User Report

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Learning Modern Algebra

From Early Attempts to Prove

Fermat’s Last Theorem

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c 2013 by The Mathematical Association of America (Incorporated)

Library of Congress Control Number: 2013940990

Print ISBN: 978-1-93951-201-7

Electronic ISBN: 978-1-61444-612-5

Printed in the United States of America

Current Printing (last digit):

10 9 8 7 6 5 4 3 2 1

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Learning Modern Algebra

From Early Attempts to Prove

Fermat’s Last Theorem

Al Cuoco

EDC, Waltham MA

and

Joseph J. Rotman

University of Illinois at Urbana–Champaign

Published and distributed by

The Mathematical Association of America

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Committee on Books

Frank Farris, Chair

MAA Textbooks Editorial Board

Zaven A. Karian, Editor

Matthias Beck Richard E. Bedient

Thomas A. Garrity

Charles R. Hampton

John Lorch

Susan F. Pustejovsky Elsa J. Schaefer

Stanley E. Seltzer

Kay B. Somers

MAA TEXTBOOKS

Bridge to Abstract Mathematics, Ralph W. Oberste-Vorth, Aristides Mouzakitis, and

Bonita A. Lawrence

Calculus Deconstructed: A Second Course in First-Year Calculus, Zbigniew H. Nitecki

Combinatorics: A Guided Tour, David R. Mazur

Combinatorics: A Problem Oriented Approach, Daniel A. Marcus

Complex Numbers and Geometry, Liang-shin Hahn

A Course in Mathematical Modeling, Douglas Mooney and Randall Swift

Cryptological Mathematics, Robert Edward Lewand

Differential Geometry and its Applications, John Oprea

Elementary Cryptanalysis, Abraham Sinkov

Elementary Mathematical Models, Dan Kalman

An Episodic History of Mathematics: Mathematical Culture Through Problem Solving,

Steven G. Krantz

Essentials of Mathematics, Margie Hale

Field Theory and its Classical Problems, Charles Hadlock

Fourier Series, Rajendra Bhatia

Game Theory and Strategy, Philip D. Straffin

Geometry Revisited, H. S. M. Coxeter and S. L. Greitzer

Graph Theory: A Problem Oriented Approach, Daniel Marcus

Knot Theory, Charles Livingston

Learning Modern Algebra: From Early Attempts to Prove Fermat’s Last Theorem, Al

Cuoco and and Joseph J. Rotman

Lie Groups: A Problem-Oriented Introduction via Matrix Groups, Harriet Pollatsek

Mathematical Connections: A Companion for Teachers and Others, Al Cuoco

Mathematical Interest Theory, Second Edition, Leslie Jane Federer Vaaler and James

W. Daniel

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Mathematical Modeling in the Environment, Charles Hadlock

Mathematics for Business Decisions Part 1: Probability and Simulation (electronic text-

book), Richard B. Thompson and Christopher G. Lamoureux

Mathematics for Business Decisions Part 2: Calculus and Optimization (electronic text-

book), Richard B. Thompson and Christopher G. Lamoureux

Mathematics for Secondary School Teachers, Elizabeth G. Bremigan, Ralph J. Bremi-

gan, and John D. Lorch

The Mathematics of Choice, Ivan Niven

The Mathematics of Games and Gambling, Edward Packel

Math Through the Ages, William Berlinghoff and Fernando Gouvea

Noncommutative Rings, I. N. Herstein

Non-Euclidean Geometry, H. S. M. Coxeter

Number Theory Through Inquiry, David C. Marshall, Edward Odell, and Michael Star-

bird

A Primer of Real Functions, Ralph P. Boas

A Radical Approach to Lebesgue’s Theory of Integration, David M. Bressoud

A Radical Approach to Real Analysis, 2nd edition, David M. Bressoud

Real Infinite Series, Daniel D. Bonar and Michael Khoury, Jr.

Topology Now!, Robert Messer and Philip Straffin

Understanding our Quantitative World, Janet Andersen and Todd Swanson

MAA Service Center

P.O. Box 91112

Washington, DC 20090-1112

1-800-331-1MAA FAX: 1-301-206-9789

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vii

Per Micky: Tutto quello che faccio, lo faccio per te.

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Contents

Preface xiii

Some Features of This Book . . . . . . . . . . . . . . . . . . . . . xiv

A Note to Students . . . . . . . . . . . . . . . . . . . . . . . . . . xv

A Note to Instructors . . . . . . . . . . . . . . . . . . . . . . . . . xv

Notation xvii

1 Early Number Theory 1

1.1 Ancient Mathematics . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Diophantus . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Geometry and Pythagorean Triples . . . . . . . . . . . . . 8

The Method of Diophantus . . . . . . . . . . . . . . . . . 11

Fermat’s Last Theorem . . . . . . . . . . . . . . . . . . . 14

Connections: Congruent Numbers . . . . . . . . . . . . . . 16

1.3 Euclid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Greek Number Theory . . . . . . . . . . . . . . . . . . . . 21

Division and Remainders . . . . . . . . . . . . . . . . . . 22

Linear Combinations and Euclid’s Lemma . . . . . . . . . 24

Euclidean Algorithm . . . . . . . . . . . . . . . . . . . . . 30

1.4 Nine Fundamental Properties . . . . . . . . . . . . . . . . . . 36

1.5 Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . 41

Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2 Induction 45

2.1 Induction and Applications . . . . . . . . . . . . . . . . . . . 45

Unique Factorization . . . . . . . . . . . . . . . . . . . . . 53

Strong Induction . . . . . . . . . . . . . . . . . . . . . . . 57

Differential Equations . . . . . . . . . . . . . . . . . . . . 60

2.2 Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . 63

Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . 69

2.3 Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

An Approach to Induction . . . . . . . . . . . . . . . . . . 73

Fibonacci Sequence . . . . . . . . . . . . . . . . . . . . . 75

3 Renaissance 81

3.1 Classical Formulas . . . . . . . . . . . . . . . . . . . . . . . 82

3.2 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . 91

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x Contents

Algebraic Operations . . . . . . . . . . . . . . . . . . . . 92

Absolute Value and Direction . . . . . . . . . . . . . . . . 99

The Geometry Behind Multiplication . . . . . . . . . . . . 101

3.3 Roots and Powers . . . . . . . . . . . . . . . . . . . . . . . . 106

3.4 Connections: Designing Good Problems . . . . . . . . . . . . 116

Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

Pippins and Cheese . . . . . . . . . . . . . . . . . . . . . 118

Gaussian Integers: Pythagorean Triples Revisited . . . . . . 119

Eisenstein Triples and Diophantus . . . . . . . . . . . . . . 122

Nice Boxes . . . . . . . . . . . . . . . . . . . . . . . . . . 123

Nice Functions for Calculus Problems . . . . . . . . . . . 124

Lattice Point Triangles . . . . . . . . . . . . . . . . . . . . 126

4 Modular Arithmetic 131

4.1 Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

4.2 Public Key Codes . . . . . . . . . . . . . . . . . . . . . . . . 149

4.3 Commutative Rings . . . . . . . . . . . . . . . . . . . . . . . 154

Units and Fields . . . . . . . . . . . . . . . . . . . . . . . 160

Subrings and Subfields . . . . . . . . . . . . . . . . . . . . 166

4.4 Connections: Julius and Gregory . . . . . . . . . . . . . . . . 169

4.5 Connections: Patterns in Decimal Expansions . . . . . . . . . 177

Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . 177

Decimal Expansions of Rationals . . . . . . . . . . . . . . 179

Periods and Blocks . . . . . . . . . . . . . . . . . . . . . . 182

5 Abstract Algebra 191

5.1 Domains and Fraction Fields . . . . . . . . . . . . . . . . . . 192

5.2 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 196

Polynomial Functions . . . . . . . . . . . . . . . . . . . . 204

5.3 Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . 206

Extensions of Homomorphisms . . . . . . . . . . . . . . . 213

Kernel, Image, and Ideals . . . . . . . . . . . . . . . . . . 216

5.4 Connections: Boolean Things . . . . . . . . . . . . . . . . . . 221

Inclusion-Exclusion . . . . . . . . . . . . . . . . . . . . . 227

6 Arithmetic of Polynomials 233

6.1 Parallels to Z . . . . . . . . . . . . . . . . . . . . . . . . . . 233

Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . 233

Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

Greatest Common Divisors . . . . . . . . . . . . . . . . . 243

Unique Factorization . . . . . . . . . . . . . . . . . . . . . 248

Principal Ideal Domains . . . . . . . . . . . . . . . . . . . 255

6.2 Irreducibility . . . . . . . . . . . . . . . . . . . . . . . . . . 259

Roots of Unity . . . . . . . . . . . . . . . . . . . . . . . . 264

6.3 Connections: Lagrange Interpolation . . . . . . . . . . . . . . 270

7 Quotients, Fields, and Classical Problems 277

7.1 Quotient Rings . . . . . . . . . . . . . . . . . . . . . . . . . 277

7.2 Field Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

Characteristics . . . . . . . . . . . . . . . . . . . . . . . . 287

Extension Fields . . . . . . . . . . . . . . . . . . . . . . . 289

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Contents xi

Algebraic Extensions . . . . . . . . . . . . . . . . . . . . 293

Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . 300

Classification of Finite Fields . . . . . . . . . . . . . . . . 305 7.3 Connections: Ruler–Compass Constructions . . . . . . . . . . 308

Constructing Regular n-gons . . . . . . . . . . . . . . . . 320

Gauss’s construction of the 17-gon . . . . . . . . . . . . . 322

8 Cyclotomic Integers 329

8.1 Arithmetic in Gaussian and Eisenstein Integers . . . . . . . . 330

Euclidean Domains . . . . . . . . . . . . . . . . . . . . . 333

8.2 Primes Upstairs and Primes Downstairs . . . . . . . . . . . . 337

Laws of Decomposition . . . . . . . . . . . . . . . . . . . 339 8.3 Fermat’s Last Theorem for Exponent 3 . . . . . . . . . . . . 349

Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . 350

The First Case . . . . . . . . . . . . . . . . . . . . . . . . 351

Gauss’s Proof of the Second Case . . . . . . . . . . . . . . 354

8.4 Approaches to the General Case . . . . . . . . . . . . . . . . 359 Cyclotomic integers . . . . . . . . . . . . . . . . . . . . . 360

Kummer, Ideal Numbers, and Dedekind . . . . . . . . . . . 365

8.5 Connections: Counting Sums of Squares . . . . . . . . . . . . 371

A Proof of Fermat’s Theorem on Divisors . . . . . . . . . 373

9 Epilog 379

9.1 Abel and Galois . . . . . . . . . . . . . . . . . . . . . . . . . 379

9.2 Solvability by Radicals . . . . . . . . . . . . . . . . . . . . . 381

9.3 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 9.4 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389

9.5 Wiles and Fermat’s Last Theorem . . . . . . . . . . . . . . . 396

Elliptic Integrals and Elliptic Functions . . . . . . . . . . . 397

Congruent Numbers Revisited . . . . . . . . . . . . . . . . 400

Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . 404

A Appendices 409

A.1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409

A.2 Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . 420 A.3 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 424

Bases and Dimension . . . . . . . . . . . . . . . . . . . . 427

Linear Transformations . . . . . . . . . . . . . . . . . . . 435

A.4 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 441

A.5 Generalized Associativity . . . . . . . . . . . . . . . . . . . . 442

A.6 A Cyclotomic Integer Calculator . . . . . . . . . . . . . . . . 444 Eisenstein Integers . . . . . . . . . . . . . . . . . . . . . . 445

Symmetric Polynomials . . . . . . . . . . . . . . . . . . . 446

Algebra with Periods . . . . . . . . . . . . . . . . . . . . . 446

References 449

Index 451

About the Authors 459

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Preface

This book is designed for college students who want to teach mathematics in

high school, but it can serve as a text for standard abstract algebra courses as

well. First courses in abstract algebra usually cover number theory, groups, and commutative rings. We have found that the first encounter with groups is

not only inadequate for future teachers of high school mathematics, it is also

unsatisfying for other mathematics students. Hence, we focus here on number

theory, polynomials, and commutative rings. We introduce groups in our last

chapter, for the earlier discussion of commutative rings allows us to explain how groups are used to prove Abel’s Theorem: there is no generalization of the

quadratic, cubic, and quartic formulas giving the roots of the general quintic

polynomial. A modest proposal: undergraduate abstract algebra should be a

sequence of two courses, with number theory and commutative rings in the

first course, and groups and linear algebra (with scalars in arbitrary fields) in

the second. We invoke an historically accurate organizing principle: Fermat’s Last The-

orem (in Victorian times, the title of this book would have been Learning Mod-

ern Algebra by Studying Early Attempts, Especially Those in the Nineteenth

Century, that Tried to Prove Fermat’s Last Theorem Using Elementary Meth-

ods). To be sure, another important problem at that time that contributed to modern algebra was the search for formulas giving the roots of polynomials.

This search is intertwined with the algebra involved in Fermat’s Last Theo-

rem, and we do treat this part of algebra as well. The difference between our

approach and the standard approach is one of emphasis: the natural direction

for us is towards algebraic number theory, whereas the usual direction is to- wards Galois theory.

Four thousand years ago, the quadratic formula and the Pythagorean The-

orem were seen to be very useful. To teach them to new generations, it was

best to avoid square roots (which, at the time, were complicated to compute),

and so problems were designed to have integer solutions. This led to Pythag-

orean triples: positive integers a; b; c satisfying a2 C b2 D c2. Two thousand years ago, all such triples were found and, when studying them in the seven-

teenth century, Fermat wondered whether there are positive integer solutions

to an C bn D cn for n > 2. He claimed in a famous marginal note that there are no solutions, but only his proof of the case n D 4 is known. This problem, called Fermat’s Last Theorem, intrigued many of the finest mathematicians, but it long resisted all attempts to solve it. Finally, using sophisticated tech-

niques of algebraic geometry developed at the end of the twentieth century,

Andrew Wiles proved Fermat’s Last Theorem in 1995.

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xiv Preface

Before its solution, Fermat’s Last Theorem was a challenge to mathemati-

cians (as climbing Mount Everest was a challenge to mountaineers). There are

no dramatic applications of the result, but it is yet another triumph of human in- tellect. What is true is that, over the course of 350 years, much of contemporary

mathematics was invented and developed in trying to deal with it. The num-

ber theory recorded in Euclid was shown to have similarities with the behavior

of polynomials, and generalizations of prime numbers and unique factoriza-

tion owe their initial study to attempts at proving Fermat’s Last Theorem. But these topics are also intimately related to what is actually taught in high school.

Thus, abstract algebra is not merely beautiful and interesting, but it is also a

valuable, perhaps essential, topic for understanding high school mathematics.

Some Features of This Book

We include sections in every chapter, called Connections, in which we explic- itly show how the material up to that point can help the reader understand and

implement the mathematics that high school teachers use in their profession.

This may include the many ways that results in abstract algebra connect with

core high school ideas, such as solving equations or factoring. But it may also

include mathematics for teachers themselves, that may or may not end up “on the blackboard;” things like the use of abstract algebra to make up good prob-

lems, to understand the foundations of topics in the curriculum, and to place

the topics in the larger landscape of mathematics as a scientific discipline.

Many students studying abstract algebra have problems understanding

proofs; even though they can follow each step of a proof, they wonder how

anyone could have discovered its argument in the first place. To address such problems, we have tried to strike a balance between giving a logical develop-

ment of results (so the reader can see how everything fits together in a coherent

package) and discussing the messier kinds of thinking that lead to discovery

and proofs. A nice aspect of this sort of presentation is that readers participate

in doing mathematics as they learn it. One way we implement this balance is our use of several design features,

such as the Connections sections described above. Here are some others.

� Sidenotes provide advice, comments, and pointers to other parts of the text related to the topic at hand. What could be more fitting for a book related to

Fermat’s Last Theorem than to have large margins? � Interspersed in the text are boxed “callouts,” such as How to Think About

It, which suggest how ideas in the text may have been conceived in the first place, how we view the ideas, and what we guess underlies the formal

exposition. Some other callouts are:

Historical Note, which provides some historical background. It often helps

to understand mathematical ideas if they are placed in historical con-

text; besides, it’s interesting. The biographies are based on those in the MacTutor History of Mathematics Archive of the School of Mathemat-

ics and Statistics, University of St. Andrews, Scotland. It can be found

on the internet: its URL is

www-history.mcs.st-andrews.ac.uk

Etymology, which traces out the origin of some mathematical terms. We

believe that knowing the etymology of terms often helps to understand

the ideas they name.

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Preface xv

Etymology. The word mathematics comes from classical Greek; it

means “knowledge,” “something learned.” But in ancient Rome through

the thirteenth century, it meant “astronomy” and “astrology.” From the

Middle Ages, it acquired its present meaning.

The word arithmetic comes from the Greek word meaning “the art of

counting.” The word geometry, in classical Greek, meant “science of measuring;” it arose from an earlier term meaning “land survey.”

It is a pleasure to acknowledge those who have contributed valuable com-

ments, suggestions, ideas, and help. We thank Don Albers, Carol Baxter, Bruce

Berndt, Peter Braunfeld, Keith Conrad, Victoria Corkery, Don DeLand, Ben Conrad’s website www.math.uconn.edu/

˜kconrad/blurbs/

is full of beautiful ideas.

Fischer, Andrew Granville, Heini Halberstam, Zaven Karian, Tsit-Yuen Lam,

Paul Monsky, Beverly Ruedi, Glenn Stevens, and Stephen Ullom.

A Note to Students

The heart of a mathematics course lies in its problems. We have tried to or-

chestrate them to help you build a solid understanding of the mathematics in

the sections. Everything afterward will make much more sense if you work through as many exercises as you can, especially those that appear difficult.

Quite often, you will learn something valuable from an exercise even if you

don’t solve it completely. For example, a problem you can’t solve may show

that you haven’t fully understood an idea you thought you knew; or it may

force you to discover a fact that needs to be established to finish the solution.

There are two special kinds of exercises.

� Those labeled Preview may seem to have little to do with the section at hand; they are designed to foreshadow upcoming topics, often with numerical ex-

periments.

� Those labeled Take it Further develop interesting ideas that are connected to the main themes of the text, but are somewhat off the beaten path. They

are not essential for understanding what comes later in the text.

An exercise marked with an asterisk, such as 1.8*, means that it is either used in some proof or it is referred to elsewhere in the text. For ease of finding

such exercises, all references to them have the form “Exercise 1.8 on page 6”

giving both its number and the number of the page on which it occurs.

A Note to Instructors

We recommend giving reading assignments to preview upcoming material.

This contributes to balancing experience and formality as described above, and it saves time. Many important pages can be read and understood by students,

and they should be discussed in class only if students ask questions about them.

It is possible to use this book as a text for a three hour one-semester course,

but we strongly recommend that it be taught four hours per week.

—Al Cuoco and Joe Rotman

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Notation

.a; b; c/ 4 triangle with sides of lengths a; b; c

ABC 4 triangle with vertices A;B; C

N 21 natural numbers

Z 21 integers

a j b 21 a is a divisor of b gcd.a; b/ 24 greatest common divisor

bxc 29 greatest integer in x Q 36 rational numbers

R 36 real numbers

) 46 implies lcm.a; b/ 55 least common multiple�

n r

� 63 binomial coefficient

<.z/ 92 real part of complex number z =.z/ 92 imaginary part of complex number z C 92 complex numbers

��! PQ 93 arrow from P to Q

z 96 conjugate of z

jzj 99 modulus of z arg.z/ 100 argument of z

ez 108 complex exponential

�.n/ 111 Euler �-function

N.z/ 116 norm of z

ZŒi  119 Gaussian integers

ZŒ!/ 120 Eisenstein integers

a � b mod m 132 a is congruent to b modulom m1 � � �cmi � � �mr 147 expression withmi deleted

Œa 154 congruence class of integer a

Zm 154 integers mod m

ZŒ� 157 cyclotomic integers

RS 157 ring of functionsR ! S C.X/ 157 ring of continuous functionsX ! R

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xviii Notation

Fun.R/ 157 ring of functionsR ! R F4 165 field with 4 elements

2X 167 Boolean ring of subsets of set X

j.m/ 172 calendar month function

Frac.D/ 194 fraction field of domain D

a=b 195 element of Frac.D/

deg.f / 198 degree of polynomial f

RŒŒx 198 all power series over R

RŒx 198 all polynomials over R

x 200 indeterminate inRŒx

f 0.x/ 202 derivative of f .x/ 2 RŒx f # 204 associated polynomial function of f

Poly.R/ 204 all polynomials functions over R

k.x/ 205 field of rational functions over k

Fq 205 finite field with exactly q elements

RŒx1; : : : ; xn 205 polynomials in several variables over R

D.x1; : : : ; xn/ 206 rational functions in several variables over

domain D

R Š S 207 ringsR and S are isomorphic ker' 217 kernel of homomorphism '

im' 217 image of homomorphism '

.b1; : : : ; bn/ 218 ideal generated by b1; : : : ; bn

.a/ 218 principal ideal generated by a

.0/ 219 zero ideal D f0g IJ 220 product of ideals I and J

I C J 220 sum of ideals I and J R � S 221 direct product of rings R and S a _ b 223 binary operation in Boolean ring jAj 227 number of elements in finite set A PID 255 principal ideal domain

UFD 258 unique factorization domain

ˆd .x/ 265 cyclotomic polynomial

aC I 278 coset of element a mod ideal I a � b mod I 279 congruent mod ideal I

R=I 280 quotient ringR mod I˝ X ˛

293 subfield generated by subset X

ŒK W k 291 degree of extension field K=k k.z1 ; : : : ; zn/ 294 extension field adjoining z1; : : : ; zn to k

irr.z; k/ 296 minimal polynomial of z over k

PQ 310 line segment with endpoints P;Q

PQ 310 length of segment PQ

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Notation xix

L.P;Q/ 309 line determined by points P;Q

C.P;Q/ 309 circle with center P , radius PQ

@ 333 size function on Euclidean domain

� 348 � D 1 � ! � 350 valuation

r.n/ 371 number of non-associate z 2 ZŒi  of norm n Q1 372 first quadrant

�.s/ 374 Riemann zeta function

�.n/ 375 a multiplicative function on ZŒi 

Gal.f / 386 Galois group of polynomial f

Gal.E=k/ 387 Galois group of field extension E=k

Sn 389 symmetric group on n letters

G=N 392 quotient group

a 2 A 409 a is an element of set A 1X 411 identity function on set X

f W a 7! b 411 f .a/ D b U � V 410 U is a subset of set V U ¨ V 410 U is a proper subset of V

¿ 410 empty set

g ı f 414 composite f followed by g Œa 421 equivalence class of element a

SpanhXi 427 subspace spanned by subset X dim.V / 433 dimension of vector space V

V � 437 dual space of vector space V

A> 438 transpose of matrix A

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1 Early Number Theory Algebra, geometry, and number theory have been used for millennia. Of course,

numbers are involved in counting and measuring, enabling commerce and ar-

chitecture. But reckoning was also involved in life and death matters such as astronomy, which was necessary for navigation on the high seas (naval com-

merce flourished four thousand years ago) as well as to predict the seasons,

to apprise farmers when to plant and when to harvest. Ancient texts that have

survived from Babylon, China, Egypt, Greece, and India provide evidence for

this. For example, the Nile River was the source of life in ancient Egypt, for its banks were the only arable land in the midst of desert. Mathematics was

used by the priestly class to predict flooding as well as to calculate area (taxes

were assessed according to the area of land, which changed after flood waters

subsided). And their temples and pyramids are marvels of engineering.

1.1 Ancient Mathematics

The quadratic formula was an important mathematical tool, and so it was

taught to younger generations training to be royal scribes. Here is a problem from an old Babylonian cuneiform text dating from about 1700 BCE. We quote

from van der Waerden [35], p. 61 (but we write numbers in base 10 instead

of in base 60, as did the Babylonians). We also use modern algebraic notation

that dates from the fifteenth and sixteenth centuries (see Cajori [6]).

I have subtracted the side of the square from the area, and it is 870. What

is the side of my square?

The text rewrites the data as the quadratic equation x2 � x D 870; it then gives a series of steps showing how to find the solution, illustrating that the

Babylonians knew the quadratic formula.

Historians say that teaching played an important role in ancient mathe- matics (see van der Waerden [35], pp. 32–33). To illustrate, the coefficients

of the quadratic equation were chosen wisely: the discriminant b2 � 4ac D 1 � 4.�870/ D 3481 D 592 is a perfect square. Were the discriminant not a The number 59 may have

been chosen because

the Babylonians wrote

numbers in base 60, and

59 D 60 � 1.

perfect square, the problem would have been much harder, for finding square

roots was not routine in those days. Thus, the quadratic in the text is well- chosen for teaching the quadratic formula; a good teaching prize would not be

awarded for x2 � 47x D 210. The Babylonians were not afraid of cubics. Another of their problems from

about the same time is

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2 Chapter 1 Early Number Theory

Solve 12x3 D 3630,

and the answer was given. The solution was, most likely, obtained by using tables of approximations of cube roots.

A standard proof of the quadratic formula is by “completing the square.”

This phrase can be taken literally. Given a quadratic x2C bx D c with b and c positive, we can view x2 C bx as the shaded area in Figure 1.1. Complete the

x

x

Figure 1.1. Completing the Square.

figure to a square by attaching the corner square having area 1 2 b � 1

2 b D 1

4 b2;

the new square has area

c C 1 4 b2 D x2 C bx C 1

4 b2 D .x C 1

2 b/2:

Thus, x C 1 2 b D

q c C 1

4 b2, which simplifies to the usual formula giving

the roots of x2 C bx � c. The algebraic proof of the validity of the quadratic

In [35], pp. 26–35, van

der Waerden considers

the origin of proofs in

mathematics, suggesting

that they arose in Europe

and Asia in Neolithic

(late Stone Age) times,

4500 BCE–2000 BCE.

formula works without assuming that b and c are positive, but the idea of the

proof is geometric.

a2

b2

a

b

a

b

a b

a

b

c2

Figure 1.2. Pythagorean Theorem.

The Babylonians were aware of the Pythagorean Theorem. Although they

believed it, there is no evidence that the Babylonians had proved the Pythag-

orean Theorem; indeed, no evidence exists that they even saw a need for a

proof. Tradition attributes the first proof of this theorem to Pythagoras, who Exercise 1.4 on page 5

asks you to show that the

rhombus in Figure 1.2

with sides of length c is a

square.

lived around 500 BCE, but no primary documents extant support this. An ele-

gant proof of the Pythagorean Theorem is given on page 354 of Heath’s 1926

translation [16] of Euclid’s The Elements; the theorem follows from equality

of the areas of the two squares in Figure 1.2.

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1.1 Ancient Mathematics 3

Here is an ancient application of the Pythagorean Theorem. Aristarchus

(ca. 310 BCE–250 BCE) saw that the Moon and the Sun appear to be about

the same size, and he wondered how far away they are. His idea was that at the time of the half-moon, the Earth E , Moon M , and Sun S form a right

triangle with right angle †M (that is, looking up at the Moon, the line of sight seems to be perpendicular to the Sun’s rays). The Pythagorean Theorem gives

a

S M

E

Figure 1.3. Earth, Moon, and Sun.

jSEj2 D jSM j2 C jMEj2. Thus, the Earth is farther from the Sun than it is from the Moon. Indeed, at sunset, ˛ D †E seems to be very close to 90ı: if we are looking at the Moon and we wish to watch the Sun dip below the horizon,

we must turn our head all the way to the left. Aristarchus knew trigonometry;

he reckoned that cos˛ was small, and he concluded that the Sun is very much

further from the Earth than is the Moon.

Example 1.1. Next, we present a geometric problem from a Chinese collec-

tion of mathematical problems, Nine Chapters on the Mathematical Art, writ- ten during the Han Dynasty about two thousand years ago. Variations of this

problem still occur in present day calculus books!

There is a door whose height and width are unknown, and a pole whose There are similar problems from the Babylonians and

other ancient cultures. length p is also unknown. Carried horizontally, the pole does not fit by 4

ch’ihI vertically, it does not fit by 2 ch’ihI slantwise, it fits exactly. What are the height, width, and diagonal of the door?

p p – 2

p – 4

Figure 1.4. Door Problem.

The data give a right triangle with sides p � 4, p � 2, and p, and the Py- thagorean Theorem gives the equation .p � 4/2 C .p � 2/2 D p2, which simplifies to p2�12pC20 D 0. The discriminant b2�4ac is 144�80 D 64, a perfect square, so that p D 10 and the door has height 8 and width 6 (the other root of the quadratic is p D 2, which does not fit the physical data). The sides of the right triangle are 6, 8, 10, and it is similar to the triangle with

sides 3; 4; 5. Again, the numbers have been chosen wisely. The idea is to teach

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4 Chapter 1 Early Number Theory

students how to use the Pythagorean Theorem and the quadratic formula. As

we have already remarked, computing square roots was then quite difficult, so

that the same problem for a pole of length p D 12 would not have been very bright because there is no right triangle with sides of integral length that hasThe word hypotenuse

comes from the Greek verb

meaning to stretch. hypotenuse 12. N

Are there right triangles whose three sides have integral length that are not

similar to the 3; 4; 5 triangle? You are probably familiar with the 5; 12; 13 tri-

angle. Let’s use 4.a; b; c/ (lower case letters) to denote the triangle whose sides have length a, b, and c; if 4.a; b; c/ is a right triangle, then c denotes the length of its hypotenuse, while a and b are its legs. Thus, the right trian-

gle with side-lengths 5, 12, 13 is denoted by 4.5; 12; 13/. (We use the usual notation, 4ABC , to denote a triangle whose vertices are A;B; C .)

Definition. A triple .a; b; c/ of positive integers with a2 C b2 D c2 is called a Pythagorean triple.

If .a; b; c/ is a Pythagorean triple, then the triangles 4.a; b; c/ and 4.b; a; c/ are the same. Thus, we declare that the Pythagorean triples .a; b; c/ and .b; a; c/ are the same.

Historical Note. Pythagorean triples are the good choices for problems teach-

ing the Pythagorean Theorem. There are many of them: Figure 1.5 shows a

Babylonian cuneiform tablet dating from the dynasty of Hammurabi, about

1800 BCE, whose museum name is Plimpton 322, which displays fifteen Pythagorean triples (translated into our number system).

b a c

120 119 169

3456 3367 4825

4800 4601 6649

13500 12709 18541

72 65 97

360 319 481

2700 2291 3541

960 799 1249

600 481 769

6480 4961 8161

60 45 75

2400 1679 2929

240 161 289

2700 1771 3229

90 56 106

Figure 1.5. Plimpton 322.

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1.1 Ancient Mathematics 5

It is plain that the Babylonians had a way to generate large Pythagorean

triples. Here is one technique they might have used. Write

a2 D c2 � b2 D .c C b/.c � b/:

If there are integers m and n with

c C b D m2

c � b D n2;

then

a D p .c C b/.c � b/ D mn: (1.1)

We can also solve for b and c:

b D 1 2

� m2 � n2

� (1.2)

c D 1 2

� m2 C n2

� : (1.3)

Summarizing, here is what we call the Babylonian method. Choose odd num-

bers m and n (forcing m2 C n2 and m2 � n2 to be even, so that b and c are integers), and define a, b, and c by Eqs. (1.1), (1.2), and (1.3). For example, if m D 7 and n D 5, we obtain 35, 12, 37. If we choose m D 179 and n D 71, we obtain 13500, 12709, 18541, the largest triple on Plimpton 322.

The Babylonian method does not give all Pythagorean triples. For example,

.6; 8; 10/ is a Pythagorean triple, but there are no odd numbers m > n with

6 D mn or 8 D mn. Of course, .6; 8; 10/ is not signifcantly different from .3; 4; 5/, which arises from 3 > 1. In the next section, we will show, follow-

ing Diophantus, ca. 250 CE, how to find all Pythagorean triples. But now we

should recognize that practical problems involving applications of pure math-

ematics (e.g., surveying) led to efforts to teach this mathematics effectively,

which led to more pure mathematics (Pythagorean triples) that seems at first to After all, what practi-

cal application does

the Pythagorean triple

.13500; 12709; 18541/

have?

have no application outside of teaching. The remarkable, empirical, fact is that pure mathematics yields new and valuable applications. For example, we shall

see in the next section that classifying Pythagorean triples leads to simplifying

the verification of some trigonometric identities as well as the solution of cer-

tain integration problems (for example, we will see a natural way to integrate

sec x).

Exercises

1.1 Prove the quadratic formula for the roots of ax2Cbx Cc D 0 whose coefficients a, b, and c may not be positive.

1.2 Give a geometric proof that .a C b/2 D a2 C 2ab C b2 for a; b positive. 1.3 * Let f .x/ D ax2C bx C c be a quadratic whose coefficients a; b; c are rational.

Prove that if f .x/ has one rational root, then its other root is also rational.

1.4 *

(i) Prove that the rhombus with side lengths c in the left square of Figure 1.2 is The book by Loomis [20] contains 370 different

proofs of the Pythagorean

Theorem, by the author’s

count.

a square.

(ii) Prove the Pythagorean Theorem in a way suggested by Figure 1.2.

(iii) Give a proof of the Pythagorean Theorem different from the one suggested

by Figure 1.2.

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6 Chapter 1 Early Number Theory

1.5 Here is another problem from Nine Chapters on the Mathematical Art. A pond is

10 ch’ih square. A reed grows at its center and extends 1 ch’ih out of the water.

If the reed is pulled to the side of the pond, it reaches the side precisely. What are

the depth of the water and the length of the reed?

Answer: Depth = 12 ch’ih and length = 13 ch’ih.

1.6 *

(i) Establish the algebraic identity

� a C b

2

�2 � �

a � b 2

�2 D ab:

(ii) Use (i) to establish the Arithmetic–Geometric Mean Inequality: if a and b

are positive reals, then

p ab � 12 .a C b/:

When is there equality?

(iii) Show how to dissect an a � b rectangle so that it fits inside a square with side-length .a C b/=2. How much is “left over?”

Hint: Try it with numbers. Cut an 8 � 14 rectangle to fit inside an 11 � 11 square.

(iv) Show that a rectangle of maximum area with fixed perimeter is a square.

(v) The hyperbolic cosine is defined by

cosh x D 12 .e x C e�x/:

Prove that cosh x � 1 for all real numbers x, while coshx D 1 if and only if x D 0.

(vi) Use Figure 1.6 to give another proof of the Arithmetic-Geometric Mean In-

equality.

a b

Figure 1.6. Arithmetic–Geometric Mean Inequality.

1.7 * Prove that there is no Pythagorean triple .a; b; c/ with c D 12.

1.8 * Let .a; b; c/ be a Pythagorean triple.

(i) Prove that the legs a and b cannot both be odd.

(ii) Show that the area of 4.a; b; c/ is an integer.

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1.2 Diophantus 7

1.9 * Show that 5 is not the area of a triangle whose side-lengths form a Pythagorean

triple.

1.10 * Let .a; b; c/ be a Pythagorean triple. If m is a positive integer, prove that

.ma; mb; mc/ is also a Pythagorean triple.

1.11 .Converse of Pythagorean Theorem/: * Let 4 D 4.a; b; c/ be a triangle with sides of lengths a; b; c (positive real numbers, not necessarily integers). Prove that

if a2 C b2 D c2, then 4 is a right triangle.

Hint: Construct a right triangle 40 with legs of lengths a; b, and prove that 40 is congruent to 4 by side-side-side.

1.12 * Prove that every Pythagorean triple .a; b; c/ arises from a right triangle 4.a; b; c/ having sides of lengths a; b; c.

1.13 If P D .a; b; c/ is a Pythagorean triple, define r.P / D c=a. If we label the Py- thagorean triples on Plimpton 322 as P1; : : : ; P15 , show that r.Pi / is decreasing:

r.Pi / > r.PiC1/ for all i � 14.

1.14 * If .a; b; c/ is a Pythagorean triple, show that .a=c; b=c/ is a point on the graph

of x2 C y2 D 1. What is the graph of x2 C y2 D 1?

1.15 Preview: Let L be the line through .�1; 0/ with slope t . (i) If t D 12 , find all the points where L intersects the graph of x

2 C y2 D 1.

Answer: .35 ; 4 5 /.

(ii) If t D 32 , find all the points where L intersects the graph of x2 C y2 D 1.

Answer: . �513 ; 12 13

/.

(iii) Pick a rational number t , not 12 or 3 2 , and find all the points where L intersects

the graph of x2 C y2 D 1. (iv) Suppose ` is a line that contains .�1; 0/ with slope r . If r is a rational number,

show that ` intersects the graph of x2 C y2 D 1 in two points, each of which has rational number coordinates.

1.16 Preview: A Gaussian integer is a complex number a C bi where both a and b are integers. Pick six Gaussian integers r C si with r > s > 0 and square them. State something interesting that you see in your results.

1.17 Preview: Consider a complex number z D q C ip, where q > p are positive integers. Prove that

.q2 � p2; 2qp; q2 C p2/

is a Pythagorean triple by showing that jz2j D jzj2.

If z is a complex number,

say z D aC bi , then we define jzj D

p a2 C b2.

1.18 Preview: Show, for all real numbers m and n, that

h 1 2 .m C n/ C

1 2 .m � n/i

i2 D mn C 12 .m

2 � n2/i:

1.2 Diophantus

We are going to classify Pythagorean triples using a geometric method of Dio-

phantus that describes all Pythagorean triples.

Historical Note. We know very little about the life of Diophantus. He was

a mathematician who lived in Alexandria, Egypt, but his precise dates are

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8 Chapter 1 Early Number Theory

unknown; most historians believe he lived around 250 CE. His extant work

shows systematic algebraic procedures and notation, but his leaps of intuition

strongly suggest that he was thinking geometrically; indeed, Newton called Diophantus’s discussion of Pythagorean triples the chord method (see Fig-

ure 1.7). Thus, geometry (the Pythagorean Theorem) and applied problems

(teaching) suggested an algebraic problem (find all Pythagorean triples), and

we now return to geometry to solve it. Here is evidence that the distinction

between algebra and geometry is an artificial one; both are parts of the same subject.

Geometry and Pythagorean Triples

Before we get into the technicalities of Diophantus’s classification of Pythag-

orean triples, let’s note that geometry is lurking nearby. Exercise 1.14 above

makes a natural observation: if .a; b; c/ is a Pythagorean triple, then

�a c

�2 C � b

c

�2 D 1;

a point on the unit circle, the circle having radius 1, center the origin, and

equation x2 C y2 D 1. Dividing through by c2 is a good idea. For example, .6; 8; 10/ is a “duplicate” of .3; 4; 5/, and both of these Pythagorean triples

determine the same point, .3=5; 4=5/, on the unit circle. Here is the main idea of Diophantus. Even though those points arising from

Pythagorean triples are special (for example, they lie in the first quadrant and

both their coordinates are rational numbers), let’s parametrize all the pointsP

on the unit circle. Choose a point on the unit circle “far away” from the first

quadrant; the simplest is .�1; 0/, and let ` D `.P / be the line joining it to P . We shall see that the slopes of such lines parametrize all the points on the

unit circle. In more detail, any line ` through .�1; 0/ (other than the tangent)

(–1,0)

P x, y= ( )

slope = t

l

Figure 1.7. Geometric Idea of Diophantus.

intersects the unit circle in a unique second point, P D .x; y/; let t be the slope of `. As t varies through all real numbers, �1 < t < 1, the intersection points P of ` and the unit circle trace out the entire circle (except for .�1; 0/).

Proposition 1.2. The points P on the unit circle .other than .�1; 0// are parametrized as

P D � 1 � t2 1C t2 ;

2t

1C t2

� ; where � 1 < t < 1:

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1.2 Diophantus 9

Proof. The line through points .a; b/ and .c; d / has equation y�b D t.x�a/, where t D .d � b/=.c � a/, so the line ` through .�1; 0/ and a point P D .x; y/ on the unit circle has an equation of the form y D t.x C 1/, so that x D .y � t/=t . Thus, .x; y/ is a solution of the system

y D t.x C 1/ x2 C y2 D 1:

An obvious solution of this system is .�1; 0/, because this point lies on both the line and the circle. Let’s find x and y in terms of t . If the slope t D 0, then ` is the x-axis and the other solution is .1; 0/. To find the solutions when

t ¤ 0, eliminate x: the equations

y � t t

D x and x2 C y2 D 1

give

� y � t t

�2 C y2 D 1:

Expanding and simplifying, we obtain

y � .1 C t2/y � 2t

� D 0:

We knew at the outset that y D 0 makes this true. If y ¤ 0, then canceling gives

y D 2t 1C t2 ;

and solving for x gives

x D y � t t

D 2t 1Ct2 � t

t D 1 � t

2

1C t2 :

In Exercise 1.12 on page 7, we saw that every Pythagorean triple .a; b; c/

arises from a right triangle 4.a; b; c/ having sides of integral lengths a; b; c. Conversely, the Pythagorean Theorem says that every right triangle 4.a; b; c/ whose sides have integral length gives the Pythagorean triple .a; b; c/. Thus,

Pythagorean triples and certain right triangles are merely two ways to view the

same idea, one algebraic, one geometric. At any given time, we will adopt that

viewpoint which is most convenient.

We have already run across distinct Pythagorean triples that are essentially

the same; Exercise 1.10 on page 7 shows that if .a; b; c/ is a Pythagorean triple, then so is .ma;mb;mc/, where m is a positive integer. The right trian-

gles 4.a; b; c/ and 4.ma;mb;mc/ determined by these Pythagorean triples are similar, for their sides are proportional. More generally, the Pythagorean

triples .6; 8; 10/ and .9; 12; 15/ are not really different, for each arises from

.3; 4; 5/; however, neither .6; 8; 10/ nor .9; 12; 15/ is obtained from the other by multiplying its terms by some integer m.

Definition. Two Pythagorean triples .a; b; c/ and .u; v; z/ are similar if their

right triangles 4.a; b; c/ and 4.u; v; z/ are similar triangles.

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10 Chapter 1 Early Number Theory

The method of Diophantus will give a formula for certain special Pythago-

rean triples, and it will then show that every Pythagorean triple is similar to a

special one.

Definition. A point .x; y/ in the plane is a rational point if both x and y are rational numbers.

A Pythagorean point is a rational point in the first quadrant, lying on the

unit circle, and above the diagonal line with equation y D x.

Remember that we regard Pythagorean triples .a; b; c/ and .b; a; c/ as the

same. Recall some analytic geometry: if a ¤ b, P D .a; b/, and Q D .b; a/, then the diagonal is the perpendicular bisector of the segment PQ. (The line

through P and Q has equation y D �x C a C b; it is perpendicular to the diagonal for the product of their slopes is �1; the line intersects the diagonal in the point .aCb

2 ; aCb 2 /, which is equidistant from P and Q. If a � b, then P

is above the diagonal and Q is below.)

Proposition 1.3. A triple .a; b; c/ of integers is a Pythagorean triple if and

only if .a=c; b=c/ is a Pythagorean point.

Proof. Let .a; b; c/ be a Pythagorean triple. Dividing both sides of the defining

equation a2 C b2 D c2 by c2 gives .a=c/2 C .b=c/2 D 1;

so that the triple gives an ordered pair of positive rational numbers .x; y/ D .a=c; b=c/ with x2 C y2 D 1. Thus, the rational point P D .x; y/ lies in the first quadrant. As both .a; b; c/ and .b; a; c/ are the same Pythagorean triple,

we may assume that

x D a=c � b=c D y;

so that .x; y/ lies above the diagonal line with equation y D x. Hence, .x; y/ is a Pythagorean point.

Conversely, let’s now see that a Pythagorean point .x; y/ gives rise to a

Pythagorean triple. Write the rational numbers x � y with the same denomi- nator, say, x D a=c and y D b=c, where a; b; and c are positive integers and a � b � c. Now

1 D x2 C y2 D a 2

c2 C b

2

c2 ;

so that a2 C b2 D c2 and hence .a; b; c/ is a Pythagorean triple. In summary, the problem of finding all Pythagorean triples corresponds to

the problem of finding all Pythagorean points. This is exactly what the geo-

metric idea of Diophantus does. In fact, a Pythagorean point .x; y/ gives rise

to infinitely many Pythagorean triples. Write the coordinates with another de-

nominator, say x D u=z and y D v=z. The calculation at the end of the proof of Proposition 1.3 shows that .u; v; z/ is another Pythagorean triple arising from .x; y/.

Etymology. Here are sources of some common words of mathematics.

� Proposition: From Latin, meaning a statement or something pictured in the mind.

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1.2 Diophantus 11

� Theorem: From the Greek word meaning “spectacle” or “contemplate.” Re- lated words are “theory” and “theater.” Theorems are important propositions.

� Corollary:From the Latin word meaning “flower.” In ancient Rome, it meant a “gratuity;” flowers were left as tips. In mathematics, corollaries follow eas-

ily from theorems; they are gifts bequeathed to us.

� Lemma: From Greek; it meant something taken for granted. In mathematics nowadays, it is usually a technical result, a minor theorem, which can be

used in the course of proving a more important theorem.

� Proof: From Medieval French, meaning an argument from evidence estab- lishing the truth. The adage, “The exception proves the rule,” uses the word in the sense of testing: it originally meant a kind of indirect proof. We test

whether a rule is true by checking whether an exception to it leads to a con-

tradiction. Nowadays, this adage seems to have lost its meaning.

The Method of Diophantus

Proposition 1.2 parametrizes all the points P on the unit circle other than

.�1; 0/. We are now going to see which values of t produce Pythagorean points: rational points on the unit circle lying in the first quadrant above the

diagonal line with equation y D x.

Theorem 1.4. Let P D .x; y/ ¤ .�1; 0/ be a point on the unit circle, and let t be the slope of the line ` joining .�1; 0/ and P .

(i) The slope t is a rational number if and only if P is a rational point.

(ii) The point P is a Pythagorean point if and only if t is a rational number

satisfying p 2 � 1 < t < 1.

Proof. (i) The parametrization P D .x; y/ gives a pair of equations:

x D 1 � t 2

1C t2 and y D 2t

1C t2 :

Clearly, if t is rational, then both x and y are rational. Conversely, if

P D .x; y/ is a rational point, then the slope t of ` is t D y�0 x�.�1/ D

y xC1 ,

and so t is a rational number.

(ii) Pythagorean points correspond to rational points on the unit circle that lie

in the first quadrant above the line y D x. Points on the circle lying in the first quadrant arise from lines having slope t with 0 < t < 1. The point The slope of the line joining

.�1; 0/ to .0; 1/ is 1.in the first quadrant that is the intersection of the unit circle and the line y D x is .

p 2 2 ;

p 2 2 /, and the slope of the line joining .�1; 0/ to .

p 2 2 ;

p 2 2 /

is

t D

p 2 2

1C p 2 2

D p 2 � 1 � :414:

Therefore, Pythagorean points correspond to the lines ` through .�1; 0/ having rational slope t satisfying

p 2 � 1 < t < 1.

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12 Chapter 1 Early Number Theory

Let’s look at this more closely. If t D p=q is a rational number betweenp 2� 1 and 1, then the Pythagorean point it gives can be expressed in terms of

p and q:

� 1 � t2 1C t2 ;

2t

1C t2

� D

0 B@ 1 �

� p q

�2

1C � p q

�2 ; 2 � p q

1C � p q

�2

1 CA

D � q2 � p2 q2 C p2 ;

2qp

q2 C p2

� : (1.4)

Theorem 1.5 (Diophantus). Every Pythagorean triple .a; b; c/ is similar to a

Pythagorean triple of the form

.2qp; q2 � p2; q2 C p2/;

where p and q are positive integers with q > p > p 2 � 1.

Proof. Since .a; b; c/ is a Pythagorean triple, P D .a=c; b=c/ is a Pythago- rean point. By Eq. (1.4),

�a c ; b

c

� D � 1 � t2 1C t2 ;

2t

1C t2 �

D � q2 � p2 q2 C p2 ;

2qp

q2 C p2 � :

It follows that 4.a; b; c/ is similar to 4.2pq; q2�p2; q2Cp2/, because their sides are proportional. Therefore, the Pythagorean triple .a; b; c/ is similar to .2qp; q2 � p2; q2 C p2/, as claimed.

How to Think About It. The strategy of Diophantus is quite elegant. The problem of determining all Pythagorean triples is reduced from finding three

unknowns, a; b; and c, to two unknowns, x D a=c and y D b=c, to one unknown, t D p=q. In effect, all Pythagorean triples are parametrized by t ; that is, as t varies over all rational numbers between

p 2�1 and 1, the formulas

involving t vary over all Pythagorean points and hence over all Pythagorean triples.

We can now show that the Babylonians had, in fact, found all Pythagorean

triples. We are tacitly using a

technique of proof called

Infinite Descent. If, for a

given positive integer n

with certain properties,

there always exists a

strictly smaller positive

integer n1 having the

same properties, then

there are infinitely many

such integers. But this is

impossible; there are only

finitely many integers with

n > n1 > n2 > � � � > 0.

Corollary 1.6. Every Pythagorean triple is similar to one arising from the

Babylonian method.

Proof. By Theorem 1.5, every Pythagorean triple is similar to one of the form

.2qp; q2�p2; q2Cp2/, where q > p are positive integers. If both q and p are even, then we can replace q > p by 1

2 q > 1

2 p, obtaining a Pythagorean triple

.1 4 2qp; 1

4 .q2�p2/; 1

4 .q2Cp2// similar to the original one. If both parameters

of the new triple are still even, replace 1 2 q > 1

2 p by 1

4 q > 1

4 p. Eventually, we

arrive at a Pythagorean triple .2rs; r2 � s2; r2 C s2/, similar to the original triple, that arises from parameters r > s, at least one of which is odd.

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1.2 Diophantus 13

There are two possibilities. If r and s have different parity, definem D rCs and n D r � s. Both m and n are odd, and the Pythagorean triple given by the Babylonian method fromm > n is

B D � mn; 1

2 .m2 � n2/; 1

2 .m2 C n2/

� :

Substitute:

mn D .rCs/.r�s/ D r2�s2; 1 2 .m2�n2/ D 2rs; and 1

2 .m2�n2/ D r2Cs2:

Thus, the Pythagorean tripleB is similar to .2rs; r2 � s2; r2 C s2/. If both r and s are odd, then the Pythagorean triple given by the Babylo-

nian method from r > s is .rs; 1 2 .r2 � s2/; 1

2 .r2 C s2// which is similar to

.2rs; r2 � s2; r2 C s2/.

Not every Pythagorean triple .a; b; c/ is equal to .2qp; q2 � p2; q2 C p2/ for some q > p, nor does the theorem say that it is; the theorem asserts only

that .a; b; c/ is similar to a Pythagorean triple arising from the formula. For example, let us show that .9; 12; 15/ is not of this form. Since the leg 9 is odd,

the even leg 12 must be 2qp, so that qp D 6, and the only possible parameters are 3 > 2 or 6 > 1. But 3 > 2 gives .5; 12; 13/ and 6 > 1 gives .12; 35; 37/,

neither of which is similar to .9; 12; 15/. However, .9; 12; 15/ is similar to

.3; 4; 5/, and .3; 4; 5/ arises from 2 > 1. A Pythagorean triple .a; b; c/ is primitive if there is no integer d > 1 that

is a divisor of a; b, and c. Thus, .3; 4; 5/ is primitive but .9; 12; 15/ is not.

In Theorem 1.25, we’ll give a rigorous proof that every Pythagorean triple is

similar to exactly one primitive Pythagorean triple.

Exercises

1.19 Find q and p in Theorem 1.5 for each of the following Pythagorean triples.

(i) .7; 24; 25/.

Answer: q D 5 and p D 3. (ii) .129396;261547;291805/:

Answer: q D 526 and p D 123.

1.20 * Show that every Pythagorean triple .x; y; z/ with x; y; z having no common

factor d > 1 is of the form

.r2 � s2; 2rs; r2 C s2/

for positive integers r > s having no common factor > 1; that is,

x D r2 � s2; y D 2rs; z D r2 C s2:

1.21 A line in the plane with equation y D mx C c is called a rational line if m and c are rational numbers. If P and Q are distinct rational points, prove that the line

joining them is a rational line.

1.22 A lattice point is a point in the plane whose coordinates are integers. Let P D .x; y/ be a Pythagorean point and ` the line through P and the origin. Prove that

if Q D .a; b/ is a lattice point on ` and c is the distance from Q to the origin, then .a; b; c/ is a Pythagorean triple.

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14 Chapter 1 Early Number Theory

1.23 * Let P D .x0; y0/ be a Pythagorean point and L the line joining P and the origin (so the equation of L is y D mx, where m D y0=x0). Show that if .a=c; b=c/ is a rational point on L, then .a; b; c/ is a Pythagorean triple.

1.24 Does every rational point in the plane correspond to a Pythagorean point? If so,

prove it. If not, characterize the ones that do.

Answer: No. For example, .12 ; 1 2 / does not correspond.

1.25 * Prove the identity � x2 C y2

�2 D � x2 � y2

�2 C .2xy/2: 1.26 *

(i) Show that the same number can occur as a leg in two nonsimilar Pythagorean

triangles.

(ii) Prove that the area of 4.a; b; c/, a right triangle with integer side lengths, is an integer.

(iii) A Heron triangle is a triangle with integer side lengths and area. Find a Heron

triangle that is not a right triangle.

Hint: Use parts (i) and (ii).

1.27 Show that every integer n � 3 occurs as a leg of some Pythagorean triple.

Hint: The cases n even and n odd should be done separately.

1.28 Distinct Pythagorean triples can have the same hypotenuse: both .33;56; 65/ and

.16; 63; 65/ are Pythagorean triples. Find another pair of distinct Pythagorean

triples having the same hypotenuse.

1.29 * If .cos �; sin �/ is a rational point, prove that both cos.� C30ı/ and sin.� C30ı/ are irrational.

Fermat’s Last Theorem

About fourteen centuries after Diophantus, Fermat (1601–1665) proved that

there are no positive integers a; b; c with a4 C b4 D c4. He was studying his copy of Diophantus’s Arithmetica, published in 1621, and he wrote in its

margin,

: : : it is impossible for a cube to be written as a sum of two cubes or a

fourth power to be written as a sum of two fourth powers or, in general,

for any number which is a power greater than the second to be written

as a sum of two like powers. I have discovered a truly marvelous demon-

stration of this proposition which this margin is too narrow to contain.

Fermat never returned to this problem (at least, not publicly) except for hisFermat was not the first mathematician to write a

marginal note in a copy

of Diophantus. Next to

the same problem, the

Byzantine mathematician

Maximus Planudes wrote,

Thy soul, Diophantus, be

with Satan because of the

difficulty of your theorems.

proof of the case n D 4, which we give below. The statement: If n > 2, there are no positive integers a; b; c with an C bn D cn, was called Fer- mat’s Last Theorem, perhaps in jest. The original text in which Fermat wrote

his famous marginal note is lost today. Fermat’s son edited the next edition

of Diophantus, published in 1670; this version contains Fermat’s annotations,

including his famous “Last Theorem;” it contained other unproved assertions as well, most true, some not. By the early 1800s, only Fermat’s Last Theorem

remained undecided. It became a famous problem, resisting the attempts of

mathematicians of the highest order for 350 years, until it was finally proved,

in 1995, by Wiles. His proof is very sophisticated, and most mathematicians

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1.2 Diophantus 15

believe that Fermat did not have a correct proof. The quest for a proof of Fer-

mat’s Last Theorem generated much beautiful mathematics. In particular, it led

to an understanding of complex numbers, factorization, and polynomials. We’ll see, in the Epilog, that extending the method of Diophantus from quadratics to

cubics involves elliptic curves, the study of which is the setting for Wiles’

proof of Fermat’s Last Theorem.

Fermat proved the next theorem (which implies the case n D 4 of Fermat’s Last Theorem) because he was interested in the geometric problem of deter- mining which right triangles having all sides of rational length have integer

area (we’ll soon discuss this problem in more detail).

Theorem 1.7 (Fermat). There is no triple .x; y; z/ of positive integers with

x4 C y4 D z2: (1.5) This proof is not difficult,

but it uses several elemen-

tary divisibility results we’ll

prove later. Since we feel

that this is the appropriate

place for this theorem, we’ll

just refer to the needed

things.

Proof. The proof will be by infinite descent (Fermat invented infinite descent

for this very problem). Given a triple of positive integers .x; y; z/ satisfying Eq. (1.5), we’ll show there is another triple .u; v; w) of the same sort with

w < z, and so repeating this process leads to a contradiction.

Let’s say that integers x and y are relatively prime if there is no integer

d > 1 dividing both of them; that is, it’s not true that x D da and y D db. We can assume that x and y are relatively prime, for otherwise a common

factor of x and y would also be a factor of z, and we could divide it out. It

follows (and we’ll prove it in the next chapter) that x2 and y2 are also relatively

prime. And note that x4 C y4 D z2 implies that � x2 �2 C

� y2 �2 D z2

so that .x2; y2; z/ form a Pythagorean triple.

We also observe that x2 and y2 can’t both be odd; if x2 D 2k C 1 and y2 D 2j C 1, then

.2k C 1/2 C .2j C 1/2 D z2:

Expanding and collecting terms gives z2 D 4hC2 for some integer h. But you can check that the square of any integer is either of the form 4h or 4hC 1.

We can now assume that .x2; y2; z/ is a Pythagorean triple in which x and

y are relatively prime, x is odd, and y is even. By Exercise 1.20 on page 13,

there are relatively prime integers r and s with r > s > 0 such that

x2 D r2 � s2; y2 D 2rs; and z D r2 C s2:

The first equation says that x2 C s2 D r2; that is, .x; s; r/ is another Pythago- rean triple with x odd. Moreover, x and s have no common factor (why?), so

that Exercise 1.20 gives relatively prime integers a and b such that

x D a2 � b2; s D 2ab; and r D a2 C b2:

Now,

y2 D 2rs D 2.a2 C b2/.2ab/ D 4ab.a2 C b2/:

Since y is even, we have an equation in integers:

�y 2

�2 D ab.a2 C b2/: (1.6)

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16 Chapter 1 Early Number Theory

As a and b are relatively prime (no common factor d > 1), each pair from

the three factors on the right-hand side of Eq. (1.6) is relatively prime. Since

the left-hand side .y=2/2 is a square, each factor on the right is a square (Ex- ercise 2.12 on page 59). In other words, there are integers u, v, and w such

that

a D u2; b D v2; and a2 C b2 D w2:

And, since a and b are relatively prime, so, too, are u and v relatively prime.

Hence, we have

u4 C v4 D w2:

This is our “smaller” solution to Eq. (1.5), for

0 < w < w2 D a2 C b2 D r < r2 < r2 C s2 D z:

We can now repeat this process on .u; v; w/. By infinite descent, there is no

solution to Eq. (1.5).

Corollary 1.8 (Fermat’s Last Theorem for Exponent 4). There are no pos-

itive integers x; y; z with

x4 C y4 D z4:

Proof. If such a triple existed, we’d have

x4 C y4 D � z2 �2 ;

and that’s impossible, by Theorem 1.7.

Call an integer n � 2 good if there are no positive integers a; b; c with an C bn D cn. If n is good, then so is any multiple nk of it. Otherwise, there are positive integers r; s; t with rnk C snk D tnk , and this gives the contradic- tion an C bn D cn, where a D rk , b D sk , and c D tk . For example, Corol- lary 1.8 shows that that any positive integer of the form 4k is good. Since every

n � 2 is a product of primes, it follows that Fermat’s Last Theorem would be true if every odd prime is good.

Connections: Congruent Numbers

Fermat’s motivation for Theorem 1.7 came, not from a desire to prove there

are no non-trivial integer solutions to x4 C y4 D z4, but from a problem in the intersection of arithmetic and geometry. In more detail, suppose that

4 D 4.a; b; c/ is the right triangle arising from a Pythagorean triple .a; b; c/. Since 4 is a right triangle, the leg a is an altitude and the area of 4.a; b; c/ is 1 2 ab; since .a; b; c/ is a Pythagorean triple, the area is an integer (Exercise 1.8

on page 6). Tipping this statement on its head, we ask which integers are areas

of right triangles having integer side-lengths. Certainly 6 is, because it’s the

area of 4.3; 4; 5/. But 5 is not the area of such a triangle (Exercise 1.9 on page 7).

However, we claim that 5 is the area of a right triangle whose side-lengths

are rational numbers. Consider the Pythagorean triple .9; 40; 41/; its right tri-

angle 4 D 4.9; 40; 41/ has area 1 2 .9 � 40/ D 180. Now 180 D 36 � 5. Scaling

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1.2 Diophantus 17

the side-lengths of 4 by 1 6

scales the area by 1 36

, so that 4 � 3 2 ; 20 3 ; 41 6

� has area

180=36 D 5. So, the question arises: “Is every integer the area of a right triangle with

rational side-lengths?” Fermat showed that 1 and 2 are not, and his proof for 2

involved Eq. (1.5).

Theorem 1.9. There is no right triangle with rational side-lengths and area 2.

Proof. Suppose, on the contrary, that the rational numbers r; s; t are the lengths

of the sides of a right triangle with area 2. Then we have two equations:

r2 C s2 D t2 1 2 rs D 2:

Multiply the first equation by r2 to obtain

r4 C .rs/2 D .rt/2; so that (since rs D 4),

r4 C 24 D .rt/2:

Write the rational numbers r and t as fractions with the same denominator:

r D a=c and t D b=c. When we clear denominators, we get a4 C z4c4 D t2, an equation in integers x; y; z of the form

x4 C y4 D z2:

This is Eq. (1.5), and Theorem 1.7 says that this cannot occur.

So, not every positive integer is the area of a right triangle with rational

side-lengths.

Definition. A congruent number is a positive integer n that is the area of a

right triangle having rational side-lengths.

Theorem 1.9 says that 2 is not a congruent number. Using similar ideas,

Fermat showed that 1 is not a congruent number (Exercise 1.31 below).

One way to generate congruent numbers is to scale a Pythagorean triple using the largest perfect square that divides its area. For example, the area of We have already used this

method on the Pythag-

orean triple .9; 40; 41/

when we showed that 5 is

a congruent number.

4.7; 24; 25/ is 84 D 22 � 21. Since 4 D 22 is the largest perfect square in 84, scaling the sides by 2 will produce a triangle of area 21, so that 21 is the area of

4.7 2 ; 12; 25

2 / and, hence, 21 is a congruent number. More generally, we have

Proposition 1.10. Let .a; b; c/ be a Pythagorean triple. If its right triangle

4.a; b; c/ has area m2n, where n is squarefree, then n is a congruent number. Moreover, every squarefree congruent number is obtained in this way.

Proof. Since .a; b; c/ is a Pythagorean triple, 4 D 4.a; b; c/ is a right trian- gle. Now area.4/ D m2n D 1

2 ab, so that

area � 4. a

m ; b m ; c m / �

D 1 2

� a m b m

� D m2n

m2 D n;

and so n is a congruent number.

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18 Chapter 1 Early Number Theory

Conversely, if n is a square-free congruent number, then there are rational

numbers r , s, and t so that

r2 C s2 D t2 1 2 rs D n:

Clearing denominators, we find integers a, b, c, and m so that

a2 C b2 D c2 1 2 ab D m2n:

The first few congruent numbers are

5; 6; 7; 13; 14; 15; 20; 21; 22; 23:

In light of Exercise 1.33 on page 20, we now have a method for determin-

ing all congruent numbers: generate the areas of all Pythagorean triangles (we

know how to do that), and then divide out its largest perfect square factor: case closed.

Not quite. The trouble with this method is that you have no idea how many

triangle areas to calculate before (if ever) you get to an area of m2n for a

particular n. For some congruent numbers, it takes a long time. For example,

157 is a congruent number, but the smallest rational right triangle with area

157 has side lengths

This triangle was found

by Don Zagier, using

sophisticated techniques

investigating elliptic curves,

and using a substantial

amount of computer power

(see [19] for more details).

224403517704336969924557513090674863160948472041

8912332268928859588025535178967163570016480830 ;

6803294847826435051217540

411340519227716149383203 ;

411340519227716149383203

21666555693714761309610 :

A method for effectively determining whether or not an integer is a congru-

ent number is an unsolved problem (this problem is at least a thousand years

old, for historians have found it in manuscripts dating from the late tenth cen- tury). A detailed discussion of the Congruent Number Problem is in [19].A readable account of

the congruent prob-

lem, with more exam-

ples than we provide

here, can be found at

www.math.uconn.edu/

˜kconrad/blurbs/

How to Think About It. Proposition 1.10 shows that every squarefree con-

gruent number n is the area of a scaled Pythagorean triangle. But there might

be more than one Pythagorean triangle whose area has n as its squarefree

part. The search for more than one rational right triangle with the same area

leads to some fantastic calculations. For example, we saw that 5 is the area of

4 � 3 2 ; 20 3 ; 41 6

� , which comes from the Pythagorean triangle 4.9; 40; 41/whose

area is 5 � 62. But 5 is also the area of

4 � 1519

492 ; 4920

1519 ; 3344161

747348

� ;

and this comes from the Pythagorean triangle 4.2420640; 2307361; 3344161/ whose area is 5 � 7473482.

As usual, this isn’t magic; in Chapter 9, we’ll show how to find infinitely

many rational right triangles with the same congruent number as area.

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1.2 Diophantus 19

There’s a surprising connection between congruent numbers and 3-term

arithmetic sequences of perfect squares of rational numbers: positive rationals

s2 < t2 < u2 with u2 � t2 D t2 � s2, like 1; 25; 49 (common difference 24) and

961

36 ;

1681

36 ;

2401

36

(common difference 20). Note that 24 D 4 � 6 and 20 D 4 � 5. So, for these examples, at least, the common difference is 4 times a congruent number. This

suggests that something’s going on. One approach is due to Fibonacci.

Our two equations

a2 C b2 D c2

ab D 2n

might lead us to think that we could find a and b by finding their sum and product, for this would lead to a quadratic equation whose roots are a and b.

Well, we know ab, and

.aC b/2 D a2 C b2 C 2ab D c2 C 4n:

So, aCb D p c2 C 4n (why can we take the positive square root?), and hence

a and b are roots of the quadratic equation

x2 � p c2 C 4n x C 2n:

The quadratic formula gives us a and b:

a D p c2 C 4nC

p .c2 C 4n/� 4.2n/ 2

D p c2 C 4nC

p c2 � 4n

2

and

b D p c2 C 4n�

p .c2 C 4n/ � 4.2n/ 2

D p c2 C 4n�

p c2 � 4n

2 :

But we want a and b to be rational, so we want c2˙4n to be perfect squares. That produces an arithmetic sequence of three perfect squares:

c2 � 4n; c2; c2 C 4n:

There are details to settle, but that’s the gist of the proof of the following theo-

rem.

Theorem 1.11. An integer n is a congruent number if and only of there is a

3-term arithmetic sequence of perfect squares whose common difference is 4n.

Exercises

1.30 * Show that 1 is not a congruent number.

1.31 Show that there are no positive rational numbers x and y so that

x4 ˙ 1 D y2:

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20 Chapter 1 Early Number Theory

1.32 Show that if n is a congruent number and m is an integer, then m2n is also a

congruent number.

1.33 Show that there are no right triangles with rational side-lengths whose area is a

perfect square or twice a perfect square.

1.34 Show that 7 and 14 are congruent numbers.

1.35 Take It Further. Show that 13 is a congruent number.

1.36 * Prove Theorem 1.11.

1.3 Euclid

Euclid of Alexandria (ca. 325 BCE–ca.265 BCE) is one of the most prominent

mathematicians of antiquity. He is best known for The Elements, his treatise consisting of thirteen books: six on plane geometry, four on number theory,

and three on solid geometry. The Elements has been used for over two thou-

sand years, which must make Euclid the leading mathematics teacher of all

time. We do not know much about Euclid himself other than that he taught in

Alexandria, Egypt around 270 BCE. We quote from Sir Thomas Heath [16],

the great translator and commentator on The Elements.

It is most probable that Euclid received his mathematical training in

Athens from the students of PlatoI for most of the geometers who could have taught him were of that school ... Pappus says ... such was .Euclid’s/Pappus (ca. 290 CE–

ca. 350 CE), was one

of the last great classic

geometers.

scrupulous fairness and his exemplary kindliness towards all who could

advance mathematical science to however small an extentI .he was/ in no wise contentious and, though exact, yet no braggart.

Eight hundred years after Euclid, Proclus (412 CE–485 CE) wrote:

Not much younger than these .pupils of Plato/ is Euclid, who put to-

gether The Elements, collecting many of Eudoxus’s theorems, perfecting

many of Theaetetus’s, and also bringing to irrefragable demonstration

the things which were only somewhat loosely proved by his predecessors.

This man lived in the time of the first Ptolemy .323 BCE � 283 BCE/. For Archimedes, who came immediately after the first Ptolemy makes men-

tion of EuclidI and further they say that Ptolemy once asked him if there were a shorter way to study geometry than The Elements, to which he

replied that there was no royal road to geometry. He is therefore younger

than Plato’s circle, but older than Eratosthenes and ArchimedesI for these were contemporaries, as Eratosthenes somewhere says.

The Elements is remarkable for the clarity with which its theorems are stated

and proved. The standard of rigor was a goal (rarely achieved!) for the inven-

tors of calculus centuries later. As Heath writes in the preface to the second

edition of his translation [16] of The Elements,

. . . so long as mathematics is studied, mathematicians will find it neces-

sary and worthwhile to come back again and again . . . to the twenty-two-

centuries-old book which, notwithstanding its imperfections, remains the

greatest elementary textbook in mathematics that the world is privileged

to possess.

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1.3 Euclid 21

More than one thousand editions of The Elements have been published since

it was first printed in 1482. In the Encyclopedia Britannica, van der Waerden

wrote,

Almost from the time of its writing and lasting almost to the present,

The Elements has exerted a continuous and major influence on human

affairs. It was the primary source of geometric reasoning, theorems, and

methods at least until the advent of non-Euclidean geometry in the 19th century. It is sometimes said that, next to the Bible, The Elements may

be the most translated, published, and studied of all the books produced

in the Western world.

Greek Number Theory

In spite of the glowing reviews of The Elements, we must deviate a bit from

Euclid, for the Greeks, and Euclid in particular, recognized neither negative numbers nor zero.

Notation. The natural numbers is the set

N D f0; 1; 2; 3; : : :g:

The set of all integers, positive, negative, and 0, is denoted by The set of integers is denoted by Z because the

German word for numbers

is Zahlen.

Z D f˙n W n 2 Ng:

We are going to assume that the set N of natural numbers satisfies a certain

property—a generalized version of Infinite Descent.

Definition. The Least Integer Axiom (often called the Well-Ordering Axiom)

states that every nonempty collectionC of natural numbers contains a smallest Note that the set of positive rationals QC does not satisfy an analogous

property: the nonempty

subset fx 2QC W x2 > 2g contains no smallest

element.

element; that is, there is a number c0 2 C with c0 � c for all c 2 C .

This axiom is surely plausible. If 0 2 C , then c0 D 0. If 0 … C and 1 2 C , then c0 D 1. If 0; 1 … C and 2 2 C , then c0 D 2. Since C is not empty, you will eventually bump into C , and c0 is the first number you’ll meet.

We now define some familiar terms.

Definition. If a and b are integers, then a divides b, denoted by

a j b;

if there is an integer c with b D ca. We also say that a is a divisor (or a factor) of b, and that b is a multiple of a.

Example 1.12. Consider some special cases. Every number a divides itself, Note that 0 divides itself:

0 j 0 is true. Do not confuse the notation a j b, which is the relation “a is a divisor

of b” with a=b, which is a

number. In particular, we

are not saying that 0=0 is a

number.

for a D a � 1; similarly, 1 divides every number. Every number a divides 0: taking c D 0, we have 0 D a�0. On the other hand, if 0 divides b, then b D 0, for b D 0� c D 0. Note that 3 j 6, because 6 D 3� 2, but 3 − 5 (that is, 3 does not divide 5): even though 5 D 3 � 5

3 , the fraction 5

3 is not an integer. N

Lemma 1.13. If a and b are positive integers and a j b, then a � b.

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22 Chapter 1 Early Number Theory

Proof. There is a positive integer c with b D ca; note that 1 � c, for 1 isInequalities are discussed in Appendix A. 4. the smallest positive integer. Multiplying by the positive number a, we have

a � ac D b.

Every integer a has 1;�1; a;�a as divisors. A positive integer a ¤ 1 having only these divisors is called prime.

Definition. An integer a is prime if a � 2 and its only divisors are ˙1 and ˙a; if a � 2 has other divisors, then it is called composite.

The first few primes are 2, 3, 5, 7, 11, 13; : : : : We will soon see that there

are infinitely many primes.

The reason we do not consider 1 to be a prime is that theorems about primes

would then require special cases treating the behavior of 1. For example, we

will prove later that every positive integer a � 2 has exactly one factorization of the form a D p1p2 � � �pt , where p1 � p2 � � � � � pt are primes. This statement would be more complicated if we allowed 1 to be a prime.

Proposition 1.14. Every integer a � 2 is a product of primes.We allow products having only one factor; it’s okay to

say that a single prime is a

product of primes. Proof. Let C be the set of all natural numbers a � 2 that are not products of primes. If the proposition is false, then C is nonempty, and the Least Integer

Axiom gives a smallest such integer, say, c0. Since c0 2 C , it is not prime; hence, it factors, say, c0 D ab, where a; b ¤ 1. As a j c0, we have a � c0, by Lemma 1.13; but a ¤ c0, lest b D 1, so that a < c0. Therefore, a … C , for c0 is the smallest number in C , and so a is a product of primes: a D p1 � � �pm for m � 1. Similarly, b is a product of primes: b D q1 � � �qn. Therefore, c0 D ab D p1 � � �pmq1 � � �qn is a product of primes, a contradiction, and so C is empty.

Division and Remainders

Dividing an integer b by a positive integer a gives

b=a D q C r=a;

where q is an integer and 0 � r=a < 1. If we clear denominators, we get the statement b D qaC r which involves only integers. For example, 225 D 4C

2 5

becomes 22 D 4 � 5C 2.

b

a a a a r

b a r= 4 +

A A A A

B

Figure 1.8. Division Algorithm.

Euclid viewed division geometrically, as in Figure 1.8. Suppose B is a line

segment of length b, and thatA is a shorter segment of length a. Lay off copies

of A along B as long as possible. If there’s nothing left over, then a is a divisor

of b; if some segment of length, say r , is left over, then r is the remainder.

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1.3 Euclid 23

Theorem 1.15 (Division Algorithm). If a and b are positive integers, then

there are unique .i.e.,exactly one/ integers q .the quotient/ and r .the remain-

der/ with

b D qaC r and 0 � r < a:

Proof. We first prove that q and r exist; afterward, we’ll prove their unique- The hypothesis of Theo-

rem 1.15 can be weakened

to a; b 2 Z and a ¤ 0; the inequalities for the remain-

der now read 0 � r < jaj.

ness.

If b < a, set q D 0 and r D b. Now b D qa C r D 0 � a C b, while 0 � b < a. Hence, we may assume that b � a; that is, b�a � 0. Consider the sequence b > b�a > b�2a > b�3a > � � � . There must be an integer q � 1 with b�qa � 0 and b�.qC1/a < 0 (this is just Infinite Descent, described on page 12; in more down-to-earth language, there can be at most b steps before this sequence becomes negative). If we define r D b � qa, then b D qa C r . If a � b, the quotient q is

the largest multiple qa with

qa � b. This is very much the way young children are

taught to find the integer

quotient in division when a

and b are small.

We also have the desired inequalities. Clearly, 0 � r . If r D b � qa � a, then b � qa � a � 0; that is, b � .q C 1/a � 0, contradicting the definition of q.

Let’s prove uniqueness. If there are integers Q and R with b D Qa C R and 0 � R < a, then qa C r D b D Qa CR and

.Q � q/a D r �R:

If Q ¤ q, there is no loss in generality in assuming that Q > q; that is, 0 < Q � q. By Lemma 1.13, a � .Q � q/a D r � R. But r < a and R � 0 gives r �R < r < a. Therefore, a � .Q � q/a D r � R < a; that is, a < a, a contradiction. Hence Q D q. It follows that R D r , and we are done.

For example, there are only two possible remainders after dividing by 2,

namely, 0 and 1. An integer b is even if the remainder is 0; b is odd if the

remainder is 1. Thus, either b D 2q or b D 2qC 1. The equation b D qa C r is of no value at all without the restriction on

the remainder r . For example, the equations 1000 D 3 � 25C 925 and 1000 D 2 � 53C 894 are true and useless.

How to Think About It. We have been trained to regard the quotient q as

more important than the remainder; r is just the little bit left over. But our viewpoint now is just the reverse. Given a and b, the important question for us

is whether a is a divisor of b. The remainder is the obstruction: a j b if and only if r D 0. This will be a common strategy: to see whether a j b, use the Division Algorithm to get b D qaC r , and then try to show that r D 0.

The next result shows that there is no largest prime. The proof shows, given

any finite set of primes, that there always exists another one.

Corollary 1.16. There are infinitely many primes.

Proof. .Euclid/ Suppose, on the contrary, that there are only finitely many primes. If p1; p2; : : : ; pk is the complete list of all the primes, define

M D .p1 � � �pk/C 1:

By Proposition 1.14, M is a product of primes. But M has no prime divisor

pi , for dividing M by pi gives remainder 1 and not 0. For example, dividing

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24 Chapter 1 Early Number Theory

M by p1 givesM D p1.p2 � � �pk/C 1, so that the quotient and remainder are q D p2 � � �pk and r D 1; dividingM by p2 givesM D p2.p1p3 � � �pk/C 1, so that q D p1p3 � � �pk and r D 1; and so forth. The assumption that there are only finitely many primes leads to a contradiction, and so there must be an

infinite number of them.

Linear Combinations and Euclid’s Lemma

The greatest common divisor of two integers is a fundamental tool in studying

factorization.

Definition. A common divisor of integers a and b is an integer c with c j a and c j b. The greatest common divisor of a and b, denoted by gcd.a; b/ (or, more briefly, by .a; b/ ), is defined by

gcd.a; b/ D ( 0 if a D 0 D b the largest common divisor of a and b otherwise.

We saw, in Lemma 1.13, that if a and m are positive integers with a j m, then a � m. It follows, if at least one of a; b is not zero, that gcd’s exist: there are always common divisors (1 is always a common divisor), and there are

only finitely many positive common divisors � maxfjaj; jbjg.

Lemma 1.17. If p is a prime and b is an integer, then

gcd.p; b/ D ( p if p j b 1 otherwise.

Proof. A common divisor c of p and b is, in particular, a divisor of p. But the only positive divisors of p are p and 1, and so gcd.p; b/ D p or 1; it is p if p j b, and it is 1 otherwise.

If b � 0, then gcd.0; b/ D b (why?).

Definition. A linear combination of integers a and b is an integer of the form

saC tb;

where s; t 2 Z (the numbers s; t are allowed to be negative).

Example 1.18. The equation b D qa C r in the Division Algorithm displays b as a linear combination of a and r (for b D qaC 1 � r ). Note that 0 is a linear combination of any pair of integers: 0 D 0 �aC 0 �b. There are infinitely many linear combinations of 12 and 16, each of which is divisible by 4 (why?). It

follows that 5, for example, is not such a linear combination. N

The next result is one of the most useful properties of gcd’s. See Exercise 1.47 on

page 30. Theorem 1.19. If a and b are integers, then gcd.a; b/ is a linear combination

of a and b.

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1.3 Euclid 25

Proof. We may assume that at least one of a and b is not zero (otherwise,

the gcd is 0 and the result is obvious). Consider the set I of all the linear

combinations of a and b:

I D fsaC tb W s; t 2 Zg:

Both a and b are in I (take s D 1 and t D 0 or vice versa). It follows that The proof of Theorem 1.19 contains an idea that

will be used again, as in

Exercise 1.49 on page 30.

I contains positive integers (if a < 0, then �a is positive, and I contains �a D .�1/aC 0b); hence, the set C of all those positive integers lying in I is nonempty. By the Least Integer Axiom, C contains a smallest positive integer,

In other words, d is the

smallest positive linear

combination of a and b.

say, d ; we claim that d is the gcd.

Since d 2 I , it is a linear combination of a and b: there are integers s and t with

d D saC tb:

We’ll show that d is a common divisor by trying to divide each of a and b

by d . The Division Algorithm gives integers q and r with a D qd C r , where 0 � r < d . If r > 0, then

r D a � qd D a � q.saC tb/ D .1 � qs/aC .�qt/b 2 C;

contradicting d being the smallest element of C . Hence r D 0 and d j a; a similar argument shows that d j b.

Finally, if c is a common divisor of a and b, then Exercise 1.46 on page 29

shows that c divides every linear combination of a and b; in particular, c j d . By Lemma 1.13, we have c � d .

If d D gcd.a; b/ and if c is a common divisor of a and b, then c � d , by Lemma 1.13. The next corollary shows that more is true: c is a divisor of d ;

that is, c j d for every common divisor c.

Corollary 1.20. Let a and b be integers. A nonnegative common divisor d is In some treatments of num- ber theory, Corollary 1.20

is taken as the definition

of gcd. Later, we will want

to define greatest common

divisor in other algebraic

structures. It often will not

make sense to say that one

element of such a structure

is greater than another, but

it will make sense to say

that one element divides

another. Corollary 1.20

will allow us to extend the

notion of gcd.

their gcd if and only if c j d for every common divisor c of a and b.

Proof. Necessity (the implication )). We showed that any common divisor of a and b divides gcd.a; b/ at the end of the proof of Theorem 1.19.

Sufficiency (the implication (). Let d D gcd.a; b/, and let D � 0 be a common divisor of a and b with c j D for every common divisor c of a and b. Now D is a common divisor, so that d j D, by hypothesis; hence, d � D, by Lemma 1.13. But the definition of gcd (d is the greatest common divisor) givesD � d , and so D D d .

The next theorem is of great interest: not only is it very useful, but it also

characterizes prime numbers.

Theorem 1.21 (Euclid’s Lemma). If p is a prime and p j ab for integers a; b, then p j a or p j b. Conversely, if m � 2 is an integer such that m j ab always impliesm j a orm j b, thenm is a prime.

Proof. ()): Suppose that p j ab and that p − a; that is, p does not divide a; we must show that p j b. Since gcd.p; a/ D 1 (by Lemma 1.17), Theorem 1.19 gives integers s and t with 1 D sp C ta. Hence,

b D spb C tab:

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26 Chapter 1 Early Number Theory

Now p divides both expressions on the right, for p j ab, and so p j b, by Exercise 1.46 on page 29.The contrapositive of a

statement P implies

Q is “not Q” implies

“not P ”. For example, the

contrapositive of “If I live

in Chicago, then I live in

Illinois” is “If I don’t live

in Illinois, then I don’t live

in Chicago.” A statement

and its contrapositive are

either both true or both

false. Thus, to prove a

statement, it suffices to

prove its contrapositive.

((): We prove the contrapositive. If m is composite, then m D ab, where a < m and b < m. Now m j m D ab, but m − a and m − b, by Lemma 1.13. Thus, m divides a product but it divides neither factor.

To illustrate: 6 j 12 and 12 D 4� 3, but 6 − 4 and 6 − 3. Of course, 6 is not prime.

We will generalize Euclid’s Lemma in the next chapter. Theorem 2.8 says

that if p is a prime and p j a1 � � �an for integers a1; : : : ; an, where n � 2, then p j ai for some i .

Definition. Call integers a and b relatively prime if their gcd is 1.

Thus, a and b are relatively prime if their only common divisors are ˙1. For example, 2 and 3 are relatively prime, as are 8 and 15.

Here is a generalization of Euclid’s Lemma having the same proof.

Corollary 1.22. Let a, b, and c be integers. If c and a are relatively prime and

c j ab, then c j b.

Proof. Theorem 1.19 gives integers s and t with 1 D sc C ta. Hence, b D scb C tab. Now c divides both expressions on the right, for c j ab, and so c j b, by Exercise 1.46 on page 29.

How to Think About It.

We have just seen one reason why it is important to know proofs: Corol-

lary 1.22 does not follow from the statement of Euclid’s Lemma, but it does follow from its proof. See Exercise 1.54 on page 34 for another example of

this.

Proposition 1.23. Let a and b be integers.

(i) gcd.a; b/ D 1 .that is, a and b are relatively prime/ if and only if 1 is a linear combination of a and b.

(ii) If d D gcd.a; b/ ¤ 0, then the integers a=d and b=d are relatively prime.

Proof. (i) By Theorem 1.19, the gcd d is a linear combination; here, d D 1. Conversely, if 1 D sa C tb and c is a common divisor of a and b, then c j 1, by Exercise 1.46 on page 29. Hence, c D ˙1.

(ii) There are integers s and t with d D sa C tb. Divide both sides by d D gcd.a; b/:

1 D s � a d

� C t

� b

d

� D � s d

� aC

� t

d

� b:

Since d is a common divisor, both a=d and b=d are integers, and part (i)

applies.

Definition. An expression a=b for a rational number (where a and b are inte-

gers and b ¤ 0) is in lowest terms if a and b are relatively prime.

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1.3 Euclid 27

Proposition 1.24. Every nonzero rational number a=b has an expression in

lowest terms.

Proof. If d D gcd.a; b/, then a D a0d , b D b0d , and a b

D a 0d

b0d D a

0

b0 . But

a0 D a d

and b0 D b d

, so gcd.a0; b0/ D 1 by Lemma 1.23.

We can now complete our discussion of Pythagorean triples.

Definition. A Pythagorean triple .a; b; c/ is primitive if a; b; c have no com-

mon divisor d � 2; that is, there is no integer d � 2 which divides each of a; b, and c.

Theorem 1.25 (Diophantus). Every Pythagorean triple .a; b; c/ is similar to

a unique primitive Pythagorean triple.

Proof. We show first that .a; b; c/ is similar to a primitive Pythagorean triple.

If d is a common divisor of a; b; c, then a D du, b D dv, and c D dz, and .u; v; z/ is a Pythagorean triple similar to .a; b; c/ (why?). If d is the largest

common divisor of a; b; c, we claim that .u; v; z/ is primitive. Otherwise, there

is an integer e � 2 with u D eu0, v D ev0, and z D ez0; hence, a D du D deu0, b D dv D dev0, and c D dz D dez0. Thus, de > d is a common divisor of a; b; c, contradicting d being the largest such.

To prove uniqueness, suppose that .a; b; c/ is similar to two primitive Py-

thagorean triples, say .u; v; z/ and .r; s; t/. It follows that the right triangles

4.u; v; z/ and 4.r; s; t/ are similar, and so their sides are proportional, so there is some positive number h with

u D hr; v D hs; and z D ht:

Since the side lengths are integers, h is rational, say h D m=`, and we may assume that it is in lowest terms; that is, gcd.m; `/ D 1. Cross multiply:

mu D `r; mv D `s and mz D `t:

By Corollary 1.22, ` is a common divisor of u, v, and z and m is a common

divisor of r , s, and t . Since both .u; v; z/ and .r; s; t/ are primitive, ` D 1 D m, and so .u; v; z/ D .r; s; t/.

This next result is significant in the history of mathematics.

Proposition 1.26. There is no rational number a=b whose square is 2.

Proof. Suppose, on the contrary, that .a=b/2 D 2. We may assume that a=b is An indirect proof or proof by contradiction has the

following structure. We

assume that the desired

statement is false and

reach a contradiction. We

conclude that the original

statement must be true.

in lowest terms; that is, gcd.a; b/ D 1. Since a2 D 2b2, Euclid’s Lemma gives 2 j a, and so 2m D a. Hence, 4m2 D a2 D 2b2, and 2m2 D b2. Euclid’s Lemma now gives 2 j b, contradicting gcd.a; b/ D 1.

It follows that the legs of a Pythagorean triple .a; b; c/ cannot be equal, for

if a D b, then a2 C a2 D c2, which implies that 2 D .c=a/2.

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28 Chapter 1 Early Number Theory

Proposition 1.26 is often stated as “ p 2 is irrational,” which is a stronger

statement than what we’ve just proved. We can assert that p 2 is irrational only

if we further assume that there exists a number u with u2 D 2. Our proof can be made more elementary; we need assume only that at least

one of a; b is odd. Also, see Exercise 1.75 on page 41.

Historical Note. The ancient Greeks defined number to mean “positive inte- ger.” Rationals were not viewed as numbers but, rather, as ways of comparing

two lengths. They called two segments of lengths a and b commensurable

if there is a third segment of length c with a D mc and b D nc for posi- tive integers m and n. That

p 2 is irrational was a shock to the Pythagoreans

(ca. 500 BCE); given a square with sides of length 1, its diagonal and side are not commensurable; that is,

p 2 cannot be defined in terms of numbers

(positive integers) alone. Thus, there is no numerical solution to the equation

x2 D 2, but there is a geometric solution.To bridge the gap between numbers and geometric

magnitudes, Eudoxus

(408 BCE–355 BCE) intro-

duced the sophisticated

notion of proportions (this

idea, discussed in The Ele-

ments, is equivalent to our

contemporary definition of

real numbers).

By the time of Euclid, around 270 BCE, this problem had been resolved by

splitting mathematics into two disciplines: number theory and geometry.

In ancient Greece, algebra as we know it did not really exist. Euclid and the Greek mathematicians did geometric algebra. For simple ideas, e.g.,

.a C b/2 D a2 C 2ab C b2 or completing the square, geometry clarifies al- gebraic formulas (for example, see the right-hand part of Figure 1.2 on page 2

without the dashed lines). For more difficult ideas, say equations of higher de-

gree, the geometric figures involved are very complicated, so that geometry is no longer clarifying. As van der Waerden writes in [34], p. 266,

one has to be a mathematician of genius, thoroughly versed in trans-

forming proportions with the aid of geometric figures, to obtain results

by this extremely cumbersome method. Anyone can use our algebraic

notation, but only a gifted mathematician can deal with the Greek theory

of proportions and with geometric algebra.

The problem of defining number has arisen several times since the classical

Greek era. Mathematicians had to deal with negative numbers and with com-

plex numbers in the 1500s after the discovery of the Cubic Formula, because

that formula often gives real roots of a cubic polynomial, even integer roots, in

unrecognizable form (see Chapter 3). The definition of real numbers generally accepted today dates from the late 1800s. But there are echos of ancient Athens

in our time. Kronecker (1823–1891) wrote,

Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Men-

schenwerk. (God created the integers; everything else is the work of

Man.)

Even today some logicians argue for a new definition of number.

Exercises

1.37 True or false, with reasons. Of course, it is important to get the right answer, but

most attention should be paid to your reasoning.

(i) 6 j 2. Answer: False. (ii) 2 j 6. Answer: True. (iii) 6 j 0. Answer: True. (iv) 0 j 6. Answer: False. (v) 0 j 0. Answer: True.

HELIANG GAO
高亮

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1.3 Euclid 29

1.38 True or false, with reasons.

(i) gcd.n;n C 1/ D 1 for every natural number n. Answer: True. (ii) gcd.n;n C 2/ D 2 for every natural number n. Answer: False.

(iii) 113 is a sum of distinct powers of 2. Answer: True.

(iv) If a and b are natural numbers, there there are natural numbers s and t with

gcd.a; b/ D sa C tb. Answer: False. (v) If an integer m is a divisor of a product of integers ab, then m is a divisor of

either a or b (or both). Answer: False.

1.39 Prove, or disprove and salvage if possible. “Disprove” here means “give a concrete counterex-

ample.” “Salvage” means

“add a hypothesis to make

it true.”

(i) gcd.0; b/ D b (ii) gcd.a2; b2/ D gcd.a; b/2

(iii) gcd.a; b/ D gcd.a; b C ka/ for all k 2 Z (iv) gcd.a; a/ D a (v) gcd.a; b/ D gcd.b; a/

(vi) gcd.a; 1/ D 1 (vii) gcd.a; b/ D � gcd.�a; b/

(viii) gcd.a; 2b/ D 2 gcd.a; b/

1.40 * If x is a real number, let bxc denote the greatest integer n with n � x. (For example, 3 D b�c and 5 D b5c.) If q is the quotient in Theorem 1.15, show that q D bb=ac.

1.41 *

(i) Given integers a and b (possibly negative) with a ¤ 0, prove that there exist unique integers q and r with b D qa C r and 0 � r < jaj.

Hint: Use the portion of the Division Algorithm that has already been proved.

(ii) If b and a are positive integers, do b and �b have the same remainder after dividing by a? Answer: No.

1.42 For each of the following pairs a; b, find the largest nonnegative integer n with

n � b=a < n C 1. (i) a D 4 and b D 5. Answer: n D 1.

(ii) a D 5 and b D 4. Answer: n D 0. (iii) a D 16 and b D 36. Answer: n D 2. (iv) a D 36 and b D 124. Answer: n D 3. (v) a D 124 and b D 1028. Answer: n D 7.

1.43 Let p1; p2; p3; : : : be the list of the primes in ascending order: p1 D 2, p2 D 3, p3 D 5, and so forth. Define fk D 1 C p1p2 � � � pk for k � 1. Find the smallest k for which fk is not a prime.

Hint: 19 j f7, but 7 is not the smallest k.

1.44 What can you say about two integers a and b with the property that a j b and b j a? What if both a and b are positive?

1.45 * Show that if a is positive and a j b, then gcd.a; b/ D a. Why do we assume that a is positive?

1.46 *.Two Out of Three/: Suppose that m, n, and q are integers and m D n C q. If c is an integer that divides any two of m; n; q, show that c divides the third one as

well.

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30 Chapter 1 Early Number Theory

1.47 *

(i) For each a and b, give the smallest positive integer d that can be written as

sa C tb for integers s and t :Allow for positive and negative values of s and t.

� a D 12 and b D 16. Answer: d D 4. � a D 12 and b D 17. Answer: d D 1. � a D 12 and b D 36. Answer: d D 12. � a D 0 and b D 4. Answer: d D 4. � a D 4 and b D 16. Answer: d D 4. � a D 16 and b D 36. Answer: d D 4. � a D 36 and b D 124. Answer: d D 4. � a D 124 and b D 1028. Answer: d D 4.

(ii) How is “smallest positive integer d expressible as saC tb” related to a and b in each case? Is d a divisor of both a and b?

1.48 * Show that the set of all linear combinations of two integers is precisely the set

of all multiples of their gcd.

1.49 * Let I be a subset of Z such that

(i) 0 2 I (ii) if a; b 2 I , then a � b 2 I

(iii) if a 2 I and q 2 Z, then qa 2 I . Prove that there is a nonnegative integer d 2 I with I consisting precisely of all the multiples of d .

1.50 How might one define the gcd.a;b; c/ of three integers? When applied to a prim-

itive Pythagorean triple .a; b; c/, your definition should say that gcd.a;b; c/ D 1.

Euclidean Algorithm

Our discussion of gcd’s is incomplete. What is gcd.12327; 2409/? To ask the question another way, is the expression 2409/12327 in lowest terms? The next

result enables us to compute gcd’s efficiently. We first prove another lemma

from Greek times.

Lemma 1.27. Let a and b be integers.

(i) If b D qa C r , then gcd.a; b/ D gcd.r; a/. (ii) If b � a, then gcd.a; b/ D gcd.b � a; a/.

Proof. (i) In light of Corollary 1.20, it suffices to show that an integer c is a

common divisor of a and b if and only if it is a common divisor of a and r . Since b D qaC r , this follows from Exercise 1.46 on page 29.

(ii) This follows from part (i) because b D 1 � aC .b � a/.

The hypothesis b � a in part (ii) of Lemma 1.27 is not necessary; it is there only to put you in the mood to accept the next example showing a method

the Greeks probably used to compute gcd’s. This method of computation is

nowadays called the Euclidean Algorithm; it is Theorem 1.29.

Example 1.28. In this example, we will abbreviate gcd.b; a/ to .b; a/. Com-

puting .b; a/ is simple when a and b are small. If b � a, then Lemma 1.27

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1.3 Euclid 31

allows us to replace .b; a/ by .b � a; a/; indeed, we can continue replacing numbers, .b�2a; a/, .b�3a; a/; : : : ; .b�qa; a/ as long as b�qa > 0. Since the natural numbers b�a; b�2a; : : : ; b�qa are strictly decreasing, the Least Integer Axiom (or Infinite Descent) says that they must reach a smallest such

integer: r D b�qa; that is, 0 < r < a. Now .b; a/ D .r; a/. (We see the proof of the Division Algorithm in this discussion.) Since .r; a/ D .a; r/ and a > r , they could continue replacing numbers: .a; r/ D .a�r; r/ D .a�2r; r/ D � � � (remember that the Greeks did not recognize negative numbers, so it was nat- ural for them to reverse direction). This process eventually ends, computing

gcd’s; we call it the Euclidean Algorithm. The Greek term for this method is

antanairesis, a free translation of which is “back and forth subtraction.” Let us

implement this idea before we state and prove the Euclidean Algorithm.

Antanairesis computes gcd.326; 78/ as follows:

.326; 78/ D .248; 78/ D .170; 78/ D .92; 78/ D .14; 78/:

So far, we have been subtracting 78 from the other larger numbers. At this

point, we now start subtracting 14 (this is the reciprocal, direction-changing, aspect of antanairesis), for 78 > 14.

.78; 14/ D .64; 14/ D .50; 14/ D .36; 14/ D .22; 14/ D .8; 14/:

Again we change direction:

.14; 8/ D .6; 8/:

Change direction once again to get .8; 6/ D .2; 6/, and change direction one last time to get

.6; 2/ D .4; 2/ D .2; 2/ D .0; 2/ D 2:

Thus, gcd .326; 78/ D 2. The Division Algorithm and Lemma 1.27(i) give a more efficient way of

performing antanairesis. There are four subtractions in the passage from

.326; 78/ to .14; 78/; the Division Algorithm expresses this as

326 D 4 � 78C 14:

There are then five subtractions in the passage from .78; 14/ to .8; 14/; the

Division Algorithm expresses this as

78 D 5 � 14C 8:

There is one subtraction in the passage from .14; 8/ to .6; 8/:

14 D 1 � 8C 6:

There is one subtraction in the passage from .8; 6/ to .2; 6/:

8 D 1 � 6C 2;

and there are three subtractions from .6; 2/ to .0; 2/ D 2:

6 D 3 � 2: N

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32 Chapter 1 Early Number Theory

Theorem 1.29 (Euclidean Algorithm I). If a and b are positive integers,

there is an algorithm computing gcd.a; b/.

Proof. Let us set b D r0 and a D r1, so that the equation b D qa C r reads r0 D q1aC r2. There are integers qi and positive integers ri such that

The beginning of the proof

of the theorem gives the

algorithm.

b D r0 D q1a C r2; r2 < a a D r1 D q2r2 C r3; r3 < r2

r2 D q3r3 C r4; r4 < r3 :::

:::

rn�3 D qn�2rn�2 C rn�1; rn�1 < rn�2 rn�2 D qn�1rn�1 C rn; rn < rn�1 rn�1 D qnrn

(remember that all qj and rj are explicitly known from the Division Algo-

rithm). There is a last remainder: the procedure stops (by Infinite Descent!)

because the remainders form a strictly decreasing sequence of nonnegative in-

tegers (indeed, the number of steps needed is less than a).

Lamé (1795–1870) proved

that the number of steps

in the Euclidean Algorithm

cannot exceed 5 times

the number of digits in the

smaller number (see [26],

p. 49).

We now show that the last remainder rn is the gcd.

b D q1aC r2 ) gcd.a; b/ D gcd.a; r2/ a D q2r2 C r3 ) gcd.a; r2/ D gcd.r2; r3/ r2 D q3r3 C r4 ) gcd.r2; r3/ D gcd.r3; r4/

:::

rn�2 D qn�1rn�1 C rn ) gcd.rn�2; rn�1/ D gcd.rn�1; rn/ rn�1 D qnrn ) gcd.rn�1; rn/ D rn:

All the implications except the last follow from Lemma 1.27. The last one

follows from Exercise 1.45 on page 29.

Let’s rewrite the previous example in the notation of the proof of Theo- rem 1.29. The passage from one line to the line below it involves moving the

boldface numbers “southwest.”

326 D 4 � 78 C 14 (1.7) 78 D 5 � 14 C 8 (1.8) 14 D 1 � 8 C 6 (1.9) 8 D 1 � 6 C 2 (1.10) 6 D 3 � 2:

The Euclidean Algorithm also allows us to find a pair of integers s and t

expressing the gcd as a linear combination.

Theorem 1.30 (Euclidean Algorithm II). If a and b are positive integers,

there is an algorithm computing a pair of integers s and t with gcd.a; b/ D saC tb.

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1.3 Euclid 33

Proof. It suffices to show, given equations

b D qaC r a D q0r C r 0

r D q00r 0 C r 00;

how to write r 00 as a linear combination of b and a (why?). Start at the bottom, and write

r 00 D r � q00r 0:

Now rewrite the middle equation as r 0 D a � q0r , and substitute:

r 00 D r � q00r 0 D r � q00.a � q0r/ D .1 � q00q0/r � q00a:

Now rewrite the top equation as r D b � qa, and substitute:

r 00 D .1 � q00q0/r � q00a D .1 � q00q0/.b � qa/ � q00a:

Thus, r 00 is a linear combination of b and a.

We use the equations to find coefficients s and t expressing 2 as a linear

combination of 326 and 78. Work from the bottom up.

2 D 8 � 1 � 6 by Eq. (1.10) D 8 � 1 � .14 � 1 � 8/ by Eq. (1.9) D 2 � 8 � 1 � 14 D 2 � .78 � 5 � 14/ � 1 � 14 by Eq. (1.8) D 2 � 78 � 11 � 14 D 2 � 78 � 11 � .326 � 4 � 78/ by Eq. (1.7) D 46 � 78 � 11 � 326:

Thus, s D 46 and t D �11.

How to Think About It. The algorithm produces one pair of coefficients that

works. However, it’s not the only pair. For example, consider gcd.2; 3/ D 1. A moment’s thought gives s D �1 and t D 1; but another moment’s thought gives s D 2 and t D �1 (see Exercise 1.57 on page 35). However, the Eu- clidean Algorithm always produces a specific pair of coefficients; assuming

that no mistakes in arithmetic are made, two people using the algorithm al-

ways come up with the same s and t .

Students usually encounter greatest common divisors in elementary school, sometimes as early as the fifth grade, when they learn how to add fractions

and put the sum in lowest terms. As we have seen, putting a fraction in lowest Putting a fraction in lowest terms is not always wise.

For example,

2 3 C 1

5 D 10

15 C 3

15

D 13 15

:

terms involves the gcd of numerator and denominator. The preferred method

of finding gcd’s in early grades involves prime factorization, for if integers a

and b are small, then it is easy to factor them into primes: after several cancel- lations, the expression a=b is in lowest terms. Pedagogically, this may be the

right choice, but finding gcd’s using prime factorization is practical only when

numbers are small; can you put the fraction 167291=223377 in lowest terms

using prime factorization?

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34 Chapter 1 Early Number Theory

How to Think About It. In calculating gcd’s with the Euclidean Algorithm,

many students get confused keeping track of the divisors and remainders. We

illustrate one way to organize the steps that has been effective with high school

students. Arrange the steps computing gcd.124; 1028/ as on the left:

8

124

� 1028

992 3

36

� 124

108 2

16

� 36

32 4

4

� 16

16

0

4 D 36 � 2 � 16 &

D 36 � 2 � .124 � 3 � 36/ D �2 � 124C 7 � 36

& D �2 � 124C 7 � .1028 � 8 � 124/ D 7 � 1028� 58 � 124

The last nonzero remainder is the gcd, so gcd.124; 1028/ D 4. This arrange- ment can be used to read off coefficients s and t so that 4 D 124sC128t . Start at the next to last division and solve for each remainder.

Exercises

1.51 If a and b are positive integers, then gcd.a; b/ D sa C tb. Prove that either s or t is negative.

1.52 * Use Infinite Descent to prove that every positive integer a has a factorization

a D 2km, where k � 0 and m is odd. Now prove that p

2 is irrational using this

fact instead of Euclid’s Lemma.

1.53 Prove that if n is squarefree (i.e., n > 1 and n is not divisible by the square of

any prime), then there is no rational number x with x2 D n. Hint: Adapt the proof of Proposition 1.26.

1.54 * Assuming there is a real number x with x3 D 2, prove that x is irrational.

1.55 (i) Find d D gcd.326; 78/, find integers s and t with d D 326s C 78t , and put the expression 326/78 in lowest terms.

Answer: d D 2, s D �11, t D 46, and 163 39

.

(ii) Find d D gcd.12327;2409/, find integers s and t with d D 12327s C 2409t , and put the expression 2409/12327 in lowest terms.

Answer: d D 3, s D 299, t D �1530, and 8034109 . (iii) Find d D gcd.7563;526/, and express d as a linear combination of 7563 and

526.

Answer: d D 1, s D �37, t D 532. (iv) Find d D gcd.73122; 7404621/ and express d as a linear combination of

73122 and 7404621.

Answer: d D 21, s D 34531, t D �7404621.

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1.3 Euclid 35

1.56 * Prove that if gcd.r; m/ D 1 and gcd.r 0 ; m/ D 1, then gcd.rr 0 ; m/ D 1. Con- clude that if both r and r 0 are relatively prime to m, then so is their product rr 0 .

Hint: If ar C bm D 1 and sr 0 C tm D 1, consider .ar C bm/.sr 0 C tm/.

1.57 * Let a; b, and d be integers. If d D sa C tb, where s and t are integers, find infinitely many pairs of integers .sk ; tk/ with d D ska C tkb.

Hint: If 2s C 3t D 1, then 2.s C 3/ C 3.t � 2/ D 1.

1.58 * If a and b are relatively prime and each divides an integer n, prove that their

product ab also divides n.

Hint: Use Corollary 1.22.

1.59 If m > 0, prove that m gcd.b; c/ D gcd.mb; mc/. (We must assume that m > 0 lest m gcd.b; c/ be negative.)

Hint: Show that if k is a common divisor of mb and mc, then k j m gcd.b; c/.

1.60 Write d D gcd.a; b/ as a linear combination of a and b. (i) a D 4 and b D 16.

Answer: d D 4 D 5 � 4 C .�1/ � 16 (or, 4 D 1 � 4 C 0 � 16). (ii) a D 16 and b D 36.

Answer: d D 4 D .�2/ � 16 C 1 � 36. (iii) a D 36 and b D 124.

Answer: d D 4 D 7 � 36 C .�2/ � 124. (iv) a D 124 and b D 1028.

Answer: d D 4 D .�58/ � 124 C 7 � 1028.

1.61 Given integers a, b, and c with c j a and c j b, prove that c divides every linear combination sa C tb.

1.62 Is anything wrong with this calculation? Explain your answer.

4

7 �

37 28

9

1.63 Given integers b, c, d , and e satisfying b D 7c C 2 and d D 7e C 4, (i) What’s the remainder when b C d is divided by 7?

Answer: 6.

(ii) What’s the remainder when bd is divided by 7?

Answer: 1.

(iii) Explain your answers.

1.64 A lattice point is a point .x; y/ in the plane with both x and y integers.

(i) Which lattice points are on the line whose equation is 4x C 6y D 24? (ii) Which lattice points are on the line whose equation is 3x C 6y D 24?

(iii) Find a line whose equation has integer coordinates but that never passes

through a lattice point.

(iv) Explain how to tell whether the line with equation y D axCb contains lattice points.

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36 Chapter 1 Early Number Theory

1.65 Consider the calculation of gcd.124; 1028/ on page 34. Show that the integer pairs

.124; 1028/; .36; 124/; .16; 36/; .4; 16/; .0; 4/

have the same greatest common divisor.

1.66 Most calculators have functions computing quotients and remainders. Let r.b; a/

denote the remainder when b is divided by a, and let q.b; a/ denote the quotient.

Find r.b; a/ and q.b; a/ if

(i) a D 12, b D 16. Answer: q.16;12/ D 1; r.16; 12/ D 4. (ii) a D 16, b D 12. Answer: q.12; 16/ D 0; r.12; 16/ D 12.

(iii) a D 124, b D 1028. Answer: q.1028;124/ D 8; r.1028; 24/ D 36. (iv) a D 78, b D 326. Answer: q.326;78/ D 4; r.326; 78/ D 14.

1.67 Preview. Using the notation in Exercise 1.66, consider the pair of recursivelyThe functions can be pro- grammed into a calculator. defined functions on N:

s.a; b/ D (

0 a D 0 t .r.b; a/; a/ � q.b; a/ � s .r.b; a/; a/ a > 0

t.a; b/ D (

1 a D 0 s .r.b; a/; a/ a > 0:

Find s.a; b/ and t .a; b/ if

(i) a D 124; b D 1028. Answer: s.124; 1028/ D �58; t .124; 1028/ D 7. (ii) a D 36; b D 124. Answer: s.36; 124/ D 7; t .36;124/ D �2.

(iii) a D 78; b D 326. Answer: s.78; 326/ D 46; t .78; 326/ D �11. (iv) a D 12327;b D 2409. Answer: s.1237;2409/ D 1186; t .1237;2409/ D

�609. (v) a D 7563; b D 526. Answer: s.7563;526/ D �37; t .7563;526/ D 532.

(vi) a D 167291;b D 223377. Answer: s.167291;223377/ D �4, t .167291;223377/ D 3.

1.4 Nine Fundamental Properties

We now focus on a small number (nine) of properties of arithmetic, for it turns

out that many of the usual rules follow from them. This obviously simplifies

things, making explicit what we are allowed to assume. But we have an ulterior

motive. The properties will eventually be treated as axioms that will describe

addition and multiplication in other systems, such as complex numbers, poly- nomials, and modular arithmetic; these systems lead naturally to their common

generalization, commutative rings.

Notation. The set of all rational numbers is denoted by Q, and the set of all

real numbers is denoted by R.

We begin by stating some basic properties of real numbers (of course, inte-

gers and rationals are special cases). These properties undergird a great deal of high school algebra; they are essential for the rest of this book and, indeed, for

abstract algebra.

Addition and multiplication are functions R � R ! R, namely, .a; b/ 7!Functions are discussed in Appendix A.1. a C b and .a; b/ 7! ab. The Laws of Substitution say that if a; a0; b; b0 are

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1.4 Nine Fundamental Properties 37

real numbers with a D a0 and b D b0, then

a C b D a0 C b0 and ab D a0b0:

The Laws of Substitution are used extensively (usually tacitly) when solving

equations or transforming expressions, and they merely say that addition and

multiplication are single-valued. For example, since �5C 5 D 0, we have

.�5C 5/ � .�1/ D 0 � .�1/ D 0:

Here are the properties of addition we are emphasizing.

Addition W For all real numbers a; b, and c,

(i) Commutativity: aC b D b C a, (ii) 0C a D a,

(iii) there is a number �a, called the negative of a (or its additive inverse), with �a C a D 0,

(iv) Associativity: a C .b C c/ D .a C b/C c.

Let’s say a bit more about associativity. Addition is defined as an operation

performed on two numbers at a time, but it’s often necessary to add three or

more numbers. Associativity says that, when evaluating, say 2 C 5 C 3, we Given associativity for the sum or product of

3 numbers, generalized

associativity is also true:

we don’t need parentheses

for the sum or product of

n � 3 numbers. A proof is in Appendix A.5.

can first add 2 and 5, giving 7 C 3 D 10, or we can first add 5 and 3, giving 2 C 8 D 10. In other words, we don’t need parentheses: writing 2 C 5 C 3 is unambiguous because .2 C 5/ C 3 D 2 C .5 C 3/. This is not the case with subtraction. What is 8� 3� 2? If we first subtract 8 � 3, then the answer is 5 � 2 D 3. However, if we evaluate 8 � .3 � 2/ D 8 � 1, we obtain a different answer. Thus, subtraction R2 ! R, defined by .a; b/ 7! a�b, is not associative, and we do need parentheses for it.

Here are the properties of multiplication that we are emphasizing; note that

they are, formally, the same as those for addition: just replace “plus” by “times”

(we usually denote the product of numbers a and b by ab, although we will

occasionally write a � b or a � b).

Multiplication W For all real numbers a; b, and c,

(i) Commutativity: ab D ba, (ii) 1 � a D a,

(iii) If a ¤ 0, there is a number a�1, called its (multiplicative) inverse (or its reciprocal) with a � a�1 D 1, Why do we assume that

a ¤ 0? Read on. (iv) Associativity: a.bc/ D .ab/c.

Finally, we highlight a property involving both addition and multiplication.

Distributivity: a.b C c/ D ac C ab:

Reading from left to right, distributivity says that we can “multiply a through;”

reading from right to left, distributivity says that we can “factor a out.”

Aside from the two Laws of Substitution, one for addition and one for mul-

tiplication, we have now listed nine properties of addition and multiplication.

Subtraction and division are defined as follows.

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38 Chapter 1 Early Number Theory

Definition. If a and b are numbers, define subtraction by

b � a D b C .�a/;

where �a is the negative of a; that is, �a is the number which, when added to a, gives 0.

Quotient (or division) is defined similarly.

Definition. If a and b are numbers with b ¤ 0, then the quotient of a by b is ab�1, where b�1 is the number which, when multiplied by b, gives 1. We often denote ab�1 by a=b.

The word quotient is used here in a different way than in the Division Algo-

rithm, where it is bb=ac, the integer part of b=a (see Exercise 1.40 on page 29).

How to Think About It. Almost all the properties just listed for the set R

of real numbers also hold for the set Z of integers—these properties are “in-

herited” from R because integers are real numbers. The only property that Z

doesn’t inherit is the existence of multiplicative inverses. While every nonzero integer does have an inverse in R, it may not be an integer; in fact, the only

nonzero integers whose inverses also lie in Z are 1 and �1. There are other familiar algebraic systems that are more like Z than R in the sense that multi-

plicative inverses may not exist in the system. For example, all polynomials in

one variable with rational coefficients form such a system, but the multiplica-

tive inverse 1=x of x is not a polynomial.

Other familiar “rules” of arithmetic are easy consequences of these fundamen-

tal ones. Here are some of them.

Proposition 1.31. For every number a, we have 0 � a D 0.

Proof. By Addition Rule (ii), we have 0C 0 D 0. Therefore,

0 � a D .0C 0/a D .0 � a/C .0 � a/:

Now subtract 0 � a from both sides to obtain 0 D 0 � a.

What is the meaning of division by 5? When we say that 20 � 5 D 4, we mean that 20=5 is a number (namely 4), and that .20=5/ � 5 D 20. Dividing is the “opposite” of multiplying: dividing by 5 undoes multiplying by 5. This agrees with our formal definition. The inverse of 5 is 5�1, and 20 � 5�1 D 20=5 D 4.Multiplication by 5 is a

bijection, and we are

saying that division by 5

is its inverse function.

See Example A.10 in

Appendix A. 1.

Can we divide by zero? If so, then 1=0would be a number with 0� .1=0/ D 1. But we have just seen that 0 � a D 0 for any number a. In particular, 0 � .1=0/ D 0, giving the contradiction 1 D 0. It follows that 1=0 D 0�1 is not a number; we cannot divide by 0.

Here is another familiar consequence of the nine fundamental properties.

Proposition 1.32. For a number a, we have

.�a/ � .�1/ D a:

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1.4 Nine Fundamental Properties 39

In particular,

.�1/ � .�1/ D 1:

Proof. The distributive law gives The Law of Substitution allows us to replace 0 by

�1C 1.0 D 0 � .�a/ D .�1 C 1/.�a/ D .�1/ � .�a/C .�a/:

Now, add a to both sides to get a D .�1/.�a/.

How to Think About It.

Even though its proof is very simple, Proposition 1.32 is often presented to

high school students as something mysterious and almost magical. We can only guess at a reason. From Euclid’s time until the 1500s, numbers were always

positive; either negative numbers were not recognized at all or, if they did

appear, they were regarded with suspicion, as not being bona fide (the complex

numbers, which came on the scene around the same time, were also suspected

of witchcraft). In the proof of Proposition 1.32, we treated negative numbers

without prejudice, and we assumed that they obey the same elementary rules as positive numbers do. And we have reaped a reward for clear thinking.

Addition Rule (iii) states that every real number has a negative, an additive

inverse. Can a number a have more than one negative? Intuition tells us no,

and this can be proved using the nine fundamental properties.

Proposition 1.33. Negatives in R are uniqueI that is, for a 2 R, there is exactly one number b in R with b C a D 0.

Multiplicative inverses of nonzero real numbers are uniqueI that is, for nonzero c 2 R, there is exactly one real number d with cd D 1.

Proof. Suppose b is a number with aC b D 0. Add �a to both sides:

�aC .a C b/ D �a:

We can now use associativity to calculate, like this.

�aC .a C b/ D �a .�a C a/C b D �a

0C b D �a b D �a:

This argument can be adapted to prove uniqueness of multiplicative in- verses; merely replace + by � and “additive inverse” by “multiplicative in- verse.”

Uniqueness theorems like Proposition 1.33 are useful because they show

that certain objects are characterized by their behavior. For example, to show

that a number b is equal to �a, add b to a and see if you get 0. This is the strategy in the next proof.

Corollary 1.34. For every real number a, we have �a D .�1/a. Similarly, if b ¤ 0, then .b�1/�1 D b.

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40 Chapter 1 Early Number Theory

Proof. We add .�1/a to a and see if we get 0.

.�1/a C a D .�1/a C 1 � a D a.�1C 1/ D a � 0 D 0:

We do get 0, and so Proposition 1.33 guarantees that �a D .�1/a. To prove the second statement, interpret the equation bb�1 D 1 as saying

that b is an element which, when multiplied by b�1, gives 1.

We can now prove the distributive law for subtraction.

Corollary 1.35. If a; b; c are real numbers, then a.b � c/ D ab � ac.

Proof. By definition, b � c D b C .�c/. But b � c D b C .�1/c, by Corol- lary 1.34. Therefore, distributivity gives

a.b � c/ D a � b C .�1/c

D ab C a.�1/c D ab C .�1/.ac/ D ab � ac:

We have just displayed some properties of addition and multiplication of real numbers following from the nine fundamental properties. The proofs fol-

low only from the nine properties; we did not use any other properties of R,

such as decimal expansions or inequalities. Hence, if we show, for example,

that addition and multiplication of complex numbers or of polynomials satisfy

the nine properties, then each of these systems satisfy the “other properties,”

Propositions 1.31, 1.32, and 1.33, as well.

Exercises

1.68 (i) Prove the additive cancellation law using only the nine properties: if a; b; c

are real numbers with a C c D b C c, then a D b. (ii) Prove the multiplicative cancellation law for real numbers using only the nine

properties: if a; b; c are real numbers with ac D bc and c ¤ 0, then a D b.

1.69 Suppose that b ¤ 0. Show that a=b is the unique real number whose product with b is a.

1.70 (i) Prove that a real number a is a square if and only if a � 0. (ii) Prove that every complex number is a square.

1.71 * Let a; b; c be numbers.

(i) Prove that �ac, the negative of ac, is equal to .�a/c; that is, acC.�a/c D 0. (ii) In the proof of Corollary 1.35, we stated that

ab C a.�1/c D ab � ac:

Prove this.

Hint: Evaluate a.0 C 0/ in two ways.

1.72 * Suppose that e and f are integers and let m D minfe; f g and M D maxfe; f g. Show that

m C M D e C f:

HELIANG GAO
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1.5 Connections 41

1.73 *

(i) If a is a positive real number such that an D 1 for an integer n � 1, prove that a D 1.

(ii) If a is a real number such that an D 1 for an integer n � 1, prove that a D ˙1.

1.74 The Post Office has only 5 and 8 cent stamps today. Which denominations of

postage can you buy?

1.75 * Later in this book, we’ll prove Theorem 2.10: every integer can be factored into

primes in essentially only one way. You may use this theorem here.

(i) If a 2 Z, prove that every prime p that divides a2 shows up with even expo- nent; that is, if p j a2, then p2 j a2.

(ii) Show that there are no integers a and b so that 2a2 D b2. (iii) Use part(ii) to show that there is no rational number x with x2 D 2.

1.76 Use Euclid’s idea of a geometric Division Algorithm (see Figure 1.8 on page 22)

to give a geometric version of the Euclidean Algorithm that uses repeated geo-

metric division. Apply your geometric algorithm to

(i) two segments of length 12 and 90.

(ii) the diagonal and the side of a square.

1.5 Connections

This section applies the method of Diophantus to trigonometry and to calculus.

Trigonometry

The formulas x D .1 � t2/=.1 C t2/ and y D 2t=.1 C t2/, where t is a real number, parametrize all the points on the unit circle except .�1; 0/. But we know that if A D .x; y/ is a point on the unit circle, then x D cos � and y D sin � , where � D †DOA (see Figure 1.9).

O D

A = (cos , sin )q q

1

q

Figure 1.9. Cosine and Sine.

If � D 30o, then .cos �; sin �/ D . p 3 2 ; 1 2 /; one coordinate is irrational and

one is rational. Are there any acute angles � with both cos � and sin � rational?

If .x; y/ is the Pythagorean point arising from .3; 4; 5/, then x D cos � D 3 5

and y D sin � D 4 5

. With a little more work, we can prove that there are

infinitely many angles � with both cos � and sin � rational (is it obvious that

Pythagorean triples arising from distinct Pythagorean points are not similar?)

HELIANG GAO
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42 Chapter 1 Early Number Theory

and also infinitely many angles with both cosine and sine irrational (see Exer-

cise 1.29 on page14).

The parametrization of the unit circle in Proposition 1.2,

cos � D 1 � t 2

1C t2 and sin � D 2t

1C t2 ; �1 < t < 1;

enables us to prove some trigonometric identities. For example, let’s prove the

identity

1C cos � C sin � 1C cos � � sin � D sec � C tan �:

First, rewrite everything in terms of sin � and cos � . The left-hand side is fine; the right-hand side is .1= cos �/ C .sin �= cos �/. Now replace these by their formulas in t . The left-hand side is

1C 1�t2 1Ct2 C

2t 1Ct2

1C 1�t2 1Ct2 �

2t 1Ct2

;

and this simplifies to a rational function of t (that is, a quotient of two poly-

nomials). Similarly, the right-hand side is also a rational function of t , for

sec � D 1 cos �

D 1C t 2

1 � t2 and tan � D 2t

1 � t2 . Thus, verifying whether the trigonometric identity is true is the same thing as verifying whether one ra-

tional expression is equal to another. This problem involves no ingenuity at

all. Just cross multiply and check whether the polynomials on either side are

equal; that is, check whether the monomials on either side having the same

degree have the same coefficients.

Integration

The parametrization of the unit circle is useful for certain integration problems.

In Figure 1.10, we see that 4AOB is isosceles, for two sides are radii; thus, the

q

O

B = (cos , sin )q q

(–1,0) = A

Figure 1.10. Tangent Half-Angle.

We denote the line de-

termined by points A

and B by L.A; B/, but

the notation ! AB is the

convention in geometry

and precalculus books.

base angles are equal. But the exterior angle � is their sum, and so †BAO D �=2. Therefore,

t D sin � 1C cos � D slope L.A; B/ D tan.�=2/I

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1.5 Connections 43

t D tan.�=2/ is called the tangent half-angle formula. Now

� D 2 arctan t and d� D 2 dt 1C t2

:

Let’s apply this substitution. In most calculus courses, the indefinite integralR sec � d� D log j sec � C tan � j is found by some unmotivated trick, but this

integration is quite natural when we use the method of Diophantus.

Z sec � d� D

Z d�

cos � D Z 1C t2 1 � t2

� 2 dt 1C t2

D Z

2 dt

1 � t2 :

Since

2

1 � t2 D 1 1C t

C 1 1 � t

;

we have Z

2 dt

1 � t2 D Z

dt

1C t C Z

dt

1 � t D log j1C t j � log j1� t j:

The hard part is now done;

log j1C t j � log j1 � t j D log ˇ̌1C t 1 � t

ˇ̌ ;

and it is cosmetic to rewrite, using the formula relating t and �

1C t 1 � t D

.1 C t/2 1 � t2 D

1C 2t C t2 1 � t2 D

1C t2 1 � t2 C

2t

1 � t2 D sec � C tan �:

Other integrands can also be integrated using the tangent half-angle formula

(see Exercise 1.78 below). Similar parametrizations of other conic sections also

lead to integration formulas (see Exercises 1.80–1.82 below and [28, pp. 86–

97]).

Exercises

1.77 Verify the following trigonometric identities.

(i) 1 C csc � D cos � cot � 1 � sin �

.

(ii) 1

csc � � cot � � 1

csc � C cot � D 2 cot � .

(iii) cot4 � C cot2 � D csc4 � � csc2 �:

1.78 Integrate the following using the tangent half-angle formula.

(i)

Z sin �

2 C cos � d� .

Answer: ln ˇ̌ ˇ 1Ct2 3Ct2

ˇ̌ ˇ, where t D sin �

1Ccos � .

(ii)

Z sin � � cos � sin � C cos � d� .

Answer: ln ˇ̌ ˇ 1Ct2 1C2tDt2

ˇ̌ ˇ, which leads to � ln j cos � C sin � j.

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44 Chapter 1 Early Number Theory

1.79 * Preview.

(i) Sketch the graph of x2 � xy C y2 D 1. (ii) Find a “sweeping lines” parametrization for the points on the graph of

x2 � xy C y2 D 1. (iii) Find a scalene triangle with integer side lengths and a 60ı angle.

1.80 Take It Further.

(i) Find a “sweeping lines” parametrization for the points on the graph of the

parabola x D y2, using lines joining A D .0; 0/ to points P D .x; y/ on the parabola.

(ii) Use this parametrization to evaluate

Z dx

1 C p

x .

1.81 Take It Further. Show that a “sweeping lines” parametrization for the points

on the ellipse x2=a2 C y2=b2 D 1, using lines joining A D .�a; 0/ to points P D .x; y/ on the ellipse, is

x D a.b 2 � a2t2/

b2 C a2t2 and

2ab2t

b2 C a2t2 :

1.82 Take It Further. Show that a “sweeping lines” parametrization for the points on

the hyperbola x2=a2 � y2=b2 D 1, using lines joining A D .�a; 0/ to points P D .x; y/ on the hyperbola, is

x D a.b 2 C a2t2/

b2 � a2t2 and y D 2ab

2t

b2 � a2t2 :

1.83 * Take It Further. Most high school texts derive the quadratic formula by “com-

pleting the square,” a method we’ll discuss and generalize in Chapter 3. Here’s

another way to derive the formula.

(i) Show that if r and s are the roots of x2 C bx C c D 0, then

r C s D �b and rs D c

(ii) If r C s D �b and rs D c, show that

.r � s/2 D b2 � 4c;

so that r � s D ˙ p

b2 � 4c. (iii) Solve the system

r C s D �b

r � s D ˙ p

b2 � 4c

for r and s.

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2 Induction In Chapter 1, we proved some basic theorems of ordinary arithmetic: Division

Algorithm; Euclidean Algorithm; prime factorization. We are now going to

prove the Fundamental Theorem of Arithmetic: any two people writing an in- teger as a product of primes always get the same factors. We need a very useful We shall see later that

many interesting number

systems do not have

unique factorization.

Indeed, not recognizing

this fact is probably

responsible for many false

“proofs” of Fermat’s Last

Theorem.

tool in order to do this, and so we interrupt our historical account to introduce

mathematical induction, a method of proof that finds application throughout

mathematics. We’ll go on here to use induction to discuss the Binomial Theo-

rem and some combinatorics.

2.1 Induction and Applications

The term induction has two meanings. The most popular one is inductive rea-

soning: the process of inferring a general law from the observation of partic-

ular instances. For example, we say that the Sun will rise tomorrow morning

because, from the dawn of time, the Sun has risen every morning. Although this notion of induction is used frequently in everyday life, it is not adequate

for mathematical proofs, as we now show.

Consider the assertion: “f .n/ D n2 � n C 41 is prime for every positive integer n.” Evaluating f .n/ for n D 1; 2; 3; : : : ; 40 gives the numbers

41; 43; 47; 53; 61; 71; 83; 97; 113; 131;

151; 173; 197; 223; 251; 281; 313; 347; 383; 421;

461; 503; 547; 593; 641; 691; 743; 797; 853; 911;

971; 1033; 1097; 1163; 1231; 1301; 1373; 1447; 1523; 1601:

It is tedious, but not very difficult (see Exercise 2.2 on page 52), to show that every one of these numbers is prime. Inductive reasoning leads you to expect

that all numbers of the form f .n/ are prime. But the next number, f .41/ D 1681, is not prime, for f .41/ D 412 � 41 C 41 D 412, which is obviously composite.

An even more spectacular example of the failure of inductive reasoning is given by the harmonic series 1 C 1

2 C 1

3 C � � � C 1

n C � � � , which diverges

(first proved by Oresme (ca.1320–1382)), and so its partial sums get arbitrarily

large. Given a number N , there is a partial sum

†m D mX

nD1

1

n D 1C 1

2 C � � � C 1

m

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46 Chapter 2 Induction

with †m > N . A high school student, unaware of this, playing with his cal-

culator and seeing that†315 � 6:33137, would probably make the reasonable guess that †m < 100 for all m. But he’s wrong; the series diverges! It is known that if m < 1:5 � 1043, then †m < 100. The most generous estimate of the age of the Earth is ten billion (10,000,000,000) years, or 3:65 � 1012 days, a number insignificant when compared to 1:5� 1043. Therefore, starting from the Earth’s very first day, if the statement †m < 100 was verified on the

mth day, then there would be today as much evidence of the general truth of these statements as there is that the Sun will rise tomorrow morning. And yet

most statements †m < 100 are false!

Inductive reasoning is valuable in mathematics, as it is in natural science,

because seeing patterns in data often helps us guess what may be true in general

(see Exercise 2.1 on page 52, for example). However, merely checking whether the first few (or first few trillion) statements are true is not enough. We have

just seen that checking the first 1:5� 1043 statements is inadequate to establish a general rule.

Let’s now discuss mathematical induction. Suppose we are given a se-

quence of statements

S.1/; S.2/; S.3/; : : : ; S.n/; : : : :

For example, the formula 2n > n for all n � 1 can be viewed as the sequence of statements

21 > 1; 22 > 2; 23 > 3; : : : ; 2n > n; : : : :

Mathematical induction is a technique for proving that all the statements are

true.

The key idea is just this. Imagine a stairway to the sky. We claim that if

its bottom step is white and the next step above any white step is also white,

then all the steps of the stairway are white. Here’s our reasoning. If some steps aren’t white, walk up to the first non-white step; call it Fido. Now Fido can’t

be at the bottom, for the bottom step is white, and so there is a step just below

Fido. This lower step must be white, because Fido is the first non-white one.

But Fido, being the next step above a white step, must also be white. This is a

contradiction; there is no Fido. All the steps are white.

To sum up, given a list of statements, we are claiming that if

(i) the first statement is true, and

(ii) whenever a statement is true, so is the next one,

then all the statements on the list are true.

Let’s apply this idea to the list of inequalities S.n/W 2n > n. Now S.1/ is true, for 21 D 2 > 1. Suppose we believe, for every n > 1, that the implication 2n�1 > n � 1 ) 2n > n is true. Since S.1/ is true and S.1/ ) S.2/ is true,The symbol) means

implies. we have S.2/ true; that is, if 21 > 1 and 21 > 1 ) 22 > 2 are both true, then 22 > 2. Since 22 > 2 is true and 22 > 2 ) 23 > 3 is true, we have 23 > 3; since 23 > 3 is true and 23 > 3 ) 24 > 4 is true, we have 24 > 4; and so forth. Mathematical induction replaces the phrase and so forth with statement (ii), which guarantees, for every n, that there is never an obstruction

in the passage from the truth of any statement S.n � 1/ to the truth of the next one S.n/. We will prove 2n > n for all n � 1 in Proposition 2.2.

Here is the formal statement of mathematical induction.

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2.1 Induction and Applications 47

Theorem 2.1 (Mathematical Induction). Let k be an integer. If S.k/;

S.k C 1/; S.k C 2/; : : : is a sequence of statements such that

(i) Base StepW S.k/ is true, and (ii) Inductive StepW If, for n > k, S.n � 1/ being true implies S.n/ true,

then the statements S.n/ are true for all n � k. Many people prefer to

write the inductive step as

S.n/) S.nC 1/ instead of S.n � 1/ ) S.n/ as we do. The difference is

cosmetic; the important

thing is the passage from

one statement to the next

one.

We’ll prove this in Theorem 2.17 (you’ll see then that the proof is our story about Fido), but let’s use the theorem now to prove some interesting results.

We start by completing our argument that 2n > n.

Proposition 2.2. 2n > n for all n � 1.

Proof. Here k D 1, and the statements are

S.1/ W 21 > 1; : : : ; S.n � 1/ W 2n�1 > n � 1; S.n/ W 2n > n; : : :

Base Step: If n D 1, then 21 D 2 > 1, so S.1/ is true. Inductive Step: We need to show that if n > 1 and S.n� 1/ is true, then S.n/ is true. It is always a good idea to write the statements out so that we can see

what needs to be proved. Here, we must show that if the inductive hypothesis

S.n � 1/ W 2n�1 > n � 1

is true, then so is S.n/; that is, 2n�1 > n � 1 implies 2n > n. Multiply both sides of the inequality S.n � 1/ by 2: if 2n�1 > n� 1, then

2n D 2 � 2n�1 > 2.n� 1/ D .n � 1/C .n � 1/ � .n � 1/C 1 D n

(the last inequality holds because n > 1 implies n � 1 � 1). Thus, if 2n�1 > n� 1 is true, then 2n > n is also true.

Since both the base step and the inductive step hold, Theorem 2.1 says that

all the statements are true: 2n > n for all n � 1.

Etymology. The word induction comes from the Latin word meaning to lead

into or to influence. It is used here because, as we have just seen, the truth of the nth statement arises from the truth of the previous statement.

Usually the base step in an inductive proof occurs when k D 1, although Define S.0/ W 20 > 0. Suppose we had taken

the base step in Propo-

sition 2.2 at k D 0. Can you write out a proof that

S.0/) S.1/?

many proofs occur when k D 0 (see Exercise 2.4 on page 52). Here is an example of an induction whose base step occurs when k D 5. Consider the statements

S.n/ W 2n > n2:

This is not true for small values of n: if n D 2 or 4, then there is equality, not inequality; if n D 3, the left side, 8, is smaller than the right side, 9. However, S.5/ is true: 32 > 25.

Proposition 2.3. 2n > n2 for all integers n � 5.

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48 Chapter 2 Induction

Proof. We have just checked the base step S.5/. Suppose that n > 5 and that

2n�1 > .n � 1/2: (2.1)

Can we use this to show that 2n > n2? Multiply both sides of inequality (2.1)

by 2 to obtain

2n > 2.n� 1/2:

We’ll be done if we show, for n > 5, that 2.n� 1/2 > n2. Now

2.n� 1/2 D .n � 1/2 C .n � 1/.n � 1/ > .n � 1/2 C 4.n � 1/ since n� 1 > 4 � � n2 � 2nC 1

� C 4.n � 1/

D n2 C 2n� 3:

But 2n� 3 is positive, because n > 5, and so n2 C 2n� 3 > n2.

We now use induction to prove a geometric result.

Definition. A polygon P in the plane is convex if, for every pair of distinct

pointsA;B on its perimeter, the line segment AB lies inside of P .

For example, every triangle is convex, but there are quadrilaterals that are

not convex. For example, the shaded quadrilateral in Figure 2.1 is not convex,

for the line segment joining boundary pointsA and B is not wholly inside it.

A B

Figure 2.1. Non-convex polygon.

Proposition 2.4. Let P be a convex polygon with vertices V1; : : : ; Vn. If �i is

the .interior/ angle at Vi , then

�1 C � � � C �n D .n� 2/180ı:

Proof. The proof is by induction on n � 3. For the base step n D 3, the polygon is a triangle, and it is well known that the sum of the interior angles

is 180ı. For the inductive step n > 3, let P be a convex polygon with vertices V1; : : : ; Vn. Since P is convex, the segment joining V1 and Vn�1 lies wholly inside P ; it divides P into the triangle D V1VnVn�1 and the polygon P 0 having vertices V1; : : : ; Vn�1. Now P 0 is convex (why?), so that the inductive hypothesis says that the sum of its interior angles is .n � 3/180ı. Figure 2.2 shows that the sum of the interior angles of P is the sum of the angles of 

and those of P 0, which is 180ı C .n � 3/180ı D .n� 2/180ı.

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2.1 Induction and Applications 49

D

Vn

Vn–1

Vn–2

V1 V2

V3

qn–2

q3

q2 q1

qn–1

qn

Figure 2.2. Convex polygon.

In any proof by induction, we must verify both the base step and the induc-

tive step; verification of only one of them is insufficient. For example, consider the statements S.n/Wn2 D n. The base step, S.1/, is true, but the inductive step is false; of course, these statements S.n/ are false for all n > 1. Another

example is given by the statements S.n/Wn D nC 1. It is easy to see that the inductive step is true: if S.n�1/ is true, i.e., if n�1 D .n�1/C1, then adding 1 to both sides gives n D .n � 1/C 2 D n C 1, which is the next statement, S.n/. But the base step is false; of course, all these statements S.n/ are false.

How to Think About It. When first seeing induction, many people sus-

pect that the inductive step is circular reasoning. Why are you allowed to use

statement S.n � 1/, which you don’t know is true, to prove that S.n/ is true? Isn’t the truth of S.n � 1/ essentially what you are supposed to be proving? A closer analysis shows that this is not at all what is happening. The inductive step, by itself, does not prove that S.n/ is true. Rather, it says that if S.n � 1/ is true, then S.n/ is also true. In other words, the inductive step proves that the

implication “If S.n � 1/ is true, then S.n/ is true” is correct. The truth of this implication is not the same thing as the truth of its conclusion. For example,

consider the two statements: “Your grade on every exam is 100%” and “Your

grade for the course is A.” The implication “If all your exams are perfect, then you will get the highest grade for the course” is true. Unfortunately, this does

not say it is inevitable that your grade for the course will be A. Here is a math-

ematical example: the implication “If n � 1 D n, then n D nC 1” is true, but the conclusion “n D nC 1” is false.

From now on, we usually abbreviate mathematical induction to induction.

Here is the first example of a proof by induction often given in most texts.

Proposition 2.5. For every integer n � 1, we have

1C 2C � � � C n D 1 2 n.nC 1/:

Proof. The proof is by induction on n � 1. Base step. If n D 1, then the left-hand side is 1 and the right-hand side is

1 2 1.1C 1/ D 1, as desired.

Inductive step. The .n � 1/st statement is

S.n � 1/W 1C 2C � � � C .n � 1/ D 1 2 .n � 1/n;

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50 Chapter 2 Induction

and we must show

1C 2C � � � C n D � 1C 2C � � � C .n � 1/

� C n:

By the inductive hypothesis, the right-hand side is

1 2 .n � 1/nC n:

But 1 2 .n � 1/n C n D 1

2 n.n C 1/. By induction, the formula holds for all

n � 1.

Historical Note. Here is one version of a popular story. As a 7-year old

prodigy, Gauss was examined by two mathematicians to evaluate his mathe-

matical ability. When asked to add up all the numbers from 1 to 100, he thought

a moment and then said the answer was 5050. Gauss let s denote the sum of all the numbers from 1 to 100: s D 1 C 2 C � � � C 99 C 100. Of course, s D 100C 99C � � � C 2C 1. Arrange these nicely

s D 1C 2C � � � C 99C 100 s D 100C 99C � � � C 2C 1

and add

2s D 101C 101C � � � C 101C 101;

the sum 101 occurring 100 times. We now solve: s D 1 2 .100 � 101/ D 5050.

This argument is valid for any number n in place of 100 (and there is no obvi-

ous use of induction!). Not only does this give a new proof of Proposition 2.5,

it shows how the formula could have been discovered.

Example 2.6. Another proof of the formula in Proposition 2.5 comes from an

analysis of the square in Figure 2.3.

Figure 2.3. Pn kD1 k.

Imagine an .nC 1/ � .nC 1/ square. It contains .nC 1/2 small unit squares.We have n D 7 in Fig- ure 2.3. Since there are nC 1 unit squares on the diagonal, there are

.nC 1/2 � .nC 1/ D n2 C n

unit squares off the diagonal. Half of them, 1 2 .n2 Cn/, are above the diagonal.

But, if you count by rows, there are

1C 2C � � � C n

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2.1 Induction and Applications 51

unit squares above the diagonal. Hence

1C 2C � � � C n D 1 2 .n2 C n/: N

How to Think About It. Proposition 2.5 illustrates a common problem stu- dents have when learning induction. Induction itself is a technique of proof

(involving just two steps), but it is not a method of discovery. However, the

two notions of proof and discovery are often intertwined. For example, merely

applying mathematical induction, as we did in the proof of Proposition 2.5,

is straightforward. But many beginning students get confused because, at the same time as they are following the steps of the proof, they are also wonder-

ing where the formula for the sum comes from. In contrast, neither Gauss’s

proof nor the proof using the .n C 1/ � .n C 1/ square is confusing, for the ideas of these proofs and their techniques of proof are separate. In Section 2.3,

we’ll describe a method for introducing mathematical induction to high school

students that usually minimizes this confusion.

Aside from proving statements, induction can also be used to define terms.

For example, here is an inductive definition of factorial. Inductive definitions are often called recursive

definitions.Definition. Define 0Š D 1 and, if n > 0, define nŠ D n � .n � 1/Š. In other words, nŠ is defined by Defining 0Š D 1 is con-

venient, as we shall see

in the next section when

we discuss the Binomial

Theorem.

nŠ D (

1 if n D 0 n � .n � 1/Š if n > 0:

Induction allows us to define the powers of a number.

Definition. If a 2 R, define the powers of a, for n � 0, by induction:

an D (

1 if n D 0 an�1a if n > 0:

If a D 0, we have defined 00 D 1.

Etymology. The terminology x square and x cube for x2 and x3 is, of

course, geometric in origin. Usage of the term power in this context arises

from a mistranslation of the Greek dunamis (from which the word dynamo

comes) as used by Euclid. The standard European rendition of dunamis was

“power;” for example, the first English translation of Euclid’s Elements by

H. Billingsley in 1570, renders a sentence of Euclid as “The power of a line is the square on the same line” (which doesn’t make much sense to us). However,

contemporaries of Euclid, e.g., Plato and Aristotle, used dunamis to mean “am-

plification.” This seems to be a more appropriate translation, for Euclid was

probably thinking that a one-dimensional line segment can sweep out a two-

dimensional square. We thank Donna Shalev for informing us of the classical usage of dunamis.

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52 Chapter 2 Induction

Proposition 2.7 (Laws of Exponents). Let a 2 R andm; n � 0 be integers. (i) amCn D aman.

(ii) .am/n D amn.

Proof. (i) The statement is true for all n � 0 when m D 0. We prove amCn D aman is true for all n by induction on m � 1. The base step says that aan D anC1, which is just the definition of powers. For the inductive step,

amCn D amCn�1a definition of powers D am�1Cna D am�1ana inductive hypothesis D am�1anC1 definition of powers D am�1CnC1 inductive hypothesis D amCn:

(ii) The statement is true for all m � 0 when n D 0. We prove .am/n D amn is true for all m by induction on n � 1. The base step says that .am/1 D am1 D am, which is obvious. For the inductive step,

.am/n D .am/n�1am definition of powers D am.n�1/am inductive hypothesis D am.n�1/Cm part (i) D amn:

Historical Note. The earliest known occurrence of mathematical induction is in Sefer ha-Mispar (also called Maaseh Hoshev, whose Hebrew title means

practical and theoretical calculating), written by Levi ben Gershon in 1321

(he is also known as Gersonides or as RaLBaG, the acronym for Rabbi Levi

ben Gershon). Induction appears later in Arithmeticorum libri duo, written by

Maurolico in 1557, and also in Traité du Triangle Arithmetique, written by Pascal around 1654 (in which Pascal discusses the Binomial Theorem).

Exercises

2.1 * Guess a formula for 1 C P n jD1 j Šj , and use mathematical induction to prove

that your formula is correct.

2.2 * Prove that if m � 2 is an integer not divisible by any prime p with p � p

m,

then m is a prime. Use this to prove that the numbers n2 � n C 41 are prime for all n � 40.

2.3 * Let m1; m2; : : : ; mn be integers such that gcd.mi ; mj / D 1 for all i ¤ j . If each mi divides an integer k, prove that their product m1m2 � � � mn also divides k. Hint: Use Exercise 1.58 on page 35.

2.4 * If a is positive, give two proofs that

1 C a C a2 C � � � C an�1 D a n � 1

a � 1 ;

by induction on n � 0 and by multiplying the left-hand expression by .a � 1/.

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2.1 Induction and Applications 53

2.5 Let x > �1 be a real number. Prove that .1 C x/n � 1 C nx for all n � 1.

2.6 What is the smallest value of k so that 2n > n3 for all n � k? Why?

2.7 Assuming the product rule for derivatives, .fg/0 D f 0g C fg0, prove that

.xn/0 D nxn�1 for all integers n � 1:

2.8 In high school, nŠ is usually defined as 1 � 2 � 3 � � � � � n. Show that this agrees with the definition on page 51 for all n � 1.

2.9 (Double Induction) Let k; k0 be integers, and let S.m; n/ be a doubly indexed family of statements, one for each pair of integers m � k and n � k0. Suppose that

(i) S.k; k0/ is true,

(ii) if S.m � 1; k0/ is true, then S.m; k0/ is true, (iii) if S.m; n � 1/ is true for all m � k, then S.m; n/ is true for all m � k. Prove that S.m; n/ is true for all m � k and n � k0.

2.10 Prove that .m C 1/n > mn for all m; n � 1.

2.11 Prove the Laws of Exponents by Double Induction.

Unique Factorization

Induction is useful in number theory. As a simple example, we generalize Eu-

clid’s Lemma to more than two factors.

Theorem 2.8 (Euclid’s Lemma). If p is a prime and p j a1a2 � � �an, where n � 2, then p j ai for some i .

Proof. The proof is by induction on n � 2. The base step is Theorem 1.21. To “The proof is by induction on n � 2” not only indicates the base step, it

also tells which variable

will be changing in the

inductive step.

prove the inductive step, suppose that p j a1a2 � � �an. We may group the fac- tors on the right side together so there are only two factors: .a1a2 � � �an�1/an . By Theorem 1.21, either p j a1a2 � � �an�1 or p j an. In the first case, the in- ductive hypothesis gives p j ai for some i � n � 1 and we are done. In the second case, p j an, and we are also done.

This proof illustrates an empirical fact. It is not always the case, in an in-

ductive proof, that the base step is very simple. In fact, all possibilities can

occur: both steps can be easy, both can be difficult, or one can be harder than

the other.

Here is an amusing inductive proof (due to Peter Braunfeld) of the existence of the quotient and remainder in the Division Algorithm.

Proposition 2.9. If a and b are positive integers, then there are integers q

and r with b D qaC r and 0 � r < a.

Proof. We do induction on b � 1. The base step: b D 1. Now a � 1, because it is a positive integer. If a D 1,

choose q D 1 and r D 0; if a > 1, choose q D 0 and r D 1. Let’s prove the inductive step. The inductive hypothesis is b � 1 D qaC r ,

where 0 � r < a. It follows that b D qaCrC1. Now r < a implies rC1 � a. If r C 1 < a, we are done. If r C 1 D a, then b D qaC .r C 1/ D qaC a D .q C 1/a, and we are done in this case as well.

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54 Chapter 2 Induction

We now use induction to prove unique factorization into primes.

Theorem 2.10 (Fundamental Theorem of Arithmetic). Every integer a � 2 is a product of primes. Moreover, ifWe often use the word

product even when there

is only one factor. Thus,

a prime is a product of

primes.

a D p1 � � �pm and a D q1 � � �qn;

where the p’s and q’s are primes, then n D m and the q’s can be re-indexed so that qi D pi for all i .

Proof. The existence of a factorization is Theorem 1.14. To prove uniqueness,

we may assume that m � n, and we use induction on m � 1. The base step is obvious: if m D 1, then n D 1 and the given equations are a D p1 D q1. For the inductive step, the equation

p1 � � �pm D q1 � � �qn

gives pm j q1 � � �qn. By Euclid’s Lemma, there is some i with pm j qi . But qi , being a prime, has no positive divisors other than 1 and itself, so that

qi D pm. Re-indexing, we may assume that qn D pm. Canceling, we have p1 � � �pm�1 D q1 � � �qn�1. By the inductive hypothesis, n � 1 D m � 1 (so that n D m) and the q’s may be re-indexed so that qi D pi for all i � m.

Corollary 2.11. If a � 2 is an integer, then there are distinct primes pi and integers ei > 0 with

a D pe11 � � �penn :

Moreover, if there are distinct primes qj and integers fj > 0 with

p e1 1 � � �penn D q

f1 1 � � �qfmm ;

then m D n, qi D pi and fi D ei for all i .after re-indexing the q0s/.

Proof. Just collect like terms in a prime factorization.

The Fundamental Theorem of Arithmetic says that the exponents e1; : : : ; en in the prime factorization a D pe11 � � �p

en n are well-defined integers determined

by a. It would not make sense to speak of the exponent of q dividing a if the

Fundamental Theorem were false; if an integer a had two factorizations, say,

a D p2q5r6 and a D p2q3s7, where p; q; r; s are distinct primes, would its q-exponent be 5 or 3?

It is often convenient to allow factorizations p e1 1 � � �p

en n having some ex-

ponents ei D 0, because this allows us to use the same set of primes when factoring two numbers. For example, 168 D 233171 and 60 D 223151 may be rewritten as 168 D 23315071 and 60 D 22315170.

Lemma 2.12. Let positive integers a and b have prime factorizations

a D pe11 � � �penn and b D p f1 1 � � �pfnn ;

where p1; : : : ; pn are distinct primes and ei , fi � 0 for all i . Then a j b if and only if ei � fi for all i .

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2.1 Induction and Applications 55

Proof. If ei � fi for all i , then b D ac, where c D pf1�e11 � � �p fn�en n . Now c

is an integer, because fi � ei � 0 for all i , and so a j b. Conversely, if b D ac, let the prime factorization of c be c D pg11 � � �p

gn n ,

where gi � 0 for all i . It follows from the Fundamental Theorem of Arithmetic that ei C gi D fi for all i , and so fi � ei D gi � 0 for all i ; that is, ei � fi for all i .

Definition. A common multiple of integers a and b is an integerm with a j m and b j m. The least common multiple, denoted by

lcm.a; b/;

is the smallest positive common multiple if both a; b ¤ 0, and it is 0 otherwise.

The following proposition describes gcd’s in terms of prime factorizations. This is, in fact, the method usually taught to students in elementary school for

putting fractions into lowest terms.

Proposition 2.13. Let a D pe11 � � �p en n and b D pf11 � � �p

fn n , where p1; : : : ; pn

are distinct primes and ei � 0, fi � 0 for all i . Then

gcd.a; b/ D pm11 � � �pmnn and lcm.a; b/ D p M1 1 � � �pMnn ;

where mi D minfei ; fig, andMi D maxfei ; fi g.

Proof. Define d D pm11 � � �p mn n . Lemma 2.12 shows that d is a (positive)

common divisor of a and b; moreover, if c is any (positive) common divisor,

then c D pg11 � � �p gn n , where 0 � gi � minfei ; fig D mi for all i . Therefore,

c j d . A similar argument shows that D D pM11 � � �p

Mn n is a common multiple

that divides every other such.

Computing the gcd for small numbers a and b using their prime factoriza-

tions is more efficient than using Euclidean Algorithm I. For example, since

168 D 23315071 and 60 D 22315170, we have gcd.168; 60/ D 22315070 D 12 and lcm.168; 60/ D 23315171 D 840. However, finding the prime factor- ization of a large integer is very inefficient, even with today’s fanciest com-

puters; it is so inefficient that this empirical fact is one of the main ingredients

in public key cryptography, the basic reason you can safely submit your credit

card number when buying something online.

Corollary 2.14. If a and b are positive integers, then

lcm.a; b/ gcd.a; b/ D ab:

Proof. The result follows from Proposition 2.13 and Exercise 1.72 on page 40: Notice how a computa- tional inquiry has given a

theorem.mi CMi D ei C fi ;

where mi D minfei ; fi g and Mi D maxfei ; fig.

Since gcd’s can be computed by Euclidean Algorithm I, this corollary al- lows us to compute lcm’s:

lcm.a; b/ D ab= gcd.a; b/:

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56 Chapter 2 Induction

Example 2.15. Sudoku is a popular puzzle. One starts with a 9 � 9 grid of cells, some filled with numbers. The object is to insert numbers in the blank

cells so that every row, every column, and every heavily bordered 3 � 3 box contains the digits 1 through 9 exactly once.

96

5

3

34

1

3

8

7

5

4

5 2

8 3

7 6

3

5

8

1

1

5

3 6

9

7

6 7

Figure 2.4. Sudoku.

KenKen is a variation of Sudoku. As in Sudoku, the object is to fill an n�nKenKenr is a registered trademark of Nextoy, LLC,

2012, KenKen Puzzle LLC.

All rights reserved.

grid with digits: 1 through 4 for a 4 � 4 grid, 1 through 5 for a 5 � 5 grid, etc., so that no digit appears more than once in any row or column. That the cells

in Sudoku are filled with 1, 2, . . . , 9 is not important; one could just as well use the first nine letters a, b, . . . , i instead. In contrast, KenKen uses arithmetic.

KenKen grids are divided into heavily bordered groups of cells, called cages,

and the numbers in the cells in each cage must produce a target number when

combined using a specified mathematical operation—either addition, subtrac-

tion, multiplication or division. Here is a 5� 5KenKen puzzle and its solution.

4– 12´ 3–

12+ 1–

2¸ 4– 3

60´ 5+ 2¸

Figure 2.5. KenKenr puzzle.

4– 12´ 3–

12+ 1–

2¸ 4– 3

60´ 5+ 2¸

5 3 2 1 4

1 2 3 4 5

2 1 4 5 3

4 5 1 3 2

3 4 5 2 1

Figure 2.6. KenKenr solution.

The difficulty in solving a KenKen puzzle arises from there being too many

ways to fill in each cage. Sometimes, the Fundamental Theorem of Arithmetic can help. Let’s start solving the puzzle in Figure 2.5. We view the grid as a

5 � 5 matrix, and we’ll abbreviate “target-operation” to T-O. Consider the L- shaped cage consisting of 4 cells whose target operation is 60�. There are two possibilities: its cells are filled with an arrangement either of 2, 2, 3, 5 or of

1, 3, 4, 5. Assume the first possibility holds. Since we cannot have both 2s in the same row or the same column, one 2 is in position .4; 3/; the other 2 is

either in position .5; 1/ or .5; 2/. Suppose 2 sits in the .5; 1/ position. There is

a cage in the first column with T-O 4�; its cells must contain 1 and 5. Hence, the other cage, with T-O 2�, must contain 3 and 5; it cannot. Thus, 2 sits in

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2.1 Induction and Applications 57

position .5; 2/. There is a cage in the second column with T-O 4�, and its cells must contain 1 and 5. This says that the top two cells in the second column

contain 3 and 4. But the L-shaped cage with T-O 12� must now have 1 in position .1; 3/. This forces the column one cage, with T-O 4�, to have 5 in position .1; 1/, because it can’t be 1. Thus, the last cage in the first row cannot

involve 1 or 5. But the only ways to fill in a 2-cell cage with T-O 3� are with 1 and 4 or with 2 and 5. Conclusion: The 4-cell cage with T-O 60� must be an arrangement of 1, 3, 4, 5. The full solution is given in Figure 2.6. N

Strong Induction

Certain situations call for a variant of induction, called Strong Induction (or the Second Form of Induction).

Definition. Given integers k and n � k, the predecessors of n are the integers ` with k � ` < n, namely, k; k C 1; : : : ; n� 1 (k has no predecessor).

Theorem 2.16 (Strong Induction). Let k be an integer. If S.k/, S.k C 1/, S.k C 2/; : : : is a sequence of statements such that

(i) Base StepW S.k/ is true, and (ii) Inductive StepW If, for n > k, S.`/ being true for all predecessors ` of n

implies S.n/ true,

then the statements S.n/ are true for all n � k.

How to Think About It. Let’s compare the two forms of induction. Both

start by verifying the base step, and both have an inductive step to prove

S.n/. The inductive hypothesis in the first form is that S.n � 1/ is true; the inductive hypothesis in Strong Induction is that all the preceding statements

S.k/; : : : ; S.n � 1/ are true. Thus, Strong Induction has a stronger inductive hypothesis (actually, each of Theorems 2.1 and 2.16 implies the other).

We are going to prove Theorem 2.16 and Theorem 2.1 simultaneously (we

haven’t yet proved the latter theorem). But first we need an easy technical

remark. The Least Integer Axiom says that every nonempty subset C of the natural numbers N contains a smallest number; that is, there is some c0 2 C with c0 � c for all c 2 C . This axiom holds, not only for N, but for any subset

Nk D fn 2 Z W n � kg

as well, where k is a fixed, possibly negative, integer. If k � 0, then Nk � N, and there is nothing to prove; if k < 0, then we argue as follows. Let C � Nk be a nonempty subset. If C contains no negative integers, then C � N, and the Least Integer Axiom applies; otherwise, keep asking, in turn, whether

k; k C 1; : : : ;�1 are in C , and define c0 to be the first one that lies in C . We have alreaady seen the basic idea of the next proof, involving Fido, on

page 46. Here are the two statements again.

Theorem 2.17 (= Theorems 2.1 and 2.16). Let k be an integer. If S.k/,

S.k C 1/, S.k C 2/; : : : is a sequence of statements such that

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58 Chapter 2 Induction

Base StepW S.k/ is true, and Inductive StepW either

(i) if, for n > k, S.n � 1/ being true implies S.n/ true, or

(ii) if, for n > k, S.`/ being true for all predecessors ` of n implies S.n/ true,

then the statements S.n/ are true for all n � k.

Proof. Let Nk D fn 2 Z W n � kg. We show that there are no integers n 2 Nk for which S.n/ is false. Otherwise, the subset C � Nk , consisting of all n for which S.n/ is false, is nonempty, and so C has a smallest element, say, c0. The

truth of the base step says that k < c0, so that c0�1 lies in Nk and hence there is a statement S.c0 � 1/.

Case 1. As S.c0/ is the first false statement, S.c0�1/must be true. Assuming inductive step (i), S.c0/ D S ..c0 � 1/C 1/ is true, and this is a contradiction. Case 2. As S.c0/ is the first false statement, all the statements S.`/, where ` is

a predecessor of c0, are true. Assuming inductive step (ii), the strong version,

we again reach the contradiction that S.c0/ is true.

In either case, C D ¿ (i.e., C is empty), which says that every S.n/ is true.

Here’s a second proof that prime factorizations exist.

Proposition 2.18 (= Proposition 1.14). Every integer n � 2 is a product of primes.

Proof. The base step S.2/ is true because 2 is a prime. We prove the inductive

step. If n � 2 is a prime, we are done. Otherwise, n D ab, where 2 � a < n and 2 � b < n. As a and b are predecessors of n, each of them is a product of primes:

a D pp0 � � � and b D qq0 � � � :

Hence, n D pp0 � � �qq0 � � � is a product of (at least two) primes.

The reason why strong induction is more convenient here is that it is more natural to use S.a/ and S.b/ than to use S.n � 1/; indeed, it is not at all clear how to use S.n � 1/.

The next result says that we can always factor out a largest power of 2 from

any integer. Of course, this follows easily from the Fundamental Theorem of

Arithmetic, but we prove the proposition to illustrate further situations in which strong induction is more appropriate than the first form.

Proposition 2.19. Every integer n � 1 has a unique factorization n D 2km, where k � 0 and m � 1 is odd.

Proof. We use strong induction on n � 1 to prove the existence of k and m. If n D 1, take k D 0 and m D 1. For the inductive step n � 1, we distinguish

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2.1 Induction and Applications 59

two cases. If n is odd, take k D 0 and m D n. If n is even, then n D 2b. Why isn’t the first form of induction convenient here?Since b < n, it is a predecessor of n, and so the inductive hypothesis allows us

to assume b D 2`m, where ` � 0 and m is odd. The desired factorization is n D 2b D 2`C1m.

To prove uniqueness (induction is not needed here), suppose that 2km D n D 2tm0, where both k and t are nonnegative and both m and m0 are odd. We may assume that k � t . If k > t , then canceling 2t from both sides gives 2k�tm D m0. Since k � t > 0, the left side is even while the right side is odd; this contradiction shows that k D t . We may thus cancel 2k from both sides, leaving m D m0.

Exercises

2.12 (i) Prove that an integer a � 2 is a perfect square if and only if whenever p is prime and p j a, then p2 j a.

(ii) Prove that if an integer z � 2 is a perfect square and d4 j z2 , then d2 j z.

2.13 Let a and b be relatively prime positive integers. If ab is a perfect square, prove

that both a and b are perfect squares.

2.14 * Let a; b; c; n be positive integers with ab D cn. Prove that if a and b are relatively prime, then both a and b are nth powers; that is, there are positive

integers k and ` with a D kn and b D `n.

2.15 * For any prime p and any positive integer n, denote the highest power of p

dividing n by Op.n/. That is, Corollary 2.11 guarantees that Op is well-defined.

Op.n/ D e;

where pe j n but peC1 − n. If m and n are positive integers, prove that (i) Op.mn/ D Op.m/ C Op.n/

(ii) Op.m C n/ � min ˚ Op.m/;Op.n/

. When does equality occur?

There is a generalization of Exercise 1.6 on page 6. Using a (tricky) inductive

proof (see FCAA [26], p. 11), we can prove the Inequality of the Means: if n � 2 and a1; : : : ; an are positive numbers, then

n p

a1 � � � an � 1n .a1 C � � � C an/:

2.16 (i) Using the Inequality of the Means for n D 3, prove, for all triangles having a given perimeter, that the equilateral triangle has the largest area.

Hint: Use Heron’s Formula for the area A of a triangle with sides of lengths

a; b; c: if the semiperimeter is s D 1 2

.a C b C c/, then

A2 D s.s � a/.s � b/.s � c/:

(ii) What conditions on a, b, and c ensure that Heron’s formula produces 0? In-

terpret geometrically.

2.17 Let a, b, and c be positive numbers with a > b > c, and let L D 13 .a Cb Cc/. (i) Show that either a > b > L > c or a > L > b > c.

(ii) Assume that a > b > L > c. Show that

L3 � .L � b/2c � .L � b/.L � c/c � .L � c/2L D abc:

(iii) Use part (ii) to prove the Inequality of the Means for three variables.

(iv) Show that a box of dimensions a � b � c can be cut up to fit inside a cube of side length L with something left over.

HELIANG GAO
高亮
HELIANG GAO
高亮

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60 Chapter 2 Induction

Differential Equations

You may have seen differential equations in other courses. If not, don’t worry;

the next example is self-contained.

Definition. A differential equation is an equation involving a function y D y.x/ and its derivatives; a solution is a function y that satisfies the equation.

Solving a differential equation generalizes indefinite integration: R f .x/ dx

is a solution to the differential equation y0 D f . There may be many solutions: for example, if y D F.x/ is an indefinite integral of

R f .x/ dx, then so is

F.x/ C c for any constant c. Assume that a differential equation has a solution y that is a power series.

Because factorials occur in the coefficients of Taylor series, let’s write a solu-

tion in the formDefining 0Š D 1 allows us to write the coefficient of

xn in Eq. (2.2) as an=nŠ

for all n � 0. y.x/ D a0 C a1x C

a2

2Š x2 C � � � C an

nŠ xn C � � � : (2.2)

We ignore questions of convergence. Of course, some power series diverge,

but we are doing algebra here!

Induction arises here because we can often find y by relating its coefficients

an�1 and an.

Example 2.20. Consider the differential equation y0 D y; that is, we seek a function equal to its own derivative (do you know such a function?). Assuming

that y is a power series, then y has an expression as in Eq. (2.2). Using term-

by-term differentiation, we see that

y0 D a1 C a2x C a3

2Š x2 C � � � C

an

.n � 1/Šx n�1 C � � � ;

so that

an

.n � 1/Šx n�1 D

an�1 .n � 1/Šx

n�1 for all n � 1I

that is, a1 D a0, a2 D a1, and, in fact, an D an�1 for all n � 1. Rewrite the equations: there is no restriction on a0 and, for small n, we see that an D a0. If this were true for all n, then

y.x/ D a0.1C x C 12x 2 C � � � C 1

nŠ xn C � � � / D a0ex:

It is true that an D a0 for all n; one proof is by induction (see Exercise 2.18 on page 62). N

Differential equations often arise with initial conditions: values of y.0/,

y0.0/, y00.0/; : : : are specified. If y is given by a power series

P .an=nŠ/x

n, then y.0/ D a0. Thus, the ini- tial condition y.0/ D 1 chooses the solutiony D ex in the preceding example.

The next example shows how strong induction can be used in solving dif-

ferential equations.

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2.1 Induction and Applications 61

Example 2.21. Consider the differential equation

y00 D y0 C 2y (2.3)

with initial conditions

y.0/ D 2 and y0.0/ D 1:

Again, let’s see if there is a power series solution

y.x/ D a0 C a1x C a2

2Š x2 C � � � :

Substituting y into Eq. (2.3) and equating like powers of x gives

an D an�1 C 2an�2:

Tabulating an for a few values shows a pattern. All the outputs an seem to be

1 away from a power of 2, either 1 more or 1 less. The first two entries record

the initial conditions.

n an

0 2

1 1

2 5 D 1C 2 � 2 3 7 D 5C 2 � 1 4 17 D 7C 2 � 5 5 31 D 17C 2 � 7 6 65

7 127

8 257

9 511

10 1025

Looking closer, the coefficients seem to satisfy an D 2nC .�1/n. Inductive reasoning suggests the conjecture, and mathematical induction is a natural way

to prove it. But there is a problem. The inductive step for an involves not only

an�1, but an�2 as well. Strong Induction to the rescue! Before dealing with the details, we show that if the formula can be proved to hold for all an, then we

can complete our discussion of the differential equation. A solution is

y.x/ D X 2n C .�1/n

nŠ xn

D X 2n

nŠ xn C

X .�1/n nŠ

xn

D X 1

nŠ .2x/n C

X 1 nŠ .�x/n

D e2x C e�x:

You can check that e2x C e�x works by substituting it into Eq. (2.3); we have solved the differential equation. N

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62 Chapter 2 Induction

The proof of the equation relating the coefficients is by Strong Induction.

Proposition 2.22. Suppose, for all n � 0, that an satisfies

an D

8 <̂

2 n D 0 1 n D 1 an�1 C 2an�2 n > 1:

Then an D 2n C .�1/n for all integers n � 0.

Proof. Because the definition has two initial values, we need to check two base

steps:

a0 D 2 D 20 C .�1/0 and a1 D 1 D 21 C .�1/1:

If n > 1 and ak D 2k C .�1/k for all the predecessors of n, 0 � k < n, then

an D an�1 C 2an�2 D � 2n�1 C .�1/n�1

� C 2

� 2n�2 C .�1/n�2

D � 2n�1 C .�1/n�1

� C � 2 � 2n�2 C 2 � .�1/n�2

D � 2n�1 C .�1/n�1

� C � 2n�1 C 2 � .�1/n�2

D � 2n�1 C 2n�1

� C � .�1/n�1 C 2 � .�1/n�2

D 2 � 2n�1 C .�1/n�2 .�1C 2/ D 2n C .�1/n�2

D 2n C .�1/n :

Exercises

2.18 * Complete the discussion in Example 2.20: show that if

y.x/ D a0 C a1x C a2

2Š x2 C � � � C an

nŠ xn C � � �

and y0 D y, then an D a0.

2.19 Assume that “term-by-term” differentiation holds for power series: if f .x/ D c0 C c1x C c2x2 C � � � C cnxn C � � � , then the power series for the derivative f 0.x/ is

f 0.x/ D c1 C 2c2x C 3c3x2 C � � � C ncnxn�1 C � � � :

(i) Prove that f .0/ D c0. (ii) Prove, for all n � 0, that if f .n/ is the nth derivative of f , then

f .n/.x/ D nŠcn C .n C 1/ŠcnC1x C x2gn.x/;

where gn.x/ is a power series.

(iii) Prove that cn D f .n/.x/.0/=nŠ for all n � 0. (This is Taylor’s formula.)Here is an instance in which it is convenient to

write 0Š D 1. This exercise shows why, in Example 2.21, that power series were denoted by a0 C a1x C .a2=2Š/x2 C .a3=3Š/x3 C � � � .

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2.2 Binomial Theorem 63

2.20 Find the solution to the differential equation

2y00 � y0 � 3y D 0:

subject to the initial conditions y.0/ D y.1/ D 1.

Answer: y D 1 5

e�x C 4 5

e3x=2 .

2.2 Binomial Theorem

We now look at a result, important enough to deserve its own section, which

involves both mathematical induction and inductive reasoning. What is the pat-

tern of the coefficients in the formulas for the powers .1C x/n of the binomial 1C x? Let

.1C x/n D c0 C c1x C c2x2 C � � � C cnxn:

Definition. The coefficients cr are called binomial coefficients: Euler introduced the notation

� n r

� , and this

symbol evolved into �

n r

� ,

which is generally used

today:

n

r

! is the coefficient cr of x

r in .1 C x/n:

The binomial coefficient � n r

� is pronounced “n choose r” because it also

arises in counting problems, as we shall soon see. Thus,

.1C x/n D nX

rD0

n

r

! xr :

For example,

.1 C x/0 D 1

.1 C x/1 D 1C 1x

.1 C x/2 D 1C 2x C 1x2

.1 C x/3 D 1C 3x C 3x2 C 1x3

.1 C x/4 D 1C 4x C 6x2 C 4x3 C 1x4:

Etymology. Binomial means aCb; trinomial means aCbCc. But monomial usually refers to a summand of a polynomial: either axe for a polynomial in

one variable, or ax e1 1 � � �x

en n for a polynomial in several variables.

The following figure, called Pascal’s triangle, displays an arrangement of the

first few coefficients.

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64 Chapter 2 Induction

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

In Pascal’s triangle, an inside number (i.e., not a 1 on the border) of the

nth row can be computed by going up to the .n � 1/st row and adding the two neighboring numbers above it. For example, the inside numbers in row 4 can

be computed from row 3 as follows:

1 3 3 1

1 4 6 4 1

(4 D 1C3, 6 D 3C3, and 4 D 3C1). Let’s prove that this observation always holds.

Lemma 2.23. For all integers n � 1 and all r with 0 � r � n, n

r

! D ( 1 if n D 0 or n D r � n�1 r�1

� C � n�1 r

� if 0 < r < n:

Proof. The nth row of Pascal’s triangle is the coefficient list for .1C x/n. The fact that the constant term and the highest degree term have coefficient 1 is Exercise 2.21 on page 67. For the inside terms, we claim that the coefficient of

xr in .1C x/n is the sum of two neighboring coefficients in .1C x/n�1. MoreYou can also prove Lemma 2.23 by induc-

tion. See Exercise 2.22 on

page 67.

precisely, we claim that if

.1C x/n�1 D c0 C c1x C c2x2 C � � � C cn�1xn�1;

and 0 < r < n, then the coefficient of xr in .1 C x/n is cr�1 C cr . We have

.1C x/n D .1C x/.1 C x/n�1 D .1 C x/n�1 C x.1C x/n�1

D .c0 C � � � C cn�1xn�1/C x.c0 C � � � C cn�1xn�1/ D .c0 C � � � C cn�1xn�1/C .c0x C c1x2 C � � � C cn�1xn/ D 1C .c0 C c1/x C .c1 C c2/x2 C � � � :

Thus � n r

� D cr�1 C cr D

� n�1 r�1 �

C � n�1 r

� .

Pascal’s triangle was known centuries before Pascal’s birth; Figure 2.7 shows a Chinese scroll from the year 1303 depicting it. Pascal’s contribution (around

1650) is a formula for the binomial coefficients.

Proposition 2.24 (Pascal). For all n � 0 and all r with 0 � r � n, n

r

! D nŠ rŠ.n� r/Š

:

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2.2 Binomial Theorem 65

Figure 2.7. Pascal’s triangle, China, 1303 CE.

Proof. The proof is by induction on n � 0. If n D 0, then

Pascal probably discovered

this formula by regarding� n r

� in a different way. We’ll

look at this in a moment. 0

0

! D 0Š 0Š0Š

D 1:

For the inductive step, note first that the formula holds when r D 0 and Here is another place showing that defining

0Š D 1 is convenient. r D n:

n

0

! D 1 D nŠ

0Š.n � 0/Š

and n

n

! D 1 D nŠ

nŠ 0Š :

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66 Chapter 2 Induction

If 0 < r < n, then n

r

! D n� 1 r � 1

! C n � 1 r

! (Lemma 2.23)

D .n � 1/Š .r � 1/Š.n� r/Š

C .n � 1/Š rŠ.n� r � 1/Š

(inductive hypothesis)

D .n � 1/Š .r � 1/Š.n� r � 1/Š

� 1 n � r C

1

r

D .n � 1/Š .r � 1/Š.n� r � 1/Š

� n r.n � r/

� D nŠ rŠ.n� r/Š :

Theorem 2.25 (Binomial Theorem). (i) For all real numbers x and all in-

tegers n � 0,

.1C x/n D nX

rD0

�n r

� xr D

nX

rD0

rŠ.n� r/Š xr :

(ii) For all real numbers a and b and all integers n � 0,

.a C b/n D nX

rD0

n

r

! an�rbr D

nX

rD0

� nŠ rŠ.n� r/Š

� an�rbr :

Proof. (i) This follows from replacing � n r

� by Pascal’s formula in Proposi-

tion 2.24.

(ii) The result is trivially true when a D 0 (we have agreed that 00 D 1). If a ¤ 0, set x D b=a in part (i), and observe that

� 1C b

a

�n D �aC b

a

�n D .a C b/

n

an :

Hence,

.a C b/n D an � 1C b

a

�n D an

nX

rD0

n

r

! br

ar D

nX

rD0

n

r

! an�rbr :

There are many beautiful connections between Pascal’s triangle and number

theory. For example, while it is not generally true that n j � n r

� (for example,

4 − 6 D � 4 2

� ), this result is true when n is prime.

Proposition 2.26. If p is a prime, then p j � p r

� for all r with 0 < r < p.

Proof. By Pascal’s Theorem,

p

r

! D

rŠ.p � r/Š D p.p � 1/ � � � .p � r C 1/

rŠ ;

and cross multiplying gives

p

r

! D p.p � 1/ � � � .p � r C 1/I

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2.2 Binomial Theorem 67

that is, p j rŠ � p r

� . But each factor of rŠ is strictly less than p, because r < p,

so that p is not a divisor of any of them. Therefore, Euclid’s Lemma says that

p − rŠ and, hence, that p must divide � p r

� .

Example 2.27. The Binomial Theorem can be used to express the sum of the

nth powers of two variables a and b in terms of the “elementary symmetric

functions” a C b and ab. Here are some examples for n D 2; 3; 4; from

.a C b/2 D a2 C 2abC b2

we have

a2 C b2 D .a C b/2 � 2ab:

From

.aC b/3 D a3 C 3a2b C 3ab2 C b3

we conclude

a3 C b3 D .a C b/3 � 3ab.aC b/:

For n D 4,

.a C b/4 D a4 C 4a3b C 6a2b2 C 4ab3 C b4

D .a4 C b4/C 4ab.a2 C b2/ � 6.ab/2:

Hence,

a4 C b4 D .a C b/4 � 4ab.a2 C b2/C 6.ab/2

We can now replace a2Cb2 by the already computed expression .aCb/2�2ab, collect like terms, and have an expression for a4 C b4 in terms of a C b and ab.

We could proceed inductively, expressing anC bn in terms of aC b and ab for n � 5. Try a few more examples; you’ll get the sense that there’s a general method expressing anCbn in terms of aCb, ab, and other terms like akCbk with k < n. N

Exercises

2.21 * Show, without using the Binomial Theorem, that, if n � 0 is an integer, then (i) the degree of .1 C x/n is n

(ii) the leading coefficient of .1 C x/n is 1 (iii) the constant term of .1 C x/n is 1.

2.22 Prove Lemma 2.23 by induction on n � 0.

2.23 Prove that the binomial coefficients are symmetric: for all r with 0 � r � n,

n

r

! D

n

n � r

! :

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68 Chapter 2 Induction

2.24 Find a formula for the sum of the entries in the nth row of Pascal’s triangle and

prove your assertion.

2.25 If n � 1, find a formula for the alternating sum of the binomial coefficients in the nth row of Pascal’s triangle:

n

0

! �

n

1

! C

n

2

! � � � � C .�1/n

n

n

! :

Prove what you say.

Hint: Consider f .x/ D .1 C x/n when x D �1.

2.26 If n � 1, find a formula for the sum of the squares of the binomial coefficients in the nth row of Pascal’s triangle:

n

0

!2 C

n

1

!2 C

n

2

!2 � � � C

n

n

!2 :

Prove what you say.

2.27 Prove, for a given n � 1, that the sum of all the binomial coefficients �n r

� with r

even is equal to the sum of all those �n r

� with r odd.

2.28 The triangular numbers count the number of squares in a staircase of height n.

Figure 2.8 displays the staircases of height n for 1 � n � 5.

Figure 2.8. Triangular numbers.

(i) Find a formula for the nth triangular number in terms of binomial coefficients,

and prove your assertion. Compare this exercise with the discussion of the

.n C 1/ � .n C 1/ square in Example 2.6. (ii) Show that the sum of two consecutive triangular numbers is a perfect square.

2.29 Take It Further. Using the notation of Example 2.27, use the Binomial Theorem

and induction to show that an C bn can be expressed in terms of a C b and ab.

2.30 Pascal’s triangle enjoys a sort of hockey stick property: if you start at the end of

any row and draw a hockey stick along a diagonal, as in Figure 2.9, the sum of the

entries on the handle of the stick is the entry at the tip of the blade. Express the

hockey stick property as an identity involving binomial coefficients and prove the

identity.

2.31 (Leibniz) A function f W R ! R is called a C1-function if it has an nth deriva- tive f .n/.x/ for every n � 1. Prove that if f and g are C1-functions, then

.fg/.n/.x/ D nX

kD0

n

k

! f .k/.x/ � g.n�k/.x/:

In spite of the strong resemblance, there is no routine derivation of the Leibniz

formula from the Binomial Theorem (there is a derivation using an idea from

hypergeometric series).

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2.2 Binomial Theorem 69

6435 64351 15 105 455 1365 5005 11510545513655005

1 14 1

1

1

1

1

13

12

11

10

1

1

1

1

1

1

11

1

1 1

1

1

11

1

1

1

1

14

13

12

11

10

2

3 3

4 4

5 5

6 6

7 7

8 8

9 9

3432

1716 1716

924

462 462

252

126 126

70

35 35

20

2121

56 5628 28

10 10

6

15 15

1 1

1 1

84 368436

210 210 120 4512045

30033003

91

78

66

55

91

78

66

55

300330032002 2002 10011001364 364

792 792 495495

12871287 715 286715286

220220

330330 165165

Figure 2.9. Hockey sticks.

2.32 * If p is a prime and a and b are integers, prove that there is an integer c with

.a C b/p D ap C bp C pc:

Combinatorics

Binomial coefficients have a combinatorial interpretation. Given a set X with n elements, define an r-subset of X to be a subset having exactly r elements.

How many r -subsets of X are there?

Example 2.28. There are ten 3-element subsets of the 5-element set X D fA;B; C;D;Eg. Think of forming a 3-person committee from 5 people. A com- mittee either contains Elvis or doesn’t. The committees are

fA;B; C g fA;B;Dg fB; C;Dg fA; C;Dg fA;B;Eg fA; C; Eg fA;D;Eg fB; C; Eg fB;D;Eg fC;D;Eg

The first row consists of the 3-subsets that don’t contain Elvis (there are four

such); the second row displays the 3-subsets that do contain Elvis (there are

six of these). N

In general, if X has n elements and 0 � r � n, denote the number of its r -element subsets by

Œn; r I

that is, Œn; r  is the number of ways one can choose r things from a box of n

things. Note that:

When n D 0, items (i) and (ii) give the same

answer. Why does this

make sense?

(i) Œn; 0 D 1 (there’s only one 0-subset, the empty set ¿). (ii) Œn; n D 1 (there’s only one n-subset of X , X itself).

If 0 < r < n, you can compute Œn; r  using the committee idea in Exam-

ple 2.28. If X D fa1; a2; : : : ; ang and you want to build an r -subset, first choose a “distinguished” element ofX , say an, and call him Elvis. Either your

subset contains Elvis or it doesn’t.

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70 Chapter 2 Induction

Case 1. If Elvis is in your r -subset, then you must pick r � 1 elements from the remaining n� 1; by definition, there are Œn� 1; r � 1 ways to do this. Case 2. If Elvis is not in your r -subset, then you must pick all r elements from

the remaining n� 1; there are Œn� 1; r ways to do this.

It follows that Œn; r  D Œn�1; r�1C Œn�1; r. We have proved the following result.

Lemma 2.29. For all integers n � 1 and all r with 0 � r � n,

Œn; r  D (

1 if n D 0 or n D r Œn � 1; rC Œn� 1; r if 0 < r < n.

The similarity between Lemmas 2.23 and 2.29 inspires the next theorem. It is also the reason why the binomial coefficient

� n r

� is usually pronounced

“n choose r .”

Theorem 2.30 (Counting Subsets). If n and r are nonnegative integers with

0 � r � n, then n

r

! D Œn; r :

Proof. Use induction on n � 0. If n D 0, the inequality 0 � r � n forces r D 0, and

0

0

! D 1 D Œ0; 0:

Suppose the result is true for n� 1. If 0 � r � n, then n

r

! D n � 1 r � 1

! C n � 1 r

! Lemma 2.23

D Œn � 1; r � 1C Œn� 1; r inductive hypothesis

D Œn; r  Lemma 2.29.

Theorems can often be proved in several ways. The following discussion

gives another proof of Theorem 2.30 using Pascal’s formula (Proposition 2.24)

instead of Lemmas 2.23 and 2.29.

We first compute Œn; r  by considering a related question.

Definition. Given an “alphabet” with n (distinct) letters and an integer r with

1 � r � n, an r-anagram is a sequence of r of these letters with no repetitions.

For example, the 2-anagrams on the alphabet a, b, c are

ab; ba; ac; ca; bc; cb

(note that aa, bb, cc are not on this list). How many r -anagrams are there on an

alphabet with n letters? We count the number of such anagrams in two ways.

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2.2 Binomial Theorem 71

(i) There are n choices for the first letter; since no letter is repeated, there are

only n � 1 choices for the second letter, only n � 2 choices for the third letter, and so forth. Thus, the number of r -anagrams is

n.n � 1/.n � 2/ � � � .n � .r � 1// D n.n � 1/.n � 2/ � � � .n� r C 1/:

In the special case n D r , the number of n-anagrams on n letters is nŠ. (ii) Here is a second way to count the anagrams. First choose an r -subset of

the alphabet (consisting of r letters); there are Œn; r  ways to do this, for

this is exactly what the symbol Œn; r  means. For each chosen r -subset,

there are rŠ ways to arrange the r letters in it (this is the special case of

our first count when n D r ). The number of r -anagrams is thus

rŠ Œn; r :

We conclude that

rŠ Œn; r D n.n � 1/.n � 2/ � � � .n � r C 1/;

from which it follows that

Œn; r  D n.n � 1/.n � 2/ � � � .n � r C 1/ rŠ

D nŠ .n � r/Š rŠ:

Therefore, Pascal’s formula gives

Œn; r  D n

r

! :

If you piece together the results of this section, you’ll see that we have

shown that the following ways to define binomial coefficients are all equiva-

lent: starting from any one of them, you can derive the others.

Algebraic: � n r

� is the coefficient of xr in the expansion of .1C x/n.

Pascal:

n

r

! D nŠ rŠ .n� r/Š

:

Combinatorial: � n r

� is the number of r -element subsets of an n-element set.

Inductive:

n

r

! D ( 1 if n D 0 or n D r� n�1 r�1 �

C � n�1 r

� if 0 < r < n.

Example 2.31. If you replace the symbols by their definition, Theorem 2.30

says something that is far from obvious: the coefficient of xr in .1 C x/n is the same as the number of r -element subsets of an n-element set. The proof by

induction of Theorem 2.30 establishes this, but many people are left wondering

if there is a more intuitive reason why the expansion of .1 C x/n contains all the information about subsets of various sizes from an n-element set.

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72 Chapter 2 Induction

If you were going to multiply out .1 C x/5 by hand, you could view the calculation like this:

.1 C x/5 D .1C x/.1 C x/.1 C x/.1C x/.1 C x/

The expansion is carried out by taking one term (1 or x) from each binomial

factor 1C x, multiplying them together, and then collecting like powers of x. For example, you could take a “1” from each of the first three binomials and

an x from the last two. That would produce 1 � 1 � 1 � x � x D x2. But that’s not the only way to get an x2. You could have taken an x from the first and third binomials and 1 from the rest. Or an x from the first two binomials and 1

from the last three. Do this in every possible way; the coefficient of x2 in the

expansion will be the number of ways you can pick two binomials from the set

of five to be “x terms.” And there are precisely 10 D Œ5; 2 ways to do this. Generalizing, view .1 C x/n as a product of n binomials:

.1 C x/n D .1C x/.1 C x/.1 C x/ : : : .1 C x/:„ ƒ‚ … n times:

The coefficient of xr in this product is the number of ways you can choose r of

the binomials to be “x terms” (and the rest to be 1). This number is precisely

Œn; r . Hence

.1 C x/n D nX

rD0 Œn; r xr :

When combined with the definition of binomial coefficients on page 63, this

gives another proof that � n r

� D Œn; r . N

Exercises

2.33 *

(i) For each value of r , 0 � r � 4, how many r-element subsets of the set fA; B; C; Dg are there?

(ii) For each value of r , 0 � r � 5, how many r-element subsets of the set fA; B; C; D; Eg are there?

2.34 How many subsets (of any size) are there in an n-element set? Prove your asser-

tion.

2.35 Show that

2n

r

! D

rX

kD0

n

k

! n

r � k

! :

Hint: Split a 2n-element set into two equal pieces.

2.36 Show that

3n

r

! D

rX

kD0

n

k

! 2n

r � k

! :

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2.3 Connections 73

2.37 If m, n, and r are nonnegative integers, prove Vandermonde’s Identity:

m C n

r

! D

rX

kD0

m

k

! n

r � k

! :

Hint: .1 C x/mCn D .1 C x/m.1 C x/n.

2.38 Show that nX

kD0 k

n

k

! D n2n�1:

2.39 How many ways can you choose two hats from a closet containing 14 different

hats? (One of our friends does not like the phrasing of this exercise. After all,

you can choose two hats with your left hand, with your right hand, with your

teeth, : : : , but we continue the evil tradition.)

2.40 Let D be a collection of ten different dogs, and let C be a collection of ten dif-

ferent cats. Prove that there are the same number of quartets of dogs as there are

sextets of cats.

2.41 (i) What is the coefficient of x16 in .1 C x/20? (ii) How many ways are there to choose 4 colors from a palette containing paints

of 20 different colors?

2.42 A weekly lottery asks you to select 5 different numbers between 1 and 45. At the

week’s end, 5 such numbers are drawn at random, and you win the jackpot if all

your numbers match the drawn numbers. What is your chance of winning?

The number of selections of 5 numbers is “45 choose 5”, which is �45 5

� D

1; 221; 759. The odds against your winning are more than a million to one.

2.3 Connections

An Approach to Induction

Teaching mathematical induction to high school students is often tough. In

particular, many students fall into the trap we described on page 49: in spite

of all our explanations to the contrary, they think that the inductive hypothesis

assumes what it is they are supposed to be proving. In this section, we look at

a well-tested method that avoids this trap.

Suppose you ask a class to come up with a function that agrees with the table

Input Output

0 4

1 7

2 10

3 13

4 16

5 19

We’ve found that about half a high school class (beginning algebra, say) comes

up with a closed form definition, something likef .n/ D 3nC4;while the other

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74 Chapter 2 Induction

half comes up with an inductive definition—something like “start with 4, and

each output is 3 more than the previous one.” This inductive definition can be

written more formally:

g.n/ D ( 4 if n D 0 g.n � 1/C 3 if n > 0.

That inductive definitions (or recursive definitions, as they are often called)

Of course, beginning

students don’t usually

write the definition of g

using this case notation.

Technology can help get

them used to it. seem to be, in some sense, natural for students, can be exploited to help stu-

dents understand proof by induction.

The part of g ’s definition

involving n, the equation

g.n/ D g.n � 1/ C 3, is called a recurrence.

Computer algebra systems (CAS) let you model the two definitions. Build-

ing such computational models allows students to experiment with both func- tions, and it also provides an opportunity to launch some important ideas. For

example, a teacher can use the models to discuss the domain of each function—

f accepts any real number, but g will accept only nonnegative integers.

Here’s what’s most germane to this section. If you try some values in a

spreadsheet or calculator, it seems, for a while, that f and g produce the same output when given the same input. But at some point (exactly where depends

on the system), f outputs an integer but g surrenders. Suppose this happens,

for example, at 255: both functions return 766 at 254, but f .255/ D 769 while g returns an error. Is this because f and g really are not equal at 255?

Or is it because of the limitations of the technology? Many students will say

immediately that the functions are equal when n D 255; it’s just that the com- puter can’t compute the value of g there. Tell the students, “I believe you that

f .254/ D g.254/. Convince me that they are also equal at 255.” After some polishing and a little help, their argument usually goes some-

thing like this.

g.255/ D g.254/C 3 (this is how g is defined) D f .254/C 3 (the calculator said so—they both output 766) D .3 � 254C 4/C 3 (this is how f is defined) D .3 � 254C 3/C 4 (algebra) D 3.254C 1/C 4 (more algebra) D 3 � 255C 4 (arithmetic) D f .255/ (this is how f is defined).

There’s nothing special about 255 here. If you had a more powerful calculator,

one that handled inputs to g up to, say, 567 (and then crashed at 568), you

could argue that f and g were equal at 568, too.

g.568/ D g.567/C 3 (this is how g is defined) D f .567/C 3 (the powerful calculator said so) D .3 � 567C 4/C 3 (this is how f is defined) D .3 � 567C 3/C 4 (algebra) D 3.567C 1/C 4 (more algebra) D 3 � 568C 4 (arithmetic) D f .568/ (this is how f is defined).

Computing g.568/ is the same as computing g.255/. Now imagine that you

had a virtual calculator, one that showed that f and g agreed up to some integer

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2.3 Connections 75

n � 1, but then crashed when you asked for g.n/. You could show that f and g are equal at n by the same argument:

g.n/ D g.n � 1/C 3 (this is how g is defined) D f .n � 1/C 3 (the virtual calculator said so) D .3 � .n � 1/C 4/C 3 (this is how f is defined) D 3nC 4 (algebra) D f .n/ (this is how f is defined).

So, every time f and g are equal at one integer, they are equal at the next

one. Since f and g are equal at 0 (in fact, since they are equal at every integer

between 0 and 254), they are equal at every nonnegative integer.

This argument is the essence of mathematical induction. In the example, it

shows that if two functions f and g are equal at one integer, then they are equal at the next one. Coupled with the fact that they are equal at 0, it makes

sense that they are equal for all integers greater than or equal to 0; that is,

f .n/ D g.n/ for all nonnegative integers n. We have seen that induction applies in much more general situations than

this one. But this simple context is quite effective in starting students onto a

path that helps them understand induction.

Fibonacci Sequence

Many interesting investigations in high school center around the following se- quence, which describes a pattern frequently found in nature and in art.

Definition. The Fibonacci sequence is defined by:

Fn D

€ 0 n D 0 1 n D 1 Fn�1 C Fn�2 n > 1:

There are two base steps in the definition: n D 0 and n D 1. The Fibonacci sequence begins: 0; 1; 1; 2; 3; 5; 8; 13; : : :.

Historical Note. The Fibonacci sequence is related to the golden ratio, a

number mentioned in Euclid, Book 6, Proposition 30. It is said that the ancient Greeks thought that a rectangular figure is most pleasing to the eye (such rect-

angles can be seen in the Parthenon in Athens) if its edges a and b are in the

proportion

a

b D

b

aC b :

In this case, a.aCb/ D b2, so that b2�ab�a2 D 0; that is, .b=a/2�.b=a/�1 D 0. The quadratic formula gives b=a D 1

2 .1 ˙

p 5/. Therefore,

b=a D D 1 2 .1 C

p 5/ or b=a D ı D 1

2 .1 �

p 5/:

But ı is negative, and so we must have

b=a D D 1 2 .1 C

p 5/:

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76 Chapter 2 Induction

Figure 2.10. Golden rectangle.

The number D 1:61803 : : :, is called the golden ratio. Since both and ı are roots of x2 � x � 1, we have

2 D C 1 and ı2 D ı C 1: (2.4)

So, what’s the connection of the golden ratio to the Fibonacci sequence?

We discovered the closed form for the sequence cn in Proposition 2.22 by

tabulating the first few terms of the sequence and looking for regularity—it

seemed “almost” exponential, off by 1 from a power of 2. Let’s tabulate the first few ratios of consecutive terms Fn=Fn�1 of the Fi-

bonacci sequence.

F1=F0 D 1=1 D 1 F2=F1 D 2=1 D 2 F3=F2 D 3=2 D 1:5 F4=F3 D 5=3 D 1:666 F5=F4 D 8=5 D 1:6 F6=F5 D 13=8 D 1:625:

If you tabulate a few more ratios (try it), a conjecture emerges—it appears that

the ratio of two consecutive terms in the Fibonacci sequence might converge to the golden ratio � 1:61803 (if the ratios were actually constant, Fn would be a geometric sequence (why?)). This is, in fact, the case, and you’ll see, in

Exercise 2.50 on page 78, how to refine the conjecture into the statement of the

following theorem (the exercise will also help you develop a method that will

let you find closed forms for many 2-term recurrences).

Theorem 2.32. For all n � 0, the nth term of the Fibonacci sequence satisfies

Fn D 1p 5 . n � ın/;

where D 1 2 .1 C

p 5/ and ı D 1

2 .1 �

p 5/:

Proof. We use strong induction because the inductive step involves the formu-

las for both Fn�1 and Fn�2. The base steps S.0/ and S.1/ are true:

1p 5 . 0 � ı0/ D 0 D F0 1p 5 . � ı/ D 1p

5

� 1 2 .1C

p 5/ � 1

2 .1 �

p 5/ �

D 1 D F1:

If n � 2, then

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2.3 Connections 77

Fn D Fn�1 C Fn�2 D 1p

5 . n�1 � ın�1/C 1p

5 . n�2 � ın�2/

D 1p 5

� . n�1 C n�2/� .ın�1 C ın�2/

D 1p 5

� n�2. C 1/� ın�2.ı C 1/

D 1p 5

� n�2. 2/ � ın�2.ı2/

� by Eq. (2.4)

D 1p 5 . n � ın/:

Isn’t it curious that the integers Fn are expressed in terms of the irrational

number p 5?

If n D 2, then F2 D 1 D 0, and so there is equality,

not inequality. Corollary 2.33. Fn >

n�2 for all integers n � 3, where D 1 2

� 1C

p 5 � .

Proof. The proof is by induction on n � 3. The base step S.3/ is true, for F3 D 2 > � 1:618. For the inductive step, we must show that FnC1 > n�1. By the inductive hypothesis,

FnC1 D Fn C Fn�1 > n�2 C n�3

D n�3. C 1/ D n�3 2 D n�1:

Exercises

2.43 Show that the following functions agree on all natural numbers.

f .n/ D 3n C 5 and g.n/ D (

5 if n D 0 g.n � 1/ C 3 if n > 0.

2.44 Show that the following two functions agree on all natural numbers.

f .n/ D 4n and g.n/ D (

4 if n D 0 4g.n � 1/ if n > 0.

2.45 Define the function h inductively:

h.n/ D (

4 if n D 0 h.n � 1/ C 2n if n > 0

Find a polynomial function p that agrees with h on all natural numbers, and prove

that your functions are equal on N .

Answer: n2 C n C 4.

2.46 Define the function m inductively:

m.n/ D (

0 if n D 0 m.n � 1/ C n2 if n > 0

Find a polynomial function s that agrees with m on all natural numbers, and prove

that your functions are equal on N .

Answer: 2n 3C3n2Cn 6 D

n.nC1/.2nC1/ 6 .

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78 Chapter 2 Induction

2.47 Consider the two functions f and g:

f .x/ D x4 � 6x3 C 14x2 � 6x C 2

and

g.x/ D (

2 if x D 0 s.x � 1/ C 6x � 3 if x > 0.

Are f and g equal on N?

2.48 Find a formula for 02 C 12 C 22 C � � � C .n � 1/2 as a function of n, and prove your assertion.

2.49 You saw, on page 76, that the ratio of two consecutive terms seems to converge to

the golden ratio. Using only the recurrence

Fn D Fn�1 C Fn�2 for all n � 2;

show that

lim n!1

Fn

Fn�1 D 12 .1 C

p 5/:

2.50 * You saw, on page 76, that the Fibonacci sequence seems to be “almost” expo-

nential.

(i) Suppose the Fibonacci sequence actually was exponential: Fn D rn. Show that r would have to be either

D 1 C p

5

2 or ı D 1 �

p 5

2 :

(ii) Show that the sequences n and ın satisfy the recurrence

fn D fn�1 C fn�2:

(iii) If a and b are any real numbers, show that a nC bın satisfies the recurrence

fn D fn�1 C fn�2:

(iv) Without using Theorem 2.32, find a and b so that

a n C bın D Fn:

2.51 Ms. D’Amato likes to take a different route to work every day. She will quit her

job the day she has to repeat her route. Her home and work are pictured in the grid

of streets in Figure 2.11. If she never backtracks (she only travels north or east),

how many days will she work at her job?

A valid trip Another valid trip Not a valid trip

Figure 2.11. Ms. D’Amato.

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2.3 Connections 79

2.52 Find a closed form for each of the following functions and prove your assertions.

(i) f .n/ D (

4 n D 0 f .n � 1/ C 3 n > 0:

Answer: f .n/ D 3n C 4.

(ii) f .n/ D (

4 n D 0 3f .n � 1/ n > 0:

Answer: f .n/ D 4 � 3n.

(iii) f .n/ D

˚ 2 n D 0 4 n D 1 4f .n � 1/ � 3f .n � 2/ n > 1:

Answer: f .n/ D 3n C 1.

(iv) f .n/ D

˚ 4 n D 0 4 n D 1 4f .n � 1/ � 3f .n � 2/ n > 1:

Answer: f .n/ D 4.

2.53 Find a closed form for the following function and prove your assertion.

f .n/ D

„ 3 n D 0 4 n D 1 14 n D 2 4f .n � 1/ � f .n � 2/ � 6f .n � 3/ n > 2:

2.54 Take It Further. Find or generate a copy of the first 30 rows of Pascal’s triangle.

Color the odd numbers red and the even numbers black. Explain any patterns that

you see. (Alternatively, you can use a spreadsheet to generate the triangle of 0s

and 1s that are the remainders when each entry is divided by 2.) For more on this

exercise, see

ecademy.agnesscott.edu/˜lriddle/ifs/siertri/Pascalmath.htm

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3 Renaissance For centuries, the Western World believed that the high point of civilization

took place from the Greek and Roman eras through the beginning of Chris-

tianity. But this worldview began to change dramatically about five hundred years ago. The printing press was invented around 1450, by Johannes Guten-

berg, Christopher Columbus landed in North America in 1492, Martin Luther

began the Reformation in 1517, and Nicolas Copernicus published De Revolu-

tionibus in 1530.

Mathematics was also developing. A formula giving the roots of certain cubic polynomials, similar to the quadratic formula, was discovered by Scipi-

one del Ferro around 1515; by 1545, it was extended to all cubics by Fontana

(Tartaglia) and Cardano. The cubic formula contributed to the change in world-

view that was the essence of the Renaissance, for it was one of the first math-

ematical results not known to the ancients. But its impact on contemporary

mathematics was much deeper, for it introduced complex numbers. As we shall see, the cubic formula is not as useful for numerical computations as we’d like,

because it often gives roots in unrecognizable form. Its importance, however,

lies in the ideas it generated. Trying to understand the formula, searching for

generalizations of it, and studying questions naturally arising from such en-

deavors, were driving forces in the development of abstract algebra. In many high school algebra courses today, the complex numbers, usually

denoted by C, are introduced to find the roots of ax2CbxCcwhen b2�4ac < 0. That’s not how it happened. Square roots of negative real numbers occur in

the cubic formula, but not as roots; indeed, in the 16th century, complex roots

would have been considered useless. But complex numbers arose in the middle of calculations, eventually producing real numbers (we will see this explicitly

in the next section). To understand this phenomenon, mathematicians were

forced to investigate the meaning of number; are complex numbers bona fide

numbers? Are negative numbers bona fide numbers?

Section 3.1 discusses the classical formulas giving the roots of cubic and

quartic polynomials. We will look more carefully at the complex numbers themselves in Section 3.2. Although initially used in purely algebraic contexts,

C has a rich geometric and analytic structure that, when taken together with its

algebraic properties, can tie together many of the ideas in high school mathe-

matics. Indeed, C finds applications all over mathematics. Section 3.4 uses C

to solve some problems that are especially useful for teachers (and interesting for all mathematicians). Just as the method of Diophantus was used to create

Pythagorean triples, C can be used to invent problems whose solutions “come

out nice.”

81

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82 Chapter 3 Renaissance

3.1 Classical Formulas

As Europe emerged from the Dark Ages, a major open problem in mathemat-

ics was finding solutions to polynomial equations. The quadratic formula had

been known for about four thousand years and, arising from a tradition of pub-

lic mathematical contests in Pisa and Venice, formulas for the roots of cubics

and quartics had been found in the early 1500s. Let’s look at these formulas

in modern algebraic notation; we will assume for now that the complex num- bers obey the usual laws of arithmetic (neither of these simplifying steps was

available to mathematicians of the 16th century).

Historical Note. Modern arithmetic notation was introduced in the late

1500s, but it was not generally agreed upon in Europe until after the influ-

ential book of Descartes, La Géométrie, was published in 1637 (before then,

words and abbreviations were used as well as various competing notations).

The symbols +, �, and p, as well as the symbol / for division, as in 4=5, were introduced by Widman in 1486. The equality sign, D, was invented by Recorde in 1557. Designating variables by letters was invented by Viète in

1591, who used consonants to denote constants and vowels to denote vari-

ables; the modern notation, using the letters a; b; c; : : : to denote constants and

the letters x; y; z at the end of the alphabet to denote variables, was introduced

by Descartes in 1637. The exponential notation 22; 23; 24; : : : was essentially invented by Hume in 1636, who wrote 2ii, 2iii, 2iv; : : : : The symbol � for mul- tiplication was introduced by Oughtred in 1631; the symbol � for division was introduced by Rahn in 1659. See Cajori [6].

Cubics

The following familiar fact (to be proved in Chapter 6) was known and used

by Renaissance mathematicians, and we will use it in this section.

Proposition 6.15. If r is a root of a polynomial f .x/, then x � r is a factor of f .x/I that is, f .x/ D .x � r/g.x/ for some polynomial g.x/.

One of the simplest cubics is f .x/ D x3�1. Obviously, 1 is root of f , and so x3 � 1 D .x � 1/g.x/, where

g.x/ D .x3 � 1/=.x � 1/ D x2 C x C 1:

The roots of g (and, hence, also of f ) are

! D 1 2 .�1 C i

p 3/ and ! D 1

2 .�1 � i

p 3/;

by the quadratic formula. Both ! and ! are called cube roots of unity, for

!3 D 1 D !3. Note that ! D !2 D 1=!. We know that a positive number a has two square roots. By convention,

p a

denotes the positive square root, so that the two square roots are ˙ p a. Any

real number a has three cube roots. By convention, 3 p a denotes the real cube

root, so that the three cube roots are 3 p a; ! 3

p a; !2 3

p a. Thus, cube roots of

unity generalize ˙. The general cubic equation aX3 C bX2 C cX C d D 0 can be simplified

by dividing both sides by a; this procedure does not affect the roots, and so

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3.1 Classical Formulas 83

we may assume that a D 1. Thus, we seek the roots of the cubic polynomial F.X/ D X3 C bX2 C cX C d , where b; c; d 2 R. The change of variable

X D x � 1 3 b

yields a simpler polynomial, f .x/ D F.x � 1 3 b/ D x3 C qx C r , where q

and r are expressions in b, c, and d . We call f the reduced polynomial arising

from F .

Lemma 3.1. Let f .x/ D x3CqxC r be the reduced polynomial arising from F.X/ D X3 C bX2 C cX C d D 0. If u is a root of f , then u � 1

3 b is a root

Exercise 3.5 on page 89

asks you to check that

the coefficient of x2 in

F .x� 13 b/ D .x� 13 b/3C b.x� 1

3 b/2Cc.x� 1

3 b/Cd

is zero. of F .

Proof. Since f .x/ D F.x � 1 3 b/ for all x, we have 0 D f .u/ D F.u � 1

3 b/;

that is, u� 1 3 b is a root of F.X/.

We will use the following consequence of the quadratic formula.

Lemma 3.2. Given a pair of numbersM andN , there are .possibly complex/

numbers g and h with g C h D M and gh D N . In fact, g and h are roots of x2 �Mx CN .

Proof. We have

.x � g/.x � h/ D x2 � .g C h/x C gh:

Thus, the roots g; h of f .x/ D x2 � Mx C N (which exist, thanks to the quadratic formula) satisfy the given equations g C h D M and gh D N .

Let’s try to find a general method for solving cubic equations—a method that doesn’t depend on the specific values of the coefficients—by first solving

a numerical equation.

Consider the polynomial f .x/ D x3 � 18x � 35. Since the constant term 35 D 5 � 7, we check whether ˙1;˙5;˙7 are roots. It turns out that 5 is a root and, dividing by x � 5, we can find the other two roots by solving the You can check that the

other two roots are

complex. Renaissance

mathematicians would

have dismissed these as

meaningless. But stay

tuned—we’ll soon see that

they, too, can be generated

by the emerging method.

quadratic f .x/=.x � 5/ D x2 C 5x C 7. But we are looking for a general method applicable to other cubics, so let’s pretend we don’t know that 5 is a

root.

It’s natural to look for a polynomial identity having the same form as the

equation we are trying to solve. Example 2.27 provides one. From

a3 C b3 D .a C b/3 � 3ab.aC b/;

we have the identity

.a C b/3 � 3ab.aC b/� � a3 C b3

� D 0:

Thinking of a C b as a single “chunk,” say, x D a C b, the correspondence looks like this:

.a C b/3 �3ab � .a C b/ �.a3 C b3/ D 0 # # # #

�3 �3ab � � �.a3 C b3/ D 0 " " " " x3 �18 � x �35 D 0:

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84 Chapter 3 Renaissance

So, if we can find two numbers a and b such that

�3ab D �18 and a3 C b3 D 35;

then a C b will be a root of the cubic. Hence we want a and b so that ab D 6 and a3Cb3 D 35. There’s an obvious solution here, namely, a D 3 and b D 2, but we’re looking for a general method. Cubing both sides of ab D 6, we get

a3b3 D 216 and a3 C b3 D 35:

By Lemma 3.2, a3 and b3 are roots of the quadratic equation

x2 � 35x C 216 D 0:

The roots of this are 27 and 8. So we can take a3 D 27, b3 D 8; surprise! a D 3 and b D 2. Hence, 3C 2 D 5 is a root of our original cubic.

The next theorem is usually attributed to Scipione del Ferro; we’ll use com-

plex numbers and modern notation in its statement and proof, neither of which

was available at the time. In light of Lemma 3.1, we may assume that cubics

are reduced.

Theorem 3.3 (Cubic Formula). The roots of f .x/ D x3 C qx C r are

g C h; !g C !2h; and !2g C !h;

where ! D 1 2

� �1C i

p 3 �

is a cube root of unity,

g3 D �r C

p R

2 ; h D �

q

3g ; and R D r2 C

4q3

27 :

Proof. Let u be a root of f .x/ D x3CqxC r and, as in the discussion above, we try

u D g C h:

We are led to

g3 C h3 D �r gh D �1

3 q:

Cube gh D �1 3 q, obtaining the pair of equations

g3 C h3 D �r g3h3 D � 1

27 q3:

Lemma 3.2 gives a quadratic equation whose roots are g3 and h3:

x2 C rx � 1 27 q3 D 0: (3.1)

The quadratic formula gives

g3 D 1 2

� �r C

q r2 C 4

27 q3 �

D 1 2

� �r C

p R �

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3.1 Classical Formulas 85

and

h3 D 1 2

� �r �

q r2 C 4

27 q3 �

D 1 2

� �r �

p R � :

Now there are three cube roots of g3, namely, g, !g, and !2g. Because

of the constraint gh D �1 3 q, each has a “mate,” namely, �q=.3g/ D h,

�q=.3!g/ D !2h, and �q=.3!2g/ D !h. Thus, the roots of f are

g C h; !g C its mate; !2g C its mateI

that is, the roots of f are g C h, !g C !2h, and !2g C !h.

Example 3.4 (Good Example). If f .x/ D x3 � 15x � 126, then q D �15, r D �126, R D 15376, and

p R D 124. Hence, g3 D 125, so we can take

g D 5. Thus, h D �q=3g D 1. Therefore, the roots of f are

Alternatively, having found

one root to be 6, the

other two roots can be

found as the roots of the

quadratic f.x/=.x� 6/ D x2C 6x C 21.

6; 5! C !2 D �3C 2i p 3; 5!2 C ! D �3 � 2i

p 3:

For Renaissance mathematicians, this cubic would have only one root—they would have ignored the complex roots. N

But things don’t always work out as we expect, as the next surprising ex-

ample shows.

Example 3.5 (Bad Example). The cubic formula may give the roots in unrec-

ognizable form. Let

f .x/ D .x � 1/.x � 2/.x C 3/ D x3 � 7x C 6I

the roots of f are, obviously, 1; 2, and �3. But the cubic formula gives

g3 D 1 2

� �6C

q �400 27

� and h3 D 1

2

� �6 �

q �400 27

� :

It is not at all obvious that gC h is a real number, let alone an integer! N

Imagine yourself, standing in Piazza San Marco in Venice in 1520, partici-

pating in a contest. Your opponent challenges you to find a root of f .x/ D x3 � 7x C 6 (he invented the cubic, so he knows that it comes from .x � 1/.x � 2/.x C 3/). Still, you are a clever rascal; your mentor taught you the cubic formula. You do as you were taught, and triumphantly announce that

a root is gC h, where g3; h3 are the awful expressions above. Most likely, the judges would agree that your opponent, who says that 1 is a root, has defeated you. After all, f .1/ D 1 � 7C 6 D 0, so that 1 is, indeed, a root. The judges even snickered when they asked you to evaluate f .g C h/.

With head hung low, you return home. Can you simplify your answer? Why

is g C h equal to 1? Let’s pretend you have modern notation. Well,

g3 D 1 2

� �6C

q �400 27

� D �3C i 10

p 3

9 :

The first question is how to compute cube roots of “numbers” of the form

aC bi , where i2 D �1. Specifically, we want uC iv with

.uC iv/3 D �3C i 10 p 3

9 :

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86 Chapter 3 Renaissance

Hmm! Perhaps it’s smart to separate terms involving i from honest numbers.

.uC iv/3 D u3 C 3u2iv C 3u.iv/2 C .iv/3

D u3 C 3u2iv � 3uv2 � iv3

D u3 � 3uv2 C i.3u2v � v3/:

Let’s see if the separation pays off. We want numbers u; v with u3 � 3uv2 DBut see Example 3.36, or try to solve the system

u3 � 3uv2 D �3 and 3u2v � v3 D 10

p 3

9 with a

computer.

�3 and 3u2v � v3 D 10 p 3

9 . These equations are intractible! Sigh.

Cube roots are tough. Let’s simplify things; perhaps solving a simpler prob-

lem, say, finding square roots, can give a clue to finding cube roots. And this

we can do.

Proposition 3.6. Every complex number a C bi has a square root.

Proof. If b D 0, then aC ib D a. If a � 0, then p a is well-known; if a < 0,

then a D �c, where c > 0, and p a D i

p c. We can now assume that b ¤ 0,

and our task is to find uC iv with

.uC iv/2 D u2 C 2iuv � v2 D a C ibI

that is, we seek numbers u; v such that

u2 � v2 D a (3.2) and

2uv D b: (3.3)

Since b ¤ 0, Eq. (3.3) gives u ¤ 0; define v D b=2u. Substituting into Eq. (3.2), we have

u2 � .b=2u/2 D aI rewriting,

4u4 � 4au2 � b2 D 0:

This is a quadratic in u2, and the quadratic formula gives

u2 D 1 8

� 4a˙

p 16a2 C 16b2

D 1 2

� a˙

p a2 C b2

� :

Since a2Cb2 > 0, it has a real square root. Now 1 2

� aC

p a2 C b2

� is positive

(because b2 > 0 implies a < p a2 C b2); hence, we can find its (real) square

root u as well as v D b=2u.

For example, our proof gives a method finding a square root of i . Set a D 0 and b D 1 to obtain

i D � 1p 2 .1 C i/

�2 :

Alas, this square root success doesn’t lead to a cube root success, although it We’ll see how to find the

roots of a complex number

in Section 3.3. does give us some confidence that our manipulations may be legitimate.

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3.1 Classical Formulas 87

You can now appreciate the confusion produced by the cubic formula; a

cloud enveloped our ancestors. First of all, what are these “numbers” a C ib? Sometimes they can help. Can we trust them to always give us the truth? Is it true that we can separate terms involving i from those that don’t? When are

two complex numbers equal? Does it make sense to do arithmetic with these

guys? Do they obey the nine properties of arithmetic on page 40 that familiar

numbers do? It took mathematicians about 100 years to become comfortable

with complex numbers, and another 100 years until all was set on a firm foun- dation.

Quartics

A method for solving fourth degree equations was found by Lodovico Ferrari

in the 1540s, but we present the version given by Descartes in 1637. Consider the quartic F.X/ D X4 C bX3 C cX2 C dX C e. The change

of variable X D x � 1 4 b yields a simpler polynomial, f .x/ D F.x � 1

4 b/ D

x4 C qx2 C rx C s, whose roots give the roots of F : if u is a root of f , then See Exercise 3.6 on page 89.u� 1

4 b is a root of F . Write f as a product of two quadratics:

f .x/ D x4 C qx2 C rx C s D .x2 C jx C `/.x2 � jx Cm/;

and determine j , `, and m (note that the coefficients of the linear terms in the

quadratic factors are j and �j because f has no cubic term). Expanding and equating like coefficients gives the equations

`Cm� j 2 D q; j.m� `/ D r;

`m D s:

The first two equations give Since j.m � `/ D r , we have �`Cm D r=j .

2m D j 2 C q C r j ;

2` D j 2 C q � r j :

Substituting these values for m and ` into the third equation and simplifying

yield a degree 6 polynomial which is a cubic in j 2 (called the resolvent cubic):

.j 2/3 C 2q.j 2/2 C .q2 � 4s/j 2 � r2:

The cubic formula gives a root j 2, from which we can determinem and ` and,

hence, the roots of the quartic.

This process is an algorithm that can easily be encoded in a computer al- gebra system; it is known as the quartic formula. The quartic formula has the

same disadvantage as the cubic formula: even though it gives correct answers,

the values it gives for the roots are usually unrecognizable. But there are some

good examples.

Example 3.7. Let’s find the roots of

f .x/ D x4 � 10x2 C 1:

First, factor f :

x4 � 10x2 C 1 D .x2 C jx C `/.x2 � jx Cm/I

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88 Chapter 3 Renaissance

in our earlier notation, q D �10, r D 0, and s D 1. The quartic formula showsWhat’s going on? Why is there a choice for ` and

j 2? If the roots of f are

˛i for 1 � i � 4, then f.x/ D

Q i.x � ˛i /. A

factorization of f into

quadratics arises from

grouping these four factors

into pairs, and there is

no reason why different

groupings should give

the same quadratics.

Of course, any such

factorization gives the

same roots of f .

us how to find j; `; m. Since r D 0, we have 2` D j 2 � 10 D 2m; hence, ` D m. But `m D 1, so that either ` D 1 and j 2 D 12 or ` D �1 and j 2 D 8. Taking ` D 1 and j 2 D 12 gives

f .x/ D .x2 C p 12x C 1/.x2 �

p 12x C 1/;

and the quadratic formula gives the four roots of f :

˛ D p 2C

p 3; ˇ D �

p 2C

p 3; D

p 2 �

p 3; ı D �

p 2 �

p 3: N

The quadratic formula can be derived in a way similar to the derivations

of the cubic and quartic formulas (in Chapter 1, we derived the formula by

completing the square). The change of variable X D x � 12b replaces the polynomialF.X/ D X2CbXCc with the simpler polynomialf .x/ D x2Cq, where q D c � 1

4 b2; the roots u D ˙p�q of f .x/ give the roots u � 1

2 b of

F . Since the roots of f are

u D ˙p�q D ˙ q

�.c � 1 4 b2/ D ˙ 1

2

p b2 � 4c;

the roots of F are our old friends

˙ 12 p b2 � 4c � 1

2 b D 1

2

� � b ˙

p b2 � 4c

� :

It is now tempting, as it was for our ancestors, to try to find the roots of the

general quinticF.X/ D X5 C bX4 C cX3 C dX2 C eX C f and to express them in a form similar to those for quadratic, cubic, and quartic polynomials;

that is, using only extraction of roots, addition, subtraction, multiplication, and division (of course, our ancestors hoped to find roots of polynomials of any

degree). They began with the change of variable X D x � 1 5 b to eliminate

the X4 term. It was natural to expect that some further ingenious substitution

together with the formulas for roots of polynomials of lower degree, analogous

to the resolvent cubic, would yield the roots of F . For almost 300 years, no

such formula was found. But, in 1824, Abel proved that there is no such quintic formula.

How to Think About It. Abel’s theorem is often misquoted. It says: there is

no formula involving only extraction of roots and the four basic operations of arithmetic that expresses the roots of the general quintic polynomial in terms

of its coefficients. Succinctly, the general quintic is not solvable by radicals.

But there are other kinds of formulas giving roots of polynomials. For exam-

ple, here is a formula, due to Viète, giving the roots in terms of trigonometric

functions. If f .x/ D x3CqxC r has three real roots, then its roots are t cos � , t cos.�C120ı/, t cos.�C240ı/, where t D

p �4q=3 and cos.3�/ D �4r=t3

(there are variations using cosh and sinh when f has complex roots ([26],

p. 445–447)). You may recall Newton’s method giving the roots as limn!1 xn, where xnC1 D xn � f .xn/=f 0.xn/. Now some quintic polynomials are solv- able by radicals; for example, we’ll see in Section 3.3 that x5 � 1 is one such. Another theorem of Abel gives a class of polynomials, of any degree, which are solvable by radicals. Galois, the young wizard who was killed before hisAbel (1802–1829) also

died young. 21st birthday, characterized all the polynomials which are solvable by radi-

cals, greatly generalizing Abel’s theorem. We will look at this more closely in

Chapter 9.

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3.1 Classical Formulas 89

Exercises

3.1 For each equation, find all roots in R and in C

(i) x2 � 2x D 15 (ii) x2 � 2x D 16 (iii) x2 � 2x D �16 (iv) 6x2 C x D 15 (v) 6x2 C x D 16 (vi) 6x2 C x D �16

(vii) x2 D 1 (viii) x3 D 1 (ix) x4 D 1 (x) x3 D 8

3.2 * We know that i satisfies x2 C 1 D 0 in C (is there another solution?). (i) Show, for all n 2 Z, that the value of in is one of 1; i; �1; �i .

(ii) Use the Division Algorithm to decide which of the four values i247 will have.

3.3 Let ! D 12 .�1 C i p

3/ be a cube root of unity.

(i) Show, for every integer n, the value of !n is one of 1; !; !2 .

(ii) Use the Division Algorithm to decide, for any fixed n, which of the three

values !n will have.

3.4 Find two numbers whose

(i) sum is 5 and product is 6. (ii) sum is 0 and product is �2. (iii) sum is 3 and product is 3. (iv) sum is �1 and product is 1. (v) sum is b and product is c (in terms of b and c).(vi)

3.5 * If F.X/ D X3CbX2CcX Cd , show that the change of variable X D x � 13b produces a polynomial f with no quadratic term,

f .x/ D F.x � 1 3

b/ D x3 C qx C r:

Express q and r in terms of b, c, and d .

3.6 *

(i) Suppose that F.X/ D X4 C bX3 C cX2 C dX C e. (a) Show that the change of variable X D x � 14b produces a polynomial f

with no cubic term,

f .x/ D F.x � 14b/ D x 4 C qx2 C rx C s:

Express q, r , and s in terms of b, c, d , and e.

(b) Show that if u is a root of f , then u � 1 4

b is a root of F .

(ii) In general, let

F.X/ D Xn C an�1Xn�1 C an�2Xn�2 C � � � C a0

be a polynomial of degree n.

(a) Show that the change of variable X D x� 1 n

an�1 produces a polynomial f with no term of degree n � 1,

f .x/ D F.x � 1nan�1/ D x n C qn�2xn�2 C � � � C q0:

(b) Show that if u is a root of f , then u � 1nan�1 is a root of F:

3.7 Take It Further. Suppose that g and h are complex numbers and

! D 1 2

� �1 C i

p 3 �

:

Show that

g3 C h3 D .g C h/.!g C !2h/.!2g C !h/:

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90 Chapter 3 Renaissance

3.8 In Example 3.7, we found the roots of x4 � 10x2 C 1 by factoring it into two quadratics (which came from taking, in the notation of page 87, ` D 1 and j 2 D 12). Another choice was ` D �1 and j 2 D 8.

(i) Using the alternate choice, get a different factorization of the quartic into

quadratic factors.

(ii) Show that the two factorizations produce the same linear factors.

3.9 The following problem, from an old Chinese text, was solved by Qin Jiushao

(Ch’in Chiu-shao) in 1247. There is a circular castle (see Figure 3.1) whose di-

ameter is unknown; it is provided with four gates, and two lengths out of the

north gate there is a large tree, which is visible from a point six lengths east of the

south gate. What is the length of the diameter? (The answer is a root of a cubic

polynomial.)

S E

N C

O

T

2

r

r

r

a

6

Figure 3.1. Castle problem.

3.10 Show that there is no real number whose square is �1.

3.11 (i) Find the roots of x3 � 3x C 1 D 0. (ii) Find the roots of x4 � 2x2 C 8x � 3 D 0.

3.12 Find a complex number s so that s3 D 9 � 46i .

3.13 Find the roots of x3 � 21x C 20. (i) by finding a root and reducing the cubic to a quadratic.

(ii) by the cubic formula.

(iii) Verify that the answers are the same.

3.14 Suppose that ˛ and ˇ are roots of the quadratic equation x2 C bx C c D 0. Find expressions in terms of b and c for

(i) ˛ C ˇ (ii) ˛2 C ˇ2

(iii) ˛3 C ˇ3

(iv) .˛ � ˇ/2

(v) Use parts (i) and (iv) to derive the quadratic formula.

3.15 Suppose that ˛, ˇ, and are roots of the cubic equation x3C bx2C cx C d D 0. Show that

(i) ˛ C ˇ C D �b (ii) ˛ˇ C ˛ C ˇ D c

(iii) ˛ˇ D �d .

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3.2 Complex Numbers 91

3.16 (i) Show that a rectangle is determined by its perimeter and area.

(ii) Is a rectangular box determined by its volume and surface area? Explain.

3.17 Suppose that ˛, ˇ, and are roots of the cubic equation x3C bx2C cx C d D 0. Exercise 3.17 can be done without the cubic formula.Find, in terms of b, c, and d ,

(i) ˛2 C ˇ2 C 2

(ii) ˛3 C ˇ3 C 3

(iii) ˛2ˇ2 C ˛2ˇ 2 C ˛ˇ2 2

3.18 Take It Further.

(i) Suppose that ˛, ˇ, and are three numbers whose sum is 0. Show that

� .˛ � ˇ/.˛ � /.ˇ � /

�2 C 4.˛ˇ C ˛ C ˇ /3 C 27 � ˛ˇ

�3 D 0:

(ii) Suppose that ˛, ˇ, and are roots of x3 C qx C r . Show that

� � .˛ � ˇ/.˛ � /.ˇ � /

�2 D 27r2 C 4q3:

3.19 Take It Further. When finding the roots of x3Cqx C r with the cubic formula, you are led to Eq. (3.1): x2 C rx � 1

27 q3, whose roots are g3 and h3.

(i) Show that the discriminant ı of this quadratic is

ı D r2 C 427q 3

(ii) If ı > 0, show that the cubic has one real root and two complex conjugate

roots. The discriminant of the cubic x3CqxCr is defined to be D �4q3 � 27r2.(iii) If ı D 0, show that the cubic has two real roots, one of them with multiplic-

ity 2.

(iv) If ı < 0, show that the cubic has three distinct real roots.

3.2 Complex Numbers

Before the cubic formula, mathematicians had no difficulty in ignoring neg- ative numbers or square roots of negative numbers. For example, consider

the problem of finding the sides x and y of a rectangle having area A and

perimeter p. The equations xy D A and 2x C 2y D p give the equation 2x2 � px C 2A D 0, and the quadratic formula gives

x D 1 4

� p˙

p p2 � 16A

� :

If p2 � 16A � 0, the problem is solved. If p2 � 16A < 0, people didn’t in- vent fantastic rectangles whose sides involve square roots of negative numbers.

Instead, they merely said that there is no rectangle whose area and perimeter are so related. But the cubic formula doesn’t allow us to avoid “imaginary”

numbers, for we have just seen, in Example 3.5, that an “honest” real and posi-

tive root can appear in terms of such expressions. Complex numbers arose, not

as an attempt to get roots of equations involving square roots of negative real

numbers, but as a device to solve cubic equations having real coefficients and real roots.

The cubic formula was revolutionary. For the next 100 years, mathemati-

cians were forced to reconsider the meaning of number, calculating with strange

objects of the form a C ib (where a and b are real numbers) as if they were

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92 Chapter 3 Renaissance

actual numbers enjoying the simplification rule i2 D �1. It was during thisIn Chapter 7, using ideas of abstract algebra, we’ll

see that the naive way of

thinking about complex

numbers, as polynomials in

i obeying the rule i2 D �1, can be made precise.

time that the terms real and imaginary arose. In this section and the next, we’ll

develop complex numbers in a more careful and formal way, and we’ll see that complex numbers are as real as real numbers!

The Complex Plane

When considering expressions of the form aC bi , it is natural to separate the two summands. Geometry rears its head.

Definition. A complex number is an ordered pair z D .a; b/ of real numbers, denoted by z D aC bi . We call a the real part of z, denoting it by <.z/ D a,Right now, aC bi is just

an “alias” for .a;b/ but, in

your previous experience

with C, the summand bi

denoted the product of b

and i . We’ll soon recover

this notion.

and b the imaginary part of z, denoting it by =.z/ D b.

Both the real and the imaginary parts of a complex number are real numbers.

Moreover, equality of ordered pairs says that complex numbers z D a C bi and z0 D a0 C b0i are equal if and only if <.z/ D <.z0/ and =.z/ D =.z0/; that is, a D a0 and b D b0. Thus, one equation of complex numbers is the same as two equations of real numbers.

There is an immediate geometric interpretation of complex numbers: they

can be viewed as points in the plane. Real numbers are complex numbers z

with =.z/ D 0; that is, they correspond to points .a; 0/ on the x-axis (which is called the real axis in this context). We usually abbreviate .a; 0/ to a; thus, the

set of real numbers R is a subset of C. We denote the complex number .0; 1/You hear the following message when you call

one of our friends. “The

number you have reached

is imaginary; please rotate

your phone 90 degrees.”

by i , so that the purely imaginary complex numbers z, those with <.z/ D 0, correspond to points on the y-axis (which is called the imaginary axis in this

context). When we view points as complex numbers, the plane R2 is called the

complex plane, and it is denoted by C.

Thus, an ordered pair .a; b/ of real numbers has two interpretations: alge-

braic, as the complex number z D a C bi , and geometric, as the point P in the plane R2 having coordinates a and b. We will use both interpretations, al-

gebraic and geometric, depending on which is more convenient for the context

in which we are working.

Historical Note. Surprisingly, it took a very long time for people to embrace

the idea of representing the elements of C as points in the plane. It wasn’t until

Wessel presented a paper in 1797 to the Royal Danish Academy of Sciences, A bijection is a one-to-

one correspondence. See

Appendix A.1, page 416,

for the precise definition.

entitled On the Analytic Representation of Direction: An Attempt, did this rep- resentation crystallize. Wessel’s discovery was not adopted immediately but,

by 1830, most mathematicians routinely used the bijection a C bi $ .a; b/ between complex numbers and points of the plane. The complex plane has

gone by other names in its history: for example, Argand Diagram and Gaus-

sian Plane.

Algebraic Operations

In Section 3.1, you saw that mathematicians were forced to add and multiply complex numbers. However, without precise definitions of the operations or of

the complex numbers themselves, they could not trust many of their results.

The complex plane allows us to resolve the many doubts our ancestors had

about the algebra of complex numbers.

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3.2 Complex Numbers 93

In a linear algebra course, R2 is often viewed as a vector space with real

scalars; we continue using these operations in the complex plane.

Definition. Define addition C � C ! C by

.a C bi/C .c C di/ D .a C c/C .b C d/i:

In terms of ordered pairs, .a; b/C .c; d / D .a C c; bC d/. Define scalar multiplication R � C ! C by

r.aC bi/ D raC rb i;

where r 2 R. In terms of ordered pairs, r.a; b/ D .ra; rb/.

As in linear algebra, it is useful to look at each point in the plane (and,

hence, each complex number) as an arrow with tail at the origin (sometimes

we say vector instead of arrow). For example, we’ll think of z D 3C 2i either as the point P D .3; 2/ or as the arrow ��!OP [where O D .0; 0/]. The context will make it clear which interpretation we are using.

Addition is illustrated by the parallelogram law (see Figure 3.2). If P D .a; b/ and Q D .c; d /, then R D .a C c; b C d/. Of course, this needs a geometric proof, especially when points don’t lie in the first quadrant or they

are collinear; see Exercise 3.33 on page 98.

O

P a,b= ( )

R a+c,b+d= ( )

Q c,d= ( )

Figure 3.2. Parallelogram Law.

Scalar multiplication of complex numbers has the same geometric interpre- tation as scalar multiplication of vectors. View a complex number z D a C ib as

��! OP , where P D .a; b/. If r 2 R, then we may view rz as the vector r��!OP ;

that is, if r � 0, then it’s an arrow in the same direction as ��!OP whose length has been stretched by a factor of r (if r > 1) or shrunk by a factor of r (if

r < 1/; if r < 0, then r ��! OP is the arrow in the reverse direction whose length

has been changed by a factor of jr j. The eight properties listed in the next proposition are precisely the defining

properties of a vector space with scalars in R.

Proposition 3.8. Let z D a C bi , w D c C di , and u D e C f i be complex numbers, and suppose that r; s 2 R.

(i) z C w D w C z (ii) z C .w C u/ D .z C w/C u

(iii) z C 0 D z

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94 Chapter 3 Renaissance

(iv) There is a complex number �z such that z C .�z/ D 0In Exercise 3.21 on page 98, you’ll show that

negatives are unique. (v) r.sz/ D .rs/z

(vi) 1z D z (vii) r.z C w/ D rz C rw

(viii) .r C s/z D rz C sz

Proof. The proofs are routine, just reducing each to a familiar statement about

real numbers, and so we’ll only prove the longest such: associativity of addi-

tion. It is clearer if we use ordered pairs.

z C .w C u/ D .a; b/C Œ.c; d /C .e; f / D .a; b/C .c C e; d C f / D .a C .c C e/; b C .d C f // D ..a C c/C e; .bC d/C f / D .a C c; bC d/C .e; f / D Œ.a; b/C .c; d /C .e; f / D .z C w/C u:

In linear algebra, every vector .a; b/ has a decomposition into components

ae1 C be2 with respect to the standard basis e1 D .1; 0/; e2 D .0; 1/:

aC bi D .a; b/ D .a; 0/C .0; b/ D a.1; 0/C b.0; 1/:

It follows that the + in the notation aC bi really does mean add and that bi is the product of b and i ; that is, bi D b.0; 1/ D .0; b/.

The set C of complex numbers has more algebraic structure: any two com-

plex numbers can be multiplied, not just when one of them is real. The defini-

tion arises from pretending that

.a C bi/.c C di/ D ac C adi C bci C bdi2

D .ac � bd/C i.ad C bc/;

where we have set i2 D �1. This is precisely what our ancestors did, which motivates the formal definition. But our definition involves no pretending.

Definition. Define multiplication C � C ! C by

.a C ib/.c C id / D .ac � bd/C i.ad C bc/:

In terms of ordered pairs, .a; b/.c; d / D .ac � bd; ad C bc/.

Notice that

i2 D .0; 1/.0; 1/ D .�1; 0/ D �1;

for ac D 0 D ad D bc. We are now obliged to prove that the familiar properties of multiplication

actually do hold for complex multiplication.

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3.2 Complex Numbers 95

Proposition 3.9. Let z D a C bi , w D c C di , and u D e C f i be complex numbers.

(i) zw D wz (ii) z.wu/ D .zw/u

(iii) 1z D z (iv) z.w C u/ D zw C zu.

Proof. Again, the proofs are routine, for each reduces to a familiar statement

about real numbers. We’ll only prove associativity. As in the proof of Proposi- tion 3.8, it is clearer if we use ordered pairs.

z.wu/ D .a; b/ Œ.c; d /.e; f / D .a; b/.ce � df; deC cf // D .a.ce � df /� b.de C cf /; b.ce � df /C a.de C cf // D .ace � adf � bde � bcf; bce � bdf C ade C acf /:

On the other hand,

.zw/u D Œ.a; b/.c; d / .e; f / D .ac � bd; bc C ad/.e; f / D .ac � bd/e � .bc C ad/f; .bc C ad/e C .ac � bd/f / D .ace � bde � bcf � adf; bceC ade C acf � bdf /:

Hence, z.wu/ D .zw/u.

The operations of addition and multiplication in C extend the definitions

in R; for example, if r and s are real numbers, then their sum rC s is the same, whether you think of doing the addition in R or in C. See Exercise 3.22 on page 98.

In Section 1.4, we displayed nine properties of addition and multiplication

in R, and we have just seen that eight of them also hold in C. We could now

define subtraction, as we did there, and prove results like z.w�u/ D zw�zu, z �0 D 0 for all complex numbers z, and the Binomial Theorem. We don’t have to repeat all of this. As we said then, once this is established, the proofs of other properties of addition and multiplication, such as 0 � z D 0 and .�z/.�w/ D zw, go through verbatim.

The ninth property describes reciprocals: If z D a C bi ¤ 0, there is a number z�1 such that z � z�1 D 1 (Exercise 3.21 on page 98 shows that such a number z�1, if it exists, is unique.) Here is an explicit formula for z�1 D 1=z when z D a C bi ¤ 0. If b D 0, then z D a is a nonzero real number, and we know it has a reciprocal 1=a. If b ¤ 0, we can easily find x C yi so that .x C yi/.a C bi/ D 1. Multiply and equate real and imaginary parts:

xa � yb D 1 and ya C xb D 0:

The second equation gives x D �ay=b; substitute this into the first equation and obtain

y D � b a2 C b2 and x D

a

a2 C b2 :

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96 Chapter 3 Renaissance

In either case (b D 0 or b ¤ 0),

z�1 D a a2 C b2

� i b a2 C b2

:

There is a more elegant derivation of this formula. The denominator a2Cb2 can be factored in C:

Hence, a2 C b2 D zz. a2 C b2 D .a C bi/.a � bi/:

This leads to the useful notion of complex conjugate.

Definition. The complex conjugate z of a complex number z D a C bi is defined to be

z D a � bi:

The function C ! C, given by z 7! z, is called complex conjugation.

If z D .a; b/, then z D .a;�b/, so, geometrically, z is obtained from z by reflection in the real axis.

Complex conjugation interacts well with addition and multiplication.

Proposition 3.10. If z D aC bi and w D c C di are complex numbers, then

(i) z C w D z Cw (ii) zw D z w

(iii) z 2 R if and only if z D z (iv) z D z

Proof. We’ll prove (i), leaving the rest to Exercise 3.25 on page 98.

z C w D .aC c/C .b C d/i D .aC c/ � .b C d/i D .a � bi/C .c � di/ D z C w:

Using induction, the first two statements in the proposition can be general- ized:

z1 C � � � C zn D z1 C � � � C zn z1 � � � zn D z1 � � � zn:

The formula for the multiplicative inverse of a complex number can be writ-

ten in terms of conjugates. Informally, cancel z to see that z=zz D 1=z.

Proposition 3.11. Every nonzero complex number z D aC bi has an inverseW

z�1 D a � bi a2 C b2 D

z

z z :

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3.2 Complex Numbers 97

Proof. It’s enough to see that you get 1 if you multiply z by .a�bi/=.a2Cb2/. And so it is:

z

� 1

z z z

� D � 1

z z

� z z D 1:

It wouldn’t be worth introducing the new term complex conjugation if our

only use of it was to give a neat proof of the formula for reciprocals. The notion has many other uses as well. For example, if f .x/ D ax2 C bx C c has real coefficients, then the quadratic formula implies that whenever z is a complex

root of f , then so is z. In fact, this is true for polynomials of any degree, and Is this true if z is real?

the proof depends only on Proposition 3.10.

Theorem 3.12. If f .x/ is a polynomial with real coefficients and a complex

number z is a root of f , then so is z.

Proof. Suppose that

f .x/ D a0 C a1x C � � � C aixi C � � � C anxn;

where each ai 2 R. Saying that z is a root means that

0 D a0 C a1z C � � � C aizi C � � � C anzn:

Hence,

0 D 0 D a0 C a1z C � � � C aizi C � � � C anzn

D a0 C a1z C � � � C aizi C � � � C anzn

D a0 C a1 z C � � � C ai zi C � � � C an zn

D a0 C a1z C � � � C aizi C � � � C anzn (because all ai are real) D f .z/ :

Therefore, z is a root of f .

Exercises

3.20 In Appendix A.4, we considered the subset P of all (strictly) positive real num-

bers; it satisfies: If we define a < b to mean b � a 2 P , then we can prove all the familiar

properties of inequality.

For example, if a < b and

c < 0, then bc < ac. See

page 441.

� if a; b 2 P , then a C b 2 P and ab 2 P ; � if r 2 R, then exactly one of the following is true:

r 2 P; r D 0; or � r 2 P:

(i) Using only the two properties of P , prove that if a 2 R, then either a D 0 or a2 2 P .

(ii) Prove that there is no subset Q � C, closed under addition and multiplica- tion, such that if z 2 C, then exactly one of the following is true:

z 2 Q; z D 0; or � z 2 Q:

Conclude that it’s impossible to order the complex numbers in a way that

preserves the basic rules for inequality listed in Proposition A.51.

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98 Chapter 3 Renaissance

3.21 * Suppose that z is a complex number. Generalize Propositions 1.33 and 1.32.

(i) Show that �z is unique. (ii) Show that �z D .�1/z.

(iii) If z ¤ 0, show that z�1 is unique.

3.22 *

(i) We may think of real numbers r and s as complex numbers. Show that their

sum r C s and their product rs in C are the same as their sum and product in R.

(ii) If z is a complex number and r is a real number, you can think of the complex

number rz in two ways: as the product of scalar multiplication, or as the

product of two complex numbers r C0i and z. Show that the two calculations give the same result.

3.23 If z 2 C, show that z C z D 2.<z/ and z � z D 2.=z/.

3.24 Find a complex number z such that z C z D 14 and zz D 49.

3.25 * Finish the proof of Proposition 3.10. If z and w are complex numbers, prove

(i) zw D z w (ii) z D z

(iii) z D z if and only if z is a real number

3.26 If z is a complex number and n is a natural number, show that

zn D .z/n :

Is this equation true if z ¤ 0 and n is a negative integer?

3.27 * Let z be a complex number and r a real number. Show how to locate rz in the

complex plane in terms of z.

Hint: r.a; b/ D .ra; rb/.

3.28 Solve the following equation for z.

.3 C 2i /z D �3 C 11i:

3.29 Find real numbers a and b such that

(i) a C bi D .8 C i /=.3 C 2i /

(ii) a C bi D .8 C i /=.3 C i /

(iii) .a C bi /2 D �5 C 12i

(iv) .a C bi /2 D 1 C i .

3.30 What’s wrong with this “proof” that 6 D �6?

6 D p

36 D p

.�9/.�4/ D p

�9 p

�4 D 3i � 2i D 6i2 D �6:

3.31 Establish the identity

.a2 C b2/.c2 C d2/ D .ac � bd/2 C .bc C ad/2

for all complex numbers a; b; c; d .

3.32 Use Theorem 3.12 to prove that every cubic polynomial with real coefficients has

a real root.

3.33 * Let z D a C bi and w D c C di . Show that in the complex plane z C w is the fourth vertex of the (possibly degenerate) parallelogram whose other vertices are

0, z, and w.

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3.2 Complex Numbers 99

Absolute Value and Direction

We’ve already seen that addition can be viewed as the parallelogram law, multi-

plication by a real number can be viewed as scalar multiplication, and conjuga- tion can be viewed as reflection in the real axis. There is a beautiful geometric

interpretation of complex multiplication; it is best understood if, first, we con-

sider a different way to describe an arrow in the complex plane using absolute

value and direction.

Definition. The absolute value (or length or modulus) of z D aC bi is

jzj D p a2 C b2:

The absolute value of a real number is its distance to the origin, and so we

have just extended the notion of absolute value from R to C. Thus, if z D a C bi , then jzj is the distance from the point P D .a; b/ to the origin O ; equivalently, it is the length of the arrow

��! OP . Because z z D a2 C b2, we can

write Does this equation hold if

z is a real number?jzj D p z z:

Proposition 3.13. Let z D aC bi and w D c C di . (i) jzj � 0, and jzj D 0 if and only if z D 0.

(ii) .Triangle Inequality/: jz C wj � jzj C jwj. (iii) jzwj D jzj jwj.

Proof. (i) Both statements follow from the definition jzj D p a2 C b2, be-

cause a2 C b2 D 0 if and only if a D 0 D b. (ii) If P D .a; b/ andQ D .c; d /, then z is the arrow ��!OP and w is the arrow

��! OQ. As in Figure 3.2, zCw D ��!OR. The inequality we want is the usual triangle inequality, which follows from the length of a line segment being

the shortest distance between its endpoints.

Make sure you can justify

each step in the proof.

Would this proof work if

either z or w (or both) is

real?

jzwj D p .zw/ .zw/

D p .zw/ .z w/

D p .z z/ .w w/

D p .z z/

p .w w/

D jzj jwj:

What do we mean by direction? The most natural way to indicate direction in the plane is to point: “He went thataway!”—an arrow shows the way. The

arrow may as well have its tail at the origin and, since the length of the arrow

doesn’t affect the direction, we may as well assume it is a unit vector; that is,

it has length 1. If we denote the tip of the arrow by P D .a; b/, then P lies on the unit circle. There are various geometric ways to describe P . One way Why did our ancestors

divide a circle into 360

parts? We can only guess

why. Perhaps it was related

to calendars, for a year has

about 360 days.

is to consider the angle � between the x-axis and ��! OP ; hence, P D .a; b/ D

.cos �; sin �/. This angle can be described with degrees; the ancients divided

the circle into 360 equal degrees. The angle can also be described with radians;

the circumference of the unit circle is 2� , and � is the length of the arc from

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100 Chapter 3 Renaissance

.1; 0/ to P . When we view the point P D .cos �; sin �/ on the unit circle as a complex number, it is equal to cos � C i sin � .

Figure 3.3 shows z D a C bi as the tip of an arrow of length jzj D r . The direction of this arrow is the same as the direction of the unit vector

��! OP having

the same direction as z. If � is the angle between the x-axis and ��! OP , then the

coordinates of P are jOAj D cos � and jAP j D sin � .

(1, 0)

r |z|=

q O A

P = (cos , sin )q q

z a, b= ( )

Figure 3.3. Absolute value and argument.

Definition. If z is a nonzero complex number, then its argument, denoted by

arg.z/;

is the counterclockwise angle � from the positive real axis to ��! OP .

In Figure 3.3, we see that the coordinates of P are cos � and sin � ; that is,Finding arg.z/ requires some way of comput-

ing values of inverse

trigonometric functions.

Nowadays, we use com-

puters; in earlier times,

tables of values of cosine

and sine were used. Fairly

accurate trigonometric

tables were known over

two thousand years ago.

if P D .a; b/, then a D cos � and b D sin � . Thus, for any nonzero complex number z D a C bi , not necessarily of absolute value 1, the definitions of cosine and sine (in terms of right triangles) give arg.z/ D � , where cos � D a=jzj and sin � D b=jzj. Note that

z D jzj � a

jzj C i b

jzj

� D jzj .cos � C i sin �/ : (3.4)

How to Think About It. Technically, the argument of a complex number is only determined up to a multiple of 360 (if we measure in degrees) or 2�

(if we measure in radians). For example, arg.1 C i/ is 45ı or � 4

radians, and

this is the same direction as 405ı or 9� 4

radians. There is a fussy way to make statements like “arg.1 C i/ D 45ı” precise (introduce a suitable equivalence relation), but we prefer, as do most people, to be a bit sloppy here; the cure is

worse than the disease.

Actually, the trig functions do make angles precise, for cos � and sin � have

the same values when � is replaced by either � C 360n degrees or � C 2�n radians.

The polar form of z is

z D jzj .cos � C i sin �/ ;

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3.2 Complex Numbers 101

where � D arg.z/. Just as a complex number z is determined by its real and imaginary parts, so too is it determined by its polar form: its absolute value

and argument.

Proposition 3.14 (Polar Form). Every complex number z has a polar formW

z D r.cos � C i sin �/;

where r � 0 and 0 � � < 2� . If z ¤ 0, then this expression is unique.

We may paraphrase

uniqueness of polar forms:

two vectors are equal if

and only if they have the

same length and the same

direction.

Proof. Existence is given by Eq. (3.4). Uniqueness is almost obvious. Suppose

that z D r .cos � C i sin �/ for some r > 0 and � . Then

jzj D jr .cos � C i sin �/j D jr j jcos � C i sin � j D jr j D r;

since r � 0 and jcos � C i sin � j D cos2 � C sin2 � D 1; thus, r D jzj. But then

cos � C i sin � D 1 r z D 1jzjz D

a

jzj C i b

jzj ;

so that

cos � D ajzj and sin � D a

jzj and � D arg.z/.

Example 3.15. If z D 3 C 4i , the Pythagorean Theorem gives jzj2 D 32 C 42 D 25, so that jzj D 5; your favorite computer gives arg.z/ D � D cos�1.3

5 / � 53:13ı. Thus, the polar form of z is

z D cos � C i sin � � 5 � cos 53:13ı C i sin 53:13ı

� : N

The Geometry Behind Multiplication

We’re ready to give a geometric interpretation of complex multiplication. Propo-

sition 3.13(iii) tells part of the story: the absolute value of a product is the prod-

uct of the absolute values. To finish the geometric analysis of multiplication, We know that zw sits on a circle of radius jzjjwj, centered at the origin. But

where?

we need to know how arg.zw/ is related to arg.z/ and arg.w/. We may as well

assume that z and w are unit vectors (why?), so that z D cos ˛ C i sin ˛ and w D cosˇ C i sinˇ. Multiply them together and collect real and imaginary parts.

zw D .cos ˛ C i sin˛/.cos ˇ C i sinˇ/ (3.5) D .cos ˛ cosˇ � sin˛ sinˇ/C i.cos ˛ sinˇ C sin ˛ cosˇ/: (3.6)

Do <.zw/ and =.zw/ look familiar? They are the addition formulas for sine and cosine:

cos.˛ C ˇ/ D cos ˛ cosˇ � sin ˛ sinˇ

and

sin.˛ C ˇ/ D cos ˛ sinˇ C sin ˛ cosˇ:

These formulas will give a beautiful characterization of the product of two

complex numbers. We now prove them, beginning with a familiar lemma that

uses Figure 3.4.

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102 Chapter 3 Renaissance

W r, s= ( )

U c, d= ( )

O

Figure 3.4. Orthogonality and dot product.

Lemma 3.16. If W D .r; s/ and U D .c; d /, then the arrows ��!OW and ��!OUWe are proving that per- pendicularity is equivalent

to the dot product being 0. are perpendicular if and only if rc C sd D 0.

Proof. We use the Pythagorean Theorem and its converse, Exercise 1.11 on

page 7: ��! OU ? ��!OW if and only if jUW j2 D jOU j2CjOW j2. Let h D jUW j;

then

h2 D .r � c/2 C .s � d/2

D r2 � 2rc C c2 C s2 � 2sd C d 2

D r2 C s2 C c2 C d 2 � 2.rc C sd/:

But jOW j2 D r2 C s2 and jOU j2 D c2 C d 2. Hence,

jOW j2 C jOU j2 D .r2 C s2/C .c2 C d 2/

and

h2 D r2 C s2 C c2 C d 2 � 2.rc C sd/:

Therefore, h2 D jOU j2 C jOW j2 if and only if rc C sd D 0.

In Figure 3.4, the coor-

dinates of U are labeled

.c; d/. This lemma shows

that .c;d/ D .�s; r/.

Now for the addition formulas.

Theorem 3.17 (Addition Theorem). Let ˛ and ˇ be angles.

(i) cos.˛ C ˇ/ D cos ˛ cosˇ � sin˛ sinˇ: (ii) sin.˛ C ˇ/ D cos˛ sinˇ C sin ˛ cosˇ:

Proof. In Figure 3.5, we have a picture of the unit circle. Let Z D .a; b/ DWe are looking at points here as elements of R2, al-

though we’ll soon interpret

this diagram in the complex

plane.

.cos ˛; sin˛/ and W D .r; s/ D .cos ˇ; sinˇ/. Rotate 4OQZ counterclock- wise through †ˇ to get 4OQ0Z0, so that 4OQZ and 4OQ0Z0 are congru- ent. Thus, Z0 D .cos.˛ C ˇ/; sin.˛ C ˇ// : Our task is to find the coordinates of Z0 in terms of r; s; a, and b.

Define U D .�s; r/. Since W D .r; s/ is on the unit circle, we have r2 C s2 D 1, and so U D .�s; r/ is also on the unit circle. Moreover, since .�s/r C rs D 0, Lemma 3.16 says that ��!OU is orthogonal to ��!OW . Therefore, O Q0 Z0M is a rectangle.

Decompose ��! OZ0 as the sum of two vectors:

��! OZ0 D

��! OQ0 C ��!OM;

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3.2 Complex Numbers 103

a

a b

Q a= ( , 0)

O

M

Z a, b= ( )

W r, s= ( )

U –s, r= ( )

Figure 3.5. Addition Theorem.

where ��! OM is the projection of

���! Q0Z0 onto

��! OU . We can get explicit expres-

sions for Q0 and M . First, ��! OQ0 is a scalar multiple of

��! OW and, because

jOQj D a, we know the scalar:

Q0 D a.r; s/ D .ar; as/:

Second, ��! OM is a scalar multiple of

��! OU , where U D .�s; r/; as jOM j D

jQZj D b, we know the scalar:

M D b.�s; r/ D .�bs; br/:

Therefore,

��! OZ0 D

��! OQ0 C ��!OM D .ar; as/C .�bs; br/ D .ar � bs; as C br/:

Making the appropriate substitutions for a, b, r , and s, we have the desired

result:

cos.˛ C ˇ/ D cos ˛ cosˇ � sin ˛ sinˇ and

sin.˛ C ˇ/ D cos˛ sinˇ C sin˛ cosˇ:

Here is the result we have been seeking.

Theorem 3.18 (The Geometry of Multiplication.). If z and w are complex In words, the length of a product is the product

of the lengths, and the

argument of a product is

the sum of the arguments.

The equality in Theo-

rem 3.18(ii) holds up to a

multiple of 2� .

numbers, then

(i) jzwj D jzj jwj, and (ii) arg.zw/ D arg.z/C arg.w/.

Proof. The first statement is Proposition 3.13, and the second follows from

Theorem 3.17 and Eq. (3.5) on page 101.

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104 Chapter 3 Renaissance

If we set arg.z/ D ˛ and arg.w/ D ˇ, then Theorem 3.18 has an especially pleasing restatement in polar form.

Corollary 3.19. If z D jzj.cos ˛ C sin˛/ and w D jwj.cosˇ C sinˇ/, then

z � w D jzwj .cos.˛ C ˇ/C i sin.˛ C ˇ// :

Proof. Both sides equal jzj .cos˛ C i sin˛/ � jwj .cosˇ C i sinˇ/.

It follows easily, by induction on k � 1, that if z is a complex number and k 2 Z , then

ˇ̌ zk ˇ̌

D jzjk and arg � zk �

D k arg.z/:

How to Think About It. There’s a way to see, without using trigonometry, that angles add in the product of two complex numbers. Essentially, we recast

the proof of Theorem 3.17 in terms of complex numbers. Given z D a C bi and w D r C si , we want to determine arg.zw/ in terms of arg.z/ D ˛ and arg.w/ D ˇ. We can assume that z and w are unit vectors; this implies that zw is also a unit vector, by Proposition 3.13. The key insight is that

zw D .aC bi/w D aw C .bi/w D aw C b.iw/:

You know the geometric effect of scalar multiplication (Exercise 3.27 on page 98),This approach does not seem to be very well

known. It appears in [22]

and Kerins, B. “Gauss,

Pythagoras, and Heron”

(Mathematics Teacher,

96:5, 2003), but we can’t

find any older sources.

you know how to add geometrically (parallelogram law), and you know

iw D i.r C si/ D �s C ri I

using Lemma 3.16, it follows that iw is obtained from w by counterclock-

wise rotation by 90ı. Figure 3.6 below is almost the same as Figure 3.5; the difference is that points are now labeled as complex numbers.

a

a b

O a

aw

b iw( )

u iw –s ri= = +

z a bi= +

w r si= +zw aw b iw= + ( )

Figure 3.6. Complex multiplication again.

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3.2 Complex Numbers 105

Let’s put this all together. Triangle OZ0Q0 is congruent to triangle OZQ, so that †Z0OW D †ZOQ D ˛; hence, arg.zw/ D ˛ C ˇ. We have shown that Theorem 3.18(ii) follows without any mention of trigonometry.

We have just used plane geometry to derive the geometric interpretation of

complex multiplication, avoiding the trigonometric addition formulas. Aside from proving these ideas for students who haven’t yet seen the addition for-

mulas, complex numbers can now be used to derive these formulas. That’s an

additional bonus, especially for a precalculus class; it allows us to use complex

numbers when trying to establish other trigonometric identities that depend on

the addition formulas. For example, to get a formula for cos � � 4

C � � , calculate

like this:

cos � � 4

C � �

C i sin � � 4

C � �

D � cos �

4 C i sin �

4

� .cos � C i sin �/

D 1p 2 .1C i/ .cos � C i sin �/

D 1p 2

� .cos � � sin �/C i.cos � C sin �/

� :

Hence

cos � � 4

C � �

D 1p 2 .cos � � sin �/

and, as a bonus,

sin � � 4

C � �

D 1p 2 .cos � C sin �/:

Over the next two centuries, people became comfortable with the fact that

polynomial equations with real coefficients can have complex solutions. It was eventually proved that every polynomial f .x/ D xnCcn�1xn�1C� � �Cc1xC c0 with real coefficients has a factorization

f .x/ D .x � ˛1/ � � � .x � ˛n/;

where ˛1; : : : ; ˛n are complex numbers. This amazing result holds for any

nonzero polynomial f with complex coefficients; it is known as the Funda- mental Theorem of Algebra. We won’t prove this result here because, in spite

of its name, it is a theorem of analysis, not of algebra; you can find a readable

account in [4], pp. 142–152.

Exercises

3.34 If z D a C bi , prove that the arrow corresponding to z, namely ��!OP , where P D .a; b/, is perpendicular to the arrow corresponding to iz.

3.35 If z and w are complex numbers with w ¤ 0, show that

arg.z=w/ D arg.z/ � arg.w/:

3.36 (i) Prove that the quadratic formula holds for polynomials with complex coeffi-

cients (use Proposition 3.6).

(ii) Find the roots of x2 C 2ix � 1. Why aren’t these roots conjugate?

3.37 If z and w are complex numbers, find a necessary and sufficient condition that

jz C wj to equal jzj C jwj.

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106 Chapter 3 Renaissance

3.38 * If z D cos ˛ C i sin ˛, show that

z D 1=z D cos.�˛/ C i sin.�˛/:

3.39 Let n � 0 is an integer and � D cos.2�n / C i sin. 2� n /. If z is a complex number,

give a geometric description of how �z is located with respect to z on the complex

plane.

3.40 Preview. Plot the roots in the complex plane for each of the polynomials x2 � 1, x3 � 1, x4 � 1, x6 � 1, and x12 � 1.

3.41 Preview. Let � D cos.45ı/ C i sin.45ı/. (i) Show that �8 D 1

(ii) Show that the distinct roots of x8 � 1 are precisely 1; �; �2; : : : ; �7. (iii) Plot these roots in the complex plane.

(iv) Show that �147 is a root of x8 � 1. (v) Show that �147 is equal to one of the roots in part (ii). Which one?

3.42 Preview. Let n be a nonnegative integer and � D cos.2� n

/ C i sin.2� n

/.

(i) Let � D cos.2�n / C i sin. 2� n /. Show that �

n D 1.

(ii) If k is any nonnegative integer, show that �

�k �n

D 1.

(iii) Give a geometric description of the subset f�k W k � 0g of the complex plane.

3.43 Take It Further. If f .x/ is a polynomial with complex coefficients, define f to

be the polynomial you get by replacing each coefficient in f by its conjugate.

Prove the following statements.

(i) f C g D f C g: (ii) fg D f g:

(iii) f D f if and only if f .x/ has real coefficients. (iv) f f has real coefficients.

(v) f .z/ D f .z/:

3.44 Take It Further. Suppose f .x/ is a polynomial with coefficients in C. If a

complex number z is a root of f , show that z is a root of f .

3.45 Take It Further. Suppose f .x/ is a polynomial with coefficients in C. Then

by Exercise 3.43 on page 106, if we define g.x/ D f .x/f .x/, then g.x/ has coefficients in R. Show that if g.z/ D 0, either f .z/ D 0 or f .z/ D 0. Hence conclude that if every polynomial with real coefficients and degree at least 1 has

a root in C, then every polynomial with complex coefficients and degree at least

1 has a root in C. (The Fundamental Theorem of Algebra says that every polyno-

mial with complex coefficients has all its roots in C. This exercise shows that it’s

enough to prove this for polynomials with real coefficients.)

3.3 Roots and Powers

We saw in the previous section that every point z on the unit circle can be written as z D cos � C i sin � for some angle � . Theorem 3.18 tells us that arguments add when complex numbers are multiplied. In particular,

.cos � C i sin �/2 D .cos � C i sin �/ .cos � C i sin �/ D cos.� C �/C i sin.� C �/ D cos.2�/C i sin.2�/:

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3.3 Roots and Powers 107

On the other hand, complex multiplication gives

.cos � C i sin �/2 D � cos2 � � sin2 �

� C i2 cos � sin �:

Equating real parts and imaginary parts gives the double angle formulas:

cos.2�/ D cos2 � � sin2 � sin.2�/ D 2 cos � sin �:

We now generalize this to any positive integer power.

Theorem 3.20 (De Moivre). For every angle � and all integers n � 0,

.cos � C i sin �/n D cos.n�/C i sin.n�/:

Proof. We prove equality by induction on n � 0. The theorem is true when n D 0, for cos 0 D 1 and sin 0 D 0. Here is the inductive step.

.cos � C i sin �/n D .cos � C i sin �/n�1 .cos � C i sin �/

D �

cos � .n � 1/�

� C i sin

� .n � 1/�

�� .cos � C i sin �/

D cos � .n � 1/� C �

� C i sin

� .n � 1/� C �

D cos.n�/ C i sin.n�/:

Example 3.21. (i)

.cos 3ı C i sin 3ı/40 D cos 120ı C i sin 120ı D �1 2

C i p 3 2 :

(ii) Let z D cos.45ı/C i sin.45ı/ D 1p 2 .1C i/. We compute z6 in two ways:

� With the Binomial Theorem:

z6 D � 1p 2 .1C i/

�6

D � 1p 2

�6 .1C i/6

D 1 8

� 1C 6i C 15i2 C 20i3 C 15i4 C 6i5 C i6

D 1 8 .1C 6i � 15 � 20i C 15C 6i � 1/

D �i:

� With De Moivre’s Theorem:

z6 D � cos.45ı/C i sin.45ı/

�6 D cos.6 � 45ı/C i sin.6 � 45ı/ D cos.270ı/C i sin.270ı/ D �i: N

Polar Decomposition and De Moivre’s Theorem combine to give a nice

formula for computing powers of any complex number.

Corollary 3.22. If z D r � cos ˛ C i sin˛

� is a complex number, then

zn D rn � cos.n˛/C i sin.n˛/

� :

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108 Chapter 3 Renaissance

We were unable to find a cube root of z D a C ib earlier, but there’s no problem now.

Corollary 3.23. Let r.cos �Ci sin �/ be the polar form of a complex number z. If n � 1 is an integer, then

h n p r �

cos � �=n

� C i sin

� �=n

��in D r.cos � C i sin �/ D z:

Example 3.24. In Example 3.15, we saw that the polar form of z D 3C 4i isOf course, we must find the polar form of z,

which involves finding

� D cos�1.a=jzj/.

approximately 5 .cos.53:13ı/C i sin.53:13ı//. Now .53:13/=3 D 17:71, and so a cube root of z is approximately

3 p 5 .cos.17:71ı/C i sin.17:71ı//. Our

calculator says that

� 3 p 5 � cos 17:71ı C i sin 17:71ı

��3 D 3:000001C 3:99999i: N

We are now going to describe a beautiful formula discovered by Euler. Re- call some power series formulas from calculus. For every real number x,

ex D 1C x C x 2

2Š C � � � C x

n

nŠ C � � � ;

cos x D 1 � x 2

2Š C x

4

4Š � � � � C .�1/

nx2n

.2n/Š C � � � ;

and

sin x D x � x 3

3Š C x

5

5Š � � � � C .�1/

n�1x2nC1

.2nC 1/Š C � � � :

We can define convergence of power series P1 nD0 cnz

n for z and cn com-

plex numbers, and we can then show that the series

1C z C z 2

2Š C � � � C z

n

nŠ C � � �

converges for every complex number z. The complex exponential ez is defined

to be the sum of this series. In particular, the series for eix converges for all

real numbers x, and

eix D 1C ix C .ix/ 2

2Š C � � � C .ix/

n

nŠ C � � � :

Theorem 3.25 (Euler). For all real numbers x,It is said that Euler was delighted by the special

case

ei� C 1 D 0;

for it contains five important

constants in one equation.

eix D cos x C i sinx:

Sketch of proof. We will not discuss necessary arguments involving conver- gence. As n varies over 0; 1; 2; 3; 4; 5; : : :, the powers of i repeat every four

steps: that is, the sequence

1; i; i2; i3; i4; i5; i6; i7; i8; i9; i10; i11; : : :

is actually

1; i; �1; �i; 1; i; �1; �i; 1; i; �1; �i; : : : I

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3.3 Roots and Powers 109

the even powers of i are all real, whereas the odd powers all involve i . It

follows, for every real number x, that .ix/n D inxn takes values

1; ix; �x2; �ix3; x4; ix5; �x6; �ix7; x8; ix9; �x10; �ix11; : : :

Thus, in the definition of the complex exponential,

eix D 1C ix C .ix/ 2

2Š C � � � C .ix/

n

nŠ C � � � ;

the even powers of ix do not involve i , whereas the odd powers do. Collecting

terms, one has eix D even terms C odd terms. But

even terms D 1C .ix/ 2

2Š C .ix/

4

4Š C � � �

D 1 � x 2

2Š C x

4

4Š � � � �

and

odd terms D ix C .ix/ 3

3Š C .ix/

5

5Š C � � � :

D i.x � x 3

3Š C x

5

5Š � � � � /:

Therefore, eix D cos x C i sin x. � As a consequence of Euler’s Theorem, the polar decomposition can be

rewritten in exponential form: every complex number z has a factorization

z D rei� ;

where r � 0 and 0 � � < 2� . We have chosen to denote

P1 nD0

.ix/n

nŠ by eix, but we cannot assert,

merely as a consequence of our notation, that the law of exponents, eixeiy D ei.xCy/, is valid. But this is precisely what Corollary 3.19 says once it is trans- lated into exponential notation.

Theorem 3.26 (Exponential Addition Theorem). For all real numbers x and

y,

eixeiy D ei.xCy/:

Proof. According to Corollary 3.19,

eixeiy D .cos x C i sinx/.cos y C i siny/ D cos.x C y/C i sin.x C y/ D ei.xCy/:

We can also translate De Moivre’s Theorem into exponential notation.

Corollary 3.27 (Exponential De Moivre). For every real number x and all

integers n � 1,

.eix/n D einx :

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110 Chapter 3 Renaissance

Proof. According to De Moivre’s Theorem,

.eix/n D .cos x C i sinx/n D cos.nx/C i sin.nx/ D einx :

It is easier to remember the trigonometric addition formulas in complex

form. For example, let’s find the triple angle formulas. On the one hand, De

Moivre’s Theorem gives

ei3x D cos.3x/C i sin.3x/:

On the other hand,

ei3x D .eix/3

D � cos x C i sinx

�3

D cos3 x C 3i cos2 x sin x C 3i2 cos x sin2 x C i3 sin3 x D cos3 x � 3 cos x sin2 x C i

� 3 cos2 x sin x � sin3 x

� :

Equating real and imaginary parts, we have

cos.3x/ D cos3 x � 3 cos x sin2 x

and

sin.3x/ D 3 cos2 x sinx � sin3 x:

Roots of Unity

De Moivre’s Theorem can be used to find the roots of an important family of

polynomials: those of the form xn � 1.

Theorem 3.28. The distinct roots of xn � 1 are

1; �; �2; � � � ; �n�1;

where � D �n D cos.2�=n/C i sin.2�=n/. These numbers are equally spaced on the unit circle and are the vertices of a regular polygon, called the unit

n-gon.

Proof. By Corollary 3.23, �n D 1, so that � is a root of xn � 1. Furthermore, for any nonnegative integer k, we have .�k/n D .�n/k D 1, so that all �k are also roots of xn � 1. But there are repetitions on the list 1; �; �2; � � � . By the Division Algorithm, for any j , we have j D qn C r , where 0 � r � n � 1. Hence,

�j D �qnCr D �qn �r D �r ;

because �qn D 1. On the other hand, all the �k , for 0 � k � n � 1, are distinct. After all, by

De Moivre’s Theorem,

�k D cos.2�k=n/C i sin.2�k=n/;

and Proposition 3.14, uniqueness of polar forms, applies, for 0 � 2�k=n < 2� are n distinct angles. Therefore, we have displayed n distinct roots of xn � 1.

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3.3 Roots and Powers 111

These are all the roots of xn � 1, for a polynomial of degree n can have at most n distinct roots. We’ll give a proof of this later (see Theorem 6.16) but,

since we haven’t yet proved this result, we now proceed in a different way. If z 2 C is a root of xn � 1, then 1 D jznj D jzjn, so that jzj D 1,

by Exercise 1.73 on page 41, and z D cos � C i sin � for some � . By De Moivre’s Theorem, 1 D zn D cos.n�/ C i sin.n�/, so that n� D 2�k for some integer k; hence, � D 2�k=n. Write k D qnC r , where 0 � r < n, and

z D cos.2�k=n/C i sin.2�k=n/ D cos.2�r=n/C i sin.2�r=n/:

Thus, z is equal to the root �r already displayed. Finally, since arg

� �k �

D k arg .�/ D 2�k=n, the roots �k are equally spaced around the circle and, hence, they are the vertices of a regular n-gon.

(See Figure 3.7 for the case n D 8.)

Definition. The roots of xn � 1 are called the nth roots of unity. An nth root of unity � is a primitive nth root of unity if n is the smallest positive integer

for which �n D 1.

For every n � 1, we see that � D e2�i=n is a primitive nth root of unity, for if 1 � m < n, then �m D cos.2�m=n/ C i sin.2�m=n/ ¤ 1. In particular, i D cos.2�=4/ C i sin.2�=4/ is a primitive fourth root of unity, and ! D 1 2 .�1C

p 3/ D cos.2�=3/C i sin.2�=3/ is a primitive cube root of unity.

Corollary 3.29. Let �k D cos.2�k=n/Ci sin.2�k=n/ be an nth root of unity. (i) �k is a primitive nth root of unity if and only if gcd.k; n/ D 1.

(ii) If �k is a primitive nth root of unity, then every nth root of unity is a power

of �k .

Proof. (i) Suppose that �k is a primitiventh root of unity. If d D gcd.k; n/ > 1, then n=d < n, and

.�k/n=d D .�n/k=d D 1:

This contradicts n being the smallest positive integer with .�k/n D 1. Suppose that �k is not primitive; that is, .�k/m D 1 for some m < n.

Since, by hypothesis, gcd.k; n/ D 1, there are integers s and t with 1 D sk C tn; hence, m D msk Cmtn. But now

�m D �mskCmtn D �msk�mtn D 1;

which contradicts � being a primitive nth root of unity.

(ii) Every nth root of unity is equal to �j for some j . If gcd.k; n/ D 1, then there are integers s and t with 1 D sk C tn. Hence,

�j D �jskCjtn D �jsk�jtn D .�k/js :

Definition. For every integer n � 1, define the Euler �-function �.n/ by

�.n/ D number of k with 1 � k � n and gcd.k; n/ D 1:

For example, �.1/ D 1 and, if p is prime, �.p/ D p � 1.

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112 Chapter 3 Renaissance

Corollary 3.30. For every positive integer n, there are exactly �.n/ primitive

nth roots of unity.

Proof. This follows at once from Corollary 3.29(i).

Example 3.31. The complex number cos.2� n / C i sin.2�

n / is a primitive nth

root of unity, by Theorem 3.28. The 8th roots of unity (shown in Figure 3.7)

are

cos.2� 8 /C i sin.2�

8 /; cos.6�

8 /C i sin.6�

8 /;

cos.10� 8 /C i sin.10�

8 /; cos.14�

8 /C i sin.14�

8 /I

that is, the primitive 8th roots of unity are all those cos.2k� 8 /C i sin.2k�

8 / for

which gcd.k; 8/ D 1. N

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Figure 3.7. 8th roots of unity.

The nth roots of unity enjoy some remarkable properties that we’ll use in

upcoming chapters; here are some of them. (See Exercise 3.51 on page 115

and Proposition 6.63 for some other interesting properties.)

Theorem 3.32. Let � be an nth root of unity.

(i) 1C �2 C �3 C � � � C �n�1 D 0. (ii) �k D 1=�k for every nonnegative integer k.

(iii) If k D qnC r , then �k D �r .

Proof. (i) We have xn � 1 D .x � 1/q.x/, and we find q by long division:

.x � 1/ � 1C x C x2 C � � � C xn�1

� D xn � 1:

Now set x D � to see that

.� � 1/.1 C � C �2 C � � � C �n�1/ D �n � 1 D 0:

But � � 1 ¤ 0, and the result follows. (ii) This follows from Exercise 3.38 on page 106.

(iii) This follows from �qn D 1.

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3.3 Roots and Powers 113

Theorem 3.28 establishes an intimate connection between the nth roots of

unity and the geometry of the unit n-gon. The next examples illustrate this

connection for small n.

Example 3.33. The vertices of unit n-gons for small values of n can be calcu- lated with plane geometry.

See Exercise 3.49 on

page 115 for more exam-

ples.

� The vertices of the unit 3-gon are 1; 1 2 .�1 C i

p 3/; 1

2 .�1 � i

p 3/:

� The vertices of the unit 4-gon are 1; i; �1; �i . � The vertices of the unit 6-gon are

1; 1 2 .1Ci

p 3/; 12 .�1Ci

p 3/; �1; 12 .�1�i

p 3/; 12 .1�i

p 3/: N

Example 3.34 (Regular Pentagon). Since a primitive 5th root of unity is � D cos.2�=5/C i sin.2�=5/, by Theorem 3.28, the vertices of the unit 5-gon are

� D cos.2�=5/C i sin.2�=5/ �2 D cos.4�=5/C i sin.4�=5/ �3 D cos.6�=5/C i sin.6�=5/ �4 D cos.8�=5/C i sin.8�=5/ �5 D cos.10�=5/C i sin.10�=5/ D 1 D �0:

1

Figure 3.8. Unit 5-gon.

Can we find explicit expressions for these vertices that don’t involve trigonom-

etry? We’ll obtain such an expression for cos.2�=5/, but we’ll leave the rest of the details for you (Exercise 3.49(i) on page 115); after all, you can evaluate,

say, cos.8�=5/.

We have � D �4, by Theorem 3.32(ii). Inspired by Lemma 3.2, we define

g D � C �4 and h D �2 C �3:

Now

g D � C �4 D 2 cos.2�=5/;

a real number that is twice the number we are after. Similarly,

h D �2 C �3 D 2 cos.4�=5/;

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114 Chapter 3 Renaissance

another real number. By Theorem 3.32(i),

g C h D � C �4 C �2 C �3

D �1:

Thus, we know that g C h D �1. Do we also know gh? Let’s see.

gh D � � C �4

� .�2 C �3/

D �3 C �4 C �6 C �7

D �3 C �4 C � C �2 by Theorem 3.32(iii) D �1:

Hence, g C h D �1 D gh, and so g and h are the roots of

x2 C x � 1:

Now x2 C x � 1 has a positive root and a negative one. Since g > h (why?), the positive root is g, and so

cos.2�=5/ D 1 2 g

D 1 4 .�1C

p 5/: N

An ancient problem, going back to the Greeks, is to determine which regular

n-gons can be constructed with ruler and compass. As we’ll see in Chapter 7,

the problem comes down to finding an expression for cos.2�=n/ that doesn’t

mention trigonometry, only the operations of arithmetic and iterated square

roots (as in Example 3.34). We’ve essentially shown that the regular pentagon

is so constructible. This argument, grouping the �k into convenient subclusters, was greatly generalized and refined by Gauss (when he was only 17 years old!)

to show that the vertices of the unit 17-gon can be constructed (Euclid did not

know this!). (This and much more is in Gauss’s masterpiece, Disquisitiones

Arithmeticae.) Gauss requested that his tombstone portray a regular 17-gon,

but the stonemason was unable to carve it, saying it would look more like a circle than a polygon.

Exercises

3.46 Is De Moivre’s Theorem true for negative integer exponents? Explain.

3.47 Let z D cos � C i sin � . Show, for all nonnegative integers n, that

zn C .z/n D 2 cos n� and zn � .z/n D 2 sin n�:

3.48 This exercise shows that there’s something special about a 72ı angle: there’s only one isosceles triangle (up to similarity) whose base angle is twice the vertex an-

gle, namely, the “72-72-36 triangle.” Let the equal sides of such a triangle have

length 1, and let q denote the length of the base.

(i) Bisect one of the base angles of the triangle.

(ii) Show that the small triangle is similar to the whole triangle.

(iii) Use (ii) to show that 1 q

D q 1�q ; and solve for q.

(iv) Show that q=2 D cos 72ı.

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3.3 Roots and Powers 115

1 1

q

72°72°

36°

Figure 3.9. 72-72-36 triangle.

1

1 – q

q

q

q

Figure 3.10. Its construction.

3.49 * Find explicit formulas (i.e., without trigonometry) for the vertices of the unit

(i) pentagon. (ii) decagon. (iii) 20-gon.

3.50 * Let n be a positive integer and let � D e2�i=n. (i) Establish the identity

xn � 1 D .x � 1/.x � �/ � x � �2

� � � � �

x � �n�1 �

:

(ii) If x and y are integers, show that

xn � yn D .x � y/.x � �y/ �

x � �2y �

� � � � x � �n�1y

� :

(iii) If x and y are integers and n is odd , show that

xn C yn D .x C y/.x C �y/ �

x C �2y �

� � � � x C �n�1y

� :

3.51 Take It Further. We saw, on page 111, that �.p/ D p � 1, where p is prime and � is the Euler-� function. Note that if n D p1p2,

then .p1 � 1/.p2 � 1/ D n� n

p1 � n

p2 C n

p1p2 :

(i) Suppose n is the product of two primes, n D p1p2 . Show that

�.n/ D .p1 � 1/.p2 � 1/:

(ii) Suppose n is the product of two primes powers, n D pe11 p e2 2 . Show that

�.n/ D n � n p1

� n p2

C n p1p2

D n �

1 � 1 p1

�� 1 � 1

p2

� :

(iii) Generalize to show that, if n D pe11 p e2 2 : : : p

en n , then

�.n/ D n nY

kD1

� 1 �

1

pk

� :

3.52 Prove or disprove and salvage if possible. If a and b are positive integers,

�.ab/ D �.a/�.b/:

3.53 Find explicit formulas (i.e., without trigonometry) for the vertices of the unit n-

gon if

(i) n D 3 (ii) n D 4 (iii) n D 6 (iv) n D 8 (v) n D 12 (vi) n D 16

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116 Chapter 3 Renaissance

3.54 For all integers n between 3 and 9, find all the primitive nth roots of unity.

3.55 Find a primitive 12th root of unity �. Is � unique?

3.56 Suppose � D cos.2� 7

/ C i sin.2� 7

/.

(i) Plot the roots of x7 � 1 in the complex plane. (ii) Show that ˛ D �C�6, ˇ D �2C�5, and D �3C�4 are real numbers.

(iii) Find a cubic equation satisfied by 2 cos.2�7 / by finding the values of ˛CˇC , ˛ˇ C ˛ C ˇ , and ˛ˇ .

3.57 If �n D cos.2�n / C i sin. 2� n

/, evaluate Pn�1 kD0 �

k n :

3.58 Show that cos.2�5 / C cos. 4� 5 / D �

1 2 :

3.59 Take It Further. If n is a nonnegative integer, how many irreducible factors over

Z does xn � 1 have? In other words, we’re looking for a pattern in the outputs of the function n 7! # of factors of xn � 1 over Z: (Use a computer).

n Number of Factors of xn � 1 1

2

3

4

5

6

7

8

9

10

11

12

3.4 Connections: Designing Good Problems

This section will use complex numbers to help create mathematics problems

that “come out nice.” When launching a new topic, you want to start with

examples which focus on the new idea; there shouldn’t be any distractions—

for example, numbers should be simple integers or rationals. Indeed, this is why the Babylonians introduced Pythagorean triples.

Norms

We begin by introducing a function C ! R, called the norm, that is closely related to absolute value. It will be an important tool for our applications; it will also be very useful in Chapter 8 when we do some algebraic number theory.

Definition. The norm of a complex number z D aC bi is

N.z/ D zz D a2 C b2:

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3.4 Connections: Designing Good Problems 117

Here are some basic properties.

Proposition 3.35. Let z D aC ib and w be complex numbers. (i) N.z/ is a nonnegative real number, andN.z/ D 0 if and only if z D 0.

(ii) N.z/ D jzj2. (iii) N.zw/ D N.z/N.w/.

Proof. (i) This follows at once from N.aC bi/ D a2 C b2. (ii) This follows at once from jzj D jaC bi j D

p a2 C b2.

(iii) N.zw/ D zw zw D zw z w D zz ww D N.z/N.w/:

It follows from Proposition 3.35(iii) that

N.zk/ D N.z/k

for all z and all k � 0. Here is an application of the norm.

Example 3.36. Let’s revisit Example 3.5, the “bad example,” in which the

cubic formula gives the roots of x3 � 7x C 6 D .x � 1/.x � 2/.x C 3/ in unrecognizable form.

Imagine again that you have just left the contest in Piazza San Marco, think-

ing about how g C h could possibly equal 1, where

g3 D �3C i 10 9

p 3 and h3 D �3 � i 10

9

p 3:

Had you known about conjugates, you’d have seen that g3 D h3. It would have been natural to guess that the cube roots g and h are also complex conjugates (you’d have guessed right: see Exercise 3.64 on page 127); thus, g D a C ib and h D a� ib. Now if gC h D 1, as your opponent loudly proclaimed, then .a C ib/C .a � ib/ D 2a D 1; that is,

g D 1 2

C ib and h D 1 2

� ib:

You really want to find g and h now—what is b? Using the norm function, you

see that

N.g/3 D N.g3/ D .�3/2 C � 10

p 3

9

�2 D 343

27 :

Since norms are always real numbers, you conclude that

N.g/ D 3 q 343 27

D 7 3

(the other cube roots are complex; they are 7 3 ! and 7

3 !2, where ! is a primitive

cube root of unity). But if g D 1 2

C ib, thenN.g/ D 1 4

Cb2. Hence, 1 4

Cb2 D 7 3

, and b D ˙ 5 2

p 3

. Thus,

g D 1 2

C i 5 2

p 3

and h D 1 2

� i 5 2

p 3 :

Bingo! For these “values” of g and h, we have g C h D 1. You were right! To find the other two roots, see Exercise 3.65 on

page 127. Elated, you run back to the square to show off g and h, but everyone has gone home. N

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118 Chapter 3 Renaissance

Pippins and Cheese

We call this subsection Pippins and Cheese, a phrase borrowed from Shake-

speare’s Merry Wives of Windsor, which describes delicious desserts.

Here are five problems. Solve them now; they are not difficult, but the an- swers may surprise you.

(i) A triangle has vertices .�18; 49/; .15;�7/, and .30;�15/: How long are its sides?

(ii) In Figure 3.11, the side lengths of 4QSU are as marked. How big is †Q?

U

Q S

5 7

8

Figure 3.11. Side lengths.

(iii) An open box is formed by cutting out squares from a 7� 15 rectangle and folding up the sides (see Figure 3.12). What size cut-out x maximizes the

volume of the box?

7

15

Figure 3.12. Making boxes.

(iv) Find the zeros, extrema, and inflection points of the function

f .x/ D 140� 144 x C 3 x2 C x3:

(v) Find the area of the triangle with sides of lengths 13, 14, and 15.

A meta-problem is a problem that asks how to design “nice” exercises of

a particular genre, such as “How do you construct integer-sided scalene trian-

gles having a 60ı angle?” As we mentioned earlier, finding Pythagorean triples was one of the first meta-problems; it was invented by teachers who wanted to

study and apply side-lengths of right triangles. In Section 1.2, we developed

the method of Diophantus for this purpose—rational points on the unit circle

correspond to Pythagorean triples. In this section, we’ll consider two types

of meta-problems: two ways of creating exercises like the five listed above. One meta-problem uses the norm function; the other generalizes Diophantus’sSee Exercises 1.79—1.82

on page 44. chord method of “sweeping lines” by replacing circles with other conic sec-

tions. (There are many other kinds of meta-problems, ranging in topic from

exponential equations to algebra word problems to trigonometry.)

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3.4 Connections: Designing Good Problems 119

Gaussian Integers: Pythagorean Triples Revisited

In Chapter 1, we saw that the Pythagorean equation a2C b2 D c2 corresponds to a rational point

� a c ; b c

� on the unit circle. But, given what you’ve just been

studying, the Pythagorean equation might conjure up another rewrite in your

mind, namely

.a C bi/.a � bi/ D c2;

or even

N.a C bi/ D c2:

So, we’re looking for complex numbers z D a C bi whose norms are perfect squares of integers. The Pythagorean equation now looks like

N.z/ D c2:

For example, N.3C 4i/ D 52, N.5C 12i/ D 132, and N.8C 15i/ D 172. This idea doesn’t work for every complex number. What’s needed are com-

plex numbers whose real and imaginary parts are integers (and, besides, whose norms are perfect squares). We’d like the real and imaginary parts to be posi-

tive integers, but any integers will do, because changing the sign of the real or

imaginary part of a complex number doesn’t change its norm (why?).

Definition. The Gaussian integers is the set ZŒi  of all complex numbers

whose real and imaginary parts are integers. In symbols,

ZŒi  D faC bi 2 C W a 2 Z and b 2 Zg :

Proposition 3.37. (i) The set ZŒi  of Gaussian integers is closed under ad-

dition and multiplicationW If aC bi; c C di 2 ZŒi , then

.a C bi/C .c C di/ D .a C c/C .b C d/i 2 ZŒi  .a C bi/.c C di/ D .ac � bd/C .ad C bc/i 2 ZŒi :

(ii) If z D aC bi , then

N.z/ D a2 C b2:

Proof. The formula for addition is clear; for multiplication, use the fact that

i2 D �1. Of course, part (ii) is just the definition of the norm.

We’ll investigate the Gaussian integers in more detail in Chapter 8. Let’s return to the norm equation N.z/ D c2 arising from Pythagorean

triples, but with z a Gaussian integer. Our question is now “Which Gaussian

integers have perfect squares as norms?” The answer comes from Proposi-

tion 3.35(iii): if z and w are complex numbers, thenN.zw/ D N.z/N.w/: In particular (letting z D w),

N � z2

� D N.z/2:

The left-hand side of this equation is the norm of a Gaussian integer: if z D aC ib, then z2 D .a2�b2/C i2ab; moreover, N.z2/ is a sum of two nonzero

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120 Chapter 3 Renaissance

perfect squares if a > 0, b > 0, and a ¤ b. Now the right-hand side is the square of an integer, namely, N.z/2, which produces a Pythagorean triple. For

example, if z D 3 C 2i , then N.z/ D 13 and z2 D 5 C 12i , and we get the Pythagorean triple .5; 12; 13/, for

52 C 122 D N � .3C 2i/2

� D N.3C 2i/2 D 132:

We now have a quick way to generate Pythagorean triples (by hand or with

a computer; one of our colleagues uses this method to amaze friends at parties). Pick a Gaussian integer r C si (with r > 0, s > 0, and r ¤ s), and square it.

The r; s entry in the following table is Œ.r C is/2; N.r C is/. For exam- ple, the top entry in the first column, arising from r D 2 and s D 1, is Œ.2 C i/2; N.2 C i/ D Œ3 C 4i; 5; the corresponding Pythagorean triple is .3; 4; 5/.

s D 1 s D 2 s D 3 s D 4 r D 2 3C 4i; 5 r D 3 8C 6i; 10 5C 12i; 13 r D 4 15C 8i; 17 12C 16i; 20 7C 24i; 25 r D 5 24C 10i; 26 21C 20i; 29 16C 30i; 34 9C 40i; 41 r D 6 35C 12i; 37 32C 24i; 40 27C 36i; 45 20C 48i; 52

Eisenstein Integers.

Let’s now look at the meta-problem of creating triangles with integer side-

lengths and a 60ı angle. Let †C D 60ı in Figure 3.13, so that cos.†C/ D 12 . By the Law of

Cosines,

c2 D a2 C b2 � 2ab cos †C D a2 C b2 � ab:

What’s important here is that the right-hand side of the equation, a2�abCb2, is the norm of a C b!, where ! D 1

2 .�1 C i

p 3/ is a primitive cube root of

unity (Exercise 3.72 on page 128). This leads to the following definition.

b

a c

C

B

A

Figure 3.13. †C D 60ı.

Definition. The Eisenstein integers is the set ZŒ! of all complex numbers of

Eisenstein did extensive

research on complex

numbers of the form

a C b�, where � is a primitive nth root of unity.

Note that ! is a primitive

cube root of unity.

the form a C b!, where ! D 1 2 .�1 C i

p 3/ is a primitive cube root of unity

and a; b are integers. In symbols,

ZŒ! D faC b! 2 C W a 2 Z and b 2 Zg :

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3.4 Connections: Designing Good Problems 121

Here are some properties of Eisenstein integers.

Proposition 3.38. (i) The set ZŒ! of Eisenstein integers is closed under ad-

dition and multiplication: If .a C b!/; .c C d!/ 2 ZŒ!, then

InZŒi, i2 D �1. In ZŒ!; !2 D �1� !.

.aC b!/C .c C d!/ D .a C c/C .b C d/! 2 ZŒ! .a C b!/.c C d!/ D .ac � bd/C .bc C ad � bd/! 2 ZŒ!:

(ii) If z D aC b!, then In ZŒi, i2 D �1. In ZŒ!; !2 D �1� !.

N.z/ D a2 � ab C b2:

Proof. (i) The formula for addition is clear. For multiplication,

.a C b!/.c C d!/ D ac C .bc C ad/! C bd!2:

Since !2 C ! C 1 D 0; we have !2 D �1 � !, and

.a C b!/.c C d!/ D .ac � bd/C .bc C ad � bd/!:

(ii) As we said above, this is Exercise 3.72 on page 128.

Definition. An Eisenstein triple is a triple of positive integers .a; b; c/ such

that

a2 � ab C b2 D c2:

The same idea that produces Pythagorean triples from norms of squares of

Gaussian integers applies to produce Eisenstein triples from norms of squares

of Eisenstein integers. If z is an Eisenstein integer, then

N.z2/ D N.z/2:

The left-hand side of this equation, being the norm of an Eisenstein integer,

is of the form a2 � ab C b2. And the right-hand side is the square of the integer N.z/. Hence a2 � ab C b2 is a perfect square, and we have produced an Eisenstein triple.

Example 3.39. If z D 3C 2!, thenN.z/ D 32 � 3 � 2C 22 D 7, and we have

z2 D 9C 12! C 4!2

D 9C 12! C 4.�1 � !/ D 5C 8!:

Hence, 52 � 5 � 8C 82 D N � z2 �

D N.z/2 D 72; and .5; 8; 7/ is an Eisenstein triple. N

In Figure 3.14, we have †Q D 60ı. We have found a quick way to generate Eisenstein triples (by hand or with

a computer). Pick an Eisenstein integer r C s! (with r > 0, s > 0, and r ¤ s) and square it.

The r; s entry in the following table is .r C s!/2; N.r C s!/. For exam- ple, the top entry in the first column, which arises from r D 2 and s D 1,

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122 Chapter 3 Renaissance

U

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5 7

8

Figure 3.14. All sides have integer length.

is � .2 C !/2; N.2C !/

� D .3C 3!; 3/; the corresponding Eisenstein triple

gives .3; 3; 3/, which is an equilateral triangle. One of our friends calls this

table a “candy store of patterns.” Which entries give equilateral triangles?

s D 1 s D 2 s D 3 s D 4 r D 2 3C 3!; 3 r D 3 8C 5!; 7 5C 8!; 7 r D 4 15C 7!; 13 12C 12!; 12 7C 15!; 13 r D 5 24C 9!; 21 21C 16!; 19 16C 21!; 19 9C 24!; 21 r D 6 35C 11!; 31 32C 20!; 28 27C 27!; 27 20C 32!; 28 r D 7 48C 13!; 43 45C 24!; 39 40C 33!; 37 33C 40!; 37 r D 8 63C 15!; 57 60C 28!; 52 55C 39!; 49 48C 48!; 48 r D 9 80C 17!; 73 77C 32!; 67 72C 45!; 63 65C 56!; 61 r D 10 99C 19!; 91 96C 36!; 84 91C 51!; 79 84C 64!; 76

Eisenstein Triples and Diophantus

There’s another, geometric, way to generate Eisenstein triples, using the same

idea as the method of Diophantus in Chapter 1. If .a; b; c/ is an Eisenstein

triple, so thatSee Exercise 1.79 on page 44.

a2 � ab C b2 D c2;

then dividing by c2 gives

.a=c/2 � .a=c/ .b=c/C .b=c/2 D 1:

Thus, .a=c; b=c/ is a rational point on the ellipse with equation

x2 � xy C y2 D 1:

(See Figure 3.15.) As with the unit circle, the graph contains .�1; 0/, and we can use the chord method idea of Diophantus.See Exercise 3.66 on

page 128.

Proposition 3.40. Let ` be a line through .�1; 0/ which intersects the ellipse with equation x2 � xy C y2 D 1 in a point P . If ` has rational slope, then P has rational coordinates, P D .a=c; b=c/, and

a2 � ab C b2 D c2:

If P D .a=c; b=c/ is in the first quadrant, then .a; b; c/ is an Eisenstein triple.

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3.4 Connections: Designing Good Problems 123

y

x 0.4 0.8 1.2 1.6

0.4

0.8

1.2

–0.4

–0.8

–1.2

–0.4–0.8–1.2–1.6

Figure 3.15. The graph of x2 � xy C y2 D 1.

Proof. The proof is almost identical to the proof of Proposition 1.2. We leave it to you to fill in the details.

For example, if ` has slope 1 4

and equation y D 1 4 .x C 1/, then ` intersects

the ellipse in � 15 13 ; 7 13

� , and .15; 7; 13/ is an Eisenstein triple. So, the triangle

whose side lengths are 15, 7, and 13 has a 60ı angle. Which angle is it?

Nice Boxes

Our next application is to a “box problem.” In an a � b rectangle, cut out little squares at the corners, and then fold up the sides to form an open-top box (see

Figure 3.16). What size cut-out maximizes the volume of the box? For most rectangles, the best cut-out has irrational side length. The meta-problem:

How can we find a and b to make the optimal cut-out a rational number?

a

b

x

Figure 3.16. Box problem.

As we tell our students, let the size of the cut-out be x. Then the volume of

the box is a function of x:

V.x/ D .a � 2x/.b � 2x/x D 4x3 � 2.aC b/x2 C abx;

and its derivative is

V 0.x/ D 12x2 � 4.a C b/x C ab:

We want V 0.x/ to have rational zeros, and so its discriminant

16.aC b/2 � 48ab

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124 Chapter 3 Renaissance

should be a perfect square. But 16 is a perfect square, and so

.a C b/2 � 3ab D a2 � ab C b2

should be a perfect square. This will be so if a and b are the legs of an Eisen-

stein triple .a; b; c/.

For example, from the Eisenstein triple .7; 15; 13/, we get a 7�15 rectangle that can be used to create a box whose maximum volume occurs at a rational-

length cut-out. The volume of the resulting box is

V.x/ D .7 � 2x/.15 � 2x/x D 4x3 � 443x2 C 105x:

So, V 0.x/ D 12x2 � 88x C 105. The roots of V 0.x/ are 3 2

and 35 6

. Both are

rational, but only 3 2

fits the context and maximizes V . (Why doesn’t 35 6

fit the context? What significance does it have? Also, see Exercise 3.69 on page 128.)

Nice Functions for Calculus Problems

Our next meta-problem is one that has occupied faculty room discussions about

calculus teaching for years.

How do you find cubic polynomials f .x/ with integer coefficients and

rational roots, whose extrema and inflection points have rational coordi-

nates?

No cheating: we want the extrema points and inflection points to be distinct. We’ll actually create cubics in which all these points have integer coordinates.

Using the notation of Theorem 3.3, we can first assume that the cubic f is reduced; that is, it has form

f .x/ D x3 C qx C r:

This immediately guarantees that f 00.x/ D 6x has an integer root, namely 0 (the inflection point of the graph is on the y-axis). Next, if we replace q by

�3p2 for some integer p, then f 0.x/ D 3x2�3p2, and f 0.x/ also has integer roots. So, our cubic now looks like f .x/ D x3 � 3p2 x C r . This will have rational extrema and inflection points (what are they?), so all we have to do is ensure that it has three rational roots.

If f has two rational roots, it has three (why?), and so it’s enough to make

two roots, say �˛ and ˇ, rational (we use �˛ instead of ˛ because we’ve experimented a bit and found that this makes the calculations come out nicer).

But if f .�˛/ D f .ˇ/ D 0, we have

�˛3 C 3p2˛ D ˇ3 � 3p2ˇ

or

ˇ3 C ˛3 D 3p2.˛ C ˇ/:

We can divide both sides by ˛ C ˇ, for ˛ C ˇ ¤ 0 (lest �˛ D ˇ; remember that we want our roots distinct); we obtain

˛2 � ˛ˇ C ˇ2 D 3p2:

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3.4 Connections: Designing Good Problems 125

Eisenstein integers again. This is the same as

N.˛ C ˇ!/ D 3p2:

This time we want an Eisenstein integer whose norm is 3 times a square.

We’re in luck: the equation a2 � abC b2 D 3 has several integer solutions, including .1; 2/. So, 3 D N.1 � !/. Hence we just need to take ˛ C ˇ! to be 1 � ! times the square of an Eisenstein integer. Indeed, if

˛ C ˇ! D .1 � !/.r C s!/2;

then

˛2 � ˛ˇ C ˇ2 D N.˛ C ˇ!/ D N � .1 � !/.r C s!/2

D N .1 � !/N � .r C s!/2

D 3N.r C s!/2;

which is 3 times the square of an integer.

Example 3.41. Let’s take s D 1 and r D 3. Then we have

˛ C ˇ! D .1 � !/.3 C !/2

D 13C 2!:

This tells us several things:

(i) Since N.13C 2!/ D 147, our cubic is

f .x/ D x3 � 147x C r:

(ii) But because 147 D 3 � 72 (so, p D 7), f 0.x/ D 3x3 � 3 � 49 will have rational roots: ˙7.

(iii) Since two roots of our cubic are �˛ and ˇ, two roots are �13 and 2. This lets us find r . Since

23 � 147 � 2C r D 0;

we have r D 286. Hence our cubic is

f .x/ D x3 � 147x C 286:

You can check that the third root is 11 and that the extrema and inflection

points are rational. N

Creating examples like this is not hard by hand, but a computer algebra

system makes it automatic. The next table was generated by a CAS, and it

shows the results of our algorithm for small values of r and s.

s D 1 s D 2 s D 3 r D 2 54 � 27x C x3 �128 � 48x C x3

r D 3 286 � 147x C x3 286 � 147x C x3 �1458 � 243x C x3

r D 4 �506 � 507x C x3 3456 � 432x C x3 �506 � 507x C x3

r D 5 �7722 � 1323x C x3 10582 � 1083x C x3 10582 � 1083x C x3

r D 6 �35282 � 2883x C x3 18304 � 2352x C x3 39366 � 2187x C x3

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126 Chapter 3 Renaissance

All these cubics have coefficient of x2 equal to 0. If you’d like examples

where this is not the case, just replace x by, say, x C 1 and simplify. Again, aSee Exercise 3.70 on page 128. CAS makes this easy.

Lattice Point Triangles

Our last meta-problem arises when illustrating the distance formula.A lattice point is a point with integer coordinates.

How can you find three lattice points A, B , and C in the plane so that

the distance between any two of them is an integer?

Clearly, solutions are invariant under translation by a lattice point; that is,

if A, B , and C form a lattice point solution and U is any lattice point, then

A� U , B � U , and C � U form another solution: since d.A � U;B � U / D d.A; B/ (where d.P;Q/ is the distance between points P and Q), we have d.A; B/ D jA� Bj. Hence, we can assume that one of the points, say C , is at the origin.

Now view the plane as the complex plane, so that lattice points are Gaussian

integers. Thus, we want Gaussian integers z and w such that jzj, jwj, and jz �wj are integers. But if z D aC bi , then

jzj D p a2 C b2 D

p N.z/:

Hence, to make the length an integer, make the norm a perfect square and, to

make the norm a perfect square, make the Gaussian integer a perfect square in

ZŒi . That is, we want Gaussian integers z and w so that z, w, and z � w are perfect squares in ZŒi . Hence, we choose z and w so that

z D ˛2 for some ˛ 2 ZŒi  w D ˇ2 for some ˇ 2 ZŒi 

z �w D 2 for some 2 ZŒi :

In other words, we want Gaussian integers ˛, ˇ, and so that

˛2 � ˇ2 D 2

or

˛2 D ˇ2 C 2:

The punchline is that one of our favorite identities,

� x2 C y2

�2 D � x2 � y2

�2 C .2xy/2 ;

which holds in any commutative ring, holds, in particular, in ZŒi . So, the trick

is to pick Gaussian integers x and y, set

˛ D x2 C y2

ˇ D x2 � y2;

and then let

See Exercise 1.25 on

page 14.

z D ˛2

w D ˇ2:

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3.4 Connections: Designing Good Problems 127

Example 3.42. Pick x D 2C i and y D 3C 2i . Then

˛ D x2 C y2 D 8C 16i and ˇ D x2 � y2 D �2 � 8i:

Now put

z D ˛2 D �192C 256i and w D ˇ2 D �60C 32i:

Hence, .0; 0/, .�192; 256/, and .�60; 32/ are vertices of an integer-sided tri- angle. Moreover, adding a lattice point to each vertex produces another such triangle with no vertex at the origin. Once again, a CAS can be used to generate

many more. N

This is just the beginning; many research problems are generalizations of

meta-problems. Fermat’s Last Theorem started as a search for integer solutions

to equations like the Pythagorean equation but with larger exponents.

There are many other meta-problems that yield to these two methods: norms from ZŒi  or ZŒ!; rational points on the unit circle or on the graph of x2 � xyCy2 D 1. Still others can be solved with norms from other number systems or from rational points on other curves. In Chapter 9, we will see that congruent

numbers lead to rational points on certain cubic curves.

Exercises

3.60 For each integer n between 3 and 9, find a polynomial of smallest degree with

integer coefficients whose roots are the primitive nth roots of unity.

3.61 * Let a and b be real numbers, and let z be a complex number.

(i) Show that a C bz D a C b z: (ii) Show that N.a C bz/ D a2 C 2<.z/ab C b2N.z/.

3.62 * If z and w are complex numbers, show that N.z/ < N.w/ if and only if

jzj < jwj.

3.63 Let  be an isosceles triangle with side lengths 13, 13, and 10.

(i) Show that the altitude to the base has length 12, and that it divides  into two

5, 12, 13 triangles.

(ii) Show that the altitude to one of the sides of length 13 divides  into two right

triangles whose side lengths are rational.

(iii) Each of the side lengths can thus be scaled to get a Pythagorean triple. Show

that one triple is similar to .5; 12; 13/ and that other comes from .5 C 12i /2. (iv) Generalize this result to any isosceles triangle formed by two copies of a

Pythagorean triple, joined along a leg.

3.64 * Let g and h be complex numbers such that g3 D h3. (i) Show that g is equal to either h, !h, or !2h, where ! D 12 .�1 C i

p 3/.

(ii) If gh is also real, show that g D h.

3.65 * Suppose that g D 12 C 5

2 p 3

i , h D g, and ! D cos.2�3 / C i sin. 2� 3 / (see

Example 3.36). Find the value of

(i) g! C h!2

(ii) g!2 C h!

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128 Chapter 3 Renaissance

3.66 * Sketch the graph of x2 � ax C y2 D 1 for (i) a D �1

(ii) a D 1 (iii) a D 2 (iv) a D 3 (v) a D 1

2

3.67 In Theorem 1.5, we saw that every Pythagorean triple is similar to one of the form

.2xy; x2 � y2; x2 C y2/:

Show how this can be obtained via the “norm from ZŒi ” method.

3.68 Obtain a formula for Eisenstein triples analgous to the one for Gaussian inte-

gers in Theorem 1.5 using norms from ZŒ! and rational points on the graph of

x2 � xy C y2 D 1.

3.69 * Assume that the square of the Eisenstein integer r C s! is used to generate an Eisenstein triple, and that the triple is used to create a “nice box,” as on page 123.

Express the volume of the box in terms of r and s.

3.70 * Replace x by x C 1 in several of the cubics in the table on page 125 to produceReplacing x by x C 1 just translates the graph by one

unit. Which way? nice cubics whose coefficient of x2 is nonzero. Show that your cubics are indeed

nice.

3.71 Describe where Gaussian integers are situated in the complex plane.

3.72 * Suppose that a and b are real numbers and

! D cos.2�3 / C i sin. 2� 3 / D

1 2 .�1 C i

p 3/:

Show that

N.a C b!/ D a2 � ab C b2:

3.73 Describe where Eisenstein integers are situated in the complex plane.

3.74 Find an integer-sided triangle one of whose angles has cosine equal to 3=5.

Hint: Let � D cos.35 / C i sin. 3 5 / and consider norms from ZŒ�. What conic

would help here?

3.75 A Heron triangle is a triangle with integer side lengths and integer area. In Ex-

ercise 1.26 on page 14, you found a Heron triangle by joining two Pythagorean

triangles together along a common leg. Show that the following method also pro-

duces Heron triangles.

Pick a rational point .cos �; sin �/ on the unit circle, where 0 < � < � , and let

˛ D � cos � C i sin � . Then pick any number z of the form r C s˛, where r and s are rational numbers and r > s > 0.

(i) What is the norm of r C s˛? (ii) Show that

˛2 C 2˛ cos � C 1 D 0:

(iii) Show that if z2 D a C b˛, then the triangle with side lengths a and b and included angle � will have a rational number, say c, as its third side length

and a rational number as an area. (This triangle can be then scaled to produce

a Heron triangle.) Use this method to generate a few Heron triangles.

3.76 Show that a triangle with lattice point vertices and integer side-lengths is a Heron

triangle.

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3.4 Connections: Designing Good Problems 129

3.77 Take It Further. Here’s a typical current problem, taken from B. Kerins, Gauss,

Pythagoras, and Heron, Mathematics Teacher, 2003, 350-357:

A boat is making a round trip, 135 miles in each direction. Without a cur-

rent, the boat’s speed would be 32 miles per hour. However, there is a con-

stant current that increases the boat’s speed in one direction and decreases

it in the other. If the round trip takes exactly 9 hours, what is the speed of

the current?

(i) Solve the problem.

(ii) Solve the corresponding meta-problem: find a method for generating current

problems that come out nice.

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4 Modular Arithmetic Theorems about integers can be generalized to other interesting contexts. For

example, an early attack on Fermat’s Last Theorem was to factor xn C yn (n odd) as in Exercise 3.50 on page 115:

xn C yn D .x C y/.x C �y/ � � � .x C �n�1y/;

where � D e2�i=n is an nth root of unity. It turns out that the most fruitful way to understand this factorization is within the system ZŒ� of cyclotomic integers, the collection of all polynomials in � with coefficients in Z (a com- We’ll discuss cyclotomic

integers in Chapter 8.mon generalization of Gaussian integers ZŒi  and Eisenstein integers ZŒ!).

Numbers in these systems can be added and multiplied, and they satisfy all but

one of the nine fundamental properties that ordinary numbers do (reciprocals

of cyclotomic integers need not be such); we will call such systems commuta-

tive rings. But for some roots of unity �, the commutative ring ZŒ� does not enjoy the unique factorization property that Z, ZŒi , and ZŒ! have, and this

caused early “proofs” of Fermat’s Last Theorem to be false. Dealing with the

lack of unique factorization was one important problem that led naturally to

the modern way of studying algebra.

In Section 4.1, we shall see that the distinction between even and odd can be generalized, using congruences: studying remainders in the Division Algo-

rithm. It turns out, as we’ll see in Section 4.3, that, for any fixed positive integer

m, the set of its remainders, 0; 1; :::; m � 1, can be viewed as a commutative ring, as can cyclotomic integers, and they behave in many, but not all, ways It turns out that many of

the “number systems”

studied in high school are

commutative rings.

as do ordinary integers. Finally, in Section 4.5, we’ll apply these results to an analysis of decimal expansions of rational numbers.

4.1 Congruence

It is often useful to know the parity of an integer n; that is, whether n is even

or odd (why else would these words be in the language?). But n being even or

odd is equivalent to whether its remainder after dividing by 2 is 0 or 1. Modular arithmetic, introduced by Euler around 1750, studies the generalization of par-

ity arising from considering remainders after dividing by any positive integer.

At a low level, it will help us answer questions of the following sort:

� London time is 6 hours ahead of Chicago time; if it is now 9:00 AM in Chicago, what time is it in London?

� If April 12 falls on a Thursday this year, on what day of the week is May 26?

131

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132 Chapter 4 Modular Arithmetic

At a more sophisticated level, it will allow us to solve some difficult number

theoretic problems.

Definition. Let m � 0 be an integer. If a; b 2 Z, then a is congruent to b modulo m, denoted by

Note that a � 0 mod m if and only if m j a. a � b mod m;

if m j .a � b/.

Etymology. The number m in the expression a � b mod m is called the modulus, the Latin word meaning a standard unit of measurement. The term

modular unit is used today in architecture: a fixed length m is chosen, say,

m D 1 foot, and plans are drawn so that the dimensions of every window, door, wall, etc., are integral multiples of m.

We claim that integers a and b have the same parity if and only if a � b mod 2. Assume that a and b have the same parity. If both are even, then

a D 2a0 and b D 2b0. Hence, a � b D 2.a0 � b0/, 2 j .a � b/, and a � b mod 2. Similarly, if both are odd, then a D 2a0 C 1 and b D 2b0 C 1. Hence, a � b D .2a0 C 1/ � .2b0 C 1/ D 2.a0 � b0/, 2 j .a � b/, and a � b mod 2 in this case as well. Conversely, suppose that a � b mod 2. If a and b have different parity, then one is even, the other is odd, and so their difference is

odd. Hence, 2 − .a � b/, and a 6� b mod 2. Having proved the contrapositive, we may now assert that a and b have the same parity.

Example 4.1. If a � r mod m, then r is obtained from a by throwing out a multiple of m. For example, let’s compute the time of day using a 12-hour

clock. When adding 6 hours to 9:00, the answer, 3:00, is obtained by takingModular arithmetic is called “clock arithmetic” in some

introductory texts. 9C 6 D 15 � 3 mod 12 (i.e., we throw away 12). In more detail, let 0 denote 12:00, 1 denote 1:00, : : : ; 11 denote 11:00. Three hours after 9:00 is 12:00;

that is, 9 C 3 D 12 � 0 mod 12; 4 hours after 9:00 is 1:00; that is, 9 C 4 D 13 � 1 mod 12, and 6 hours after 9:00 is 3:00; that is, 9C6 D 15 � 3 mod 12.

The same idea applies to calendars. Let 0 denote Sunday, 1 denote Mon-

day, : : :, 6 denote Saturday.

Sun Mon Tues Wed Thurs Fri Sat

0 1 2 3 4 5 6

If today is Tuesday, what day of the week is 90 days from now? Since 2C90 D 92 � 1 mod 7, the answer is Monday.

Let’s now answer the question: if April 12 falls on Thursday this year, on

what day of the week is May 26? There are 18 days to April 30, so there are

18C 26 D 44 days until May 26 (for April has only 30 days). Now Thursday corresponds to 4, so that May 26 corresponds to 4 C 44 D 48 � 6 mod 7; therefore, May 26 falls on Saturday. N

There are at least two ways to state the solutions of Exercises 3.2 and 3.3

on page 89. We expected you to say then that in D im if and only if n and m leave the same remainder when divided by 4 and, if ! is a primitive cube root

of unity, that !n D !m if and only if n and m leave the same remainder when

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4.1 Congruence 133

divided by 3. In light of the next proposition, we can also say that in D im if and only if 4 j .n � m/; that is, n � m mod 4, and !n D !m if and only if 3 j .n �m/; that is, n � m mod 3.

Proposition 4.2. Let m � 2 and a; b 2 Z.

(i) If a D qmC r , then a � r mod m. (ii) a � b mod m if and only if each of a and b have the same remainder

after dividing by m.

Proof. (i) Since a � r D qm, we have m j .a � r/; that is, a � r mod m. (ii) Assume a � b mod m. Let r; r 0 be the remainders after dividing a; b, re-

spectively, bym; that is, a D qmCr and b D q0mCr 0, where 0 � r < m and 0 � r 0 < m. We want to show that r 0 D r . If not, suppose that r 0 < r (the argument is the same if r < r 0). Then a � b D m.q � q0/C .r � r 0/ with 0 < r � r 0 < m. Now Exercise 1.46 on page 29 gives m j .r � r 0/. Hence, m � r � r 0, by Lemma 1.13, contradicting r � r 0 < m.

Conversely, if a D qmC r , b D q0mC r 0, and r D r 0, then a � b D m.q � q0/ and a � b mod m.

Notice that Proposition 4.2 generalizes the fact that integers a and b have

the same parity if and only if a � b mod 2. We are now going to see that congruence modulo m behaves very much

like ordinary equality; more precisely, it is an equivalence relation (see Ap- pendix A.2): it is reflexive, symmetric, and transitive.

Proposition 4.3. Let m � 0. For all integers a; b; c, we have

(i) a � a mod mI (ii) if a � b mod m, then b � a mod mI

(iii) if a � b mod m and b � c mod m, then a � c mod m.

Proof. All are easy to check. We have a � a mod m, because a � a D 0 and m j 0 is always true (even whenm D 0). Since b�a D �.a�b/, ifm j .a�b/, thenm j .b � a/. Finally, .a� b/� .b � c/ D a� c, so that ifm j .a� b/ and m j .b � c/, then m j .a � c/.

How to Think About It. Congruence mod 1 makes sense, but it is not very

interesting, for a � b mod 1 if and only if 1 j .a� b/. But this latter condition is always true, for 1 is a divisor of every integer. Thus, every two integers are congruent mod 1. Similarly, congruence mod 0 makes sense, but it, too, is not

very interesting, for 0 j c if and only if c D 0. Thus, a � b mod 0 if and only If 0 j c, then there is some k with c D 0 � k D 0; that is, c D 0.

if 0 j .a � b/; that is, a � b mod 0 if and only if a D b, and so congruence mod 0 is just ordinary equality. You should not be surprised that we usually

assume that m � 2.

Corollary 4.4. If m � 2, then every integer a is congruent mod m to exactly See Exercise 4.5 on page 140 for a generaliza-

tion. one integer on the list

0; 1; : : : ; m� 1:

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134 Chapter 4 Modular Arithmetic

Proof. By the Division Algorithm, we have a D qmC r , where 0 � r < m; that is, a � r mod m.

If a were congruent to two integers on the list, say, r < r 0, then r � r 0 mod m by transitivity, so that m j .r 0 � r/. Since 0 < r 0 � r < m, this would contradict Lemma 1.13.

Congruence gets along well with addition and multiplication.

Proposition 4.5. Let m � 0. (i) If a � a0 mod m and b � b0 mod m, then

aC b � a0 C b0 mod m:

More generally, if ai � a0i mod m for i D 1; : : : ; k, then

a1 C � � � C ak � a01 C � � � C a0k mod m:

(ii) If a � a0 mod m and b � b0 mod m, then

ab � a0b0 mod m:

More generally, if ai � a0i mod m for i D 1; : : : ; k, then

a1 � � �ak � a01 � � �a0k mod m:

(iii) If a � b mod m, then

ak � bk mod m for all k � 1:

Proof. (i) If m j .a � a0/ and m j .b � b0/, then m j .a C b/ � .a0 C b0/, because .aC b/� .a0 C b0/ D .a� a0/C .b � b0/. The generalization to k summands follows by induction on k � 2.

(ii) We must show that ifm j .a� a0/ and m j .b � b0/, thenm j .ab � a0b0/. This follows from the identity

ab � a0b0 D ab � ab0 C ab0 � a0b0 D a.b � b0/C .a � a0/b0:

The generalization to k factors follows by induction on k � 2. (iii) This is the special case of part (ii) in which all ai D a and all a0i D b.

How to Think About It. The key idea in calculating with congruences

mod m is that every number can be replaced by its remainder after dividing

by m, for this is precisely what Proposition 4.5 permits; it allows you to “re-

duce as you go” in calculations, as the next example shows.

Example 4.6. The last (units) digit of a positive integer is the remainder when

it is divided by 10. What is the last digit of

103243 C 2348 � 5267‹

We could do this by brute force: cube 10324, multiply 2348 and 5267, add, and

look at the last digit. But, as one of our friends says, why should the calculator

have all the fun? You can do this more cleverly using congruence.

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4.1 Congruence 135

� To compute 103243, first look at 10324.

10324 � 4 mod 10; so that 103243 � 43 mod 10:

Now 43 D 64 � 4 mod 10, so that 103243 � 4 mod 10, and the last digit of 103243 is 4.

� To multiply 2348 and 5267, note that 2348 � 8 mod 10 and 5267 � More simply, think of multiplying 2348 and 5267

by hand. What’s the last

digit? This is what most

middle school students

would do. We just want

to illustrate the general

principle here.

7 mod 10. Hence,

2348 � 5267 � 8 � 7 D 56 � 6 mod 10: � Thus,

103243 C 2348 � 5267 � 4C 6 D 10 mod 10;

and 0 is the last digit; 103243 C 2348 � 5267 is divisible by 10.

Now you try one: what is the last digit of 752843 C 10988 � 310767? N

The next example uses congruence to solve more difficult problems.

Example 4.7. (i) If a 2 Z, then a2 � 0, 1, or 4 mod 8. If a is an integer, then a � r mod 8, where 0 � r � 7; moreover,

by Proposition 4.5 (iii), a2 � r2 mod 8, and so it suffices to look at the squares of the remainders. We see in Figure 4.1 that only 0, 1, or 4 can be

a remainder after dividing a perfect square by 8.

r 0 1 2 3 4 5 6 7

r2 0 1 4 9 16 25 36 49

r2 mod 8 0 1 4 1 0 1 4 1

Figure 4.1. Squares mod 8.

(ii) n D 1003456789 is not a perfect square. Since 1000 D 8 � 125, we have 1000 � 0 mod 8, and so

1003456789D 1003456 � 1000C 789 � 789 mod 8:

Dividing 789 by 8 leaves remainder 5; that is, n � 5 mod 8. But if nwere a perfect square, then n � 0; 1, or 4 mod 8.

(iii) There are no perfect squares of the form 3m C 3nC 1, wherem and n are positive integers.

Again, let’s look at remainders mod 8. Now 32 D 9 � 1 mod 8, and so we can evaluate 3m mod 8 as follows: if m D 2k, then 3m D 32k D 9k � 1 mod 8; if m D 2k C 1, then 3m D 32kC1 D 32k � 3 � 3 mod 8. Thus,

3m � ( 1 mod 8 if m is even

3 mod 8 if m is odd:

Replacing numbers by their remainders after dividing by 8, we have the

following possibilities for the remainder of 3m C 3n C 1, depending on

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136 Chapter 4 Modular Arithmetic

the parities of m and n:

3C 1C 1 � 5 mod 8 3C 3C 1 � 7 mod 8 1C 1C 1 � 3 mod 8 1C 3C 1 � 5 mod 8:

In no case is the remainder 0, 1, or 4, and so no number of the form

3m C 3n C 1 can be a perfect square, by part (i). N

Many beginning algebra students wish that .aCb/p D apCbp in Z; if only .a C b/2 D a2 C b2! The next proposition (which paraphrases Exercise 7.27 on page 293) would delight them. If theorems were movies, Proposition 4.8 would be X-rated: only adults would be allowed to see it.

Proposition 4.8. If p is a prime and a; b are integers, then

.a C b/p � ap C bp mod p:

Proof. The Binomial Theorem gives

.a C b/p D pX

rD0

p

r

! ap�rbr :

But � p r

� � 0 mod p for all r with 0 < r < p, by Proposition 2.26. The result

now follows from Proposition 4.5(i).

The next theorem (sometimes called the Little Fermat Theorem to distin-

guish it from Fermat’s Last Theorem) turns out to be very useful.

Theorem 4.9 (Fermat). Let p be a prime and a 2 Z.See Corollary 4.67 for another proof.

(i) ap � a mod p. (ii) ap

n � a mod p for all n � 1. (iii) If p − a, then ap�1 � 1 mod p.

Proof. (i) We first prove the statement when a � 0, by induction on a. The base step a D 0 is obviously true. For the inductive step, the inductive hypothesis is ap � a mod p. Hence, Proposition 4.8 gives

.a C 1/p � ap C 1 � a C 1 mod p:

To complete the proof, consider �a, where a > 0; now

.�a/p D .�1/pap � .�1/pa mod p:

If p is an odd prime (indeed, if p is odd), then .�1/p D �1, and .�1/pa D �a, as desired. If p D 2, then .�a/2 D a2 � a mod 2, and we are fin- ished in this case as well.

(ii) The proof is by induction on n � 1: the base step is part (i), while the inductive step follows from the identity ap

n D � ap

n�1�p .

(iii) By part (i), p j .ap � a/; that is, p j a.ap�1 � 1/. Since p − a, Euclid’s Lemma gives p j .ap�1 � 1/; that is, ap�1 � 1 mod p.

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4.1 Congruence 137

Later in this chapter, we will use the next corollary to construct codes that

are extremely difficult for spies to decode.

Corollary 4.10. If p is a prime andm � 1 mod .p�1/, then am � a mod p for all a 2 Z.

Proof. If a � 0 mod p, then am � 0 mod p, and so am � a mod p. Assume now that a 6� 0 mod p; that is, p − a. By hypothesis,m � 1 D k.p � 1/ for some integer k, and so m D 1C .p � 1/k. Therefore,

am D a1C.p�1/k D aa.p�1/k D a � ap�1

�k � a mod p;

for ap�1 � 1 mod p, by Theorem 4.9(iii)

We can now explain a well-known divisibility test.

Proposition 4.11. A positive integer a is divisible by 3 if and only if the sum

of its (decimal) digits is divisible by 3.

Proof. The decimal notation for a is dk : : : d1d0; that is,

a D dk10k C � � � C d110C d0;

where 0 � di < 10 for all i . Now 10 � 1 mod 3, and Proposition 4.5(iii) gives 10i � 1i D 1 mod 3 for all i ; thus parts (i) and (ii) of Proposition 4.5 give a � dk C � � � C d1 C d0 mod 3. Therefore, a is divisible by 3 if and only if a � 0 mod 3 if and only if dk C � � � C d1 C d0 � 0 mod 3.

How to Think About It. The proof of Proposition 4.11 shows more than its

statement claims: the sum of the (decimal) digits of any positive integer a is

congruent to a mod 3, whether or not a is divisible by 3. For example,

172 � 1C 7C 2 mod 3I

that is, both 172 and 10 (the sum of its digits) are � 1 mod 3.

Since 10 � 1 mod 9, Proposition 4.11 holds if we replace 3 by 9 (it is often called casting out 9s): A positive integer a is divisible by 9 if and only if the Is the sum of the decimal

digits of an integer a

congruent mod 9 to a

itself?

sum of its digits,†.a/, is divisible by 9.

Define two operations on the decimal digits of a positive integer a.

(i) Delete all 9s (if any) and delete any group of digits whose sum is 9

(ii) Add up all the digits.

It is easy to see that repeated applications of these operations to a positive integer a yields a single digit; call it r.a/. For example,

5261934 ! 526134 ! 561 (for 2C 3C 4 D 9) ! 12 ! 3:

(It is now clear why this procedure is called casting out 9s.) In light of a � †.a/ mod 9, we have †.a/ � r.a/ mod 9, so that r.a/, which seems to de- pend on a choice of operations (i) and (ii), depends only on a, for the variation

of Proposition 4.11 for 9 says that†.a/ is the remainder after dividing a by 9.

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138 Chapter 4 Modular Arithmetic

Before today’s calculators, casting out 9s was used by bookkeepers to detect

errors in calculations (alas, it could not detect all errors). For example, suppose

the end of a calculation gave the equation

.22345C 5261934/1776D 9347119504:

Casting out 9s from each number gives

.7 C 3/3 D 8;

for r.22345/ D 7, r.5261934/ D 3, r.1776/ D 3, and r.9347119504/ D 8. But .7C 3/3 D 30 � 3 mod 9, not 8 mod 9, and so there was a mistake in the calculation.

The word “bookkeeper” is unusual in that it has three consecutive double

letters: oo, kk, ee. This reminds us of a silly story about a word having six

consecutive double letters. A zoo discovered that one of its animals, Ricky the raccoon, was quite remarkable. Ricky was a born showman: he could do

somersaults, hang by his tail, and give wonderful soft-shoe dances whenever

spectators sang. As his fame spread, the zoo provided him with a special cage

containing a private corner where he could unwind after popular performances.

Crowds came from far and wide came to see him. Indeed, Ricky became so

famous that the zoo was forced to hire attendants to take care of his needs. In particular, someone was sought to maintain Ricky’s corner; the job description:

raccoonnookkeeper.

The usual decimal notation for the integer 5754 is an abbreviation of

5 � 103 C 7 � 102 C 5 � 10C 4:

But there is nothing special about the number 10.

Example 4.12. Let’s write 12345 in “base 7.” Repeated use of the Division Algorithm gives

12345 D 1763 � 7C 4 1763 D 251 � 7C 6 251 D 35 � 7C 6 35 D 5 � 7C 0 5 D 0 � 7C 5:

Back substituting (i.e., working from the bottom up),

0 � 7C 5 D 5 5 � 7C 0 D 35

.0 � 7C 5/ � 7C 0 D 35 35 � 7C 6 D 251

..0 � 7C 5/ � 7C 0/ � 7C 6 D 251 251 � 7C 6 D 1763

...0 � 7C 5/ � 7C 0/ � 7C 6/ � 7C 6 D 1763 1763 � 7C 4 D 12345

....0 � 7C 5/ � 7C 0/ � 7C 6/ � 7C 6/ � 7C 4 D 12345:

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4.1 Congruence 139

Expanding and collecting terms gives

5 � 74 C 0 � 73 C 6 � 72 C 6 � 7C 4 D 12005C 0C 294C 42C 4 D 12345: N

This idea works for any integer b � 2.

Proposition 4.13. If b � 2 is an integer, then every positive integer h has an expression in base bW there are unique integers di with 0 � di < b such that

h D dkbk C dk�1bk�1 C � � � C d0:

Proof. We first prove the existence of such an expression, by induction on h.

By the Division Algorithm, h D qb C r , where 0 � r < b. Since b � 2, we have h D qb C r � qb � 2q. It follows that q < h: otherwise, q � h, giving the contradiction h � 2q � 2h. By the inductive hypothesis,

h D qb C r D .d 0kb k C � � � C d 00/b C r D d 0kb

kC1 C � � � C d 00b C r:

We prove uniqueness by induction on h. Suppose that

h D dkbk C � � � C d1b C d0 D embm C � � � C e1b C e0;

where 0 � ej < b for all j . that is, h D .dkbk�1 C � � � C d1/b C d0 and h D .embm�1 C � � �C e1/bC e0. By the uniqueness of quotient and remainder in the Division Algorithm, we have

dkb k�1 C � � � C d1 D embm�1 C � � � C e1 and d0 D e0:

The inductive hypothesis gives k D m and di D ei for all i > 0.

Definition. If h D dkbk C dk�1bk�1 C � � � C d0, where 0 � di < b for all i , then the numbers dk; : : : ; d0 are called the b-adic digits of h. Example 4.12 shows that

the 7-adic digits of 12345

are 50664.That every positive integer h has a unique expression in base 2 says that there is exactly one way to write h as a sum of distinct powers of 2 (for the

only binary digits are 0 and 1).

Example 4.14. Let’s calculate the 13-adic digits of 441. The only complica-

tion here is that we need 13 digits d (for 0 � d < 13), and so we augment 0 through 9 with three new symbols

t D 10; e D 11; and w D 12: Now

441 D 33 � 13C 12 33 D 2 � 13C 7 2 D 0 � 13C 2:

So, 441 D 2 � 132 C 7 � 13C 12, and the 13-adic expansion for 441 is

27w:

Note that the expansion for 33 is just 27. N

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140 Chapter 4 Modular Arithmetic

The most popular bases are b D 10 (giving everyday decimal digits), b D 2 (giving binary digits, useful because a computer can interpret 1 as “on” and 0

as “off”), and b D 16 (hexadecimal, also for computers). The Babylonians preferred base 60 (giving sexagesimal digits).

Fermat’s Theorem enables us to compute np k

mod p for every prime p

and exponent pk; it says that np k � n mod p. We now generalize this result

to compute nh mod p for any exponent h.

Lemma 4.15. Let p be a prime and let n be a positive integer. If h � 0, then

nh � n†.h/ mod p;

where †.h/ is the sum of the p-adic digits of h.This lemma generalizes Fermat’s Theorem, for if

h D pk , then †.h/ D 1; see Exercise 4.10 on

page 141.

Proof. Let h D dkpk C � � � C d1p C d0 be the expression of h in base p. By Fermat’s Theorem, np

i � n mod p for all i ; thus, ndipi D .ndi /pi � ndi mod p. Therefore,

nh D ndkpkC���Cd1pCd0

D ndkpkndk�1pk�1 � � � nd1pnd0

D � np

k �dk �npk�1 �dk�1 � � �

� np �d1nd0

� ndkndk�1 � � � nd1nd0 mod p � ndkC���Cd1Cd0 mod p � n†.h/ mod p:

Example 4.16. What is the remainder after dividing 312345 by 7? By Exam-

ple 4.12, the 7-adic digits of 12345 are 50664. Therefore, 312345 � 321 mod 7 (because 5 C 0 C 6 C 6 C 4 D 21). The 7-adic digits of 21 are 30 (because 21 D 3 � 7+0), and so 321 � 33 mod 7 (because 2 + 1 = 3). We conclude that 312345 � 33 D 27 � 6 mod 7. N

Exercises

4.1 Show that if integers a and b are congruent mod m to the same thing, say r , then

they are congruent to each other.

4.2 We saw in Exercise 1.41 on page 29 that an integer b and its negative �b can have different remainders, say r and s, after dividing by some nonzero a. Prove that

s � �r mod a.

4.3 Show that if a � b mod n and m j n, then a � b mod m.

4.4 A googol is 10100; that is, 1 followed by 100 zeros. Compute the remainder mod 7

of a googol.

4.5 *

(i) If m � 2, show that every integer a is (ii) congruent mod m to exactly one integer on the list

1; 2; : : : ; m:

(iii) Generalize Corollary 4.4 by showing that if m � 2, every integer a is congru- ent mod m to exactly one integer on any list of m consecutive integers.

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4.1 Congruence 141

4.6 (i) Show that every nonnegative integer is congruent mod 6 to the sum of its

7-adic digits.

(ii) Show that every nonnegative integer is congruent mod 3 to the sum of its

7-adic digits.

(iii) Suppose b and n are nonnegative integers. If n j .b � 1/, show that every integer is congruent mod n to the sum of its b-adic digits.

4.7 (i) Show that every nonnegative integer is congruent mod 11 to the alternating

sum of its decimal digits.

(ii) Show that every nonnegative integer is congruent mod b C1 to the alternating sum of its b-adic digits.

4.8 Let a nonnegative integer n have decimal expansion n D Pk iD0 di10

i . Define

t .n/ D n�d010 � 4d0. (i) Show that n is divisible by 41 if and only if t .n/ is.

(ii) Is n � t .n/ mod 41 for all nonnegative n?

4.9 Find the b-adic digits of 1000 for b D 2; 3; 4; 5; and 20. You will have to invent

symbols for some 20-adic

digits. 4.10 (i) Find the 11-adic digits of 115.

(ii) What is the b-adic expansion for bk (k a nonnegative integer)?

4.11 Let a be a positive integer, and let a0 be obtained from a by rearranging its (dec- imal) digits (e.g., a D 12345 and a0 D 52314). Prove that a � a0 is a multiple of 9.

4.12 Prove that there are no positive integers a; b; c with

a2 C b2 C c2 D 999:

4.13 Prove that there is no perfect square whose last two decimal digits are 35.

4.14 Using Fermat’s Theorem 4.9, prove that if apCbp D cp, then aCb � c mod p.

Linear congruences

We are now going to solve linear congruences; that is, we’ll find all the inte-

gers x, if any, satisfying

ax � b mod m:

Later, we will consider several linear congruences in one unknown with dis- tinct moduli (see Theorems 4.21, 4.25, and 4.27). And we’ll even consider two

linear congruences in more than one unknown (see Theorem 4.44).

Theorem 4.17. If gcd.a; m/ D 1, then, for every integer b, the congruence

ax � b mod m

can be solved for xI in fact, x D sb, where as C mt D 1. Moreover, any two solutions are congruent mod m.

Proof. Since gcd.a; m/ D 1, there are integers s and t with as C mt D 1; that is, as � 1 mod m. Multiplying both sides by b, Proposition 4.5(ii) gives asb � b mod m, so that x D sb is a solution. If y is another solution, then ax � ay mod m, and so m j a.x � y/. Since gcd.m; a/ D 1, Corollary 1.22 givesm j .x � y/; that is, x � y mod m.

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142 Chapter 4 Modular Arithmetic

Corollary 4.18. If p is prime and p − a .i.e., p does not divide a/, then the congruence ax � b mod p is always solvable.

Proof. Since p is a prime, p − a implies gcd.a; p/ D 1.

Example 4.19. When gcd.a; m/ D 1, Theorem 4.17 says that the set of solu- tions of ax � b mod m is

fsbC km W where k 2 Z and sa � 1 mod mg:

Now sa C tm D 1 for some integer t , so that s can always be found by Euclidean Algorithm II. When m is small and you are working by hand, it is

easier to find such an integer s by trying each of ra D 2a; 3a; : : : , .m� 1/a in turn, at each step checking whether ra � 1 mod m.

For example, let’s find all the solutions to

2x � 9 mod 13:

Considering each of the products 2 � 2; 3 � 2; 4 � 2; : : : mod 13 quickly leads to 7 � 2 D 14 � 1 mod 13; that is, s D 7. By Theorem 4.17, x D 7 � 9 D 63 � 11 mod 13. Therefore,

x � 11 mod 13;

and the solutions are : : : ;�15; �2; 11; 24; 37; : : : : N

Example 4.20. Find all the solutions to 51x � 10 mod 94. Since 94 is large, seeking an integer s with 51s � 1 mod 94, as in Exam-

ple 4.19, is tedious. Euclidean Algorithm II gives 1 D �35 � 51 C 19 � 94, and so s D �35. (The formulas in Exercise 1.67 on page 36 implement Eu- clidean Algorithm II, and they can be programmed on a calculator to produce the value of s. In fact, a CAS can solve specific congruences, but it can’t (yet)

solve them in general.) Therefore, the set of solutions consists of all integers x

with x � �35 � 10 mod 94; that is, all numbers of the form �350C 94k. If you prefer s to be positive, just replace �35 by 59, for 59 � �35 mod 94.

The solutions are now written as all integers x with x � 59 � 10 mod 94; that is, numbers of the form 590C 94k. N

There are problems solved in ancient Chinese manuscripts, arising from

studying calendars, that involve simultaneous congruences with relatively prime moduli.

Theorem 4.21 (Chinese Remainder Theorem). If m and m0 are relatively prime, then the two congruences

x � b mod m x � b0 mod m0

have a common solution. Moreover, any two solutions are congruent mod Theorem 4.27 will gener-

alize Theorem 4.21 to any

number of moduli. mm0 .

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4.1 Congruence 143

Proof. Every solution of the first congruence has the form x D b C km for some integer k; hence, we must find k such that bC km � b0 mod m0; that is, km � b0 � b mod m0. Since gcd.m;m0/ D 1, however, Theorem 4.17 applies at once to show that such an integer k does exist.

If y is another common solution, then both m and m0 divide x � y; by Exercise 1.58 on page 35, mm0 j .x � y/, and so x � y mod mm0 .

Example 4.22. Let’s find all the solutions to the simultaneous congruences

x � 5 mod 8 x � 11 mod 15:

Every solution to the first congruence has the form

x D 5C 8k;

for some integer k. Substituting, x D 5C 8k � 11 mod 15, so that

8k � 6 mod 15:

But 2 � 8 D 16 � 1 mod 15, so that multiplying by 2 gives

16k � k � 12 mod 15:

We conclude that x D 5C8�12 D 101 is a solution, and the Chinese Remainder Theorem (which applies because 8 and 15 are relatively prime) says that every solution has the form 101C 120n for n 2 Z (because 120 D 8 � 15). N

Example 4.23. We solve the simultaneous congruences

x � �6 mod 13 x � 8 mod 20:

Now gcd.13; 20/ D 1, so that we can solve this system as in the proof of the Chinese Remainder Theorem. The first congruence gives

x D 13k � 6;

for k 2 Z, and substituting into the second congruence gives

13k � 6 � 8 mod 20I

that is,

13k � 14 mod 20:

Since 13 �17 D 221 � 1 mod 20, multiplying by 17 gives k � 17 �14 mod 20, One finds 17 either by try- ing each number between

1 and 19 or by using the

Euclidean Algorithm.

that is,

k � 18 mod 20:

By the Chinese Remainder Theorem, all the simultaneous solutions x have the form

x D 13k � 6 � .13 � 18/� 6 � 228 mod 260I

that is, the solutions are

: : : ;�32; 228; 488; : : : : N

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144 Chapter 4 Modular Arithmetic

Example 4.24 (A Mayan Calendar). A congruence arises whenever there is

cyclic behavior. For example, suppose we choose some particular Sunday as

time zero and enumerate all the days according to the time elapsed since then. Every date now corresponds to some integer, which is negative if it occurred

before time zero. Given two dates t1 and t2, we ask for the number x D t2 � t1 of days from one to the other. If, for example, t1 falls on a Thursday and t2

Remember that 0 denotes

Sunday, : : :, 6 denotes

Saturday. falls on a Tuesday, then t1 � 4 mod 7 and t2 � 2 mod 7, and so x D t2� t1 D �2 � 5 mod 7. Thus, x D 7k C 5 for some k and, incidentally, x falls on a Friday.

About 2500 years ago, the Maya of Central America and Mexico developed

three calendars (each having a different use). Their religious calendar, called

tzolkin, consisted of 20 “months,” each having 13 days (so that the tzolkin

“year” had 260 days). The months were

1. Imix 6. Cimi 11. Chuen 16. Cib

2. Ik 7. Manik 12. Eb 17. Caban

3. Akbal 8. Lamat 13. Ben 18. Etznab

4. Kan 9. Muluc 14. Ix 19. Cauac

5. Chicchan 10. Oc 15. Men 20. Ahau

Let us describe a tzolkin date by an ordered pair

Œm; d ;

where 1 � m � 20 and 1 � d � 13; thus, m denotes the month and d denotes the day. Instead of enumerating as we do (so that Imix 1 is followed

by Imix 2, then Imix 3, and so forth), the Maya let both month and day cycle

simultaneously; that is, the days proceed as follows:

Imix 1, Ik 2, Akbal 3,. . . , Ben 13, Ix 1, Men 2,. . . ,

Cauac 6, Ahau 7, Imix 8, Ik 9,. . . .

We now ask how many days have elapsed between Oc 11 and Etznab 5.

More generally, let x be the number of days from tzolkin Œm; d  to tzolkin

Œm0; d 0. As we remarked at the beginning of this example, the cyclic behavior of the days gives the congruence

x � d 0 � d mod 13;

while the cyclic behavior of the months gives the congruence

x � m0 �m mod 20:

To answer the original question, Oc 11 corresponds to the ordered pair Œ10; 11

and Etznab 5 corresponds to Œ18; 5. Since 5 � 11 D �6 and 18 � 10 D 8, the simultaneous congruences are

x � �6 mod 13 x � 8 mod 20:

In the previous example, we found the solutions:

x � 228 mod 260:

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4.1 Congruence 145

It is not clear whether Oc 11 precedes Etznab 5 in a given year (one must

look). If it does, then there are 228 days between them; otherwise, there are

260� 228 D 32 days between them (the truth is 228). N

If we do not assume that the moduli m and m0 are relatively prime, then there may be no solutions to a linear system. For example, if m D m0 > 1, then uniqueness of the remainder in the Division Algorithm shows that there

is no solution to

x � 0 mod m x � 1 mod m:

Theorem 4.25. Let d D gcd.m;m0/. The system

x � b mod m x � b0 mod m0

has a solution if and only if b � b0 mod d . Exercise 4.19 on page 148 gives a condition guar-

anteeing uniqueness of

solutions. Proof. If h � b mod m and h � b0 mod m0, thenm j .h�b/ andm0 j .h�b0/. Since d is a common divisor ofm andm0, we have d j .h�b/ and d j .h�b0/. Therefore, d j .b � b0/, because .h � b0/ � .h � b/ D b � b0, and so b � b0 mod d .

Conversely, assume that b � b0 mod d , so that there is an integer k with b0 D b C kd . If m D dc and m0 D dc0, then gcd.c; c0/ D 1, by Propo- sition 1.23. Hence, there are integers s and t with 1 D sc C tc0. Define h D b0sc C btc0. Now

h D b0sc C btc0

D .b C kd/sc C btc0

D b.sc C tc0/C kdsc D b C ksm � b mod m:

A similar argument, replacing b by b0 � kd , shows that h � b0 mod m0.

Example 4.26. Solve the linear system

x � 1 mod 6 x � 4 mod 15:

Here, b D 1 and b0 D 4, while m D 6;m0 D 15, and d D 3; hence, c D 2 and c0 D 5 (for 6 D 3 � 2 and 15 D 3 � 5). Now s D 3, and t D �1 (for 1 D 3 � 1C .�1/ � 4). Theorem 4.25 applies, for 1 � 4 mod 3. Define

h D 4 � 3 � 2C 1 � .�1/ � 5 D 19:

We check that 19 � 1 mod 6 and 19 � 4 mod 15. Since lcm.6; 15/ D 30, the solutions are : : : ;�41;�11; 19; 49; 79; : : :. N

We are now going to generalize the Chinese Remainder Theorem for any

number of linear congruences whose moduli are pairwise relatively prime. We

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146 Chapter 4 Modular Arithmetic

shall see in Chapter 6 that this new version, whose solutions are given more

explicitly, can be used to reveal a connection with Lagrange Interpolation, a

method for finding a polynomial that agrees with a finite set of data. Consider the following problem, adapted from Qin Jiushao, Nine Chapters

on the Mathematical Art, 1247 CE.

Three farmers equally divide the rice that they have grown. One goes to

a market where an 83-pound weight is used, another to a market that

uses a 112-pound weight, and the third to a market using a 135-pound

weight. Each farmer sells as many full measures as possible, and when

the three return home, the first has 32 pounds of rice left, the second 70

pounds, and the third 30 pounds. Find the total amount of rice they took

to market.

We can model the situation in the problem with three congruences:

x � 32 mod 83 x � 70 mod 112 (4.1) x � 30 mod 135:

Now, you could solve this system using the same method we used in Exam-

ple 4.22: just write out each congruence in terms of its corresponding divisi-

bility tests, and work from there. There’s another technique for solving Eqs. (4.1) that works in more general

settings. The idea is to “localize” a solution x, where “localize” means consid-

ering only one modulus at a time, ignoring the other two; that is, making the

other two congruent to zero. Suppose we can find integers u; v; w such that

u � 32 mod 83 v � 0 mod 83 w � 0 mod 83 u � 0 mod 112 v � 70 mod 112 w � 0 mod 112 u � 0 mod 135 v � 0 mod 135 w � 30 mod 135:

Now take x to be u C v C w. Thanks to Proposition 4.5, we can find the remainder when u C v C w is divided by 83 by first finding the remainders when each of u, v, and w is divided by 83, and then adding the answers:

x D uC v C w � 32C 0C 0 mod 83:

Similarly, x � 70 mod 112 and x � 30 mod 135. So, how do we find such u, v, and w? Let’s look at what we want u to do:

u � 32 mod 83 u � 0 mod 112 u � 0 mod 135:

It’s easy to make u congruent to 0 mod 112 and 0 mod 135: just let it be a

multiple of 112 � 135 D 15120. So, we want u to look like

u D k � 112 � 135 D 15120k

for some integer k. And we choose k to meet the local condition that u wants

to be 32 modulo 83:

15120k � 14k � 32 mod 83 (4.2)

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4.1 Congruence 147

Now comes the important step: since 112 and 135 are relatively prime to 83,

so is their product (Exercise 1.56 on page 35). Hence, 14 (which is the same as

112�135modulo 83) is also relatively prime to 83, and so Theorem 4.17 implies that we can solve Eq. (4.2) for k. There is an integer s with 14s � 1 mod 83, and multiplying both sides of 14k � 32 mod 83 by s gives

k D 32s mod 83:

There are several methods for finding s (since 83 is not so small, the Euclidean

Algorithm is probably the most efficient); in fact, s D 6, and so k satisfies

6 � 32 D 192 � 26 mod 83:

Hence,

u D 26 � 112 � 135 D 393120:

To get a feel for this method, it’s a good idea to go through it twice more,

finding v and w. In fact, that’s Exercise 4.22 on page 149.

The method just developed generalizes to a proof of the extended Chinese

Remainder Theorem. Let’s first introduce some notation.

Notation. Given numbersm1; m2; : : : ; mr , define

Mi D m1m2 � � �bmi � � �mr D m1 � � �mi�1miC1 � � �mr I

that is,Mi is the product of all mj other than mi .

Theorem 4.27 (Chinese Remainder Theorem Redux). If m1; m2; : : : ; mr are pairwise relatively prime integers, then the simultaneous congruences

x � b1 mod m1 x � b2 mod m2 :::

:::

x � br mod mr

have an explicit solution, namely

x D b1 .s1M1/C b2 .s2M2/C � � � C br .srMr / ; where

Mi D m1m2 � � �bmi � � �mr and siMi � 1 mod mi for 1 � i � r:

Furthermore, any solution to this system is congruent to x mod m1m2 � � �mr .

Proof. Use our discussion on the previous page as a model for the proof.

That the specified x works is a consequence of Proposition 4.5. That all solu-

tions are congruent modm1m2 : : : mr is a consequence of Exercise 1.58 on

page 35.

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148 Chapter 4 Modular Arithmetic

Exercises

4.15 * Complete the proof of Theorem 4.27.

4.16 (i) Solve

x � 5 mod 7 x � 2 mod 11:

(ii) In the year 2000, the remainder after dividing my age by 3 was 2, and the

remainder after dividing by 8 was 3. If I was a child when people first walked

on the Moon, how old was I in 2000?

(iii) Solve

x7 � 5 mod 7 x11 � 2 mod 11:

4.17 (i) Find a solution v to

v � 3 mod 17 v � 0 mod 11:

Answer: v � 88 mod 187. (ii) Find a solution w to

w � 0 mod 17 w � 9 mod 11:

Answer: w � 119 mod 187. (iii) Using your v and w from (i) and (ii), show that v C w is a solution to the

system

x � 3 mod 17 x � 9 mod 11:

4.18 Solve

x � 32 .mod 83/ x � 70 .mod 112/ x � 30 .mod 135/:

4.19 * Theorem 4.25 says that if d D gcd.m; m0/, then the system

x � b mod m x � b0 mod m0

has a solution if and only if b � b0 mod d . Prove that any two solutions are congruent mod `, where ` D lcm.m; m0/.

4.20 How many days are there between Akbal 13 and Muluc 8 in the Mayan tzolkin

calendar?

4.21 On a desert island, five men and a monkey gather coconuts all day, then sleep. The

first man awakens and decides to take his share. He divides the coconuts into five

equal shares, with one coconut left over. He gives the extra one to the monkey,

hides his share, and goes to sleep. Later, the second man awakens and takes his

fifth from the remaining pile; he, too, finds one extra and gives it to the monkey.

Each of the remaining three men does likewise in turn. Find the minimum number

of coconuts originally present.

Hint: Try �4 coconuts.

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4.2 Public Key Codes 149

4.22 * Finish the calculations solving Qin Jiushao’s problem on page 146 by first find-

ing s and t , and then finding the smallest positive solution.

4.23 A band of 17 pirates stole a sack of gold coins. When the coins were divided

equally, there were three left over. So, one pirate was made to walk the plank.

Again the sack was divided equally; this time there were 10 gold coins left over.

So, another unlucky member of the crew took a walk. Now, the gold coins could

be distributed evenly with none left over. How many gold coins were in the sack?

4.24 .Bhaskara I; ca: 650 C.E./: If eggs in a basket are taken out 2, 3, 4, 5, and 6 at a

time, there are 1, 2, 3, 4, and 5 eggs left over, respectively. If they are taken out 7

at a time, there are no eggs left over. What is the least number of eggs that can be

in the basket?

4.2 Public Key Codes

A thief who knows your name and credit card number can use this information

to steal your money. So why isn’t it risky to buy something online, and pay

for it by sending your credit card data? After all, thieves can read the message

you are sending. Here’s why: the online company’s software encodes your

information before it is transmitted; the company can decode it, but the thieves cannot. And the reason the thieves cannot decode your message is that codes

are constructed in a clever way using number theory.

It is no problem to convert a message in English into a number. Make a list

of the 52 English letters (lower case and upper case) together with a space and

the 11 punctuation marks

. , ; : ! ? - ’ " ( )

In all, there are 64 symbols. Assign a two-digit number to each symbol. For example,

a 7! 01; : : : ; z 7! 26;A 7! 27; : : : ;Z 7! 52 space 7! 53; : 7! 54; ; 7! 55; : : : ; . 7! 63; / 7! 64

(we could add more symbols if we wished: say, $, +, �, =, !, 0, 1, . . . , 9). A cipher is a code in which distinct letters in the original message are replaced by

distinct symbols. It is not difficult to decode a cipher; indeed, many newspapers print daily cryptograms to entertain their readers. In the cipher we have just

described, “I love you!” is encoded

I love you! D 3553121522055325152158

Notice that any message coded in this cipher has an even number of digits, and

so decoding, converting the number into English, is a simple matter. Thus,

.35/.53/.12/.15/.22/.05/.53/.25/.15/.21/.58/ D I love you!

What makes a good code? If a message is a natural number x (and this is no loss in generality, as we have just seen), we need a way to encode x (in a fairly

routine way so as to avoid introducing any errors into the coded message),

and we need a (fairly routine) method for the recipient to decode the message.

Of utmost importance is security: an unauthorized reader of a coded message

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150 Chapter 4 Modular Arithmetic

should not be able to decode it. An ingenious way to find a code with these

properties, now called an RSA code, was found in 1978 by Rivest, Shamir,

and Adleman; they received the 2002 Turing Award for their discovery. The following terms describe two basic ingredients of RSA codes.

Definition. A public key is an ordered pair .N; e/, whereN D pq is a product of distinct primes p and q, and e is a positive integer with gcd.e; p � 1/ D 1Why these conditions

on e? Read on. and gcd.e; q � 1/ D 1.

The numbers N and e are public—they are published on the web—but the

primes p and q are kept secret. In practice, the primes p and q are very large.

If x is a message, encoded by assigning natural numbers to its letters as

discussed above, then the encoded message sent is

xe mod N:

Definition. Given a public key .N; e/, a private key is a number d such that

xed � x mod N for all x 2 Z:

A private key essentially decodes the sent message, for

xed D .xe/d � x mod N:

Only the intended recipients know the private key d . To find d , we’ll see

that you need to factor N , and that’s very hard. Indeed, the modulus N being

a product of two very big primes—each having hundreds of digits—is what

makes factoring N so difficult. Since breaking the code requires knowing p

and q, this is the reason RSA codes are secure. Now for the details.

Ease of Encoding and Decoding

Given a public key .N; e/ and a private key d , we encode x as xe, and we send the congruence class xe mod N . A recipient who knows the number d

can decode this, because

.xe/d D xed � x mod N:

There is a minor problem here, for decoding isn’t yet complete: we know the

congruence class of the original message x but not x itself; that is, we know

x C kN for some k 2 Z but not x. There is a routine way used to get around this; one encodes long blocks of text, not just letters (see [18], pp. 88–91).

Given any positive integer m, an efficient computation of xm mod N is based on the fact that computing x2 mod N is an easy task for a computer.

Since computing x2 i

is just computing i squares, this, too, is an easy task. Now write the exponent m in base 2:

m D 2i C 2j C � � � C 2z :

Computing 2m is the same as multiplying several squares:Note that x4 D .x2/2, x8 D .x4/2 D ..x2/2/2, etc. xm D x2i C2j C���C2z D x2ix2j � � �x2z :

In particular, after writing e in base 2, computers can easily encode a message

x mod N as xe mod N and, after writing ed in base 2, they can easily decode

xed mod N . Since xed � x mod N , this congruence essentially recaptures x.

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4.2 Public Key Codes 151

Finding a Private Key

Let .N; e/ be a public key, where N D pq. We want to find a private key; that is, a number d so that xed � x mod N for all x 2 Z. More generally, let’s find conditions on any integer m so that xm � x mod pq. By Corollary 4.10, we have xm � x mod p if m � 1 mod .p � 1/; similarly, xm � x mod q if m � 1 mod .q � 1/. Now ifm satisfies both congruence conditions, then

p j .xm � x/ and q j .xm � x/:

As p and q are distinct primes, they are relatively prime, and so pq j .xm � x/, by Exercise 2.20 on page 33. Hence, xm � x mod pq for all x; that is, xm � x mod N for all x 2 Z.

Return now to the special case m D ed ; can we find a private key d so that ed � 1 mod .p�1/ and ed � 1 mod .q�1/? By hypothesis, gcd.e; p�1/ D 1 D gcd.e; q � 1/; by Exercise 1.56 on page 35, gcd

� e; .p � 1/.q � 1/

� D 1.

We can now find d with Proposition 4.17, which shows how to construct an

integer d such that

ed � 1 mod .p � 1/.q � 1/:

We have constructed an RSA code.

Example 4.28. Let’s create a public key and a private key using p D 11 and q D 13. (This is just for the sake of illustration; in practice, both p and q need to be extremely large primes.)

The modulus is N D pq D 11 � 13 D 143, and so p � 1 D 10 and q � 1 D 12. Let’s choose e D 7 (note that gcd.7; 10 � 12/ D 1). Hence the public key is

.N; e/ D .143; 7/: If x is a message in cipher (i.e., a natural number), then the encoded message is

the congruence class x7 mod 143. To find the private key, we need a number d

so that 7d � 1 mod 120. Using Euclidean Algorithm II or a CAS, we find a private key

d D 103; for 7 � 103 D 721 D 6 � 120C 1.

Let’s encode and decode the word “dog”: d D 4; o D 15; g D 7. Thus, the cipher for dog is 041507. In the real world, the encoding is .41507/7, and the message sent out is the congruence class .41507/7 mod 143. Decoding

involves computing .41507/721 mod 143. As we said earlier, decoding is not

finished by finding this congruence class; the numbers in this class are of the

form .41507/721 C 143k, and only one of these must be determined. As we said above, the method used in actual RSA transmissions encodes blocks of letters to get around this ambiguity. For this example, however, we’ll use a

simpler method—we’ll send each letter separately, so that “dog” is sent as as

three codes

047; 157; 077:

This eliminates the ambiguity of recovering a congruence class rather than

an integer, because the each letter will correspond to a (unique) integer less

than 143.

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152 Chapter 4 Modular Arithmetic

The encoding is calculated like this:

d: 47 � 82 mod 143 o: 157 � 115 mod 143 g: 77 � 6 mod 143:

To decode these messages, apply the private key:

82103 � 4 mod 143 115103 � 15 mod 143 6103 � 7 mod 143:

We get 4 $ d, 15 $ o, and 7 $ g: “dog,” which was the original message. N

How to Think About It. A CAS can easily tell you that 82103 � 4 mod 143, but it’s interesting to see how the theorems developed in this chapter can

allow you to do the computation by hand. Start with the fact that the reduction

of 82103 mod 143 is equivalent to two calculations, since 143 is 11 � 13:

82103 mod 11

82103 mod 13:

The computations of the remainders when 82103 is divided by a prime are

made easy via Fermat’s Little Theorem and the “reduce as you go” idea:

82103 � 5103 mod 11 (because 82 � 5 mod 11/ 82103 � 4103 mod 13 (because 82 � 4 mod 13).

Now work on the exponents:

5103 D 510�10C3 D � 510 �10

53 � 53 mod 11 (Little Fermat) D 125 mod 11 � 4 mod 11

and

4103 D 412�8C7 D � 412 �10

47 � 47 mod 13 (Little Fermat) D 43 � 4.43/ D 64 � 4.64/ � .�1/ � 4.�1/ mod 13 � 4 mod 13:

Constructing Secure RSA Codes

Let’s construct a specific type of public key. Choose distinct primes p � 2 mod 3 and q � 2 mod 3. Now p � 1 � 1 mod 3, so that gcd.3; p � 1/ D 1, and q � 1 � 1 mod 3, so that gcd.3; q � 1/ D 1. Therefore, .N; 3/ is a public key, where N D pq. The reason that these RSA codes are so secure is that the factorization of a product N D pq of two very large primes is very difficult. Thieves may know the transmitted message x3 mod N , and they may even know N , but without knowing the factorization of N D pq, they don’t know p � 1 and q � 1, hence, they don’t know d (for 3d � 1 mod .p � 1/.q � 1/), and they can’t decode. Indeed, if both p and q have about 200 digits (and,

for technical reasons, they are not too close together), then the fastest existing

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4.2 Public Key Codes 153

computers need two or three months to factor N . A theorem of Dirichlet ([5],

p. 339) says that if gcd.a; b/ D 1, then the arithmetic progression a C bn, where n � 0, contains infinitely many primes. In particular, there are infinitely many primes of the form 2 C 3n; that is, there are infinitely many primes p with p � 2 mod 3. Hence, we may choose a different pair of primes p and q every month, say, thereby stymying the crooks.

RSA codes have been refined and made even more secure over the years.

Some of these refinements make use of elliptic curves, which we’ll touch on in Chapter 9. The book In Code [13] is a readable account of how a high school

student contributed to other refinements.

Exercises

4.25 For this exercise, use the primes p D 5 and q D 17 to create public and private keys.

(i) What will be the modulus N for the public key?

(ii) The exponent e for the public key must have no common factors with p � 1 and q � 1. List the five smallest numbers relatively prime to .p � 1/.q � 1/.

(iii) There are many possibilities for e; for now, use e D 3. To encode letters (a The public key reveals e D 3 and N D pq, but p and q are not revealed.

(Why not?)

computer would do blocks of letters), use the rule x 7! x3 mod 85. (iv) Encode the phrase “cell phones” using this method.

(v) The private key d satisfies

ed � 1 mod .p � 1/.q � 1/:

Find d , decode your message using the private key, and verify that it is, in-

deed, what was sent.

4.26 The following message was encoded using the public key .85; 3/:

01 42 59 10 49 27 56I

decode this message. It answers the question, “What do you call a boomerang that

doesn’t come back when you throw it?”

4.27 Decode the following message encoded using the public key .91; 5/:

04 31 38 38 23 71 14 31:

4.28 Let m and r be nonnegative integers, and p be a prime. If m � r mod .p � 1/, show that xm � xr for all integers x.

4.29 Take It Further. (Electronic Signatures) Consider this scenario: Elvis receives

an email, encoded with his public key, from his abstract algebra instructor Mr. Jag-

ger, which says that algebra is a waste of time and Elvis should spend all his time

watching TV. Elvis suspects that the message didn’t really come from Mr. J., but

how can he be sure?

Suppose both Elvis and Mr. Jagger have private keys, and each knows the

other’s public keys. They can communicate in total privacy, with no one able to

read their messages. Here’s how: if Elvis wants to send a message to Mr. J., he

follows these steps:

� Write the message to get x1 .

� Encode the message with his private key to get x2 . � Encode x2 with Mr. J.’s public key to get x3.

� Send x3.

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154 Chapter 4 Modular Arithmetic

When Mr. J. receives the message, he can follow a procedure to get the original

message back.

(i) What is the procedure?

(ii) Explain why no one besides Mr. Jagger could read the message from Elvis.

4.30 Take It Further. Elvis is home sick with the flu. He decides to send a message

to Mr. Jagger, using the method from Exercise 4.29. Suppose Elvis’s public keyThese public keys are not realistic. In reality, public

keys use much larger

primes.

is .253; 7/ and Mr. J.’s public key is .203;5/. Elvis sends the message

FIDO ATE MY HOMEWORK.

What is the encoded message that Mr. J. receives? Show how Elvis encodes it and

how Mr. J. decodes it.

4.3 Commutative Rings

We begin this section by showing, for an integer m � 2, that we can add and multiply the remainders 0; 1; : : : ; m� 1 in such a way that the new operations behave very much like ordinary addition and multiplication in Z. Once this is

done, we will be able to revisit congruences and understand what “makes them tick.”

It is shown in Appendix A.2 that if � is an equivalence relation on a set X , then the equivalence class of an element a 2 X is

Œa D fx 2 X W x � ag:

Now Proposition 4.3 says that congruence mod m is an equivalence relation

on Z; the equivalence class of an integer a is called its congruence class

mod m.

Definition. The congruence class mod m of an integer a isThe congruence class Œa does depend on m,

but it is standard practice

not to make m part of

the notation. In fact, we’ll

eventually write a instead

of Œa.

Œa Dfk 2 Z W k � a mod mg Df: : : ; a� 2m; a �m; a; a Cm; aC 2m; : : :g:

The integers mod m is the set of all congruences classes:

Zm D fŒ0; Œ1; : : : ; Œm� 1g:

Corollary 4.4 says that the list Œ0; Œ1; : : : ; Œm�1 is complete; that is, there are no other congruence classes mod m.

For example, Z2, the integers mod 2, is the set fŒ0; Œ1g; we may think of Œ0 as even (for Œ0 D fa 2 Z W a � 0 mod 2g is the set of all even integers) and Œ1 as odd (for Œ1 is the set of all odds).

Here is the “theological reason” for introducing congruence classes. We

could continue to deal with integers and congruence; this is, after all, what Gauss did. We saw in Proposition 4.5 that + and � are compatible with con- gruence: if a � b mod m and a0 � b0 mod m, then a C b � a0 C b0 mod m and ab � a0b0 mod m. But wouldn’t life be simpler if we could replace � by =; that is, if we could replace congruence by equality? We state the following

special case of Lemma A.16 in Appendix A.2 explicitly:

a � b mod m if and only if Œa D Œb in Zm:

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4.3 Commutative Rings 155

We often say “odd + odd = even,” which does replace � by = at the cost of replacing integers by their congruence classes. Thus, we should define addition

of these congruence classes so that Œ1C Œ1 D Œ0. Addition and multiplication of evens and odds leads to the following tables.

C even odd even even odd

odd odd even

� even odd even even even

odd even odd

Rewrite these tables using congruence classes mod 2.

C Œ0 Œ1 Œ0 Œ0 Œ1

Œ1 Œ1 Œ0

� Œ0 Œ1 Œ0 Œ0 Œ0

Œ1 Œ0 Œ1

We saw above that Œ1 C Œ1 D Œ0 says that “odd C odd D even;” note that Œ1 � Œ1 D Œ1 says “odd � odd D odd.” The table above on the left de- A binary operation on a set

R is a function R �R! R (in particular, R is closed

under f : if a and b are in

R, then f.a; b/ is in R).

Can you prove associativity

of the binary operations ˛

and � when R D Z2?

fines addition ˛W Z2 � Z2 ! Z2; the table on the right defines multiplication �W Z2 � Z2 ! Z2. As usual, we view congruence as generalizing parity, and we now extend the definitions to give addition and multiplication of congru-

ence classes mod m for allm � 2.

Definition. Ifm � 2, addition and multiplication Zm�Zm ! Zm are defined by

Œr C Œs D Œr C s and Œr Œs D Œrs:

The definitions are simple and natural.However, we are adding and multi-

plying congruence classes, not remainders. After all, remainders are integers

between 0 andm�1, but the sum and product of remainders can exceed m�1, and hence are not remainders.

Lemma 4.29. Addition and multiplication Zm � Zm ! Zm are well-defined functions.

Proof. To see that addition is well-defined, we must show that if Œr  D Œr 0 and Œs D Œs0, then Œr C s D Œr 0 C s0. But this is precisely what was proved in Proposition 4.5. A similar argument shows that multiplication is

well-defined.

Binary operations f WR � R ! R, being functions, are single-valued. This is usually called the Law of Substitution in this context: If .r; s/ D .r 0; s0/, then f .r; s/ D f .r 0; s0/. In particular, if f W Zm � Zm ! Zm is addition or multiplication, then Œr  D Œr 0 and Œs D Œs0 imply Œr C Œs D Œr 0 C Œs0 and Œr Œs D Œr 0Œs0.

We are now going to show that these binary operations on Zm enjoy eight

of the nine fundamental properties of ordinary arithmetic on page 37. We have

already seen several number systems in which addition and multiplication sat-

isfy these familiar properties, so let’s make these properties into a definition.

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156 Chapter 4 Modular Arithmetic

Definition. A commutative ring is a nonempty set R having two binary oper-

ations: addition R � R ! R, denoted by .r; s/ 7! r C s, and multiplication R � R ! R, denoted by .r; s/ 7! rs, which satisfy the following axioms for all a; b; c 2 R:

(i) aC b D b C a; (ii) there is 0 2 R with aC 0 D a for all a 2 R;

(iii) for each a 2 R, there is �a 2 R, called its negative, such thatNegatives are often called additive inverses. �aC a D 0;

(iv) .Associativity of Addition/ aC .b C c/ D .aC b/C cI (v) .Commutativity of Multiplication/ ab D baI

(vi) there is 1 2 R, called its identity, with 1 � a D a for all a 2 RI (vii) .Associativity of Multiplication/ a.bc/ D .ab/cI

(viii) .Distributivity/ a.b C c/ D ab C ac:

How to Think About It. There are more general (non-commutative) rings in which (v), commutativity of multiplication, is not assumed, while (vi) is

modified to say that 1 � a D a D a � 1 and (viii) is modified to say a.b C c/ D ab C ac and .b C c/a D ba C bc. A good example is the ring of all 2 � 2 matrices with entries in R, with identity element

� 1 0 0 1

� , and binary operations

ordinary matrix addition and multiplication:

" a b

c d

# C " a0 b0

c0 d 0

# D " a C a0 b C b0

c C c0 d C d 0

#

and " a b

c d

#" a0 b0

c0 d 0

# D " aa0 C bc0 ab0 C bd 0

ca0 C dc0 cb0 C dd 0

# :

Since all rings in this book are commutative, we will often abuse language and

abbreviate “commutative ring” to “ring.”

The ninth fundamental property of real numbers is: If a ¤ 0, there is a real number a�1, called its (multiplicative) inverse, such that a � a�1 D 1. We will soon consider commutative rings, called fields, which enjoy this property as

well.

How to Think About It. The notion of commutative ring wasn’t conceived

in a vacuum. Mathematicians noticed that several useful systems shared the

basic algebraic properties listed in the definition. Definitions usually emerge

in this way, distilling common features of different interesting examples.

Precise definitions are valuable; we couldn’t prove anything without them.

For example, political discourse is often vapid because terms are not defined:

what is a liberal; what is a conservative? A mathematician who asserts that

there are infinitely many primes can be believed. But can you believe a politi-

cian who says his opponent is a fool because he’s a liberal (or she’s a conser- vative)?

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4.3 Commutative Rings 157

Example 4.30. (i) Z, Q, and R are commutative rings. The ninth funda-

mental property, reciprocals, does not hold in Z; for example, 2�1 D 12 does not lie in Z.

(ii) Propositions3.8 and 3.9 show that C is a commutative ring, while Propo-

sition 3.11 shows that every nonzero complex number has an inverse.

(iii) The set of even integers does not form a commutative ring, for it has no

identity.

(iv) The Gaussian integers ZŒi  form a commutative ring (see Exercise 4.64

on page 168).

(v) The Eisenstein integers ZŒ!, where ! is a primitive cube root of unity,

form a commutative ring (see Exercise 4.64).

(vi) More generally, the cyclotomic integers ZŒ�, where � is any primitive

root of unity, form a commutative ring (see Exercise 4.65 on page 168).

(vii) The next theorem shows that Zm is a commutative ring for every integer

m � 2. (viii) We’ll see, in the next chapter, that all polynomials whose coefficients lie

in a commutative ring (e.g., all polynomials with coefficients in Z) is itself a commutative ring with the usual addition and multiplication. N

Example 4.31. (i) If R is a commutative ring and S is a set, let RS be the

set of all functions S ! R. Define uWS ! R to be the constant function with value 1, where 1 is the identity element of R: that is, u.s/ D 1 for all s 2 S . Define the sum and product of f; g 2 RS , for all s 2 S , by

f C gW s 7! f .s/C g.s/ and

fgW s 7! f .s/g.s/I

these operations are called pointwise addition and pointwise multipli-

cation. We leave the straightforward checking that RS is a commuta-

tive ring as Exercise 4.34. An important special case of this example is If R D R, then Fun.R/ D RR arises in calculus.

After all, what are the

functions x C cos x and x cos x?

Fun.R/ D RR, the ring of all functions from a commutative ring R to itself.

(ii) If X D Œa; b is an interval on the line, then

C.X/ D ff WX ! R W f is continuousg

is a commutative ring under pointwise operations. If both f; g 2 C.X/ are continuous, then it is shown in calculus that both f C g and fg are also continuous. The constant function e with e.t/ D 1 for all t 2 X is continuous; we let the reader prove that the other axioms in the definition

of commutative ring hold. N

Etymology. The word ring was probably coined by Hilbert in 1897 when

he wrote Zahlring. One of the meanings of the word ring, in German as in

English, is “collection,” as in the phrase “a ring of thieves.” It has also been

suggested that Hilbert used this term because, for a commutative ring such as the Gaussian integers ZŒi , powers of some elements “cycle back” to being a

linear combination of smaller powers (for example, i; i2; i3; i4 D 1, i5 D i ).

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158 Chapter 4 Modular Arithmetic

Theorem 4.32. Zm is a commutative ring for every integerm � 2.

Proof. The proof of each of the eight statements is routine; in essence, they

are inherited from the analogous statement in Z (the inheritance is made pos-

sible by Proposition 4.5). We prove only statements (i), (vii), and (viii) in the

definition of commutative ring; the other proofs are left to Exercise 4.31 below.

Convince yourself that

each step in these proofs

is legitimate by supplying a

reason.

(i) ŒaC Œb D ŒaC b D Œb C a D ŒbC Œa. (vii) .Associativity of Multiplication/W

Œa � ŒbŒc

� D ŒaŒbc D Œa.bc/ D Œ.ab/c D ŒabŒc D

� ŒaŒb

� Œc:

(viii) .Distributivity/W

Œa.ŒbC Œc/ D ŒaŒb C c D Œa.b C c/ D Œab C ac D ŒabC Œac D ŒaŒbC ŒaŒc:

A commutative ring is an algebraic system we view as a generalization of

ordinary arithmetic. One remarkable feature of the integers mod m is that an

integer a is divisible by m if and only if Œa D Œ0 in Zm (for m j a if and only if a � 0 mod m); that is, we have converted a statement about divisibility into an equation.

Exercises

4.31 * Prove the remaining parts of Theorem 4.32

4.32 Prove that every commutative ring R has a unique identity 1.

4.33 (i) Prove that subtraction in Z is not an associative operation.

(ii) Give an example of a commutative ring in which subtraction is associative.

4.34 * If R is a commutative ring and S is a set, verify that RS is a commutative ring

under pointwise operations. (See Example 4.31.)

4.35 * Define the weird integers W as the integers with the usual addition, but with

multiplication � defined by

a � b D (

ab if a or b is odd

�ab if both a and b are even:

Prove that W is a commutative ring.

Hint: It is clear that 1 is the identity and that * is commutative; only associativity

of * and distributivity must be checked.

4.36 For each integer a between 1 and 11, find all solutions to Œax D Œ9 in Z12. (There may be no solutions for some a.)

4.37 In Z8, find all values of x so that .x � 1/.x C 1/ D 0.

4.38 Solve the equation x2 C 3x � 3 D 0 in Z5.

4.39 How many roots does the polynomial x2 C 1 D 0 have in each of the following commutative rings?

(i) Z5 (ii) Z7 (iii) Z11

(iv) Z101 (v) Z13

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4.3 Commutative Rings 159

Properties of Commutative Rings

One advantage of precise definitions is that they are economical: proving a the-

orem for general commutative rings automatically proves it for each particular commutative ring. For example, we need not prove that .�1/.�1/ D 1 holds in the Gaussian integers ZŒi  because we prove below that it holds in all com-

mutative rings. The nice thing here is that some general proofs can be copied

verbatim from those in Chapter 1. Alas, this is not always so. For example, the

generalization of the Chinese Remainder Theorem does hold in ZŒi , but its proof requires more than merely copying, mutatis mutandis, its proof in Z.

Proposition 4.33. For every a in a commutative ring R, we have a � 0 D 0.

Proof. Identical to the proof of Proposition 1.31.

Can 1 D 0 in a commutative ringR? The answer is “yes,” but not really. If 1 D 0 in R, then a D 1a D 0a D 0 for all a 2 R, by Proposition 4.33; that is, R consists of only one element, namely, 0. So, 1 ¤ 0 in any commutative ring having more than one element. Commutative rings with only one element are

called zero rings; they are not very interesting, although they do arise every

once in a while. For example, Theorem 4.32 says that Zm is a commutative

ring for every integer m � 2. Actually, Zm is a commutative ring for m � 0: we have Z0 D Z, and Z1 the zero ring. Since zero rings arise rarely, we declare that 1 ¤ 0 for all commutative rings in this book unless we say otherwise.

Proposition 4.34. For any a in a commutative ring R, we have

.�a/.�1/ D a: In particular,

.�1/.�1/ D 1:

Proof. Identical to the proof of Proposition 1.32.

Can an element a in a commutative ringR have more than one negative?

Proposition 4.35. Let R be a commutative ring. Negatives in R are uniqueI that is, for each a 2 R, there is exactly one a0 2 R with aC a0 D 0.

Multiplicative inverses, when they exist, are uniqueI that is, for each b 2 R, there is at most one b0 2 R with bb0 D 1.

Proof. Identical to the proof of Proposition 1.33. As usual, the negative of a is

denoted by �a, and the inverse of b, when it exists, is denoted by b�1.

Corollary 4.36. For every a in a commutative ring R, we have �a D .�1/a. Moreover, if an element b has an inverse, then .b�1/�1 D b.

Proof. Identical to the proof of Corollary 1.34.

The distributive law for subtraction holds, where b � c is defined as b C .�1/c.

Corollary 4.37. If a; b; c lie in a commutative ringR, then a.b�c/ D ab�ac.

Proof. Identical to the proof of Corollary 1.35.

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160 Chapter 4 Modular Arithmetic

Definition. Let R be a commutative ring. If a 2 R, define its powers by induction on n � 0. Set a0 D 1 and, if n � 0, then anC1 D aan.

We have defined a0 D 1 for all a 2 R; in particular, 00 D 1. The notation an is a hybrid: a is an element ofR while n is an integer. Here

is the additive version of this notation.

Definition. If R is a commutative ring and k > 0 is an integer, define ka D a C � � � C a, the sum of a with itself k times. If k D 0, define ka D 0a D 0, where the 0 on the right is the zero element ofR. If k < 0, then �k D jkj > 0, and we define ka D .�k/.�a/; that is, ka is the sum of �a with itself jkj times.

The hybrid ka can be viewed as the product of two elements in the com-

mutative ring R. If e D 1 (the identity element in R), then ke 2 R and ka D .ke/a. For example, if k > 0, then

ka D aC a C � � � C a D .e C e C � � � C e/a D .ke/a:

We note that we could have defined ka, for k � 0, by induction. Set 0a D 0 and, if k � 0, then .k C 1/a D a C ka.

The Binomial Theorem holds in every commutative ring R. Since we have

defined ka whenever k is an integer and a 2 R, the notation � n j

� a makes sense.

Theorem 4.38 (Binomial Theorem). Let R be a commutative ring.

(i) For all x 2 R and all integers n � 0,

.1 C x/n D nX

jD0

n

j

! xj D

nX

jD0

j Š.n � j /Š xj :

(ii) For all a; b 2 R and all integers n � 0,

.a C b/n D nX

jD0

n

j

! an�j bj D

nX

jD0

� nŠ j Š.n � j /Š

� an�j bj :

Proof. Identical to the proof of Theorem 2.25.

Units and Fields

Let’s return to the ninth fundamental property of ordinary arithmetic. A nonzero

element in a commutative ring may not have an inverse. For example, Œ2 ¤ Œ0 in Z4, but there is no Œa 2 Z4 with Œ2Œa D Œ1: the products Œ2Œa are

Œ2Œ0 D Œ0; Œ2Œ1 D Œ2; Œ2Œ2 D Œ4 D Œ0; Œ2Œ3 D Œ6 D Œ2I

none of these is Œ1.

If m � 2, which nonzero elements in Zm have multiplicative inverses?

Proposition 4.39. Let m � 2. An element Œa 2 Zm has an inverse if and only if gcd.a; m/ D 1.

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4.3 Commutative Rings 161

Proof. Since gcd.a; m/ D 1, Theorem 4.17 says that there is an integer s so that sa � 1 mod m. Translating this congruence to Zm (using the definition of multiplication in Zm), we have

ŒsŒa D Œsa D Œ1I

thus, Œs is the inverse of Œa.

Conversely, if ŒsŒa D Œ1 in Zm, then Œsa D Œ1 and sa � 1 mod m. Therefore, m j .sa � 1/, so that sa � 1 D tm for some integer t , and gcd.a; m/ D 1, by Proposition 1.23.

If a and m are relatively prime, then the coefficients s and t displaying 1 as

a linear combination are not unique (see Exercise 1.57 on page 35). However,

Proposition 4.35 shows that the congruence class of s mod m is unique: if

also 1 D s0a C t 0m, then Œs0 D Œs, for both equal Œa�1 in Zm; that is, s0 � s mod m, for inverses are unique when they exist.

Dividing by an element a 2 R means multiplying by a�1. Thus, dividing by zero requires an element 0�1 2 R with 0�1 � 0 D 1. But we saw, in Proposition 4.33, that a � 0 D 0 for all a 2 R; in particular, 0�1 � 0 D 0. It follows that if 1 ¤ 0 in R; that is, if R has more than one element, then 0�1 does not exist; therefore, we cannot divide by 0.

How to Think About It. There is a strong analogy between the method for

solving linear equations in elementary algebra and the proof of Theorem 4.17.

When solving an equation like 3x D 4 in first-year algebra, you multiply both sides by the number u with u3 D 1, namely, u D 1

3 :

3x D 4 1 3 .3x/ D 1

3 4

� 1 3 3 � x D 4

3

x D 4 3 :

Now look at a congruence like 3x � 4 mod 7 as an equation in Z7,

Œ3x D Œ4;

and go through the same steps as above, using the fact that Œ5 � Œ3 D Œ1:

Œ3x D Œ4 Œ5.Œ3x/D Œ5 � Œ4

.Œ5 � Œ3/ x D Œ6 x D Œ6:

As we remarked on page 158, the notion of commutative ring allows us to turn

congruences into equations that obey the usual rules of elementary algebra.

Notation: The various Zm are important examples of commutative rings. It

is getting cumbersome, as in the above calculation, to decorate elements of

Zm with brackets. From now on, we will usually drop the brackets, letting

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162 Chapter 4 Modular Arithmetic

the context make things clear. For example, the calculation in Z7 above will

usually be written

3x D 4 5.3x/ D 5 � 4

.5 � 3/ x D 6 x D 6:

Definition. An element u in a commutative ring R is a unit if it has a multi-

plicative inverse inR; that is, there is v 2 R with uv D 1.

Note that v must be in R in order that u be a unit in R. For example, 2 is

not a unit in Z because 1 2

is not in Z; of course, 2 is a unit in Q.

Knowledge of the units in a commutative ring R tells us a great deal about

how much elementary algebra carries over toR. For example, knowing whether or not a is a unit inR tells us whether or not we can solve the equation ax D b in R by dividing both sides by a.

Example 4.40. (i) The only units in Z are ˙1. (ii) Proposition 4.39 describes all the units in Zm. It says that Œa is a unit in

Zm if and only if gcd.a; m/ D 1. (iii) Every nonzero element of Q, R, and C is a unit. N

What are the units in ZŒi ? Our work in Chapter 3 lets us find the answer.

Every nonzero Gaussian integer z has an inverse in C, but that inverse may not be in ZŒi . Proposition 3.11 shows, in C, that

z�1 D z z z :

The denominator on the right-hand side is none other thanN.z/, the norm of z,

and this suggests the following proposition.

Proposition 4.41. A Gaussian integer z is a unit in ZŒi  if and only ifN.z/ D 1.

Proof. If N.z/ D 1, the formula z�1 D z=.z z/ D z=N.z/ shows that z�1 D z, a Gaussian integer, and so z is a unit in ZŒi .

Conversely, if z is a unit in ZŒi , then there is a Gaussian integer w with

zw D 1. Take the norm of both sides; Proposition 3.35(iii) gives N.z/N.w/ D 1:

This is an equation in Z saying that a product of two integers is 1. The only way

this can happen is for each factor to be ˙1. But norms are always nonnegative, by Proposition 3.35, and so N.z/ (and also N.w/) is equal to 1.

Proposition 4.41 leads to the question “Which Gaussian integers have norm

1?” If z D aCbi is a Gaussian integer andN.z/ D 1, then a and b are integers with a2 C b2 D 1. Using the fact that .a; b/ is a lattice point on the unit circle, its distance to the origin is 1, and we see that the only .a; b/ satisfying the equation are

.1; 0/; .0; 1/; .�1; 0/; .0;�1/: Hence, we have

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4.3 Commutative Rings 163

Proposition 4.42. There are exactly four units in ZŒi ; namely

1; i ;�1; �i:

We know that 0 is never a unit in a nonzero commutative ring R; what if

every nonzero element in R is a unit?

Definition. A field is a nonzero commutative ring F in which every nonzero

a 2 F is a unit; that is, there is b 2 F with ab D 1.

Familiar examples of fields are Q, R, and C; here is a new example.

Theorem 4.43. Ifm � 2, then Zm is a field if and only if m is a prime.

Proof. Ifm is prime and 0 < a < m, then gcd.a; m/ D 1, and Proposition 4.39 says that a is a unit in Zm. Hence, Zm is a field. We have removed the

brackets from the notation

for elements of Zm. Conversely, suppose that m is not prime; that is, m D ab, where 0 < a,

b < m. In Zm, both a and b are nonzero, and ab D 0. If a has an inverse in Zm, say, s, then sa D 1,which gives the contradiction:

0 D s0 D s.ab/ D .sa/b D 1b D b:

Who would have thought that a field could have a finite number of elements?

When one of us was a graduate student, a fellow student was tutoring a 10-year

old prodigy. To illustrate the boy’s talent, he described teaching him how to

multiply 2� 2matrices. As soon as he was shown that the 2� 2 identity matrix I satisfies IA D A for all matrices A, the boy immediately began writing; after a few minutes he smiled, for he had just discovered that A D

� a b c d

� has

an inverse if and only if ad � bc ¤ 0! Later, when this boy was told the definition of a field, he smiled as the usual examples of Q, R, and C were

trotted out. But when he was shown Z2, he threw a temper tantrum and ended the lesson.

In Theorem 4.17, we considered linear congruences in one variable. We

now consider linear systems in two variables.

Theorem 4.44. If p is a prime, then the system

ax C by � u mod p cx C dy � v mod p

has a solution .x; y/ if and only if the determinant ad � bc 6� 0 mod p.

Proof. Since p is a prime, we know that Zp is a field. Now the system of

congruences can be considered as a system of equations in Zp .

ax C by D u cx C dy D v:

You can now complete the proof just as in linear algebra.

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164 Chapter 4 Modular Arithmetic

Example 4.45. Find the solution in Z7 of the system

4x � 5y D �2 2x C 3y D 5:

We proceed as in linear algebra. The determinant is 4 �3�.�5/ �2 D 22 ¤ 0 in Z7, and so there is a solution. Now 4

�1 D 2 in Z7 (for 4 �2 D 8 � 1 mod 7), so the top congruence can be rewritten as x � 10y � �4. Since �10 D �3, we have

x D 3y � 4:

Substituting into the bottom equation gives 2.3y � 4/C 3y D 5; that is, 9y D 13; rewrite this as 2y D 6. Multiply by 4 D 2�1 to obtain y D 24 D 3. Finally, x D 3y� 4 D 9� 4 D 5. Therefore, the solution is .5; 3/. Let’s check this. If x D 5 and y D 3, then

4 � 5 � 5 � 3 D 4 � �2 mod 7 2 � 5C 3 � 3 D 19 � 5 mod 7: N

How to Think About It. Had you mimicked the method in the example

when proving Theorem 4.44, you would have found Cramer’s Rule, a generic

formula for the solution to the system

ax C by D u cx C dy D v:

The solution is .x; y/, where

x D det

" u b

v d

#

det

" a b

c d

# and y D det

" a u

c v

#

det

" a b

c d

# :

Thus, Cramer’s Rule holds, giving us an easily remembered formula for solv-

ing 2� 2 systems of equations in any field. Most linear algebra courses present a more general Cramer’s Rule for n � n systems.

Exercises

4.40 Give an example of a commutative ring R containing an element a with a ¤ 0, a ¤ 1, and a2 D a.

4.41 * The notation in this exercise is that of Example 4.31.

(i) Find all the units in Fun.R/ D RR. (ii) Prove that a continuous function uW X ! R is a unit in C.X/ if and only if

u.t/ ¤ 0 for all t 2 X .

4.42 Let R D ZŒ p

3 D fa C b p

3 W a; b 2 Zg: (i) Show, with the usual addition and multiplication of real numbers, that R is a

commutative ring.

(ii) Show that u D 2 C p

3 is a unit in R.

(iii) Show that R has infinitely many units.

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4.3 Commutative Rings 165

4.43 If p is a prime, show that a quadratic polynomial with coefficients in Zp has at

most two roots in Zp .

4.44 * Prove or give a counterexample. Let R be a commutative ring.

(i) The product of two units in R is a unit.

(ii) The sum of two units in R is a unit.

4.45 * Describe all the units in the Eisenstein integers ZŒ!.

4.46 * Just as in C, a root of unity in a ring R is an element a 2 R with an D 1 for some positive integer n.

Find all roots of unity in Zm for all integers m between 5 and 12.

4.47 * Show that Zm contains exactly �.m/ units, where � is the Euler �-function.

4.48 * Show that an element u 2 Zm is a unit if and only if u is a root of unity.

4.49 * If u is a unit in Zm, then Exercise 4.48 says there is some positive integer n

with un D 1; the smallest such n is called the order of u in Zm. For each integer m between 5 and 12, make a table that shows the units and

their orders. Any conjectures about which integers can be orders of units?

4.50 State and prove Cramer’s Rule for a 3 � 3 system of linear equations in a field.

4.51 Solve the system of congruences

3x � 2y C z � 1 mod 7 x C y � 2z � 0 mod 7

�x C 2y C z � 4 mod 7:

4.52 For what values of m will the system

2x C 5y D 7 x C 4y D 9

have a unique solution in Zm?

4.53 Find a system of two linear equations in two unknowns that has a unique solution

in Zm for all m � 2.

4.54 (i) Show that

M2 D ("

a b

�b a

# W a; b 2 Z

)

is a commutative ring under matrix addition and multiplication.

(ii) What are the units in M2?

4.55 *

(i) Show that

F4 D ("

a b

b a C b

# W a; b 2 Z2

) ;

with binary operations matrix addition and multiplication, is a field having

exactly four elements.

(ii) Write out addition and multiplication tables for F4 .

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166 Chapter 4 Modular Arithmetic

Subrings and Subfields

Sometimes, as with Z and Q, one ring sits inside another ring.

More precisely, if ˛ is the

addition on R, then its

restriction ˛j.S � S/ has image in S, and it is the

addition on S. Similarly for

multiplication.

Definition. A subring of a commutative ring R is a commutative ring S con-

tained inR that has the same 1, the same addition, and the same multiplication as R; that is, 1 2 S and if a; b 2 S , then a C b 2 S and ab 2 S .

Each commutative ring on the list Z � Q � R � C is a subring of the next one. Example A.20 in Appendix A.3 says that if R is a commutative ring andQuery: Is Zm a subring of

Z? k � R is a subring that is a field, then R is a vector space over k. Thus, C is a vector space over R (and also over Q), and R is a vector space over Q.

Proposition 4.46. A subset S of a commutative ringR is a subring ofR if and

only if

(i) 1 2 S I (ii) if a; b 2 S , then aC b 2 S I

(iii) if a; b 2 S , then ab 2 S .

Proof. If S is a subring of R, then the three properties clearly hold.

Conversely, if S satisfies the three properties, then S contains 1, and so it

only remains to show that S is a commutative ring. Items (ii) and (iii) (closure

under addition and multiplication) show that the (restrictions of) addition and multiplication are binary operations on S . All the other items in the definition

of commutative ring are inherited from R. For example, the distributive law

holds: since a.bCc/ D abCac holds for all a; b; c 2 R, it holds, in particular, for all a; b; c 2 S � R.

Proposition 4.46 is more powerful than it looks. A subset S of a commuta- tive ring R is a subring if, using the same operations as those in R, it satisfies

all the conditions in the definition of commutative ring. But there’s no need

to check all the properties; you need check only three of them. For example,

Exercise 4.64 on page 168 asks you to prove that the Gaussian integers ZŒi 

and the Eisenstein integers ZŒ! are commutative rings. This could be tedious:

there are ten things in the definition of commutative ring to check: addition and multiplication are binary operations and the eight axioms. However, if we

know that C is a commutative ring and ZŒi  and ZŒ! are subrings of C (facts

that can be established via Proposition 4.46), then both ZŒi  and ZŒ! are com-

mutative rings in their own right.

Example 4.47. Here is an example of a commutative ring arising from set

theory. If A and B are subsets of a set X , then their symmetric difference is

AC B D .A [ B/ � .A \ B/

(see Figure 4.2). If U and V are subsets of a set X , then

U � V D fx 2 X W x 2 U and x … V g:

Let X be a set, let 2X denote the set of all the subsets of X , define addition

ACB is exclusive or; that is, all x 2 X lying in either A or B but not in both. In

terms of Venn diagrams,

this pictures the statement:

Take it or leave it!

Recall that BA is the family

of all functions from a set

A to a set B . Why is this

ring denoted by 2X ? We’ll

see why in Example 5.16.

on 2X to be symmetric difference, and define multiplication on 2X to be inter-

section. Exercises 4.68 through 4.74 on page 169 essentially show that 2X is a

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4.3 Commutative Rings 167

A B

Figure 4.2. Symmetric difference.

commutative ring. The empty set ¿ is the zero element, for AC ¿ D A, while each subset A is its own negative, for AC A D ¿. These exercises also show that symmetric difference is associative and that the distributive law holds. Fi-

nally, X itself is the identity element, for X \ A D A for every subset A. We call 2X a Boolean ring.

Suppose now that Y ¨ X is a proper subset of X ; is 2Y a subring of 2X? If A and B are subsets of Y , then A C B and A \ B are also subsets of Y ; that is, 2Y is closed under the addition and multiplication on 2X . However, the

identity element in 2Y is Y , not X , and so 2Y is not a subring of 2X . N

The example of 2X may have surprised you. It was natural for us to in-

troduce the notion of commutative ring, for we had already seen many ex-

amples of numbers or of functions in which addition and multiplication make sense and obey the usual rules. But the elements of 2X are neither numbers nor

functions. And even though we call their binary operations addition and mul-

tiplication, they are operations from set theory. This is a happy circumstance,

which we will exploit in the next chapter. It’s not really important what we

call addition and multiplication; what is important is that the operations satisfy eight fundamental properties; that is, the axioms in the definition of commuta-

tive ring.

Just as the notion of a subring of a commutative ring is useful, so too is the

notion of a subfield of a field.

Definition. If F is a field, then a subfield of F is a subring k � F that is also a field.

For example, Q is a subfield of R, and both Q and R are subfields of C.

There is a shortcut for showing that a subset is a subfield.

Proposition 4.48. A subring k of a field F is a subfield of F if and only if

a�1 2 k for all nonzero a 2 k.

Proof. This is Exercise 4.57 below.

Exercises

4.56 Give an example of a subring of a field that is not a field.

4.57 * Prove Proposition 4.48.

4.58 (i) Show that f0; 2g � Z4 has the same addition and multiplication tables as Z2 . (ii) Is Z2 a subring of Z4?

(iii) Is f0; 2; 4; 6g a subring of Z8?

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168 Chapter 4 Modular Arithmetic

4.59 Let R D ZŒ p

�3 D fa C b p

�3 W a; b 2 Zg. (i) Show that R is a subring of the Eisenstein integers.

(ii) What are the units in R?

4.60 (i) If S and T are subrings of a ring R, show that S \ T is also a subring of R. (ii) Show that the intersection of the Gaussian and Eisenstein integers is Z.

4.61 *

(i) If .Si /i2I is a family of subrings of a commutative ring R, prove that their intersection

T i2I Si is also a subring of R.

(ii) If X is a subset of a commutative ring R, define G.X/, the subring generated

by X , to be the intersection of all the subrings of R that contain X .

Prove that G.X/ is the smallest subring containing X in the following

sense: if S is any subring of R containing X , then G.X/ � S . (iii) Let .Si /i2I be a family of subrings of a commutative ring R, each of which

is a field. Prove that the subring T i2I Si is a field. Conclude that the inter-

section of a family of subfields of a field is a subfield.

4.62 Let p be a prime and let Ap be the set of all fractions with denominator a power

of p.

(i) Show, with the usual operations of addition and multiplication, that Ap is a

subring of Q.

(ii) Describe the smallest subring of Q that contains both A2 and A5.

4.63 Let p be a prime and let Qp be the set of rational numbers whose denominator

(when written in lowest terms) is not divisible by p.

(i) Show, with the usual operations of addition and multiplication, that Qp is a

subring of Q.

(ii) Show that Q2 \ Q5 is a subring of Q. (iii) Is Qp a field? Explain.

(iv) What is Qp \ Ap, where Ap is defined in Exercise 4.62?

4.64 *

(i) Prove that ZŒi  D fa C bi W i2 D �1 and a; b 2 Zg, the Gaussian integers, is a commutative ring.

(ii) Prove that ZŒ! D fa C b! W !3 D 1 and a; b 2 Zg, the Eisenstein integers, is a commutative ring.

4.65 * Prove that ZŒ� D fa C b�i W 0 � i < n and a; b 2 Zg is a commutative ring, where � is a primitive nth root of unity.

4.66 * It may seem more natural to define addition in 2X as union rather than symmet-

ric difference. Is 2X a commutative ring if addition A ˚ B is defined as A [ B and AB is defined as A \ B?

4.67 If X is a finite set with exactly n elements, how many elements are in 2X?

4.68 * If A and B are subsets of a set X , prove that A � B if and only if A D A \ B .

4.69 * Recall that if A is a subset of a set X , then its complement is

Ac D fx 2 X W x … Ag:

Prove, in the commutative ring 2X , that Ac D X C A.

4.70 * Let A be a subset of a set X . If S � X , prove that Ac D S if and only if A [ S D X and A \ S D ¿.

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4.4 Connections: Julius and Gregory 169

4.71 Let A; B; C be subsets of a set X .

(i) Prove that A [ .B \ C/ D .A [ B/ \ .A [ C/. (ii) Prove that A \ .B [ C/ D .A \ B/ [ .A \ C/.

4.72 If A and B are subsets of a set X , then A � B D fx 2 A W x … Bg. Prove that A � B D A \ Bc . In particular, X � B D Bc , the complement of B .

4.73 * Let A and B be subsets of a set X . Prove the De Morgan laws:

.A [ B/c D Ac \ Bc and .A \ B/c D Ac [ Bc ;

where Ac denotes the complement of A.

4.74 * If A and B are subsets of a set X , define their symmetric difference by ACB D .A � B/ [ .B � A/ (see Figure 4.2). (i) Prove that A C B D .A [ B/ � .A \ B/.

(ii) Prove that .A C B/ [ .A \ B/ D A [ B . (iii) Prove that A C A D ¿. (iv) Prove that A C ¿ D A. (v) Prove that A C .B C C/ D .A C B/ C C .

Hint: Show that each of A C .B C C/ and .A C B/ C C is described by Figure 4.3.

(vi) Prove that the Boolean ring 2X is not a field if X has at least two elements.

A B

C

Figure 4.3. Associativity.

4.75 Prove that A \ .B C C/ D .A \ B/ C .A \ C/.

4.4 Connections: Julius and Gregory

On what day of the week was July 4, 1776? We’ll use congruence to answer

this question. In fact, we’ll answer in two ways: with an exact formula com-

puting the day, and with a faster refinement, due to Conway. Let’s begin by seeing why our calendar is complicated. A year is the amount

of time it takes the Earth to make one complete orbit around the Sun; a day is

the amount of time it takes the Earth to make a complete rotation about the axis

through its north and south poles. There is no reason why the number of days

in a year should be an integer, and it isn’t; a year is approximately 365.2422 days long. In 46 BCE, Julius Caesar (and his scientific advisors) changed the

old Roman calendar, creating the Julian calendar containing a leap year every

four years; that is, every fourth year has an extra day, namely, February 29, and

so it contains 366 days (a common year is a year that is not a leap year). This

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170 Chapter 4 Modular Arithmetic

would be fine if the year were exactly 365.25 days long, but it has the effect of

making the year 365:25 � 365:2422 D :0078 days (about 11 minutes and 14 seconds) too long. After 128 years, a full day was added to the calendar; that is, the Julian calendar overcounted the number of days. In the year 1582, the

vernal equinox (the Spring day on which there are exactly 12 hours of daylight

and 12 hours of night) occurred on March 11 instead of on March 21. Pope

Gregory XIII (and his scientific advisors) then installed the Gregorian calen-

dar by erasing 10 days that year; the day after October 4, 1582 was October 15, 1582. This caused confusion and fear among the people; they thought their

lives had been shortened by ten days.

The Gregorian calendar modified the Julian calendar as follows. Call a

year y ending in 00 a century year. If a year y is not a century year, then

it is a leap year if it is divisible by 4; if y is a century year, it is a leap year only if it is divisible by 400. For example, 1900 is not a leap year, but 2000 is a leap

year. The Gregorian calendar is the one in common use today, but it was not

uniformly adopted throughout Europe. For example, the British empire didn’t

accept it until 1752, when 11 days were erased, and the Russians didn’t ac-

cept it until 1918, when 13 days were erased (thus, Trotsky called the Russian

revolution, which occurred in 1917, the October Revolution, even though it occurred in November of the Gregorian calendar).

The true number of days in 400 years is about

400 � 365:2422 D 146096:88 days:

In this period, the Julian calendar has

400� 365C 100 D 146; 100 days;

while the Gregorian calendar, which eliminates three leap years from this time period, has 146,097 days. Thus, the Julian calendar gains about 3.12 days every

400 years, while the Gregorian calendar gains only 0.12 days (about 2 hours

and 53 minutes.

Historical Note. There are 1628 years from 46 BCE to 1582 CE. The Julian

calendar overcounts one day every 128 years, and so it overcounted 12 days in this period (for 12� 128 D 1536). Why didn’t Gregory have to erase 12 days? The Council of Nicaea, meeting in the year 325 CE, defined Easter as the first

Sunday strictly after the Paschal full moon, which is the first full moon on or

after the vernal equinox (now you know why Pope Gregory was interested in

the calendar). The vernal equinox in 325 CE fell on March 21, and the Synod

of Whitby, in 664 CE, officially defined the vernal equinox to be March 21. The discrepancy observed in 1582 was thus the result of only 1257 D 1582� 325 years of the Julian calendar: approximately 10 days.

We now seek a calendar formula. For easier calculation, choose 0000 as our

reference year, even though there was no year zero! Assign a number to each

day of the week, according to the scheme

Sun Mon Tues Wed Thurs Fri Sat

0 1 2 3 4 5 6 .

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4.4 Connections: Julius and Gregory 171

In particular, March 1, 0000, has some number a0, where 0 � a0 � 6. In the next year 0001, March 1 has number a0C1 (mod 7), for 365 days have elapsed from March 1, 0000, to March 1, 0001, and

365 D 52 � 7C 1 � 1 mod 7:

Similarly, March 1, 0002, has number a0 C 2, and March 1, 0003, has number a0 C 3. However, March 1, 0004, has number a0 C 5, for February 29, 0004, fell between March 1, 0003, and March 1, 0004, and so 366 � 2 mod 7 days had elapsed since the previous March 1. We see, therefore, that every common

year adds 1 to the previous number for March 1, while each leap year adds 2. Thus, if March 1, 0000, has number a0, then the number a

0 of March 1, year y, is

a0 � a0 C y C L mod 7;

where L is the number of leap years from year 0001 to year y. To compute L,

count all those years divisible by 4, then throw away all the century years, and

then put back those century years that are leap years. Thus,

L D by=4c � by=100c C by=400c;

where bxc denotes the greatest integer in x. Therefore, we have

a0 � a0 C y C L � a0 C y C by=4c � by=100c C by=400c mod 7:

We can actually find a0 by looking at a calendar. Since March 1, 2012, fell

on a Thursday,

4 � a0 C 2012C b2012=4c � b2012=100c C b2012=400c � a0 C 2012C 503 � 20C 5 mod 7;

and so

a0 � �2496 � �4 � 3 mod 7

(that is, March 1, 0000 fell on Wednesday). We can now determine the day of

the week a0 on which March 1 will fall in any year y > 0, for

a0 � 3C y C by=4c � by=100c C by=400c mod 7:

Historical Note. There is a reason we have been discussing March 1, for

it was the first day of the year in the old Roman calendar (753 BCE). There

were only ten months: Martius, . . . , Iunius, Quintilis, Sextilis, Septembris,

. . . , Decembris (which explains why September is so named; originally, it was month 7). In 713 BCE, Numa added January and February, and the Julian cal-

endar changed the names of Quintilis and Sextilis to July and August.

Let us now analyze February 28. For example, suppose that February 28, 1600, has number b. As 1600 is a leap year, February 29, 1600, occurs between

February 28, 1600, and February 28, 1601; hence, 366 days have elapsed be-

tween these two February 28s, so that February 28, 1601, has number b C 2. February 28, 1602, has number b C 3, February 28, 1603, has number b C 4,

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172 Chapter 4 Modular Arithmetic

February 28, 1604, has number bC5, but February 28, 1605, has number bC7 (for there was a February 29 in 1604).

Let us compare the pattern of behavior of February 28, 1600, namely, b; bC2; bC3; bC4; bC5; bC7; : : : ; with that of some date in 1599. If May 26, 1599, has number c, then May 26, 1600, has number c C 2, for February 29, 1600, comes between these two May 26s, and so there are 366 � 2 mod 7 in- tervening days. The numbers of the next few May 26s, beginning with May 26,

1601, are c C 3, c C 4, c C 5, c C 7. We see that the pattern of the days for February 28, starting in 1600, is exactly the same as the pattern of the days

for May 26, starting in 1599; indeed, the same is true for any date in January

or February. Thus, the pattern of the days for any date in January or February

of a year y is the same as the pattern for a date occurring in the preceding

year y � 1: a year preceding a leap year adds 2 to the number for such a date, whereas all other years add 1. Therefore, we pretend we have reverted to the

ancient calendar by making New Year’s Day fall on March 1, so that any date

in January or February is treated as if it had occurred in the previous year.

Historical Note. George Washington’s birthday, in the Gregorian calendar,

is February 22, 1732. But the Gregorian calendar was not introduced in the

British colonies until 1752. Thus, his original birthday was February 11. But

New Year’s Day was also changed; before 1752, England and its colonies cele-

brated New Year’s Day on March 25; hence, February, which had been in 1731, was regarded, after the calendar change, as being in 1732. George Washington

used to joke that not only did his birthday change, but so did his birth year. See

Exercise 4.80 on page 176.

How do we find the day corresponding to a date other than March 1? Since March 1, 0000, has number 3 (as we have seen above), April 1, 0000, has

number 6, for March has 31 days and 3 C 31 � 6 mod 7. Since April has 30 days, May 1, 0000, has number 6 C 30 � 1 mod 7. Figure 4.4 is the table giving the number of the first day of each month in year 0000.

Remember that we are pretending that March is month 1, April is month 2, and so on. Let us denote these numbers by 1 C j.m/, where j.m/, for m D 1; 2; : : : ; 12, is defined by

j.m/ W 2, 5, 0, 3, 5, 1, 4, 6, 2, 4, 0, 3.

It follows that month m, day 1, year y, has number

1C j.m/C g.y/ mod 7;

where

g.y/ D y C by=4c � by=100c C by=400c:

Note that a0 D 1Cj.1/, so that the values of j.m/ depend on our knowing a0. Here’s a formula for j.m/:

j.m/ D b2:6m � 0:2c; where 1 � m � 12I

the values are displayed in Figure 4.4. This formula is not quite accurate. For

example, this number for December, that is, form D 10, is b2:6m�0:2c D 25; but j.10/ D 4. However, 25 � 4 mod 7, and so the formula for j.m/ really gives the congruence class mod 7.

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4.4 Connections: Julius and Gregory 173

Date Number Date Number Date Number

March 1 2 July 1 5 November 1 2

April 1 5 August 1 1 December 1 4

May 1 0 September 1 4 January 1 0

June 1 3 October 1 6 February 1 3

Figure 4.4. Values of j.m/.

Theorem 4.49 (Calendar Formula). The date with month m, day d , year y The word calendar comes from the Greek “to call,”

which evolved into the

Latin word for the first day

of a month (when accounts

were due).

has number

d C j.m/C g.y/ mod 7;

where j.m/ is given in Figure 4.4,

g.y/ D y C by=4c � by=100c C by=400c;

and dates in January and February are treated as having occurred in the pre-

vious year.

Proof. The number mod 7 corresponding to monthm, day 1, year y, is

1C j.m/C g.y/:

It follows that 2C j.m/ C g.y/ corresponds to month m, day 2, year y, and, more generally, d C j.m/Cg.y/ corresponds to monthm, day d , year y.

Let’s find the day of the week on which July 4, 1776 fell; here m D 5, d D 4, and y D 1776. Substituting in the formula, we obtain the number

4C 5C 1776C 444 � 17C 4 D 2216 � 4 mod 7I

therefore, July 4, 1776, fell on a Thursday.

Example 4.50. Does every year y contain a Friday 13? We have

5 � 13C j.m/C g.y/ mod 7:

The question is answered positively if the numbers j.m/, as m varies from 1

through 12, give all the remainders 0 through 6 mod 7. And this is what hap-

pens. The sequence of remainders mod 7 is

2; 5; [0; 3; 5; 1; 4; 6; 2]; 4; 0; 3:

Indeed, we see that there must be a Friday 13 occurring between May and

November. No number occurs three times on the list, but it is possible that

there are three Friday 13s in a year because January and February are viewed as having occurred in the previous year; for example, there were three Friday 13s

in 1987 (see Exercise 4.79 on page 176). Of course, we may replace Friday by

any other day of the week, and we may replace 13 by any number between 1

and 28. N

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174 Chapter 4 Modular Arithmetic

Most of us need paper and pencil (or a calculator) to use the calendar for-

mula in the theorem, but here’s a way to simplify the formula so you can do

the calculation in your head and amaze your friends. A mnemonic for j.m/ is the sentence

My Uncle Charles has eaten a cold supper; he eats nothing hot.

2 5 (7 � 0) 3 5 1 4 6 2 4 (7 � 0) 3

Corollary 4.51. The date with month m, day d , year y D 100C CN , where 0 � N � 99, has number

d C j.m/CN C bN=4c C bC=4c � 2C mod 7;

provided that dates in January and February are treated as having occurred in

the previous year.

Proof. If we write a year y D 100C CN , where 0 � N � 99, then

y D 100C CN � 2C CN mod 7; by=4c D 25C C bN=4c � 4C C bN=4c mod 7;

by=100c D C; and by=400c D bC=4c:

Therefore,

y C by=4c � by=100c C by=400c � N C 5C C bN=4c C bC=4c mod 7 � N C bN=4c C bC=4c � 2C mod 7:

This formula is simpler than the first one. For example, the number corre-

sponding to July 4, 1776 is now obtained as

4C 5C 76C 19C 4 � 34 D 74 � 4 mod 7;

agreeing with our calculation above. The reader may now compute the day of his or her birth.

Example 4.52. The birthday of Rose, the grandmother of Danny and Ella, was

January 1, 1909; on what day of the week was she born?January is counted as belonging to the previous

year 1908. We use Corollary 4.51. If A is the number of the day, then j.m/ D 0 (for January corresponds to month 11), and

A � 1C 0C 8C b8=4c C b19=4c � 38 � �23 mod 7 � 5 mod 7:

Rose was born on a Friday. N

J. H. Conway found an even simpler calendar formula. The day of the week on which the last day of February occurs is called the doomsday of the year.

We can compute doomsdays using Corollary 4.51.

Knowing the doomsday D of a century year 100C finds the doomsday D0

of any other year y D 100C C N in that century. Since 100C is a century

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4.4 Connections: Julius and Gregory 175

year, the number of leap years from 100C to y does not involve the Gregorian

alteration. Thus,

D0 � D CN C bN=4c mod 7:

For example, since doomsday 1900 is Wednesday = 3, we see that doomsday

1994 is Monday = 1, for

3C 94C 23 D 120 � 1 mod 7:

February 29, 1600 2 Tuesday

February 28, 1700 0 Sunday

February 28, 1800 5 Friday

February 28, 1900 3 Wednesday

February 29, 2000 2 Tuesday

Figure 4.5. Recent doomsdays.

Proposition 4.53 (Conway). Let D be doomsday 100C , and let 0 � N � 99. If N D 12q C r , where 0 � r < 12, then D0, doomsday 100C CN , is given by

D C q C r C br=4c mod 7:

Proof.

D0 � D CN C bN=4c � D C 12q C r C b.12q C r/=4c � D C 15q C r C br=4c � D C q C r C br=4c mod 7:

For example, what isD0 D doomsday 1994? NowN D 94 D 12� 7C 10, so that q D 7 and r D 10. Thus, D0 D 3 C 7 C 10 C 2 � 1 mod 7; that is, doomsday 1994 is Monday, as we saw above.

Once we know doomsday of a particular year, we can use various tricks (e.g., Uncle Charles) to pass from doomsday to any other day in the year. Con-

way observed that some other dates falling on the same day of the week as the

doomsday are

April 4, June 6, August 8, October 10, December 12,

May 9, July 11, September 5, and November 7:

If we return to the everyday listing beginning with January as the first month,

then it is easier to remember these dates using the notation month/day:

4=4; 6=6; 8=8; 10=10; 12=12;

5=9; 7=11; 9=5; 11=7:

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176 Chapter 4 Modular Arithmetic

Since doomsday corresponds to the last day of February, we are now within

a few weeks of any date in the year, and we can easily interpolate to find the

desired day. For example, let’s use this method for July 4, 1776. Notice that July 4 occurs on the same day of the week as July 11, and so we need only find

doomsday 1776. By Proposition 4.53,

D0 � 0C 76C b76=4c D 95 � 4 mod 7:

We see again that July 4, 1776 fell on a Thursday.

Example 4.54. Let’s use Conway’s method to compute Rose’s birthday again (recall Example 4.52: Rose was born on January 1, 1909). Since Conway’s

method applies within a given century, there is no need to pretend that Jan-

uary and February live in the preceding year; we can work within 1909. Now

doomsday 1900 is 3, so that Proposition 4.53 gives doomsday 1909 D 0; that is, Sunday. By definition, doomsday is the number corresponding to the last date in February, which is here February 28 (for 1909 is not a leap year). Thus,

we interpolate that 3 is the number for 1=31, 1=24, 1=3; that is, January 3 fell

on Sunday, and so January 1 fell on Friday (which agrees with what we saw in

Example 4.52). N

Exercises

4.76 A suspect said that he had spent the Easter holiday April 21, 1893, with his ailing

mother; Sherlock Holmes challenged his veracity at once. How could the great

detective have been so certain?

Hint: Easter always falls on Sunday. (There is a Jewish variation of this problem,

for Yom Kippur must fall on either Monday, Wednesday, Thursday, or Saturday;

secular variants can involve Thanksgiving Day, which always falls on a Thursday,

or Election Day in the US, which always falls on a Tuesday.)

4.77 How many times in 1900 did the first day of a month fall on a Tuesday?

Hint: The year y D 1900 was not a leap year.

4.78 On what day of the week did February 29, 1896 fall?

Hint: On what day did March 1, 1896, fall? Conclude from your method of solu-

tion that no extra fuss is needed to find leap days.

4.79 *

(i) Show that 1987 had three Friday 13s.

Hint: See Example 4.50.

(ii) Show, for any year y > 0, that g.y/ � g.y � 1/ D 1 or 2, where g.y/ D y C b y=4c � by=100c C b y=400c.

(iii) Can there be a year with exactly one Friday 13?

Hint: Either use congruences or scan the 14 possible calendars: there are 7

possible common years and 7 possible leap years, for January 1 can fall on

any of the 7 days of the week.

4.80 * JJR’s Uncle Ben was born in Pogrebishte, a village near Kiev, and he claimed

that his birthday was February 29, 1900. JJR told him that this could not be, for

1900 was not a leap year. Why was JJR wrong?

Hint: When did Russia adopt the Gregorian calendar?

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4.5 Connections: Patterns in Decimal Expansions 177

4.5 Connections: Patterns in

Decimal Expansions

One of the most beautiful applications of modular arithmetic is to the classi-

fication of decimal expansions of rational numbers, a circle of ideas that runs

throughout precollege mathematics.

We now ask what we can infer from knowing the decimal expansion of a real number x. You probably know Proposition 4.58: x is rational if and only

if its decimal expansion either terminates or eventually repeats. Is there any

nice way to tell ahead of time which fractions terminate? Can you be sure

that the ones that don’t terminate really do repeat? And, for fractions whose

decimals repeat, can you predict the period (the number of digits in its block, the repeating part) as well as the actual sequence of digits in it?

Many conjectures about fractions and decimals come from a careful anal-

ysis of numerical calculations. In this section, we (and you) will perform a

great many calculations, looking at patterns you’ll observe, with the goal of

analyzing them, and seeing how they are explained by “how the calculation

goes.”

Real Numbers

We assume that every real number x has a decimal expansion; for example,

�� D �3:14159 : : : . This follows from identifying each real number x with a “point on a number line” having signed distance from a fixed origin on a

coordinatized line. In particular, rational numbers have decimal expansions,

which you can find by long division.

The term expansion will be used in a nonstandard way: we restrict the ter-

minology so that, from now on, the decimal expansion of a real number is the sequence of digits after the decimal point. With this usage, for example, the

decimal expansion of �� is :14159 : : : . We are going to see that decimal expansions of real numbers are unique,

with one possible exception: if there is an infinite string of all 9s. For example,

:328 D :327999 : : : :

This is explained using the geometric series.

Lemma 4.55. If r is a real number with jr j < 1, then 1X

nD0 rn D 1C r C r2 C � � � D 1

1 � r :

Proof. For every positive integer n, the identity

1 � rn D .1 � r/ � 1C r C r2 C � � � C rn�1

gives the equation

1C r C r2 C � � � C rn�1 D 1 � r n

1 � r D 1

1 � r � rn

1 � r

for every real number r ¤ 1. Since jr j < 1, we have lim n!1

rn=.1� r/ D 0.

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178 Chapter 4 Modular Arithmetic

For example, taking r D 1=10, we have

:999 : : : D 9 10

C 9 102

C 9 103

C � � �

D 9 10

� 1C 1

10 C 1 102

C 1 103

C � � � �

D 9 10

� 1 1 � 1=10

D 9 10

� 10 9

D 1:

Hence,

:327999 : : : D :327C :000999 : : : D :327C 1 103

:999 : : :

D :327C 1 103

D :327C :001 D :328:

We’ll resolve this ambiguity by choosing, once for all, to avoid infinite strings

of 9s. Indeed, we’ll soon see that the choice :328 comes from long division. If we disregard “all nines from some point on,” then we can show that every

real number has a unique decimal expansion. For this, we need the following

corollary to Lemma 4.55.

Corollary 4.56. If x D :d1d2 : : : and dj < 9 for some j > 1, then

x < d1 C 1 10

:

Proof. Each digit di is at most 9, and there is some j > 1 with dj strictly less than 9. So, writing x as a series, we have

x D d1 10

C d2 102

C d3 103

C � � � C dj 10j

C � � �

< d1

10 C d2 102

C d3 103

C � � � C 9 10j

C � � �

� d1 10

C 9 102

C 9 103

C � � � C 9 10j

C � � �

D d1 10

C 9 102

� 1C 1

10 C � � � C 1

10j�2 C � � �

D d1 10

C � 9

102

�� 10

9

D d1 10

C 1 10

D d1 C 1 10

:

Proposition 4.57. Every real number x has a unique decimal expansion that

does not end with infinitely many consecutive 9s.

Proof. Suppose that

:d1 : : : dk : : : D x D :e1 : : : ek : : : ;

where d1 D e1; : : : ; dk�1 D ek�1, but that dk ¤ ek . We may assume that dk < ek , so that dk C 1 � ek.

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4.5 Connections: Patterns in Decimal Expansions 179

Multiplying by a power of 10, we see that

:dkdkC1 : : : D :ekekC1 : : : :

Because there’s not an infinite string of 9s in our expansions, we can apply Corollary 4.56 to find that

:dkdkC1 : : : < dk C 1 10

� ek 10

� :ekekC1 : : : ;

contradicting the fact that the extreme left-hand and right-hand expressions are

equal.

Decimal Expansions of Rationals

Let’s now focus on rational numbers. Some decimal expansions of rationals

terminate; for example,

1

10 D :1; 1

4 D :25; 3

8 D :375; 1

40 D :025:

And there are some fractions whose decimal expansions repeat (after a possible

initial string of digits):

You can think of

.32142857142857. . . as

the 10-adic expansion

of 9=28, using negative

powers of 10.

1

3 D :333 : : : ; 1

7 D :142857142857 : : : ; 9

28 D :32142857142857 : : :

Definition. Let a real number x have decimal expansion

r D :d1d2d3 : : : I

that is, x D k:d1d2d3 � � � D k C r for some k 2 Z.

(i) We say that x terminates if there exists an integer N so that di D 0 for all i > N .

(ii) We say that x repeats with period m � 1 if (a) it doesn’t terminate

(b) there exist positive integersN andm so that di D diCm for all i > N (c) m is the smallest such integer.

If x repeats, then its block is the first occurrence of its repeating part

didiC1 : : : diCm�1:

We could say that “terminating” and “repeating” decimals are not really different, for terminating rationals have decimal expansions that repeat with

period 1 and with block having the single digit 0, but it’s convenient and natural

to distinguish such rationals from those having infinitely many nonzero digits,

as you’ll see in Proposition 4.59.

The way to get the decimal expansion for 1=7 is to divide 7 into 1 via long division, as in Figure 4.6. Each of the remainders 1 through 6 shows up exactly

once in this calculation, in the order 3; 2; 6; 4; 5; 1. Once you get a remainder

of 1, the process will start over again, and the digits in the quotient, namely,

1; 4; 2; 8; 5; 7, will repeat. The block of 1=7 is 142857. However, even though

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180 Chapter 4 Modular Arithmetic

0.142857. . .

7 �

1.000000. . .

7

30

28

20

14

60

56

40

35

50

49

1

Figure 4.6. 1=7 D :142857142857142857: : : :

0.153846. . .

13 �

2.000000. . .

1 3

70

65

50

39

110

104

60

52

80

78

2

Figure 4.7. 2=13 D :153846 : : : :

142857142857 also repeats, it is not a block because it is too long: 1=7 has

period 6, not 12.

Consider a second example: the calculation of 2=13 in Figure 4.7. It too has

period 6.

Next, we’ll see that every rational number terminates or repeats; that is, the two types in the definition are the only possibilities.

Proposition 4.58. A real number x is rational if and only if it either terminates

or repeats. Moreover, if x D a=b is rational, then it has period at most b.

Proof. The arguments for 1=7 and 2=13 generalize. Imagine expressing a frac-

tion a=b (with a; b > 0) as a decimal by dividing b into a via long division.

There are at most b possible remainders in this process (integers between 0

and b � 1), so after at most b steps a remainder appears that has shown up before. After that, the process repeats.

Conversely, let’s see that if a real number x terminates or repeats, then x is

rational. A terminating decimal is just a fraction whose denominator is a power

of 10, while a repeating decimal is made up of such a fraction plus the sum of

a convergent geometric series. An example is sufficient to see what’s going on.Middle school students practice another method

for doing this (for days on

end). See Exercise 4.84 on

page 181.

:1323232 : : :D :1 C :0323232 : : :

D :1 C 32 103

C 32 105

C 32 107

C : : :

D :1 C 32 103

� 1C 1

102 C 1 104

C : : : �

D 1 10

C 4 25

1

1 � 1 102

! : (by Lemma 4.55)

The last expression is clearly a rational number. The general proof is a generic

version of this idea; it is left as Exercise 4.81 below.

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4.5 Connections: Patterns in Decimal Expansions 181

Which rationals terminate? Certainly, any rational a=b whose denomina-

tor b is a power of 10 terminates. But some other rationals can also be put in

this form; for example,

5

8 D 5 � 125 8 � 125 D

625

1000 D :625:

The basic idea is to take a denominator of the form 2u5v , and multiply top and Theorem 4.61 below gives a necessary and sufficient

condition for a=b to repeat. bottom of the fraction to produce a denominator that’s a power of 10.

Proposition 4.59. Let x D a=b be rational, written in lowest terms. Then x terminates if and only if the only prime factors of b are 2 and 5.

Proof. If x terminates, say, x D k:d1d2 : : : dm, then x D kCD=10m for some k 2 Z, where D is the integer with digits d1d2 : : : dm; thus, x is a fraction whose denominator is divisible only by 2 and 5. Conversely, if x D k C r D k C a=2u5v, then

r D a 2u5v

D � 2v5u

2v5u

�� a 2u5v

� D 2

v5u a

10uCv

a fraction whose denominator is a power of 10. Hence, x terminates.

How to Think About It. Exercise 4.83 shows that if r is a rational number

and 5mr or 2`r terminates, then r also terminates. However, if kr terminates (for some integer k), then r need not terminate; for example, r D :271333 : : : does not terminate, but 3r D :814 does terminate.

Exercises

4.81 * Complete the proof of Proposition 4.58 that a decimal that eventually repeats is

the decimal expansion of a rational number.

4.82 * Let r D a=b be rational. (i) If r terminates, then kr terminates for every integer k.

(ii) If gcd.a; b/ D 1, prove that a=b terminates if and only if 1=b terminates. Hint: 1=b D .sa C tb/=b D sa=b C t .

4.83 * If ` � 0, m � 0, and 2`5mr terminates, prove that r terminates.

4.84 * Here’s a method used by many precollege texts for converting repeating dec-

imals to fractions. Suppose that you want to convert :324324 : : : to a fraction.

Calculate like this: If x D :324324 : : :, then 1000x D 324:324324 : : : , and

1000x � x D 999x D 324:

Hence, x D 324=999: (i) There is a hidden assumption about geometric series in this method. Where is

it?

(ii) Try this method with the following decimal expansions:

(a) :356356 : : : (b) :5353 : : :

(c) :2222 : : : (d) :07593 : : :

(e) :0123563563: : :

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182 Chapter 4 Modular Arithmetic

4.85 What’s wrong with the following calculation? Let x D 1 C 2 C 22 C : : : . Then

x D 1 C 2 C 22 C : : : 2x D 2 C 22 C 23 C : : : :

Subtract the top equation from the bottom to obtain x D �1.

4.86 Calculate decimal expansions for the followings fractions using long division. For

each one, what other fractions-to-decimal expansions (if any) do you get for free?

(i) 1

3 (ii)

1

6 (iii)

1

9

(iv) 1

15 (v)

4

15 (vi)

7

15

(vii) 1

8 (viii)

1

13 (ix)

1

20

(x) 1

19 (xi)

1

31 (xii)

1

37

Periods and Blocks

What is the period of a “unit fraction” 1=n? Our result will come from taking aCorollary 4.62 says that the period of a=n is equal

to the period of 1=n if

gcd.a; n/ D 1.

closer look at how decimal expansions are calculated; the analysis generalizes

to the decimal expansion of any rational number.

0.076923. . .

13 �

1.000000. . .

0 0

1 00

91

90

78

120

117

30

26

40

39

1

0.153846. . .

13 �

2.000000. . .

1 3

70

65

50

39

110

104

60

52

80

78

2

Figure 4.8. Decimal expansions of 1=13 and 2=13.

An analysis of the calculation for 1=13 yields another insight (see Figure

4.8). Pretend that the decimal point isn’t there, so we are dividing 1;000;000 D 106 by 13. Since 1 appears as a remainder, the initial sequence of remainders

will repeat, and the period of 1=13 is 6. Thus, the period of 1=13 is the smallest power of 10 congruent to 1 mod 13. In other words, the period of 1=13 is the

order of 10 in Z13 (see Exercise 4.49 on page 165).

We will generalize this observation in Theorem 4.61: the period of any frac-

tion 1=n is the order of 10 in Zn as long as there is some positive integer e with

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4.5 Connections: Patterns in Decimal Expansions 183

10e � 1 mod n. But, by Exercise 4.48 on page 165, some power of 10 is con- gruent to 1 mod n if and only if 10 is a unit in Zn. Now this condition is

equivalent to gcd.10; n/ D 1; that is, if and only if n is not of the form 2u5v . Thus, Proposition 4.59 shows why the dichotomy of terminating and repeating

rationals is so natural.

To prove the general result for 1=n, we just need to make sure that the first

remainder that shows up twice is, in fact, 1. That’s the content of the next

lemma.

Lemma 4.60. If gcd.10; n/ D 1, then 1 occurs as a remainder in the long di- vision of 1 by nI moreover, there cannot be two identical remainders occurring before 1 occurs.

Proof. As we saw above, because gcd.10; n/ D 1, a remainder of 1 will first appear in the long division after e steps where e is the order of 10 in Zn.

We must prove that there is no repeat of some other remainder before that

remainder of 1 shows up. First of all, there can’t be an earlier 1 (why?). Next,

suppose you see the same remainder, say c, occurring earlier, say, at steps e1 < e2 < e. Then we’d have

10e1 � 10e2 � c mod n:

Since 10 is a unit in Zn, this would imply that

1 � 10e2�e1 mod n:

And since e2 � e1 < e, this would contradict the fact that e is the order of 10 in Zm.

Putting it all together, we have a refinement of Propositions 4.58 and 4.59:

Theorem 4.61. If n > 0 is an integer, then 1=n either terminates or repeats.

(i) 1=n terminates if and only if n D 2u5v for nonnegative integers u and v. (ii) If gcd.n; 10/ D 1, then 1=n repeats with period m, where m is the order

of 10 in Zn.

Proof. Part (i) was proved in Proposition 4.59. The essence of the proof of part (ii) lies in the discussion on page 182 about the decimal expansion of 1=13:

the expansion for 1=n repeats after e steps, where e is the order of 10 in Zn;

that is, the first occurrence of remainder 1 occurs at the eth step of the long

division. And Lemma 4.60 shows that there can be no earlier occurrences.

So, if gcd.n; 10/ D 1, then 1=n repeats, and we know that its period is the order of 10 in Zn. What about fractions of the form a=n? The next corollary

shows that the same thing is true, as long as the fraction is in lowest terms.

Corollary 4.62. If gcd.a; n/ D 1 and a < n, then the period of a=n is the same as that of 1=n, namely the order of 10 in Zn.

Proof. Suppose the period of a=n is `. Then, arguing as in Lemma 4.60, the

expansion will repeat only after the remainder a occurs in the long division

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184 Chapter 4 Modular Arithmetic

p period for 1p

3 1

7 6

11 2

13 6

17 16

19 18

23 22

29 28

31 15

37 3

41 5

43 21

47 46

53 13

59 58

61 60

67 33

71 35

73 8

79 13

83 41

89 44

p period for 1p

97 96

101 4

103 34

107 53

109 108

113 112

127 42

131 130

137 8

139 46

149 148

151 75

157 78

163 81

167 166

173 43

179 178

181 180

191 95

193 192

197 98

199 99

Figure 4.9. Periods of 1=p for small primes p.

of a by n (see Exercise 4.88 on page 190). But this implies that ` is the smallest

positive integer such that

a 10` � a mod n:

Since a is a unit in Zn, multiplying by a �1 gives

10` � 1 mod n:

It follows that ` D m, the order of 10 in Zn.

Theorem 4.61 doesn’t answer every question about the periods of 1=p,

where p is a prime other than 2 or 5. Sometimes the period is p � 1, as when p D 7, but this not always so, for 1=13 has period 6, not 12. In all the entries in Figure 4.9, we see that periods of 1=p are divisors of p � 1. This turns out to be always true, and you’ll prove it soon. What about non-prime denominators?

Perhaps the length of the period of the expansion of 1=n is a factor of n � 1? No such luck: 1=21 D :047619047619 : : : has period 6 which is not a divisorBut stay tuned—we’ll

return to the period of 1=n

shortly. of 20.

Historical Note. In Disquisitiones Arithmeticae [14], Gauss conjectured

that there are infinitely many primes p that have the property that the deci-

mal expansion for 1=p has period p � 1. Gauss’s conjecture can be restated as

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4.5 Connections: Patterns in Decimal Expansions 185

follows: there are infinitely many primes p for which the order of 10 in Zp is

p � 1. E. Artin generalized Gauss’s conjecture. He claimed that if b is a posi- See Exercises 4.92 and 4.93 on page 190.tive integer that is not a perfect square, then there are infinitely many primes p

for which the b-adic expansion of 1=p has period p � 1. These are still con- jectures (as Gauss’s conjecture above), and very celebrated ones at that. Many

seemingly simple questions in arithmetic are extremely hard to answer.

But some things are known. For example, Gauss proved in Disquisitiones

that for any prime p, there is always at least one number (not necessarily 10) whose order in Zp is p � 1. Such a number is called a primitive root mod p.

We now know that the period of 1=n, where gcd.10; n/ D 1, is the order of 10 in Zn. In Exercise 4.49 on page 165, you did some calculations of orders of

units. We can now say a little more.

Theorem 4.63. If u is a unit in Zn, then

u�.n/ D 1

where � is the Euler �-function.

Proof. By Proposition 4.39, there are �.n/ units in Zn. Suppose we list them

all:

u1; u2; : : : ; u�.n/:

One of these units is u. Now multiply all these units by u; you get

uu1; uu2 : : : ; uu�.n/:

All these elements are units (Exercise 4.44 on page 165), and they are distinct

(Exercise 4.89 on page 190). This means that the second list contains all the

units, perhaps in a different order (they are distinct units, and there are �.n/ of them). Now multiply all the units together, first using the original order, and

then using the permuted order:

�.n/Y

iD1 ui D

�.n/Y

iD1 uui D u�.n/

�.n/Y

iD1 ui :

But Q�.n/ iD1 ui is a unit (Exercise 4.44 again), so you can divide both sides by

it, and the result follows.

Corollary 4.64. The order of a unit in Zn is a factor of �.n/.

Proof. Suppose that u is a unit in Zn with order e. Divide �.n/ by e to get a quotient and remainder:

�.n/ D qe C r 0 � r < e:

Then

u�.n/ D uqeCr D .ue/q ur :

Now use Theorem 4.63 and the fact that e is the minimal positive exponent

such that ue D 1 to conclude that r D 0.

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186 Chapter 4 Modular Arithmetic

Specializing to u D 10, we have

Theorem 4.65. If n is relatively prime to 10 .that is, if 1=n repeats/, then the

period of 1=n is a divisor of �.n/.

This greatly reduces the number of possibilities. For example, all we could

say about the length of the period of 1=231 before is that it is at most 230. Now

we can say it is a factor of �.231/ D 120. Which one is it?

Example 4.66. We saw earlier, on page 184, that 1=21 D :047619047619 : : : , so that the period of 1=21 is 6. Now �.21/ D 12 and, of course, 6 is a divisor of 12. N

Proposition 4.63 gives us an added bonus: another proof of Fermat’s Little

Theorem.

Corollary 4.67. If p is a prime, then ap D a in Zp for all integers a.

Proof. As in the proof of Theorem 4.9, we have two cases. If p j a, then

ap � a � 0 mod p;

and ap D a in Zp . If gcd.a; p/ D 1, then a is a unit in Zp , and Proposition 4.63 gives

a�.p/ D 1

in Zp . But �.p/ D p � 1, because p is prime. Hence,

ap�1 D 1:

Multiplying both sides by a gives ap D a in Zp .

As another application, we know that the period of 1=n is at most n � 1. When is it as large as possible?

Corollary 4.68. If the period of 1=n is n � 1, then n is prime.

Proof. If n is not prime, then �.n/ < n � 1, and the period of 1=n is not n � 1.

How to Think About It. The converse of Corollary 4.68 is not true, as the

example of 1=13 shows. As we said on page 184, it’s still an open question

about which primes p have the property that the decimal expansion for 1=p

has maximal period. All we can say is that the decimal expansion is a divi- sor of �.p/ D p � 1, providing an explanation for the evidence gathered in Figure 4.9.

We have discovered information about periods of repeating rationals; let’s

now look a bit at their blocks. Before continuing, it’s worth working out some

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4.5 Connections: Patterns in Decimal Expansions 187

other decimal expansions to look for interesting patterns. For example, calcu-

late the decimal expansions of

1

8 ;

2

3 ;

1

15 ;

1

19 ;

2

19 ;

1

13 ;

1

20

to see whether you can come up with some conjectures for connections be-

tween the integers a and b and the blocks in the decimal expansion of a=b.

Figure 4.10 displays the digits in the blocks of k=7 for 1 � k � 6. Is there a way to explain where each rearrangement starts?

1

7 D :142857142857 : : :

2

7 D :285714285714 : : :

3

7 D :428571428571 : : :

4

7 D :571428571428 : : :

5

7 D :714285714285 : : :

6

7 D :857142857142 : : :

Figure 4.10. The expansions of k=7 for 1 � k < 7.

There are quite a few patterns here. For example, each block consists of six repeating digits—some “cyclic” permutation of 142857:

142857; 285714; 428571; 571428; 714285; 857142

It’s the sequence of remainders that explains the various decimal expansions

of k=7—what they are and why they are in a particular order. For example, in

calculating 6=7, you look down the remainder list and see where you get a 6.

The process for 6=7 will start there, as in Figure 4.11.

The point of Figure 4.11 is that you can “pick up” the calculation at any step in the process—in a way, the calculation of 6=7 is embedded in the calculation

of 1=7. So are the calculations for all the other k=7 for 2 � k � 5. So, the sequence of remainders in a long division provides the key to which

decimal expansions can be obtained from the same long division. For the ra-

tionals k=7, there were six remainders before things started to repeat, so we

get all the expansions 1=7; 2=7; : : : ; 6=7 from one calculation. But it isn’t al- ways the case that you get all the expansions for k=n (where 1 � k < n) from the calculation of 1=n. That only happens when the period for the decimal ex-

pansion of 1=n has the maximal length n � 1 (implying that n is prime). For example, for the various k=13, you need two calculations, because the period

of the expansion for 1=13 is 6, not 12. Earlier, on page 187, we listed the blocks for the various k=7, noting that

there seemed to be no apparent pattern to where each block starts. In fact, a

closer analysis of the long division gives us a way to calculate the digits in each

block. Consider again the calculation of the expansion for 6=7. As before, if

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188 Chapter 4 Modular Arithmetic

0. 142857. . .

7 1. 000000. . .

7

30

28

20

14

60

56

40

35

50

49

1

0 .857142. . .

7 6 . 000000. . .

56

40

35

50

49

10

7

30

28

20

14

6

Figure 4.11. The expansion of 6=7 from that of 1=7.

we “forget” the decimal point, each new remainder gives the remainder when 6

times a power of 10 is divided by 7. Referring to Figure 4.11, we have

6 � 6 � 1 mod 7 4 � 6 � 10 mod 7 5 � 6 � 100 mod 7 1 � 6 � 1000 mod 7 3 � 6 � 10000 mod 7 2 � 6 � 100000 mod 7 6 � 6 � 1000000 mod 7:

Now, these are the remainders, not the digits in the block. Still, we have an

interesting preliminary result.

Lemma 4.69. Let 1 � a < n, and suppose that gcd.10; n/ D gcd.a; n/ D 1. If e is the order of 10 in Zn, then the j th remainder in the long division calcu-

lation of a=n, where 0 � j � e, is the solution cj of the congruence

cj � a � 10j mod n

with 0 � cj < n.

Proof. Imagine dividing n into a with long division, up to j places. Suppose

that the remainder is cj :

.q1q2q3q4. . . qj

n � a. 0 0 0 0. . . 0

cj .

This says that

Look at the example of

2=13:

2 D .13 � :1/C 10�1 � 7 2 D .13 � :15/C 10�2 � 5 2 D .13� :13/C10�3 �11 and so on. a D .n � :q1 q2 q3 q4 : : : qj /C 10�j cj :

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4.5 Connections: Patterns in Decimal Expansions 189

Multiply both sides of the equation by 10j to find

a10j D .n � q1 q2 q3 q4 : : : qj /C cj :

This says that

cj � a � 10j mod n:

What about the digits in the blocks? As in Lemma 4.69, let gcd.10; n/ D gcd.a; n/ D 1 and e be the order of 10 in Zn. Then we know that the eth remainder is a, where

a � 10e � a mod n: (4.3)

What is the block? Our old friend the Division Algorithm gives the answer:

Theorem 4.70. Let gcd.10; n/ D gcd.a; n/ D 1. If e is the order of 10 in Zn, then the sequence of digits in the block of the decimal expansion for a=n is

a.10e � 1/ n

:

Proof. The above discussion shows that the block is the partial quotient up to a remainder of a in the division. Rewrite Eq. (4.3) as:

a � 10e D qnC a

Solving for q, we have the desired result.

Example 4.71. For the various k=7,

1

7 D :142857 : : : and 1.106 � 1/=7 D 142857

2

7 D :285714 : : : and 2.106 � 1/=7 D 285714

3

7 D :428571 : : : and 3.106 � 1/=7 D 428571

4

7 D :571428 : : : and 4.106 � 1/=7 D 571428

5

7 D :714285 : : : and 5.106 � 1/=7 D 714285

6

7 D :857142 : : : and 6.106 � 1/=7 D 857142:

Figure 4.12. The blocks of k=7 for 1 � k < 7.

It’s an interesting calculation to go through the same process for the vari-

ous k=13. N

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190 Chapter 4 Modular Arithmetic

Exercises

4.87 Find the order of 10 modulo n (if it exists) for each value of n, and verify that the

decimal expansion of 1=n has period equal to the order.

(i) 7 (ii) 9 (iii) 3 (iv) 6 (v) 8

(vi) 11 (vii) 13 (viii) 39 (ix) 22 (x) 41

(xi) 73 (xii) 79 (xiii) 123 (xiv) 71 (xv) 61

4.88 * Finish the proof of Corollary 4.62 by showing that the decimal expansion of

1=n, where 1 � a < n and gcd.a; n/ D 1, will repeat only after a remainder of a occurs in the long division of a by n.

4.89 If

L D fu1; u2; : : : ; u�.n/g

is the list of units in Zn and u is any unit, show that the elements of

uL D fuu1; uu2; : : : ; uu�.n/g

are all distinct.

4.90 Theorem 4.70 says that if gcd.10; n/ D gcd.a; n/ D 1 and e is the order of 10 in Zn, then the block in the decimal expansion of a=n is a.10

e � 1/=n. Why is this latter fraction an integer?

4.91 Suppose that gcd.10; n/ D gcd.a; n/ D 1, e is the order of 10 in Zn, and cj is the remainder when a � 10j is divided by n. If the block of a=n is :a1a2 : : : ae , show,Let a0 D a. for 1 � j � e, that

aj D 10cj�1 � cj

n :

4.92 Just as there are b-adic expansions of integers, there are also such expansions

for rational numbers. For example, if we are working in base 5, then 1=5 D :1, 1=52 D :01, 1=53 D :001, and so on. Find rational numbers (written as a=b in the usual way) that are equal to each 5-adic expansion.

(i) :2 (ii) :03 (iii) .1111. . .

(iv) :171717 : : : (v) :001001001: : :

4.93 Find the 5-adic expansion of each rational number

(i) 25 (ii) 4 25 (iii)

1 4

(iv) 124 (v) 17 24 (vi)

20 24

4.94 Show that a positive rational number has a terminating b-adic expansion for some

positive base b.

4.95 (i) What is the decimal expansion of 1=9801?

(ii) What is the period of this expansion?

Hint:

10000

9801 D 1�

1 � 1100 �2

4.96 (i) What is the decimal expansion of 1=9899?

(ii) What is the period of this expansion?

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5 Abstract Algebra Why do mathematics? The answer is simple: we want to understand a corner

of our universe. But we are surrounded by so many different things that it

makes sense to organize and classify, thereby imposing some order. Naturally, we draw on our experience, so we can decide what we think is important and

what is less interesting.

Numbers and calculations have been very useful for thousands of years, and

we have chosen to study them. In particular, we have seen that certain arith-

metic and geometric ideas help us understand how numbers behave. Some- times the connections are quite surprising: for example, the relation between

Pythagorean triples and the method of Diophantus. We have also developed

several tools to facilitate our work: an efficient notation and mathematical in-

duction; the complex numbers and congruences have also enhanced our view.

There are unexpected consequences. As we investigate, we find that even

when we find a satisfying answer, new, interesting questions arise. Even though the method of Diophantus explains almost every question we might have about

Pythagorean triples, it also suggests that we replace the unit circle by other

conic sections, thereby giving insight into some calculus.

It is now time to organize the number theory we have studied. The main

idea is to abstract common features of integers, rational numbers, complex numbers, and congruences, as we did when we introduced the definition of

commutative ring. This will further our understanding of number theory itself

as well as other important topics, such as polynomials.

This chapter continues this adventure. In Section 5.1, we study domains,

an important class of commutative rings. In Section 5.2, we study polynomi- als, one of the most important examples of commutative rings. We will show,

in particular, that any commutative ring can serve as coefficients in a ring of

polynomials. Section 5.3 introduces homomorphisms, which allows us to com-

pare and contrast commutative rings, as well as to make precise the idea that

two rings have structural similarities.

The rest of the chapter is devoted to the structure of rings of polynomi- als. Using the results developed in Section 5.3, we’ll see how the two main

rings in high school mathematics—Z and polynomials in one variable with

coefficients in a field—share many structural similarities. For example, every

polynomial has a unique factorization as a product of primes (primes here are

called irreducible polynomials). And we’ll also revisit many of the theorems from advanced high school algebra, like the factor theorem and the fact that

polynomials of degree n have at most n roots, putting these results in a more

general setting.

191

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192 Chapter 5 Abstract Algebra

5.1 Domains and Fraction Fields

We now introduce a class of commutative rings that satisfy a property enjoyed

by our favorite rings: any product of nonzero integers is nonzero. On the other

hand, there are commutative rings in which a product of nonzero elements is 0.

For example, 2� 3 D 0 in Z6, even though both 2 and 3 are nonzero. We now promote this property to a definition, for there are interesting examples (e.g., polynomials) where it occurs.

Definition. A domain D is a nonzero commutative ring in which every prod-Many texts say integral domain instead of domain. uct of nonzero elements is nonzero.

A nonzero element a in a commutative ring R is called a zero divisor if

there is a nonzero b 2 R with ab D 0. Using this language, we can describe a domain as a commutative ring without zero divisors.

The commutative ring of integers Z is a domain, but Zm is not a domain when m is composite: if m D ab for 0 < a � b < m, then a ¤ 0 and b ¤ 0, but ab D m D 0. Recall the Boolean ring 2X in Example 4.47: its elements are all the subsets of a set X , and its operations are symmetric difference and

intersection. If X has at least two elements, then there are nonempty disjoint

subsets A and B; that is, A\ B D ¿. Thus, A and B are nonzero elements of 2X whose product AB D 0, and so 2X is not a domain.

How to Think About It. Everyone believes that Z is a domain—the product

of two nonzero integers is nonzero—but a proof from first principles is surpris-

ingly involved. If you grant that Z sits inside R, a fact that is a cornerstone of elementary school arithmetic using the “number line representation” of R, and

if you grant the fact that R is a field, then there is a simple proof (see Proposi-

tion 5.3). But that’s a fair amount of “granting.” We’ll simply assume that Z is

a domain.

Proposition 5.1. A nonzero commutative ring D is a domain if and only if it

satisfies the cancellation law: If ab D ac and a ¤ 0, then b D c.

Proof. Assume thatD is a domain. If ab D ac and a ¤ 0, then 0 D ab�ac D a.b � c/. Since a ¤ 0, we must have b � c D 0. Hence, b D c and the cancellation law holds.

Conversely, suppose that ab D 0, where both a and b are nonzero. Rewrite this as ab D a0. Since a ¤ 0 and the cancellation law holds, we have b D 0, a contradiction. Hence, D is a domain.

Corollary 5.2. Every field F is a domain.

Proof. The cancellation law holds: if a 2 F is nonzero and ab D ac, then a�1ab D a�1ac and b D c.

Proposition 5.3. Every subring S of a field F is a domain.

Proof. By Corollary 5.2, F is a domain. If a; b 2 S are nonzero, then their product (in F , and hence in S ) is also nonzero. Hence, S is a domain.

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5.1 Domains and Fraction Fields 193

For example, if we assume that R is a field and Z is a subring of R, then Z is

a domain. The proof of Proposition 5.3 shows more: every subring of a domain

is a domain.

Fraction Fields

The converse of Proposition 5.3—every domain is a subring of a field—is much more interesting than the proposition. Just as the domain Z is a sub-

ring of the field Q, so, too, is any domain a subring of its fraction field. We’ll

construct such a field containing a given domain using the construction of Q

from Z as inspiration. This is not mere generalization for generalization’s sake;

we shall see, for example, that it will show that certain polynomial rings are subrings of fields of rational functions.

How to Think About It. Warning! Over the years, school curricula have

tried using the coming discussion to teach fractions to precollege students,

even to fourth graders. This is a very bad idea. Experience should precede

formalism and, in this particular case, introducing rational numbers as ordered pairs of integers was a pedagogical disaster.

Elementary school teachers often say that 2 4

D 3 6

because 2 � 6 D 4 � 3. Sure enough, both products are 12, but isn’t this a non sequitur? Does it make

sense? Why should cross multiplication give equality? Teachers usually con-

tinue: suppose you have two pizzas of the same diameter, the first cut into four pieces of the same size, the second into six pieces of the same size; eating two

slices of the first pizza is just as filling as eating three slices of the second. This

makes more sense, and it tastes better, too. But wouldn’t it have been best had

the teacher said that if a=b D c=d , then multiplying both sides by bd gives ad D bc; and, conversely, if ad D bc, multiplying both sides by d�1b�1 gives a=b D c=d?

What is 1 2

? What is a fraction? A fraction is determined by a pair of integers–

its numerator and denominator—and so we start with ordered pairs. Let X be

the set of all ordered pairs .a; b/ of integers with b ¤ 0 (informally, we are thinking of a=b when we write .a; b/). Define cross multiplication to be the

relation on X

.a; b/ � .c; d / if ad D bc:

This is an equivalence relation. It is reflexive: .a; b/ � .a; b/ because ab D ba. It is symmetric: if .a; b/ � .c; d /, then .c; d / � .a; b/ because ad D bc implies cb D da. We claim it is transitive: if .a; b/ � .c; d / and .c; d / � .e; f /, then .a; b/ � .e; f /. Since .a; b/ � .c; d /. we have ad D bc, so that adf D bcf ; similarly, .c; d / � .e; f / gives cf D de, so that bcf D bde. Thus,

adf D bcf D bde:

Hence, adf D bde and, canceling d (which is not 0), gives af D be; that is, .a; b/ � .e; f /.

How to Think About It. One reason cross multiplication is important is that

it converts many problems about fractions into problems about integers.

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194 Chapter 5 Abstract Algebra

Lemma 5.4. If D is a domain and X is the set of all .a; b/ 2 D � D with b ¤ 0, then cross multiplication is an equivalence relation on X .

Proof. The argument given above for Z is valid forD. The assumption thatD

is a domain is present so that we can use the cancellation law to prove trans- itivity.

Notation. If D is a domain, the equivalence class of .a; b/ 2 X � D �D is denoted by

Œa; b:

Specialize Lemma A.16 in Appendix A.2 to the relation � on X : Œa; b D Œc; d  if and only if .a; b/ � .c; d /; that is, Œa; b D Œc; d  if and only if ad D bc.

Let’s finish the story in the context of arbitrary domains.

Definition. The fraction field of a domain D isFraction field? Stay tuned.

Frac.D/ D fŒa; b W a; b 2 D and b ¤ 0g:

How to Think About It. In the back of our minds, we think of Œa; b as

the fraction a=b. But, in everyday experience, fractions (especially rational numbers) are used in calculations—they can be added, multiplied, subtracted,

and divided. The next theorem equips Frac.D/ with binary operations that will

look familiar to you if you keep thinking that Œa; b stands for a=b.

Theorem 5.5. Let D be a domain.

(i) Frac.D/ is a field if we define

Œa; bC Œc; d  D Œad C bc; bd  and Œa; bŒc; d D Œac; bd :

(ii) The subset D0 of Frac.D/, defined by

D0 D fŒa; 1 W a 2 Dg;

is a subring of Frac.D/.

(iii) Every h 2 Frac.D/ has the form uv�1, where u; v 2 D0.

Proof. (i) Define addition and multiplication on F D Frac.D/ as in the statement. The symbols ŒadCbc; bd and Œac; bd  in the definitions make sense, for b ¤ 0 and d ¤ 0 imply bd ¤ 0, because D is a domain. The proof that F is a field is now a series of routine steps.

We show that addition F � F ! F is well-defined (i.e., single- valued): if Œa; b D Œa0; b0 and Œc; d  D Œc0; d 0, then Œad C bc; bd  D Œa0d 0 C b0c0; b0d 0. Now ab0 D a0b and cd 0 D c0d . Hence,

.ad C bc/b0d 0 D adb0d 0 C bcb0d 0 D .ab0/dd 0 C bb0.cd 0/ D a0bdd 0 C bb0c0d D .a0d 0 C b0c0/bd I

that is, .ad C bc; bd/ � .a0d 0 C b0c0; b0d 0/, as desired. A similar com- putation shows that multiplicationF � F ! F is well-defined.

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5.1 Domains and Fraction Fields 195

The verification that F is a commutative ring is also routine, and it is

left as Exercise 5.5 below, with the hints that the zero element is Œ0; 1,

the identity is Œ1; 1, and the negative of Œa; b is Œ�a; b. To see that F is a field, observe first that if Œa; b ¤ 0, then a ¤ 0

(for the zero element of F is Œ0; 1 D Œ0; b). We claim that the inverse of Œa; b is Œb; a, for Œa; bŒb; a D Œab; ab D Œ1; 1. Therefore, every nonzero element of F has an inverse in F .

(ii) We show thatD0 is a subring of F :

Œ1; 1 2 D0

Œa; 1C Œc; 1 D ŒaC c; 1 2 D0

Œa; 1Œc; 1D Œac; 1 2 D0:

(iii) If h D Œa; b, where b ¤ 0, then

h D Œa; 1Œ1; bD Œa; 1Œb; 1�1:

Notation. From now on, we use standard notation: If D is a domain, then the But be careful: for arbitrary fraction fields, the notation

a=b is just an alias for

Œa; b. For Q, the notation

is loaded with all kinds of

extra meanings that don’t

carry over to the general

setting (for example, as a

number having a decimal

expansion obtained by

dividing a by b).

element Œa; b in Frac.D/ will be denoted by

a=b:

Of course, Q D Frac.Z/. Not surprisingly, elementary school teachers are correct: it is, indeed, true that a=b D c=d if and only if ad D bc.

We started this section with two goals: to show that every domain is a sub-

ring of a field, and to make precise the notion of “fraction.” We’ve done the

second, but we didn’t quite show that a domain D is a subring of Frac.D/;

instead, we showed that D0 is a subring of Frac.D/, where D0 consists of all Œa; 1 for a 2 D. Now D and D0 do bear a strong resemblance to each other. If we identify each a in D with Œa; 1 in D0 (which is reminiscent of identify- ing an integer m with the fractionm=1), then not only do elements correspond

nicely, but so, too, do the operations: a C b corresponds to ŒaC b; 1:

Œa; 1C Œb; 1 D Œa � 1C 1 � b; 1 � 1 D ŒaC b; 1I

similarly, ab corresponds to Œab; 1 D Œa; 1Œb; 1. In Section 5.3, we will dis- cuss the important idea of isomorphism which will make our identification here

precise. For the moment, you may regard D and D0 as algebraically the same.

Exercises

5.1 Let R be a domain. If a 2 R and a2 D a, prove that a D 0 or a D 1. Compare with Exercise 4.40 on page 164.

5.2 Prove that the Gaussian integers ZŒi  and the Eisenstein integers ZŒ! are do-

mains.

5.3 * Prove that Zm is a domain if and only if Zm is a field. Conclude, using Theo-

rem 4.43, that Zm is a domain if and only if m is prime.

5.4 Prove that every finite domain D (i.e., jDj < 1) is a field.

Hint: Use the Pigeonhole Principle, Exercise A.11 on page 419.

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196 Chapter 5 Abstract Algebra

5.5 * Complete the proof of Theorem 5.5.

5.6 Let Q.i / D fr C si W r; s 2 Qg be the set of complex numbers whose real and imaginary parts are rational.

(i) Show that Q.i / is a field.

(ii) True or false? Frac .ZŒi / D fŒr C si;1 W r C si 2 Q.i /g :

5.7 *

(i) Show that Q.!/ D fr C s! W r; s 2 Qg is a field, where ! D e2�i=3 is a cube root of unity.

(ii) True or false? Frac .ZŒ!/ D fŒr C s!; 1 W r C s! 2 Q.!/g : Why?

5.2 Polynomials

You are surely familiar with polynomials; since they can be added and mul-

tiplied, it is not surprising that they form commutative rings. However, there are some basic questions about them whose answers may be less familiar. Is a

polynomial a function? Is x a variable? If not, just what is x? After all, we first

encounter polynomials as real-valued functions having simple formulas; for

example, f .x/ D x3 � 2x2 C 7 is viewed as the function f #W R ! R defined by f #.a/ D a3�2a2C7 for every a 2 R. But some polynomials have complex coefficients. Is it legitimate to consider polynomials whose coefficients lie in

any commutative ringR? When are two polynomials equal? Every high school

algebra student would say that the functions defined by f .x/ D x7 C 2x � 1 and g.x/ D 3x C 6 are not the same, because they are defined by different polynomials. But these two functions are, in fact, equal when viewed as func- tions Z7 ! Z7, a fact that you can check by direct calculation. Here’s another example. Is it legitimate to treat 2x C 1 as a polynomial whose coefficients lie in Z4? If so, then .2x C 1/2 D 4x2 C 4x C 1 D 1 (for 4 D 0 in Z4); that is, the square of this linear polynomial is a constant! Sometimes polynomials

are treated as formal expressions in which x is just a symbol, as, for example,

when you factor x6 � 1 or expand .x C 1/5. And sometimes polynomials are treated as functions that can be graphed or composed. Both of these perspec-

tives are important and useful, but they are clearly different.

We now introduce polynomials rigorously, for this will enable us to answer

these questions. In this section, we’ll first study polynomials from the formal

viewpoint, after which we’ll consider polynomial functions. In the next sec- tion, we will see that the notion of homomorphism will link the formal and the

function viewpoints, revealing their intimate connection.

How to Think About It. As we said on page 193 in the context of fractions,

rigorous developments should not be points of entry. One goal of this section is

to put polynomials on a firm footing. This will prepare you for any future work

you do with beginning algebra students, but it is in no way meant to take the

place of all of the informal experience that’s necessary before the formalities can be appreciated and understood.

We investigate polynomials in a very formal way,beginning with the allied

notion of power series. A key observation is that one should pay attention to

where the coefficients of polynomials live.

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5.2 Polynomials 197

Definition. If R is a commutative ring, then a formal power series over R is a

sequence

� D .s0; s1; s2; : : : ; si ; : : : /I

the entries si 2 R are called the coefficients of � .

Be patient. The reason for this terminology will be apparent in a few pages. In linear algebra, you may have seen the example of

the vector space V of all

polynomials of degree, say,

3 or less, with coefficients

in R. As a vector space,

V can be thought of as

R4, where the 4-tuple

.5; 6; 8; 9/ corresponds to

the polynomial 5C 6x C 8x2C 9x3.

In the meantime, pretend that .s0; s1; s2; : : : ; si ; : : : / is really s0 C s1x C s2x

2 C � � � C sixi C � � � . A formal power series � over R is a sequence, but a sequence is just a

function � W N ! R (where N is the set of natural numbers) with �.i/ D si for all i � 0. By Proposition A.2 in Appendix A.1, two sequences � and � are equal if and only if �.i/ D �.i/ for all i 2 N. So, formal power series are equal if and only if they are equal “coefficient by coefficient.”

Proposition 5.6. Formal power series � D .s0; s1; s2; : : : ; si ; : : : / and � D .t0; t1; t2; : : : ; ti ; : : : / over a commutative ringR are equal if and only if si D ti for all i � 0.

How to Think About It. Discussions of power series in calculus usually involve questions asking about those values of x for which s0 C s1xC s2x2 C � � � converges. In most commutative rings, however, limits are not defined, and so, in general, convergence of formal power series does not even make

sense. Now the definition of formal power series is not very complicated, while

limits are a genuinely new and subtle idea (it took mathematicians around 200 years to agree on a proper definition). Since power series are usually introduced

at the same time as limits, however, most calculus students (and ex-calculus

students!) are not comfortable with them; the simple notion of power series is

entangled with the sophisticated notion of limit.

Today’s calculus classes do not follow the historical development. Calcu-

lus was invented to answer a practical need; in fact, the word calculus arose because it described a branch of mathematics involving or leading to calcu-

lations. In the 1600s, navigation on the high seas was a matter of life and

death, and practical tools were necessary for the safety of boats crossing the

oceans. One such tool was calculus, which is needed in astronomical calcula-

tions. Newton realized that his definition of integral was complicated; telling a navigator that the integral of a function is some fancy limit of approxima-

tions and fluxions would be foolish. To make calculus useful, he introduced

power series (Newton discovered the usual power series for sin x and cos x),

he assumed that most integrands occurring in applications have a power series

expansion, and he further assumed that term-by-term integration was valid for them. Thus, power series were actually introduced as “long polynomials” in

order to simplify using calculus in applications.

Polynomials are special power series.

Definition. A polynomial over a commutative ringR is a formal power series

� D .s0; s1; : : : ; si ; : : : / overR for which there exists some integer n � 0with si D 0 for all i > n; that is,

� D .s0; s1; : : : ; sn; 0; 0; : : : /:

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198 Chapter 5 Abstract Algebra

The zero polynomial, denoted by � D 0, is the sequence � D .0; 0; 0; : : : /.

A polynomial has only finitely many nonzero coefficients; that is, it is a

“short power series.”

Definition. If � D .s0; s1; : : : ; sn; 0; 0; : : : / is a nonzero polynomial, thenSome authors define the degree of the zero

polynomial 0 to be �1, where �1C n D �1 for every integer n 2 N (this is sometimes convenient).

We choose not to assign

a degree to 0 because,

in proofs, it often must

be treated differently than

other polynomials.

there is n � 0 with sn ¤ 0 and si D 0 for all i > n. We call sn the leading coefficient of � , we call n the degree of � , and we denote the degree by n D deg.�/.

The zero polynomial 0 does not have a degree because it has no nonzero

coefficients.

Etymology. The word degree comes from the Latin word meaning “step.”

Each term six i (in the usual notation s0 C s1x C s2x2 C � � � C sixi C � � � ) has

degree i , and so the degrees suggest a staircase.

The word coefficient means “acting together to some single end.” Here, co-

efficients collectively give one formal power series or one polynomial.

Notation. If R is a commutative ring, then

RŒŒx

denotes the set of all formal power series over R, and

RŒx � RŒŒx

denotes the set of all polynomials over R.

We want to make RŒŒx into a commutative ring, and so we define addition

and multiplication of formal power series. Suppose that

� D .s0; s1; : : : ; si ; : : : / and � D .t0; t1; : : : ; ti ; : : : /:

Define their sum by adding term by term:

� C � D .s0 C t0; s1 C t1; : : : ; si C ti ; : : : /:

What about multiplication? The product of two power series is also computed

term by term; multiply formally and collect like powers of x:

.s0 C s1x C s2x2 C � � � C sixi C � � � /.t0 C t1x C t2x2 C � � � C tjxj C � � � / D s0.t0 C t1x C t2x2 C � � � /C s1x.t0 C t1x C t2x2 C � � � /C � � � D .s0t0 C s0t1x C s0t2x2 C � � � /C .s1t0x C s1t1x2 C s1t2x3 C � � � /C � � � D s0t0 C .s1t0 C s0t1/x C .s0t2 C s1t1 C s2t0/x2 C � � � :

Motivated by this, we define multiplication of formal power series by

�� D .s0t0; s0t1 C s1t0; s0t2 C s1t1 C s2t0; : : : /I

more precisely,

�� D .c0; c1; : : : ; ck; : : : /;

where ck D P iCjDk si tj D

P k iD0 si tk�i .

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5.2 Polynomials 199

Proposition 5.7. If R is a commutative ring, then RŒŒx, together with the

operations of addition and multiplication defined above, is a commutative ring.

Proof. Addition and multiplication are operations onRŒŒx: the sum and prod-

uct of two formal power series are also formal power series. Define zero to be

the zero polynomial, define the identity to be the polynomial .1; 0; 0; : : : /, and

define the negative of .s0; s1; : : : ; si ; : : : / to be .�s0;�s1; : : : ;�si ; : : : /. Veri- fications of the axioms of a commutative ring are routine, and we leave them as

Exercise 5.8 on page 202. The only difficulty that might arise is proving the as- sociativity of multiplication. Hint: if � D .r0; r1; : : : ; ri ; : : : /, then the `th co- ordinate of the polynomial �.��/ turns out to be

P iCjCkD` ri .sj tk/, while the

`th coordinate of the power series .��/� turns out to be P iCjCkD`.ri sj /tk ;

these are equal because associativity of multiplication in R gives ri .sj tk/ D .ri sj /tk for all i; j; k.

We’ll see in a moment that the subset RŒx of polynomials is a subring of

the commutative ring of formal power series RŒŒx.

Lemma 5.8. Let R be a commutative ring and � , � 2 RŒx be nonzero poly- nomials.

(i) Either �� D 0 or deg.��/ � deg.�/ C deg.�/. (ii) If R is a domain, then �� ¤ 0 and

deg.��/ D deg.�/C deg.�/:

Proof. Let � D .s0; s1; : : : / have degree m, let � D .t0; t1; : : : / have degree n, and let �� D .c0; c1; : : : /.

(i) It suffices to prove that ck D 0 for all k > mC n. By definition,

ck D s0tk C � � � C smtk�m C smC1tk�m�1 C � � � skt0:

All terms to the right of smtk�m are 0, because deg.�/ D m, and so si D 0 for all i � m C 1. Now smtk�m, as well as all the terms to its left, are 0, because deg.�/ D n, and so tj D 0 for all j � k �m > n.

(ii) We claim that cmCn D smtn, the product of the leading coefficients of � and � . Now

cmCn D X

iCjDmCn si tj

D s0tmCn C � � � C sm�1tnC1 C smtn C smC1tn�1 C � � � :

We show that every term si tj in cmCn, other than smtn, is 0. If i < m, then m � i > 0; hence, j D m � i C n > n, and so tj D 0; that is, each term to the left of smtn is 0. If i > m, then si D 0, and each term to the right of smtn is 0. Therefore,

cmCn D smtn:

If R is a domain, then sm ¤ 0 and tn ¤ 0 imply smtn ¤ 0; hence, cmCn D smtn ¤ 0, �� ¤ 0, and deg.��/ D mC n:

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200 Chapter 5 Abstract Algebra

Corollary 5.9. (i) If R is a commutative ring, thenRŒx andR are subrings

of RŒŒx.

(ii) If R is a domain, thenRŒx is a domain.Exercise 5.22 on page 203 shows that if R is a

domain, then RŒŒx is a

domain. Proof. (i) Let �; � 2 RŒx. Now � C � is a polynomial, for either � C � D 0

or deg.� C �/ � maxfdeg.�/; deg.�/g. By Lemma 5.8(i), the product of two polynomials is also a polynomial. Finally, 1 D .1; 0; 0; : : :/ is a polynomial, and so RŒx is a subring of RŒŒx.

It is easy to check that R0 D f.r; 0; 0; : : : / W r 2 Rg is a subring of RŒx, and we may view R0 as R by identifying r 2 R with .r; 0; 0; : : : /.See Exercise 5.9 on

page 202. (ii) If � and � are nonzero polynomials, then Lemma 5.8(ii) shows that �� ¤

0. Therefore, RŒx is a domain.

We remark that R can’t be a subring of RŒx or of RŒŒx because it’s not

even a subset of these rings. This is why we have introduced the subring R0. A similar thing happened when we couldn’t view a domain D as a subring

of its fraction field Frac.D/. We shall return to this point when we discuss isomorphisms.

From now on, we view RŒx and RŒŒx as rings, not merely as sets.

Definition. IfR is a commutative ring, then RŒx is called the ring of polyno-

mials over R, and RŒŒx is called the ring of formal power series over R.

Here is the link between this discussion and the usual notation.

Definition. The indeterminate x is the element

x D .0; 1; 0; 0; : : : / 2 RŒx:

How to Think About It. Thus, x is neither “the unknown” nor a variable; it is a specific element in the commutative ring RŒx, namely, the polynomial

.a0; a1; a2; : : : / with a1 D 1 and all other ai D 0; it is a polynomial of de- gree 1.

Note that we need the unit 1 in a commutative ring R in order to define the

indeterminate inRŒx.

Lemma 5.10. Let R be a commutative ring.

(i) If � D .s0; s1; : : : ; sj ; : : : / 2 RŒŒx, then

x� D .0; s0; s1; : : : ; sj ; : : : /I

that is, multiplying by x shifts each coefficient one step to the right.

(ii) If n � 0, then xn is the polynomial having 0 everywhere except for 1 in the nth coordinate.

(iii) If r 2 R and .s0; s1; : : : ; sj ; : : : / 2 RŒŒx, then

.r; 0; 0; : : : /.s0; s1; : : : ; sj ; : : : / D .rs0; rs1; : : : ; rsj ; : : : /:

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5.2 Polynomials 201

Proof. (i) Write x D .a0; a1; : : : ; ai ; : : : /, where a1 D 1 and all other ai D 0, and let x� D .c0; c1; : : : ; ck; : : : /. Now c0 D a0s0 D 0, because a0 D 0. If k � 1, then the only nonzero term in the sum ck D

P iCjDk ai sj

is a1sk�1 D sk�1, because ai D 0 for i ¤ 1; thus, for k � 1, the kth coordinate ck of x� is sk�1, and x� D .0; s0; s1; : : : ; si ; : : : /.

(ii) Use induction and part (i).

(iii) This follows from the definition of multiplication.

If we identify .r; 0; 0; : : : / with r , as in the proof of Corollary 5.9, then See Exercise 5.11 on page 202.Lemma 5.10(iii) reads

r.s0; s1; : : : ; si ; : : : / D .rs0; rs1; : : : ; rsi; : : : /:

We can now recapture the usual polynomial notation.

Proposition 5.11. LetR be a commutative ring. If � D .s0; s1; : : : ; sn; 0; 0; : : : / 2 RŒx has degree n, then

� D s0 C s1x C s2x2 C � � � C snxn;

where each element s 2 R is identified with the polynomial .s; 0; 0; : : : /. More- over, if � D t0 C t1x C t2x2 C � � � C tmxm, then � D � if and only if n D m So, two polynomials are

equal in RŒx if and only

if they are equal “term by

term.”

and si D ti for all i � 0.

Proof.

� D .s0; s1; : : : ; sn; 0; 0; : : : / D .s0; 0; 0; : : : /C .0; s1; 0; : : : /C � � � C .0; 0; : : : ; 0; sn; 0; : : : / D s0.1; 0; 0; : : : /C s1.0; 1; 0; : : : /C � � � C sn.0; 0; : : : ; 0; 1; 0; : : :/ D s0 C s1x C s2x2 C � � � C snxn:

The second statement merely rephrases Proposition 5.6, equality of polyno-

mials, in terms of the usual notation.

We shall use this familiar (and standard) notation from now on. As is cus-

tomary, we shall write

f .x/ D s0 C s1x C s2x2 C � � � C snxn

instead of � D .s0; s1; : : : ; sn; 0; 0; : : : /.

Corollary 5.12. If R is a commutative ring, then the polynomial ring RŒx is

infinite.

Proof. By Proposition 5.11, xi ¤ xj if i ¤ j .

If f .x/ D s0C s1xC s2x2C � � �C snxn, where sn ¤ 0, then s0 is called its constant term and, as we have already said, sn is called its leading coefficient.

If its leading coefficient sn D 1, then f .x/ is called monic. Every polynomial other than the zero polynomial 0 (having all coefficients 0) has a degree. A constant polynomial is either the zero polynomial or a polynomial of degree

0. Polynomials of degree 1, namely a C bx with b ¤ 0, are called linear, polynomials of degree 2 are quadratic, degree 3s are cubic, then quartics,

quintics, and so on.

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202 Chapter 5 Abstract Algebra

Etymology. Quadratic polynomials are so called because the particular qua-

dratic x2 gives the area of a square (quadratic comes from the Latin word

meaning four, which reminds us of the 4-sided figure); similarly, cubic polyno-

mials are so called because x3 gives the volume of a cube. Linear polynomials

are so called because the graph of a linear polynomial axC b in RŒx is a line.

Exercises

5.8 * Fill in the details and complete the proof of Proposition 5.7.

5.9 * Suppose that R is a commutative ring. In the proof of Corollary 5.9(i), we de-

fined R0 as the set of all power series of the form .r; 0; 0; 0; : : : / where r 2 R, and we said “we may view R0 as R by identifying r 2 R with .r; 0; 0; : : : /.” Show, if r; s 2 R, that (i) r C s is identified with .r; 0; 0; 0; : : : / C .s; 0; 0; 0; : : : /

(ii) rs is identified with .r; 0; 0; 0; : : : /.s; 0; 0; 0; : : : /.

5.10 If .t0; t1; t2; : : : / is a power series over R and r 2 R, show that

.r; 0; 0; 0; : : : /.t0; t1; t2; : : : / D .rt0; rt1; rt2; : : : /:

5.11 * Suppose that F is a field. Show that F ŒŒx is a vector space over F where

addition is defined as addition of power series and scalar multiplication is definedVector spaces over arbi- trary fields are discussed

in Appendix A.3. by

r.s0; s1; s2; : : : / D .rs0; rs1; rs2; : : : /:

5.12 If R is the zero ring, what are RŒx and RŒŒx? Why?

5.13 Prove that if R is a commutative ring, then RŒx is never a field.

Hint: If x�1 exists, what is its degree?

5.14 (i) Let R be a domain. Prove that if a polynomial in RŒx is a unit, then it is a

nonzero constant (the converse is true if R is a field).

Hint: Compute degrees.

(ii) Show that .2x C 1/2 D 1 in Z4Œx. Conclude that 2x C 1 is a unit in Z4Œx, and that the hypothesis in part (i) that R be a domain is necessary.

5.15 * If R is a commutative ring and

f .x/ D s0 C s1x C s2x2 C � � � C snxn 2 RŒx

has degree n � 1, define its formal derivative f 0.x/ 2 RŒx by

f 0.x/ D s1 C 2s2x C 3s3x2 C � � � C nsnxn�1I

if f is a constant polynomial, define its derivative to be the zero polynomial.

Prove that the usual rules of calculus hold for derivatives in RŒx:

.f C g/0 D f 0 C g0

.rf /0 D r.f 0/ if r 2 R .fg/0 D fg0 C f 0g

Œ1pt .f n/0 D nf n�1f 0 for all n � 1:

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5.2 Polynomials 203

5.16 Take It Further. Define R

W QŒx ! QŒx by Z

f D a0x C 12a1x 2 C � � � C 1nC1anx

nC1 2 QŒx;

where f .x/ D a0 C a1x C � � � C anxn 2 QŒx. (i) Prove that

R .f C g/ D

R f C

R g.

(ii) If D is the derivative, prove that D R

D 1QŒx, but that R

D ¤ 1QŒx.

1QŒx denotes the identity

function on the set QŒx.

Why didn’t we defineR WRŒx ! RŒx for any

commutative ring R? 5.17 *Preview. Let R be a commutative ring, let f .x/ 2 RŒx, and let f 0.x/ be its

derivative.

(i) Prove that if .x � a/2 is a divisor of f in RŒx, then x � a is a divisor of f 0 in RŒx.

(ii) Prove that if x � a is a divisor of both f and f 0, then .x � a/2 is a divisor of f .

5.18 (i) If f .x/ D ax2p C bxp C c 2 Zp Œx, prove that f 0.x/ D 0. (ii) Prove that a polynomial f .x/ 2 Zp Œx has f 0.x/ D 0 if and only if there is a

polynomial g.x/ D P

anx n with f .x/ D g.xp/; that is, f .x/ D

P anx

np .

5.19 If p is a prime, show, in Zp Œx, that

.x C 1/p D xp C 1:

5.20 *

(i) If R is a domain and � D 1 C x C x2 C � � � C xn C � � � 2 RŒŒx, prove that � is a unit in RŒŒx; in fact, .1 � x/� D 1.

(ii) Show that .1 � x/2 is a unit in QŒŒx, and express 1=.1 � x/2 as a power series.

Hint: See Exercise 5.22 below.

5.21 Show that 1 � x � x2 is a unit in QŒŒx, and express 1=.1 � x � x2/ as a power series.

5.22 *

(i) Prove that if R is a domain, then RŒŒx is a domain.

Hint: If � D .s0; s1; : : : / 2 RŒŒx is nonzero, define the order of � , denoted by ord.�/, to be the smallest n � 0 for which sn ¤ 0. If R is a domain and �; � 2 RŒŒx are nonzero, prove that ord.��/ � ord.�/ C ord.�/, and use this to conclude that �� ¤ 0.

(ii) Let k be a field. Prove that a formal power series � 2 kŒŒx is a unit if and only if its constant term is nonzero; that is, ord.�/ D 0.

(iii) Prove that if � 2 kŒŒx and ord.�/ D n, then � D xnu, where u is a unit in kŒŒx.

5.23 *

(i) Prove that Frac.ZŒx/ D Q.x/. (ii) Let D be a domain with K D Frac.D/. Prove that Frac.DŒx/ D K.x/.

5.24 (i) Expand .C2 C S2 � 1/.S2 C 2CS � C2/, where C and S are elements in some commutative ring.

(ii) Establish the trigonometric identity

cos2 x C 2 cos3 x sin x C 2 cos x sin3 x C sin4 x D cos4 x C 2 cos x sin x C sin2 x:

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204 Chapter 5 Abstract Algebra

5.25 Preview. Suppose p is a prime and

fp.x/ D xp � 1 x � 1 :

(i) Show that fp.x/ D xp�1 C xp�2 C � � � C 1. (ii) Show that fp.x C 1/ D xp in Zp Œx.

Polynomial Functions

Let’s now pass to viewing polynomials as functions. Each polynomial f .x/ D s0Cs1xCs2x2C� � �Csnxn 2 RŒx defines its associated polynomial function f #WR ! R by evaluation:

f #.a/ D s0 C s1aC s2a2 C � � � C snan 2 R;

where a 2 R (in this way, we can view the indeterminate x as a variable). But polynomials and polynomial functions are different things. For example,In Proposition 6.18, we will

see that there’s a bijection

between polynomials and

their associated polynomial

functions if R is an infinite

field.

Corollary 5.12 says, for every commutative ring R, that there are infinitely

many polynomials in RŒx. On the other hand, if R is finite (e.g., R D Zm), then there are only finitely many functions from R to itself, and so there are

only finitely many polynomial functions. Fermat’s Theorem (ap � a mod p for every prime p) gives a concrete example of distinct polynomials defining

the same polynomial function; f .x/ D xp�x is a nonzero polynomial, yet its associated polynomial function f #W Zp ! Zp is the constant function zero.

Recall Example 4.31: if R is a commutative ring, then Fun.R/ D RR, the set of all functions from R to itself, is a commutative ring under pointwise

operations. We have seen that every polynomial f .x/ 2 RŒx has an associated polynomial function f # 2 Fun.R/, and we claim that

Poly.R/ D ff # W f .x/ 2 RŒxg

is a subring of Fun.R/ (we admit that we are being very pedantic, but you

will see in the next section that there’s a good reason for this fussiness). The identity u of RR is the constant function with value 1, where 1 is the identity

element of R; that is, u D 1#, where 1 is the constant polynomial. We claim that if f .x/; g.x/ 2 RŒx, then

f # C g# D .f C g/# and f #g# D .fg/# :

(In the equation f # C g# D .f C g/#, the plus sign on the left means addition of functions, while the plus sign on the right means the usual addition of poly-

nomials in RŒx; a similar remark holds for multiplication.) The proof of these

equations is left as Exercise 5.27 on page 206.

Etymology. In spite of the difference between polynomials and polynomial

functions, RŒx is often called the ring of all polynomials in one variable over R.

Since kŒx is a domain when k is a field, by Corollary 5.9(ii), it has a frac-

tion field.

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5.2 Polynomials 205

Definition. If k is a field, then the fraction field Frac.kŒx/ of kŒx, denoted We can define R.x/ for arbitrary domains R. See

Exercise 5.23 on page 203. by

k.x/;

is called the field of rational functions over k.

How to Think About It. By convention, the elements of k.x/ are called

rational “functions” but they are simply elements of the fraction field for kŒx.

Of course, a rational function can be viewed as an actual function via evalu- ation at elements of k, in the same way that a polynomial in kŒx gives rise

to its associated polynomial function defined on k. But the domain of such a

rational function may not be all of k (why?).

We’ll use the standard notation for elements in fraction fields (introduced

on page 195) for rational functions over a field: Œf; g will be denoted by f=g.

Proposition 5.13. If p is prime, then the field of rational functions Zp.x/ is

an infinite field containing Zp as a subfield.

Proof. By Corollary 5.12, Zp Œx is infinite, because the powers x n, for n 2 N,

are distinct. Thus, its fraction field, Zp.x/, is an infinite field containing Zp Œx

as a subring. But Zp Œx contains Zp as a subring, by Corollary 5.9. Well, Zp .x/ contains a domain with a “strong

resemblance” to ZpŒx

(see page 195). We’ll make

this precise in Section 5.3.

Notation. We’ve been using Zp to stand for the integers mod p, and we know

that it is a field. There are other finite fields that do not have a prime number

of elements; you met one, F4, in Exercise 4.55 on page 165. It’s customary to

denote a field with q elements by

Fq:

It turns out that q D pn for some prime p and some n � 1. Moreover, there exists essentially only one field with q elements, a fact we’ll prove in Chap-

ter 7. In particular, there is only one field with exactly p elements and so, from

now on, we’ll use the notations

Zp D Fp interchangeably (we’ll use Fp when we’re viewing it as a field).

Let’s now consider polynomials overR in two variables x and y. A quadratic

polynomial ax2 C bxy C cy2 C dx C ey C f can be rewritten as

ax2 C .by C d/x C .cy2 C ey C f /I

that is, it is a polynomial in x with coefficients inRŒy. If we writeA D RŒy, then it is clear that AŒx is a commutative ring.

Definition. IfR is a commutative ring, thenRŒx; y D AŒx, whereA D RŒy, is the ring of all polynomials over R in two variables.

By induction, we can form the commutative ring RŒx1; x2; : : : ; xn of all

polynomials in n variables over R:

RŒx1; x2; : : : ; xnC1 D � RŒx1; x2; : : : ; xn

� ŒxnC1:

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206 Chapter 5 Abstract Algebra

Corollary 5.9 can now be generalized, by induction on n, to say that if D

is a domain, then so is DŒx1; x2; : : : ; xn; we call Frac.DŒx1; x2; : : : ; xn/ the

ring of rational functions in n variables. Exercise 5.23 on page 203 can be generalized to several variables: ifK D Frac.D/, then

Frac.DŒx1; x2; : : : ; xn/ D K.x1; x2; : : : ; xn/I

its elements have the form f=g, where f; g 2 KŒx1; x2; : : : ; xn and g ¤ 0.

Exercises

5.26 Let R be a commutative ring. Show that if two polynomials f .x/; g.x/ 2 RŒx are equal, then their associated polynomial functions are equal; that is, f # D g#.

5.27 * If R is a commutative ring, prove that Poly.R/ is a subring of Fun.R/ D RR.

5.28 True or false, with reasons:

(i) .x2 � 9/=.x2 � 2x � 3/ D .x C 3/=.x C 1/ in Q.x/. (ii) What are the domains of the functions x 7! .x2 � 9/=.x2 � 2x � 3/ and

x 7! .x C 3/=.x C 1/? Are the functions equal?

5.3 Homomorphisms

The question whether two given commutative rings R and S are somehow the

same has already arisen, at least twice.

(i) On page 195 we said

For the moment, you may regard D and D0 as algebraically the same.

(ii) And on page 201 we said

If we identify .r; 0; 0; : : : / with r , then Lemma 5.10(iii) reads

r.s0; s1; : : : ; si ; : : : / D .rs0; rs1; : : : ; rsi; : : : /:

What does “the same” mean in statement (i)? What does “identify” mean in

statement (ii)? More important, ifR is a commutative ring, we wish to compare the (formal) polynomial ring RŒx with the ring Poly.R/ of all polynomial

functions on R.

We begin our discussion by considering the ring Z2; it has two elements, the

congruence classes 0; 1, and the following addition and multiplication tables.

C 0 1 0 0 1

1 1 0

� 0 1 0 0 0

1 0 1

The two words even, odd also form a commutative ring, call it P; its addi-

tion and multiplication are pictured in the following tables.

C even odd even even odd

odd odd even

� even odd even even even

odd even odd

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5.3 Homomorphisms 207

Thus, oddCodd D even and odd�odd D odd. It is clear that the commutative rings Z2 and P are distinct; on the other hand, it is equally clear that there

is no significant difference between them. The elements of Z2 are given in terms of numbers; those of P in terms of words. We may think of P as a trans-

lation of Z2 into another language. And more than just a correspondence of

elements, the operations of addition and multiplication (that is, the two tables)

get translated, too.

A reasonable way to compare two systems is to set up a function between

them that preserves certain essential structural properties (we hinted at this

idea earlier when we noted that a ring R is essentially a subring of RŒx). The notions of homomorphism and isomorphism will make this intuitive idea

precise. Here are the definitions; we will discuss what they mean afterward.

Definition. Let R and S be commutative rings. A homomorphism is a func-

tion 'WR ! S such that, for all a; b 2 R, (i) '.a C b/ D '.a/C '.b/,

(ii) '.ab/ D '.a/'.b/, (iii) '.1/ D 1 (the 1 on the left-hand side is the identity of R; the 1 on the

right-hand side is the identity of S ).

If ' is also a bijection, then ' is called an isomorphism. Two commutative

rings R and S are called isomorphic, denoted by R Š S , if there exists an isomorphism 'WR ! S between them.

In the definition of a homomorphism 'WR ! S , the + on the left-hand side is addition in R, while the + on the right-hand side is addition in S ; similarly

for products. A more complete notation for a commutative ring would display its addition, multiplication, and unit: instead ofR, we could write .R;C;�; 1/. Similarly, a more complete notation for S is .S;˚;˝; e/. The definition of homomorphism can now be stated more precisely

(i) '.a C b/ D '.a/˚ '.b/, (ii) '.a � b/ D '.a/˝ '.b/,

(iii) '.1/ D e.

Etymology. The word homomorphism comes from the Greek homo, mean-

ing “same,” and morph, meaning “shape” or “form.” Thus, a homomorphism

carries a commutative ring to another commutative ring of similar form. The

word isomorphism involves the Greek iso, meaning “equal,” and isomorphic rings have identical form.

Consider the two simple examples above of addition tables arising from the rings Z2 and P (the symbol P stands for “parity.”). The rings Z2 and P

Z2W 0 1

1 0 I PW

even odd

odd even :

Figure 5.1. Addition tables.

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208 Chapter 5 Abstract Algebra

are isomorphic, for the function 'W Z2 ! P, defined by '.0/ D even and '.1/ D odd, is an isomorphism, as the reader can quickly check (of course, you must also check the multiplication tables).

Let a1; a2; : : : ; aj ; : : : be a list with no repetitions of all the elements of a

ringR. An addition table forR is a matrix whose ij entry is ai C aj .

C a1 � � � aj � � � a1 a1 C a1 � � � a1 C aj � � �

ai ai C a1 � � � ai C aj � � �

A multiplication table for R is defined similarly.

The addition and multiplication tables for a ring R depend on the listing of its elements, so that a ring has many tables. Let a1; a2; : : : ; aj ; : : : be a list of

all the elements of a ringR with no repetitions. If S is a ring and 'WR ! S is a bijection, then '.a1/; '.a2/; : : : ; '.aj /; : : : is a list of all the elements of S

with no repetitions, and so this latter list determines addition and multiplication

tables for S . That ' is an isomorphism says that if we superimpose the tables

forR (determined by a1; a2; : : : ; aj ; : : : ) upon the tables for S (determined by '.a1/; '.a2/; : : : ; '.aj /; : : : ), then the tables match. In more detail, if ai C aj is the ij entry in the given addition table of R, then '.ai / C '.aj / is the ij entry of the addition table of S . But '.ai /C'.aj / D '.ai Caj /, because ' is an isomorphism. In this sense, isomorphic rings have the same addition tables

and the same multiplication tables (see Figure 5.1). Informally, we say thatDoes a homomorphism “preserve 0”? See

Lemma 5.17. a homomorphism preserves addition, multiplication, and 1. Thus, isomorphic

rings are essentially the same, differing only in the notation for their elements

and their operations.

Here are two interesting examples of homomorphisms: the first will be used

often in this book; the second compares the two different ways we view poly- nomials.

Example 5.14. (i) Reduction mod m:

We didn’t have the language to say it at the time, but Proposition 4.5

sets up a homomorphism rmW Z ! Zm for any nonnegative integer m, namely, rmWn 7! Œn. It’s not an isomorphism because Z is infinite and Zm is finite (so there can’t be any bijection between them). Another rea- son is that rm.m/ D 0 D rm.2m/, so rm can’t be injective. Is it surjective?

(ii) Form to Function:

In Example 4.31, we saw that every f .x/ 2 RŒx, where R is a com- mutative ring, determines its associated polynomial function f #WR ! R. The function 'Wf 7! f # is a homomorphism RŒx ! Fun.R/ D RR: as we saw on page 204, addition of polynomials corresponds to pointwise

addition of polynomial functions .f C g/# D f # C g#, and multiplica- tion of polynomials corresponds to pointwise multiplication of polyno- mial functions .fg/# D f #g#. Is ' an isomorphism? No, because it’s not surjective—not every function on R is a polynomial function. Is '

an isomorphism between RŒx and the subring im' D Poly.R/ of RR consisting of all polynomial functions? That depends on R. We’ve seen,

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5.3 Homomorphisms 209

for example, that f .x/ D x7 C 2x � 1 and g.x/ D 3x C 6 give the same function (f # D g#) when R D Z7. But we’ll see, in Theorem 6.20, that if R is an infinite field (as it almost always is in high school), then 'WRŒx ! Poly.R/ is an isomorphism. N

As with all important definitions in mathematics, the idea of homomor- phism existed long before the name. Here are some examples that you’ve en-

countered so far.

Example 5.15. (i) Complex conjugation z D a C ib 7! z D a � ib is a homomorphism C ! C, because 1 D 1; z C w D zCw, and zw D z w; it is a bijection because z D z (so that it is its own inverse) and, therefore, complex conjugation is an isomorphism.

(ii) Let D be a domain with fraction field F D Frac.D/. In Theorem 5.5, we proved thatD0 D fŒa; 1 W a 2 Dg is a subring of F . We can now identify D0 with D, for the function 'WD ! D0, given by '.a/ D Œa; 1 D a=1, is an isomorphism.

(iii) In the proof of Corollary 5.9, we “identified” an element r in a com-

mutative ring R with the constant polynomial .r; 0; 0; : : : /. We said that

R is a subring of RŒx, but that is not the truth. The function 'WR ! RŒx, defined by '.r/ D .r; 0; 0; : : : /, is a homomorphism, and R0 D f.r; 0; 0; : : : / W r 2 Rg is a subring of RŒx isomorphic to R, and 'WR ! R0 is an isomorphism.

(iv) If S is a subring of a commutative ring R, then the inclusion i WS ! R is a ring homomorphism (this is one reason why we insist that the identity

of R lie in S ).

(v) Recall Example 4.47: if X is a set, then 2X is the Boolean ring of all the subsets of X , where addition is symmetric difference, multiplication is

intersection, and the identity is X . If Y is a proper subset ofX , then 2Y is

not a subring of 2X , for the identity of 2Y is Y , notX . Thus, the inclusion

i W 2Y ! 2X is not a homomorphism, even though i.aC b/ D i.a/C i.b/ and i.ab/ D i.a/i.b/. Therefore, the part of the definition of homomor- phism requiring identity elements be preserved is not redundant. N

Example 5.16. Example 4.31 on page 157 shows, for a commutative ring R

and a set X , that the family RX of all functions from X to R, equipped with pointwise addition and multiplication, is a commutative ring. We’ve also used

the notation 2X in Example 4.47 on page 166 to stand for the Boolean ring of

all subsets of a set X . The goal of this example is to prove that 2X and .Z2/ X

This example is rather

dense. It’s a good idea

to pick a concrete set,

say X D f1; 2; 3g, and work out the characteristic

function for each of the 8

subsets of X .

are isomorphic rings.

The basic idea is to associate every subset A � X with its characteristic function fA 2 .Z2/X , defined by

fA.x/ D ( 1 if x 2 A 0 if x 62 A:

We claim that ' W 2X ! .Z2/X , defined by The characteristic function fA is sometimes called the

indicator function, for it tells

you whether an element

x 2 X is or is not in A.

'.A/ D fA;

is an isomorphism.

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210 Chapter 5 Abstract Algebra

First, ' is a bijection:

(i) ' is injective: if A;B � X and '.A/ D '.B/, then fA D fB : for any x 2 X , we have fA.x/ D 1 if and only if fB .x/ D 1. Thus, x 2 A if and only if x 2 B; that is, A D B .

(ii) ' is surjective: given a function g W X ! .Z2/X , define AŒg � X by

AŒg D fx 2 X W g.x/ D 1g:

It is easy to check that '.AŒg/ D fAŒg D g. Finally, ' is a homomorphism:

(i) ' maps the identity to the identity: the (multiplicative) identity of 2X isX ,

and '.X/ D fX . Now fX.x/ D 1 for all x 2 X , because every element of X lies in X! Hence, fX is the (multiplicative) identity in .Z2/

X .

(ii) ' preserves addition: we must show that Recall the different mean-

ings of C in this equation: ACB is symmetric differ- ence, and '.A/C '.B/ is pointwise addition.

'.AC B/ D '.A/C '.B/

for all A;B � X . Consider the following table:

fA.x/ fB .x/ fA.x/C fB .x/ 1 1 0

1 0 1

0 1 1

0 0 0

It follows that fA C fB D fACB , for each of the functions fA C fB and fACB has value 1 if x 2 .A [ B/ � .A \ B/ and value 0 otherwise. Therefore,

'.AC B/ D fACB D fA C fB D '.A/C '.B/:

(iii) ' preserves multiplication: we must show that '.AB/ D '.A/'.B/ for all A;B � X . The proof is similar to that in part (ii), using a table for fAfB to prove that fAfB D fAB ; you will supply the details in Exer- cise 5.39 on page 212.

We conclude that 2X and .Z2/ X are isomorphic. In Exercise 5.40 on page 212,

we will see that if jX j D n, then .Z2/X Š .Z2/n, the ring of all n-tuples having coordinates in Z2 with pointwise operations. N

How to Think About It. There are two strategies in trying to show that a

homomorphism 'WR ! S is an isomorphism. One way is to show that ' is a bijection; that is, it is injective and surjective. A second way is to show that

the inverse function '�1WS ! R exists (see Exercise 5.30 on page 211 and Exercise 5.39(ii) on page 212).

Here are some properties of homomorphisms.

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5.3 Homomorphisms 211

Lemma 5.17. Let R and S be commutative rings, let 'WR ! S be a homo- morphism, and let a 2 R.

(i) '.0/ D 0. (ii) '.�a/ D �'.a/.

(iii) '.na/ D n'.a/ for all n 2 Z. (iv) '.an/ D '.a/n for all n 2 N. (v) If a is a unit in R, then '.a/ is a unit in S , and '.a�1/ D '.a/�1.

Proof. (i) Since 0C 0 D 0, we have '.0C 0/ D '.0/C '.0/ D '.0/. Now subtract '.0/ from both sides.

(ii) Since 0 D �a C a, we have 0 D '.0/ D '.�a/ C '.a/. But Propo- sition 4.35 says that negatives are unique: there is exactly one s 2 S , namely, s D �'.a/, with s C '.a/ D 0. Hence, '.�a/ D �'.a/.

(iii) If n � 0, use induction to prove that '.na/ D n'.a/. Now use (ii) : '.�na/ D �'.na/ D �n'.a/.

(iv) Use induction to show that '.an/ D '.a/n for all n � 0. (v) By Proposition 4.35, there is exactly one b 2 R, namely, b D a�1, with

ab D 1. Similarly in S ; since '.a/'.b/ D '.ab/ D '.1/ D 1, we have '.a�1/ D '.a/�1.

Example 5.18. If 'WA ! B is a bijection between finite sets A and B , then they have the same number of elements. In particular, two finite isomorphic

commutative rings have the same number of elements. We now show that the

converse is false: there are finite commutative rings with the same number of

elements that are not isomorphic.

Recall Exercise 4.55 on page 165: there is a field, F4, having exactly four elements. If a 2 F4 and a ¤ 0, then a2 ¤ 0, for F4 is a domain (even a field), and so the product of nonzero elements is nonzero. Suppose there

were an isomorphism 'W F4 ! Z4. Since ' is surjective, there is a 2 F4 with '.a/ D 2. Hence, '.a2/ D '.a/2 D 22 D 0. This contradicts ' being injective, for a2 ¤ 0 and '.a2/ D 0 D '.0/. N

Exercises

5.29 Let R and S be commutative rings, and let 'W R ! S be an isomorphism. (i) If R is a field, prove that S is a field.

(ii) If R is a domain, prove that S is a domain.

5.30 *

(i) If ' is an isomorphism, prove that its inverse function '�1W S ! R is also an isomorphism.

(ii) Show that ' is an isomorphism if and only if ' has an inverse function '�1.

5.31 (i) Show that the composite of two homomorphisms (isomorphisms) is again a

homomorphism (an isomorphism).

(ii) Show that R Š S defines an equivalence relation on the class of all commu- tative rings.

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212 Chapter 5 Abstract Algebra

5.32 Prove that the weird integers W (see Exercise 4.35 on page 158) is not isomorphic

to Z.

5.33 Recall that ZŒ! D fa C b! W a; b 2 Zg, where ! D �1 2

C i p 3 2

. Show that

' W ZŒ! ! ZŒ!, defined by

'W a C b! 7! a C b!2;

is a homomorphism. Is ' an isomorphism?

5.34 If R is a commutative ring and a 2 R, is the function 'W R ! R, defined by 'W r 7! ar , a homomorphism? Why?

5.35 Prove that two fields having exactly four elements are isomorphic.

Hint: First prove that 1 C 1 D 0.

5.36 * Let k be a field that contains Zp as a subfield (e.g., k D Zp.x/). For every integer n > 0, show that the function 'nW k ! k, given by 'n.a/ D ap

n , is an

injective homomorphism. If k is finite, show that 'n is an isomorphism.

5.37 * If R is a field, show that R Š Frac.R/. More precisely, show that the homo- morphism 'W R ! Frac.R/, given by 'W r 7! Œr; 1, is an isomorphism.

5.38 * Recall, when we constructed the field Frac.D/ of a domain D, that Œa; b de-

noted the equivalence class of .a; b/, and that we then reverted to the usual nota-

tion: Œa; b D a=b. (i) If R and S are domains and 'W R ! S is an isomorphism, prove that

Œa; b 7! Œ'.a/; '.b/

is an isomorphism Frac.R/ ! Frac.S/. (ii) Prove that a field k containing an isomorphic copy of Z as a subring must

contain an isomorphic copy of Q.

(iii) Let R be a domain and let 'W R ! k be an injective homomorphism, where k is a field. Prove that there exists a unique homomorphism ˆW Frac.R/ ! k extending '; that is, ˆjR D '.

5.39 * In Example 5.16, we proved that if X is a set, then the function 'W 2X ! .Z2/X , given by '.A/ D fA, the characteristic function of A, is an isomorphism. (i) Complete the proof in Example 5.16 by showing that '.AB/ D '.A/ '.B/

for all A; B 2 2X . (ii) Give another proof that 2X Š .Z2/X by showing that '�1 exists.

5.40 * If n is a positive integer, define .Z2/ n to be the set of all n-tuples .a1; : : : ; an/

with ai 2 Z2 for all i (such n-tuples are called bitstrings). (i) Prove that .Z2/

n is a commutative ring with pointwise operations

.a1; : : : ; an/ C .b1; : : : ; bn/ D .a1 C b1; : : : ; an C bn/

and

.a1; : : : ; an/.b1; : : : ; bn/ D .a1b1; : : : ; anbn/:

(ii) If X is a finite set with jX j D n, prove that .Z2/X Š .Z2/n. Conclude, in this case, that 2X Š .Z2/n (see Example 5.16).

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5.3 Homomorphisms 213

Extensions of Homomorphisms

Suppose that a ring R is a subring of a commutative ring E with inclusion

i WR ! E . Given a homomorphism ' W R ! S , an extension ˆ of ' is a homomorphismˆWE ! S with restrictionˆjR D ˆi D '. If ˆWU ! Y is any

function, then its restriction

ˆjX to a subset X � U is equal to the composite

ˆi , where iWX ! U is the inclusion.

E

ˆ

R

i

' S:

Some obvious questions about extensions are

(i) Can we extend 'WR ! S to a homomorphism ˆWE ! S? (ii) Can we extend 'WR ! SŒx to a homomorphismˆWE ! SŒx?

Theorems 5.19 and 5.20 below answer the first question when E D RŒx and E D RŒx1; : : : ; xn; Corollary 5.22 answers the second question when E D RŒx. The basic idea is to let ' handle the elements of R, then specify what happens to x, and then use the definition of homomorphism to make sure that the extension preserves addition and multiplication.

Even though the coming proof is routine, we give full details because of the

importance of the result.

Theorem 5.19. Let R and S be commutative rings, and let 'WR ! S be a homomorphism. If s 2 S , then there exists a unique homomorphism

ˆWRŒx ! S

withˆ.x/ D s and ˆ.r/ D '.r/ for all r 2 R.

Proof. If f .x/ D P i rix

i D r0 C r1xC � � � C rnxn, define ˆ W RŒx ! S by

ˆ.f / D '.r0/C '.r1/s C � � � C '.rn/sn:

Proposition 5.6, uniqueness of coefficients, shows that ˆ is a well-defined

function, and the formula shows that ˆ.x/ D s and ˆ.r/ D '.r/ for all r 2 R.

We now prove thatˆ is a homomorphism. First,ˆ.1/ D '.1/ D 1, because ' is a homomorphism.

Second, if g.x/ D a0 C a1x C � � � C amxm, then

ˆ.f C g/ D ˆ X

i

.ri C ai /xi !

D X

i

'.ri C ai /si

D X

i

� '.ri /C '.ai /

� si

D X

i

'.ri /s i C

X

i

'.ai /s i

D ˆ.f /Cˆ.g/:

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214 Chapter 5 Abstract Algebra

Third, let f .x/g.x/ D P k ckx

k , where ck D P iCjDk riaj . Then

ˆ.fg/ D ˆ X

k

ckx k

!

D X

k

'.ck/s k

D X

k

'

0 @ X

iCjDk riaj

1 A sk

D X

k

0 @ X

iCjDk '.ri /'.aj /

1 A sk:

On the other hand,

ˆ.f /ˆ.g/ D X

i

'.ri /s i

!0 @X

j

'.aj /s j

1 A D

X

k

0 @ X

iCjDk '.ri /'.aj /

1 A sk:

Uniqueness ofˆ is easy: if‚WRŒx ! S is a homomorphism with‚.x/ D s and ‚.r/ D '.r/ for all r 2 R, then

‚ � r0 C r1x C � � � C rdxd

� D '.r0/C '.r1/s C � � � C '.rd /sd

D ˆ � r0 C r1x C � � � C rdxd

� :

This theorem generalizes to polynomial rings in several variables.

Theorem 5.20. Let R and S be commutative rings and 'WR ! S a homo- morphism. If s1; : : : ; sn 2 S , then there exists a unique homomorphism

ˆWRŒx1; : : : ; xn ! S

withˆ.xi / D si for all i andˆ.r/ D '.r/ for all r 2 R.

Proof. The proof is by induction on n � 1. The base step is Theorem 5.19. For the inductive step, let n > 1 and define A D RŒx1; : : : ; xn�1. The inductive hypothesis gives a homomorphism WA ! S with .xi / D si for all i � n � 1 and .r/ D '.r/ for all r 2 R. The base step gives a homomorphism ‰WAŒxn ! S with ‰.xn/ D sn and ‰.a/ D .a/ for all a 2 A. The result follows, because RŒx1; : : : ; xn D AŒxn, ‰.xi / D .xi / D si for all i � n� 1, ‰.xn/ D .xn/ D sn, and‰.r/ D .r/ D '.r/ for all r 2 R.

How to Think About It. There is an analogy between Theorem 5.20 and

an important theorem of linear algebra, Theorem A.43 in Appendix A.4: Let

V and W be vector spaces over a field k; if v1; : : : ; vn is a basis of V and

w1; : : : ; wn 2 W , then there exists a unique linear transformation T WV !W with T .vi / D wi for all i (linear transformations are homomorphisms of vec- tor spaces). The theorem is actually the reason why matrices can describe linear

transformations.

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5.3 Homomorphisms 215

Here is a familiar special case of Theorem 5.19.

Definition. If R is a commutative ring and a 2 R, then evaluation at a is the If f # is the polynomial function determined by f ,

then ea.f / D f #.a/. function eaWRŒx ! R given by ea.f / D f .a/; that is,

ea

X

i

rix i

! D X

i

ria i :

So, in this language, we have:

Corollary 5.21. If R is a commutative ring and a 2 R, then eaWRŒx ! R, evaluation at a, is a homomorphism.

Proof. In the notation of Theorem 5.19, set R D S , ' D 1R (the identity function R ! R), and s D a 2 R. The homomorphism ˆWRŒx ! R is ea, which sends

P i rix

i into P i ria

i .

As an illustration of Corollary 5.21, if f; g 2 RŒx and h D fg, then

h.a/ D ea.h/ D ea.f /ea.g/ D f .a/g.a/:

In other words, we get the same element of R if we first multiply polynomials

in RŒx and then substitute a for x, or if we first substitute a for x and then

multiply the elements f .a/ and g.a/. For example, if R is a commutative

ring and a 2 R, then f .x/ D q.x/g.x/ C r.x/ in RŒx implies f .a/ D q.a/g.a/ C r.a/ inR.

Let’s return to question (ii) on page 213. Given a homomorphism

' W R ! S , can we extend it to a homomorphism RŒx ! SŒx? The basic idea? Let ' handle the coefficients and send x to x.

Corollary 5.22. If R and S are commutative rings and 'WR ! S is a ho- momorphism, then there is a unique homomorphism '�WRŒx ! SŒx given by

'�W r0 C r1x C r2x2 C � � � 7! '.r0/C '.r1/x C '.r2/x2 C � � � :

Moreover, '� is an isomorphism if ' is.

Proof. The existence of the homomorphism '� is a special case of Theo- rem 5.19. More precisely, consider the following diagram in which �WR ! RŒx and �WS ! SŒx are the usual inclusions viewing elements of R and of S as constant polynomials. The role of 'WR ! S is now played by the composite �'WR ! SŒx, namely, r 7! .'.r/; 0; 0: : : : /.

RŒx

ˆ

R '

S � SŒx:

If ' is an isomorphism, then ˆ�1 is the inverse of the extension of '�1.

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216 Chapter 5 Abstract Algebra

Example 5.23. If rmW Z ! Zm is reduction mod m, that is, rm.a/ D Œa, then the homomorphism r�mW ZŒx ! ZmŒx reduces each coefficient of a polyno- mial mod m:

r�mW a0 C a1x C a2x2 C � � � 7! Œa0C Œa1x C Œa2x2 C � � � :

We will usually write a instead of Œa when using r�m. N

Example 5.24. Complex conjugation extends to an isomorphism CŒx ! CŒx in which every polynomial is mapped to the polynomial obtained by taking the

complex conjugate of each coefficient. (We have already used this construction

in Theorem 3.12 and in Exercises 3.43–3.45 on page 106.) N

Exercises

5.41 If R is a commutative ring, prove that RŒx; y Š RŒy; x. In fact, prove that there is an isomorphism ˆ with ˆ.x/ D y, ˆ.y/ D x, and ˆ.r/ D r for all r 2 R.

Hint: Use Theorem 5.20.

If you look very carefully at the definitions, you’ll see that RŒx; y and RŒy; x

are different rings. Recall that elements a in a ring A correspond to .a; 0; : : :/

in AŒx. In particular, the element x 2 RŒx corresponds to .x; 0; 0; : : :/ in RŒxŒy; that is, we have x D

� .0; 1; 0; : : : /; 0; 0; : : :

� so that in RŒxŒy the element x has

.0; 1; 0; : : : / in coordinate 1. This is not the same element as x in RŒyŒx, which

has 1 sitting in coordinate 1. However, this exercise allows you to relax and regard

these polynomials rings as the same.

5.42 *

(i) If R is a commutative ring and c 2 R, prove that there is a homomorphism 'W RŒx ! RŒx with '.x/ D x C c and '.r/ D r for all r 2 R; that is, '. P i rix

i / D P i ri .x C c/i . Is ' an isomorphism?

(ii) If deg.f / D n, show thatNo calculus is needed for this exercise.

'.f / D f .c/ C f 0.c/.x C c/ C f 00.c/ 2Š

.x C c/2 C � � � C f .n/.c/

nŠ .x C c/n;

where f 0.x/ is the formal derivative of f defined in Exercise 5.15 on page 202.

Kernel, Image, and Ideals

There’s a great deal of talk about “modeling” in high school mathematics: we

wonder whether a given statement is true, and the idea is to see whether it

holds in some model. A homomorphism R ! S is a good illustration of this idea; it transports the structure of R to the structure of S , so that we may test

whether a statement in R is true by asking whether its analog in the “model”

S is true. For example, is �1 a square in Z; is there k 2 Z with �1 D k2? Now we can list all the squares in Z3: 0

2 D 0; 12 D 1; 22 D 4 D 1, and we see that �1 D 2 is not a square. But if �1 D k2 in Z, then reduction mod 3, the homomorphism r3W Z ! Z3 taking a 7! Œa (see Example 5.14(i)), would give

r3.�1/ D r3.k2/ D r3.k/2 ;

contradicting �1 not being a square in Z3.

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5.3 Homomorphisms 217

We should be cautious when viewing a homomorphism 'WR ! S as mod- eling a ringR. First, ' may not give a faithful model ofR, thereby losing some

information; for example, ' might take different elements of R to the same el- ement of S . Also, some information might get missed: there may be elements

of S that don’t get “hit” by an element of R. The information that’s lost is

called the kernel of '; the information that’s hit is called its image.

Definition. If 'WR ! S is a homomorphism, then its kernel is

ker' D fa 2 R with '.a/ D 0g � R;

and its image is

im' D fs 2 S W s D '.a/ for some a 2 Rg � S:

Here are the first properties of these subsets. Note that Lemma 5.17 says, for every homomorphism 'WR ! S , that 0 2 ker' and 0 2 im'. In general, im' is a subset of S but, as for any function, ' is surjective if and only if

im' D S .

Proposition 5.25. Let R and S be rings and 'WR ! S a homomorphism. (i) im' D f'.r/ W r 2 Rg is a subring of S .

(ii) If a; b 2 ker ', then aC b 2 ker '. (iii) If a 2 ker' and r 2 R, then ra 2 ker '.

Proof. (i) To see that im' is a subring of S , note first that 1 2 im', because '.1/ D 1. If s; t 2 im', then there are a; b 2 R with s D '.a/ and t D '.b/. Hence, s C t D '.a/ C '.b/ D '.a C b/ 2 im', and st D '.a/'.b/ D '.ab/ 2 im'. Therefore, im' is a subring of S .

(ii) If a; b 2 ker', then '.a/ D 0 D '.b/. Hence, '.aCb/ D '.a/C'.b/ D 0C 0 D 0, and aC b 2 ker '.

(iii) If a 2 ker', then '.a/ D 0. Hence, '.ra/ D '.r/'.a/ D '.r/ � 0 D 0, and so ra 2 ker '.

Here are some examples of kernels and images.

Example 5.26. (i) If 'WR ! S is an isomorphism, then ker ' D f0g and im' D S .

(ii) If ' is injective, then ker' D f0g, for if r ¤ 0, then '.r/ ¤ '.0/ D 0. We will soon see that the converse is true, so that ' is injective if and only

ker' D f0g. (iii) If rmW Z ! Zm is reduction mod m, then ker rm consists of all the multi-

ples of m.

(iv) Let k be a commutative ring, let a 2 k, and let eaW kŒx ! k be the evaluation homomorphism f .x/ 7! f .a/. Now ea is always surjective: if b 2 k, then b D ea.f /, where f .x/ D x � a C b. By definition, ker ea consists of all those polynomials g for which g.a/ D 0.

In particular, let 'W RŒx ! C be defined by '.x/ D i and '.a/ D a for all a 2 R. Then ker ' is the set of all polynomials f .x/ 2 RŒx having i as a root. For example, x2 C 1 2 ker'. N

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218 Chapter 5 Abstract Algebra

Proposition 5.25 suggests that ker' is a subring of R but, in fact, it almost

never is because it usually doesn’t contain 1. The definition of homomorphism

says that '.1/ D 1. If 1 2 ker ', then '.1/ D 0, and so 1 D 0 in S ; that is, S is the zero ring. We conclude that if S has more than one element, then ker'

is not a subring of R. However, kernels are always ideals.

Definition. An ideal in a commutative ringR is a subset I of R such that

(i) 0 2 I (ii) if a, b 2 I , then a C b 2 I

(iii) if a 2 I and r 2 R, then ra 2 I .

An ideal I ¤ R is called a proper ideal.

The ring R itself and f0g, the subset of R consisting of 0 alone, are always ideals in a commutative ring R. Proposition 5.25 says that the kernel of a ho-

momorphism 'WR ! S is always an ideal in R; it is a proper ideal if S is not the zero ring because 1 … ker '.

We have seen ideals in a completely different context. Theorem 1.19, which

says that gcd.a; b/ is a linear combination of a; b, involved showing that the set

of all linear combinations is an ideal in Z. Indeed, Exercise 1.49 on page 30 makes this explicit (of course, we had not introduced the term ideal at that

time).

Etymology. As we said on page 131, a natural attempt to prove Fermat’s Last Theorem involves factoring xp C yp in the ring ZŒ�p of cyclotomic integers, where �p is a pth root of unity. In Chapter 8, we shall sketch the ideas that

show that if this ring has unique factorization into primes, that is, if the analog

of the Fundamental Theorem of Arithmetic holds in ZŒ�p, then there are no

positive integers a; b; c with apCbp D cp . For some primes p, such an analog is true but, alas, there are primes for which it is false. In his investigation of Fermat’s Last Theorem, Kummer invented ideal numbers in order to restore

unique factorization. His definition was later recast by Dedekind as the ideals

we have just defined, and this is why ideals are so called.

Here is a construction of ideals that generalizes that which arose when we

studied gcd’s. Recall that a linear combination of elements b1; b2; : : : ; bn in a commutative ringR is an element ofR of the form

r1b1 C r2b2 C � � � C rnbn; where ri 2 R for all i .

Definition. If b1; b2; : : : ; bn lie in a commutative ring R, then the set of allIt is very easy to check that .b1; b2; : : : ; bn/ is an

ideal. linear combinations, denoted by

.b1; b2; : : : ; bn/;

is an ideal in R, called the ideal generated by b1; b2; : : : ; bn. In particular, if

n D 1, then .b/ D frb W r 2 Rg

consists of all the multiples of b; it is called the principal ideal generated by b. The principal ideal .b/ is

sometimes denoted by Rb.

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5.3 Homomorphisms 219

Both R and f0g are ideals; indeed, both are principal ideals, for R D .1/ and, obviously, f0g D .0/ is generated by 0. Henceforth, we will denote the zero ideal f0g by .0/.

Example 5.27. (i) The even integers comprise an ideal in Z, namely, .2/.

(ii) Proposition 5.25 says that if 'WR ! S is a homomorphism, then ker' is an ideal in R. In particular, we can generalize part (i): if rmW Z ! Zm is reduction mod m, then ker rm D .m/.

(iii) If I and J are ideals in a commutative ring R, then it is routine to check

that I \ J is also an ideal in R. More generally, if .Ij /j2J is a family of ideals in a commutative ring R, then

T j2J Ij is an ideal in R (see

Exercise 5.53 below).

(iv) By Example 5.26(iv), the set I , consisting of all polynomials f .x/ in

RŒx having i as a root, is an ideal in RŒx containing x2 C 1 (it is the kernel of the evaluation ei ). We shall see, in Corollary 6.26, that I D .x2 C 1/. N

Example 5.28. Let R be a commutative ring. For a subset A of R, define

I D I.A/ D ff .x/ 2 RŒx W f .a/ D 0 for all a 2 Ag:

It is easy to check that I is an ideal inRŒx. Clearly, 0 2 I . If f 2 I and r 2 R, then .rf /# D rf #, and so .rf /.a/ D r .f .a// D 0 for all a 2 A. Finally, if f; g 2 I , then .f C g/# D f # C g# , so that .f C g/# W a 7! f .a/C g.a/ D 0 for all a 2 A, and f C g 2 I . Therefore, I is an ideal. (Alternatively, show that I.A/ D

T a2A ker ea, where ea is evaluation at a, and use Exercise 5.53

below that says the intersection is an ideal.)

In the special case when R is a field, then I.A/ is a principal ideal. If A is

finite, can you find a monic d.x/ with I.A/ D .d/? What if A is infinite? N

Theorem 5.29. Every ideal I in Z is a principal ideal.

Proof. If I D .0/, then I is the principal ideal with generator 0. If I ¤ .0/, then there are nonzero integers in I ; since a 2 I implies �a 2 I , there are positive integers in I ; let d 2 I be the smallest such. Clearly, .d/ � I . For the reverse inclusion, let b 2 I . The Division Algorithm gives q; r 2 Z with b D qd C r , where 0 � r < d . But r D b � qd 2 I . If r ¤ 0, then its existence contradicts d being the smallest positive integer in I . Hence, r D 0, d j b, b 2 I , and I � .d/. Therefore, I D .d/.

We’ll see in the next chapter that there are commutative rings having ideals

that are not principal ideals.

Example 5.30. (i) If an ideal I in a commutative ring R contains 1, then

I D R, for now I contains r D r1 for every r 2 R. Indeed, if I contains a unit u, then I D R, for then I contains u�1u D 1.

(ii) It follows from (i) that if R is a field, then the only ideals I in R are .0/ and R itself: if I ¤ .0/, it contains some nonzero element, and every nonzero element in a field is a unit.

Conversely, assume that R is a nonzero commutative ring whose only

ideals are R itself and .0/. If a 2 R and a ¤ 0, then .a/ D fra W r 2 Rg

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220 Chapter 5 Abstract Algebra

is a nonzero ideal, and so .a/ D R; but 1 2 R D .a/. Thus, there is r 2 R with 1 D ra; that is, a has an inverse in R, and so R is a field. N

Proposition 5.31. A homomorphism 'WR ! S is an injection if and only if ker ' D .0/.

Proof. If ' is an injection, then a ¤ 0 implies '.a/ ¤ '.0/ D 0. Hence, ker ' D .0/. Conversely, assume that ker' D .0/. If '.a/ D '.b/, then '.a � b/ D '.a/ � '.b/ D 0; that is, a � b 2 ker' D .0/. Therefore, a D b and ' is an injection.

Corollary 5.32. If k is a field and 'W k ! S is a homomorphism, where S is a nonzero commutative ring, then ' is an injection.

Proof. The only proper ideal in k is .0/, by Example 5.30; now apply Propo-

sition 5.31.

Exercises

5.43 Construct a homomorphism from ZŒi  ! ZŒi  that has i in its kernel. What is the entire kernel?

5.44 Find the kernel of the homomorphism QŒx ! QŒ p

2 defined by f 7! f . p

2/,

where QŒ p

2 D fa C b p

2 W a; b 2 Qg.

5.45 Show that the kernel of the evaluation homomorphism ea in Corollary 5.21 is the

set of polynomials in RŒx that have a as a root.

5.46 Consider the set I of polynomials in RŒx that vanish on the set f3˙ p

5; 5˙ p

7g. Show that I is a principal ideal in RŒx.

5.47 * Find three ideals .a/ in Z with the property that

The notation ¨ means “is a proper subset of”

(in contrast to � which indicates a subset which

may or may not be proper). .24/ ¨ .a/:

5.48 * Suppose a and b are integers. Show that a j b if and only if .b/ � .a/.

5.49 * If a; b 2 Z, prove that .a/ \ .b/ D .m/, where m D lcm.a; b/.

5.50 * Define the sum of ideals I and J in a commutative ring R by

I C J D fu C v W u 2 I and v 2 J g:

(i) Prove that I C J is an ideal. (ii) If a; b 2 Z, prove that .a/ C .b/ D .a; b/ D .d/, where d D gcd.a; b/.

5.51 * Define the product of ideals I and J in a commutative ring R by

IJ D fa1b1 C � � � C anbn W ai 2 I; bi 2 J; n � 1g:

(i) Prove that IJ is an ideal in R.

(ii) Prove that if I and J are principal ideals, then IJ is principal. More precisely,

if I D .a/ and J D .b/, then IJ D .ab/. (iii) If I D .a1; : : : ; as/ and J D .b1; : : : ; bt/, prove that

IJ D .aibj W 1 � i � s and 1 � j � t /:

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5.4 Connections: Boolean Things 221

5.52 * Let I; J , and Q be ideals in a commutative ring R.

(i) Prove that IJ D JI . (ii) Prove that RI D I .

(iii) Prove that I.JQ/ D .IJ /Q.

5.53 * If .Ij /j2J is a family of ideals in a commutative ring R, prove that T j2J Ij

is an ideal in R.

5.54 *

(i) If R and S are commutative rings, show that their direct product R�S is also a commutative ring, where addition and multiplication in R � S are defined coordinatewise:

.r; s/ C .r 0; s0/ D .r C r 0; s C s0/

and

.r; s/.r 0; s0/ D .rr 0; ss0/:

This construction generalizes that of .Z2/ n in Exercise 5.40 on page 212.

(ii) Show that R � S is not a domain. (iii) Show that R � .0/ is an ideal in R � S . (iv) Show that R � .0/ is a ring isomorphic to R, but it is not a subring of R � S . (v) Prove that Z6 Š Z2 � Z3.

(vi) Show that Z4 6Š Z2 � Z2 . (vii) Prove that Zmn Š Zm � Zn if m and n are relatively prime.

Hint: Use the Chinese Remainder Theorem.

5.55 If R1; : : : ; Rn are commutative rings, define their direct product R1 � � � � � Rn by induction on n � 2 (it is the set of all n-tuples .r1; : : : ; rn/ with ri 2 Ri for all i ). Prove that the ring .Z/X in Example 5.16, where X is a set with jX j D n, is the direct product of n copies of Z2.

5.56 (i) Give an example of a commutative ring R with nonzero ideals I and J such

that I \ J D .0/. (ii) If I and J are nonzero ideals in a domain R, prove that I \ J ¤ .0/.

5.57 Let F be the set of all 2 � 2 real matrices of the form

A D "

a b

�b a

# :

(i) Prove that F is a field (with operations matrix addition and matrix multipli-

cation).

(ii) Prove that 'W F ! C, defined by '.A/ D a C ib, is an isomorphism.

5.4 Connections: Boolean Things

In some high school programs, Boolean Algebra is called the “algebra of sets;”

it usually focuses on establishing set-theoretic identities like

A\ .B [ C/ D .A\ B/[ .A\ C/

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222 Chapter 5 Abstract Algebra

for subsets A, B , and C of a set X . Such formulas are proved by showing that

an element lies in the left-hand side if and only it lies in the right-hand side.

Exercises 4.68 through 4.74 on page 169 gave you practice in doing this sort of thing, but they actually showed more. Recall Example 4.47: if 2X is

the family of all the subsets of a set X , then 2X is a commutative ring with

addition defined as symmetric difference,

AC B D .A � B/ [ .B �A/ D A[ B � .A \ B/;

and multiplication defined as intersection,Recall: if U; V are subsets of X , then U � V D fx 2 X W x 2 U and x … V g.

AB D A\ B:

It follows, for all subsets A of X , that

A2 D A; AC ¿ D A; AC A D ¿; and AX D AI

the identity element 1 is the subsetX itself. It follows fromACA D ¿ that ev- ery A 2 2X is its own additive inverse; that is, A D �A. Indeed, Exercise 5.58 on page 226 says that 1 D �1 in 2X . Since we often pass back and forth be- tween the commutative ring 2X and set theory, we say out loud that a minus

sign will be used in set theory, as in the definition of symmetric difference, but

it shall never be used when we are working in 2X viewed as a ring.

We are going to show that calculations in the ring 2X give more satifisfy-

ing proofs of set-theoretic identities; thus, regarding all subsets as forming a commutative ring is a definite advantage. Another goal is to use the calcula-

tions to establish the inclusion-exclusion principle, a very useful technique in

counting problems.

Venn diagrams are visual representations in the plane of relationships among

subsets in X . They convert words into pictures. For example, symmetric dif-

ference and intersection are illustrated by the Venn diagram in Figure 5.2.

Some standard words occurring in set theory, actually in logic, are NOT,

AND, OR, and EXCLUSIVE OR. If we picture a statement a as the inside of a region A in the plane, then the Venn diagram of “NOT a” is the outside of A;

it is the complement

Ac D fx 2 X W x … Ag: Exercise 4.69 on page 168 says that Ac D X C A. If a and b are statements, then the Venn diagram of the statement “a AND b” is the intersection A\ B , while the diagram of “a OR b” is the union A [ B . EXCLUSIVE OR is the symmetric difference AC B; it pictures the statement “a OR b but not both” (as in the statement “Take it or leave it!”).

The next result is Exercise 4.73 on page 169; you probably solved this ex-

In Figure 5.2, ACB is the shaded region, AB is the

unshaded region.

ercise then using elements, as we now do.

A B

Figure 5.2. A C B and AB .

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5.4 Connections: Boolean Things 223

Proposition 5.33 (De Morgan). If A and B are subsets of a set X , then

.A [ B/c D Ac \ Bc:

Proof. We first show that .A [ B/c � Ac \ Bc . If x 2 .A [ B/c , then x … A [ B . But A [ B consists of all elements in A or in B . So, x … A [ B implies x … A and x … B; that is, x 2 Ac \ Bc .

For the reverse inclusion, take x 2 Ac \ Bc . Hence, x 2 Ac and x 2 Bc ; that is, x … A and x … B . Thus, x … A[ B , and x 2 .A[ B/c .

This proof is not very difficult, but it’s also not very satisfying. The reason- ing very much depends on the meanings of the connectives NOT, AND, and

OR, as do the definitions of union and intersection. If feels as if we are just

playing with words.

We are going to give a second proof, more in the spirit of commutative

rings, that uses the binary operations in a special kind of commutative ring that distills the distinguishing feature of 2X into one property.

Definition. A Boolean ring is a commutative ring R in which a2 D a for all a 2 R.

Example 5.34. (i) The ring 2X of subsets of a set X , in Example 4.47, is a

Boolean ring.

(ii) If X is a set, possibly infinite, then the family R of all finite subsets of X

together withX itself is a Boolean ring with operations symmetric differ-

ence and intersection. N

Let’s extend familiar facts in 2X to arbitrary Boolean ringsR. Some of these

calculations might look strange; just keep 2X and Venn diagrams in the back of your mind as you work through them. For example, the following definitions,

inspired by the particular Boolean ring 2X , make sense in any Boolean ring.

Complement: a0 D 1C a (see Exercise 4.69 on page 168) Union: a _ b D aC b C ab (see Exercise 4.74(ii) on page 169) Disjoint: ab D 0

Lemma 5.35. Suppose that R is a Boolean ring and a 2 R. Then:

(i) aC a D 0 (ii) aa0 D 0

(iii) a0 C a D 1.

Proof. (i)

a C a D .a C a/.a C a/ D a2 C a2 C a2 C a2

D a C aC aC a:

Now subtract aC a from both sides to obtain aC a D 0. (ii) aa0 D a.1C a/ D a C a2 D aC a D 0:

(iii) a0 C a D .1C a/C a D 1C .aC a/ D 1C 0 D 1.

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224 Chapter 5 Abstract Algebra

Proposition 5.36. Let R be a Boolean ring and a; b 2 R.

(i) aC b D ab0 C a0b, and the summands ab0 and a0b are disjoint. (ii) a_b D ab0 Ca0bCab, and the summands ab0, a0b, and ab are pairwise

disjoint.

Proof. (i) For all x 2 R, x C x D 0, xx0 D 0, and x C x0 D 1. Hence,

aC b D a.b C b0/C b.aC a0/ D ab C ab0 C ab C a0b D ab0 C a0b:

The summands are disjoint, because .ab0/.a0b/ D 0. (ii)

a _ b D aC b C ab D ab0 C a0b C ab:

The summands are disjoint, for part (i) shows that ab0 and a0b are disjoint,

Work though these proofs

for yourself, justifying

each step. Notice how

the particulars of 2X are

fading into the background.

while .ab0/ab D 0 D .a0b/ab.

Let’s now see how working in an arbitrary Boolean ring reduces the proofs about facts in specific such rings like 2X to algebraic calculations. Compare

the set-theoretic proof of Proposition 5.33 with the following proof.

Proposition 5.37 (De Morgan = Proposition 5.33). If A andB are subsets of

a set X , then

.A[ B/c D Ac \ Bc :

Proof. We first work in a Boolean ring R and then pass to 2X . If a; b 2 R, we want to show that .a _ b/0 D a0b0. ButThe proof that

1 C .a C b C ab/ D .1Ca/.1C b/ could be an exercise in any first-year

high school algebra text.

.a _ b/0 D 1C a _ b D 1C .a C b C ab/;

which is equal to a0b0 D .1C a/.1 C b/. Now interpret this general result in R in the particular Boolean ring 2X ,

using the translationsA _ B D A[ B , AB D A\ B , and 1C A D Ac .

There’s another De Morgan law in Exercise 4.73 on page 169. Algebra to

the rescue.

Proposition 5.38 (De Morgan). If A and B are subsets of a set X , then

.A\ B/c D Ac [ Bc :

Proof. Let R be a Boolean ring and a; b 2 R. We want to show that

.ab/0 D a0 _ b0:

The idea again is to first use “pure algebra,” reducing everything to statements about addition and multiplication in R, and then translate the result into the

language of 2X . Now .ab/0 D 1C ab, and

a0 _ b0 D a0 C b0 C a0b0 D .1C a/C .1C b/C .1C a/.1 C b/:

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5.4 Connections: Boolean Things 225

Calculate:

.1C a/C .1 C b/C .1C a/.1 C b/ D 1C aC 1C b C 1C aC b C ab D .1 C 1/C .a C a/

C .b C b/C .1C ab/ D 1C ab D .ab/0:

We now solve an earlier exercise using this point of view.

Proposition 5.39 (= Exercise 4.70). Let A; S be subsets of a set X . Then

S D Ac if and only if A\ S D ¿ and A[ S D X .

Proof. It suffices to work in a Boolean ring and then to see what it says in the

particular Boolean ring 2X .

Assume that s D a0 D 1C a. Then

as D a.1 C a/ D aC a2 D aC a D 0;

and

a _ s D a _ .1C a/ D aC .1C a/C a.1C a/ D aC 1C aC aC a2 D 1:

Conversely, if as D 0 and a C s C as D 1, then a C s D 1. But �1 D 1 in every Boolean ring, by Exercise 5.58 on page 226, and so s D 1 C a D a0.

The usual distributive law in a commutative ring is a.bCc/ D abCac. The proof that the equation holds in 2X essentially follows from the set-theoretic

identity

A\ .B [ C/ D .A \ B/ [ .A \ C/:

We are now going to show that interchanging \ and [ gives another valid identity.

Proposition 5.40. If A;B and C are subsets of a set X , then

A[ .B \ C/ D .A [ B/ \ .A [ C/:

Proof. We must show that a _ bc D .a _ b/.a _ c/; that is,

a C bc C abc D .aC b C ab/.a C c C ac/:

Expand the right-hand side, remembering that x2 D x and xCx D 0 for all x:

.a C b C ab/.a C c C ac/ D a2 C ac C a2c C ab C bc C abc C a2b C abc C a2bc

D aC bc C abc:

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226 Chapter 5 Abstract Algebra

Exercises

5.58 * Prove that �1 D 1 in every Boolean ring.

5.59 * Proposition 5.40 proves that if A; B , and C are subsets of a set X , then

A [ .B \ C/ D .A [ B/ \ .A [ C/:

Give another proof using set theory.

5.60 * If A, B1, B2, and B3 are subsets of a set X , show that

A \ .B1 \ B2 \ B3/ D .A \ B1/ [ .A \ B2/ [ .A \ B3/:

Generalize to A \ �Sn

iD1Bi � .

5.61 Let R be a ring in which multiplication is not assumed to be commutative (see

the callout on page 156). If a2 D a for every a 2 R, prove that R must be a commutative ring.

5.62 (i) If A; B are subsets of a set X , prove that B � A D B \ Ac . (ii) In any Boolean ring R, prove that b C a D b.1 C a/ C a.1 C b/.

5.63 Suppose that R is a Boolean ring and a; b 2 R. Show thatIn Exercises 5.66 and 5.67, we use minus signs

in a Boolean algebra.

Since �1 D C1, all these signs are really +, but

this notation invites you to

compare these formulas

with the statement of

Inclusion-Exclusion.

a0b0 D 1 � a � b C ab;

where a0 D 1 C a.

5.64 Suppose that R is a Boolean ring and a; b 2 R. Show that

1 � a0b0 D a _ b:

5.65 Suppose that R is a Boolean ring and a; b 2 R. Show that

a _ .b _ c/ D .a _ b/ _ c:

5.66 Suppose that R is a Boolean ring and � ai �n iD1 is a collection of n elements in R.

Show that

1 � nY

iD1 a0i D

n_

iD1 ai :

5.67 Suppose that R is a Boolean ring and � ai �n iD1 is a collection of n elements in R.

Show that

nY

iD1 a0i D 1 �

X

1�i�n ai C

X

1�i<j�n aiaj � � � � C .�1/na1a2 : : : an:

Hint: a0 D 1 � a.

5.68 Suppose that R is a Boolean ring and � ai �n iD1 is a collection of n elements in R.

Show that

n_

iD1 ai D

X

1�i�n ai �

X

1�i<j�n aiaj C � � � C .�1/n�1a1a2 : : : an:

5.69 In a Boolean ring, define a � b to mean a D ab. Viewing 2X as a Boolean ring, prove that A � B in 2X if and only if A � B .

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5.4 Connections: Boolean Things 227

5.70 An atom in a Boolean ring R is a nonzero element a 2 R with x � a if and only if x D 0 or x D a. If R is a finite Boolean ring, prove that every x 2 R is a sum of atoms.

5.71 (i) If R is a finite Boolean ring, prove that R Š 2X , where X is the set of all atoms in R.

(ii) Take It Further. Let R be the Boolean ring of all finite subsets of an infinite

set X (see Example 5.34(ii)). Prove that R 6Š 2Y for any set Y .

Hint: The simplest solution involves some set theory we have not discussed.

If X is countable, then R is countable; however, if Y is any infinite set, then

2Y is uncountable. Hence, there is no bijection R ! 2Y .

Inclusion-Exclusion

Suppose you have a class of students, all of whom take either French or Span-

ish, but none of whom take both. If 15 students are studying French and 12

students are studying Spanish, you have 15C 12 D 27 students in your class. Denote the number of elements in a finite set A by

jAj:

Then one way to state the above fact is that if F is the set of students studying French and S is the set of students studying Spanish, then

jF [ S j D jF j C jS j:

We make the above counting principle explicit.

Addition Principle. If A and B are disjoint finite subsets of a set X , then

jA[ Bj D jAj C jBj:

The Addition Principle extends, by induction, to any number of finite sets.

Lemma 5.41. If .Ai / n iD1 is a family of pairwise disjoint finite sets, then

ˇ̌ ˇ n[

iD1 Ai

ˇ̌ ˇ D

nX

iD1 jAi j:

Proof. The proof is by induction on n � 2. The base step is the Addition Principle. For the inductive step,

n[

iD1 Ai D

n�1[

iD1 Ai

! [ An:

Now �Sn�1

iD1 Ai �

\An D ¿: Exercise 5.60 on page 226 gives

n�1[

iD1 Ai

! \ An D .A1 \An/[ � � � [ .An�1 \An/;

and each Ai \An D ¿ because the subsets are pairwise disjoint.

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228 Chapter 5 Abstract Algebra

Let’s return to your class of students, 15 of whom are studying French and

12 of whom are studying Spanish. What if 4 of them are studying both French

and Spanish? You’d then have fewer than 27 students in the class because of double counting. A Venn diagram can help you figure out how to calculate

the actual number. The goal of this subsection is to develop a general method

of calculating the number of elements in the union of a finite collection of

possibly overlapping finite sets.

As a Venn diagram illustrates, the Addition Principle no longer holds if A and B overlap, for elements in A\ B are counted twice in jAj C jBj. What is the formula giving a precise count of jA [ Bj? The number of things that get counted twice must be subtracted once.

Lemma 5.42. If A and B are finite subsets of a set X , then

jA[ Bj D jAj C jBj � jA\ Bj:

Proof. First note that A[ B is the disjoint union

A[ B D .A � B/ [ .B �A/ [ .A \ B/;

so that Lemma 5.41 gives

jA[ Bj D jA\ Bcj C jAc \ Bj C jA\ Bj: (5.1)

As usual, we first compute in a Boolean ring R, after which we specialize

to 2X . Recall Proposition 5.36(ii): if a; b 2 R, then

a _ b D ab0 C a0b C ab;

where the summands on the right-hand side are pairwise disjoint. Hence, there

are two more equations: factor out a to get

a _ b D a.b0 C b/C a0b D aC a0b;

or factor out b to get

a _ b D ab0 C b.a0 C a/ D ab0 C b:

Since the summands on the right-hand side of each of the equations are pair-

wise disjoint, we can pass back to 2X to obtain

jA[ Bj D jAj C jAc \ Bj

and

jA[ Bj D jBj C jA\ Bcj:

Add the equations:

2jA[ Bj D jAj C jBj C jAc \ Bj C jA\ Bc j: (5.2)

Now Eq. (5.1) says that the last two terms on the right-hand side of Eq. (5.2)

can be replaced by jA[ Bj � jA\ Bj, giving

2jA[ Bj D jAj C jBj C jA[ Bj � jA\ Bj:

Subtracting jA[ Bj from both sides gives the desired result.

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5.4 Connections: Boolean Things 229

Example 5.43. How many positive integers < 1000 are there that are not di-

visible by 5 or by 7? If the number of positive integers that are divisible by 5

or 7 isD, then the answer is 999�D. We computeD using Lemma 5.42. Let

A D fn 2 Z W 5 j n and 0 < n < 1000g and

B D fn 2 Z W 7 j n and 0 < n < 1000g:

The Division Algorithm gives jAj D 199, because 999 D 199 �5C4; similarly, jBj D 142 and jA\ Bj D 28, where A \ B D fn 2 Z W 35 j n and 0 < n < 1000g. Hence,

jA [ Bj D jAj C jBj � jA\ Bj D 199C 142� 28 D 313:

Therefore, there are exactly 999 � 313 D 686 positive numbers < 1000 that are not divisible by 5 or by 7. N

How to Think About It. You could probably convince yourself of the result

in Lemma 5.42 with a Venn diagram accompanied by a few examples. While

diagrams and examples can motivate insight, they are not substitutes for rig-

orous proof. The reason is that a picture can be misleading. For example, if

you aren’t careful about drawing a Venn diagram for the union of four or more regions, then some possible intersections might be overlooked.

Example 5.44. Let’s look at the case of three finite subsets A, B , and C of a

set X . Before reading on, what do you think the formula should be? The basic

idea is to apply Lemma 5.42 twice.

jA[ B [ C j D j.A [ B/ [ C j D jA[ Bj C jC j � j.A [ B/ \ C j D jAj C jBj � jA\ Bj C jC j � j.A \ C/ [ .B \ C/j D jAj C jBj C jC j � jA\ Bj

� � jA\ C j C jB \ C j � jA\ B \ C j

D jAj C jBj C jC j � � jA\ Bj C jA\ C j C jB \ C j

C jA\ B \ C j:

So, the number of elements in the union of three sets is the sum of the number

of elements in each, minus the sum of the number of elements in the pairwise

intersections, plus the number of elements that are common to all three. N

We want to generalize the formula in Example 5.44 to count the number of

elements in a union of finitely many subsets. The difficulty in deriving such

a formula by a brutal assault is that we must be careful that an element in the union is not counted several times, for it may occur in the intersection of

several of the Ai . To illustrate, consider Figures 5.3 and 5.4, Venn diagrams

depicting the various intersections obtained from three subsets and from four

subsets.

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230 Chapter 5 Abstract Algebra

A B

C

Figure 5.3. Subsets of three regions. Figure 5.4. Subsets of four regions.

To count the number of elements in a union

A1 [ � � � [Ar ;

we proceed by induction, using the idea of Example 5.44 by shearing off Ar and treating the union of the rest as one set. The details are technical, so

sharpen a pencil and follow along.

Given finite subsets A1; : : : ; Ar of a set X , let us write

Aij D Ai \ Aj ;where 1 � i < j � r; Aijk D Ai \ Aj \Ak;where 1 � i < j < k � r;

::: :::

Ai1i2���iq D Ai1 \Ai2 \ � � � \A1q ;where 1 � i1 < i2 < � � � iq � r; :::

:::

A12���r D A1 \ � � � \ Ar :

Theorem 5.45 (Inclusion–Exclusion). Given finite subsets A1; : : : ; Ar of a

set X , we have

jA1[� � �[Ar j D X

i�r jAi j�

X

i<j�r jAij jC

X

i<j<k�r jAijkj�� � �C.�1/r�1jA12���r j:

Proof. The proof is by induction on r � 2. The base step is Lemma 5.42. For the inductive step, the same Lemma gives

jA1[� � �[Ar�1/[Ar j D jA1[� � �[Ar�1jC jArj� j.A1[� � �[Ar�1/\Ar j:

Now

j.A1 [ � � � [Ar�1/\ Ar j D j.A1 \ Ar/ [ � � � [ .Ar�1 \Ar/j D jA1r [ � � � [Ar�1 r j;

and the inductive hypothesis gives

jA1 [ � � � [Ar�1j D X

i�r�1 jAi j �

X

i<j�r�1 jAij j C � � � C .�1/r�2jA12���.r�1/j

as well as

jA1r [ � � � [Ar�1 r j D X

i<r

jAir j � X

i<j<r

jAijrj C � � � C .�1/r�2jA12���r j:

Finally, collect terms, realizing that �.�1/r�2 D .�1/r�1.

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5.4 Connections: Boolean Things 231

In Exercise 5.76 below, you will use Inclusion-Exclusion to give a formula

computing the Euler �-function �.n/.

Here is an interesting special case of Theorem 5.45 that applies when all

intersectionsAi1 \� � �\Aiq , for each q � r , have the same number of elements.

Corollary 5.46 (Uniform Inclusion-Exclusion). If A1; : : : ; Ar are finite sub-

sets of a set X such that, for each q � r , there is an integer sq with

jAi1 \ � � � \ Aiq j D sq;

then

jA1 [ � � � [Ar j D rs1 � r

2

! s2 C

r

3

! s3 � � � � C .�1/r�1sr :

Proof. By hypothesis, jAi j D s1 for all i , and so P i jAi j D rs1. How many

terms are there in the sum P 1�i1<:::<iq�r jAi1���iq j? If q D 2, there is one term

jAij j D jAi \ Aj j for each pair of distinct Ai , Aj in fA1; : : : ; Arg; that is, there’s one term for each choice of 2 of the r subsets. If q D 3, there is one term jAijkj D jAi \Aj \Akj for each triple of distinctAi , Aj ,Ak in fA1; : : : ; Arg; that is, there’s one term for each choice of 3 of the r subsets. In general, there

are r choose q terms in the sum P 1�i1<i2<���<iq�r jAi1i2���iq j; thus, there are�

r q

� terms of the form jAi1i2���iq j.Therefore, the sum

P 1�i1<:::<iq�r jAi1���iq j in

the Inclusion-Exclusion formula is here equal to � r q

� sq .

Example 5.47. Social Security numbers are 9-digit numbers of the form xxx-

xx-xxxx (there are some constraints on the digits, but let’s not worry about

them here). How many Social Security numbers are there that contain all the odd digits?

As usual, it is easier to compute the size of the complement of a union. Let

X be the set of all 9-digit numbers and, for i D 1; 3; 5; 7; 9, let

Ri D fn 2 X W i is not a digit in ng:

Thus, R1[R3[R5 [R7[R9 consists of all 9-digit numbers missing at least one odd digit. There are 109 Social Security numbers. For each i , we have jRi j D 99 (for i does not occur). If i < j , then jRi \ Rj j D 89 (for i and j do not occur); if i < j < k, then jRi \ Rj \ Rk j D 79, and so forth. By Corollary 5.46,

jR1 [R3 [ R5 [R7 [R9j D 5 � 99 � 5

2

! 89 C

5

3

! 79 �

5

4

! 69 C 59:

Therefore, the answer is 109 � jR1 [R3 [R5 [ R7 [R9j. N You can compute this number explicitly if you

really care to know it.

Exercises

5.72 There is a class of students, all of whom are taking French or Spanish. If 15

students are studying French, 12 are studying Spanish, and 4 are studying both,

how many students are in the class? Notice that “or” is not “exclusive or.”

Answer: 23.

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232 Chapter 5 Abstract Algebra

5.73 There is a class of students, all of whom are taking either French, German, or

Spanish. Suppose that 15 students are studying French, 12 students are studying

German, and 10 students are studying Spanish; moreover, 4 students are study-

ing French and German, 5 are studying German and Spanish, and 3 are studying

French and Spanish. One brave soul is studying all three at once. How many stu-

dents are in the class?

Answer: 26.

5.74 Is “Inclusion-Exclusion” an appropriate name for Theorem 5.45? Why?

5.75 Elvis is playing a game in which he tosses a fair coin and rolls a fair die. He wins

if either the coin comes up heads or the die rolls a multiple of 3. What is the

probability that Elvis wins the game?

5.76 * Recall that if p is a prime and � is the Euler-� function, then �.p/ D p � 1 (see page 111).

(i) Suppose n D pe11 p e2 2 p

e3 3 is a product of three prime powers. Show that

�.n/ D n � n p1

� n p2

� n p3

C n p1p2

C n p1p3

C n p2p3

� n p1p2p3

:

(ii) Generalize to show that if n D pe11 p e2 2 : : : p

en n , where p1; : : : ; pn are distinct

primes, then

�.n/ D n

0 @1 �

X

i

1

pi C X

i;j

1

pipj � X

i;j;l

1

pipjpl C � � � C .�1/k 1

p1 � � � pk

1 A :

(iii) Using the notation of part (ii), show that

�.n/ D n nY

kD1

� 1 � 1

pk

� :

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6 Arithmetic of Polynomials The two most important rings appearing in precollege mathematics are Z and

kŒx (where k is usually Q, R, or C). The goal of this chapter is to show that

these rings share some basic structural properties: both are domains, each has a division algorithm, and non-units in each are products, in essentially only one

way, of irreducibles (primes in Z, polynomials in kŒx having no nontrivial

factorizations); there are numerous other parallels as well. Our program is to

take familiar results about Z and investigate their analogs in kŒx. Sometimes a

translation from Z to kŒx is quite simple—not only is the analog of a theorem in Chapter 1 true, but so is its proof, mutatis mutandis; in other cases, however,

some modifications in proofs are necessary.

6.1 Parallels to Z

Divisibility

Let’s begin with a discussion of divisibility.

Definition. If R is a commutative ring and a; b 2 R, then a is a divisor of b, denoted by

a j b;

if there is r 2 R with b D ar . We continue using the usual synonyms: a divides b or b is a multiple of a.

The next result, analogous to Lemma 1.13, will be very useful in what fol-

lows. It allows us to use degree in kŒx as a proxy for absolute value in Z.

Lemma 6.1. Let k be a field and let f .x/; g.x/ 2 kŒx. If f ¤ 0 and f j g, then

deg.f / � deg.g/:

Proof. If g D f q, where q.x/ 2 kŒx, then Lemma 5.8(ii) gives deg.g/ D deg.f q/ D deg.f /Cdeg.q/. Since deg.q/ � 0, we have deg.f / � deg.g/.

Recall that a unit in a commutative ringR is an element that has a multiplic-

itive inverse in R. The only units in Z are ˙1, but the polynomial ring kŒx, where k is a field, has many units, as the next proposition shows.

Proposition 6.2. If k is a field, then u.x/ 2 kŒx is a unit if and only if u is a

Sometimes, we’ll denote a

polynomial by f.x/; other

times, we’ll simply write

f . Both conventions are

commonly used in algebra.

nonzero polynomial of degree 0I that is, u is a nonzero constant. 233

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234 Chapter 6 Arithmetic of Polynomials

Proof. If u is a unit, then there is a polynomialv.x/ 2 kŒxwithuv D 1. Thus, u j 1 and, by Lemma 6.1, we have deg.u/ � deg.1/ D 0. Hence, deg.u/ D 0.

Conversely, if deg.u/ D 0, then u 2 k. Since k is a field and u ¤ 0, there is an inverse u�1 in k; that is, u is a unit in k. A fortiori,u is a unit in kŒx.

Describing the units in kŒxwhen k is not a field is much more complicated.

For example, a nonzero constant need not be a unit: 5 is not a unit in ZŒx. And

a unit need not be a constant: .2xC 1/2 D 4x2C 4xC 1 D 1 in Z4Œx, so that 2x C 1 is a unit in Z4Œx (it is its own inverse).

Multiplying an element of a commutative ring by a unit doesn’t change any of its essential algebraic properties. It’s convenient to give a name to elements

that are so related.

Definition. An associate of an element a in a commutative ring R is an ele-

ment of the form ua for some unit u 2 R.

Example 6.3. (i) Since the only units in Z are ˙1, the associates of an inte- ger m are ˙m.

(ii) There are only four units in the Gaussian integers ZŒi , by Proposition 4.42: namely ˙1 and ˙i . Hence, every nonzero Gaussian integer z has four associates: z;�z; iz;�iz.

(iii) There are exactly six units in the Eisenstein integers ZŒ!, where ! D 1 2

� �1C i

p 3 �

, by Exercise 4.45 on page 165. Hence, every Eisenstein

integer z has exactly six associates: ˙z;˙!z;˙!2z: (iv) If k is a field, Proposition 6.2 says that the associates of f .x/ 2 kŒx are

nonzero multiples uf for u 2 k. N

Proposition 6.4. If k is a field, every nonzero polynomial in kŒx has a monic

associate.

Proof. If the leading coefficient of f is c, then c, being a nonzero element of

k, is a unit, and so f is associate to c�1f .

In a commutative ring R, every element a 2 R is divisible by units u (for a D u.u�1a/) and associates ua [for a D u�1.ua/]. An element having only these obvious divisors is called irreducible.

Definition. An element a in a commutative ring R is irreducible in R if it is

neither zero nor a unit and its only divisors are units and associates.

An integer n is irreducible in Z if and only if n D ˙p for some prime p;The definition of prime on page 22 says that primes

are positive. that is, n is an associate of a prime. When k is a field, Proposition 6.2 implies

that every associate uf of a polynomial f .x/ 2 kŒx has the same degree as f , and it is easy to see that if f is irreducible, then uf is also irreducible.

How to Think About It. We have defined irreducible in R, not irreducible, for irreducibility depends on the ambient ring. In particular, irreducibility of a

polynomial in kŒx depends on the coefficient ring k, hence on R D kŒx. For example, the polynomial x2 C 1, when viewed as lying in RŒx, is irreducible. On the other hand, when x2 C 1 is viewed as lying in the larger ring CŒx,

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6.1 Parallels to Z 235

it is not irreducible, for x2 C 1 D .x C i/.x � i/ and neither factor is a unit. Similarly, a primep may factor in some larger commutative ring containing Z.

For example, in the Gaussian integers ZŒi , the prime 5 in Z factors: 5 D .2C i/.2� i/. Since the only units in ZŒi  are ˙1 and ˙i , by Example 6.3(iii), the factors are neither units nor associates of 5 in ZŒi .

In general, testing a polynomial for irreducibility is hard. Here is a crite-

rion for irreducibility of polynomials over fields that uses degree to narrow the

kinds of polynomials that need to be tested as factors.

Proposition 6.5. Let k be a field and let f .x/ 2 kŒx be a nonconstant polyno- mial. Then f is irreducible in kŒx if and only if it has no factorizationf D gh in kŒx with both factors having degree < deg.f /.

Proof. If f is irreducible in kŒx and f D gh, then one factor, say g, is a unit (why?). By Proposition 6.2, we have deg.g/ D 0 < deg.f /, for f is nonconstant.

Conversely, if f D gh and f is not a product of polynomials of smaller degree, then one factor, say g, must have degree 0, hence it is a unit. Therefore And the other factor is

h D g�1f , an associate of f .

f is irreducible in kŒx.

Every linear polynomial a.x/ D rx C s 2 kŒx, where k is a field, is ir- reducible in kŒx: if a D fg, then 1 D deg.a/ D deg.f / C deg.g/. Hence, deg.f /; deg.g/ 2 f0; 1g. It follows that one degree is 0 and the other is 1, and so a is irreducible, by Proposition 6.5. There are fields k whose only ir- See Corollary 6.15.

reducible polynomials are linear; for example, the Fundamental Theorem of

Algebra says that C is such a field.

Proposition 6.5 need not be true if the ring of coefficients is not a field. Indeed, linear polynomials need not be irreducible. For example, 5x C 5 D 5.x C 1/ is not irreducible in ZŒx, even though one factor has degree 0 and the other degree 1, for 5 is not a unit in ZŒx.

Proposition 6.6. Let R be a domain and let a; b 2 R.

(i) a j b and b j a if and only if a and b are associates. (ii) Let k be a field and a; b 2 R D kŒx be monic polynomials. If a j b and

b j a, then a D b.

Proof. (i) If a j b and b j a, there are r; s 2 R with b D ra and a D sb, and so b D ra D rsb. If b D 0, then a D 0 (because b j a); if b ¤ 0, then we may cancel it (R is a domain) to obtain 1 D rs. Hence, r and s are units, and a and b are associates. The converse is obvious (and it does not

need the hypothesis that R be a domain).

(ii) Corollary 5.9 tells us that R is a domain, so, by part (i), there is a unit u 2 kŒx with a D ub. Now u is a nonzero constant, by Proposition 6.2. Because a j b and b j a, a and b have the same degree (by Lemma 6.1), say m. Since they are monic, the leading coefficient of ub is u and the

leading coefficient of a is 1. Hence u D 1 and a D b.

The next example shows that we need the hypothesis in Proposition 6.6 that

R be a domain.

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236 Chapter 6 Arithmetic of Polynomials

Example 6.7 (Kaplansky). Let X be the interval Œ0; 3. We claim that there

are elements a; b 2 C.X/ (see Example 4.31(ii)) each of which divides the other yet which are not associates. Define

a.t/ D 1 � t D b.t/ for all t 2 Œ0; 1 a.t/ D 0 D b.t/ for all t 2 Œ1; 2 a.t/ D t � 2 for all t 2 Œ2; 3 b.t/ D �t C 2 for all t 2 Œ2; 3:

If v 2 C.X/ satisfies v.t/ D 1 for all t 2 Œ0; 1 and v.t/ D �1 for all t 2 Œ2; 3, then it is easy to see that b D av and a D bv (same v); hence, a and b divide each other.

Suppose a and b are associates: there is a unit u 2 C.X/ with b D au. As for v above, u.t/ D 1 for all t 2 Œ0; 1 and u.t/ D �1 for all t 2 Œ2; 3; in particular, u.1/ D 1 and u.2/ D �1. Since u is continuous, the Intermediate Value Theorem of calculus says that u.t/ D 0 for some t 2 Œ1; 2. But this contradicts Exercise 4.41(ii) on page 164, which says that units in C.X/ are never 0. N

The next result shows that irreducible polynomials over a field behave like

primes in Z; they are “building blocks” in the sense that every nonconstant

polynomial can be expressed in terms of them.

Proposition 6.8. If k is a field, then every nonconstant polynomial in kŒx is

a product of irreducibles.

Proof. If the proposition is false, then the setWe continue to use the term product as we have in

earlier chapters: a product

can have only one factor.

Thus, it’s okay to say that

a single irreducible is a

product of irreducibles.

C D fa.x/ 2 kŒx W a is neither a constant nor a product of irreduciblesg

is nonempty. Let h.x/ 2 C have least degree (the Least Integer Axiom guaran- tees h exists). Since h 2 C , it is not a unit, and so 0 < deg.h/; since h is not ir- reducible, h D fg, where neither f nor g is a unit, and so, by Proposition 6.2, neither f nor g is constant. Hence, Lemma 6.1 gives 0 < deg.f / < deg.h/

and 0 < deg.g/ < deg.h/. It follows that f … C and g … C , for their degrees are too small (h has the smallest degree of polynomials in C ). Thus, both f

and g are products of irreducibles and, hence, h D fg is a product of irre- ducibles, contradicting h 2 C . Therefore, C is empty, and the proposition is true.

Corollary 6.9. If k is a field, then every nonconstant f .x/ 2 kŒx has a fac- torization

f .x/ D ap1.x/ � � �pn.x/;

where a is a nonzero constant and the pi are monic irreducibles.

Proof. Apply the result of Exercise 6.8 on page 243 to a factorization of f as

in the proposition.

We continue showing that polynomials over fields behave very much like

integers. Let’s first do some long division.

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6.1 Parallels to Z 237

4x3�14x2

x2 C 3x � 2 � 4x5� 2x4 C x3 � � � 4x5C12x4�8x3

�14x4C9x3 :::

This process can be completed until we get 0 or a remainder of degree < 2

(which is it?). Generalizing, there is a Division Algorithm for RŒx, where R

is any commutative ring: if a.x/; b.x/ 2 RŒx and a is monic, then there are q.x/; r.x/ 2 RŒx with b D qa C r , where r D 0 or deg.r/ < deg.a/. The basic idea is to mimic what we’ve just done.

Proposition 6.10. Let R be a commutative ring and f .x/; g.x/ 2 RŒx. If f is monic, then there exist q.x/; r.x/ 2 RŒx with

g D qf C r;

where r D 0 or deg.r/ < deg.f /.

Proof. Let

f D xn C an�1xn�1 C � � � C a0 and g D bmxm C bm�1xm�1 C � � � C b0:

If m D deg.g/ < deg.f / D n, then take q D 0 and r D g. In Z, if b < a, then b D 0aC b; for example, 27 D 0 � 35C 27. Similarly for polynomials: x2C 1 D 0.x3Cx2�1/C .x2C1/.

If m � n, the quotient begins with bmxm�n multiplied by f ; now subtract, getting a polynomial of degree less thanm. The rest of the proof is by induction

on m D deg.g/ � n. If

G.x/ D g � bmxm�nf;

then either G D 0 or deg.G/ < m D deg.g/. If G D 0, we are done: set q D bmxm�n and r D 0. IfG ¤ 0, the inductive hypothesis gives polynomials q0 and r withG D q0f Cr , where either r D 0 or deg.r/ < deg.f /. Therefore, g � bmxm�nf D q0f C r , and so

g D � bmx

m�n C q0 � f C r:

When R is a field, we can divide by every nonzero polynomial, not merely

by monic ones; moreover, the quotient and remainder are unique.

Theorem 6.11 (Division Algorithm). Let k be a field and f .x/; g.x/ 2 kŒx. If f ¤ 0, then there exist unique q.x/; r.x/ 2 kŒx with

g D qf C r;

where r D 0 or deg.r/ < deg.f /.

Proof. We first prove the existence of q and r . Now f D anxn C � � � C a0, where an ¤ 0. Since k is a field, it contains the inverse a�1n . Hence, a�1n f is monic, and Proposition 6.10 gives q0.x/; r.x/ 2 kŒx with

g D q0.a�1n f /C r;

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238 Chapter 6 Arithmetic of Polynomials

where either r D 0 or deg.r/ < deg.a�1n f / D deg.f /. Therefore,

g D qf C r;

where q D q0a�1n . To prove uniqueness of q and r , assume that g D Qf C R, where R D 0

or deg.R/ < deg.f /. Then qf C r D g D Qf CR, and

.q �Q/f D R � r:

If R ¤ r , then each side, being nonzero, has a degree. Since k is a field, kŒx is a domain (Lemma 5.8), and so

deg � .q �Q/f

� D deg.q �Q/C deg.f / � deg.f /;

while deg.R � r/ � maxfdeg.R/; deg.r/g < deg.f /, a contradiction. Hence, R D r and .q �Q/f D 0. As f ¤ 0, it can be canceled: thus, q �Q D 0 and q D Q.

By Exercise 6.5 on page 243, Theorem 6.11 remains true if we weaken the

hypothesis so that k is only a domain.

There is a two-step strategy to determine whether one integer divides an-

other: first, use the Division Algorithm; then show that the remainder is zero. This same strategy can now be used for polynomials.

Example 6.12. This example shows that quotients and remainders may not

be unique when the coefficients do not lie in a domain. In Z4Œx, let b.x/ D 2x3 C 3 and a.x/ D 2x2 C 2x C 1. Then

2x3 C 3 D .x C 1/.2x2 C 2x C 1/C .x C 2/ D .x C 3/.2x2 C 2x C 1/C x:

The quotient and remainder in the first equation are xC 1 and xC 2, while the quotient and remainder in the second equation are x C 3 and x. Note that both x C 2 and x are linear, and hence

deg.x C 2/ D deg.x/ D 1 < deg.a/

D 2: N

In forthcoming investigations into roots of unity, we’ll need to know whether

xm � 1 divides xn � 1. Certainly this is true when m j n because, if n D mq,

xn � 1 D xmq � 1 D .xm/q � 1

D .xm � 1/ � .xm/

q�1 C .xm/q�2 C � � � C � � � .xm/2 C xm C 1 � :

The converse is also true, and the proof uses the Division Algorithms in Z and

in kŒx.

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6.1 Parallels to Z 239

Proposition 6.13. If k is a field, then xm�1 divides xn�1 in kŒx if and only ifm j n.

Proof. We’ve seen above that xm � 1 divides xn � 1 ifm j n. Conversely, suppose that xm � 1 divides xn � 1. If n D mq C r where

0 � r < m, then

xn � 1 D xmqCr � 1 D xmqCr � xr C xr � 1 D xr .xmq � 1/C .xr � 1/ :

We’re assuming that xm � 1 divides xn � 1 and, as in the discussion just preceding this proposition, xm � 1 divides xmq � 1. Hence, by the 2 out of 3 property for polynomials (Exercise 6.7 on page 243), xm � 1 divides xr � 1. Since r < m, we must have r D 0 (why?).

Roots

We are going to apply the preceding results to roots of polynomials. We’ve

been using the word “root” all along; let’s begin with a formal definition.

Definition. If f .x/ 2 kŒx, where k is a field, then a root of f in k is an element a 2 k with f .a/ D 0.

How to Think About It. We have just defined “root in k,” not “root.” Often,

a root of a polynomial f .x/ 2 kŒx may live in a larger field K containing k, but we still call it a root of f . For example, f .x/ D x2 � 2 has its coefficients in Q, but we usually say that

p 2 is a root of f even though

p 2 is irrational;

that is, p 2 … Q.

Etymology. Why is a root so called? Just as the Greeks called the bottom

side of a triangle its base (as in the area formula 1 2

altitude � base), they also called the bottom side of a square its base. A natural question for the Greeks

was: given a square of area A, what is the length of its side? Of course, the answer is

p A. Were we inventing a word for

p A, we might have called it the

base ofA or the side ofA. Similarly, consider the analogous three-dimensional

question: given a cube of volume V , what is the length of its edge? The answer 3 p V might be called the cube base of V , and

p A might then be called the

square base of A. Why, then, do we call these numbers cube root and square

root? What has any of this to do with plants? Since tracing the etymology of words is not a simple matter, we only sug-

gest the following explanation. Through 400 CE, most mathematics was written

in Greek, but, by the fifth century, India had become a center of mathematics,

and important mathematical texts were also written in Sanskrit. The Sanskrit

term for square root is pada. Both Sanskrit and Greek are Indo-European lan- guages, and the Sanskrit word pada is a cognate of the Greek word podos;

both mean base in the sense of the foot of a pillar or, as above, the bottom of a

square. In both languages, however, there is a secondary meaning “the root of a

plant.” In translating from Sanskrit, Arab mathematicians chose the secondary

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240 Chapter 6 Arithmetic of Polynomials

meaning, perhaps in error (Arabic is not an Indo-European language), perhaps

for some unknown reason. For example, the influential book by al-Khwarizmi,

Al-jabr w’al muqabala, which appeared in the year 830 CE, used the ArabicThis title can be translated from Arabic, but the

words already had a

technical meaning: both

jabr and muqabala refer

to operations akin to

subtracting the same

number from both sides of

an equation.

word jidhr, meaning root of a plant. (The word “algebra” is a European ver-

sion of the first word in the title of this book; the author’s name has also come

into the English language as the word “algorithm.”) This mistranslation has

since been handed down through the centuries; the term jidhr became standard

in Arabic mathematical writings, and European translations from Arabic into Latin used the word radix (meaning root, as in radish or radical). The notation

r2 for p 2 occurs in European writings from about the twelfth century (but the

square root symbol did not arise from the letter r ; it evolved from an old dot

notation). However, there was a competing notation in use at the same time,

for some scholars who translated directly from the Greek denoted p 2 by l2,

where l abbreviates the Latin word latus, meaning “side.” Finally, with the in-

vention of logarithms in the 1500s, r won out over l , for the notation l2 was

then commonly used to denote log2. The passage from square root to cube

root to the root of a polynomial equation other than x2 � a and x3 � a is a natural enough generalization. Thus, as pleasant as it would be, there seems to

be no botanical connection with roots of equations.

Proposition 6.14 (Remainder Theorem). Let f .x/ 2 kŒx, where k is a field. If u 2 k, then there is q.x/ 2 kŒx with

f .x/ D q.x/.x � u/C f .u/:

Proof. The Division Algorithm gives

f .x/ D q.x/.x � u/C r;

where either r D 0 or deg.r/ < deg.x � u/ D 1; hence, the remainder r is a constant. By Corollary 5.21, evaluation at u is a homomorphism; hence,

f .u/ D q.u/.u � u/C r , and so f .u/ D r .

Proposition 6.14 is often paraphrased to say that f .u/ is the remainder after dividing f .x/ by x � u.

Here is a connection between roots and factoring.

Corollary 6.15 (Factor Theorem). Let f .x/ 2 kŒx, where k is a field, and let a 2 k. Then a is a root of f in k if and only if x � a divides f .

Proof. If a is a root of f in k, then f .a/ D 0, and Proposition 6.14 gives f .x/ D q.x/.x � a/. Conversely, if f .x/ D g.x/.x � a/, then evaluating at a gives f .a/ D g.a/.a � a/ D 0; that is, a is a root of f in k.

The next result turns out to be very important.

Theorem 6.16. Let k be a field. If f .x/ 2 kŒx has degree n, then f has at most n roots in k.

Proof. We prove the statement by induction on n � 0. If n D 0, then f is a nonzero constant, and the number of its roots in k is zero. Now let n > 0. If f

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6.1 Parallels to Z 241

has no roots in k, we are done, for 0 � n. Otherwise, we may assume that f has a root a in k. By Corollary 6.15,

f .x/ D q.x/.x � a/I

moreover, q.x/ 2 kŒx has degree n� 1. If there is another root of f in k, say b, where b ¤ a, then evaluating at b gives

0 D f .b/ D q.b/.b � a/:

Since b � a ¤ 0, we have q.b/ D 0 (for k is a field, hence a domain); that is, b is a root of q. But deg.q/ D n� 1, so that the inductive hypothesis says that q has at most n � 1 roots in k. Therefore, f has at most n roots in k, namely a and the roots of q.

Example 6.17. Theorem 6.16 is not true for polynomials with coefficients in

an arbitrary commutative ring. For example, the quadratic polynomial

x2 � 1 in Z8Œx has four roots in Z8, namely 1; 3; 5, and 7. On the other hand, Exercise 6.14 on page 247 says that Theorem 6.16 remains true if we assume

that the coefficient ring is only a domain. N

Recall that every polynomial f .x/ 2 kŒx determines the polynomial func- tion f # 2 Poly.k/, where f #W k ! k is defined by a 7! f .a/ for all a 2 k. On page 204, however, we saw that the nonzero polynomial f .x/ D xp � x 2 FpŒx determines the constant function zero; different polynomials can deter-

mine the same polynomial function. This pathology vanishes when the field k

is infinite.

Proposition 6.18. Let k be an infinite field and f .x/; g.x/ 2 kŒx. If f and g determine the same polynomial function .that is, f # D g#, so that f .a/ D g.a/ for all a 2 k/, then f D g.

Proof. If f ¤ g, then the polynomial h D f �g, being nonzero, has a degree, say n. But every element of k is a root of h; since k is infinite, h has more than

n roots, and this contradicts Theorem 6.16.

This proof yields a more general result.

Corollary 6.19. Let k be a .possibly finite/ field, and let f .x/; g.x/ 2 kŒx, where deg.f / � deg.g/ D n. If f .a/ D g.a/ for nC 1 elements a 2 k, then f D g.

Proof. If f ¤ g, then deg.f � g/ is defined; but deg.f � g/ � n, and so f � g has too many roots.

We can now show that kŒx and Poly.k/ are structurally the same for the

most familiar fields k.

Theorem 6.20. If k is an infinite field, then

kŒx Š Poly.k/:

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242 Chapter 6 Arithmetic of Polynomials

Proof. In Example 5.14(ii), we saw that 'W kŒx ! Poly.k/, sending f 7! f #, is a surjective homomorphism. Since k is infinite, Proposition 6.18 applies to

show that ' is injective. Therefore, ' is an isomorphism.

We now generalize Proposition 6.18 to polynomials in several variables.

Denote the n-tuple .x1; : : : ; xn/ by X .

Proposition 6.21. Let k be an infinite field.

(i) If f .X/ 2 kŒX D kŒx1; : : : ; xn is nonzero, then there are a1; : : : ; an 2 k with f .a1; : : : ; an/ ¤ 0.

(ii) If f .X/; g.X/ 2 kŒX and

f .a1; : : : ; an/ D g.a1 ; : : : ; an/ for all .a1; : : : ; an/ 2 kn;

then f D g.

Proof. (i) The proof is by induction on n � 1. If n D 1, then the result is Proposition 6.18, because f .a/ D 0 for all a 2 k implies f D 0. For the inductive step, assume that

f .x1; : : : ; xn/ D B0 C B1xn C B2x2n C � � � C Brxrn;

where all Bi 2 kŒx1; : : : ; xn�1 and Br ¤ 0. By induction, there is ˛ D .a1; : : : ; an�1/ 2 kn�1 with Br.˛/ ¤ 0. Hence, f .˛; xn/ 2 kŒxn, and

f .˛; xn/ D B0.˛/C B1.˛/xn C � � � C Br.˛/xrn ¤ 0:

By the base step, there is an 2 k with f .˛; an/ ¤ 0.Since f.˛; xn/ and g.˛;xn/ lie in kŒxn, we

have f.˛; b/; g.˛; b/ 2 k. (ii) The proof is by induction on n � 1; the base step is Proposition 6.18. For the inductive step, write

f .X; y/ D X

i

pi .X/y i and g.X; y/ D

X

i

qi .X/y i ;

where X denotes .x1; : : : ; xn�1/ (by allowing some p’s and q’s to be zero,we may assume that both sums involve the same indices i ). Suppose that f .˛; b/ D g.˛; b/ for every ˛ 2 kn�1 and every b 2 k. For fixed ˛ 2 kn�1, define F˛.y/ D

P i pi.˛/y

i andG˛.y/ D P i qi.˛/y

i . Since

both F˛.y/ and G˛.y/ are in kŒy, the base step gives pi.˛/ D qi .˛/ for all ˛ 2 kn. By the inductive hypothesis, pi .X/ D qi .X/ for all i , and hence

f .X; y/ D X

i

pi .X/y i D

X

i

qi .X/y i D g.X; y/:

Exercises

6.1 Prove that the only units in ZŒx are ˙1, and that the only associates of a polyno- mial f .x/ 2 ZŒx are ˙f .

6.2 * Let R be a domain, and let p.x/; q.x/ 2 RŒx. (i) If p and q are irreducible, prove that p j q if and only if there is a unit u with

q D up. (ii) If, in addition, both p and q are monic, prove that p j q implies u D 1 and

p D q.

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6.1 Parallels to Z 243

6.3 (i) If R is a domain, prove that the only units in RŒx are units in R.

(ii) The domain Z2 has only one unit. Give an example of an infinite domain

having only one unit.

6.4 Let R be a commutative ring and let a.x/; b.x/ 2 RŒx, where a ¤ 0. Prove that Proposition 6.10 generalizes: if the leading coefficient of a is a unit, then there

exist q.x/; r.x/ 2 RŒx with b D qaCr , where either r D 0 or deg.r/ < deg.a/.

6.5 * Let k be a domain and let a.x/; b.x/ 2 kŒx, where a ¤ 0. Prove that the uniqueness statement in the Division Algorithm generalizes: if there are q; r; Q; R

in kŒx with qa C r D b D Qa CR, where r D 0 or deg.r/ < deg.a/, and where R D 0 or deg.R/ < deg.a/, then R D r and Q D q.

6.6 Let k be a domain and let f .x/ 2 kŒx. If a.x/ is an associate of f , prove that deg.f / D deg.a/. Give an example to show that the statement may be false if k is not a domain.

6.7 * Show that there is a “2 out of 3” result for polynomials, analogous to the one for

integers: if k is a field and f; g; h 2 kŒx are polynomials such that f D g C h, then a polynomial that divides two of the three will divide the third.

6.8 * Let R be a domain and f .x/ 2 RŒx be nonzero. If f D g1 � � � gn, where gi .x/ 2 RŒx for all i , show that there exist a nonzero a 2 R and monic g0i .x/ 2 RŒx with f D ag01 � � � g0n.

6.9 (i) Let f .x/; g.x/ 2 QŒx with f monic. Write a pseudocode (or a program in a CAS) implementing the Division Algorithm with input f; g and output

q.x/; r.x/, the quotient and remainder.

(ii) Find the quotient and remainder by dividing x3 C 2x2 � 8x C 6 by x � 1 as you would in high school.

6.10 * If R is a commutative ring, define a relation � on R by a � b if they are associates. Prove that � is an equivalence relation on R.

6.11 A student claims that x � 1 is not irreducible in QŒx because there is a factoriza- tion x � 1 D .

p x C 1/.

p x � 1/. Explain the error of his ways.

6.12 * Prove that the ideal .x; y/ in kŒx; y, where k is a field, is not a principal ideal.

Greatest Common Divisors

We now introduce gcd’s of polynomials f .x/; g.x/ 2 RŒx. It doesn’t make sense to say that f � g, even when R D R, but it does make sense to say deg.f / � deg.g/. Although some of the coming definitions make sense for polynomial rings RŒx over a commutative ring R, we will focus our attention

on the rings kŒx for fields k.

Definition. Let k be a field. A common divisor of polynomials a.x/; b.x/ 2 kŒx is a polynomial c.x/ 2 kŒxwith c j a and c j b. If a and b are not both 0, define their greatest common divisor, denoted by Note the convention that

greatest common divisors

are monic. We’ll say more

about this in a moment. gcd.a; b/;

to be a monic common divisor d of a and b of largest degree. If a D 0 D b, define gcd.0; 0/ D 0.

The next proposition shows that gcd’s exist; it is true, but not obvious, that

every pair a; b 2 kŒx has a unique gcd (Corollary 6.29).

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244 Chapter 6 Arithmetic of Polynomials

Proposition 6.22. If k is a field and a.x/; b.x/ 2 kŒx, then a gcd of a; b exists.

Proof. We saw, in Lemma 6.1, that if c and a are polynomials with c j a, then deg.c/ � deg.a/. It follows that gcd’s exist, for common divisors do exist (1 is always a common divisor), and there is an upper bound on the degrees of

common divisors, namely, maxfdeg.a/; deg.b/g. Finally, a common divisor d of largest degree can be replaced by a monic associate.

Defining gcd’s of polynomials to be monic is just a normalization; after all,

when we defined gcd’s of integers, we insisted they should be positive. This will be needed to prove uniqueness of gcd’s.

Example 6.23. Here is an easy computation of a gcd, generalizing Lemma 1.17.

Let k be a field and p.x/ 2 kŒx be a monic irreducible polynomial. If b.x/ 2 kŒx, then

gcd.p; b/ D ( p if p j b 1 otherwise.

A common divisor c of p and b is, in particular, a divisor of p. But the only

monic divisors of p are p and 1, and so gcd.p; b/ D p or 1; it is p if p j b, and it is 1 otherwise. N

We are going to see that gcd’s of polynomials are linear combinations. The

proof of this fact for gcd’s of integers essentially involved ideals in Z, and so we now examine ideals in kŒx.

In any commutative ring R, associates a and b generate the same principal

ideal (the converse may be false ifR is not a domain).

Proposition 6.24. Let R be a domain and a; b 2 R. The principal ideals .a/ and .b/ are equal if and only if a and b are associates.

Proof. If .a/ D .b/, then a 2 .b/; hence, a D rb for some r 2 R, and so b j a. Similarly, b 2 .a/ implies a j b, and so Proposition 6.6 shows that a and b are associates.

Conversely, if a D ub, where u is a unit, then a 2 .b/ and .a/ � .b/. Similarly, b D u�1a implies .b/ � .a/, and so .a/ D .b/.

Ideals in general commutative rings can be quite complicated, but we have seen, in Theorem 5.29, that every ideal in Z is principal. When k is a field, all

the ideals in kŒx are also principal.

Theorem 6.25. If k is a field, then every ideal in kŒx is a principal ideal. In

fact, either I D .0/ or there is a unique monic d.x/ with I D .d/ D frd W r 2 kg.

Proof. If I D .0/, then I is a principal ideal with generator 0; that is, I D .0/. Otherwise, let a.x/ be a polynomial in I of least degree. Since a 2 kŒx is nonzero, its leading coefficient c ¤ 0; since k is a field, c�1 exists, and d D c�1a is monic. By Proposition 6.24, .a/ D .d/.

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6.1 Parallels to Z 245

Clearly, .d/ � I . For the reverse inclusion, let f .x/ 2 I . By the Division Algorithm, there are q.x/; r.x/ 2 kŒx with f D qd C r , where either r D 0 or deg.r/ < deg.d/. But r D f � qd 2 I , so that if r ¤ 0, its existence contradicts d being a polynomial in I of least degree. Hence, r D 0, d j f , and f 2 .d/. Therefore, I � .d/, and I D .d/.

To prove uniqueness, suppose that d 0.x/ 2 kŒx is a monic polynomial with .d 0/ D .d/. By Proposition 6.24, d 0 and d are associates; there is a unit u 2 kŒx with d 0 D ud . Now u 2 k, by Proposition 6.2. Since both d 0 and d are monic, we have u D 1 and d 0 D d .

Recall Example 5.27(iv): the set I consisting of all polynomials f .x/ 2 RŒx having i as a root is an ideal in RŒx containing .x2 C 1/. We can now say more.

Corollary 6.26. The ideal I � RŒx consisting of all polynomials f .x/ 2 RŒx having i as a root is equal to .x2 C 1/.

Proof. Now .x2 C 1/ � I . For the reverse inclusion, we know that I D .d/, where d is the unique monic polynomial of least degree in I . But x2 C 1 is a monic polynomial in I , and there can be no such polynomial of smaller degree

lest i be a root of a linear polynomial in RŒx.

It is not true that ideals in arbitrary commutative rings are necessarily prin-

cipal, as the next example shows.

Example 6.27. Let R D ZŒx, the commutative ring of all polynomials over Z. It is easy to see that the set I of all polynomials with even constant term is

an ideal in ZŒx. We show that I is not a principal ideal.

Suppose there is d.x/ 2 ZŒx with I D .d/. The constant 2 2 I , so that there is f .x/ 2 ZŒx with 2 D df . Since the degree of a product is the sum of the degrees of the factors, 0 D deg.2/ D deg.d/ C deg.f /. Since degrees are nonnegative, it follows that deg.d/ D 0; i.e., d is a nonzero constant. As constants here are integers, the candidates for d are ˙1 and ˙2. Suppose d D ˙2; since x 2 I , there is g.x/ 2 ZŒx with x D dg D ˙2g. But every coefficient on the right side is even, while the coefficient of x on the left side is 1. This contradiction gives d D ˙1. Thus, d is a unit and, by Example 5.30, I D .d/ D ZŒx, another contradiction. Therefore, no such d exists; that is, I But see Exercise 6.22

on page 248. There

is h.x/ 2 ZŒx with I D .2; h/

is not a principal ideal. N

Recall that if R is any commutative ring and a; b 2 R, then a linear combi- nation of a; b is an element of R of the form sa C tb, where s; t 2 R. Given a; b, the set I of all linear combinations of a; b is an ideal in R. The next

theorem parallels Theorem 1.19.

Theorem 6.28. If k is a field and f .x/, g.x/ 2 kŒx, then any gcd of f; g is a linear combination of f and gI that is, if d.x/ is a gcd, then there are s.x/; t.x/ 2 kŒx with

d D sf C tg:

Proof. The set I of all linear combinations of f and g is an ideal in kŒx; by

Theorem 6.25, there is d.x/ 2 kŒx with I D .d/. If both f and g are 0, then

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246 Chapter 6 Arithmetic of Polynomials

d D 0, and we are done; otherwise, we may assume that d is monic. We know that d D sf C tg for some s and t , because d lies in I . We claim that d is a gcd. Now d is a common divisor, for f; g 2 I D .d/. If h is a common divisor of f and g, then f D f1h and g D g1h. Hence, d D sf C tg D .sf1C tg1/h and h j d . Therefore, deg.h/ � deg.d/, and so d is a monic common divisor of largest degree.

We can characterize gcd’s in kŒx.

Corollary 6.29. Let k be a field and let f .x/, g.x/ 2 kŒx. A monic common divisor d.x/ is a gcd of f; g if and only if d is divisible by every common

divisor; that is, if h is any common divisor of f; g, then h j d .

Proof. The end of the proof of Theorem 6.28 shows that if h is a common

divisor, then h j d . Conversely, if h j d , then deg.h/ � deg.d/, and so d is a monic common divisor of largest degree.

Theorem 6.30. Let f .x/, g.x/ 2 kŒx, where k is a field, and let I D .f; g/ be the ideal of all linear combinations of f and g.

(i) If d.x/ 2 kŒx is monic, then d D gcd.f; g/ if and only if I D .d/. (ii) f and g have a unique gcd.Recall that gcd’s are re-

quired to be monic. That’s

essential to uniqueness. Proof. (i) Suppose that d D gcd.f; g/. We show that .d/ � I and I �

.d/. Theorem 6.28 shows that d 2 I ; therefore, .d/ � I (for every multiple of d is also a linear combination). For the reverse inclusion, let

h D uf Cvg 2 I . Now d j f and d j g, because d is a common divisor, and so d j h. Hence, h D rd 2 .d/; that is, I � .d/, and so I D .d/.

Conversely, suppose that I D .d/. Then d D sf C tg, and so every common divisor h of f; g is a divisor of d . Hence, Corollary 6.29 gives d D gcd.f; g/.

(ii) If d and d 0 are gcd’s of f and g, then .d/ D .d 0/, by part (i). Since both d and d 0 are monic, we must have d D d 0, by Theorem 6.25.

How to Think About It. It’s a good idea to stop and take stock of where we

are in our program of displaying parallels between integers and polynomials.

For polynomials over a field, we have, so far

� extended the notion of divisibility

� generalized “prime” to “irreducible”

� shown that factorizations into irreducibles exist

� established a division algorithm

� shown that the gcd of two polynomials exists and is unique

� shown that the gcd of two polynomials is a linear combination of them.

Thinking back to Chapter 1, what’s next? There were two main paths we took

then: one led to unique factorization—the Fundamental Theorem of Arith-

metic; one led to Euclidean Algorithms. We’ll follow both these paths for poly-

nomials.

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6.1 Parallels to Z 247

Exercises

6.13 Find the gcd of each pair .f; g/ in QŒx and write it as a linear combination of f

and g.

(i) .x3 � x2 � x � 2; x3 � 3x2 C 3x � 2/ (ii) .x6 � 1; x5 � 1/

(iii) .x3 � x2 � x � 2; 2x3 � 4x2 C 2x � 4/ (iv) .x6 � 1; x6 C x5 � 2/ (v)

� .2x C 1/.x6 � 1/; .2x C 1/.x5 � 1/

(vi) .3x6 � 3; 2x5 � 2/

6.14 * Let R be a domain. If f .x/ 2 RŒx has degree n, prove that f has at most n roots in R.

Hint: Use Frac.R/.

6.15 If k is a field in which 1 C 1 ¤ 0, prove that p

1 � x2 is not a rational function over k.

Hint: Mimic the classical proof that p

2 is irrational.

6.16 In Exercise 6.10 on page 243, we saw that the relation � on a commutative ring R, defined by a � b if they are associates, is an equivalence relation. Prove that if R is a domain, then there is a bijection from the family of all equivalence classes to

the family of all principal ideals in R.

6.17 *

(i) If f .x/ and g.x/ are relatively prime in kŒx (k a field) and each divides a

polynomial h, prove that their product fg also divides h.

(ii) If p1; p2; : : : ; pn are polynomials so that gcd.pi ; pj / D 1, and each pi di- vides a polynomial h, prove that p1p2 � � � pn also divides h.

6.18 *

(i) Find gcd.3x3 � 2x2 C 3x � 2; 3x2 C x � 2/ in CŒx. (ii) Write a pseudocode (or a program in a CAS) implementing Euclidean Algo-

rithm I.

(iii) Write a pseudocode (or a program in a CAS) implementing Euclidean Algo-

rithm II.

Hint: Model your routine after the functions in Exercise 1.67 on page 36.

6.19 * Prove the converse of Euclid’s Lemma. Let k be a field and let f .x/ 2 kŒx be a nonconstant polynomial; if, whenever f divides a product of two polynomials, it

necessarily divides one of the factors, then f is irreducible. (See Theorem 1.21.)

6.20 (i) Find two polynomials in QŒx whose associated polynomial functions agree

with this input-output table:

Input Output

1 3

4 17

5 26

(ii) Classify the set of all polynomials that agree on the table.

6.21 (i) Show that the set of polynomials in QŒx that vanish on f1; 2; 3g is an ideal in QŒx.

(ii) What is a generator of this ideal?

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248 Chapter 6 Arithmetic of Polynomials

6.22 * In Example 6.27, we saw that the ideal I in ZŒx consisting of all polynomials

with even constant term is not a principal ideal. Find a polynomial h.x/ 2 ZŒx so that I D .2; h/; that is, I consists of all the linear combinations of 2 and h.

6.23 Let k be a field and f .x/; g.x/ 2 kŒx. Generalize Exercises 5.49 and 5.50 on page 220: if d.x/ D gcd.f; g/ and m.x/ D lcm.f; g/, prove that

.f / C .g/ D .d/ and .f / \ .g/ D .m/:

6.24 Show, in Z8Œx, that x 2 � 1 has two distinct factorizations into irreducibles.

Hint: See Example 6.17.

Unique Factorization

The main result in this subsection is a generalization of the Fundamental Theo-

rem of Arithmetic to polynomials: the factorization of every nonconstant poly-

nomial over a field as a product of irreducibles is essentially unique. We begin by proving Euclid’s Lemma for polynomials. As for integers, it

shows that irreducibility is a strong assumption when dealing with divisibility.

Theorem 6.31 (Euclid’s Lemma). Let k be a field and let f .x/; g.x/ 2 kŒx. If p.x/ is an irreducible polynomial in kŒx and p j fg, then

p j f or p j g:

More generally, if p j f1 � � �fn, then p j fi for some i .

Proof. Assume thatp j fg but that p − f . Sincep is irreducible, gcd.p; f / D 1, and so 1 D spC tf for some polynomials s.x/ and t.x/. Therefore,

g D spgC tfg:

But p j fg, by hypothesis, and so Exercise 6.7 on page 243 gives p j g. The last statement follows by induction on n � 2.

The converse of Euclid’s Lemma is true; see Exercise 6.19 on page 247.

Polynomial versions of arithmetic theorems in Chapter 1 now follow.

Definition. Two polynomials f .x/; g.x/ 2 kŒx, where k is a field, are called relatively prime if their gcd is 1.

Corollary 6.32. Let f .x/; g.x/; h.x/ 2 kŒx, where k is a field, and let h and f be relatively prime. If h j fg, then h j g.

Proof. The proof of Theorem 6.31 works here. Since gcd.h; f / D 1, we have 1 D shC tf , and so g D shgC tfg. But fg D hh1 for some h1.x/ 2 kŒx, and so g D h.sg C th1/.

Definition. If k is a field, then a rational function f .x/=g.x/ 2 k.x/ is in lowest terms if f and g are relatively prime.

Proposition 6.33. If k is a field, every nonzero f .x/=g.x/ 2 k.x/ can be put in lowest terms.

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6.1 Parallels to Z 249

Proof. If f D df 0 and g D dg0, where d D gcd.f; g/, then f 0 and g0 are relatively prime, and so f 0=g0 is in lowest terms.

There is an analog of the Euclidean Algorithm in Z that can be applied to

compute gcd’s of polynomials.

Theorem 6.34 (Euclidean Algorithm I). If k is a field and f .x/; g.x/ 2 kŒx, then there is an algorithm computing gcd.f; g/.

Proof. The proof is essentially a repetition of the proof of the Euclidean Algo-

rithm in Z; just iterate the Divison Algorithm. Each line comes from the line

above it by moving some terms “southwest.”

g D q1f C r1 f D q2r1 C r2 r1 D q3r2 C r3

:::

rn�3 D qn�1rn�2 C rn�1 rn�2 D qnrn�1 C rn rn�1 D qnC1rn:

Since the degrees of the remainders are strictly decreasing, the procedure must

stop after at most deg.f / steps. The claim is that d D rn is the gcd, once it is made monic. We see that d is a common divisor of f and g by back

substitution: repeated applications of “2 out of 3,” working from the bottom

up. To see that d is the gcd, work from the top down to show that if c is any common divisor of f and g, then c j ri for every i .

The Euclidean Algorithm may not produce a monic last remainder. The gcd

is the monic associate of the last nonzero remainder.

Example 6.35 (Good Example). Let

f .x/ D 3x3 � 2x2 C 3x � 2 and g.x/ D 3x2 C x � 2I

we compute gcd.f; g/.

3x3 � 2x2 C 3x � 2 D .x � 1/.3x2 C x � 2/C .6x � 4/ 3x2 C x � 2 D

� 1 2 x C 1

2

� /.6x � 4/C 0:

Rewriting in simpler notation:

f D .x � 1/g C r g D

� 1 2 x C 1

2

� r:

The last remainder is 6x�4. As we warned, it’s not monic, and we must make it so. Thus, we need to take its monic associate (multiplying by 1

6 ):

gcd.f; g/ D x � 2 3 : N

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250 Chapter 6 Arithmetic of Polynomials

Example 6.36 (Bad Example). The Euclidean Algorithm applied to integers

is quite efficient, in terms of the number of steps it takes to get to the answer.It’s the same for ordinary long division—hand calcu-

lations for small integers

are quite simple, but they

get very messy when

dividing two large integers.

It’s the same for polynomials, but the steps get quite cumbersome when carried out by hand—the complexity comes from the computational overhead in the

hand calculations, not in the efficiency of the algorithm itself. A CAS removes

this obstacle.

The following steps calculate gcd.x4�x3�5x2C8x�4; 3x3�6x2Cx�2/ via the Euclidean Algorithm; all the quotients and remainders were calculated with a CAS:

x4 � x3 � 5x2 C 8x � 4 D � 1 3 x C 1

3

� .3x3 � 6x2 C x � 2/

C � �10 3 x2 C 25

3 x � 10

3

3x3 � 6x2 C x � 2 D � � 9 10 x � 9

20

� � �10 3 x2 C 25

3 x � 10

3

� C � 7 4 x � 7

2

�10 3 x2 C 25

3 x � 10

3 D � �40 21 x C 20

21

� � 7 4 x � 7

2

Multiplying by 4 7

produces the gcd of x � 2. N

Here is an unexpected bonus from the Euclidean Algorithm.

Corollary 6.37. Let k be a subfield of a field K, so that kŒx is a subring

of KŒx. If f .x/; g.x/ 2 kŒx, then their gcd in kŒx is equal to their gcd in KŒx.

Proof. We may assume that f ¤ 0, for gcd.0; g/ D g (actually, g’s monic associate). The Division Algorithm in KŒx gives

g D Qf CR;

whereQ;R 2 KŒx and eitherR D 0 or deg.R/ < deg.f /; since f; g 2 kŒx, the Division Algorithm in kŒx gives

g D qf C r;

where q; r 2 kŒx and either r D 0 or deg.r/ < deg.f /. But the equation g D qf C r also holds in KŒx because kŒx � KŒx, so that the uniqueness of quotient and remainder in the Division Algorithm in KŒx gives Q D q 2 kŒx and R D r 2 kŒx. Therefore, the list of equations occurring in the Euclidean Algorithm in KŒx is exactly the same as the list occurring in the

Euclidean Algorithm in the smaller ring kŒx. In particular, the gcd, being the

last remainder (made monic), is the same in both polynomial rings.

To illustrate, even though there are more divisors with complex coefficients,

the gcd of 3x3 � 2x2 C 3x � 2 and 3x2 C x � 2, computed in RŒx, is equalSee Exercise 6.18(i) on page 247. to their gcd computed in CŒx.

As in Z, the Division Algorithm in kŒx can also be used to compute coef-

ficients occurring in an expression of the gcd as a linear combination.

Theorem 6.38 (Euclidean Algorithm II). If k is a field and f .x/; g.x/ 2 kŒx, then there is an algorithm finding a pair of polynomials s.x/ and t.x/

with gcd.f; g/ D sf C tg.

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6.1 Parallels to Z 251

Proof. Let d D gcd.f; g/. To find s and t with d D sf C tg, again work from the last remainder back to f and g:

rn D rn�2 � qnrn�1 D rn�2 � qn.rn�3 � qn�1rn�2/ D .1 C qnqn�1/rn�2 � qnrn�3

:::

D sf C tg

Example 6.39. Let’s compute gcd.f; g/, where f .x/ D x3�2x2Cx�2 and g.x/ D x4 � 1.

x4 � 1 D .x C 2/.x3 � 2x2 C x � 2/C .3x2 C 3/ x3 � 2x2 C x � 2 D .x C 2/.3x2 C 3/C 15:

Rewriting in simpler notation:

g D .x C 2/f C r f D .x C 2/r C 15:

We see that the last remainder, 15, is a constant. As we warned, it need not

be monic, and we must make it so. Thus, gcd.f; g/ D 1; that is, f and g are relatively prime.

We now use Euclidean Algorithm II to find s.x/; t.x/ with d D sf C tg. Using letters,

d D f � q0r g D f � q0.g � qf /

D .1 C q0q/f � q0g:

Now set r D 3x2 C 3, q D x C 2, and q0 D 3x C 6. We have

15 D � .1 C .3x C 6/.x C 2/

� f � .x C 2/g

D .3x2 C 12x C 13/f � .x C 2/g:

Since gcd’s are monic,

1 D 1 15 .3x2 C 12x C 13/f � 1

15 .x C 2/g:

A computer can be programmed to carry out Euclidean Algorithm II (see

Exercise 6.18 on page 247). Once programmed, messy calculations are not a

problem. Indeed, using the polynomials from Example 6.36, we have

gcd.x4 � x3 � 5x2 C 8x � 4; 3x3 � 6x2 C x � 2/ D x � 2

Working as above (with the help of a CAS), we get � 9 10 x C 9

20

� .x4 � x3 � 5x2 C 8x � 4/

C � � 3 10 x2 � 9

20 x C 17

20

� .3x3 � 6x2 C x � 2/

D 7 4 x � 7

2 :

Multiplying both sides of this equation by 4 7

gives the linear combination. N

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252 Chapter 6 Arithmetic of Polynomials

The next result, an analog for polynomials of the Fundamental Theorem

of Arithmetic, shows that the factorization of a polynomial as a product of

irreducible polynomials is essentially unique.

Theorem 6.40 (Unique Factorization). If k is a field, then every noncon-

stant polynomial f .x/ 2 kŒx is a product of a nonzero constant and monic irreducibles.

Moreover, if f has two such factorizations,

f D ap1 � � �pm and f D bq1 � � �qn;

.that is, a and b are nonzero constants and the p’s and q’s are monic irredu-

cibles/, then a D b,m D n, and the q’s may be re-indexed so that qi D pi for all i .

Proof. We proved the existence of a factorization in Corollary 6.9.

To prove uniqueness, suppose that there is an equation

ap1 � � �pm D bq1 � � �qn

in which a and b are nonzero constants and the p’s and q’s are monic ir-

reducibles. We prove, by induction on M D maxfm; ng � 1, that a D b, m D n, and the q’s may be re-indexed so that qi D pi for all i . For the base step M D 1, we have ap1 D bq1. Now a is the leading coefficient, because p1 is monic, while b is the leading coefficient, because q1 is monic. Therefore, a D b, and canceling gives p1 D q1. For the inductive step, the given equation shows that pm j q1 � � �qn. By Euclid’s Lemma for polynomials, there is some i with pm j qi . But qi , being monic irreducible, has no monic divisors other than 1 and itself, so that qi D pm. Re-indexing, we may assume that qn D pm. Canceling this factor, we have ap1 � � �pm�1 D bq1 � � �qn�1. By the inductive hypothesis, a D b, m � 1 D n � 1 (hence m D n) and, after re-indexing, qi D pi for all i .

Here is another way to state uniqueness, using Proposition 6.24: after re-

indexing, the ideals .p1/; : : : ; .pm/ and .q1/; : : : ; .qm/ are the same.

Collect like factors.

Definition. Let f .x/ 2 kŒx, where k is a field. A prime factorization of f is

f .x/ D ap1.x/e1 � � �pm.x/em ;

where a is a nonzero constant, the pi ’s are distinct monic irreducible polyno- mials, and ei � 0 for all i .

Theorem 6.40 shows that every nonconstant polynomial f has prime fac- torizations; moreover, if all the exponents ei > 0, then the factors in it are

unique. Let f .x/; g.x/ 2 kŒx, where k is a field. As with integers, using zero exponents allows us to assume that the same irreducible factors occur in both

prime factorizations:

f D pa11 � � �pamm and g D p b1 1 � � �pbmm :

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6.1 Parallels to Z 253

Definition. If f and g are elements in a commutative ringR, then a common

multiple is an element h 2 R with f j h and g j h. If f and g in R are not both 0, define their least common multiple, denoted by

lcm.f; g/;

to be a monic common multiple c of them with c j h for every common multi- ple h. If f D 0 D g, define their lcm to be 0.

We now use prime factorizations having zero exponents.

Proposition 6.41. Let f .x/; g.x/ 2 kŒx, where k is a field, have prime fac- torizationsf D pa11 � � �p

an n and g D pb11 � � �p

bn n in kŒx, where ai ; bi � 0 for

all i .

(i) f j g if and only if ai � bi for all i . (ii) If mi D minfai ; big and Mi D maxfai ; big, then

gcd.f; g/ D pm11 � � �p mn n and lcm.f; g/ D p

M1 1 � � �p

Mn n :

Proof. (i) If f j g, then g D f h, where h D pc11 � � �p cn n and ci � 0 for

all i . Hence,

g D pb11 � � �p bn n D

� p a1 1 � � �p

am m

�� p c1 1 � � �p

cn n

� D pa1Cc11 � � �p

anCcn n :

By uniqueness, ai C ci D bi ; hence, ai � ai C ci D bi . Conversely, if ai � bi , then there is ci � 0 with bi D ai C ci . It follows that h D p c1 1 � � �p

cn n 2 kŒx and g D f h.

(ii) Let d D pm11 � � �p mn n . Now d is a common divisor, for mi � ai ; bi . If

D D pe11 � � �p en n is any other common divisor, then 0 � ei � minfai ; big D

mi , and so D j d . Therefore, deg.D/ � deg.d/, and d is the gcd (for it is monic). The argument for lcm is similar.

Corollary 6.42. If k is a field and f .x/; g.x/ 2 kŒx are monic polynomials, then

lcm.f; g/ gcd.f; g/ D fg:

Proof. The result follows from Proposition 6.41, for mi CMi D ai C bi .

Since the Euclidean Algorithm computes the gcd in kŒx when k is a field,

Corollary 6.42 computes the lcm.

lcm.f; g/ D fg gcd.f; g/

:

We can use roots to detect whether two polynomials are relatively prime.

Corollary 6.43. If f .x/; g.x/ 2 RŒx have no common root in C, then f; g are relatively prime in RŒx.

Proof. Assume that d D gcd.f; g/ ¤ 1, where d 2 RŒx. By the Funda- mental Theorem of Algebra, d has a complex root ˛ . By Corollary 6.37,

d D gcd.f; g/ in CŒx. Since .x � ˛/ j d in CŒx, we have .x � ˛/ j f and .x � ˛/ j g. By Corollary 6.15, ˛ is a common root of f and g.

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254 Chapter 6 Arithmetic of Polynomials

How to Think About It. There is nothing magic about R and C. It can be

proved that every field k has an algebraic closure k; that is, there is a field k

containing k as a subfield, and every f .x/ 2 kŒx is a product of linear factors. In particular, since kŒx � kŒx, every f .x/ 2 kŒx is a product of linear factors in kŒx; that is, k contains all the roots of f . Thus, Corollary 6.43 can

be generalized by replacing R and C by k and k.

We know that C can be viewed as a vector space over its subfield R, and

dimR.C/ D 2. But things are not so simple for algebraic closures k of other fields k. It is always true that k is a vector space over k, but its dimension need not be 2. In fact, dimk.k/ need not even be finite: for example, dimQ.Q/ D 1 and, if k is finite, then dimk.k/ D 1.

Let k be a field, and assume that all the roots of a polynomial f .x/ 2 kŒx lie in k: there are a; r1; : : : ; rn 2 k with

f .x/ D a nY

iD1 .x � ri/:

If r1; : : : ; rs, where s � n, are the distinct roots of f , then a prime factorization of f is

f .x/ D a.x � r1/e1.x � r2/e2 � � � .x � rs/es :

We call ej the multiplicity of the root rj . As linear polynomials are always irreducible, unique factorization shows that multiplicities of roots are well-

defined.

Exercises

6.25 Let f .x/; g.x/ 2 kŒx, where k is a field. If fg is a square, must f or g be a square? What if gcd.f; g/ D 1?

6.26 Let f .x/; g.x/ 2 kŒx, where k is a field, be relatively prime. If h.x/ 2 kŒx and h2 j fg, prove that h2 j f or h2 j g.

6.27 Let k D F2.x/. Prove that f .t/ D t2�x 2 kŒt  is an irreducible polynomial. (We shall see later that there is a field K containing k and an element u with u2 D x, so that f .t/ D .t � u/2 in KŒt .)

6.28 In Zp Œx, show that if f is an irreducible factor of x pn � x, then f 2 does not

divide xp n � x.

6.29 Determine, for each of the following polynomials in QŒx whether or not it is

irreducible in QŒx, in RŒx, or in CŒx.

(i) x2 � 7x C 6. (ii) x2 C 2x � 1.

(iii) x2 C x C 1.

6.30 * Show that f .x/ D x3 C 5x2 � 10x C 15 is irreducible in QŒx.

In Section 6.2, we will give different criteria for determining whether poly-

nomials are irreducible (in particular, we will discuss f on page 267). However,

we ask you to solve this problem now so you will appreciate the theorems to be

proved.

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6.1 Parallels to Z 255

Principal Ideal Domains

There are other classes of domains that enjoy an analog of the Fundamental Theorem of Arithmetic; one such is the following.

Definition. A domain R is a principal ideal domain, usually abbreviated by

the acronym PID, if every ideal in R is a principal ideal.

We have already seen examples of PIDs.

Example 6.44. (i) Theorem 5.29 shows that Z is a PID.

(ii) Theorem 6.25 shows that kŒx is a PID when k is a field.

(iii) Every field k is a PID, for its only ideals are k D .1/ and .0/. (iv) Not every domain is a PID. In Example 6.27, we saw that there are ideals

in ZŒx that are not principal ideals.

(v) Exercise 6.12 on page 243 shows that kŒx; y, polynomials in two vari-

ables over a field k, is not a PID. N

(vi) In Chapter 8, we shall see that the rings of Gaussian integers and of Eisen- stein integers are PIDs.

PIDs enjoy many of the properties shared by Z and kŒx (k a field). In

particular, they have a fundamental theorem of arithmetic, and the proof of

this fact parallels the program we developed for Z and kŒx. We begin by defining gcd’s in a general commutative ring R. We can’t use

� as we did in Z, nor can we use degrees as we did in kŒx, but we can use the idea in Corollaries 1.20 and 6.29.

Definition. Let R be a commutative ring. If a; b 2 R, then a gcd of a; b is a common divisor d 2 R that is divisible by every common divisor: if c j a and c j b, then c j d .

Just defining a term doesn’t guarantee it always exists—we could define unicorn if we were asked to do so, and there are rings, even domains, con-

taining elements having no gcd (see Exercise 6.33 on page 259). Even if a

gcd does exist, there is the question of uniqueness. In Z, uniqueness of a gcd

follows from our assuming, as part of the definition, that gcd’s are positive;

in kŒx, uniqueness follows from our assuming, as part of the definition, that gcd’s are monic. Neither assumption makes sense in a general commutative

ring; however, we do have a measure of uniqueness in domains.

Proposition 6.45. Let R be a domain. If d and d 0 are gcd’s of a; b in R, then d and d 0 are associates and .d/ D .d 0/.

Proof. By definition, both d and d 0 are common divisors of a; b; moreover, d j d 0 and d 0 j d . Since R is a domain, Proposition 6.6 applies, and d and d 0 are associates. The second statement follows from Proposition 6.24.

Although there are domains with elements not having a gcd, we now show

gcd’s always exist in PIDs.

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256 Chapter 6 Arithmetic of Polynomials

Theorem 6.46. Let R be a PID. If a; b 2 R, then a gcd of a; b exists and it is a linear combination of a; b.

Proof. As every ideal in R, the ideal .a; b/ D fua C vb W u; v 2 Rg is a principal ideal: there is d 2 R with .a; b/ D .d/. Of course, d is a linear combination of a; b, say, d D sa C tb for some s; t 2 R, and it suffices to prove d is a gcd. Now d is a common divisor: a 2 .a; b/ D .d/, so that a D rd for some r 2 R; hence, d j a; similarly, d j b. Finally, if c j a and c j b, then c j d D saC tb. .

We can now show that Euclid’s Lemma holds in PIDs.

Theorem 6.47 (Euclid’s Lemma). Let R be a PID and p 2 R be irreducible. If p j ab, where a; b 2 R, then p j a or p j b.

Proof. If p − a, then 1 is a gcd of p; a, for the only divisors of p are units and associates. Thus, Theorem 6.46 says that there exist s; t 2 R with 1 D spC ta.This proof should look

quite familiar to you. Hence, b D spbCtab. But ab D pr , for some r 2 R, and so p j b, as desired.

How to Think About It. To prove the unique factorization theorem in Z and in kŒx, we first proved

that every element can be factored into irreducibles. After that, we showed that

such factorizations are essentially unique. Let’s carry on with that development

for arbitrary PIDs.

To prove factorization into irreducibles in a PID, we need an abstract prop-

erty of principal ideal domains, one that was previewed in Exercises 5.47 and

5.48 on page 220. Suppose that R is a PID and that r 2 R is neither zero nor a unit. Must r be

a product of irreducibles? If not, then r is not irreducible (recall that we allow

products to have only one factor); thus, r factors: say, r D ab, where neither a nor b is a unit. If both a and b are products of irreducibles, then so is r , and

we’re done. So, suppose one of them, say a, is not a product of irreducibles. Thus, a is not irreducible, and a D cd . where neither c nor d is a unit. If both c and d are products of irreducibles, then so is a, and this is a contradiction.

We’ve got a tiger by the tail! We can keep repeating this argument ad infinitum.

Let’s rephrase these factorizations of r in terms of ideals; after all, a j r says that r D r 0a for some r 0 2 R; that is, .r/ � .a/. But this inclusion must be strict: .r/ ¨ .a/, lest r and a be associates (they’re not, because b is not a unit). The tiger tells us that there is an infinite strictly increasing sequence of

ideals.

Lemma 6.48. If R is a PID, then every ascending chain of ideals

I1 � I2 � � � � � In � InC1 � � � �

stopsI that is, there is N with In D IN for all n � N .

Proof. Suppose there is an ascending chain of ideals that does not stop. Throw-

ing away any repetitions In D InC1 if necessary, we may assume that there is

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6.1 Parallels to Z 257

a strictly ascending chain of ideals

I1 ¨ I2 ¨ � � � ¨ In ¨ InC1 ¨ � � � :

By Exercise 6.31 on page 258, J D S n�1 In is an ideal in R. And since R

is a PID, J is principal; there is d 2 J with J D .d/. Now d got into J by being in Im for some m; that is, .d/ � Im. Hence,

J D .d/ � Im ¨ ImC1 � J:

This is a contradiction.

Lemma 6.48 gives us factorization into irreducibles.

Proposition 6.49. If R is a PID, then every nonzero, non-unit r 2 R is a product of irreducibles.

Proof. If a is a divisor of r , then .r/ � .a/, as we saw above. Call a divisor a of r a proper divisor if a is neither a unit nor an associate of r . If a is a proper divisor of r , then .r/ ¨ .a/: if the inclusion is not strict, then .a/ D .r/, and this forces a and r to be associates, by Proposition 6.6.

Call a nonzero non-unit r 2 R sweet if it is a product of irreducibles; call it sour otherwise. We must show that there are no sour elements. So, suppose r

is a sour element. Now r is not irrreducible, so r D ab, where both a and b are proper divisors. But the product of sweet elements is sweet, so that at least one

of the factors, say, a, is sour. As we observed in the first paragraph, we have

.r/ ¨ .a/. Repeat this for a instead of r . It follows by induction that there exists a sequence a1 D r , a2 D a, a3; : : : ; an; : : : of sour elements with each anC1 a proper divisor of an. But this sequence yields a strictly ascending chain of ideals

.a1/ ¨ .a2/ ¨ .a3/ ¨ � � � ;

contradicting Lemma 6.48.

Proposition 6.49 gives existence. The next theorem gives a fundamental the-

orem of arithmetic for PIDs: every nonzero non-unit has a unique factorization as a product of irreducibles.

Theorem 6.50. Let R be a PID. Every r 2 R, neither 0 nor a unit, has a fac- torization as a product of irreducibles which is unique in the following senseW if

p1 � � �pn D r D q1 � � �qm;

where the p’s and q’s are irreducible, then m D n and the q’s can be re- indexed so that qi and pi are associates for all i .

Proof. Proposition 6.48(iii) shows that every r 2 R, neither 0 nor a unit, is a product of irreducibles.

To prove uniqueness, suppose that r is a nonzero non-unit and

p1 � � �pn D r D q1 � � �qm;

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258 Chapter 6 Arithmetic of Polynomials

where the p’s and q’s are irreducible. By Euclid’s Lemma, pn irreducible im-

plies pn j qi for some i . Since qi is irreducible, we have pn and qi are asso- ciates: there is a unit u 2 R with qi D upn. Re-index the q’s so that qi is now qm D upn, cancel pn from both sides, and replace q1 by uq1. Thus,

p1 � � �pn�1 D r D .uq1/ � � �qm�1:

Note that uq1 is irreducible (for it is an associate of an irreducible). The proof

is completed, as in Theorem 6.40, by induction on maxfn;mg.

So, every PID has a fundamental theorem of arithmetic. It turns out that there are other domains occurring in nature, not PIDs, which also enjoy such a

theorem.

Definition. A domainR is a unique factorizationdomain, usually abbreviated UFD, if

(i) every a 2 R that is neither 0 nor a unit is a product of irreducibles; (ii) this factorization is unique in the following sense: if

p1 � � �pn D a D q1 � � �qm;

where the p’s and q’s are irreducible, then n D m and, after re-indexing, pi and qi are associates for all i .

Further Results. We’ve just seen that every PID is a UFD. The converse is false: there are UFDs that are not principal ideal domains. A theorem of Gauss

states that if a domain A is a UFD, thenAŒx is also a UFD. For example, ZŒx

is a UFD (this is not a PID). If k is a field, then it follows by induction on

n � 1 that R D kŒx1; : : : ; xn, polynomials in several variables, is a UFD (R is not a PID if n � 2).

As we’ve mentioned earlier, the erroneous assumption that every domain

is a UFD was behind many incorrect “proofs” of Fermat’s Last Theorem. The

ring ZŒ p

�5 is not a UFD: we’ll see in Chapter 8 that

3 � 2 D 6 D .1 C p

�5/.1 � p

�5/

are two different factorizations of 6 into irreducibles in ZŒ p

�5 (and 1C p

�5 is not an associate of 2 or of 3). Another example: ZŒ�23 is not a UFD, and 23

is the smallest prime p for which ZŒ�p is not a UFD (see [23] Chapter 1, p. 7).

Exercises

6.31 *

(i) Let I and J be ideals in a commutative ring R. Prove that their union I [ J is an ideal if and only if I � J or J � I .

(ii) Let I1 � I2 � � � � � In � � � � be an ascending chain of ideals in a commu- tative ring R. Prove that

1[

n�1 In

is an ideal in R.

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6.2 Irreducibility 259

6.32 Consider ascending chains of ideals in Z:

(i) Find two different ascending chains of ideals in which I1 D .24/. (ii) Show that every ascending chain of ideals has only finitely many distinct

terms.

(iii) Find the longest strictly ascending chain of ideals that starts with .72/ (an

ascending chain of ideals is strictly ascending if all inclusions Ij � IjC1 are strict inclusions Ij ¨ IjC1).

(iv) Find the longest strictly ascending chain of ideals that starts with .101/.

6.33 * Let R be the subset of kŒx (where k is a field) consisting of all polynomials

f .x/ having no linear term; that is,

f .x/ D a0 C a2x2 C a3x3 C � � � :

(i) Prove that R is a subring of kŒx.

(ii) Prove that x5 and x6 do not have a gcd in R.

6.34 Recall that RR, the set of all real valued functions of a real variable, is a commu-

tative ring under pointwise addition and multiplication. Let n � 0 be an integer, and let In be the set of all functions in R

R vanishing on integer multiples of n.

(i) Show that In is an ideal in R R.

(ii) Find a function that is in I8 but not in I4.

(iii) Show that

I2 ¨ I4 ¨ I8 ¨ � � � ¨ I2j ¨ : : :

(iv) Show that this ascending chain of ideals does not stop.

(v) Conclude that there are ideals in RR that are not principal.

6.2 Irreducibility

Although there are some techniques to help decide whether an integer is prime, the general problem is open and is very difficult (indeed, this is precisely why

RSA public key codes are secure). Similarly, it is very difficult to determine

whether a polynomial is irreducible, but there are some useful techniques that

frequently work. Most of our attention will be on QŒx and ZŒx, but some of

the results do generalize to other rings of coefficients.

For polynomials of low degree, we have a simple and useful irreducibility criterion.

Proposition 6.51. Let k be a field and let f .x/ 2 kŒx be a quadratic or cubic polynomial. Then f is irreducible in kŒx if and only if f has no root in k.

Proof. An irreducible polynomial f of degree > 1 has no roots in k, by

Corollary 6.15, for if r 2 k is a root, then f .x/ D .x � r/g.x/ in kŒx. Conversely, if f is not irreducible, then f D gh, where neither g nor h is constant; thus, neither g nor h has degree 0. Since deg.f / D 2 or 3 and deg.f / D deg.g/Cdeg.h/, at least one of the factors has degree 1 and, hence, f has a root in k.

Proposition 6.51 is no longer true for polynomials of degree � 4; for exam- ple, f .x/ D x4 C 2x2 C 1 D .x2 C 1/.x2 C 1/ obviously factors in RŒx, so it’s not irreducible, yet f has no real roots.

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260 Chapter 6 Arithmetic of Polynomials

A polynomial f .x/ is reducible if has a linear factor x � a, and there’s a simple test for that; see whether a is a root of f . But to check whether f has

a root a, we need a candidate for a.

Theorem 6.52 (Rational Root Theorem). If f .x/ D a0Ca1xC� � �Canxn 2 ZŒx � QŒx, then every rational root of f has the form b=c, where b j a0 and c j an. In particular, if f is monic, then every rational root of f is an integer.

Proof. We may assume that a root b=c is in lowest terms; that is, gcd.b; c/ D 1. Evaluating gives 0 D f .b=c/ D a0 C a1b=c C � � � C anbn=cn, and multi- plying through by cn gives

0 D a0cn C a1bcn�1 C � � � C anbn:

Reducing this mod b shows that b j a0cn; since gcd.b; c/ D 1, Corollary 1.22 gives b j a0. Similarly, reducing mod c gives c j anbn. Since gcd.b; c/ D 1, we have c j an.

It follows from the second statement that if an integer a is not the nth power of an integer, then xn � a has no rational roots; that is, n

p a is irrational. In

particular, p 2 is irrational. Thus, Theorem 6.52 is a vast generalization of

Proposition 1.26.

Had we known Theorem 6.52 earlier, we could have easily dealt with the

“bad cubic” f .x/ D x3 � 7x C 6 in Example 3.5. Since the candidates for its rational roots are ˙1;˙2;˙3;˙6, we would have quickly found the factor- ization f .x/ D .x � 1/.x � 2/.x C 3/.

If f .x/ 2 QŒx happens to be in ZŒx, there is a useful theorem of Gauss comparing the factorizations of f in ZŒx and in QŒx that concludes that f is irreducible over Q. Our proof involves Example 5.23: the homomorphism

rpW Z ! Zp, sending j 7! Œj , gives a homomorphism r�p W ZŒx ! Zp Œx, called reduction mod p. If f .x/ D a0 C a1x C � � � C anxn 2 ZŒx, then

r�p Wf 7! f ;where f .x/ D Œa0C Œa1x C � � � C Œanxn 2 Zp Œx:

Thus, r�p merely reduces all coefficients mod p.

Theorem 6.53 (Gauss’s Lemma). Let f .x/ 2 ZŒx. If there areG.x/;H.x/ 2 QŒx with f D GH , then there are g.x/; h.x/ 2 ZŒx with deg.g/ D deg.G/, deg.h/ D deg.H/, and f D gh.

Proof. Clearing denominators in the equation f D GH , there are positive integers n0; n00 so that g D n0G and h D n00H , where both g; h lie in ZŒx. Setting n D n0n00, we have

nf D .n0G/.n00H/ D gh in ZŒx: (6.1)

Let p be a prime divisor of n, and reduce the coefficients mod p. Eq. (6.1)

becomesRecall that Fp is another notation for Zp ; we use it

when we want to regard

Zp as a field.

0 D g.x/h.x/:

But FpŒx is a domain, because Fp is a field, and so at least one of the factors,

say g, is 0; that is, all the coefficients of g are multiples of p. Therefore, we

may write g D pg0, where all the coefficients of g0 lie in Z. If n D pm, then

.pm/f D nf D gh D .pg0/h in ZŒx:

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6.2 Irreducibility 261

Cancel p, and continue canceling primes until we reach a factorization f D g�h� in ZŒx. Note that deg.g�/ D deg.g/ and deg.h�/ D deg.h/.

The contrapositive of Gauss’s Lemma is more convenient to use.

Corollary 6.54. If f .x/ 2 ZŒx is irreducible in ZŒx, then f is irreducible in QŒx.

How to Think About It. We agree that Gauss’s Lemma, though very useful,

is rather technical. Gauss saw that the ideas in the proof could be generalized to apply to polynomials in several variables over a field.

The basic use of reduction mod p was previewed on page 216 when we

proved that �1 is not a square in Z by showing that it’s not a square in Z3. Reduction mod p gives a criterion for irreducibility of f in ZŒx by testing the

irreducibility of f in FpŒx. The precise statement is:

Proposition 6.55. Let f .x/ D a0 C a1x C � � � C xn 2 ZŒx be monic. If p is prime and f 2 FpŒx is irreducible in FpŒx, then f is irreducible in ZŒx and, hence, in QŒx. The hypothesis that f is

monic can be relaxed; we

may assume instead that p

does not divide its leading

coefficient.

Proof. Suppose f factors in ZŒx; say f D gh, where 0 < deg.g/ < deg.f / and 0 < deg.h/ < deg.f /. By Exercise 6.8, we may assume that both g and

h are monic. Now f D gh (for r�p is a homomorphism), so that deg.f / D deg.g/ C deg.h/. And f , g, and h are monic, because f; g, and h are, so deg.f / D deg.f /, deg.g/ D deg.g/, and deg.h/ D deg.h/; this contradicts the irreducibility of f in FpŒx. Therefore, f is irreducible in ZŒx. Finally, f

is irreducible in QŒx, by Gauss’s Lemma.

For example, x2 C 1 is irreducible in QŒx because x2 C 1 is irreducible in Z3Œx.

Theorem 6.55 says that if one can find a prime p with f irreducible in

FpŒx, then f is irreducible in QŒx. The finiteness of Fp is a genuine advan-

tage, for there are only a finite number of polynomials in FpŒx of any given

degree. In principle, then, we can test whether a polynomial of degree n in

FpŒx is irreducible by looking at all possible factorizations of it.

The converse of Theorem 6.55 is false: x2 � 2 is irreducible in QŒx (it has no rational root), but it factors mod 2 (as x2); you can check, however, that

x2 � Œ2 is irreducible in F3Œx. But Theorem 6.55 may not apply at all: we’ll see in Example 6.67 that x4 C 1 is irreducible in QŒx, but it factors in FpŒx for every prime p (see [26], p. 304).

In order to use Theorem 6.55, we will need an arsenal of irreducible poly- nomials over finite fields.

Example 6.56. We determine the irreducible polynomials in F2Œx of small

degree.

As always, the linear polynomials x and x C 1 are irreducible. There are four quadratics: x2, x2 C x, x2 C 1, x2 C xC 1 (more generally,

there are pn monic polynomials of degree n in FpŒx, for there are p choices

for each of the n coefficients a0; : : : ; an�1). Since each of the first three has a root in F2, there is only one irreducible quadratic, namely, x

2 C x C 1.

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262 Chapter 6 Arithmetic of Polynomials

There are eight cubics, of which four are reducible because their constant

term is 0 (so that x is a factor). The remaining polynomials are

x3 C 1; x3 C x C 1; x3 C x2 C 1; x3 C x2 C x C 1: Since 1 is a root of the first and fourth, the middle two are the only irreducible cubics. Proposition 6.51 now applies.

There are sixteen quartics, of which eight are reducible because their con-

stant term is 0. Of the eight with nonzero constant term, those having an even

number of nonzero coefficients have 1 as a root. There are now only four sur-

viving polynomials f , and each has no roots in F2; that is, they have no linear factors. The only possible factorization for any of them is f D gh, where both g and h are irreducible quadratics. But there is only one irreducible quadratic,

namely, x2 C xC 1. Therefore, x4 C x2 C 1 D .x2 CxC 1/2 factors, and the other three quartics are irreducible.

Irreducible Polynomials of Low Degree over F2

Degree 2: x2 C x C 1. Degree 3: x3 C x C 1, x3 C x2 C 1. Degree 4: x4 C x3 C 1, x4 C xC1, x4 C x3 C x2 C x C 1. N

Example 6.57. Here is a list of the monic irreducible quadratics and cubics in F3Œx. You can verify that the list is correct by first enumerating all such

polynomials; there are six monic quadratics having nonzero constant term, and

there are eighteen monic cubics having nonzero constant term. It must then beNote that �1 D 2 in Z3. checked which of these have 1 or �1 as a root, for Proposition 6.51 applies.

Monic Irreducible Quadratics and Cubics over F3

Degree 2: x2 C 1, x2 C x � 1, x2 � x � 1. Degree 3: x3 � x C 1, x3 C x2 � x C 1, x3 � x2 C 1,

x3 � x2 C x C 1, x3 � x � 1, x3 C x2 � 1, x3 C x2 C x � 1, x3 � x2 � x � 1. N

Example 6.58. Here are some applications of Theorem 6.55.

(i) The polynomial f .x/ D 3x3 � 3x C 1 is irreducible in QŒx, for f D x3 C x C 1 is irreducible in F2Œx.

(ii) We show that f .x/ D x4 � 5x3 C 2x C 3 is irreducible in QŒx. By Theorem 6.52, the only candidates for rational roots of f are ˙1 and ˙3, and you can check that none is a root. Since f is a quartic, we cannot yet

conclude that it is irreducible, for it might be a product of (irreducible) quadratics.

The criterion of Theorem 6.55 works like a charm. Since f D x4 C x3 C 1 in F2Œx is irreducible, by Example 6.56, it follows that f is irre- ducible in QŒx. (It wasn’t necessary to check that f has no rational roots;

irreducibility of f is enough to conclude irreducibility of f . In spite of this, it is a good habit to first check for rational roots.)

(iii) Let ˆ5.x/ D x4 C x3 C x2 C x C 1 2 QŒx. In Example 6.56, we saw thatˆ5.x/ D x4 C x3 C x2 C xC 1 is irreducible in F2Œx, and soˆ5 is irreducible in QŒx. N

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6.2 Irreducibility 263

Further Results.

We can count the number Nn of irrreducible polynomials of degree n in

FpŒx. In [17], pp. 83–84, it is shown that

pn D X

d jn d Nd ; (6.2)

where the sum is over the positive divisors d of n.

This equation can be solved for Nn. If m D pe11 � � �p en n , define the Möbius The definition of � seems

to come out of nowhere,

but it occurs in many prob-

lems at the intersection of

combinatorics and num-

ber theory. See A.Cuoco,

“Searching for Möbius,”

College Mathematics

Journal, 37:2, (148–153),

2009.

function by

�.m/ D

8 <̂

1 if m D 1I 0 if any ei > 1I

.�1/n if 1 D e1 D e2 D � � � D en:

It turns out that Eq. (6.2) is equivalent to

Nn D 1

n

X

d jn �.d/pn=d :

One application of this formula is that, for every n � 1, there exists an irreducible polynomial in FpŒx of degree n.

Exercises

6.35 Let f .x/ D x2 C x C 1 2 F2Œx. Prove that f is irreducible in F2Œx, but that f has a root ˛ 2 F4. Use the construction of F4 in Exercise 4.55 on page 165 to display ˛ explicitly.

6.36 Show that x4 C x C 1 is not irreducible in RŒx even though it has no roots in R.

6.37 (i) If k is a field and each of f .x/; g.x/ 2 kŒx has a root ˛ in k, show that ˛ is a root of gcd.f; g/.

(ii) How does this apply to the polynomials in Examples 6.35 and 6.36?

6.38 If p is a prime, show that, in Zp Œx,

xp � x D p�1Y

iD0 .x � i / and xp�1 � 1 D

p�1Y

iD1 .x � i /:

6.39 .Wilson0s Theorem/: Suppose that p is a prime in Z. Show that

.p � 1/Š � �1 mod p:

6.40 *

(i) Let f .x/ D .x � a1/ � � � .x � an/ 2 kŒx, where k is a field. Show that f has no repeated roots (i.e., all the ai are distinct) if and only if gcd.f; f

0/ D 1, where f 0 is the derivative of f .

Hint: Use Exercise 5.17 on page 203.

(ii) Prove that if p.x/ 2 QŒx is an irreducible polynomial, then p has no re- peated roots in C.

Hint: Use Corollary 6.37.

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264 Chapter 6 Arithmetic of Polynomials

6.41 If p is prime, prove that there are exactly 13 .p 3 � p/ monic irreducible cubic

polynomials in FpŒx.

6.42 Determine whether the following polynomials are irreducible in QŒx.

(i) f .x/ D x5 � 4x C 2. (ii) f .x/ D x4 C x2 C x C 1.

Hint: Show that f has no roots in F3 and that a factorization of f as a product

of quadratics would force impossible restrictions on the coefficients.

(iii) f .x/ D x4 � 10x2 C 1.

Hint: Show that f has no rational roots and that a factorization of f as a

product of quadratics would force impossible restrictions on the coefficients.

6.43 Is x5 C x C 1 irreducible in F2Œx?

Hint: Use Example 6.56.

6.44 Let f .x/ D .xp � 1/=.x � 1/, where p is prime. Using the identity

f .x C 1/ D xp�1 C pq.x/;

where q.x/ 2 ZŒx has constant term 1, prove that

ˆp.x pn / D xpn.p�1/ C � � � C xpn C 1

is irreducible in QŒx for all n � 0.

6.45 Let k be a field, and let f .x/ D a0Ca1x C� � �Canxn 2 kŒx have degree n and nonzero constant term a0. If f is irreducible, prove that anCan�1xC� � �Ca0xn is irreducible.

Roots of Unity

In Chapter 3, we defined an nth root of unity � to be primitive if every nth root of unity is a power of �. For example, i is a primitive 4th root of unity. Note

that i is also an 8th root of unity, for i8 D 1, but it’s not a primitive 8th root of unity;

p 2 2 .1C i/ is a primitive 8th root of unity.

Lemma 6.59. Every nth root of unity � 2 C is a primitive d th root of unity for a unique divisor d of n.

Proof. We know that �n D 1; let d be the smallest positive integer for which �d D 1. By the Division Algorithm, there are integers q and r withn D qdCr , where 0 � r < d . Now

1 D �n D �qdCr D �dq�r D �r ;

because �d D 1. But r < d and �r D 1; if r > 0, then we contradict d being the smallest positive such exponent. Therefore, r D 0 and d j n. This shows that � is a d th root of unity. Its first d powers,

1; �; �2; : : : ; �d�1 ;

are all distinct (Exercise 6.51 on page 269). Since there are exactly d d th roots

of unity, they are all powers of �, and so � is primitive.

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6.2 Irreducibility 265

Definition. If d is a positive integer, then the d th cyclotomic polynomial is

defined by

ˆd .x/ D Y .x � �/;

where � ranges over all the primitive d th roots of unity.

Proposition 6.60. Let n be a positive integer and regard xn � 1 2 ZŒx. Then This proposition sheds light on your discovery in

Exercise 3.59 on page 116.(i)

xn � 1 D Y

d jn ˆd .x/;

where d ranges over all the positive divisors d of n (in particular, both

ˆ1.x/ and ˆn.x/ are factors).

(ii) ˆn.x/ is a monic polynomial in ZŒx.

Proof. (i) For each divisor d of n, collect all terms in the equation xn � 1 DQ .x � �/ with � a primitive d th root of unity. Thus,

xn � 1 D Y

d jn hd .x/;

where hd .x/ D Q .x � �/ with � an nth root of unity that is also a prim-

itive d th root of unity. But every such � must be an nth root of unity: by

Lemma 6.59, n D dq for some integer d , and 1 D �n D �dq . Therefore, hd .x/ D ˆd .x/.

(ii) The proof is by strong induction on n � 1. The base step is true, for ˆ1.x/ D x � 1. For the inductive step n > 1, write

xn � 1 D ˆn.x/F.x/;

where F.x/ D Q ˆd .x/ with d j n and d < n. The inductive hypoth-

esis says that all the factors ˆd of F are monic polynomials in ZŒx; hence, F is a monic polynomial in ZŒx. By Proposition 6.10, ˆn.x/ D .xn � 1/=F.x/ is a monic polynomial in ZŒx, as desired.

Example 6.61. The formula in Proposition 6.60(i) can be used to calculate ˆn.x/ for any n. Indeed, solving for ˆn.x/ in

xn � 1 D Y

d jn ˆd .x/

we have

ˆn.x/ D xn � 1Y

d jn; d<n ˆd .x/

Using the fact thatˆ1.x/ D x � 1, we have a recursively defined function:

ˆn.x/ D (

x � 1 if n D 1 .xn � 1/=

Q d jn; d<nˆd .x/ if n > 1:

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266 Chapter 6 Arithmetic of Polynomials

n ˆn.x/

1 x � 1 2 x C 1 3 x2 C x C 1 4 x2 C 1 5 x4 C x3 C x2 C 1 6 x2 � x C 1 7 x6 C x5 C x4 C x3 C x2 C 1 8 x4 C 1 9 x6 C x3 C 1 10 x4 � x3 C x2 � x C 1 11 x10 C x9 C x8 C x7 C x6 C x5 C x4 C x3 C x2 C 1 12 x4 � x2 C 1

Figure 6.1. Cyclotomic polynomials.

You should verify that x12 � 1 D Q d2f1;2;3;4;6;12g ˆd .x/. The recursive def-

inition can be programmed into a CAS (see Exercise 6.55 on page 270); Fig-

ure 6.1 displays the first dozen cyclotomic polynomials. There’s no simple

pattern to these polynomials, but calculating a good number of them gives you food for thought and leads to interesting conjectures. For example, can

you conjecture anything about deg.ˆn/? All the coefficients of the cyclotomic

polynomials displayed in Figure 6.1 are 0 and ˙1, but your guess that this is always true is wrong [see Exercise 6.55(iii) on page 270]. Do any of theˆn.x/

factor in ZŒx? N

When p � 11 is prime, p̂.x/ is xp�1 C xp�2 C � � � C x2 C x C 1. We now prove this is true for every prime p.

Proposition 6.62. If p is prime,

p̂.x/ D xp�1 C xp�2 C � � � C x2 C x C 1:

Proof. By Proposition 6.60,

xp � 1 D ˆ1.x/ p̂.x/ D .x � 1/ p̂.x/;

and the Division Algorithm gives

p̂.x/ D xp � 1 x � 1

D xp�1 C xp�2 C � � � C x2 C x C 1:

Recall that the Euler �-function �.n/ is defined by

�.n/ D number of k with 1 � k � n and gcd.k; n/ D 1:

The next proposition shows that �.n/ is intimately related to ˆn.x/, and this

leads to a simple proof of a fact from number theory.

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6.2 Irreducibility 267

Proposition 6.63. (i) �.n/ D deg.ˆn/. (ii) For every integer n � 1, we have n D

P d jn �.d/. Proposition 6.63(ii) is often

proved in number theory

courses without mentioning

cyclotomic polynomials;

the resulting proof is much

more difficult.

Proof. (i) This follows at once from Corollary 3.30, which says that there

are �.n/ primitive nth roots of unity.

(ii) Immediate from Proposition 6.60(i) and part (i), for

n D X

d jn deg.ˆd / D

X

d jn �.d/:

We’ve shown that ˆn.x/ 2 ZŒx, and we’ll finish this section by showing that p̂ is irreducible in QŒx when p is prime. It turns out that ˆn is actually

irreducible in QŒx for every n, but the proof is more difficult (see [17] p.195).

As any linear polynomial over a field, the cyclotomic polynomial ˆ2 D xC1 is irreducible in QŒx;ˆ3 D x2CxC1 is irreducible in QŒx because it has no rational roots; we saw, in Example 6.58, that ˆ5 is irreducible in QŒx.

We’ll next introduce another irreducibility criterion, useful in its own right,

that will allow us to prove that p̂ is irreducible in QŒx for all primes p. An

example will motivate the criterion.

Example 6.64. Exercise 6.30 on page 254 asked you to show that f .x/ D x3C5x2�10xC15 is irreducible in ZŒx. You now have machinery that makes this easy. For example, you could invoke Theorem 6.52 (the Rational Root

Theorem) to show that f has no root in Q (or Z) and then use Proposition 6.51.

But let’s use another technique that shows the power of reducing coeffi-

cients. Suppose that f .x/; g.x/; h.x/ 2 ZŒx and f D gh, where neither g nor h is constant; reduce the coefficients mod 5. Because reduction mod 5 is

a homomorphism, we have f D g h. But all the coefficients of f (except the leading one) are divisible by 5, so we have

x3 D gh in Z5Œx:

Since x is irreducible (it’s a linear polynomial), we can apply unique factor-

ization in Z5Œx to conclude that both g and h are of the form ux m where u is

a unit in Z5. Pulling this back to Z, we see that all the coefficients of g and h,

except their leading coefficients, are divisible by 5. Hence the constant term of gh (which is the product of the constant terms of g and h) is divisible by 25.

But gh D f and the constant term of f is 15, which is not divisible by 25. Hence no non-trivial factorization of f exists. N

Theorem 6.65 (Eisenstein Criterion). Let f .x/ D a0 C a1xC � � � Canxn 2 ZŒx. If there is a prime p dividing ai for all i < n but with p − an and Usually, Kadiddlehopper

was the first to discover

Kadiddlehopper’s Theo-

rem, but not always. For

example, the Eisenstein

Criterion is in a paper of

Eisenstein of 1850, but

it appeared in a paper of

Schönemann in 1845.

p2 − a0, then f is irreducible in QŒx.

Proof. .R: Singer/: Let r�p W ZŒx ! Fp Œx be reduction mod p, and let f de- note r�p .f /. If f is not irreducible in QŒx, then Gauss’s Lemma gives polyno- mials g.x/; h.x/ 2 ZŒx with f D gh, where g.x/ D b0Cb1xC� � �Cbmxm, h.x/ D c0Cc1xC� � �Cckxk , andm; k > 0. There is thus an equation f D gh in FpŒx.

Since p − an, we have f ¤ 0; in fact, f D uxn for some unit u 2 Fp, because all its coefficients, aside from its leading coefficient, are 0. By

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268 Chapter 6 Arithmetic of Polynomials

Theorem 6.40, unique factorization in kŒx where k is a field, we must have

g D vxm and h D wxk (for units v; w in Fp), so that each of g and h has constant term 0. Thus, Œb0 D 0 D Œc0 in Fp; equivalently, p j b0 and p j c0. But a0 D b0c0, and so p2 j a0, a contradiction. Therefore, f is irreducible in QŒx.

Let’s see that p̂.x/ D xp�1 C xp�2 C � � � C x C 1 is irreducible in QŒx when p is prime. Gauss showed how to transform p̂.x/ so that the Eisenstein Criterion applies.

Lemma 6.66. Let g.x/ 2 ZŒx. If there is c 2 Z with g.x C c/ irreducible in ZŒx, then g is irreducible in QŒx.

Proof. By Exercise 5.42 on page 216, the function 'W ZŒx ! ZŒx, given by

f .x/ 7! f .x C c/;

is an isomorphism (its inverse is f .x/ 7! f .x � c/). If g.x/ D s.x/t.x/, then

'.g/ D '.st/ D '.s/'.t/: (6.3)

But '.g/ D g.xCc/, so that Eq. (6.3) is a forbidden factorization of g.xCc/. Hence, Corollary 6.54 says that g is irreducible in QŒx.

Example 6.67. Consider f .x/ D x4 C 1 2 QŒx. Now

f .x C 1/ D .x C 1/4 C 1 D x4 C 4x3 C 6x3 C 4x C 2:

The Eisenstein Criterion, using the prime p D 2, shows that f .x C 1/ is irre- ducible in QŒx, and Lemma 6.66 shows that x4C1 is irreducible in QŒx. N

Theorem 6.68 (Gauss). For every prime p, the pth cyclotomic polynomial

p̂.x/ is irreducible in QŒx.

Proof. Since p̂.x/ D .xp � 1/=.x � 1/, we have

p̂.x C 1/ D Œ.xC 1/p � 1=x D xp�1 C p

1

! xp�2 C

p

2

! xp�3 C � � � Cp:

Since p is prime, we have p j � p i

� for all i with 0 < i < p (Proposition 2.26);

hence, the Eisenstein Criterion applies, and p̂.x C 1/ is irreducible in QŒx.It’s not true that xn�1 C xn�2 C � � � C x C 1 is irreducible when n is not

prime. For example, when

nD 4, x3C x2CxC 1D .xC 1/.x2C 1/.

By Lemma 6.66, p̂ is irreducible in QŒx.

Further results. Gauss used Theorem 6.68 to prove that a regular 17-gon can be constructed with ruler and compass (ancient Greek mathematicians did

not know this). He also constructed regular 257-gons and 65537-gons. We will

look at ruler–compass constructions in Chapter 7.

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6.2 Irreducibility 269

Exercises

6.46 * Let � D e2�i=n be a primitive nth root of unity. (i) Prove, for all n � 1, that

xn � 1 D .x � 1/.x � �/.x � �2/ � � � .x � �n�1/; and, if n is odd, that

xn C 1 D .x C 1/.x C �/.x C �2/ � � � .x C �n�1/: (ii) For numbers a and b, prove that

an � bn D .a � b/.a � �b/.a � �2b/ � � � .a � �n�1b/; and, if n is odd, that

an C bn D .a C b/.a C �b/.a C �2b/ � � � .a C �n�1b/: Hint: Set x D a=b if b ¤ 0.

6.47 * Let k be a field and a 2 k. Show that, in kŒx, xn � an D .x � a/

� xn�1 C xn�1a C xn�2a2 C : : : an�1x C an

� :

6.48 If k is a field, a 2 k, and f .x/ D cnxn C cn�1xn�1 C � � � C c0 2 kŒx, then rewrite

f .x/ � f .a/ D � cnx

n C cn�1xn�1 C � � � C c0 �

� � cna

n C cn�1an�1 C � � � C c0 �

and use Exercise 6.47 to give another proof of Corollary 6.15.

6.49 Determine whether the following polynomials are irreducible in QŒx.

(i) f .x/ D 3x2 � 7x � 5. (ii) f .x/ D 2x3 � x � 6. (iii) f .x/ D 8x3 � 6x � 1. (iv) f .x/ D x3 C 6x2 C 5x C 25. (v) f .x/ D x4 C 8x C 12. Hint: In F5Œx, f .x/ D .x C 1/g.x/, where g is irreducible.

6.50 Use the Eisenstein Criterion to prove that if a is a squarefree integer, then

xn � a is irreducible in QŒx for every n � 1. Conclude that there are irreducible polynomials in QŒx of every degree n � 1.

6.51 * In the proof of Lemma 6.59, we claimed that the first d powers of � are distinct,

where � is an nth root of unity and d is the smallest positive integer with �d D 1. Prove this claim.

6.52 * Let � be an nth root of unity. Lemma 6.59 shows that � is a primitive d th root

of unity for some divisor d of n. Show that the divisor is unique.

6.53 Consider a finite table of data:

Input Output

a1 b1

a2 b2

a3 b3 :::

:::

an bn

Show that two polynomial functions (defined over Q) agree on the table if and

only if their difference is divisible by

nY

iD1 .x � ai /:

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270 Chapter 6 Arithmetic of Polynomials

6.54 (i) Show that the set of polynomials in QŒx that vanish on f˛1; : : : ; ˛ng is an ideal in QŒx.

(ii) What is a generator of this ideal?

6.55 *

(i) Implement the recursively defined function for ˆn.x/ given in Example 6.61

in a CAS.

(ii) Use it to generate ˆn.x/ for, say, 1 � n � 50. (iii) Use the CAS to find the smallest value of n for which a coefficient of ˆn.x/

is not 0, 1, or �1.

6.3 Connections: Lagrange Interpolation

A popular activity in high school mathematics is finding a polynomial that

agrees with a table of data. For example, students are often asked to find a polynomial agreeing with a table like this:There are several methods

for fitting polynomial

functions to data; see

Chapter 1 of [7]. Input Output

�3 12 2 22

3 72

�4 �26

On the surface, the problem of fitting data seems to have little to do with the

ideas in this chapter. However, by placing it in a more abstract setting, we’ll

see that it yields Lagrange Interpolation, a result useful in its own right. Thus,

this problem fits right into the theory of commutative rings; in fact, it’s really the Chinese Remainder Theorem! But isn’t the Chinese Remainder Theorem

about solving some congruences? Well, yes, and we’ll see that we can make

the notion of congruence apply here. But first, let’s find a polynomial f .x/ by

hand that fits the table.

f .�3/ D 12 , the remainder when f .x/ is divided by .x C 3/ is 12

f .2/ D 22 , the remainder when f .x/ is divided by .x � 2/ is 22

f .3/ D 72 , the remainder when f .x/ is divided by .x � 3/ is 72

f .�4/ D �26 , the remainder when f .x/ is divided by .x C 4/ is �26.

The statement about remainders looks structurally similar to the solution we

constructed to the problem from Qin Jiushao on page 146. We now make this

similarity precise. Recall that a � b mod m in Z means that m j .a � b/. We can define

congruence in kŒx, where k is a field: givenm.x/ 2 kŒx, then f .x/; g.x/ are congruent mod m, denoted by

f � g mod m;

if m j .f � g/.

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6.3 Connections: Lagrange Interpolation 271

Take It Further. Throwing caution to the winds, here’s a fantastic general-

ization. Rephrase congruence modm in Z in terms of ideals: sincem j .a�b/ if and only if a � b 2 .m/, we have a � b mod m if and only if a � b 2 .m/. Let R be a commutative ring and I be any, not necessarily principal, ideal inR.

If a; b 2 R, define

a � b mod I

to mean a � b 2 I (we’ll actually use this generalization when we discuss quotient rings in Chapter 7).

We can now rephrase the Division Algorithm in kŒx using congruence. The

statement: given m.x/; f .x/ 2 kŒx with m.x/ ¤ 0, there exist q.x/; r.x/ 2 kŒx with f D qmC r , where r D 0 or deg.r/ < deg.m/, can be rewritten to say

f � r mod m:

And Proposition 6.14, the Remainder Theorem, says that if m.x/ D x � a, then f � f .a/ mod .x � a/. Thus, the constraints on f can be rewritten

f � 12 mod .x C 3/ f � 22 mod .x � 2/ f � 72 mod .x � 3/ f � �26 mod .x C 4/:

Notice that the four linear polynomials are pairwise relatively prime.

Let’s push the similarity a little further, using the localization idea on page 146.

Suppose we can find polynomials g, h, k, and ` satisfying

g.�3/ D 1 h.�3/ D 0 k.�3/ D 0 `.�3/ D 0 g.2/ D 0 h.2/ D 1 k.2/ D 0 `.2/ D 0 g.3/ D 0 h.3/ D 0 k.3/ D 1 `.3/ D 0

g.�4/ D 0 h.�4/ D 0 k.�4/ D 0 `.�4/ D 1:

Setting f D 12g C 22h C 72k � 26`, we have a polynomial that fits the original table (why?). Now Proposition 6.15, the Factor Theorem, shows that g is divisible by the linear polynomials x � 2, x � 3, and xC 4. Since they are irreducible in QŒx, they are pairwise relatively prime; hence, Exercise 6.17(ii)

on page 247 says that g is divisible by their product: there is A.x/ such that

g.x/ D A.x � 2/.x � 3/.x C 4/:

In fact, we can choose A to be a constant and have g.�3/ D 1: set

1 D g.�3/ D A.�3C 2/.�3 � 3/.�3 C 4/I

that is, A D 1=30 and g.x/ D 1 30 .x � 2/.x � 3/.x C 4/. Similarly,

� h.x/ D B.x C 3/.x � 3/.x C 4/ and h.2/ D 1 impliesB D �1=30, so

h.x/ D � 1 30 .x C 3/.x � 3/.x C 4/

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272 Chapter 6 Arithmetic of Polynomials

� k.x/ D C.x C 3/.x � 2/.x C 4/ and k.3/ D 1 implies C D 1=42, so

k.x/ D 1 42 .x C 3/.x � 2/.x C 4/

� `.x/ D D.x C 3/.x � 2/.x � 3/ and `.�4/ D 1 impliesD D �1=42, so

`.x/ D � 1 42 .x C 3/.x � 2/.x � 3/:

Now puttingf D 12gC22hC72k�26`, we have, after simplification (carried out by a CAS):

f .x/ D 2x3 C 4x2 � 8x C 6:

You can check that f matches the table.

This method is called Lagrange Interpolation, and it applies to any finite

set of input-output pairs. As we just saw, it’s the same method used in the

proof of the Chinese Remainder Theorem, but applied to polynomials rather

than integers. Since our goal is merely to display connections, we leave the proof to the reader.

Theorem 6.69 (Chinese Remainder Theorem for Polynomials). Let k be

a field. If m1; : : : ; mr 2 kŒx are pairwise relatively prime and b1; : : : ; br 2 kŒx, then the simultaneous congruences

f � b1 mod m1 f � b2 mod m2 :::

:::

f � br mod mr have an explicit solution, namely,

f D b1 .s1M1/C b2 .s2M2/C � � � C br .srMr/ ;

where

Mi D m1m2 � � �bmi � � �mr and siMi � 1 mod mi for 1 � i � r: Furthermore, any solution to this system is congruent to f mod m1m2 � � �mr .

Proof. The proof of Theorem 4.27 can be easily adapted to prove this.

Example 6.70. The calculations we have just made can be used to illustrate

the theorem. The table gives

b1 D 12; b2 D 22; b3 D 72; b4 D �26I

applying the Remainder Theorem to the table entries,

m1 D x C 3; m2 D x � 2; m3 D x � 3; m4 D x C 4:

Then

M1 D .x � 2/.x � 3/.x C 4/; M2 D .x C 3/.x � 3/.x C 4/; M3 D .x C 3/.x � 2/.x C 4/; M4 D .x C 3/.x � 2/.x � 3/;

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6.3 Connections: Lagrange Interpolation 273

and

s1 D 1=30; s2 D �1=30; s3 D 1=42; s4 D �1=42:

Note, for example, that

s1M1 � 1 D 130 .x � 2/.x � 3/.x C 4/� 1

D x 3 � x2 � 14x C 24

30 � 1

D x 3 � x2 � 14x � 6

30

D .x C 3/x 2 � 4x � 2 30

;

so that s1M1 � 1 mod m1. This is not magic; if you look carefully at how s1 is calculated (we called it A on the previous page), you’ll see that it is none

other than

1=M1.�3/:

And the Remainder Theorem (again) says that the remainder when g.x/ D M1.x/=M1.�3/ is divided by x C 3 is

g.�3/ D M1.�3/=M1.�3/ D 1:

Similarly, we have siMi � 1 mod mi for the other values of i . Finally, the statement about any other solution to the system follows from

Exercise 6.53 on page 269. N

Compare the statement of Theorem 6.69 to the more typical statement of

Lagrange Interpolation.

Theorem 6.71 (Lagrange Interpolation). Let k be a field. An explicit way of

writing the polynomial f .x/ 2 kŒx of minimal degree that takes the values bi at distinct points ai , for 1 � i � r , is

f .x/ D b1 M1.x/

M1.a1/ C b2

M2.x/

M2.a2/ C b3

M3.x/

M3.a3/ C � � � C br

Mr .x/

Mr .ar / ;

where Mi .x/ is the polynomial defined by

Mi .x/ D .x � a1/.x � a2/ � � � .x � ai�1/3.x � ai /.x � aiC1/ � � � .x � ar /:

Some things to note:

(i) Theorem 6.69 is more general than Theorem 6.71, for it allows the moduli

mi to be any finite set of relatively prime polynomials (Lagrange Interpo-

lation only considers moduli of the form x � a). (ii) On the other hand, Theorem 6.71 is more explicit: it implies that si (in

the statement of the Chinese Remainder Theorem) is 1=Mi.ai / (in the

statement of Lagrange Interpolation).

(iii) The statement of Lagrange Interpolation goes on to say that the poly- nomial obtained by this method is the one of lowest degree that fits the

conditions.

You’ll verify these last two items in Exercises 6.60 and 6.63 below.

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274 Chapter 6 Arithmetic of Polynomials

Exercises

6.56 Find a polynomial that agrees with the table

Input Output

0 3

1 4

2 7

3 48

4 211

6.57 A radio show offered a prize to the first caller who could predict the next term in

the sequence

1; 2; 4; 8; 16:

(i) What would you get if you used “common sense?”

(ii) What would you get if you used Lagrange Interpolation?

6.58 Another radio show offered a prize to the first caller who could predict the next

term in the sequence

14; 3; 26; 8; 30:

After no one got it for a few days, the host announced that these are the first

five numbers that were retired from the Mudville Sluggers baseball team. Use

Lagrange Interpolation to predict the next number that was retired in Mudville.

6.59 The following table fits the quadratic f .x/ D x2 � 3x C 5; that is, f .0/ D 5, f .1/ D 3, etc. Now forget about f and use Lagrange Interpolation to find a polynomial that fits the table.The result of Exercise 6.53

allows you to fool many

standardized tests. Input Output

0 5

1 3

2 3

3 5

4 9

It seems that this table should fool Lagrange Interpolation, which produces a de-

gree 4 polynomial. Does it?

6.60 * Show that Lagrange Interpolation produces a polynomial of smallest degree that

agrees with a given input-output table.

6.61 (i) Find a polynomial g.x/ that agrees with the table

Input Output

4 24

5 60

6 120

7 210

8 336

(ii) Factor g into irreducibles.

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6.3 Connections: Lagrange Interpolation 275

6.62 It is known that there’s a cubic polynomial function f .x/ 2 QŒx such that, for positive integers n,

f .n/ D n�1X

kD0 k2:

(i) Find f .

Hint: A cubic is determined by four inputs.

(ii) Prove that f .n/ D Pn�1 kD0 k

2 for all positive integers n.

6.63 * Using the notation of Theorem 6.71, show that

Mi .x/=Mi .ai / � 1 mod .x � ai /:

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7 Quotients, Fields, and ClassicalProblems In Chapter 4, we introduced the idea of congruence modulo an integer m as a

way to “ignore” multiples ofm in calculations by concentrating on remainders.

This led to an arithmetic of congruence classes and the construction of the commutative ring Zm in which multiples of m are set equal to 0. You now

know that the multiples ofm form an ideal .m/ in Z, and so Zm can be thought

of as a commutative ring obtained from Z in which all the elements of .m/ are

set equal to 0.

In this chapter, we introduce quotient rings, a generalization of this con- struction. Given a commutative ring R and an ideal I � R, we will produce a new commutative ringR=I that forces all the elements of I to be 0.

In particular, beginning with the commutative ring kŒx and the ideal .f /,

where k is a field and f .x/ 2 kŒx, we shall see that identifying .f / with 0 produces an element ˛ in the quotient ring kŒx=.f / that is a root of f : if

f .x/ D c0 C c1x C � � � C cnxn, then f .˛/ D c0 C c1˛ C � � � C cn˛n D 0. Moreover, the complex number field is a special case: C is the quotient ring

arising from RŒx and the ideal .x2 C 1/. Another byproduct of the quotient ring construction is the existence and classification of all finite fields (there are

others beside Fp and F4).

In the last section, we will apply fields to settle classical geometric prob- lems that arose over two millenia ago: using only ruler and compass, can we

duplicate the cube, trisect an angle, square the circle, or construct regular n-

gons?

7.1 Quotient Rings

In Chapter 3, we said that the approach of many Renaissance mathematicians

to the newly invented complex numbers was to consider them as polynomi-

als or rational functions in i , where calculations are carried out as usual with

the extra simplification rule i2 D �1. In constructing Zm from Z, we ig- nored multiples of m (that is, we set them all equal to 0), and we saw that this

idea is compatible with addition and multiplication. Let’s see if we can mimic

this idea, starting with RŒx, and apply it to C, as our Renaissance ancestors

wished. Can we replace the symbol x in a polynomial f .x/ by a new sym-

bol i that satisfies i2 D �1? Well, if we make x2 D �1, then we are setting x2 C 1 D 0. So, as with constructing Zm, let’s set all the multiples of x2 C 1 in RŒx equal to 0.

This analogy with Zm looks promising: let the commutative ring RŒx cor-

respond to Z, and let the (principal) ideal .x2 C 1/ in RŒx correspond to 277

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278 Chapter 7 Quotients, Fields, and Classical Problems

the ideal .m/ in Z. Push this analogy further. Elements in Zm are congruence

classes Œa, where a 2 Z. Let’s invent new elements Œf  corresponding to poly- nomials f .x/. In more detail, Œa denotes fa C qm W q 2 Zg, and we forced qm D 0 (actually, using the Division Algorithm in Z, this allowed us to focus on remainders after dividing by m). Defining Œf  D ff .x/C q.x/.x2 C 1/ W q.x/ 2 RŒxg would allow us to focus on the remainder after dividing f .x/ by x2 C 1. Indeed, the Division Algorithm in RŒx writes

f .x/ D q.x/.x2 C 1/C r.x/;

where r.x/ D 0 or deg.r/ < 2. In other words, we could write Œf  D Œr , where r.x/ D a C bx for a; b 2 R. Hold it! If the bracket notation makes x2 C 1 D 0, then x2 D �1, and we may as well write i instead of Œx. Looks a lot like C to us! Now it turns out that this idea is also compatible with addition and multiplication, as we shall see when we introduce quotient rings precisely.

The construction makes sense for any commutative ring R and any ideal I

in R; moreover, it constructs not only C but many other important systems

as well.

Definition. Let I be an ideal in a commutative ring R. We say that a; b 2 R are congruent mod I , written

a � b mod I; if a � b 2 I .

This does generalize our earlier definition of congruence when R D Z, m � 0, and I D .m/. If a; b 2 Z and a � b mod .m/, then a � b 2 .m/. But a � b 2 .m/ if and only if m j a � b; that is, a � b mod m in the old sense.

We now note that congruence mod I is an equivalence relation on R.

Proposition 7.1. Let a; b; c be elements in a commutative ring R. If I is an

ideal inR, then

(i) a � a mod I (ii) if a � b mod I , then b � a mod I

(iii) if a � b mod I and b � c mod I , then a � c mod I .

Proof. Just modify the proof of Proposition 4.3.

The next result shows that the new notion of congruence is compatible with

addition and multiplication of elements in R.

Proposition 7.2. Let I be an ideal in a commutative ring R.

(i) If a � a0 mod I and b � b0 mod I , then aC b � a0 C b0 mod I:

More generally, if ai � a0i mod I for i D 1; : : : ; k, then a1 C � � � C ak � a01 C � � � C a0k mod I:

(ii) If a � a0 mod I and b � b0 mod I , then ab � a0b0 mod I:

More generally, if ai � a0i mod I for i D 1; : : : ; k, then a1 � � �ak � a01 � � �a0k mod I:

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7.1 Quotient Rings 279

(iii) If a � b mod I , then ak � bk mod I for all k � 1:

Proof. This is a straightforward modification of the proof of Proposition 4.5.

For example, here is the proof of Proposition 4.5(i). If m j .a � a0/ and m j .b�b0/, thenm j .aCb/�.a0 Cb0/, because .aCb/�.a0 Cb0/ D .a�a0/C .b � b0/. Rewrite this here by changing “m j .a � a0/” to “a � a0 2 I .”

We now mimic the construction of the commutative rings Zm by first gen-

eralizing the idea of a congruence class.

Definition. Let I be an ideal in a commutative ringR. If a 2 R, then the coset aC I is the subset

aC I D faC z W z 2 I g � R:

Thus, the coset a C I is the set of all those elements in R that are congru- ent to a mod I . Cosets generalize the notion of congruence class and so, by

analogy, the coset aC I is often called a mod I .

Proposition 7.3. If R D Z, I D .m/, and a 2 Z, then the coset

aC I D aC .m/ D faC km W k 2 Zg

is equal to the congruence class Œa D fn 2 Z W n � a mod mg.

Proof. If u 2 aC.m/, then u D aCkm for some k 2 Z. Hence, u�a D km, m j .u � a/, u � a mod m, and u 2 Œa; that is, aC .m/ � Œa.

For the reverse inclusion, if v 2 Œa, then v � a mod m, m j .v � a/, v � a D `m for some ` 2 Z, and v D a C `m 2 a C .m/. Therefore, Œa � aC .m/, and so aC .m/ D Œa.

In Proposition 7.1, we saw that congruence mod I is an equivalence rela-

tion on R; in Exercise 7.6 on page 285, you will prove that if a 2 R, then its equivalence class is the coset a C I . It follows that the family of all cosets is a partition of R (see Proposition A.17); that is, cosets are nonempty, R is the

union of the cosets, and distinct cosets are disjoint: if a C I ¤ b C I , then .a C I /\ .b C I / D ¿.

When are two cosets mod I the same? In Proposition 4.2, we answered this question by proving that a � b mod m if and only if each of a and b has the same remainder after dividing by m.

Proposition 7.4. Let I be an ideal in a commutative ringR. If a; b 2 R, then a C I D b C I if and only if a � b mod I . In particular, a C I D I if and only if a 2 I .

Proof. Note first that a 2 aC I , for 0 2 I and a D aC 0. If aC I D b C I , then a 2 b C I ; hence, a D b C i for some i 2 I , and so a � b 2 I and a � b mod I .

Conversely, assume that a� b 2 I ; say a� b D i . To see whether aC I � bC I , we must show that if aC i 0 2 aC I , where i 0 2 I , then aC i 0 2 bC I . But aC i 0 D .bC i/C i 0 D bC .i C i 0/ 2 bC I (for ideals are closed under addition). The reverse inclusion, bCI � aCI , is proved similarly. Therefore, aC I D b C I .

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280 Chapter 7 Quotients, Fields, and Classical Problems

We have now generalized congruence mod m to congruence mod an ideal

and congruence classes to cosets. The next step is to assemble the cosets and

make a commutative ring with them.

Definition. If I is an ideal in a commutative ring R, we denote the set of all its cosets byR=I :

R=I D faC I W a 2 Rg:

Once the set Zm was built, we equipped it with the structure of a commu-

tative ring by defining addition and multiplication of congruence classes. We

carry out that program now forR=I .

Definition. Let I be an ideal in a commutative ring R.

Define addition ˛WR=I � R=I ! R=I by

˛W .aC I; b C I / 7! a C b C I

and define multiplication�WR=I �R=I ! R=I by

�W .a C I; b C I / 7! ab C I:

Example 7.5. Suppose thatR D ZŒx and I is the principal ideal .x2CxC1/. If a D 3C 2x and b D 4C 3x, then

.aC I /.b C I / D abC I D .3C 2x/.4C 3x/C I D 12C 17x C 6x2 C I:

But, by Exercise 7.4(ii) on page 285, 12C 17x C 6x2 � 6 C 11x mod I (in fact, .12C 17x C 6x2/ � .6C 11x/ D 6.x2 C x C 1/), so that

.3 C 2x C I /.4 C 3x C I / D 6C 11x C I: N

Lemma 7.6. Addition and multiplicationR=I�R=I ! R=I are well-defined functions.

Proof. Let a C I D a0 C I and b C I D b0 C I ; that is, a � a0 2 I and b � b0 2 I .

To see that addition is well-defined, we must show that a0 C b0 C I D aC b C I . This is true:

.a C b/� .a0 C b0/ D .a � a0/C .b � b0/ 2 I:

To see that multiplication R=I � R=I ! R=I is well-defined, we must show that .a0 C I /.b0 C I / D a0b0 C I D abC I ; that is, ab � a0b0 2 I . But this is true:

ab � a0b0 D ab � a0b C a0b � a0b0 D .a � a0/b C a0.b � b0/ 2 I:

The proof of Theorem 4.32, which shows that Zm is a commutative ring,

generalizes to show that R=I is a commutative ring. Here are the details.

Theorem 7.7. If I is an ideal in a commutative ring R, then R=I is a com-

mutative ring.

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7.1 Quotient Rings 281

Proof. Each of the eight axioms in the definition of commutative ring must be

verified; all the proofs are routine, for they are inherited from the correspond-

ing property in R. If a; b; c 2 R, then we have (i) Commutativity of addition:

.a C I /C .b C I / D a C bC I D b C aC I D .b C I /C .a C I /:

(ii) The zero element is I D 0C I , for I C .a C I / D 0C aC I D aC I . (iii) The negative of aC I is �aC I , for .aC I /C .�aC I / D 0C I D I . (iv) Associativity of addition:

Œ.aC I /C .b C I / C .c C I / D .aC b C I /C .c C I / D Œ.aC b/C cC I D ŒaC .b C c/ C I D .a C I /C .b C c C I / D .a C I /C Œ.b C I /C .c C I /:

(v) Commutativity of multiplication:

.a C I /.b C I / D ab C I D baC I D .b C I /.a C I /:

(vi) The multiplicative identity is 1CI , for .1CI /.aCI / D 1aCI D aCI . (vii) Associativity of multiplication:

Œ.a C I /.b C I /.c C I / D .ab C I /.c C I / D Œ.ab/cC I D Œa.bc/C I D .a C I /.bc C I / D .a C I /Œ.b C I /.c C I /:

(viii) Distributivity:

.a C I / � .b C I /C .c C I /

� D .a C I /.b C c C I / D Œa.b C c/ C I D .ab C ac/C I D .ab C I /C .ac C I / D .a C I /.b C I /C .aC I /.c C I /:

Definition. The commutative ring R=I constructed in Theorem 7.7 is called the quotient ring of R modulo I (it is usually pronounced “R mod I ”).

We said that quotient rings generalize the construction of Zm. Let’s show

that the commutative rings Z=.m/ and Zm are not merely isomorphic, they are

identical.

We have already seen, in Proposition 7.3, that they have the same elements: for every a 2 Z, the coset a C .m/ and the congruence class Œa are subsets of Z, and they are equal. But the operations coincide as well. They have the

same addition: � aC .m/

� C � b C .m/

� D aC b C .m/ D ŒaC b D ŒaC Œb

and they have the same multiplication: � aC .m/

�� b C .m/

� D ab C .m/ D Œab D ŒaŒb:

Thus, quotient rings truly generalize the integers mod m.

If I D R, then R=I consists of only one coset, and so R=I is the zero ring (in Chapter 4, we said that the zero ring does arise occasionally). Since the

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282 Chapter 7 Quotients, Fields, and Classical Problems

zero ring is not very interesting, we usually assume, when forming quotient

rings, that ideals are proper ideals. Recall, in constructing Zm, that we usually

assumed that m � 2.

The definitions of addition and multiplication in R=I involve an interplay between reducing modulo the ideal I and the operations of addition and multi-

plication inR=I . In the special case of Zm, we called this interplay “reduce as

you go.” But that’s just an informal way of describing a homomorphism. More

precisely, if we define a function � WR ! R=I by � W a 7! a C I , then we can rewrite aC bC I D .aC I /C .bC I / as �.aC b/ D �.a/C�.b/; similarly, ab C I D .a C I /.b C I / can be rewritten as �.ab/ D �.a/�.b/.

Definition. If I is an ideal in a commutative ring R, then the natural map isThe word “map” is often used as a synonym for

function or homomor-

phism.

the function � WR ! R=I given by

a 7! aC I I

that is, �.a/ D aC I .

Proposition 7.8. If I is an ideal in a commutative ring R, then the natural

map � WR ! R=I is a surjective homomorphism, and ker � D I .

Proof. We have just seen that �.aCb/ D �.a/C�.b/ and�.ab/ D �.a/�.b/. Since �.1/ D 1C I , the multiplicative identity inR=I , we see that � is a ho- momorphism.

Now � is surjective: if a C I 2 R=I , then a C I D �.a/. Finally, by definition, ker� D fa 2 R j �.a/ D 0 C I g. But �.a/ D a C I , and aC I D 0C I if and only if a 2 I (Proposition 7.4). The result follows.

Here is the converse of Proposition 5.25: Every ideal is the kernel of some

homomorphism.

Corollary 7.9. Given an ideal I in a commutative ringR, there exists a com-

mutative ring A and a homomorphism 'WR ! A with I D ker '.

Proof. If we set A D R=I , then the natural map � WR ! R=I is a homomor- phism with I D ker � .

We know that isomorphic commutative rings are essentially the same, being “translations” of one another; that is, if 'WR ! S is an isomorphism, we may think of r 2 R as being in English while '.r/ 2 S is in French. The next theorem shows that quotient rings are essentially images of homomorphisms.

It also shows how to modify a homomorphism to make it an isomorphism.

Theorem 7.10 (First Isomorphism Theorem). Let R and A be commutative

rings. If 'WR ! A is a homomorphism, then ker ' is an ideal in R, im' is aThere are second and third isomorphism theorems,

but they are less useful

(see Exercise 7.15 on

page 286).

subring of A, and

R= ker' Š im':

Proof. Let I D ker'. We have already seen, in Proposition 5.25, that I is an ideal in R and im' is a subring of A.

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7.1 Quotient Rings 283

Definee'WR=I ! im' by e'.r C I / D '.r/:

We claim thate' is an isomorphism. First,e' is well-defined. If r C I D s C I , then r � s 2 I D ker ', '.r � s/ D 0, and '.r/ D '.s/. Hence

e'.r C I / D '.r/ D '.s/ D e'.s C I /:

Next,e' is a homomorphism because ' is.

e' � .r C I /C .s C I /

� D e'.r C s C I / D '.r C s/ D '.r/C '.s/ D e'.r C I /Ce'.s C I /:

Similarly,e' � .r C I /.sC I /

� De'.rC I /e'.sC I / (Exercise 7.7 on page 285).

Ase'.1 C I / D '.1/ D 1, we see thate' a homomorphism. We show thate' is surjective. If a 2 im', then there is r 2 Rwith a D '.r/;

plainly, a D '.r/ D e'.r C I /. Finally, we show that e' is injective. If e'.r C I / D 0, then '.r/ D 0, and

r 2 ker ' D I . Hence, r C I D I ; that is, kere' D fI g and e' is injective, by Proposition 5.31. Therefore,e' is an isomorphism.

We can illustrate this last proof with a picture; such a picture is often called

a commutative diagram if composites of maps having same domain and same target are equal. Here, i W im' ! A is the inclusion, and ' D ie'� .

R '

A

R=I e' im' i

Here’s a trivial example. If R is a commutative ring, then .0/ is an ideal.

The identity 1RWR ! R is a surjective homomorphism with ker 1R D .0/, so that the First Isomorphism Theorem gives the isomorphisme1RWR=.0/ ! R; that is, R=.0/ Š R.

Theorem 7.10 has more interesting applications than showing thatR=.0/ Š R. For example, it gives us the tools needed to tighten up the discussion of the al-

ternate construction of C that began this section.

Theorem 7.11. The quotient ring RŒx=.x2 C 1/ is a field isomorphic to the Hence, the high school approach to complex

numbers contains the

germ of a correct idea.

complex numbers C.

Proof. Consider the evaluation 'W RŒx ! C (as in Corollary 5.21) with '.x/ D i and '.a/ D a for all a 2 R; that is,

'Wf .x/ D a0 C a1x C a2x2 C � � � 7! f .i/ D a0 C a1i C a2i2 C � � � :

Now ' is surjective, for a C ib D '.a C bx/, and so the First Isomorphism Theorem gives an isomorphisme'W RŒx= ker' ! C, namely f .x/C ker' 7! f .i/. But Corollary 6.26 gives ker' D .x2C1/; therefore, RŒx=.x2C1/ Š C as commutative rings, by the First Isomorphism Theorem. We know that C is

a field, and any commutative ring isomorphic to a field must, itself, be a field.

Thus, the quotient ring RŒx=.x2 C 1/ is another construction of C.

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284 Chapter 7 Quotients, Fields, and Classical Problems

How to Think About It. Because every element of RŒx is congruent to a

linear polynomial aC bx mod .x2 C 1/, every element of RŒx=.x2 C 1/ can be written as aC bx C .x2 C 1/ for some real numbers a and b.

Example 7.12. (i) Since RŒx=.x2 C 1/ is a field, every nonzero element in it has a multiplicative inverse. Let’s find the inverse of an element a C bx C .x2 C 1/ by “pulling back” the formula in C,

1

aC bi D a � bi a2 C b2

;

to RŒx=.x2 C 1/, using the inverse of the isomorphism ' in the proof of Theorem 7.11. Now '�1.a C bi/ D aC bx C .x2 C 1/; so that

1

a C bx C .x2 C 1/ D a � bx a2 C b2

C .x2 C 1/:

(ii) Euclidean Algorithm II gives another way of finding the inverse, writing

gcd.a C bx; x2 C 1/ as a linear combination of aC bx and x2 C 1. The algorithms in Exercise 6.18(iii) produce the linear combination in RŒx:

� a � bx b2

� .a C bx/C 1.x2 C 1/ D a

2 C b2 b2

:

Dividing both sides by .a2 C b2/=b2, we have

a � bx a2 C b2 .a C bx/C

b2

a2 C b2 .x 2 C 1/ D 1:

Moving to RŒx=.x2 C 1/, this implies again that � a C bx C .x2 C 1/

��1 D a � bx a2 C b2

C .x2 C 1/ N

We end this section with a generalization of Theorem 7.11. If you chase

back the arguments to their source, you’ll see that all we needed is that RŒx is

a PID and x2 C 1 is an irreducible element in RŒx.

Proposition 7.13. If R is a PID and p is an irreducible element in R, then

R=.p/ is a field.

Proof. It suffices to show that every nonzero element a C .p/ in the commu- tative ring R=.p/ has a multiplicative inverse. Since a C .p/ ¤ 0, we have a … .p/; that is, p − a. Since R is a PID, Theorem 6.46 says that gcd’s exist and are linear combinations. In particular, gcd.a; p/ D 1, so there are s; t 2 R with saC tp D 1. Thus,

1C .p/ D saC .p/ D � saC .p/

� D � s C .p/

�� aC .p/

in R=.p/, and � aC .p/

��1 D s C .p/. Therefore, R=.p/ is a field.

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7.1 Quotient Rings 285

Exercises

7.1 Are any cosets of .5/ in Z ideals?

7.2 Prove Proposition 7.1.

7.3 Prove Proposition 7.2.

7.4 * In QŒx=.x2 C x C 1/, write each term in the form a C bx with a; b 2 Q. (i) .3 C 2x/.4 C 3x/ (ii) 12 C 17x C 6x2

(iii) x2 (iv) x3

(v) .1 � x/2 (vi) .1 � x/.1 � x2/ (vii) .a C bx/.a C bx2/ (viii) .a C bx/2.

7.5 In QŒx=.x4Cx3Cx2Cx C1/, write each term in the form aCbx Ccx2Cdx3 with a; b; c; d rational numbers.

(i) x5

(ii) .1 � x/.1 � x2/.1 � x3/.1 � x4/ (iii) .1 C x/.1 C x2/.1 C x3/.1 C x4/.

7.6 * In Proposition 7.1, we saw that if I is an ideal in a commutative ring R, then

congruence mod I is an equivalence relation on R. Prove that the equivalence

classes are the cosets mod I .

7.7 In the notation of Theorem 7.10, show that

e'�.r C I /.s C I /� De'.r C I /e'.s C I /: 7.8 * Let 'W R ! S be an isomorphism of commutative rings. Assume that I � R

and J � S are ideals and that '.I / D J , where '.I / D f'.a/ W a 2 I g. Prove thate'W R=I ! S=J , given bye'W r C I 7! '.r/ C J , is an isomorphism.

7.9 Let I be an ideal in a commutative ring R.

(i) If S is a subring of R and I � S , prove that

S=I D fr C I W r 2 Sg

is a subring of R=I .

(ii) If J is an ideal in R and I � J , prove that

J=I D fr C I W r 2 J g

is an ideal in R=I .

7.10 Show that the subring ZŒx=.x2 C 1/ of RŒx=.x2 C 1/ is isomorphic to the Gaussian integers ZŒi .

7.11 Show that there is an isomorphism of fields:

RŒx=.x2 C 1/ Š RŒx=.x2 C x C 1/

Hint: Both are isomorphic to C.

7.12 Show that

QŒx=.x2 C x C 1/ Š QŒ! D fu C v! W u; v 2 Qg;

where ! D 12 �

�1 C i p

3 �

.

7.13 Show that the subring ZŒx=.x2C x C 1/ of RŒx=.x2C x C 1/ is isomorphic to the Eisenstein integers ZŒ!.

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286 Chapter 7 Quotients, Fields, and Classical Problems

7.14 For each element of QŒx=.x2 C x C 1/, find the multiplicative inverse. (i) 3 C 2x C .x2 C x C 1/

(ii) 5 � x C .x2 C x C 1/ (iii) 15 C 7x � 2x2 C .x2 C x C 1/ (iv) a C bx C .x2 C x C 1/ (in terms of a and b).

7.15 * Prove the Third Isomorphism Theorem: If R is a commutative ring having

ideals I � J , then J=I is an ideal in R=I , and there is an isomorphism .R=I /=.J=I / Š R=J .

Hint: Show that the function 'W R=I ! R=J , given by a C I 7! a C J , is a homomorphism, and apply the First Isomorphism Theorem.

7.16 For every commutative ring R, prove that RŒx=.x/ Š R.

7.17 An ideal I in a commutative ring R is called a prime ideal if I is a proper ideal

such that ab 2 I implies a 2 I or b 2 I . (i) If p is a prime number, prove that .p/ is a prime ideal in Z.

Hint: Euclid’s Lemma.

(ii) Prove that if an ideal .m/ in Z is a prime ideal, then m D 0 or jmj is a prime number.

7.18 Let I be a proper ideal in kŒx, where k is a field.

(i) If p is an irreducible polynomial, prove that .p/ is a prime ideal in kŒx.

(ii) Prove that if an ideal .f / in kŒx is a prime ideal, then f D 0 or f is an irreducible polynomial.

7.19 Let I be a proper ideal in a commutative ring R.

(i) Prove that .0/ is a prime ideal in R if and only if R is a domain.

(ii) Prove that I is a prime ideal if and only if a … I and b … I imply ab … I . (iii) Prove that I is a prime ideal if and only if R=I is a domain.

7.20 Prove that .x/ is a prime ideal in ZŒx.

Hint: Is ZŒx=.x/ a domain?

7.21 An ideal I in a commutative ring R is called a maximal ideal if I is a proper ideal

for which there is no proper ideal J with I ¨ J .

(i) If p is a prime number, prove that .p/ is a maximal ideal in Z.

(ii) Prove that if an ideal .m/ in Z is a maximal ideal, then jmj is a prime number.

7.22 Let I be a proper ideal in kŒx, where k is a field.

(i) If p is an irreducible polynomial, prove that .p/ is a maximal ideal in kŒx.

(ii) Prove that if an ideal .f / in kŒx is a maximal ideal, then f is an irreducible

polynomial.

7.23 * Let I be a proper ideal in a commutative ring R.

(i) Prove that .0/ is a maximal ideal in R if and only if R is a field.

(ii) Prove that I is a maximal ideal if and only if R=I is a field. Conclude that if

k is a field and p.x/ 2 kŒx is irreducible, then kŒx=.p/ is a field. (iii) Prove that every maximal ideal is a prime ideal.

7.24 (i) Prove that J is a maximal ideal in ZŒx, where J consists of all polynomials

with even constant term.

Hint: Prove that ZŒx=J Š F2 . (ii) Prove that the prime ideal .x/ in ZŒx is not a maximal ideal.

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7.2 Field Theory 287

7.2 Field Theory

General results about quotient rings R=I have a special character when R

enjoys extra hypotheses. In this section, we investigate properties of fields with an eye to using the ideas behind the isomorphism

RŒx=.x2 C 1/ Š C:

We are going to apply quotient rings to prove some interesting results: for every polynomial f .x/ 2 kŒx, where k is a field, there exists a field extensionE=k containing all the roots of f ; we will also be able to prove the existence of

finite fields other than Fp.

Characteristics

Contemplating “any field” seems quite daunting, and so it makes sense for us to

begin classifying fields. First of all, fields come in two types: those that contain

a subfield isomorphic to Q, and those that contain a subfield isomorphic to Fp for some prime p.

Recall the definition of na on page 160, where n 2 Z and a is an element of a commutative ringR. For example, 3a means aC aC a and .�3/a means �a � a � a. More generally, if n is a nonnegative integer, then na means

aC aC � � � C a„ ƒ‚ … n times

;

0a D 0, and �na is the sum of jnj copies of �a. Note that na is a hybrid in the sense that it is the product of an integer and an element ofR, not the product of

two ring elements. However, na can be viewed as the product of two elements

in R, for if e is the multiplicative identity inR, then ne 2 R and na D .ne/a. In particular, 3a D a C a C a D .e C e C e/a D .3e/a. This “action” of Z on R is really a homomorphism.

Lemma 7.14. If R is a commutative ring with multiplicative identity e; then

the function �W Z ! R, given by

�.n/ D ne;

is a homomorphism.

Proof. Exercise 7.25 on page 293.

Proposition 7.15. If k is a field and �W Z ! k is the map �Wn 7! ne, where e is the multiplicative identity in R, then either im� Š Z or im� Š Fp for some prime p.

Proof. Since every ideal in Z is principal, ker� D .m/ for some integer m � 0. If m D 0, then � is an injection, and im� Š Z. If m ¤ 0, the First Isomorphism Theorem gives Zm D Z=.m/ Š im� � k. Since k is a field, im� is a domain, and so m is prime (Exercise 5.3 on page 195). Writing p

instead of m, we have im� Š Zp D Fp.

Corollary 7.16. Every field k contains a subfield isomorphic to either Q or Fp .

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288 Chapter 7 Quotients, Fields, and Classical Problems

Proof. Proposition 7.15 shows that k contains a subring isomorphic to Z or

to Fp for some prime p. If the subring is Z, then, because the field k contains

multiplicative inverses for all of its non-zero elements, it contains an isomor- phic copy of Q D Frac.Z/. More precisely, Exercise 5.38(ii) on page 212 says that a field containg an isomorphic copy of Z as a subring must contain an

isomorphic copy of Q.

By Exercise 7.28(i) on page 293, k can’t contain an isomorphic copy of

both Q and Fp; by Exercise 7.28(ii) on page 293, k can’t contain copies of Fp and Fq for distinct primes p and q.

Definition. A field has characteristic 0 if ker� D .0/; it has characteristic p if ker� D .p/ for some prime p.

This distinction is the first step in classifying different types of fields. The fields Q, R, C, and C.x/ have characteristic 0, as do any of their sub-

fields. Every finite field has characteristic p for some prime p (after all, if

ker� D .0/, then im� Š Z is infinite); Fp.x/, the field of all rational func- tions over Fp, is an infinite field of characteristic p.

Proposition 7.17. Let k be a field of characteristic p > 0.

(i) pa D 0 for all a 2 k. (ii) If q D pn, then .a C b/q D aq C bq for all a; b 2 k.Exercise 5.36 on page 212

proves a congruence

version of (ii) for a; b 2 Z. (iii) If k is finite, then 'W k ! k, given by

'W a 7! ap ;

is an isomorphism.

Proof. (i) Since k has characteristic p, we have ker.�/ D .p/; that is, �.p/ D p1 D 0 (we have reverted to our usual notation, so that 1 denotes the mul- tiplicative identity). But the hybrid product pa can be viewed as a product

of two ring elements: pa D .p1/a D 0a D 0. (ii) Expand .a C b/p by the Binomial Theorem, and note that p j

� p j

� for

all 1 � j � p � 1. By (i), all the inside terms vanish. The argument is completed by induction on n � 1.

(iii) It is obvious that '.1/ D 1 and

'.ab/ D .ab/p D apbp D '.a/'.b/:

By (ii), '.aCb/ D '.a/C'.b/. Therefore, ' is a homomorphism. Since ker ' is a proper ideal in k (for 1 … ker '), we have ker ' D .0/, because k is a field, and so ' is an injection. Finally, since k is finite, the Pigeonhole

Principle applies, and ' is an isomorphism.

We have seen finite fields Fp with p elements, for every prime p, and in

Exercise 4.55 on page 165, we saw a field F4 with exactly four elements. The next result shows that the number of elements in a finite field must be a prime

power; there is no field having exactly 15 elements. Theorem 7.38 will show,

for every prime p and every integer n � 1, that there exists a field having exactly pn elements.

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7.2 Field Theory 289

Example A.20(iv) in the Appendices shows that if a commutative ring R

contains a subring k that is a field, then R is a vector space over k: vectors are

elements r 2 R, while scalar multiplication by a 2 k is the given multiplica- tion ar of elements inR. The vector space axioms are just some of the axioms

in the definition of commutative ring.

Proposition 7.18. If K is a finite field, then jKj D pn for some prime p and some n � 1.

Proof. The prime field of K is isomorphic to Fp for some prime p, by Propo- If K is a vector space over k, its dimension is

denoted by dimk.K/ or,

more briefly, by dim.K/.

sition 7.15. As we remarked, K is a vector space over Fp ; as K is finite, it is

obviously finite-dimensional. If dimFp .K/ D n, then jKj D pn, by Corol- lary A.34 in the Appendix.

Extension Fields

The Fundamental Theorem of Algebra states that every nonconstant polyno-

mial in CŒx is a product of linear polynomials in CŒx; that is, C contains all

the roots of every polynomial in CŒx. Using ideas similar to those allowing us

to view C as a quotient ring, we’ll prove Kronecker’s Theorem, a local ana- log of the Fundamental Theorem of Algebra for polynomials over an arbitrary

field k: given f .x/ 2 kŒx, there is some fieldE containing k as a subfield that also contains all the roots of f . (We call this a local analog, for even though

the larger fieldE contains all the roots of the polynomial f , it may not contain

roots of some other polynomials in kŒx.) In fact, we’ll see how to construct

such an E making basic use of quotient rings of the form kŒx=I , where k is a field. Theorem 4.43 says that

Zm is a field if and only

if m is a prime in Z;

Proposition 7.19 is the

analog for kŒx.

Proposition 7.19. If k is a field and I D .f /, where f .x/ 2 kŒx is noncon- stant, then the following are equivalentW

(i) f is irreducible

(ii) kŒx=I is a field

(iii) kŒx=I is a domain.

Proof. (i) ) (ii) Since kŒx is a PID, this follows at once from Proposi- tion 7.13.

(ii) ) (iii) Every field is a domain. (iii) ) (i) Assume that kŒx=I is a domain. If f is not irreducible, then

there are g.x/; h.x/ 2 kŒx with f D gh, where deg.g/ < deg.f / and deg.h/ < deg.f /. Recall that the zero in kŒx=I is 0 C I D I . Thus, if gCI D I , then g 2 I D .f / and f j g, contradicting deg.g/ < deg.f /. Similarly, hCI ¤ I . However, the product .gCI /.hCI / D f CI D I is zero in the quotient ring, which contradicts kŒx=I being a domain.

Therefore, f is irreducible.

The structure of general quotient rings R=I can be complicated, but for This section will be using various facts about dimen-

sion, and you may wish

to look in Appendix A.3 to

refresh your memory.

special choices ofR and I , the commutative ringR=I can be easily described.

For example, when k is a field and p.x/ 2 kŒx is an irreducible polynomial, the following proposition gives a complete description of the field R=I D kŒx=.p/, and it shows how to build a field K in which p.x/ has a root.

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290 Chapter 7 Quotients, Fields, and Classical Problems

Proposition 7.20. Let k be a field andK D kŒx=.p/, where p.x/ 2 kŒx is a monic irreducible polynomial of degree d and I D .p/:

(i) K is a field, and

k0 D faC I W a 2 kg

is a subfield ofK isomorphic to k. If k0 is identified with k via a 7! aCI , then k is a subfield of K.

(ii) z D x C I is a root of p in K.If we view k as a subfield of K, then it makes sense

to speak of a root of p in

K.

(iii) If g.x/ 2 kŒx and z is a root of g in K, then p j g in kŒx. (iv) p is the unique monic irreducible polynomial in kŒx having z as a root.

(v) K is a vector space over k, the list 1; z; z2; : : : ; zd�1 is a basis, and

dimk.K/ D d:

Proof. (i) Since p is irreducible, Proposition 7.19 says that the quotient ring

K D kŒx=I is a field, while Corollary 5.32 on page 220 says that the natural map a 7! aC I restricts to an isomorphism k ! k0.

(ii) Let p.x/ D a0 C a1x C � � � C ad�1xd�1 C xd , where ai 2 k for all i . In light of the identification of k and k0 in (i), we may view p.x/ asP j .aj C I /xj . Hence, since z D x C I ,

p.z/ D .a0 C I /C .a1 C I /z C � � � C .1 C I /zd

D .a0 C I /C .a1 C I /.x C I /C � � � C .1C I /.x C I /d

D .a0 C I /C .a1x C I /C � � � C .xd C I / D a0 C a1x C � � � C xd C I D p C I D I;

because I D .p/. But I D 0 C I is the zero element of K D kŒx=I ; thus, p.z/ D 0 and z is a root of p.

(iii) If p − g in kŒx, then gcd.g; p/ D 1 because p is irreducible. Therefore, there are s; t 2 kŒx with 1 D sp.x/C tg.x/. Since kŒx � KŒx, we may regard this as an equation in KŒx. Setting x D z gives the contradiction 1 D 0.

(iv) Let h.x/ 2 kŒx be a monic irreducible polynomial having z as a root. By part (iii), we have p j h. Since h is irreducible, we have h D cp for some constant c; since h and p are monic, we have c D 1 and h D p.

(v) Example A.20 in the Appendices shows that K is a vector space over k.

Every element of K has the form f .x/ C I , where f 2 kŒx. By the Division Algorithm, there are polynomials q; r 2 kŒx with f D qp C r and either r D 0 or deg.r/ < d D deg.p/. Since f � r D qp 2 I , it follows that f CI D rCI . Let r.x/ D b0Cb1xC� � �Cbd�1xd�1, where bi 2 k for all i . As in (ii), we see that rCI D b0Cb1zC� � �Cbd�1zd�1. Therefore, 1; z; z2; : : : ; zd�1 spans K.

To see that the list is linearly independent, suppose that

d�1X

iD0 ciz

i D 0 in K D kŒx=.p/I

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7.2 Field Theory 291

lifting to kŒx, this says that

d�1X

iD0 cix

i � 0 mod .p/ in kŒx;

so that p j Pd�1 iD0 cix

i in kŒx. But deg.p/ D d , so that all ci D 0.

Definition. IfK is a field containing k as a subfield, thenK is called an exten- This notation should not be confused with the notation

for a quotient ring, for a

field K has no interesting

ideals; in particular, if

k ¨ K, then k is not an ideal in K.

sion field of k, and we write “K=k is an extension field.” An extension field

K=k is a finite extension ifK is a finite-dimensional vector space over k. The dimension of K, denoted by

ŒK W k; is called the degree of K=k.

Corollary 7.21. If k.z/=k is a field extension, where z is a root of an irre-

ducible polynomial p.x/ 2 kŒx, then

Œk.z/ W k D deg.p/:

Proof. This is just a restatement of Proposition 7.20(v).

Corollary 7.21 shows why ŒK W k is called the degree ofK=k.

How to Think About It. At first glance, many people see Proposition 7.20 as a cheat: we cook up a field that contains a root of p.x/ by reducing mod p.

The root is thus a coset, not a “number.” But, just as mathematicians gradually

came to see the elements of C as numbers (through their constant use in calcu-

lations), one can develop a feel for arithmetic in kŒx=.p/ in which the cosets

become concrete objects in their own right, as in the next example.

Example 7.22. Suppose that k D Q and

p.x/ D x3 C x2 � 2x � 1:

You can check that p is irreducible (it’s a cubic without a rational root (why?), so it can’t factor). Now

K D QŒx=.x3 C x2 � 2x � 1/

is a field, ŒK W Q D 3, and a basis for K over Q is

1C .p/; x C .p/; x2 C .p/:

Hence, every element of K can be represented by a quadratic expression

aC bx C cx2 C .p/;

where a; b; c 2 Q. The expression is unique, because p is a cubic: if two Eventually, we dropped the bracket notation for

congruence classes,

abbreviating Œa to a.

quadratics f; g 2 QŒx are congruent mod .p/, then either f � g D 0 or deg.f � g/ < 2; the latter cannot occur, and so f D g. So, let’s drop the “C.p/” decoration and just represent an element ofK by the unique quadratic

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292 Chapter 7 Quotients, Fields, and Classical Problems

polynomial in its congruence class mod p. Using this convention, the elements

of K are thus named by quadratic polynomials in QŒx.

What about the arithmetic? Just as in C D RŒx=.x2C1/, calculations inK are carried out by calculating in QŒx, dividing by p, and taking the remainder.

Indeed, because x3 C x2 � 2x � 1 D 0 inK, we have an equation in K,

x3 D �x2 C 2x C 1:

Hence, to calculate in K, we calculate in QŒx with the additional simplifica-

tion rule that x3 is replaced by �x2 C 2x C 1. Beginning to sound familiar? So, for example, here are some calculations inK:

(i) Addition looks just the same as in QŒx because addition doesn’t increase

degree:

� a C bx C cx2

� C � d C ex C f x2

D .a C d/C .b C e/x C .c C f /x2:

(ii) Multiplication requires a simplification. For example, in K,

.3C 2x C 4x2/.�1 C 5x C 7x2/ D 3C 53x C 77x2;

a fact that you can verify (by hand or CAS) by expanding the left-hand

side and reducing mod .p/. In general, expand

� a C bx C cx2

� � d C ex C f x2

as

cf x4 C .bf C ce/x3 C .af C be C cd/x2 C .ae C bd/x C ad;

and then simplify, replacing occurrences of x3 by �x2 C 2x C 1:

D cf x � x3 �

C .bf C ce/x3 C .af C beC cd/x2 C .ae C bd/x C ad D cf x

� �x2 C 2x C 1

� C .bf C ce/

� �x2 C 2x C 1

C.af C beC cd/x2 C .ae C bd/x C ad D etc.

A little practice with such calculations gives you the feeling that you are indeed

CAS environments will do

all of this work for you—just

ask for the remainder when

a product is divided by p.

working with “numbers” in a system and, ifK had any use, you’d soon become

very much at home in it just as our Renaissance predecessors became at home in C. N

While we’ve constructed a field extension K=k in which p.x/ 2 kŒx has a root ˛, we have little idea about what that root is, even when k D R and ˛ is a complex number. For example, if p.x/ D x2 C 1, we can’t tell whether ˛ D i or ˛ D �i . Proposition 7.20 doesn’t give you a way to find roots— it just gives you a way to construct an extension field containing k and inYou could use Cardano’s

formula to find expressions

for the roots of p.x/. Why

not try it?

which the operations behave as if p.x/ is 0. Playing with these operations

might actually give you some ideas about the three complex numbers that make

p.x/ D x3 C x2 � 2x � 1 equal to 0 in C; see Example 7.22 above and Exercises 7.29 and 7.30 below.

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7.2 Field Theory 293

Exercises

7.25 * Prove Lemma 7.14.

7.26 * If X is a subset of a field k, then ˝ X ˛ , the subfield generated by X , is the

intersection of all the subfields containing X (by Exercise 4.61(iii) on page 168,

the intersection of any family of subfields of k is itself a subfield of k).

(i) Prove that ˝ X ˛

is the smallest such subfield in the sense that any subfield F

containing X must contain ˝ X ˛ .

(ii) Define the prime field of a field k to be the intersection of all the subfields

of k. Prove that the prime field of k is the subfield generated by 1.

(iii) Prove that the prime field of a field is isomorphic to either Q or Fp .

7.27 * If k is a field of characteristic p > 0 and a 2 k, prove that

.x C a/p D xp C ap :

Hint: Use Proposition 7.17 and the Binomial Theorem.

7.28 Let R be a commutative ring, and let p; q be distinct primes.

(i) Prove that R cannot have subrings A and B with A Š Q and B Š Fp. (ii) Prove that R cannot have subrings A and B with A Š Fp and B Š Fq .

(iii) Why doesn’t the existence of R D Fp�Fq contradict part (ii)? (Exercise 5.54 on page 221 defines the direct product of rings.)

7.29 * As in Example 7.22, let p.x/ D x3 C x2 � 2x � 1 and let K D QŒx=.p/. Let

˛ D x; ˇ D x2 � 2; and D x3 � 3x D �x2 � x C 1:

Calculate in K , writing each result as a C bx C cx2: (i) ˛ C ˇ C

(ii) ˛ˇ C ˛ C ˇ (iii) ˛ˇ .

7.30 * As in Example 7.22, let p.x/ D x3 C x2 � 2x � 1 and let K D QŒx=.p/. Show that, in K ,

p.x2 � 2/ D p.x3 � 3x/ D 0:

Hence the three roots of p in K are x C .p/, x2 � 2 C .p/, and x3 � 3x C .p/.

Algebraic Extensions

The first step in classifying fields is by their characteristics. Here is the second step: we define algebraic extensions.

Definition. Let K=k be an extension field. An element z 2 K is algebraic over k if there is a nonzero polynomial f .x/ 2 kŒx having z as a root; other- wise, z is transcendental over k. A field extension K=k is algebraic if every

z 2 K is algebraic over k.

When a real number is called transcendental, it usually means that it is

transcendental over Q. For example, it was proved by Lindemann that � is

a transcendental number (see [15], pp. 47–57 or [3], p. 5); there is no nonzero

f .x/ 2 QŒx with f .�/ D 0.

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294 Chapter 7 Quotients, Fields, and Classical Problems

Proposition 7.23. If K=k is a finite extension field, then K=k is an algebraic

extension.

Proof. By definition, K=k finite means that K has finite dimension n as a

vector space over k. Suppose that z is an element of K. By Corollary A.39 in

Appendix A.3, the list of n C 1 vectors 1; z; z2; : : : ; zn is linearly dependent: there are c0; c1; : : : ; cn 2 k, not all 0, with

P ciz

i D 0. Thus, the polynomial f .x/ D

P cix

i is not the zero polynomial, and z is a root of f . Therefore, z is algebraic over k.

The converse of the last proposition is not true; the field A of all complex

numbers algebraic over Q is an algebraic extension of Q that is not a finite

extension. (The fact that A is a field is not obvious, but it is true (see [17], Chapter 6).)

Definition. If K=k is an extension field and z 2 K, then k.z/, the subfield of K obtained by adjoining z to k, is the intersection of all those subfields of K

containing k and z. More generally, if A is a subset of K, define k.A/ to be the intersection of

all the subfields ofK containingk[A; we call k.A/ the subfield ofK obtained by adjoining A to k. In particular, if A D fz1; : : : ; zng is a finite subset, then we may denote k.A/ by k.z1 ; : : : ; zn/.

In Exercise 7.43 on page 308, you’ll show that k.A/ is the smallest subfield

ofK containing k and A; that is, ifE is any subfield of K containing k and A,

then k.A/ � E . Proposition 7.20 starts with an irreducible polynomial p.x/ 2 kŒx and

constructs an extension K=k in which p has a root. Suppose we start with the root; that is, suppose that z is algebraic over k. Can we find a polynomial p

so that k.z/ (the smallest extension of k that contains z) can be realized as

kŒx=.p/? Let’s look at an example.

Example 7.24. Suppose that K D R, k D Q, and z D p 2 C

p 3. First of

all, z is algebraic over Q. To see this, proceed as you would in high school

algebra.

z2 D 5C 2 p 6;

.z2 � 5/2 D 24; z4 � 10z2 C 1 D 0:

Hence, z is a root of h.x/ D x4 � 10x2 C 1, so it is algebraic over Q. Consider the evaluation homomorphism W QŒx ! R (provided by Theo-

rem 5.19) given by

Wf .x/ 7! f .z/:

The First Isomorphism Theorem suggests that we look at im and ker .

� im contains Q (because .a/ D a for all a 2 Q) and z (because .x/ D z). It follows that any subfield of R that contains Q and z contains im . In other

words,

im D Q.z/:

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7.2 Field Theory 295

� If I D ker , then the First Isomorphism Theorem gives

QŒx=I Š im :

� im is a subring of R, so it is a domain. And I is an ideal in QŒx, so it is principal, say I D .p/, where p.x/ 2 QŒx.

� Furthermore, since QŒx=I is a domain, p is an irreducible polynomial in QŒx, which we can take to be monic.

Thus, we have an isomorphism:

‰W QŒx=.p/ Š im ;

namely ‰Wf .x/C .p/ 7! f .z/. Since im D Q.z/, we have

QŒx=.p/ Š Q.z/:

There it is: Q.z/ is realized as a quotient of QŒx by an irreducible polynomial. And, because we started with a specific z, we can do better: we can find p.

We’ll see later, in Example 7.32, that p.x/ D h.x/. N

Example 7.24 contains most of the ideas of the general result.

Theorem 7.25. (i) If K=k is an extension field and z 2 K is algebraic over k, then there is a unique monic irreducible polynomial p.x/ 2 kŒx having z as a root. Moreover, if I D .p/, then kŒx=I Š k.z/; indeed, there exists an isomorphism

‰ W kŒx=I ! k.z/

with‰.x C I / D z and‰.c C I / D c for all c 2 k. (ii) If z0 2 K is another root of p.x/, then there is an isomorphism

� W k.z/ ! k.z0/

with �.z/ D z0 and �.c/ D c for all c 2 k.

Proof. (i) As in Example 7.24, consider the evaluation homomorphism W kŒx ! K, given by Theorem 5.19:

Wf 7! f .z/:

Now im is the subring of K consisting of all polynomials in z, that is,

all elements of the form f .z/ with f 2 kŒx, while ker is the ideal in kŒx consisting of all those g.x/ 2 kŒx having z as a root. Since every ideal in kŒx is a principal ideal, we have ker D .p/ for some monic polynomial p.x/ 2 kŒx. But the First Isomorphism Theorem says that kŒx=.p/ Š im , which is a domain, and so p is irreducible, by Proposition 7.19. The same proposition says that kŒx=.p/ is a field, and

so there is an isomorphism‰W kŒx=.p/ Š im ; namely ‰Wf .x/C I 7! .f / D f .z/. Hence, im‰ is a subfield of K containing k and z. But every such subfield of K must contain im , so that im‰ D im D k.z/. We have proved everything in the statement except the uniqueness

of p; but this follows from Proposition 7.20(iv).

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296 Chapter 7 Quotients, Fields, and Classical Problems

(ii) By (i), there are isomorphisms ‰W kŒx=I ! k.z/ and ‰0W kŒx=I ! k.z0/with‰.cCI / D c D ‰0.cCI / for all c 2 k; moreover, ‰WxCI 7! z and ‰0Wx C I 7! z0. The composite � D ‰0 ı ‰�1 is the desired isomorphism; it satisfies �.c/ D c for all c 2 k, and

� Wf .z/ 7! f .z0/;

for all f .x/ 2 kŒx.

The proof of Theorem 7.25(ii) is described by the following diagram.

kŒx=.p/

‰0

‰ k.z/

�D‰0ı‰�1

k.z0/

Definition. If K=k is an extension field and z 2 K is algebraic over k, then the unique monic irreducible polynomial p.x/ 2 kŒx having z as a root is called the minimal polynomial of z over k, and it is denoted by

p.x/ D irr.z; k/:

The minimal polynomial irr.z; k/ depends on k. For example, irr.i;R/ D x2 C 1, while irr.i;C/ D x � i .

Example 7.26. We know that i 2 C is algebraic over R, and irr.i;R/ D x2 C 1. Now �i is another root of x2 C 1. The isomorphism � W C D R.i/ ! R.�i/ D C with �.i/ D �i and �.c/ D c for all c 2 R is, of course, complex conjugation.

The adjunction of one root of irr.i;R/ also adjoins the other root, for the minimal polynomial here is quadratic. But this doesn’t always happen. For

example, z D 3 p 5 is algebraic over Q with minimal polynomial x3 � 5. The-

orem 7.25 tells us that

‰W QŒx=.x3 � 5/ ! Q.z/;

given by f .x/ C .x3 � 5/ 7! f . 3 p 5/, is an isomorphism. But the roots of

x3 � 5 are not all contained in Q.z/; indeed, the other two roots are not real, while QŒz � R; in fact, in CŒx,

x3 � 5 D .x � z/.x � !z/.x � !2z/;

where ! is our old friend 1 2

� �1C i

p 3 �

. Theorem 7.25(i) tells us that the

fields

Q.z/; Q.!z/; Q.!2z/

are all isomorphic via isomorphisms that fix Q pointwise. One of these is

� W Q.z/ ! Q.!z/;

defined as follows: every element of Q.z/ is of the form f .z/ where f .x/ C .x3 � 5/ is a coset in QŒx=.x3 � 5/. Then

� .f .z// D f .!z/:

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7.2 Field Theory 297

Again, a diagram illustrates the work just done.

QŒx=.x3 � 5/

‰0

‰ Q.z/

�D‰ı‰�1

Q.!z/ N

The following formula is quite useful, especially when proving a theorem

by induction on degree. Before reading the proof, you may want to refresh your

memory by looking at Appendix A.3 on linear algebra.

Theorem 7.27. Let k � E � K be fields, withE=k andK=E finite extension fields. Then K=k is a finite extension field, and

ŒK W k D ŒK W EŒE W k:

Proof. If A D a1; : : : ; an is a basis of E over k and B D b1; : : : ; bm is a basis of K over E , then it suffices to prove that the list X of all aibj is a basis of K

over k.

To see that X spans K, take u 2 K. Since B is a basis of K over E , there are scalars �j 2 E with u D

P j �j bj . Since A is a basis of E over k, there

are scalars �j i 2 k with �j D P i �j iai . Therefore, u D

P ij �j iaibj , and X

spansK over k. (Check that this makes sense in the special caseA D a1; a2; a3 and B D b1; b2.)

To prove thatX is linearly independent over k, assume that there are scalars

�j i 2 k with P ij �j iaibj D 0. If we define �j D

P i �j iai , then �j 2 E

and P j �j bj D 0. Since B is linearly independent over E , it follows that

0 D �j D X

i

�j iai

for all j . Since A is linearly independent over k, it follows that �j i D 0 for all j and i , as desired.

There are several classical geometric problems, such as trisecting an arbi-

trary angle with ruler and compass in which Theorem 7.27 plays a critical role (see Section 7.3).

Example 7.28. We now show how Theorem 7.27, the multiplicativity of de-

gree in a tower of extension fields, can be used to calculate degrees; we also

show, given an extension field E=k, that an explicit basis of E over k can sometimes be constructed. We urge you to work though this example care-

fully; it will help make the preceding development much more concrete, and

you will see how all these ideas come together.

Let’s return to Exercise 3.56 on page 116 (if you haven’t attempted this

exercise, you should try it now). It involves � D cos.2�=7/ C i sin.2�=7/, a primitive 7th root of unity; note that the powers of � are the vertices of a regular 7-gon in the complex plane. Using Proposition 6.62 and the language we have

since introduced, we can now say that

irr.�;Q/ D ˆ7.x/ D x6 C x5 C x4 C x3 C x2 C 1;

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298 Chapter 7 Quotients, Fields, and Classical Problems

because ˆ7.x/ is irreducible (Theorem 6.68), and so ŒQ.�/ W Q D 6. For any nonnegative integer k, we have

� �k ��1

D �k D �7�k ;

by Theorem 3.32(ii). We defined ˛, ˇ, and in Exercise 3.56 on page 116 by

˛ D � C �6 D 2 cos.2�=7/ ˇ D �2 C �5 D 2 cos.4�=7/ D �3 C �4 D 2 cos.6�=7/;

and we saw that

˛ C ˇ C D �1 ˛ˇC ˛ C ˇ D �2

˛ˇ D 1:

It follows that ˛, ˇ, and are roots of

x3 C x2 � 2x � 1:

Ah, but this is precisely the irreducible p.x/ in Example 7.22. There, you

constructed a field in which p has a root, but you didn’t know what the roots

are. Now you know: they are ˛, ˇ, and , all real numbers, determined by

expressions involving cosines. Furthermore, the construction in Theorem 7.25 gives a field isomorphic to Q.˛/. But Q.˛/ contains all the roots of p; for

example,

˛2 D � � C �6

�2

D �2 C 2�7 C �12

D �2 C 2C �5

D ˇ C 2;

and, hence,

ˇ D ˛2 � 2 2 Q.˛/: In the same way, you can expand ˛3 to see that

D ˛3 C 3˛;

so that is also an element of Q.˛/.

Since ˛ D � C ��1, we see that Q.˛/ is a subfield of Q.�/. And, since ŒQ.˛/ W Q D deg p D 3;

Theorem 7.27 gives

This “tower of fields” is

sometimes illustrated with

a diagram that displays the

degrees:

Q.�/

2

Q.˛/

3

Q

ŒQ.�/ W Q.˛/ŒQ.˛/ W Q D ŒQ.�/ W Q:

Hence, ŒQ.�/ W Q.˛/ � 3 D 6, and ŒQ.�/ W Q.˛/ D 2. Therefore, the exten- sion Q.�/=Q decomposes into a cubic extension of Q followed by a quadratic

extension of Q.˛/. This implies that

deg � irr .�;Q.˛//

� D 2I

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7.2 Field Theory 299

that is, � is a root of a quadratic polynomial with coefficients in Q.˛/. Finding

this quadratic is deceptively easy. We have

� C �6 D ˛ and � �6 D �7 D 1:

Thus, � is a root of

x2 � ˛ x C 1 2 Q.˛/Œx:

There’s another tower, Q � E � Q.�/, with E=Q a quadratic extension and Q.�/=E a cubic extension (i.e., writing 6 D 2 � 3 rather than 6 D 3 � 2). We constructed ˛; ˇ; by breaking up the roots of ˆ7 into three sums, each

a pair of complex conjugates. Instead, let’s try to break the roots up into two

sums, say ı and �, each having three terms, and each sum containing just one

member of every conjugate pair f�j ; �j D �7�j g; further, we’d like both ıC � and ı� rational. A little experimenting (and a CAS) leads to defining them like

this:

ı D � C �2 C �4

� D �6 C �5 C �3:

Note that � C ı D �1, so that � 2 Q.ı/. We can now form the tower

Q � Q.ı/ � Q.�/:

You can also check that � ı D 2, so that � and ı are the roots of the quadratic The roots of this polynomial are, by the quadratic

formula, 1 2 ˙

p 7

2 . See

Exercise 7.35 on page 300.

polynomial in QŒx:

x2 C x C 2:

Hence, ŒQ.ı/ W Q D 2 and ŒQ.�/ W Q.ı/ D 3. We now have two ways to decompose the extension Q.�/=Q, which we draw in the following diagram:

Q.�/

3

2

Q.˛/

3 Q.ı/

2

Q

Finally, � must be a root of a cubic polynomial with coefficients in Q.ı/.

Again, the calculations are deceptively simple:

� C �2 C �4 D ı; � �2 C � �4 C �2 �4 D �3 C �5 C �6 D �;

��2�4 D �7 D 1;

so that �, �2, and �4 are roots of

x3 � ıx2 C �x � 1 2 Q.ı/Œx: N

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300 Chapter 7 Quotients, Fields, and Classical Problems

Exercises

7.31 As usual, let �n D cos.2�=n/ C i sin.2�=n/. (i) Find the minimal polynomial of �n over Q for all n between 1 and 10.

(ii) What is minimal polynomial of �p over Q if p is prime?

(iii) What is minimal polynomial of �p2 over Q if p is prime?

7.32 If p is a prime, and �p D cos.2�=p/ C i sin.2�=p/, show that � Q.�p/ W Q

� D p � 1:

7.33 Show that x2 � 3 is irreducible in Q �p

2 � Œx.

7.34 Let k � K � E be fields. Prove that if E is a finite extension of k, then E is a finite extension of K , and K is a finite extension of k.

7.35 Show that

(i) cos.2�=7/ C cos.4�=7/ C cos.8�=7/ D �1 2

.

(ii) �

sin.2�=7/ C sin.4�=7/ C sin.8�=7/ �2 D 72 .

7.36 Let k � F � K be a tower of fields, and let z 2 K . Prove that if k.z/=k is finite, then ŒF.z/ W F  � Œk.z/ W k. Conclude that ŒF.z/ W F  is finite.

Hint: Use Proposition 7.20 to obtain an irreducible polynomial p.x/ 2 kŒx; the polynomial p may factor in KŒx.

7.37 Let K=k be an extension field. If A � K and u 2 k.A/, prove that there are only finitely many a1; : : : ; an 2 A with u 2 k.a1; : : : ; an/.

7.38 Let E=k be a field extension. If v 2 E is algebraic over k, prove that v�1 is algebraic over k.

Splitting Fields

We now prove a result of Kronecker that says that if f .x/ 2 kŒx is not con- stant, where k is a field, then there is some extension field K=k containing all

the roots of f .

Theorem 7.29 (Kronecker). If k is a field and f .x/ 2 kŒx is nonconstant, there exists an extension field K=k with f a product of linear polynomials

in KŒx.

Proof. The proof is by induction on deg.f / � 1. If deg.f / D 1, then f is linear and we can take K D k. If deg.f / > 1, write f D pg in kŒx, where p.x/ is irreducible. Now Proposition 7.20 provides an extension field F=k

containing a root z of p. Hence, p D .x � z/h, and so f D pg D .x � z/hg in F Œx. By induction (since deg.hg/ < deg.f /), there is an extension field

K=F (so thatK=k is also an extension field) with hg, and hence f , a product

of linear factors in KŒx.

How to Think About It. For the familiar fields Q, R, and C, Kronecker’s Theorem offers nothing new. The Fundamental Theorem of Algebra says that

every nonconstant f .x/ 2 CŒx has a root in C; it follows, by induction on the degree of f , that all the roots of f lie in C; that is, f .x/ D a.x � z1/ � � � .x � zn/, where a 2 C and zj 2 C for all j . On the other hand, if k D Fp or

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7.2 Field Theory 301

k D C.x/ D Frac.CŒx/, the Fundamental Theorem does not apply. However, Kronecker’s Theorem does apply to tell us, for any f .x/ 2 kŒx, that there is always some larger field K containing all the roots of f ; for example, there is an extension field K=C.x/ containing

p x, and there is an extension field

E=F3 containing the roots of x 2 � x � 1 2 F3Œx.

A field F is called algebraically closed if every nonconstant polynomial

f .x/ 2 F Œx has a root in F (for example, C is algebraically closed). In contrast, extension fields K=k constructed in Kronecker’s Theorem (that give roots of only one polynomial at a time) are usually not algebraically closed.

Every field k does have an algebraic closure: there is an algebraic extension

F=k that is algebraically closed (Kronecker’s Theorem is one ingredient of the

proof; see [25], p. 328).

The extension field K=k in Kronecker’s Theorem need not be unique. In-

deed, if f is a product of linear factors in K, then it is so in any extension of

K. Therefore, let’s consider the “smallest” field in which f is a product of lin- ear factors. But the lack of uniqueness is not necessarily a consequence of K

being too large, as we shall see in Example 7.33.

Definition. If K=k is an extension field and f .x/ 2 kŒx is nonconstant, then f splits over K if f .x/ D a.x � z1/ � � � .x � zn/, where z1; : : : ; zn are in K and a 2 k.

An extension fieldE=k is called a splitting field of f over k if f splits over

E , but f does not split over any extension field F=k such that k � F ¨ E .

Consider f .x/ D x2 C 1 2 QŒx. The roots of f are ˙i , and so f splits over C; that is, f .x/ D .x � i/.x C i/ is a product of linear polynomials in CŒx. However, C is not a splitting field of f over Q: there are proper subfields

of C containing Q and all the roots of f . For example, Q.i/ is such a subfield;

in fact, it is the splitting field of f over Q.

A splitting field of a polynomial g.x/ 2 kŒx depends on k as well as on g. A splitting field of x2C 1 over Q is Q.i/, while a splitting field of x2C 1 over R is R.i/ D C.

Corollary 7.30. If k is a field and f .x/ 2 kŒx, then a splitting field of f over k exists.

Proof. By Kronecker’s Theorem, there is an extension field K=k such that

f splits in KŒx; say f .x/ D a.x � z1/ � � � .x � zn/. The subfield E D k.z1; : : : ; zn/ ofK is a splitting field of f over k, because a proper subfield of

E must omit some zi .

Example 7.31. (i) Let f .x/ D xn � 1 2 QŒx, and let E=Q be a splitting field. If � D e2�i=n is a primitive nth root of unity, then Q.�/ D E is a splitting field of f , for every nth root of unity is a power of �, and

�j 2 Q.�/ for all j . (ii) There are n distinct nth roots of unity in C, but there may be fewer roots

of unity over fields of characteristic p. For example, let f .x/ D x3 � 1 2 Z3Œx. Since x

3 � 1 D .x � 1/3, by Exercise 7.27 on page 293, we see that there is only one cube root of unity here. N

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302 Chapter 7 Quotients, Fields, and Classical Problems

How to Think About It. When we defined the field k.A/ obtained from a

field k by adjoining a set A, we assumed that A � K for some extension field K=k. But suppose no larger field K is given at the outset. For example, can

the roots of f .x/ D x2 � x � 1 2 F3Œx be adjoined to F3? Yes. In light of Kronecker’s Theorem, there is some field extension K=F3 containing the

roots of f , say ˛; ˇ; now we do have the larger field, and so F3.˛; ˇ/ makes sense; we can adjoin the roots of f to F3. Such an extension field K may not

be unique, but we shall see that any two of them are isomorphic.

Example 7.32. Let’s return to Example 7.24, where we saw that

z D p 2C

p 3

is a root of h.x/ D x4 � 10x2 C 1, and that Q. p 2 C

p 3/ can be realized

as a quotient QŒx=.p/ for some irreducible monic polynomial p.x/ 2 QŒx having z as a root.

As promised, we can now do better: we can show that p D h and that E D Q

�p 2 C

p 3 �

is a splitting field of h.x/ D x4 � 10x2 C 1, as well as a splitting field of g.x/ D .x2 � 2/.x2 � 3/.

Because x4 � 10x2 C 1 is a quadratic in x2, we can apply the quadratic formula to see that if w is any root of h, then w2 D 5˙ 2

p 6. But the identityNote that x2 D w,

where w is a root of

w2 � 10w C 1; that is, w D 1

2

� 10˙

p 96 � D

5˙ 2 p

6.

�p aC

p b �2 D aC2

p abCb givesw D ˙.

p 2C

p 3/. Similarly, 5�2

p 6 D�p

2 � p 3 �2

, so that h has distinct roots, namely

z D p 2C

p 3; �

p 2�

p 3;

p 2 �

p 3; �

p 2C

p 3:

By Theorem 6.52, the only possible rational roots of h are ˙1, and so we have just proved that all these roots are irrational.

We claim that h is irreducible in QŒx (so, p D h after all). It suffices to show that h has no quadratic factor q.x/ 2 QŒx (why?). If, on the contrary, h D qq0 for two monic quadratic polynomials in QŒx, then the roots of h are paired up, two for q and two for q0. Suppose q.z/ D 0. Then the other root of q, call it z0, is one of

p 2�

p 3; �

p 2 �

p 3; �

p 2C

p 3:

Now, if q.x/ D x2CbxCc, then �b D zCz0 and c D zz0. But you can check, for each choice of z0, that either zC z0 or zz0 is irrational. Since q 2 QŒx, this is a contradiction, and so h is irreducible.

F

�2E

4 Q. p

2/

2

Q

We now know that ŒE W Q D 4. Let F D Q. p 2;

p 3/, so that we have a

tower of fields Q � E � F . Theorem 7.27 tells us that

ŒF W Q D ŒF W EŒE W Q:

On the other hand,

ŒF W Q D ŒF W Q �p 2 � ŒQ

�p 2 �

W Q:

Now ŒQ �p 2 �

W Q D 2, because p 2 is a root of the irreducible quadratic

x2 � 2 in QŒx. We claim that ŒF W Q �p 2 �  � 2. The field F arises by ad-

joining p 3 to Q

�p 2 � ; either

p 3 2 Q

�p 2 � , in which case the degree is 1,

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7.2 Field Theory 303

or x2 � 3 is irreducible in Q �p 2 � Œx, in which case the degree is 2 (by Exer-

cise 7.33 on page 300, it is 2). It follows that ŒF W Q � 4, and so the equation ŒF W Q D ŒF W EŒE W Q gives ŒF W E D 1; that is, F D E , so that F not only arises from Q by adjoining all the roots of h, but it also arises from Q by

adjoining all the roots of g.x/ D .x2 � 2/.x2 � 3/. N

A splitting field of f .x/ 2 kŒx is a smallest field extensionE=k containing all the roots of f . We say “a” splitting field instead of “the” splitting field

because splitting fields of f over k are not unique. Corollary 7.30 constructed

a splitting field of f .x/ 2 kŒx as a subfield of a field extension K=k, where f splits in K. But there may be distinct such field extensions K=k.

Example 7.33. Consider f .x/ D x2 C x C 1 2 F2Œx. Now f is irreducible (for it is a quadratic with no roots in F2), and f is a product of linear poly- nomials in K D F2Œx=.f /, by Proposition 7.20: if z D x C .f / 2 K, then f .x/ D .x � z/2 in KŒx (remember that �1 D 1 here). On the other hand, in Exercise 4.55 on page 165, we constructed a field K 0 with elements the four matrices

� a b b aCb

� (where a; b 2 F2) and operations matrix addition

and matrix multiplication. You can check that if u D � 1 1 1 0

� , then u 2 K 0 and

f .x/ D .x � u/2 in K 0Œx. N

Our next goal is to show that splitting fields are unique up to isomorphism. We paraphrase Theorem 7.25(ii).

Let K=k be an extension field, and let z; z0 2 K be roots of some ir- reducible p.x/ 2 kŒx. Then there is an isomorphism � W k.z/ ! k.z0/ with �.z/ D z0 and �.c/ D c for all c 2 k.

We need a generalization. Suppose that f .x/ 2 kŒx is a polynomial, not nec- essarily irreducible, and let E D k.z1; : : : ; zt/ and E 0 D k.z01; : : : ; z0t/ be splitting fields of f . Is there an isomorphism � WE ! E 0 that carries the roots z1; : : : ; zt to the roots z

0 1; : : : ; z

0 t and that fixes all the elements c 2 k? The

obvious way to proceed is by induction on deg.f / (making use of the fact that

f has an irreducible factor, which will let us use Theorem 7.25). Think about proving the inductive step. We’ll have an isomorphism 'W k.z1/ ! k.z01/ that we’ll want to extend to an isomorphisme'W k.z1 ; z2/ ! k.z01; z0j / for some j ; that is, e'W k.z1/.z2/ ! k.z01/.z0j /. The base fields k.z1/ and k.z01/ are no longer equal; they are only isomorphic. The upshot is that we have to compli-

cate the statement of what we are going to prove in order to take account of this.

First, recall Corollary 5.22:

If R and S are commutative rings and 'WR ! S is a homomorphism, then there is a unique homomorphism '�WRŒx ! SŒx given by

'�W r0 C r1x C r2x2 C � � � 7! '.r0/C '.r1/x C '.r2/x2 C � � � :

Moreover, '� is an isomorphism if ' is.

As we said, we are forced to complicate our earlier result.

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304 Chapter 7 Quotients, Fields, and Classical Problems

Lemma 7.34. Let 'W k ! k0 be an isomorphism of fields and '�W kŒx ! k0Œx the isomorphism of Corollary 5.22; let p.x/ 2 kŒx be irreducible, and let p0 D '�.p/.

(i) p0 is irreducible in k0Œx, and the mapˆW kŒx=.p/ ! k0Œx=.p0/, defined by ˆWf C .p/ 7! '�.f /C .p0/; is an isomorphism of fields.

(ii) Let K=k be a field extension, let z 2 K be algebraic over k, and let p.x/ D irr.z; k/. If p0 D '�.p/ 2 k0Œx and z0 is a root of p0 in some extension of k0, then ' can be extended to an isomorphism k.z/ ! k0.z0/ that maps z to z0.

Proof. (i) This is straightforward, for ' carries the ideal .p/ in kŒx onto the ideal .p0/ in k0Œx, and Exercise 7.8 on page 285 applies. Exercise 7.44 on page 308 asks you to give the details.

(ii) By (i), there are isomorphisms

W kŒx=.p/ ! k.z/ and 0 W k0Œx=.p0/ ! k0.z0/:

By Lemma 7.34, there is an isomorphism

ˆ W kŒx=.p/ ! k0.x/=.p0/;

and the composite 0 ıˆ ı �1 is the desired isomorphism.

Here is a picture of the Lemma’s proof.

kŒx=.p/

ˆ

k.z/

0ıˆı �1

k0Œx=.p0/ 0

k0.z0/

We now give the version we need.

Theorem 7.35. Let 'W k ! k0 be an isomorphism of fields and '�W kŒx ! k0Œx the isomorphism kŒx=.p/ ! k0Œx=.p0/ in Lemma 7.34. Let f .x/ 2 kŒx and f �.x/ D '�.f / 2 k0Œx. If E is a splitting field of f over k and E 0 is a splitting field of f � over k0, then there is an isomorphismˆWE ! E 0 extending '.

Proof. The proof is by induction on d D ŒE W k. If d D 1, then f is a product of linear polynomials in kŒx, and it follows easily that f � is also a product of linear polynomials in k0Œx. Therefore, E 0 D k0, and we may set ˆ D '.

For the inductive step, choose a root z of f in E that is not in k, and let

p.x/ D irr.z; k/ be the minimal polynomial of z over k. Now deg.p/ > 1, because z … k; moreover, Œk.z/ W k D deg.p/, by Proposition 7.20. Let z0 be a root of p� inE 0, so that p� D irr.z0; k0/.

By Lemma 7.34(ii), there is an isomorphisme'W k.z/ ! k0.z0/ extending ' with e'.z/ D z0. We may regard f as a polynomial with coefficients in k.z/, for k � k.z/ implies kŒx � k.z/Œx. We claim that E is a splitting field of f over k.z/; that is,

E D k.z/.z1 ; : : : ; zn/;

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7.2 Field Theory 305

where z1; : : : ; zn are the roots of f .x/=.x � z/. After all,

E D k.z; z1; : : : ; zn/ D k.z/.z1 ; : : : ; zn/:

Similarly,E 0 is a splitting field of f � over k0.z0/. But ŒE W k.z/ < ŒE W k, by Theorem 7.27, so the inductive hypothesis gives an isomorphism ˆWE ! E 0 that extendse' and, hence, '.

Corollary 7.36. If k is a field and f .x/ 2 kŒx, then any two splitting fields of f over k are isomorphic via an isomorphism that fixes k pointwise.

Proof. Let E andE 0 be splitting fields of f .x/ over k. If ' is the identity, then Theorem 7.35 applies at once.

Classification of Finite Fields

We know, thanks to Proposition 7.18, that every finite field has pn elements for

some prime p and integer n � 1. We also know that a field k with pn elements must have characteristic p, so that pa D 0 for all a 2 k, by Proposition 7.17. In this section, we show that fields with exactly pn elements exist, and that any two having the same number of elements are isomorphic.

First, we show that every nonzero element a in a finite field with q elements

is a .q � 1/st root of unity (of course, a is not a complex root of unity). We have seen the idea of the next proof in the proof of Theorem 4.63.

Lemma 7.37. Let k be a finite field having q elements. If a 2 k is nonzero, then aq�1 D 1.

Proof. Let k# D fa1; a2; : : : ; aq�1g be the nonzero elements of k. We claim, for any a 2 k#, that the function �aW ai 7! aai takes values in k#: since k is a field, it is a domain, and so aai ¤ 0. We now claim �aW k# ! k# is injective: if aai D aaj , then the cancellation law gives ai D aj . Finally, since k# is finite, the Pigeonhole Principle shows that �a is a bijection. It follows that

aa1; aa2; : : : ; aaq�1 is just a rearrangement of a1; a2; : : : ; aq�1. Hence,

a1a2 � � �aq�1 D .aa1/.aa2/ � � � .aaq�1/ D aq�1a1a2 � � �aq�1:

Now cancel a1a2 � � �aq�1 to obtain 1 D aq�1.

We now show, given a prime power q D pn, that there exists a field with pn elements. Our guess is that Galois realized that C can be constructed by

adjoining a root of x2 C 1 to R, so that it was natural for him (but not for any of his contemporaries!) to adjoin a root of a polynomial to Fp. However,

Kronecker’s Theorem was not proved until a half century after Galois’s death.

Theorem 7.38 (Galois). If p is prime and n is a positive integer, then there

exists a field having exactly pn elements.

Proof. Write q D pn. In light of Lemma 7.37, it is natural to consider roots of the polynomial

g.x/ D xq � x 2 FpŒx:

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306 Chapter 7 Quotients, Fields, and Classical Problems

By Kronecker’s Theorem, there is a field extension K=Fp with g a product of

linear factors in KŒx. Define

E D fz 2 K W g.z/ D 0gI

that is, E is the set of all the roots of g. We claim that all the roots of g are

distinct. Since the derivative g0.x/ D qxq�1 � 1 D pnxq�1 � 1 D �1 (by Proposition 7.17), we have gcd.g; g0/ D 1. By Exercise 6.40 on page 263, all the roots of g are, indeed, distinct; that is, E has exactly q D pn elements.

The theorem will follow if E is a subfield of K. Of course, 1 2 E . If a, b 2 E , then aq D a and bq D b. Hence, .ab/q D aqbq D ab, and ab 2 E . By Proposition 7.17, .a C b/q D aq C bq D a C b, so that a C b 2 E . Therefore, E is a subring of K. Finally, if a ¤ 0, then Lemma 7.37 says that aq�1 D 1, and so the inverse of a is aq�2 (which lies inE because E is closed under multiplication).

Proposition 7.39. If k is a finite field having q D pn elements, then every a 2 k is a root of xq � x.

Proof. This follows directly from Lemma 7.37.

It is remarkable that the next theorem was not proved until the 1890s, 60

years after Galois discovered finite fields.E. H. Moore was an algebraist who later did

research in geometry and

foundations of analysis. Corollary 7.40 (Moore). Any two finite fields having exactly pn elements are

isomorphic.

Proof. By Proposition 7.39, every element of E is a root of g.x/ D xq � x 2 FpŒx, and so E is a splitting field of g over Fp.

Finite fields are often called Galois fields in honor of their discoverer. In

light of Corollary 7.40, we may speak of the field with q elements, where

q D pn is a power of a prime p, and we denote it by

Fq:

The next example displays different finite fields with the same number of

elements; by Moore’s Theorem, they are isomorphic.

Example 7.41. (i) In Exercise 4.55 on page 165, we constructed the field F4 with four elements:

F4 D ˚� a b b aCb

� W a; b 2 F2

:

On the other hand, since f .x/ D x2 C x C 1 2 F2Œx is irreducible, the quotientK D F2Œx=.f / is a field. By Proposition 7.20, F consists of all a C bz, where z D x C .f / is a root of f and a; b 2 F2. Hence K also is a field with four elements.

(ii) According to the table in Example 6.57, there are three monic irreducible quadratics in F3Œx, namely

p.x/ D x2 C 1; q.x/ D x2 C x � 1; and r.x/ D x2 � x � 1I

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7.2 Field Theory 307

each gives rise to a field with 9 D 32 elements, namely quotient rings of F3Œx. Let us look at the first two in more detail. Proposition 7.20 says

that E D F3Œx=.p/ is given by

E D faC bz W where z2 C 1 D 0g:

Similarly, if F D F3Œx=.q/, then

F D fa C bu W where u2 C u � 1 D 0g:

Without Moore’s Theorem, it is not instantly obvious that the two fields

are isomorphic. You can check that the map 'WE ! F (found by trial and error), defined by '.a C bz/ D aC b.1 � u/, is an isomorphism.

Now F3Œx=.x 2 � x � 1/ is another field with nine elements; Exer-

cise 7.46 asks for an explicit isomorphism with E .

(iii) In Example 6.57, we exhibited eight monic irreducible cubics p.x/ 2 F3Œx; each gives rise to a field F3Œx=.p/ having 27 D 33 elements, and Moore’s Theorem says that they are all isomorphic to one another. N

The following result is known.

Theorem 7.42 (Primitive Element). Let K=k be a finite field extensionI that is, ŒK W k < 1. If either k has characteristic 0 orK is a finite field, then there exists ˛ 2 K such thatK D k.˛/.

Proof. [26], p. 301.

Actually, more is known whenK is finite: it can be shown that every nonzero

element ofK is a power of ˛ (not merely a linear combination of powers of ˛).

Corollary 7.43. For every integer n � 1, there exists an irreducible polyno- mial in FpŒx of degree n.

Proof. Let h.x/ D irr.˛;Fp/ be the minimal polynomial of ˛. Since h is ir- reducible, Corollary 7.21 gives dimk.K/ D deg.h/. But if jKj D pn, then dimk.K/ D n. Therefore, since there exists a finite field with exactly pn ele- ments, there exists an irreducible polynomial of degree n.

Exercises

7.39 Let f .x/; g.x/ 2 kŒx be monic polynomials, where k is a field. If g is irreducible and every root of f (in an appropriate splitting field) is also a root of g, prove that

f D gm for some integer m � 1.

Hint: Use induction on deg.h/.

7.40 Determine whether any of the following pairs of fields are isomorphic.

(i) Q.i / and Q.12 .1 C i // (ii) Q.i / and Q.

p 3/

(iii) Q. p

2/ and Q. p

3/

(iv) Q. p

2/ and Q. p

6/

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308 Chapter 7 Quotients, Fields, and Classical Problems

7.41 Let f .x/ D s0 C s1x C � � � C sn�1xn�1 C xn 2 kŒx, where k is a field, and suppose that f .x/ D .x�z1/.x�z2/ � � � .x�zn/, where the zi lie in some splitting field. Prove that sn�1 D �.z1 C z2 C � � � C zn/ and s0 D .�1/nz1z2 � � � zn. Conclude that the sum and product of all the roots of f lie in k.

7.42 (i) Show that ŒQ.cos.2�=7// W Q D 3. (ii) Find the minimal polynomial for cos 2�7 over Q.

(iii) Find all the roots of this polynomial.

Hint: See Exercise 3.56 on page 116.

7.43 Suppose that K=k is an extension field and if A � K . Show that k.A/ is the smallest subfield of K containing k and A; that is, if E is any subfield of K

containing k and A, then k.A/ � E .

7.44 Prove Lemma 7.34.

7.45 Using the setup from Example 7.41(i), show that the map ' W F4 ! K , defined by '

�h a b b aCb

i� D a C bz, is an isomorphism.

7.46 Using the setup from Example 7.41(ii), show that

F3=.x 2 C 1/ Š F3Œx=.x2 � x � 1/

without using Corollary 7.40.

7.47 Prove that F3Œx=.x 3 � x2 C 1/ Š F3Œx=.x3 � x2 C x C 1/ without using

Corollary 7.40.

7.48 Write addition and multiplication tables for the field F8 with eight elements using

an irreducible cubic over F2.

7.49 (i) Is F4 isomorphic to a subfield of F8?

(ii) For a prime p, prove that if Fpn is isomorphic to a subfield of Fpm , then n j m (the converse is also true).

Hint: View Fpm as a vector space over Fpn .

7.3 Connections:

Ruler–Compass Constructions

There are myths in several ancient civilizations in which the gods demand

precise solutions of mathematical problems in return for granting relief from catastrophes. We quote from van der Waerden [35].

In the dialogue Platonikos of Eratosthenes, a story was told about the

problem of doubling the cube. According to this story, as Theon of

Smyrna recounts it in his book Exposition of mathematical things use- ful for the reading of Plato, the Delians asked for an oracle in order to

be liberated from a plague. The god (Apollo) answered through the ora-

cle that they had to construct an altar twice as large as the existing one

without changing its shape. The Delians sent a delegation to Plato, who

referred them to the mathematicians Eudoxus and Helikon of Kyzikos.

The altar was cubical in shape, and so the problem involves constructing 3 p 2

(the volume of a cube with edges of length ` is `3). The gods were cruel,

for although there is a geometric construction of p 2 (it is the length of the

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7.3 Connections: Ruler–Compass Constructions 309

diagonal of a square with sides of length 1), we are going to prove that it is

impossible to construct 3 p 2 by the methods of Euclidean geometry — that is,

by using only ruler and compass. (Actually, the gods were not so cruel, for the Greeks did use other methods. Thus, Menaechmus constructed

3 p 2 as the

intersection of the parabolas y2 D 2x and x2 D y; this is elementary for us, but it was an ingenious feat when there was no analytic geometry and no

algebra. There was also a solution found by Nicomedes.)

There are several other geometric problems handed down from the Greeks. Can one trisect every angle? Can one construct a regular n-gon? More pre-

cisely, can one inscribe a regular n-gon in the unit circle? Can one “square the

circle;” that is, can one construct a square whose area is equal to the area of

a given circle? Since the disk with radius 1 has area � , can one construct a

square with sides of length p �?

If we are not careful, some of these problems appear ridiculously easy. For

example, a 60ı angle can be trisected using a protractor: just find 20ı and draw the angle. Thus, it is essential to state the problems carefully and to agree on

certain ground rules. The Greek problems specify that only two tools, ruler

and compass, are allowed, and each must be used in only one way. The goal

of this section is to determine exactly what can be constructed using the two “Euclidean tools.” The answer will involve some surprising applications of

ideas from this chapter.

How to Think About It. In many geometry classes, constructions are now

taught using dynamic geometry software. These environments can be used in

the same way that one uses physical rulers and compasses; the principles are

the same, and what’s possible in them is what’s possible with pencil and paper. This brings up an important point. Constructions made in dynamic geometry

environments are likely to be more accurate than those carried out with pencil

and paper, but the goal here is not approximation—we are not content with

constructing 3 p 2 correct to 100 decimal places; the goal is to find

3 p 2 exactly,

just as we can find p 2 exactly as the length of the diagonal of the unit square.

We now seek to determine just what constructions are possible, and so we must use precise definitions.

Notation. Let P andQ be points in the plane; we denote the line segment with endpoints P and Q by PQ, and we denote its length by PQ. If P and Q are

In many high school texts,

L.P; Q/ is written as ! PQ. Of course, we can’t

physically draw the infinite

line L.P; Q/, but PQ has

endpoints and L.P; Q/

does not.

points, we’ll let L.P;Q/ denote the line through P and Q, and C.P;Q/ the

circle with center P and radius PQ. We’ll also denote the circle with center P

and radius r (for a positive number r ) by C.P; r/.

The formal discussion begins with defining the tools by saying exactly what

each is allowed to do.

Definition. A ruler is a tool that can be used to draw the line L.P;Q/ deter-

mined by points P and Q.

A compass is a tool that can be used to draw circles; given two points P and Q, it can draw C.P;Q/ and C.Q; P /.

What we are calling a ruler, others call a straightedge. For them, a ruler

can be used not only to draw lines but to measure distances as well.

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310 Chapter 7 Quotients, Fields, and Classical Problems

How to Think About It. Just to show you how fussy we are, let us point

out a subtlety about what a compass cannot do. Suppose we are given three

points:P ,Q, andR. We are allowed to draw the circle C.P;Q/ with center P

and radius r D PQ. But we are not allowed to draw the circle C.R; r/ with center R and radius r . Reason: a compass is allowed to draw a circle only

if two points are given at the outset; but the circle C.R; r/ cannot be drawn (using the compass as in the definition) because only one point, namely R, is

given at the outset. Our compass is called a collapsible compass as compared

to the more versatile compass that’s allowed to draw C.R; r/. We mention this

now only because the proof of Theorem 7.48(ii) may appear more complicated

than necessary (we’ll say something more there).

Constructions with ruler and compass are carried out in the plane. Since ev-

ery construction has only a finite number of steps, we shall be able to define

constructible points inductively. Once this precise definition is given, we will

be able to show that it is impossible to double the cube or to trisect arbitrary angles using only a ruler and compass. Angles such as 90ı and 45ı can be trisected using a ruler and compass (for we can construct a 30ı angle, whichAbout 425 BCE, Hippias

of Elis was able to square

the circle by drawing a

certain curve as well as

lines and circles. We shall

see that this construction is

impossible using only ruler

and compass.

can then be bisected), but we shall see that a 60ı angle is impossible to tri- sect. When we say impossible, we mean what we say; we do not mean that it is

merely very difficult. You should ponder how anything can be proved to be im- possible. This is an important idea, and we recommend letting students spend

an evening trying to trisect a 60ı angle by themselves as one step in teaching them the difference between hard and impossible.

Given the plane, we establish a coordinate system by first choosing two dis-

tinct points, A and A0; call the line they determine the x-axis. Use a compass to draw the two circles C.A; A0/ and C.A0; A/ of radius AA0 with centers A and A0, respectively (see Figure 7.1). These two circles intersect in two points P1 and P2; the line L.P1; P2/ they determine is called the y-axis; it is the

perpendicular-bisector of AA0, and it intersects the x-axis in a pointO , called the origin. We define the distance OA to be 1. We have introduced coordi-

nates into the plane; of course, O D .0; 0/, A D .1; 0/, and A0 D .�1; 0/. Consider the point P1 in Figure 7.1. Now OAP1 is a right triangle with legs

OA and OP1. The hypotenuse AP1 has length 2 D AA0 (for this is the radius of C.A; A0/). Since OA D 1, the Pythagorean Theorem gives P1 D .0;

p 3/.

Similarly, P2 D .0;� p 3/.

Informally, we construct a new pointQ from old pointsE; F;G, andH by using the first pair E ¤ F to draw a line or circle, the second pair G ¤ H to draw a line or circle, and then obtainingQ as one of the points of intersection

O

P1

P2

AA¢

Figure 7.1. The first constructible points.

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7.3 Connections: Ruler–Compass Constructions 311

of the two lines, of the line and the circle, or of the two circles. More generally,

a point is called constructible if it is obtained fromA and A0 by a finite number of such steps.

Given a pair of constructible points, we do not assert that every point on the

line or the circle they determine is constructible. For example, we can draw the

x-axis L.A0; A/, but, as we’ll see, not every point on it is constructible.

We now begin the formal discussion. Our goal is Theorem 7.52 which gives an algebraic characterization of constructibility. Recall, given distinct points

P and Q in the plane, that L.P;Q/ is the line they determine and C.P;Q/ is

the circle with center P and radius PQ.

Definition. Let E ¤ F and G ¤ H be points in the plane. A point Q is constructible from E; F;G, and H if either

(i) Q 2 L.E; F /\ L.G;H/, where L.E; F / ¤ L.G;H/; (ii) Q 2 L.E; F /\ C.G;H/;

(iii) Q 2 C.E; F /\ C.G;H/, where C.E; F / ¤ C.G;H/.

A pointQ is constructible if Q D A, Q D A0, or there are points P1; : : : ; Pn with Q D Pn such that every point PjC1 (for 1 � j ) is constructible from pointsE; F;G;H in fA;A0; P1; : : : ; Pj g.

If L.E; F / ¤ L.G;H/ and L.E; F / is not parallel to L.G;H/, then L.E; F / \L.G;H/ is a single point (there are at most two points comprising any of these intersections).

We illustrate the formal definition of constructibility by showing that every

angle can be bisected with ruler and compass.

Lemma 7.44. (i) The perpendicular-bisector of a given line segment AB

can be drawn.

(ii) If A and B are constructible points, then the midpoint of AB is con-

structible.

(iii) If a pointP D .cos �; sin �/ is constructible, thenQ D .cos.�=2/; sin.�=2// is constructible.

Proof. (i) The construction is the same as in Figure 7.1. Here, there are two

points of intersection of the circles C.A; B/ and C.B; A/, say P1 and P2,

and L.P1; P2/ is the perpendicular-bisector of AB .

(ii) The midpoint is the intersection of AB and its perpendicular-bisector.

O q

Q

A = (1, 0)

P = (cos , sin )q q

X

Figure 7.2. Bisecting an angle.

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312 Chapter 7 Quotients, Fields, and Classical Problems

(iii) The point A D .1; 0/ is constructible. By (i), the perpendicular-bisector L.O;X/ of the chord PA can be drawn. The point Q lies in the inter-

section of L.O;X/ and the unit circle, and so Q is constructible. (See Figure 7.2.)

Here is the (tricky) constructible version of the Parallel Postulate.We know that the Parallel Postulate is not true in

non-Euclidean geometry.

What hidden hypotheses

of Euclidean geometry

are we using to make this

construction?

Lemma 7.45. If U; V; P are distinct constructible points with P … L.U; V /, then there is a constructible pointQ with L.P;Q/ parallel to L.U; V /.

Q P

U B V

L U,V( )

Figure 7.3. Parallel postulate.

Proof. The proof refers to Figure 7.3. Choose U so that L.P; U / is not per-

pendicular to L D L.U; V /. Thus, L is not tangent to the circle C.P; U /, and so C.P; U / meets L in another point, say B (of course, B is constructible).

Let Q 2 C.P; U / \ C.U; P /. Clearly, Q is constructible, and we claim that L.P;Q/ is parallel to L. Indeed, we claim that the quadrilateral PBUQ is

a rhombus and hence it is a parallelogram. Now PQ is a radius of C.P; U /,

PU is a radius of both C.P; U / and C.B; U /, and BU is a radius of both

C.B; U / and C.U; P /. Hence, PQ D PU D PB , as we want.

In high school geometry, the goal is to construct certain figures with ruler and compass. We are about to shift the focus, considering instead the notion

of constructible numbers. “Numbers?” Well, analytic geometry equips points

with coordinates, and we have seen how to regard points as complex numbers.

Definition. A complex number z D x C iy is constructible if the point .x; y/ is a constructible point.

Exercise 7.51 on page 326 shows that every element of ZŒi  is a con-

structible number. So is our old friend ! D �1 2

C i p 3 2

.

How to Think About It. We asked you earlier to contemplate how we could

prove that something is impossible. The basic strategy is an elaborate indirect

proof: assuming a certain point Y is constructible, we will reach a contra-

diction. The first step is essentially analytic geometry: replace points by their

coordinates, as we have just done by defining constructible complex numbers. The next step involves modern algebra; don’t just consider one constructible

number; consider the set K of all constructible numbers, for the totality of

them may have extra structure that we can exploit. In fact, we will see that K

is a subfield of C. Not only can we translate points to numbers, we can also

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7.3 Connections: Ruler–Compass Constructions 313

translate the definition of constructibility into algebra. If a point PnC1 is con- structible from (constructible) pointsP0; P1; : : : ; Pn, we shall see that its com-

plex brother PnC1 is algebraic over the subfield F D Q.P0; P1; : : : ; Pn/. Given that lines have linear equations and circles have quadratic equations,

it is not surprising that ŒF.PnC1/ W F  � 2. The ultimate criterion that a point Y be constructible is essentially that ŒQ.Y / W Q is a power of 2.

It follows, for example, that the classical geometric problem of duplicating

the cube corresponds to the algebraic problem of checking whether ŒQ. 3 p 2/ W

Q is a power of 2. Since this degree is 3, the assumption that we can duplicate

the cube leads to the contradiction in arithmetic that 3 D 2k for some integer k.

We continue our discussion of constructibility.

Lemma 7.46. A complex number z D aC ib is constructible if and only if its real and imaginary parts are constructible.

O

Q b= (0, ) P a b= ( , )

B b= ( , 0) A a= ( , 0)

Figure 7.4. Real and imaginary parts.

Proof. If z D aC ib is constructible, then construct lines throughP D .a; b/ parallel to each axis (Lemma 7.45). The intersection of the vertical line and the x-axis is A D .a; 0/, so that A is constructible, and hence a D aC 0i is a constructible real number. Similarly, the point Q D .0; b/, the intersection of the horizontal line and the y-axis, is constructible. It follows that B D .b; 0/ is constructible, for it is an intersection point of the x-axis and C.O;Q/. Hence,

b D b C 0i is a constructible real number. Conversely, assume that a and b are constructible real numbers; that is,

.a; 0/ and B D .b; 0/ are constructible points. The point Q D .0; b/ is con- structible, being the intersection of the y-axis and C.O; B/. By Lemma 7.45,

the vertical line through .a; 0/ and the horizontal line through .0; b/ can be

drawn, and .a; b/ is their intersection. Therefore, .a; b/ is a constructible point,

and so z D aC ib is a constructible number.

Definition. Denote the subset of C of all constructible numbers by K.

The next lemma allows us to focus on real constructible numbers.

Lemma 7.47. (i) If K \ R is a subfield of R, then K is a subfield of C. (ii) If K \ R is a subfield of R and if

p a 2 K whenever a 2 K \ R is

positive, then K is closed under square roots.

Proof. (i) If z D aC ib and w D c C id are constructible, then a; b; c; d 2 K\R, by Lemma 7.46. Hence, aCc; bCd 2 K\R, becauseK\R is a

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314 Chapter 7 Quotients, Fields, and Classical Problems

subfield, and so .aCc/Ci.bCd/ 2 K, by Lemma 7.46. Similarly, zw D .ac � bd/C i.ad C bc/ 2 K. If z ¤ 0, then z�1 D .a=zz/ � i.b=zz/. Now a; b 2 K \ R, by Lemma 7.46, so that zz D a2 C b2 2 K \ R, because K \ R is a subfield of C. Therefore, z�1 2 K.

(ii) If z D a C ib 2 K, then a; b 2 K \ R, by Lemma 7.46, and so r2 D a2 C b2 2 K \ R, as in part (i). Since r2 is nonnegative, the hypothesis gives r 2 K \ R and

p r 2 K \ R.

Now z D r .cos � C i sin �/, so that cos � C i sin � D r�1z 2 K, because K is a subfield of C by part (i). By Lemma 7.44, cos �

2 C i sin �

2

can be constructed, and hence is inK. But p z D

p r �

cos � 2

C i sin � 2

� 2

K, as desired.

Theorem 7.48. The set of all constructible numbers K is a subfield of C that

is closed under square roots and complex conjugation.

Proof. It suffices to prove that the properties of K \ R in Lemma 7.47 hold. Let a and b be constructible real numbers.

(i) �a is constructible. If P D .a; 0/ is a constructible point, then .�a; 0/ is the other inter-

section of the x-axis and C.O; P /.

(ii) aC b is constructible.

I b Q

1

a b

O P S

Figure 7.5. a C b.

Assume that a and b are positive. Let I D .0; 1/, P D .a; 0/, and Q D .b; 1/. NowQ is constructible: it is the intersection of the horizontalYou are tempted to use

a compass with center I

and radius b to draw

C.P; b/. But this is illegal.

Remember: we’re using a

collapsible compass that

requires two points given

at the outset; here, only

one is available, namely P .

line through I and the vertical line through .b; 0/, both of which can be

drawn by Lemma 7.45 (the latter point is constructible, by hypothesis).

The line throughQ parallel to IP intersects the x-axis in S D .aCb; 0/, as desired.

To construct b � a, let P D .�a; 0/ in Figure 7.5. Thus, a C b and �a C b are constructible; by part (i), �a � b and a � b are also con- structible. Thus, a C b is constructible, no matter whether a and b are both positive, both negative, or have opposite sign.

(iii) ab is constructible. By part (i), we may assume that both a and b are positive. In Fig-

ure 7.6, A D .1; 0/, B D .1 C a; 0/, and C D .0; b/. Define D to be the intersection of the y-axis and the line throughB parallel to AC . Since the

trianglesOAC and OBD are similar,

OB=OAD OD=OC I

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7.3 Connections: Ruler–Compass Constructions 315

D

C

A BO

1 a b

Figure 7.6. ab.

hence .a C 1/=1 D .b C CD/=b, and CD D ab. Therefore, b C ab is constructible. Since �b is constructible, by part (i), we have ab D .b C ab/ � b constructible, by part (ii).

(iv) If a ¤ 0, then a�1 is constructible.

AO

1

T

1

S

a

B

Figure 7.7. a�1.

Let A D .1; 0/, S D .0; a/, and T D .0; 1 C a/. Define B as the intersection of the x-axis and the line through T parallel to AS ; thus,

B D .1Cu; 0/ for some u. Similarity of the trianglesOSA andOTB gives

OT=OS D OB=OA:

Hence, .1 C a/=a D .1 C u/=1, and so u D a�1. Therefore, 1 C a�1 is constructible, and so .1C a�1/ � 1 D a�1 is constructible.

(v) If a � 0, then p a is constructible.

R

PQAO

a

Figure 7.8. p

a.

Let A D .1; 0/ and P D .1 C a; 0/; construct Q, the midpoint of OP (if U; V are constructible points, then the midpoint of the segment

UV is its intersection with the perpendicular-bisector, constructed as in

Figure 7.1). Define R as the intersection of the circle C.Q;O/ with the

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316 Chapter 7 Quotients, Fields, and Classical Problems

vertical line throughA. The (right) trianglesAOR and ARP are sim-

ilar, so that

OA=AR D AR=AP;

and hence AR D p a.

(vi) If z D aC ib 2 K; then z D a � ib is constructible.

By Lemma 7.47,K is a subfield of C. Now a; b 2 K, by Lemma 7.46, and i 2 K, as we saw on page 310. It follows that �bi 2 K, and so a � ib 2 K.

Corollary 7.49. If a, b, c are constructible, then the roots of the quadratic

ax2 C bx C c are constructible.

Proof. This follows from Theorem 7.48 and the quadratic formula.

We now consider subfields of C to enable us to prove an inductive step in

the upcoming theorem.

Lemma 7.50. Let F be a subfield of C containing i that is closed under

complex conjugation. Let z D a C ib; w D c C id 2 F , and let P D .a; b/ and Q D .c; d /.

(i) If aC ib 2 F , then a 2 F and b 2 F . (ii) If the equation of L.P;Q/ is y D mx C q, where m; q 2 R, then

m; q 2 F . (iii) If the equation ofC.P;Q/ is .x�a/2C.y�b/2 D r2, where a; b; r 2 R,

then r2 2 F .

Proof. (i) If z D aC ib 2 F , then a D 1 2 .zC z/ 2 F and ib D 1

2 .z � z/ 2

F ; since we are assuming that i 2 F , we have b D �i.ib/ 2 F . (ii) By (i), the numbers a; b; c; d lie in F . Hence, m D .d � b/=.c � a/ 2 F

and q D b �ma 2 F . (iii) The circle C.P;Q/ has equation .x � a/2 C .y � b/2 D r2, and r2 2 F

because r2 D .c � a/2 C .d � b/2.

As we said earlier, the next result is intuitively obvious, for the equation of

a line is linear and the equation of a circle is quadratic. However, the coming

proof involves some calculations.

Lemma 7.51. Let F be a subfield of C containing i and which is closed under

complex conjugation. Let P;Q;R; S be points whose coordinates lie in F ,

and let ˛ D uC iv 2 C. If either of the following is true,

˛ 2 L.P;Q/ \L.R; S/; where L.P;Q/ ¤ L.R; S/; ˛ 2 L.P;Q/ \ C.R; S/; ˛ 2 C.P;Q/ \ C.R; S/; where C.P;Q/ ¤ C.R; S/;

then ŒF.˛/ W F  � 2.

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7.3 Connections: Ruler–Compass Constructions 317

Proof. If L.P;Q/ is not vertical, then Lemma 7.50(ii) says that L.P;Q/ has

equation y D mxCb, wherem; b 2 F . IfL.P;Q/ is vertical, then its equation is x D b because P D .a; b/ 2 L.P;Q/, and so b 2 F , by Lemma 7.50(i). Similarly, L.R; S/ has equation y D nx C c or x D c, where m; b; n; c 2 F . Since these lines are not parallel, one can solve the pair of linear equations for

.u; v/, the coordinates of ˛ 2 L.P;Q/ \ L.R; S/, and they also lie in F . In this case, therefore, ŒF.˛/ W F  D 1. See Exercises 7.52

and 7.53 on page 326.Let L.P;Q/ have equation y D mx C b or x D b, and let C.R; S/ have equation .x � c/2 C .y � d/2 D r2; by Lemma 7.50, we have m; q; r2 2 F . Since ˛ D uC iv 2 L.P;Q/ \ C.R; S/,

r2 D .u � c/2 C .v � d/2

D .u � c/2 C .muC q � d/2;

so that u is a root of a quadratic polynomial with coefficients inF \ R. Hence, ŒF.u/ W F  � 2. Since v D muC q, we have v 2 F.u/, and, since i 2 F , we have ˛ 2 F.u/. Therefore, ˛ D uC iv 2 F.u/, and so ŒF.˛/ W F  � 2.

Let C.P;Q/ have equation .x�a/2C .y�b/2 D r2 and let C.R; S/ have equation .x � c/2 C .y�d/2 D s2. By Lemma 7.50, we have r2; s2 2 F \ R. Since ˛ 2 C.P;Q/ \ C.R; S/, there are equations

.u � a/2 C .v � b/2 D r2 and .u � c/2 C .v � d/2 D s2:

After expanding, both equations have the form u2 C v2 C something D 0. Setting the something’s equal gives an equation of the form tuC t 0vC t 00 D 0, where t; t 0; t 00 2 F . Coupling this with the equation of one of the circles returns us to the situation of the second paragraph.

Here is the criterion we have been seeking: an algebraic characterization of

a geometric idea; it is an exact translation from geometry into algebra.

Theorem 7.52. A complex number z is constructible if and only if there is a

tower of fields

Q D K0 � K1 � � � � � Kn � C;

where z 2 Kn and ŒKjC1 W Kj  � 2 for all j .

Proof. Let z D a C ib, and let P D .a; b/ be the corresponding point in the plane. If z is a constructible number, then P is a constructible point, and so

there is a sequence of points A;A0; P1; : : : ; Pn D P with each PjC1 obtain- able from fA;A0; P1; : : : ; Pj g; since i is constructible, we may assume that P1 D .0; 1/. Define

K1 D Q.z1/ and KjC1 D Kj .zjC1/;

where zj corresponds to the point Pj and there are pointsE; F;G;H lying in

fA;A0; P1; : : : ; Pj g with one of the following:

PjC1 2 L.E; F / \L.G;H/; PjC1 2 L.E; F / \ C.G;H/; PjC1 2 C.E; F /\ C.G;H/:

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318 Chapter 7 Quotients, Fields, and Classical Problems

We may assume, by induction on j � 1, that Kj is closed under complex conjugation, so that Lemma 7.51 applies to show that ŒKjC1 W Kj  � 2. Fi- nally, KjC1 is also closed under complex conjugation, for if zjC1 is a root of a quadratic f .x/ 2 Kj Œx, then zjC1 is the other root of f .

Conversely, given a tower of fields as in the statement, then Theorem 7.48

and Lemma 7.50 show that z is constructible.

Corollary 7.53. If a complex number z is constructible, then ŒQ.z/ W Q is a power of 2.

Proof. This follows from Theorems 7.52 and 7.27: If k � E � K are fields withE=k and K=E finite extension fields, then ŒK W k D ŒK W EŒE W k.

The converse of Corollary 7.53 is false; it can be shown that there are non-

constructible numbers z with ŒQ.z/ W Q D 4 (see [27], p. 136).

Corollary 7.54. (i) The real number cos .2�=7/ is not constructible.

(ii) The complex 7th root of unity �7 is not constructible.

Proof. (i) We saw in Example 7.28 that ŒQ .cos.2�=7// W Q D 3. (ii) �7 D cos.2�=7/C i sin.2�=7/.

We’ll soon have more to say about constructibility of roots of unity.

We can now deal with the Greek problems, two of which were solved by

Wantzel in 1837. The notion of dimension of a vector space was not known

at that time; in place of Theorem 7.52, Wantzel proved that if a number is constructible, then it is a root of an irreducible polynomial in QŒx of degree

2n for some n.

Theorem 7.55 (Wantzel). It is impossible to duplicate the cube using onlyNicomedes solved the Delian problem of doubling

the cube using a marked

ruler and compass.

ruler and compass.

Proof. The question is whether z D 3 p 2 is constructible. Since x3 � 2 is

irreducible, ŒQ.z/ W Q D 3, by Theorem 7.20; but 3 is not a power of 2.

Consider how ingenious this proof is. At the beginning of this section, you

were asked to ponder how we can prove impossibility. As we said when we outlined this argument, the constructibility of a point was translated into al-

gebra, and the existence of a geometric construction produces an arithmetic

contradiction. This is a spectacular use of the idea of modeling!

A student in one of our classes, imbued with the idea of continual progress

through technology, asked, “Will it ever be possible to duplicate the cube with ruler and compass?” Impossible here is used in its literal sense.

Theorem 7.56 (Wantzel). It is impossible to trisect a 60ı angle using only ruler and compass.

Proof. We may assume that one side of the angle is on the x-axis, and so

the question is whether z D cos.20ı/ C i sin.20ı/ is constructible. If z were constructible, then Lemma 7.46 would show that cos.20ı/ is constructible.

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7.3 Connections: Ruler–Compass Constructions 319

The triple angle formula on page 110 gives

cos.3˛/ D 4 cos3 ˛ � 3 cos˛:

Setting ˛ D 20ı, we have cos 3˛ D 1 2

, so that z D cos.20ı/ is a root of 4x3 � 3x � 1

2 ; equivalently, cos.20ı/ is a root of f .x/ D 8x3 � 6x � 1 2

ZŒx. A cubic is irreducible in QŒx if and only if it has no rational roots. By

Theorem 6.52, the only candidates for rational roots are ˙1, ˙ 1 2

, ˙ 1 4

, and

˙ 1 8

; since none of these is a root, as one easily checks, it follows that f is

irreducible. (Alternatively, we can prove irreducibility using Theorem 6.55,

for f .x/ D x3 C x � 1 is irreducible in Z7Œx.) Therefore, 3 D ŒQ.z/ W Q, by Theorem 7.20(ii), and so z D cos.20ı/ is not constructible because 3 is not a power of 2.

Theorem 7.57 (Lindemann). It is impossible to square the circle with ruler

and compass.

Proof. The problem is whether we can construct a square whose area is � ,

the area of the unit circle; If the side of the square has length z, we are asking

whether z D p � is constructible. Now Q.�/ is a subfield of Q.

p �/. We have

already mentioned that Lindemann proved that � is transcendental (over Q), so that ŒQ.�/ W Q is infinite. It follows from Corollary A.41 in Appendix A.3 that ŒQ.

p �/ W Q is also infinite. Thus, ŒQ.

p �/ W Q is surely not a power of

2, and so p � is not constructible.

Other construction tools

If a ruler is allowed not only to draw a line but to measure distance using

marks on it (as most of our rulers are used nowadays), then the added function

makes it a more powerful instrument. Both Nicomedes and Archimedes were

able to trisect arbitrary angles using a marked ruler and a compass; we present Archimedes’ proof here.

Theorem 7.58 (Archimedes). Every angle can be trisected using a marked

ruler and compass.

Proof. It is easy to construct D 30ı, D 60ı, and D 90ı. The trigonomet- ric Addition Formula shows that if z D cosˇC i sinˇ and z0 D cos C i sin can be found, so can zz0 D cos.ˇ C /C i sin.ˇ C /. Now if 3ˇ D ˛, then 3.ˇC 30ı/ D ˛C 90ı, 3.ˇC 60ı/ D ˛C 180ı, and 3.ˇC 90ı/ D ˛C 270ı.

A

E O

U V

X F

U V

U V

Figure 7.9. Sliding ruler.

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320 Chapter 7 Quotients, Fields, and Classical Problems

Thus, it suffices to trisect an acute angle ˛, for if ˛ D 3ˇ and ˇ can be found, then so can ˇ C 30ı, ˇ C 60ı, and ˇ C 90ı be found.

Draw the given angle ˛ D †AOE, where the originO is the center of the unit circle. Take a ruler on which the distance 1 has been marked; that is, there

are points U and V on the ruler with UV D 1. There is a chord through A parallel to L.E; F /; place the ruler so that the chord is AU . Since ˛ is acute,

U lies in the first quadrant. Keeping A on the sliding ruler, move the point U

down the circle; the ruler intersects the extended diameter L.E; F / in some point X with UX > 1. Continue moving U down the circle, keeping A on the

sliding ruler, until the ruler intersects L.E; F / in the point V .

A

E O

C F

B d

e

ga b

Figure 7.10. Trisecting ˛.

We claim that ˇ D †UVO D 1 3 ˛. Now

˛ D ı C ˇ;

because ˛ is an exterior angle of AOV , and hence it is the sum of the two opposite internal angles. Since OAU is isosceles .OA and OU are radii),

ı D ", and so

˛ D "C ˇ:

But " D C ˇ D 2ˇ, for it is an exterior angle of the isosceles triangle UVO ; therefore,

˛ D 2ˇC ˇ D 3ˇ:

In addition to investigating more powerful tools, one can look at what can

be accomplished with fewer tools. It was proved by Mohr in 1672 and, inde- pendently, by Mascheroni in 1797, that every geometric construction carried

out by ruler and compass can be done without the ruler. There is a short proof

of the theorem given by Hungerbühler in American Mathematical Monthly,

101 (1994), pp. 784–787.

Constructing Regular n-gons

High school geometry students are often asked to construct various regular

polygons. In light of our present discussion, we can phrase such problems more

carefully.

Which regular polygons can be inscribed in the unit circle using only

ruler and compass?

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7.3 Connections: Ruler–Compass Constructions 321

Because they can construct 90ı and 60ı angles, high school students can con- struct squares and hexagons (just make right and 60ı central angles), and they can connect every other vertex of their hexagon to inscribe an equilateral trian- gle. Also, by using the perpendicular-bisector construction, they can inscribe

a regular polygon with twice as many sides as an already constructed one, so

they can inscribe regular polygons with 3 � 2n and 4 � 2n sides for any positive integer n. Archimedes knew that � is the area of the unit circle, and he approx-

imated it by inscribing and circumscribing a regular 96-gon (he began with a regular hexagon and then doubled the number of sides four times).

This is about as far as most high school programs get, although some treat

polygons with 5 � 2n sides (using Exercise 3.48 to construct the decagon and We’ll revisit the construc- tion of the pentagon in just

a minute, putting it a more

general setting.

then connecting every other vertex). This is also about as far as Greek geome-

ters got, although they also were able to show (see Exercise 7.67) that if a regular m-gon and and a regular n-gon are inscribable in a circle (again, with

only ruler and compass), then so is a regular nm-gon; for example, a regular

15-gon can be inscribed. However, it was unknown whether all regular poly-

gons could be so inscribed.

About 2000 years later, around 1796, Gauss—still in his teens—essentially

invented the main results in this section, and he applied them to the problem of determining whether a regular polygon could be inscribed in a circle with ruler

and compass (he wrote that his main result on this problem led to his decision

to become a mathematician). We’ll develop his methods here.

Theorem 3.28 tells us that the vertices of a regular n-gon inscribed in the

unit circle can be realized in the complex plane as the set of roots to xn � 1, Given the development so far, you may already

see that the problem

can be translated to the

algebra of constructible

complex numbers, but

this was a huge leap

for mathematicians of

Gauss’s time and certainly

out of reach for Greek

geometers.

and that these roots are all powers of

�n D cos.2�=n/C i sin.2�=n/:

So, we can recast our question about inscribability and ask:

For which values of n is �n a constructible number?

Well, we can hit this question with Theorem 7.52:

Corollary 7.59. A regular n-gon can be inscribed in the unit circle with ruler

and compass if and only if there is a tower of fields

Q D K0 � K1 � � � � � Kn � C;

where �n D e2�i=n lies in Kn and ŒKjC1 W Kj  � 2 for all j .

Proof. Indeed, a regular n-gon can be so inscribed if and only if �n and, hence, all its powers, are constructible numbers.

Gauss showed how to construct such a tower when n D 17, and his method was general in principle, leading to a complete classification of inscribable

regular polygons. Before we state the main result, let’s work through two ex- amples as Gauss did (all laid out in detail by him in [14], Section VII).

Example 7.60. In Example 3.34 on page 113, we showed how to find explicit

formulas for the vertices of a regular pentagon inscribed in the unit circle.

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322 Chapter 7 Quotients, Fields, and Classical Problems

1

Figure 7.11. Unit 5-gon.

Let’s look at this from the perspective of this chapter. Write � instead of �5.

The nonreal roots of x5 � 1, namely �; �2; �3; �4; are the roots of the irre- ducible polynomial

ˆ5.x/ D x4 C x3 C x2 C x C 1:

It follows that

ŒQ.�/ W Q D 4;

so Corollary 7.53 tells us that there’s a chance that � is constructible. In Ex-

ample 3.34, without using this language, we actually constructed the tower of

quadratic extensions necessary to guarantee that � is, in fact, constructible. We

showed that if g D � C �4 D 2 cos.2�=5/ and h D �2 C �3 D cos.4�=5/, then g and h are roots of the quadratic equation x2 C x � 1, so that

ŒQ.� C �4/ W Q D 2:

By Theorem 7.27,

ŒQ.�/ W Q.� C �4/ŒQ.� C �4/ W Q D ŒQ.�/ W Q D 4;

so that Q.�/

2

Q.� C �4/

2

Q

ŒQ.�/ W Q.� C �4/ D 2;

and we have our tower of quadratic extensions:

Q � Q.� C �4/ � Q.�/ N

Gauss’s construction of the 17-gon

Stepping back a bit, we can describe what we did with the pentagon: the non-

real roots are

f�; �2; �3; �4g

(we are still writing � instead of �5). There are four roots. The first story of our

tower is Q.� C �4/, generated by the sums of pairs of the roots: � C �4 and �2 C �3. The top story, Q.�/, is generated by the individual roots themselves.

This is the basic idea behind Gauss’s insight into the 17-gon, but the situa-

tion here is more complicated. Change notation again; now let

� D �17 D cos.2�=17/C i sin.2�=17/:

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7.3 Connections: Ruler–Compass Constructions 323

Because the minimal polynomial of � over Q is

ˆ.x/ D x16 C x15 C � � � C x2 C x C 1;

we have ŒQ.�/ W Q D 16. There are sixteen roots of ˆ17: ˚ �k W 0 � k � 15

.

Together with 1, these points on the unit circle are the vertices of our regular

17-gon.

For each factorization 16 D ef , Gauss divided the roots into e sums of We’ll see what method Gauss used to partition the

roots into these sums in

just a minute.

f D 16=e roots each: �e;0; �e;1; : : : ; �e;e�1, where each �e;k is a sum of f roots. For example, he divided the sixteen roots into two sums of eight each, which we can call �2;0 and �2;1, as follows:

�2;0 D � C �9 C �13 C �15 C �16 C �8 C �4 C �2

�2;1 D �3 C �10 C �5 C �11 C �14 C �7 C �12 C �6:

There are also four sums of four each:

�4;0 D � C �13 C �16 C �4

�4;1 D �3 C �5 C �14 C �12

�4;2 D �9 C �15 C �8 C �2

�4;3 D �10 C �11 C �7 C �6:

And there are eight sums of two each: Each period of length > 1 is a real number; you can

check that if �k occurs in

�e;k , so does

�17�k D ��k D �k;

so that each period is a

sum of terms of form zCz, and hence a real number.

�8;0 D � C �16

�8;1 D �3 C �14

�8;2 D �9 C �8

�8;3 D �10 C �7

�8;4 D �13 C �4

�8;5 D �5 C �12

�8;6 D �15 C �2

�8;7 D �11 C �6:

Finally, there are sixteen “sums” of one each, namely

n �16;k D �k j 0 � k � 15

o :

Gauss called each of the �e;k a period of length f D 16=e. The plan is to show that the periods of length eight lie in a quadratic extension K1 of Q,

the periods of length four lie in a quadratic extension K2 of K1, and so on,

building a tower of quadratic extensions ending with Q.�/.

The calculations will sometimes be involved so, once again, pull out your If you use a CAS, you can perform all of these calcu-

lations in QŒx=.ˆ.x//. pencil or computer.

� Because ˆ.�/ D 0, we see that �2;0 C �2;1 D �1. � With a little patience and care (or a CAS), you can check that

�2;0�2;1 D 4.�2;0 C �2;1/ D �4:

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324 Chapter 7 Quotients, Fields, and Classical Problems

Hence �2;0 and �2;1 are roots of x 2 C x � 4, and so

ŒQ.�2;1/ W Q � 2:

The first step in our tower is Q � Q.�2;1/. Note that �2;0 D �4=�2;1, so that �2;1 2 Q.�2;0/. Next, we move up to the periods of length 4. You can check (Exercise 7.61 on page 327) that

�4;1 C �4;3 D �2;1 �4;1�4;3 D �1:

Hence, �4;1 and �4;3 are roots of x 2 � �2;1x � 1, and

You can also check that

�2 2;0 D �n;1 (Exercise 7.60

on page 327).

ŒQ.�4;1/ W Q.�2;1/ � 2:

Q.�4;1/=Q.�2;1/ is the second story in our tower.

Q � Q.�2;1/ � Q.�4;1/:

Note that �4;3 D �1=�4;1, so that �4;3 2 Q.�4;1/. Up one more story—the periods of length eight: you can check (Exer-

cise 7.62 on page 327) that

�8;1 C �8;5 D �4;1 �8;1�8;5 D �4;2:

So, �8;1 and �8;5 are roots of x 2 � �4;1x C �4;2. This says that

ŒQ.�8;1/ W Q.�4;1; �4;2/ � 2:

But, by Exercise 7.63 on page 327,Q.�17/

2

Q.�8;1/

2

Q.�4;1/

2

Q.�2;1/

2

Q

Q.�4;1; �4;2/ D Q.�4;1/;

so that

ŒQ.�8;1/ W Q.�4;1/ � 2:

Assemble what we have built:

Q � Q.�2;1/ � Q.�4;1/ � Q.�8;1/ � Q.�/:

The degree of each extension is at most 2; since

Q.�/ D 16;

all the degrees are equal to 2 (Theorem 7.27). Hence, we have constructed a tower of fields, each quadratic over the one below, starting with Q and ending

with Q.�/. We have proved that � is constructible.

Theorem 7.61. A regular 17-gon can be inscribed in the unit circle with ruler

and compass.

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7.3 Connections: Ruler–Compass Constructions 325

One detail that remains is to see what method Gauss used to assign different One of the reasons has already been mentioned:

each period should contain

a sum of terms, each

of form �k C ��k . This ensures that every story in

our tower except the last

is contained in R, so we

“save the complex step” for

the end (why is this a good

thing?).

powers of � to each period; how did Gauss decide which powers of � should

occur in each �e;k? The answer comes from Galois theory (a subject we only briefly touch on in Chapter 9). He employed an ingenious method using the

fact that 3 is a primitive element in F17; that is, every nonzero element in F17 is a power of 3.

k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

3k 1 3 9 10 13 5 15 11 16 14 8 7 4 12 2 6

Gauss used this special property of 3 to define the periods: if ef D 16, there are k periods of length f defined by

�e;k D f �1X

jD0 �3

kCje :

So, for example,

�8;4 D �3 4C0�8 C �34C1�8

D �81 C �531441

D �4 C �13:

Sufficiency of the following theorem, a feat the Greeks would have envied,

was discovered by Gauss around 1796. He claimed necessity as well, but none

of his published papers contains a complete proof of it. The first published

proof of necessity is due to Wantzel, in 1837.

Theorem 7.62 (Gauss-Wantzel). If p is an odd prime, then a regular p-gon

is constructible if and only if p is a Fermat primeI that is, a prime of the form p D 22t C 1 for some t � 0. Gauss established suffi-

ciency by generalizing the

construction of the 17-gon,

giving explicit formulas of

the �e;f for all pairs .e; f /

with ef D p � 1 (see [33], pp. 200–206).

Proof. We only prove necessity. The problem is whether z D e2�i=p is con- structible. Now z is a root of the cyclotomic polynomial p̂.x/, which is an

irreducible polynomial in QŒx of degree p � 1, by Corollary 6.68. Since z is constructible, p � 1 D 2s for some s, so that

p D 2s C 1:

We claim that s itself is a power of 2. Otherwise, there is an odd number k > 1

with s D km. Now k odd implies that �1 is a root of xk C 1; in fact, there is a factorization in ZŒx:

xk C 1 D .x C 1/.xk�1 � xk�2 C xk�3 � � � � C 1/:

Thus, setting x D 2m gives a forbidden factorization of p in Z:

p D 2s C 1 D .2m/k C 1 D Œ2m C 1Œ.2m/k�1 � .2m/k�2 C .2m/k�3 � � � � C 1:

The only known Fermat primes are 3, 5, 17, 257, and 65537. It follows from

Theorem 7.62, for example, that it is impossible to construct regular 7-gons,

11-gons, or 13-gons.

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326 Chapter 7 Quotients, Fields, and Classical Problems

Further results. Fermat conjectured that all numbers of the form Fm D 22

m C 1 are prime, but Euler factored

F5 D 232 C 1 D 4; 294; 967; 297D 641 � 6700417:

We now know that Fm is composite for 5 � m � 32, but it is unknown whether F33 is prime. The largest Fm that has been shown to be composite (as of 2011)

is F2543548, and the latest conjecture is that there are only finitely many Fermat

primes.

The strongest known result is:

Theorem 7.63. A regular n-gon is constructible with ruler and compass if and

only if n D 2kp1 � � �pt , where k � 0 and the pi ’s are distinct Fermat primes.

Proof. See [15], page 97.

Exercises

7.50 Explain how to carry out each of the following constructions with ruler and com-

pass. Prove that your method works.

(i) Copy a segment.

(ii) Copy an angle.

(iii) Construct a line parallel to a given one through a given point not on the line.

(iv) Construct a line perpendicular to a given one through a given point either on

or off the line.

7.51 *

(i) Prove that every lattice point .m; n/ in the plane is constructible. Conclude

that every Gaussian integer is constructible.

(ii) Prove that every Eisenstein integer is constructible.

7.52 Suppose that ` and `0 are lines with equations ax C by D c and dx C ey D f , and suppose that the coefficients of the equations are all in a field k.

(i) What condition on the coefficients guarantees that ` and `0 intersect in a unique point?

(ii) If ` and `0 intersect in a unique point P , show that P is point whose coordi- nates are in k.

7.53 If the quadratic polynomial ax2 C bx C c has coefficients in a field k, show that its roots are a quadratic extension of k.

7.54 Given a segment of length a, show how to construct a segment of length a=5.

7.55 Given a segment of length a, show how to construct a segment of length a=n,

where n is any positive integer.

7.56 Show how to construct segments of length

(i) p

5 (ii) 3 C p

5

(iii) p

3 C p

5 (iv)

p 3C

p 5

3

(v)

p 3C

p 5

3� p 5

.

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7.3 Connections: Ruler–Compass Constructions 327

7.57 Show how to construct the complex numbers

(i) 1 C i

(ii) p 2 2 .1 C i /

(iii) p 2 2 .1 C i /

(iv) cos 15ı C i sin 15ı

(v) cos 22:5ı C i sin 22:5ı

(vi) cos 36ı C i sin 36ı.

7.58 Show that the side of a regular decagon inscribed in the unit circle is constructible.

7.59 Show that if the side-length of a regular n-gon inscribed in the unit circle is con-

structible, so is the side-length of a regular 2n-gon inscribed in the unit circle.

7.60 * Using the notation of this section, show that

�2;0�2;1 D �4 and �22;0 D �n;1:

7.61 * Using the notation of this section, show that

�4;1 C �4;3 D �2;1 and �4;1�4;3 D �1:

7.62 * Using the notation of this section, show that

�8;1 C �8;5 D �4;1 and �8;1�8;5 D �4;2:

7.63 * Using the notation of this section, show that

�34;1 � 6�4;1 C 3 D �2�4;2;

so that �4;2 2 Q.�4;1/

7.64 Let � D �17 and let k be a nonnegative integer. Show that

�k C ��k D 2 cos.2k�=17/:

7.65 Show that 3 is a primitive element for Z5 , and apply this to Example 7.60.

7.66 Find the minimal polynomial over Q of �8;k for all 0 � k � 7.

7.67 Show that if .m; n/ D 1, and if �n and �m are constructible, so is �mn. Hint: Use Theorem 7.63.

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8 Cyclotomic Integers After proving Corollary 1.8, the special case of Fermat’s Last Theorem for

exponent n D 4, we observed that the full theorem would follow if we could prove, for every odd prime p, that there are no positive integers a; b; c with ap C bp D cp . It is natural to factor this expression as in Exercise 3.50(ii) on page 115: for odd p, we have

.a C b/.a C �b/ � � � .a C �p�1b/ D cp; (8.1)

where � D e2�i=p is a pth root of unity. We didn’t have the language of rings at the time but, later you showed, in Exercise 4.65 on page 168, that the cyclo-

tomic integers ZŒ� is a domain. How could we begin to use this observation?

Recall Exercise 2.14 on page 59: if ab D cn in Z, where n is a positive integer and a; b are relatively prime, then both a and b are nth powers. If ZŒ� behaved

like Z and the factors on the left-hand side of Eq. (8.1) are pairwise relatively

prime, then all the factors a C �j b would be pth powers; that is,

aC �j b D dpj for some dj 2 ZŒ�. For example, consider the case p D 3, so that ZŒ� D ZŒ! is the ring of Eisenstein integers. The factorization is

.a C b/.a C !b/.a C !2b/ D c3:

If the factors on the left-hand side are pairwise relatively prime, then each

of them is a cube of an Eisenstein integer. We can even say something if the

factors are not relatively prime: assuming that ZŒ! has unique factorization, any prime divisor of c must occur three times on the left-hand side (prime now

means a prime in ZŒ!; that is, an element whose only divisors are units and

associates—we will use the term prime here instead of irreducible).

But does the ring of cyclotomic integers behave as Z? To solve Exercise

2.14, we need to use the Fundamental Theorem of Arithmetic: factorization

into primes exists and is essentially unique. It turns out that some rings of cy- clotomic integers do enjoy unique factorization into primes, but some do not.

Indeed, it is known (see [36], p. 7) that ZŒ�23 does not have unique factoriza-

tion.

It’s clear that we need a more thorough investigation of the arithmetic in

rings of cyclotomic integers. In particular, we already know the units in ZŒi  and ZŒ! (Example 6.3); what are the primes?

In Section 8.1, we retrace the by-now-familiar developments in Chapters 1

and 6 to establish division algorithms in ZŒi  and ZŒ! (using norm to mea-

sure size). Even though these are the easiest rings of cyclotomic integers, this

329

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330 Chapter 8 Cyclotomic Integers

will give us a clue how to proceed with ZŒ� for other roots of unity �. There

will be a bonus: we’ll be able to prove Fermat’s Two-Square Theorem that

characterizes all primes in Z which are sums of two squares. As is our custom (because it is so useful), we’ll generalize from these and

the earlier examples of Z and kŒx, where k is a field, to define a Euclidean

domain—a domain having a generalized division algorithm. We’ll show that

every Euclidean domain is a PID, so that, by Theorem 6.50, Euclidean domains

are UFDs and thus have unique factorization. In Section 8.2, we’ll see that there are primes in ZŒi  and ZŒ! that are not

ordinary integers. On the other hand, some primes in Z remain prime in the

larger rings, while some split into non-unit factors. We’ll then make a complete

analysis of this phenomenon for ZŒi  and sketch the analogous theory for ZŒ!.

In Section 8.3, we’ll prove Fermat’s theorem for exponent 3. The fact that there are no non-trivial integer solutions to x3Cy3 D z3 is attributed to Euler; we’ll prove the result making basic use of the arithmetic of ZŒ!.

In Section 8.4, we’ll briefly sketch how the proof for exponent 3 gener-

alizes to prime exponent p when the ring of cyclotomic integers ZŒ�p is a

UFD, where � D e2�i=p. But there are primes p for which ZŒ�p  does not have unique factorization. What then? We’ll finish this section with a brief discussion about how Kummer’s construct of ideal numbers (which Dedekind

recognized as equivalent to what we now call ideals, which is why they are

so-called) could be used to restore a kind of unique factorization to ZŒ�p.

Finally, in Section 8.5 we develop the machinery to prove a lovely theorem

of Fermat that determines the number of ways a positive integer can be written as the sum of two perfect squares.

8.1 Arithmetic in Gaussian

and Eisenstein Integers

We begin by showing that ZŒi  and ZŒ! have generalized division algorithms.

Actually, we’ll show that long division is possible in these rings; that is, there are quotients and remainders. However, quotients and remainders are not nec-

essarily unique; stay tuned.

Given two Gaussian integers z and w, can we find Gaussian integers q and

r so that w D qz C r , where r is “smaller than” z? The obvious way to compare size in C is with absolute values, but it’s easier to calculate norms (squares of absolute value); by Exercise 3.62 on page 127, jr j < jzj if and only if N.r/ < N.z/. Let’s start with an example.

Example 8.1. Take z D �19 C 48i and w D �211 C 102i . Can we find q and r so thatw D qzC r , where N.r/ < N.z/? We can certainly divide z into w in QŒi ; it’s just that w=z may not be a Gaussian integer. In fact,

�211C 102i �19C 48i

D .�211C 102i/.�19 � 48i/ .�19C 48i/.�19 � 48i/

D 8905C 8190i 2665

D 137 41

C 126 41 i:

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8.1 Arithmetic in Gaussian and Eisenstein Integers 331

The idea is to take q to be the Gaussian integer closest to w=z in the complex

plane, and then to find an r that makes up the difference. Since

w

z D 137

41 C 126

41 i � 3:34C 3:07i;

we’ll take q D 3 C 3i . What about r? There’s no choice; since we want w D qz C r , set r D w � qz:

r D w � qz D .�211C 102i/� .3C 3i/.�19 C 48i/ D �10C 15i:

By construction, w D qz C r . What’s more, that q is the closest Gaussian integer to w=z implies, as we’ll see in the proof of the next proposition, that N.r/ < N.z/. Indeed,N.z/ is much bigger thatN.r/ in this example, because

w=z is so close to q.

N.r/ D .�10/2C .15/2 D 325 and N.z/ D .�19/2C .48/2 D 2665: N

This method for choosing q and r works in general.

Proposition 8.2 (Generalized Division Algorithm). If z and w are Gaussian

integers with z ¤ 0, then there exist Gaussian integers q and r such that

w D qz C r and N.r/ < N.z/:

Proof. Suppose that w=z D a C bi , where a and b are rational numbers (but not necessarily integers). As in Example 8.1, take q to be a Gaussian integer

closest to w=z in the complex plane; more precisely, choose integers m and n

so that

ja �mj � 1 2

and jb � nj � 1 2 ;

and let q D mC ni . Now define r to be the difference:

r D w � qz:

Clearly, w D qz C r , so the only thing to check is whether N.r/ < N.z/. To this end, we have

N.r/ D N.w � qz/ D N � z �w z

� q ��

D N.z/N �w z

� q � :

But w=z � q D .a �m/C .b � n/i , so that

N �w z

� q �

D .a �m/2 C .b � n/2 � 1 4

C 1 4 < 1:

It follows thatN.r/ < N.z/.

How to Think About It. The earlier statements of the division algorithms

for Z (Theorem 1.15) and for kŒx, where k is a field (Theorem 6.11), differ

from that in Proposition 8.2; the latter statement does not assert uniqueness of quotient and remainder.

In fact, the way q and r are constructed shows that there may be several

choices for q and r—locate w=z inside a unit square in the complex plane

whose vertices are Gaussian integers, and then pick a closest vertex. There

may be several of these, as the next example shows. Luckily, we won’t need uniqueness of quotients and remainders to get unique factorization into primes.

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332 Chapter 8 Cyclotomic Integers

Example 8.3. If z D 2C 4i and w D �9C 17i , then

w

z D 5

2 C 7

2 i D 2:5C 3:5i:

In contrast to Example 8.1,w=z sits smack in the middle of a unit square whose

vertices are Gaussian integers and, hence, there are four choices for q, namely

2C 3i; 3C 3i; 3C 4i; 2C 4i I

and there are four corresponding divisions, namely

�9C 17i D .2C 3i/.2 C 4i/C .�1 C 3i/I �9C 17i D .3C 3i/.2 C 4i/C .�3 � i/I �9C 17i D .3C 4i/.2 C 4i/C .1 � 3i/I �9C 17i D .2C 4i/.2 C 4i/C .3C i/:

All of these work. In fact, all the remainders have (the same) norm 10 < 20 D N.z/. Even more: the remainders are all associates. Is this an accident? See

Exercises 8.1 through 8.4 on page 336. N

There is an analogous result for the Eisenstein integers ZŒ!, and its proofAlas, there are other rings ZŒ� of cyclotomic

integers which do not have

a generalized division

algorithm.

is almost identical to that for the Gaussian integers. Recall that c C d! D c C d!2 D c C d.�1 � !/, and that

N.c C d!/ D c2 � cd C d 2:

Proposition 8.4 (Generalized Division Algorithm). If z andw are Eisenstein

integers with z ¤ 0, then there exist Eisenstein integers q and r such that

w D qz C r and N.r/ < N.z/:

Proof. Suppose that w=z D aC b!, where a and b are rational numbers (but not necessarily integers). Take q to be a Eisenstein integer closest to w=z in

the complex plane (with respect to the norm); more precisely, choose integers

m and n so that

ja �mj � 1 2

and jb � nj � 1 2 ;

and let q D mC n!. Now define r to be the difference

r D w � qz:

Clearly, w D qz C r , so the only thing to check is whether N.r/ < N.z/. To this end, we have

N.r/ D N.w � qz/ D N � z �w z

� q ��

D N.z/N �w z

� q � :

But w=z � q D .a �m/C .b � n/!, so that

N .w=z � q/ D .a �m/2 � .a �m/.b � n/C .b � n/2 � 1 4

C 1 4

C 1 4 < 1:

It follows thatN.r/ < N.z/.

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8.1 Arithmetic in Gaussian and Eisenstein Integers 333

Why can’t we modify the proof of Proposition 8.4 to prove the result for

every ring of cyclotomic integers? The short answer is that there are counter-

examples. But the reason the proof fails to generalize is that we can’t verify N.r/ < N.z/ in every ZŒ�.

Example 8.5. Let’s divide w D 91 C 84! by z D 34 C 53!. First calculate w=z in QŒ!:

91C 84! 34C 53!

D .91C 84!/.34 C 53! 2/

.34C 53!/.34 C 53!2/ D 2723� 1967!

2163

D 389 309

� 281 309

! � 1:26 � :91!:

Now set q D 1 � !: As with ZŒi, there may be several choices for q

when dividing Eisenstein

integers (see Exercise 8.5

on page 337).

r D w � zq D 4C 12!:

You can check that

N.z/ D 342�34�53C532 D 2163 and N.r/ D 42�4�12C122 D 112: N

How to Think About It. If z, w are either in ZŒi  or in ZŒ!, we could

iterate the respective generalized division algorithms, as we did in Z or kŒx,

to obtain Euclidean algorithms giving a greatest common divisor d of z and

w; moreover, d can be expressed as a linear combination of z and w ensuring, as in earlier instances of this argument, that any common divisor of z and w is

a factor of d . We’ll give an example of such a calculation shortly.

Along the way, there may be choices to be made for quotients and remain-

ders, possibly resulting in different “greatest” common divisors d . We ran into

this situation before: the Euclidean Algorithm in kŒx produces a gcd up to a

unit factor. The same is true in ZŒi  and ZŒ!, although it may not be obvious at this point because of the twists and turns that the Euclidean Algorithm might

take. This is another example in which a more abstract setting can make things

clearer (one reason for the added clarity is that abstraction casts away noise,

allowing you to focus on the heart of a problem).

Euclidean Domains

Looking at our main examples—Z; kŒx; ZŒi ; ZŒ!—we see that one key

to a division algorithm is a measure of size: absolute value for Z, degree for kŒx, norm for ZŒi  and ZŒ!. Now we generalize.

Definition. A Euclidean domain is a domainR equipped with a size function @ is defined on the nonzero elements of R and takes

nonnegative integer val-

ues. @ W R � f0g ! N

such that, for all a; b 2 R with a ¤ 0, there exist q and r inR with

b D qa C r;

where either r D 0 or @.r/ < @.a/.

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334 Chapter 8 Cyclotomic Integers

Some size functions have extra properties. For example, when R is a do-

main, then RŒx is a domain, and degree (which is a size function on RŒx)

satisfies deg.fg/ D deg.f /C deg.g/, while normN (which is a size function on ZŒi  and ZŒ!) satisfies N.˛ˇ/ D N.˛/N.ˇ/. On the other hand, if @ is a size function of a Euclidean domain R, then so is @0, where @0.a/ D @.a/C 1 for all a 2 R � f0g. It follows that a size function may have no algebraic properties; moreover, there may be no elements inR having size 0.

Euclidean domains have nice properties. The proof of the next proposition is essentially the same as that of Theorem 1.19.

Proposition 8.6. Every Euclidean domain is a principal ideal domain.

Proof. Suppose that R is a Euclidean domain with size function @. We want to show that every ideal I in R is principal. If I D f0g, then I is principal, and so we can assume that I contains nonzero elements. The set

S D [email protected]/ W z 2 I g

is a set of nonnegative integers and, hence, it has a least element; call it m.

Choose d to be any element of I of size m. We claim that I D .d/. Clearly, .d/ � I . To get the reverse inclusion, suppose that z 2 I is not 0;

we must show that z 2 .d/. Now there are q and r such that

z D qd C r;

where either r D 0 or @.r/ < @.d/. But r 2 I , because r D z � dq and both z and d are in I . But @.d/ is the smallest size among elements of I ; hence, r D 0, and z D qd 2 .d/.

Corollary 8.7 (Euclid’s Lemma). LetR be a Euclidean domain with a; b 2 R. If p 2 R is irreducible and p j ab, then p j a or p j b.

Proof. This is a direct consquence of Theorem 6.47.

Corollary 8.8. Every Euclidean domain is a unique factorization domainI that is, every nonzero non-unit has a factorization into irreducibles that is essen-

tially unique.

Proof. This is a direct consequence of Theorem 6.50: every PID is a unique

factorization domain.

Corollary 8.8 probably piques your curiosity about what primes look like in Euclidean domains. We’ll consider this question for ZŒi  and ZŒ! in the next

section.

How to Think About It. Points about the development so far.

� Euclidean domains have generalized division algorithms, but they are not necessarily algorithms in the technical sense. They are procedures for com-

puting quotients and remainders, but the division procedures, even for ZŒi 

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8.1 Arithmetic in Gaussian and Eisenstein Integers 335

and ZŒ!, are not deterministic: there is a choice about how to calculate

quotients and remainders.

� In Section 1.3, we studied a direct path in Euclid’s Elements from the Divi- sion Algorithm in Z to the Fundamental Theorem of Arithmetic. This path

can be followed in a much more general setting. We just saw that every Eu-

clidean domain enjoys Euclid’s Lemma and a fundamental theorem.

� One way to show that a domain is a PID is to show that it is Euclidean— indeed, this is one of the most important uses of this notion. On the other

hand, it’s hard from first principles to show that a domain is not Euclidean

(you have to show that no size function exists). Often, the easiest way to

show that a domain is not Euclidean is to show that it’s not a PID. So, for

example, ZŒx is not a Euclidean domain.

� There are PIDs that are not Euclidean, so that the converse of Proposition 8.6 is false. Motzkin found a property of Euclidean domains that can be de-

fined without mentioning its size function. He called an element d in an

arbitrary domain R a universal side divisor if d is not a unit and, for ev-

ery r 2 R, either d j r or there is some unit u 2 R with d j .r C u/. He then proved that every Euclidean domain contains a universal side divi- sor, namely any non-unit of smallest size. Now it was known that if ˛ D 1 2 .1 C

p �19/, then the ring ZŒ˛ is a PID. Motzkin then showed that ZŒ˛

has no universal side divisors, and he concluded that ZŒ˛ is a PID that is

not a Euclidean domain (see Wilson, A principal ideal ring that is not a

Euclidean ring. Math. Magazine 46 (1973), 34–38 and Williams, Note on non-Euclidean principal ideal domains, Math. Magazine 48 (1975),

176–177).

The fact that a Euclidean domain is a PID allows us a to talk about gcd’s,

thanks to Theorem 6.46. Using exactly the same logic as in Chapter 1, we

can iterate division, creating a Euclidean algorithm that finds a gcd for us:

just move factors on one line southwest on the next line (as in the next

example).

Example 8.9. Building on the calculation in Example 8.5, let’s find a gcd for

91C 84! and 34C 53! in ZŒ!. We’ll use the algorithm outlined in Proposi- tion 8.4 to carry out the divisions (a CAS is very useful here). There are four

equations, which we present in “southwestern style:” if a row has the form

f D qhC r , then the next row moves h and r southwest and looks like h D q0r C r 0.

91C 84! D .1 � !/.34 C 53!/C .4 C 12!/

34 C 53! D .3 � !/.4 C 12!/C .10 C 9!/

4 C 12! D .1 C !/.10 C 9!/C .3 C 2!/ 10 C 9! D .4 C !/.3 C 2!/:

Here is a second format, arranging the calculations as we did in Z in Chapter 1,

that shows more detail.

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336 Chapter 8 Cyclotomic Integers

1� ! 34C53!

� 91C84! 87C72! 3� ! 4C12!

� 34C53! 24C44! 1C ! 10C 9!

� 4C12! 1C10! 4 C ! 3C 2!

� 10C9! 10C9!

0

So, we end with 3 C 2!. Repeated application of Exercise 8.6 on page 337Recall that in a general PID, the gcd of two

elements a and b is a

generator d of the principal

ideal consisting of all linear

combinations of a and b; in

symbols, .a;b/ D .d/.

shows that there is a chain of equalities of ideals:

.91C 84!; 34C 53!/ D .34C 53!; 4C 12!/ D .4C 12!; 10C 9!/ D .10C 9!; 3C 2!/ D .3 C 2!/I

that is,

.91C 84!; 34C 53!/ D .3C 2!/:

Thus, 3C 2! is a gcd of 91C 84! and 34C 53!. While the calculations are a bit tedious, you can work the four equations

above backwards, as we did in Chapters 1 and 6, to write 3 C 2! as a linear combination of 91C 84! and 34C 53!. Using a CAS, we found thatAppendix A.6 outlines a

package for a CAS that

allows you to calculate in

ZŒ!.

3C 2! D .5 C 3!/.91 C 84!/� .9 C 2!/.34 C 53!/ N

How to Think About It. We’ve seen, in ZŒi  or ZŒ!, that there are some- times choices for quotients and corresponding remainders in the generalized

division algorithms. Hence, there may be more than one way to implement the

Euclidean Algorithm and, so, more than one end result. But, thanks to Proposi-

tion 6.45 and the fact that ZŒi  and ZŒ! are PIDs, any two gcd’s are associates.

See Exercise 8.7 below for an example.

Exercises

8.1 Prove or Disprove and Salvage if Possible. Two Gaussian integers are associates

if and only if they have the same norm.

8.2 How many possible numbers of “closest Gaussian integers” to a complex number

are there? For each number, give an example.

8.3 Let z and w be Gaussian integers, and suppose that q and q0 are Gaussian integers equidistant from w=z in the complex plane. Show that

N. w

z � q/ D N.

w

z � q0/:

8.4 Let z and w be Gaussian integers, and suppose that q and q0 are Gaussian integers equidistant from w=z in the complex plane. Are w=z�q and w=zz�q0 associates? If so, prove it; if not, give a counterexample.

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8.2 Primes Upstairs and Primes Downstairs 337

8.5 If z ¤ 0 and w are Eisenstein integers, how many possible numbers of quotients w=z are there in ZŒ! satisfying the conditions of the division algorithm? For

each number, give an example.

8.6 * Let R be a commutative ring. If a, b, c, and d are elements of R such that

b D da C c, show that there is equality of ideals .b; a/ D .a; c/. 8.7 If z D 6 C 12i and w D �13 C 74i , show that

w

z D 9

2 C 10

3 i:

(i) Show that there are two q’s with w D qz C r in the generalized division algorithm, namely q D 4 C 3i and q D 5 C 3i .

(ii) Apply the Euclidean Algorithm to find a gcd of z and w starting with q D 4 C 3i .

(iii) Apply the Euclidean Algorithm to find a gcd of z and w starting with q D 5 C 3i .

(iv) Are the two gcd’s associates in ZŒi ?

8.2 Primes Upstairs and Primes Downstairs

We saw in the last chapter that an irreducible polynomial in kŒx (for some field k) may factor inKŒx for some extension fieldK=k. For example, x2C1 is irreducible in RŒx but it factors in CŒx. A similar phenomenon occurs for

primes in Z. Every ring of cyclotomic integers R has Z as a subring, and a

prime p 2 Z may factor in R. Our goal in this section is to investigate primes in ZŒi  and in ZŒ!, and the

obvious way to begin doing this is by studying primes downstairs, that is, in Z, and look at their behavior upstairs, that is, in rings of cyclotomic integers.

Corollary 8.8, the Fundamental Theorem for Euclidean domains, tells us

that every element in such a domain has an essentially unique factorization

into primes.

Lemma 8.10. Let R D ZŒ�p  for any prime p. If u 2 R, then u is a unit in R if and only ifN.u/ D 1.

Proof. If u is a unit, there is v 2 R with uv D 1. Hence, 1 D N.uv/ D N.u/N.v/. As N.u/ and N.v/ are positive integers, we must have N.u/ D 1 D N.v/.

Conversely, suppose that N.u/ D 1. Since N.u/ D uu, we have u a unit in R (with inverse u).

The actual factorization of Gaussian integers or of Eisenstein integers into primes can be a tricky task, but here is a useful tool.

Proposition 8.11. Let R D ZŒi  or ZŒ!. If z 2 R and N.z/ is prime in Z, then z is prime inR.

Proof. We prove the contrapositive of the statement of the Proposition. If z 2 R and z D wv for non-unitsw and v, then Lemma 8.10 gives N.w/ > 1 and N.v/ > 1. Hence,

N.z/ D N.wv/ D N.w/N.v/; and N.z/ is not prime in Z.

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338 Chapter 8 Cyclotomic Integers

The converse of Proposition 8.11 is false; we’ll soon see that 7 is prime

in ZŒi , but that N.7/ D 49.

Example 8.12. (i) In Example 8.1, we divided w D �211 C 102i by z D �19C 48i and got quotient 3C 3i and remainder �10C 15i . Let’s carry out the rest of the Euclidean Algorithm to get a gcd of z and w.

We use two formats for the Euclidean Algorithmfinding the gcd. Here’s the southwestern version.

w D .3 C 3i/z C .�10C 15i/ 3C 3i D .3 � i/.�10 C 15i/C .�4 � 7i/

�10C 15i D .�1 � 2i/.�4 � 7i/:

Hence, gcd.z; w/ D 4C 7i (we take an associate of �4 � 7i ). Here is a more detailed version of this calculation.

3C 3i �19C48i

� �211C102i �201C 87i 3� i �10C 15i

� �19C48i �15C55i �1� 2i �4� 7i

� �10C15i �10C15i

0

Again we see that gcd.z; w/ D 4 C 7i . So 4 C 7i is a factor of both z and w.

(ii) Is the gcd 4C 7i prime? If not, can we factor it explicitly? Since

N.4C 7i/ D 65 D 13 � 5;

we claim that the norm of any prime factor � of 4 C 7i is either 13 or 5. If 4 C 7i D �˛, where ˛ 2 ZŒi  is not a unit, then N.˛/ > 1, by Lemma 8.10. Hence, as N.4C 7i/D N.�/N.˛/, we must have N.�/ D 13 or N.�/ D 5. Well, 5 is small enough to do a direct check, and the only Gaussian integers with norm 5 are 2 ˙ i and their associates. And we’re in luck; 2C i is a factor:

4C 7i 2C i D 3C 2i:

Note that N.3C 2i/ D 13, so that 3C 2i is prime, by Proposition 8.11. (iii) To factor w D �211C 102i , divide by the gcd:

�211C 102i 4C 7i

D �2C 29i:

Since N.�2 C 29i/ D 845 D 5 � 132, the same process as in part (ii) shows that

�2C 29i D .2C i/.5 C 12i/ D .2 C i/.3 C 2i/2:

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8.2 Primes Upstairs and Primes Downstairs 339

Putting it all together, we have the prime factorization of w: In Section 3.4, we used the fact that

5C 12i D .3C 2i/2 when we generated Pythagorean

triples with Gaussian

integers.

w D �211C 102i D .4 C 7i/.�2C 29i/ D .2 C i/.3 C 2i/.2C i/.3 C 2i/2

D .2 C i/2.3 C 2i/3:

We leave it to you to find the prime factorization of z and to show that

gcd.z; w/ lcm.z; w/ D zw. N

How to Think About It. How do we factor a positive rational integer m

into primes? First of all, there is an algorithm determining whetherm is prime.

Use the Division Algorithm to see whether 2 j m. If 2 − m, use the Division Algorithm to see whether 3 j m. And so forth. Now if d is a divisor of m, then d � m, and so there are only finitely many candidates for divisors; hence, this process must stop. Of course, if we have any extra information about m,

we may use it to cut down on the number of candidates. We must say that

this algorithm is useful only for small numbers m; after all, the difficulty in

factoring large numbers is the real reason that public key codes are secure. A variation of this algorithm can be used to factor nonzero Gaussian in-

tegers. If d; w 2 ZŒi  and d j w, then N.d/ � N.w/; hence, there are only finitely many Gaussian integers z which are candidates for being divisors of w.

If N.w/ is prime, then Proposition 8.11 says that w is prime; if N.z/ is com-

posite, we can proceed as in the last part of Example 8.12.

Laws of Decomposition

We now describe the primes in ZŒi  (there will be a similar story for ZŒ!).

The next lemma lets us concentrate on how primes downstairs in Z behave when they are viewed as elements upstairs in ZŒi .

Notation. It gets tedious to keep saying “let p be a prime in Z:” From now on,

let’s call primes in Z rational primes to distinguish them from primes in other

rings. Remember that a prime (or irreducible) element in a commutative ring is one whose only divisors are units and associates. We may also say rational

integer to distinguish an ordinary integer in Z from a Gaussian integer, an

Eisenstein integer or, more generally, a cyclotomic integer.

Lemma 8.13. Every prime � in ZŒi  divides a rational prime.

Proof. Every Gaussian integer z divides its norm in ZŒi , for N.z/ D zz. In particular, � divides a rational integer, namely its norm. Now N.�/ factors

into primes in Z:

�� D N.�/ D p1p2 : : : pk;

and so

The primes on the right-

hand side are elements of

ZŒi as well as of Z.

� j p1p2 : : : pk:

But � is a prime in ZŒi ; hence, by Euclid’s Lemma in ZŒi , it divides one of

the (prime) factors pj on the right.

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340 Chapter 8 Cyclotomic Integers

Example 8.14. We have seen that � D 3 C 2i is a prime in ZŒi ; note that � divides 13, for

N.�/ D .3C 2i/.3 � 2i/ D 13:

We sometimes say that 3C 2i lies above 13. N

As we said earlier, there are primes in Z that remain prime in ZŒi  and

others that factor into new primes; the same is true for ZŒ!. It turns out that there’s a beautiful theory, going back to Gauss, for how primes decompose in

these rings, a theory that brings together many of the ideas you’ve studied so

far. For example, here are some factorizations of rational primes when viewed

as elements in ZŒi :

5 D .2C i/.2 � i/I 13 D .3C 2i/.3 � 2i/I 29 D .5 C 2i/.5 � 2i/:

In each of these cases, the rational prime decomposes as a norm: the product

of a Gaussian integer and its conjugate. This is always the case.

Lemma 8.15. Let p be a rational prime. If p is not prime in ZŒi , then there

exists some prime z in ZŒi  with

p D zz D N.z/:

Proof. Suppose p D zw, where z and w are non-unit Gaussian integers. Then

p2 D N.p/ D N.zw/ D N.z/N.w/;

where neither N.z/ nor N.w/ is 1. But this is an equation in Z, and so uniqueAnd, in fact, z and w must be associates. factorization in Z gives p D N.z/ D zz. Finally, z is prime in ZŒi , by

Proposition 8.11, because N.z/ is a rational prime.

Lemma 8.15 narrows the situation quite a bit. It says that if a rational prime factors in ZŒi , it factors into exactly two conjugate Gaussian integers, each

prime in ZŒi . We say that such a rational prime splits in ZŒi . We can state

the result of Lemma 8.15 using only the arithmetic of Z. Since N.a C bi/ D a2 C b2, the lemma says that a prime splits if it can be written as the sum of two perfect squares. And the converse is also true.

Proposition 8.16. A rational prime p splits in ZŒi  if and only if p is a sum of

two squares in Z.

Proof. If p splits in ZŒi , then Lemma 8.15 says that there is a Gaussian integer

z D a C bi such that p D N.z/ D a2 C b2. Conversely, if p D a2Cb2, then p D .aCbi/.a�bi/. But aCbi is prime

in ZŒi , by Proposition 8.11, because its norm, N.a C bi/ D p, is a rational prime.

The question of which rational primes split in ZŒi  thus comes down to the question of which primes are sums of two squares. Not every rational prime is

a sum of two squares; for example, it’s easy to see that 11 is not. Here is a nice

(and perhaps surprising) connection to modular arithmetic. A quick example

gives the idea. The prime 29 is the sum of two squares:

29 D 22 C 52:

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8.2 Primes Upstairs and Primes Downstairs 341

As an equation in F29, this says that 2 2 C 52 D 0. Multiply both sides by 62, Recall that Fp is another

notation for Zp , the field of

integers modulo p. for 6 D 5�1 in F29:

0 D 2262 C 5262

D .2 � 6/2 C .5 � 6/2

D .2 � 6/2 C 1;

so that 2 � 6 D 12 is a root of x2 C 1 in F29. More generally, suppose that p is a prime and p D a2 C b2. We can assume that 0 < a; b < p, so that both a and b are units in Fp. We can write this as an equation: a

2 C b2 D 0 in Fp . Multiplying both sides by

� b�1

�2 , we get:

� ab�1

�2 C 1 D 0I that is, ab�1 is a root of x2 C 1 in Fp . And the converse is true as well:

Proposition 8.17. A rational prime p is a sum of two squares if and only if

x2 C 1 has a root in Fp. Another way to say this: p is the sum of two squares if

and only if �1 is a square in Fp .

Proof. We’ve just seen that an expression of p as the sum of two squares leads

to a root of x2 C 1 in Fp . Going the other way, suppose there is an integer n whose congruence class

satisfies

n2 C 1 D 0 in Fp:

Then, moving back to Z, we see that n satisfies

p j .n2 C 1/:

Now go upstairs to ZŒi . We have

p j .nC i/.n � i/:

But p divides neither n C i nor n � i in ZŒi  (otherwise, n=p ˙ i=p would be Gaussian integers). Euclid’s Lemma says that p is not prime in ZŒi  and,

hence, by Lemma 8.15, p is the norm of some Gaussian integer z—that is, p

is a sum of two squares.

Corollary 8.18. A rational prime p factors in ZŒi  if and only if x2 C 1 has a So, p factors in ZŒi if and only if x2 C 1 factors in FpŒx.

root in Fp.

Proof. Apply Proposition 8.17 and the Factor Theorem (Corollary 6.15).

Let’s summarize these various equivalent statements about a rational primep.

� p factors in ZŒi . � p D N.z/ for some z in ZŒi . � p D a2 C b2 in Z. � x2 C 1 has a root in Fp.

The last criterion may seem the most remote, but it is actually the easiest to

use—you have to check at most .p � 1/=2 possible solutions to x2 C 1 D 0 (because if ˛ is a solution, so is �˛). If you try a few numerical cases, a pat- tern begins to emerge—the primes satisfying the last criterion all seem to be

congruent to 1 mod 4. That’s quite a beautiful and elegant result, which adds

one more equivalent statement to the summarizing list above.

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342 Chapter 8 Cyclotomic Integers

Theorem 8.19. If p is an odd prime, then x2 C 1 has a root in Fp if and only if p � 1 mod 4.

Proof. Assume that p D 4k C 1. Since p is prime, Theorem 4.9 (Fermat’s Little Theorem) gives ap�1 D 1 for all nonzero a in Fp . Thus, we have the factorization in FpŒx:See Exercise 6.38 on

page 263.

xp�1 � 1 D .x � 1/.x � 2/.x � 3/ : : : .x � .p � 1// : (8.2)

Hence,

xp�1 � 1 D x4k � 1

D � x4 �k � 1

D .x4 � 1/ h� x4 �k�1 C

� x4 �k�2 C

� x4 �k�3 C � � � C 1

i

by Exercise 6.47 on page 269

D .x2 C 1/ h .x2 � 1/

� x4 �k�1 C

� x4 �k�2 C

� x4 �k�3 C � � � C 1

i

D .x2 C 1/h.x/: (8.3)

Comparing Eqs. (8.2) and (8.3), the two factorizations of xp�1 � 1, and using unique factorization in FpŒx, we see that x

2 C 1 D .x � ˛/.x � ˇ/ for some ˛; ˇ 2 Fp.

Conversely, if p is odd and p 6� 1 mod 4, then p � 3 mod 4 (it can’t be congruent to 0 or 2). But, by Proposition 8.17, if x2 C 1 has a root in Fp, then p is the sum of two squares in Z. However, the sum of two squares in Z is

never congruent to 3 mod 4: If a D 0; 1; 2; 3, then a2 � 0; 1; 0; 1mod 4, and so a2 C b2 � 0; 1; 2 mod 4.

Propostion 8.16, when combined with Theorem 8.19, gives us a nice fact of

arithmetic, first established by Gauss.

The name of a theorem

may not coincide with the

name of the first person

who proved it.

Corollary 8.20 (Fermat’s Two-Square Theorem). An odd rational prime p

is a sum of two squares if and only if p � 1 mod 4.

Theorem 8.19 tells the whole story for odd primes: primes that are congru-

ent to 1 mod 4 split into two conjugate factors, and primes that are congruent to 3 mod 4 stay prime (we call primes downstairs that stay prime upstairs inert).

There is one prime we haven’t yet considered: p D 2. Now 2 factors in ZŒi , because x2 C 1 factors in Z2Œx. In fact

2 D .1C i/.1 � i/:

But note that these two factors are associates:Are there other primes in ZŒi that are associate

to their conjugates? That

question is Exercise 8.10

on page 343.

1C i D i.1 � i/; and so

2 D i.1 � i/2:

Thus, 2 splits in a special way: it is associate to the square of a prime. We say

that 2 ramifies in ZŒi . Hence, our discussion gives a complete classification

of how rational primes decompose in the Gaussian integers.

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8.2 Primes Upstairs and Primes Downstairs 343

Theorem 8.21 (Law of Decomposition in Gaussian Integers). Every ratio-

nal prime p decomposes in ZŒi  in one of three ways.

(1) p splits into two conjugate prime factors if p � 1 mod 4 (2) p is inert if p � 3 mod 4 (3) p D 2 ramifies: 2 D i.1 � i/2.

Corollary 8.22 (Classification of Gaussian Primes). The primes � in ZŒi 

are of three types:

(1) � D a C bi , which lies above a rational prime p with p � 1 mod 4I in this case, N.�/ D a2 C b2

(2) � D p, where p is a rational prime with p � 3 mod 4I in this case, N.p/ D p2

(3) � D 1 � i and its associatesI in this case, N.1 � i/ D 2.

Proof. Now� divides some rational primep, by Lemma 8.13. If p � 1 mod 4, then � is of the first type; if p � 3 mod 4, then p is inert and � D p; if p � 2 mod 4, then p ramifies and � D 1 � i .

As we mentioned earlier, 7 is a prime in ZŒi , for 7 � 3 mod 4, but its norm 49 is not a rational prime. Thus, the converse of Proposition 8.11 is false.

How to Think About It. The fact that 2 D i.1 � i/2 can be stated in terms of ideals in Z[i]: there is equality of ideals

.2/ D � .1 � i/2

� :

In fact, if we use the definition of the product of ideals from Exercise 5.51 on

page 220, the above equation of ideals can be written as

.2/ D .1 � i/ .1 � i/ D .1 � i/2 :

Exercises

8.8 (i) In Example 8.12 we found a gcd of z D �19 C 48i and w D �211 C 102i to be 4 C 7i . Write 4 C 7i as a linear combination of z and w.

(ii) Use part (i) to find the prime factorization of z.

(iii) Show that gcd.z; w/ lcm.z; w/ D zw.

8.9 Show that if two Gaussian integers z and w have relatively prime norms in Z,

then z and w are relatively prime in ZŒi . Is the converse true?

8.10 * Which primes in ZŒi  are associate to their conjugates?

8.11 How many non-associate primes in ZŒi  lie above 5?

8.12 In ZŒi , show that every associate of a Cbi is conjugate to an associate of b Cai .

8.13 Show that every Gaussian integer is associate to one in the first quadrant of the

complex plane. (We define the first quadrant to include the nonnegative x-axis but

not the positive y-axis.)

8.14 Show that if two integers a and b can each be written as the sum of two squares,

so can ab.

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344 Chapter 8 Cyclotomic Integers

8.15 Factor each of these into primes in ZŒi .

(i) 101 (ii) 31 (iii) 37

(iv) 7 C 4i (v) 8 C i (vi) 65 (vii) 7 C 3i (viii) 40 C 42i (ix) 154 C 414i

8.16 Find the number of elements in each of the quotient rings.

(i) ZŒi =.1 � i / (ii) ZŒi =.2 C i / (iii) ZŒi =.3 C 2i / (iv) ZŒi =.5 C 12i /

8.17 Take It Further. If z is a Gaussian integer, show that

jZŒi =.z/j D N.z/:

Eisenstein Primes

The whole theory just given for ZŒi  carries over to ZŒ!. Of course, the state-

ments have to be modified slightly, but the proofs are almost identical to the

corresponding results in ZŒi . If you think about it, this shouldn’t be a surprise:

a proof using only algebraic properties of norm (for example, it is multiplica-

tive) and properties of PIDs (unique factorization and Euclid’s Lemma) should carry over mutatis mutandis.

We summarize the results for Eisenstein integers, providing sketches of

proofs where we think it’s necessary, but we leave the details to you. And

these are important exercises, because they will help you digest the ideas in

both rings.

Lemma 8.23. Every prime in ZŒ! divides a rational prime.

Proof. Mimic the proof of Lemma 8.13.

How about a law of decomposition for Eisenstein integers? Some rational

primes factor in ZŒ!; for example:Try some other primes in ZŒ!. Any conjectures

about which ones split? 7 D .3C !/.3 C !2/I 31 D .5 � !/.5 � !2/I 97 D .3 � 8!/.3 � 8!2/:

In each of these cases, the prime in Z decomposes in ZŒ! into a norm: the

product of an Eisenstein integer and its conjugate. This is always the case.

Lemma 8.24. Let p be a rational prime. If p is not prime in ZŒ!, then p D zz D N.z/ for some prime z in ZŒ!.

Proof. Mimic the proof of Lemma 8.15.

As happened in ZŒi , we can restate the result of Lemma 8.24 completely in terms of the arithmetic of Z. Since N.aC b!/ D a2 � abC b2, the lemma says that a prime splits if it can be written in this form. The converse is also

true.

Proposition 8.25. A rational prime p splits in ZŒ! if and only if p can be

expressed as a2 � ab C b2 for integers a; b 2 Z.

Proof. Mimic the proof of Lemma 8.16.

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8.2 Primes Upstairs and Primes Downstairs 345

Proposition 8.17 says that a rational prime p is a norm of some prime in

ZŒi  if and only if x2 C 1 has its roots in Fp. What might be an analog of this result for ZŒ!? Well, x2 C 1 is the minimal polynomial for i ; the minimal polynomial for ! is x2CxC1. A careful look at the proof of Proposition 8.17 shows that we can use x2 C x C 1 and modify the proof slightly to obtain another lovely result.

Proposition 8.26. A rational prime p can be expressed as a2 � ab C b2 for integers a; b 2 Z if and only if x2 C x C 1 has a root in Fp. What is the discriminant of

x2C x C 1? Proof. Suppose that p D a2 � ab C b2 for integers a and b. Then p doesn’t divide either a or b (otherwise p2 j .a2 � ab C b2)) and, hence, b is a unit in Fp. Multiply both sides by

� b�1

�2 to obtain

p � b�1

�2 D � �ab�1

�2 C � �ab�1

� C 1;

so that �a � b�1

� is a root of x2 C x C 1 in Fp.

Going the other way, suppose that there is a congruence class Œm 2 Fp with

Œm2 C ŒmC 1 D 0 in Fp : Then, in Z, we have

m2 CmC 1 � 0 mod pI that is,

p j .m2 CmC 1/:

Now move up to ZŒ!. We have Check that

.m � !/.m � !2/ D m2CmC 1.

p j .m � !/.m � !2/; or

p j .m � !/.m C 1C !/: But p doesn’t divide either m� ! or mC 1C ! in ZŒ! (otherwise m

p � 1 p !

or mC1 p

C 1 p ! would be an Eisenstein integer). Thus, by Euclid’s Lemma, p is

not prime in ZŒ! and hence by Lemma 8.24, p is the norm of an Eisenstein

integer aC b!; that is, p D a2 � ab C b2 for integers a and b.

We get the next corollary.

Corollary 8.27. A rational prime p factors in ZŒ! if and only if x2 C x C 1 So, p factors in ZŒ! if and only if x2C x C 1 factors in FpŒx.

has a root in Fp .

Proof. Apply Proposition 8.26 and the Factor Theorem (Corollary 6.15).

We summarize the chain of equivalent statements.

� p factors in ZŒ!. � p D N.z/ for some z in ZŒ!. � p D a2 � ab C b2 in Z. � x2 C x C 1 has a root in Fp.

Onward to a law of decomposition in ZŒ!. Numerical experiments (we

hope you’ll try some) suggest that if p is a rational prime and p � 1 mod 3, then x2CxC 1 has a root in Fp . The proof of Theorem 8.19 suggests a reason why.

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346 Chapter 8 Cyclotomic Integers

Proposition 8.28. If p is a prime and p � 1 mod 3, then x2 C x C 1 has a root in Fp.

Proof. Suppose that p D 3k C 1. Because p is prime, ap�1 D 1 for all non- zero a in Fp (Fermat’s Little Theorem—Theorem 4.9). Hence, in FpŒx, we

have the factorizationSee Exercise 6.38 on page 263.

xp�1 � 1 D .x � 1/.x � 2/.x � 3/ � � � .x � .p � 1// : (8.4)

But

xp�1 � 1 D x3k � 1

D � x3 �k � 1

D .x3 � 1/ h� x3 �k�1 C

� x3 �k�2 C

� x3 �k�3 C � � � C 1

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Exercise 6.47 on page 269

D .x2 C x C 1/ h .x � 1/

� x3 �k�1C

� x3 �k�2C

� x3 �k�3C � � � C 1

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D .x2 C x C 1/h.x/: (8.5)

Comparing Eqs. (8.4) and (8.5), the two factorizations of xp�1 � 1, and using unique factorization in FpŒx, we see that x

2 C x C 1 D .x � ˛/.x � ˇ/ for some ˛; ˇ 2 Fp; that is, x2 C x C 1 has a root in Fp.

What about primes that are not congruent to 1 mod 3? One case is easily

handled. Suppose that p � 2 mod 3. By Proposition 8.26, x2 C x C 1 has a root in Fp if and only if p can be written as a

2�abCb2 for a; b 2 Z. But you can check that, for any choice of a and b, a2 �abC b2 is never congruent to 2 mod 3 (just look at the possible congruence classes of a and b mod 3). Thus,

p is inert; that is, p is prime in ZŒ!. There is only one more prime, namely 3,

the prime congruent to 0 mod 3. And x2 C x C 1 certainly has roots in F3, namely 0 and 1. Therefore, 3 must split; in fact,

3 D .2 C !/.2 C !2/ D .2 C !/.1 � !/:

But the important thing is that the two factors on the right are associates. YouRecall that the units in ZŒ! are ˙1, ˙!, and ˙!2.

can check that

Are there other primes in

ZŒ! that are associate

to their conjugates? That

question is Exercise 8.18

on page 349.

�!2.1 � !/ D 2C !:

So, our factorization of 3 can be written as

3 D �!2.1 � !/2;

and 3 is a ramified prime. Putting it all together, we have the law of decompo-In terms of ideals, .3/ D� .1� !/2

� D .1� !/2. sition in ZŒ! as well as a description of all Eisenstein primes.

Theorem 8.29 (Law of Decomposition in Eisenstein Integers). Every ratio-

nal prime p decomposes in ZŒ! in one of three ways.

(1) p splits into two conjugate prime factors if p � 1 mod 3 (2) p is inert if p � 2 mod 3 (3) 3 ramifies into �!2.1 � !/2 .

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8.2 Primes Upstairs and Primes Downstairs 347

Corollary 8.30 (Classification of the Eisenstein primes). The primes � in

ZŒ! are of three typesW (1) � D aC b! which lies above a rational prime congruent to 1 mod 3I in

this case, N.�/ D a2 � ab C b2. (2) primes p in Z that are congruent to 2 mod 3I in this case, N.p/ D p2. (3) the prime 1 � ! and its associatesI in this case, N.1 � !/ D 3.

The relation between the factorization of a prime p in ZŒ! and the factor-

ization of x2 C x C 1 in FpŒx can be used to factor Eisenstein integers.

Example 8.31. (i) Let p D 31. There are two roots of x2 C x C 1 in F31, namely 5 and 25, and so x2 C x C 1 D .x � 5/.x � 25/ in F31Œx. Lift this equation to ZŒx:

x2 C x C 1 D .x � 5/.x � 25/C 31.x � 4/:

So, letting x D !, we have

.5 � !/.25 � !/ D 31.4 � !/:

Now N.5 � !/ D 31, so 5 � ! is a prime factor of 31, and the other is This example connects to Exercise 8.24 on page 349.

25 � ! 4 � ! D 6C !:

(ii) Let p D 97. There are two roots of x2 CxC 1 in F97, namely 61 and 35. In fact,

x2 C x C 1 D .x � 61/.x � 35/C 97.x � 22/:

Letting x D !,

.61 � !/.35 � !/ D 97.22� !/;

and so

.61 � !/.35 � !/ 22� !

D 97:

Now,N.22�!/ D 507 D 3�132; sinceN.1�!/ D 3 andN.4C!/ D 13, so (checking for unit factors), we have

22 � ! D .1 � !/.4 C !/2:

Some of these factors divide 61� !; the rest divide 35� !. We have

N.61 � !/ D 3 � 13 � 97 N.35 � !/ D 13 � 97:

We can cancel the factor of 13 by dividing by 4 C !; it’s easier to work with 35 � !:

35 � ! 4C ! D

.35 � !/.4 C !2/ 13

D 8 � 3!:

Bingo:N.8 � 3!/ D 97, so that

97 D .8 � 3!/.8 � 3!2/: N

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348 Chapter 8 Cyclotomic Integers

How to Think About It. Because ZŒi  and ZŒ! are commutative rings, we

can construct quotient rings. And, since both rings are PIDs, they often look

very similar to rings we have already met.

Example 8.32. We investigate the quotient ring R D ZŒ!=.�/, whereThe result of this example will be useful in the next

section. � D 1 � !

is the prime lying over the rational (and ramified) prime 3. For any Eisenstein integer z, let’s look at the remainder after dividing z by �. Proposition 8.4 gives

Eisenstein integers q and r such that

z D q�C r; with r D 0 or N.r/ < N.�/.

Now N.�/ D 3, so that N.r/ must be 0, 1, or 2. There are no Eisenstein integers of norm 2, because 2 is inert in ZŒ!. Hence N.r/ is 0 or 1. IfN.r/ D 0, then r D 0; if N.r/ D 1, then r is a unit in ZŒ!. So, aside from 0, we need only investigate the six Eisenstein units. It turns out that each of these is

congruent to 1 or �1 modulo �:

If r D 1; then z D q�C 1 and z � 1 mod �: If r D �1; then z D q� � 1 and z � �1 mod �: If r D !; then z D q�C ! D .q � 1/�C 1 and z � 1 mod �: If r D �!; then z D q� � ! D .q C 1/� � 1 and z � �1 mod �: If r D !2; then z D q�C !2 D .q � 1� !/� C 1 and z � 1 mod �: If r D �!2; then z D q� � !2 D .q C 1C !/� � 1 and z � �1 mod �: N

So, every element of ZŒ! is congruent mod � to one of 0; 1; or �1. This suggests that ZŒ!=.�/ is none other than our friend F3. And, in fact that’s

true.

Proposition 8.33. If � D 1 � !, then

ZŒ!=.�/ Š F3:

Proof. By Proposition 7.13, the quotient ring ZŒ!=.�/ is a field, while Exam- ple 8.32 shows that the field has exactly 3 elements. Therefore, ZŒ!=.�/ Š F3, for Corollary 7.40 says that two finite fields with the same number of ele-

ments are isomorphic.

The results in this section just scratch the surface, for life is more compli- cated; there are rings of cyclotomic integers that are not PIDs. We shall have

more to say about this when we discuss the work of Kummer.

Further Results. The laws of decomposition for ZŒi  (Theorem 8.21) and ZŒ! (Theorem 8.29) show that the decomposition of a rational prime depends

only on its congruence class modulo a fixed integer: 4 for ZŒi  and 3 for ZŒ!.

This theory was greatly generalized in the twentieth century to Class Field

Theory, which determines laws of decomposition of primes in rings of cy-

clotomic integers, thereby bringing together under one roof many of the main

ideas in modern algebra.

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8.3 Fermat’s Last Theorem for Exponent 3 349

Exercises

8.18 * Which primes in ZŒ! are associate to their conjugates?

8.19 For which primes p is x2 C x C 1 a perfect square in Fp Œx?

8.20 Working in ZŒ!], under what conditions are a C b! and b C a! associates?

8.21 In ZŒ!],

(i) What are all the associates of the prime 1 � !? (ii) Show that 1 � ! and 1 � !2 are associates.

(iii) Write .1 � !/.1 � !2/ as a C b!. Note that x2 C x C 1 D .x � !/.x � !2/. What happens if you put x D 1?(iv) What is the minimal polynomial of 1 � !?

8.22 If z, w, v are elements of ZŒ! (or ZŒi ), show that

(i) If z j w, then z j w. (ii) If z j w, then N.z/ j N.w/ in Z.

(iii) If z � w mod v then z � w mod v.

8.23 Show that a rational prime p splits in ZŒ! if and only if �3 is a square mod p.

8.24 Show there are isomorphisms of commutative rings,

(i) ZŒi  Š ZŒx=.x2 C 1/. (ii) ZŒ! Š ZŒx=.x2 C x C 1/.

8.25 * Find all units u in ZŒ! such that u � 1 mod 3. Note that 3 is a unit times .1� !/2.

8.26 Factor into primes in ZŒ!.

(i) 301 (ii) 307 (iii) 5 C 8! (iv) 5 C ! (v) 19 C 18! (vi) 39 C 55!

(vii) 61 � ! (viii) 62 C 149! (ix) 87 � 62!

8.27 Find the number of elements in

(i) ZŒ!=.2 C !/ (ii) ZŒ!=.4 � !/ (iii) ZŒ!=.6 C !/ (iv) ZŒ!=.31/

8.28 Take It Further. If z is an Eisenstein integer, show that

jZŒ!=.z/j D N.z/:

8.3 Fermat’s Last Theorem for Exponent 3

The goal of this section is to prove Fermat’s Last Theorem for exponent 3:

there are no positive integers x; y; z satisfying x3 C y3 D z3. The earliest proof is attributed to Euler [12] in 1770 (his proof has a gap that was eventually

closed). We develop a different proof in this section that is a nice application

of the arithmetic in ZŒ!.

How to Think About It. The development of the proof is quite technical

(we’ve polished it as much as we were able), but the essential idea is straight- forward and has already been mentioned several times. It’s based on the fac-

torization of x3 C y3 in ZŒ! (see Exercise 3.50 on page 115):

x3 C y3 D .x C y/.x C y!/.x C y!2/:

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350 Chapter 8 Cyclotomic Integers

If there are positive integers x; y; z with x3 C y3 D z3, then we’d have a factorization of z3 in ZŒ!:

z3 D .x C y/.x C y!/.x C y!2/:

The primes dividing z all show up with exponent at least 3 in z3, and the idea is

to show that this can’t happen on the right-hand side. Heuristically, if the three

factors on the right are relatively prime and none is divisible by the square of

a prime, we’re done. But it’s not so easy, mainly because of some mischief

caused by � D 1 � !, the prime lying above 3. So, pull out your pencil again and follow along.

Preliminaries

Our development will often make use of a fact about ZŒ! adapted from Corol-

lary 6.37.

Proposition 8.34. If x and y are rational integers that are relatively prime

in Z, then they are relatively prime in ZŒ!.

Proof. If � is a prime in ZŒ! dividing both x and y, then N.�/ j N.x/ and N.�/ j N.y/. That is, N.�/ j x2 and N.�/ j y2 (for both x and y lie in Z). Now N.�/ 2 Z; hence, if p is a prime factor of N.�/, then p is a commonSee Exercise 8.22 on

page 349. factor of x and y, a contradiction.

The prime � D 1�! will figure prominently in the story. In Theorem 8.29, we saw that � j 3 and, in fact,

3 D �!2�2: (8.6)

That � lies above 3 implies that a rational integer divisible by � in ZŒ! is di-

visible by 3 in Z. The next lemma explains the ubiquity of � in the forthcoming

proofs.

Lemma 8.35. If x is a rational integer, then 3 j x in Z if and only if � j x in ZŒ!.

Proof. If 3 j x in Z, then Eq. (8.6) shows that � j 3 in ZŒ!; hence, � j x in ZŒ!.

Conversely, if � j x, then Exercise 8.22 on page 349 shows that N.�/ j N.x/ in Z. But N.�/ D 3 and N.x/ D x2, so that 3 j x2 in Z and hence Euclid’s Lemma gives 3 j x in Z.

In Example 8.32, we saw that every element in ZŒ! is congruent mod �

to 0, 1, or �1. We’ll often need to know “how congruent” an Eisenstein integer ˛ is to one of these; that is, whether ˛ is divisible by a power of �. We introduce notation to capture this idea.

Definition. Define a function �W ZŒ! � f0g ! N as follows: if z 2 ZŒ! is nonzero and n � 0 is the largest integer with �n j z, then �.z/ D n. We call � the valuation.

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8.3 Fermat’s Last Theorem for Exponent 3 351

Thus, �.z/ is the exponent of the highest power of � dividing z: Some treatments define �.0/ to be 1, but we won’t do that here. Also, a

valuation can be defined

in an analogous way for

any prime q in a UFD; just

replace � by q.

�.z/ D n if and only if z D �nz0 and � − z0:

Put another way, ��.z/ j z and ��.z/C1 − z. For example, �.�/ D 1, �.!/ D 0, and �.3/ D 2; in Example 8.31, we saw

that �.61 � !/ D 1. Indeed, �.u/ D 0 for every unit u. The valuation � enjoys some properties that come from the properties of

exponentiation. The next proposition reminds us of Exercise 2.15 on page 59.

Proposition 8.36. If z; w are nonzero elements of ZŒ!, then

(i) �.zw/ D �.z/ C �.w/. (ii) If n is a nonnegative integer, then �.zn/ D n �.z/.

(iii) �.z ˙ w/ � minf�.z/; �.w/g and if �.z/ ¤ �.w/ then

�.z ˙ w/ D minf�.z/; �.w/g:

Proof. This is Exercise 8.30 on page 358.

How to Think About It. Most proofs of the theorem for exponent 3 are

broken into two parts: the first case in which 3 doesn’t divide x, y, or z, and

the second case in which 3 does divide one of them. We’ll follow this program

and treat the two cases in turn. There are many proofs in the literature; our

proof of the first case is not the easiest (see Exercise 8.31 on page 358 for a fairly simple alternative approach), but we choose it because it generalizes to

a proof of the first case for any odd prime exponent p when ZŒ�p has unique

factorization (see Chapter 1 of [36] for the details). Our proof of the second

case is based on the development in Chapter 17 of [17].

The First Case

The main result of this section is that there are no positive integers x; y; z with

gcd.x; y/ D 1 and 3 − xyz such that

x3 C y3 D z3:

Assuming x and y are relatively prime is no loss in generality: a prime factor

of x and y is a prime factor q of z, both sides can be divided by q3, preserv-

ing the relationship; hence, infinite descent would apply. The proof will be by

contradiction, and it will depend on the following lemma.

Lemma 8.37. If x, y, and z are integers such that 3 − xyz, gcd.x; y/ D 1, and x3 C y3 D z3; then the Eisenstein integers As we said on page 349,

this relatively prime condi-

tion leads fairly directly to

the desired proof. x C y; x C !y; x C !2y

are pairwise relatively prime in ZŒ!.

Proof. Suppose that � is a prime in ZŒ! that divides two of the three integers, say

� j x C !iy and � j x C !jy;

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352 Chapter 8 Cyclotomic Integers

where 0 � i < j � 2. Then � divides the difference

.x C !iy/ � .x C !j y/ D !iy � 1 � !j�i

But, by Exercise 8.21(ii) on page 349, 1 � !j�i is an associate of � D 1 � !, so that � divides

uy .1 � !/ ;

where u is some unit in ZŒ!. Hence, by Euclid’s Lemma, � j y or � D 1 � ! D �. Similarly,

� j !j � x C !iy

� � !i

� x C !j y

� D !ix

� !j�i � 1

� :

And, because !j�i�1 and 1�! are associates, we have � j x or � D 1�! D �. Hence, if � ¤ �, then � j x and � j y. This implies that x and y have a common factor in ZŒ!; thus, by Proposition 8.34, they have a common factor

in Z, contradicting the assumption that x and y are relatively prime. So � D �. Because x C y � .x C !iy/ D y.1 � !i / � 0 mod �,

x C y � x C !iy mod �:

We are assuming that x C !iy � 0 mod �; thus, Lemma 8.35 implies that x C y � 0 mod 3 in Z. But then,

z3 D x3 C y3

� x C y mod 3 (Fermat’s Little Theorem) � 0 mod 3I

that is, 3 j z, which contradicts the hypothesis 3 − xyz.

The hard work is done.

Proposition 8.38 (First Case for Exponent 3). There are no positive integers

x; y; z with gcd.x; y/ D 1 and 3 − xyz such that

x3 C y3 D z3: (8.7)

Proof. Suppose, on the contrary, that we have positive integers x; y; z as in the statement. Factoring the left-hand side of Eq. (8.7), we have

.x C y/.x C !y/.x C !2y/ D z3:

Lemma 8.37 guarantees that the three factors on the left-hand side are relatively

prime. Hence, by unique factorization in ZŒ!, each is a unit times a cube in

that ring (if ˇ is a prime divisor of any factor, say ˇ j xC!y, then ˇ j z3 and, by Euclid’s Lemma, ˇ j z. Hence, ˇ3 j .x C !y): there exists an Eisenstein integer s such that

x C !y D ˙!i s3;

where ˙!i is one of the six units in ZŒ! (i is 0, 1, or 2). We want to look at the equation mod 3. Suppose that s D a C b! with a; b 2 Z. Then

s3 D a3 C 3a2b! C 3ab2! C b3!3

D a3 C 3a2b! C 3ab2! C b3

� a3 C b3 mod 3I

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8.3 Fermat’s Last Theorem for Exponent 3 353

hence, Recall that ˛ � ˇ mod 3 means that there is

ı 2 ZŒ! with 3ı D ˛ � ˇ .x C !y � ˙!in mod 3; (8.8)

where n is a rational integer.

It follows (see Exercise 8.22 on page 349) that

x C !y � ˙!i n mod 3:

But ! D !�1, so that

x C !�1y � ˙!�in mod 3: (8.9)

Eqs. (8.8) and (8.9) can be rewritten as:

!�i .x C !y/ � ˙n mod 3 !i � x C !�1y

� � ˙n mod 3I

hence,

!�i .x C !y/ � !i � x C !�1y

� mod 3:

Multiplying by !i gives

x C !y � !2i � x C !�1y

� mod 3;

and so

x C !y � !2ix � !2i�1y � 0 mod 3: (8.10)

We claim, for each possible value of i , namely 0, 1, or 2, that Eq. (8.10) leads

to a contradiction.

(i) i D 0: Eq. (8.10) becomes

x C !y � x � !�1y � 0 mod 3I

that is, � ! � 1

!

� y � 0 mod 3:

Multiplying both sides by !2 gives

.1 � !/ y � 0 mod 3I

that is, there is some ˛ 2 ZŒ! with �y D 3˛. But 3 D �!2�2, by Eq. (8.6), so canceling � gives y D �!2�˛. Thus, � j y in ZŒ!, so that 3 j y in Z, by Lemma 8.35. This contradicts the hypothesis 3 − xyz.

(ii) i D 1: Eq. (8.10) becomes

x C !y � !2x � !y � 0 mod 3:

Thus, the !y’s drop out, and there is ˛ 2 ZŒ! with x.1 � !2/ D 3˛. But 1�!2 D i�, by Exercise 8.21 on page 349, and so xi� D �!2�2˛. Hence, � j x in ZŒ!, and Lemma 8.35 gives 3 j x in Z, another contra- diction.

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354 Chapter 8 Cyclotomic Integers

(iii) i D 2: Eq. (8.10) becomes

x C !y � !4x � !3y � 0 mod 3:

The left-hand side simplifies to x � !x D �x, because !3 D 1. As in parts (i) and (ii), this leads to 3 j xyz, which is a contradiction.

We conclude that there is no solution to x3Cy3 D z3 of the desired type.

Gauss’s Proof of the Second Case

Gauss gave an elegant proof of the second case of Fermat’s Last Theorem for

exponent 3, and we’ll present it here. It turns out to be convenient to prove a

more general result. The object of this section is to prove the following theo-

rem.

There are no Eisenstein integers u; x; y; z with xyz ¤ 0, u a unit, and 3 a factor of exactly one of x; y, z, such that

This is another example

where it’s easier to do

things in more generality.

The reason for introducing

u comes from the fact that

we allow x, y, and z to

be elements of ZŒ!, so

arithmetic statements are

true up to unit factors.

x3 C y3 D uz3:

The proof, which will use infinite descent, is a consequence of several lem-

mas and propositions. To start with, we can assume, for u; x; y; z as in the

statement, that x, y, and z are not all divisible in ZŒ! by � D 1�! (otherwise there’s a contradiction, for 3 is a divisor in Z of x, y, and z, by Lemma 8.35).

We’ll first prove the theorem in case � − xy but � j z. Since x and y are in- terchangeable in the hypothesis, the remaining case is � − yz but � j x. We’ll see that the theorem is an easy consequence of this.

Example 8.32 shows that every Eisenstein integer ˛ is congruent mod � to 0, 1, or �1. In particular, if � − ˛, then

˛ � ˙1 mod �:

Gauss’s proof requires a lemma that shows how an “extra �” sneaks into the

cube of this congruence.

Lemma 8.39. If ˛ is an Eisenstein integer for which � − ˛, then

˛3 � ˙1 mod �4:

Proof. Let’s first consider the case ˛ � 1 mod �; say

˛ D 1C �ˇ

for some ˇ 2 ZŒ!. Substitute this into the usual factorization in ZŒ!:

˛3 � 1 D .˛ � 1/.˛ � !/.˛ � !2/:

Rewrite the first factor on the right-hand side: ˛ � 1 D 1 C �ˇ � 1 D �ˇ. Next, since 1 � ! D �, we can rewrite the second factor:

˛ � ! D 1C �ˇ � ! D �C �ˇ D �.1 C ˇ/:

Now rewrite the third factor, using Exercise 8.21 on page 349, which says that 1 � !2 D �!2�:

˛ � !2 D 1C �ˇ � !2 D �ˇ � !2� D �.ˇ � !2/:

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8.3 Fermat’s Last Theorem for Exponent 3 355

Therefore,

˛3 � 1 D �3ˇ.1 C ˇ/.ˇ � !2/: (8.11)

Example 8.32 shows that ˇ is 0, �1, or 1 mod �. In each of these cases, we’ll see that there’s an extra factor of � in the expression on the right-hand

side of Eq. (8.11). If ˇ � 0 mod � for some ˇ0 2 ZŒ!, then ˇ D �ˇ0, and the expression begins �4ˇ0. If ˇ � �1C�ˇ0, then the middle factor 1Cˇ equals �ˇ0, which contributes an extra �. If ˇ D 1C�ˇ0, then Exercise 8.21 says that 1 � !2 D �!2�, and the last factor on the right-hand side becomes

ˇ � !2 D 1C �ˇ0 � !2

D 1 � !2 C �ˇ0

D �!2�C �ˇ0

D �.�!2 C ˇ0/:

Therefore, if ˛ � 1 mod �, then ˛3 � 1 is a multiple of �4; that is, ˛3 � 1 mod �4.

The remaining case ˛ � �1 mod � is now easy. We have �˛ � 1 mod �, so that

.�˛/3 � 1 mod �4;

and so ˛3 � �1 mod �4.

Gauss used infinite descent on �.z/ and showed (as we will shortly) that if

there was a solution to x3 C y3 D uz3 of the desired type, then one could find another solution .x0; y0; z0/ of the same type with �.z0/ < �.z/. Iterating this process will eventually contradict the next lemma.

Lemma 8.40. Suppose x3 C y3 D uz3 for nonzero Eisenstein integers x, y, z. If � − xy and � j z, then �2 j z.

Proof. Since � − xy, Euclid’s Lemma in ZŒ! says that � − x and � − y, and so Lemma 8.39 applies to say that both x3 and y3 are congruent to ˙1 mod �4. Hence, reducing x3 C y3 D uz3 mod �4 yields

.˙1/C .˙1/ � uz3 mod �4:

The left-hand side of these congruences is one of 0, 2, or �2. Since � j z and Note that � j z implies that �.z/ � 1.� − 2 (why?), we see that ˙2 are impossible. Thus, 0 � uz3 mod �4, so that

�4 j z3 and �.z3/ D 3�.z/ � 4. But �.z) is an integer; hence, �.z/ � 2 and �2 j z.

Here’s the main piece of the puzzle: the key step for infinite descent. Before we dig into the proof, think about why

this result, combined with

Lemma 8.40, implies

that there is no solution

to x3 C y3 D uz3 in Eisenstein integers with

� − xy and � j z.

Proposition 8.41. Suppose that u is a unit in ZŒ! and x3 C y3 D uz3 for Eisenstein integers x; y; z with � − xy and � j z. Then there exists a unit u0 and x0, y0; z0 2 ZŒ! with � − x0y0 and �.z0/ D �.z/ � 1, such that

� x0 �3 C

� y0 �3 D u0

� z0 �3 :

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356 Chapter 8 Cyclotomic Integers

Proof. Given x; y; z as in the statement, we factor x3 C y3 to get

.x C y/.x C y!/.x C y!2/ D uz3: (8.12)

Lemma 8.40 implies that �2 j z so that � � uz3

� � 6. Hence at least one

factor on the left-hand side of the above equation is divisible by �2. Because

x, y, and z are Eisenstein integers, we can replace y by y! or y!2 without

changing the equation or the claim of the proposition. Hence we can assume,

without loss of generality, that �2 j x C y; that is, �.x C y/ � 2.Recall the role of the factor x C y in the proof of the first case.

Now,

x C y! D .x C y/ � .1 � !/y D x C y � �y

and, since � − y, we have �.�y/ D 1. Hence, �.x C y!/ D 1, by Proposi- tion 8.36(iii). Similarly, �.x C y!2/ D 1. So, applying � to Eq. (8.12) and using Proposition 8.36, we have:

3�.z/ D � .x C y/C � .x C y!/ C � � x C y!2

D � .x C y/C 1C 1;

so that

� .x C y/ D 3�.z/ � 2:

For convenience, let’s call the right-hand side k:

k D 3�.z/ � 2: (8.13)

The factors on the left-hand side of Eq. (8.12) are each divisible by �. We

claim that they can’t have any other common factors. To see this, suppose that

is a prime in ZŒ!, ¤ �. If divided x C y and x C y!, then it would divide their difference, which is �y. By Euclid’s Lemma, j y, but then j x, contradicting the fact that gcd.x; y/ D 1. Hence

gcd.x C y; x C y!/ D �:

The same reasoning shows that the gcd of each of the other pairs of factors

is �. Putting all this together, we have the following equation in ZŒ!:

� x C y �k

� � x C y! �

� � x C y!2

� D u

� z ��.z/

�3 ; (8.14)

where the three factors on the left-hand side are relatively prime [remember

Eq. (8.13): 3�.z/ D k C 2]. Now invoke unique factorization in ZŒ!: the right-hand side of Eq. (8.14)

is a cube, and the left-hand side is a product of three relatively prime factors

(each having no factor �). Hence, each is a cube and, more precisely, there are

units u1; u2; u3 and Eisenstein integers z1; z2; z3 with

x C y �k

D u1 z31 and � − z1,

x C y! �

D u2 z32 and � − z2,

x C y!2 �

D u3 z33 and � − z3.

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8.3 Fermat’s Last Theorem for Exponent 3 357

Clearing fractions, it follows that

x C y D u1�kz31 where � − z1; x C y! D u2�z32 where � − z2; x C y!2 D u3�z33 where � − z3:

Multiply the second of these equations by !, the third by !2, and add them to the first to obtain

0 D u1 �k z31 C u2! �z32 C u3!2 �z33 : Divide both sides by �:

0 D u1 �k�1 z31 C u2! z32 C u3! z33 : We’re almost there. Letting v1 D �u1, v2 D u2!, and v3 D u3!2, we have Note that v1, v2, and v3

are all units. v2z

3 2 C v3z33 D v1�k�1z31 :

Recalling that k D �.z/ � 2, this can be written as

v2z 3 2 C v3z33 D v1

� ��.z/�1z1

�3 ; where � − z1z2z3. (8.15)

Divide both sides by v2, relabel everything, and we have the equation � x0 �3 C v

� y0 �3 D v0

� z0 �3 ;

where � − x0y0, v and v0 are units, and �.z0/ D �.z/ � 1. Now �2 divides the right-hand side of the equation, by Lemma 8.40, so reducing the equation

mod �2 yields

.˙1/C .˙v/ � 0 mod �2: Once again, trying all six Eisenstein units and all possible signs, you can check

that v D ˙1. Replacing y0 by �y0 if necessary, we have � x0 �3 C

� y0 �3 D v0

� z0 �3 ;

where � − x0y0 and �.z0/ D �.z/�1, and this is what we wanted to show.

Proposition 8.42. There are no Eisenstein integers u; x; y; z with u a unit,

� − xy, and � j z, such that x3 C y3 D uz3:

Proof. Suppose such elements u; x; y; z exist. Repeated use of Proposition 8.41

shows that there are elements u0; x0; y0; z0 with �.z0/ < 2. But Lemma 8.40 Once again, we use infinite descent.says that this is impossible.

It remains to settle the case where � j yz. If you’ve held on this long, there’s a relatively simple finish: Given Eisenstein integers u; x; y; z with u a

unit, � j x, and � − yz and x3 C y3 D uz3I

reduce mod �2 to obtain ˙1 � u mod �2. A check shows that u D ˙1. But See Exercise 8.25 on page 349.then

.˙z/3 C .�y/3 D x3; and we can apply Proposition 8.41.

This establishes Gauss’s Theorem.

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358 Chapter 8 Cyclotomic Integers

Theorem 8.43 (Gauss). There are no Eisenstein integers u; x; y; z with u

a unit, xyz ¤ 0, and 3 a factor of exactly one of x; y, and z, such that

x3 C y3 D uz3:

Proof. Since � j 3, the hypothesis in Proposition 8.42 that � is a factor of exactly one of x, y, and z implies that 3 is a factor of exactly one of x, y, and z.

After all this work, we have, as a simple corollary, what we wanted in the

first place.

Theorem 8.44 (Fermat’s Last Theorem for Exponent 3). There are no pos-

itive integers x, y, z such that x3 C y3 D z3.

Proof. Since � j 3, Proposition 8.38 and Theorem 8.43 (with u D 1) cover all the possible cases for x, y, z.

Proving Fermat’s Last Theorem for a given exponent n was split into two

cases, as we have just seen for n D 3; the second case was also divided into two parts. The first case for all n < 100 was proved, around 1806, by Germain.

In 1825, Legendre proved one part of the second case for n D 5, while Dirich- let proved the other part. In 1839, Lamé proved Fermat’s Last Theorem for

exponent n D 7. The level of difficulty increased with the exponent. It was not until Kummer that many exponents were completely settled simultaneously.

Exercises

8.29 Show that none of the six units u in ZŒ! is congruent mod � to 0, 2, or �2. (As usual, � D 1 � !.)

8.30 Prove Proposition 8.36.

8.31 Without using Proposition 8.38, show that there are no integers x, y, z with

3 − xyz such that x3 C y3 D z3 mod 9. This exercise gives an alternative proof of Proposition 8.38.

8.32 Show that there are no integers x, y, z with 5 − xyz such that x5 C y5 D z5 mod 25. This exercise implies Fermat’s Last Theorem for exponent 5 in the

case that 5 − xyz.

8.33 Are there any integers x; y; z with 7 − xyz such that x7 C y7 D z7 mod 49?

8.34 (i) Sketch the graph of x3 C y3 D 1. (ii) Show that the only rational points on the graph are .1; 0/ and .0; 1/.

8.35 Take It Further. Let G be the graph of x3 C y3 D 9. (i) Sketch G.

(ii) Find the equation of the line ` tangent to G at .2; 1/.

(iii) Find the intersection of ` and G.

(iv) Show that there are infinitely many triples of integers .x; y; z/ such that

x3 C y3 D 9z3:

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8.4 Approaches to the General Case 359

8.4 Approaches to the General Case

Almost all attempts to prove there are no positive integers x; y; z satisfying

xpCyp D zp , where p is an odd prime, divided the problem in half. The first case assumes that gcd.x; y/ D 1 and p − xyz; the second case assumes that exactly one of x, y, and z is divisible by p.

Our choice of proof for exponent 3 contains some of the main ingredients

of a proof of the first case for any odd prime p, provided that the ring ZŒ�p,

where �p D cos.2�=p/C i sin.2�=p/, is a UFD. Once again, this is based on the factorization in Exercise 3.50 on page 115:

xp C yp D .x C y/.x C �py/ : : : .x C �p�1p y/ (8.16)

The basic idea is to use the fact, in a UFD, that if a product of relatively prime elements is a pth power, then each of its factors is also a pth power. The proof

is more complicated for large p because, while ZŒ�3 D ZŒ! has only six units, the ring ZŒ�p for p > 3 may have infinitely many units. As we saw

in the proof of Proposition 8.38, much of the argument depends on a careful

analysis of how units enter into the calculations.

The commutative rings ZŒ�p  are called rings of cyclotomic integers, and investigating them has played an important part of the story of Fermat’s Last

Theorem, well into the 20th century. We’ll start this section with a brief sketch

of arithmetic in ZŒ�p, pointing to some major results, perhaps without proof,

that generalize results we’ve already established for ZŒ!.

After that, we’ll sketch the work of Kummer that deals with the situation when unique factorization fails. While these efforts didn’t lead him to a proof

of Fermat’s Last Theorem, they did lead to some ideas that have had real stay-

ing power in algebra. One of them is his introduction of ideals as an important

structural component of a commutative ring (Kummer called them divisors),

not merely as subsets that happen to arise, say in studying gcd’s. Another im- portant idea is that of class number, a measure of how far off ZŒ�p is from

having unique factorization.

Here is a biography of Kummer we have adapted from that given in the

history archives of the School of Mathematics and Statistics of the University

of St. Andrews in Scotland.

Ernst Eduard Kummer was born in Sorau, Prussia, in 1810. He entered the University of Halle in 1828 with the intention of studying Protestant theology,

but he received mathematics teaching as part of his degree which was designed

to provide a proper foundation to the study of philosophy. Kummer’s mathe-

matics lecturer H. F. Scherk inspired his interest in mathematics, and Kummer

soon was studying mathematics as his main subject.

In 1831 Kummer was awarded a prize for a mathematical essay he wrote on a topic set by Scherk. In the same year he was awarded his certificate enabling

him to teach in schools and, on the strength of his prize-winning essay, he

was awarded a doctorate. In 1832, Kummer was appointed to a teaching post

at the Gymnasium in Liegnitz, now Legnica in Poland. He held this post for

ten years, where he taught mathematics and physics. Some of his pupils had great ability and, conversely, they were extremely fortunate to find a school

Joachimsthal was famed

for the high quality of his

lectures. His colleagues

in Berlin included many

famous mathematicians

such as Eisenstein, Dirich-

let, Jacobi, Steiner, and

Borchardt.

teacher of Kummer’s quality and ability to inspire. His two most famous pupils

were Kronecker and Joachimsthal and, under Kummer’s guidance, they began

mathematical research while at school, as did Kummer himself. He published

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360 Chapter 8 Cyclotomic Integers

a paper on hypergeometric series in Crelle’s Journal in 1836, which he sent to

Jacobi, and this led to Jacobi, and later Dirichlet, corresponding with Kummer.

In 1839, although still a school teacher, Kummer was elected to the Berlin Academy on Dirichlet’s recommendation. Jacobi now realized that he had to

find Kummer a university professorship.

In 1842, with strong support from Jacobi and Dirichlet, Kummer was ap-

pointed a full professor at the University of Breslau, now Wroclaw in Poland,

where he began research in number theory. In 1855, Dirichlet left Berlin to succeed Gauss at Göttingen, and he recommended that Berlin offer the va-

cant chair to Kummer, which they did. The clarity and vividness of Kummer’s

presentations brought him great numbers of students—as many as 250 were

counted at his lectures. Kummer’s popularity as a professor was based not

only on the clarity of his lectures but on his charm and sense of humor as well. Moreover, he was concerned for the well-being of his students and willingly

aided them when material difficulties arose.

During Kummer’s first period of mathematics, he worked on function the-

ory. He extended Gauss’s work on hypergeometric series, giving developments

that are useful in the theory of differential equations. He was the first to com-

pute the monodromy groups of these series. In 1843 Kummer, realizing that at- tempts to prove Fermat’s Last Theorem broke down because the unique factor-

ization of integers did not extend to other rings of complex numbers, attempted

to restore the uniqueness of factorization by introducing “ideal” numbers. Not

only has his work been most fundamental in work relating to Fermat’s Last

Theorem, since all later work was based on it for many years, but the con- cept of an ideal allowed ring theory, and much of abstract algebra, to develop.

The Paris Academy of Sciences awarded Kummer the Grand Prize in 1857 for

this work. Soon after, he was elected to membership of the Paris Academy of

Sciences and then, in 1863, he was elected a Fellow of the Royal Society of

London. Kummer received numerous other honors in his long career; he died in 1893.

Cyclotomic integers

We shall assume throughout this section that p is an odd prime and � D �p D cos.2�=p/C i sin.2�=p/. Recall some facts about Q.�/.

(1) irr.�;Q/ D p̂.x/ D 1 C x C x2 C � � � C xp�2 C xp�1 (Theorem 6.68 and Exercise 7.31 on page 300).

(2) � Q.�p/ W Q

� D p � 1 (Exercise 7.32 on page 300).

(3) xp � 1 D .x � 1/.x � �/.x � �2/ : : : .x � �p�1/ (Exercise 6.46(i) on page 269).

(4) Q.�/ Š QŒx= � p̂.x/

� (Theorem 7.25(i)).

We recall Proposition 7.20(v), which we now state as a lemma for your convenience.

Lemma 8.45. Let p be an odd prime and � D �p be a pth root of unity. A basis for Q.�/ as a vector space over Q is

B D 1; �; �2; : : : ; �p�2:

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8.4 Approaches to the General Case 361

The ring ZŒ� � Q.�/ is thus the set of all linear combinations Pp�2 iD0 ai�

i

with ai 2 Z. It shares many of the algebraic properties of the Gaussian and Eisenstein integers except, alas, it is not always a UFD (more about this in the next section). But there are analogs for the laws of decomposition that we

developed in ZŒi  and in ZŒ!. Recall, for example, that there is equality of

ideals in ZŒi :

.2/ D .1 � i/2;

and also in ZŒ!,

.3/ D .1 � !/2:

It turns out that the ideal .p/ ramifies in ZŒ� in a similar way. Let’s look into

this.

Lemma 8.46. If p is an odd prime and � D �p is a pth root of unity, then

p D p�1Y

kD1 .1 � �k/:

Proof. Since xp � 1 D .x � 1/.x � �/.x � �2/ : : : .x � �p�1/, we have

xp � 1 x � 1 D

p�1Y

kD1 .x � �k/:

But .xp � 1/=.x � 1/ D p̂.x/ D 1C x C � � � C xp�1; so that

1C x C � � � C xp�1 D p�1Y

kD1 .x � �k/:

Now put x D 1.

Lemma 8.46 gives a factorization of p in ZŒ� into p factors. Our next goal

is to show that the factors are all associates.

Proposition 8.47. If s; t 2 N and p − st , then 1��s and 1��t are associates in ZŒ�.

Proof. In the field Fp, let t D r�1s, so that t r � s mod p. Then �tr D �s (why?), and so

1 � �s 1 � �t

D 1 � � tr

1 � �t

D 1 �

� �t �r

1 � �t

D 1C �t C � �t �2 C � � � C

� �t �r�1

:

Hence, .1 � �t /=.1 � �s/ 2 ZŒ�, and

.1 � �t / j .1 � �s/

in ZŒ�. A similar argument shows that .1 � �s/ j .1 � �t / in ZŒ�. It follows from Proposition 6.6 that 1 � �s and 1 � �t are associates in ZŒ�.

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362 Chapter 8 Cyclotomic Integers

As an immediate consequence, we can generalize the fact that there is equal-

ity of ideals: .3/ D .1 � !/2 . The next result says that p ramifies in ZŒ�.

Corollary 8.48. In ZŒ�, there is a unit u such that

p D u.1 � �/p�1 ;

which gives a factorization of ideals

.p/ D .1 � �/p�1 :

Proof. Lemma 8.46 shows, as elements of ZŒ�, that

p D p�1Y

kD1 .1 � �k/ D .1 � �/

� 1 � �2

� : : : � 1 � �p�1

� : (8.17)

Proposition 8.47 shows that there is a unit uk (2 � k � p � 1) so that

1 � �k D uk.1 � �/:

Factoring out the units from Eq. (8.17) and writing their product as u, we see

that

p D u.1 � �/p�1 :

Hence, we have equality of ideals in ZŒ�:

.p/ D .1 � �/p�1:

Corollary 8.49. If s; t 2 N and p − st , then

1 � �s 1 � �t

is a unit in ZŒ�.

Proof. Since 1 � �s and 1 � �t are associates, there is a unit u in ZŒ� with 1 � �s D u.1 � �t /, so .1 � �s/=.1 � �t / D u.

How to Think About It. Since �p D 1, every integer power of � occurs among

1; �; �2; : : : ; �p�1 :

In particular, if 1 � s � p � 1, then ��s D �p�s . We can calculate in ZŒ� by calculating in

ZŒx=.xp�1 C xp�2 C � � � C 1/:

This allows us to use a CAS to do calculations and then to translate to ZŒ� via

the map f .x/ 7! f .�/.

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8.4 Approaches to the General Case 363

There are other units in ZŒ� that are real numbers; Corollary 8.49 gives a

way to produce them.

Proposition 8.50. The real number

� C ��1 D 2 cos � 2�

p

is a unit in ZŒ�.

Proof. Exercise 6.46 on page 269 shows that, in ZŒ�Œx,

xp C 1 D .x C 1/.x C �/.x C �2/ � � � .x C �p�1/: Put x D ��1 to find that

� ��1

�p C 1 D .��1 C 1/.��1 C �/.��1 C �2/ � � � .��1 C �p�1/: (8.18) The second factor on the right-hand side of Eq. (8.18) is the focus of the

proposition. The left-hand side is 2 (because � ��1

�p D 1). Finally, the last factor on the right-hand side is

��1 C �p�1 D 1 �

C �p�1 D 1C � p

� D 2 �

D 2�p�1:

Hence

2 D 2�p�1.��1 C 1/.��1 C �/.��1 C �2/ � � � .��1 C �p�2/: Dividing both sides by 2, we see that ��1 C � is a unit; in fact, its inverse is:

1 D � ��1 C �

� � �p�1.��1 C 1/.��1 C �2/ � � � .��1 C �p�2/

� :

(The last equation gives us other units besides � C ��1.)

Further results. This is just the beginning.

� Corollary 8.48 is a piece of a law of decomposition in ZŒ�. In ZŒ!, rational primes either stay prime, split, or ramify (and 3 is the only ramified prime).

In ZŒ�p  for p > 3, rational primes can decompose in other ways, but it’s

still true that the way a prime decomposes depends only on its congruence It follows that if q � 1 mod p, then q “splits completely” into p � 1 factors in ZŒ�. What does

this say in ZŒ! (when

p D 3)?

class mod p. Indeed, if q is a prime and f is the smallest integer such that

qf � 1 mod p, then q spits into f prime factors in ZŒ�. This lovely theory is detailed in [5] Chapter 3 and [36] Chapter 2.

� Corollary 8.49 and Proposition 8.50 show how to build units in ZŒ�. This is a piece of a complete classification of units in cyclotomic integers: Kummer

proved that every unit in ZŒ� is a product �s� for some integer s, where

� 2 R D ZŒ� C ��1 (for a proof, see [36], p.3). Since � C ��1 D 2 cos.2�=p/ 2 R;

R is a subring of R, and every unit in ZŒ� is the product of a power of � and

a real unit of ZŒ�. For p D 3, � D ! D 1 2

� �1C i

p 3 �

and

� C ��1 D �1: Hence R D Z, and every unit is a power of ! times a unit in Z, namely ˙1; this recovers the result from Exercise 4.45 on page 165.

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364 Chapter 8 Cyclotomic Integers

The results in this section set the stage for a proof of Fermat’s Last Theorem,

along the same lines as our proof of the theorem for exponent 3, for arbitrary

prime exponents p, as long as ZŒ�p has unique factorization. Kummer did exactly this, for both cases of the theorem (a detailed historical account is

in [23]). As in the case p D 3, the key players are Eq. (8.16), the prime � D 1 � �, and the units �s� where � is a real unit in ZŒ�. We leave the story here, pointing to [5] Chapter 3 for the rest of the technical details.

Exercises

8.36 As usual, let � D cos.2�=p/ C i sin.2�=p/, where p is a rational prime. (i) Show that

� Q.�/ W Q

� � C ��1

�� D 2, and find

irr.�; Q � � C ��1

� :

(ii) What is � Q � � C ��1

� W Q

� ?

8.37 (i) Experiment with various values of p and calculate

p�1Y

iD1

� 1 C �ip

� :

(ii) Find a general formula (for any prime p) for

p�1Y

iD1

� 1 C �ip

� :

8.38 For 1 � s � p � 1 (p a prime), show that

�sp C ��sp

is a unit in ZŒ�p . Is � s p C ��sp a real number?

8.39 In ZŒx= .ˆ5.x//, calculate

(i) x4 � x4 C 1

� � x4 C x2

� � x4 C x3

(ii) � x C x4

� � x3 C x2

(iii) � x C x4

� .1 C x/.

8.40 In ZŒ�5, calculate

(i) �4 � �4 C 1

� � �4 C �2

� � �4 C �3

(ii) � � C �4

� � �3 C �2

(iii) � � C �4

� .1 C �/.

8.41 Write 1 C �5 as the product of a power of �5 and an element of Z.�5 C ��15 /.

8.42 Show that

��15 1 � �35 1 � �5

D sin.3�=5/ sin.�=5/

:

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8.4 Approaches to the General Case 365

Kummer, Ideal Numbers, and Dedekind

It is natural to think that the rings ZŒ�p  are UFDs, as evidenced by the num-

ber of mathematicians in the 17th, 18th, and 19th centuries who assumed it.

Indeed, it’s true for all primes less than 23, but ZŒ�23 does not enjoy unique factorization ([23], p.7). How could so many not know this? It may seem that

23 is not that large, but the calculations in rings of cyclotomic integers are

hefty, even with computers. Imagine the stamina required to calculate by hand

with polynomials of degree 22 in �23. Some of Kummer’s tour-de-force cal-

culations are recounted in [11] Chapter 4. The proof that unique factorization fails in ZŒ�23 is technical (again, see [11], Chapter 4, but the essential idea can

be illustrated in the ring R D ZŒ p

�5 This is a perfectly good commutative ring (of course, R is not a ring of cyclotomic integers):

R D ZŒ p

�5 Š ZŒx= � x2 C 5

� :

If we let ˛ D p

�5, then elements ofR can be written as aCb˛ with a; b 2 Z. If z D a C b˛, then its complex conjugate is, as usual, z D a � b˛. Just as in Gaussian and Eisenstein integers, we can take norms: N.z/ D z z, and we have

N.a C b˛/ D .a C b˛/.a � b˛/ D a2 C 5b2:

The usual properties of norm hold in R: it is multiplicative, the norm of a unit

is 1, and conjugates have the same norm (Exercise 8.43 below).

There are two factorizations of 6 in R:

6 D 2 � 3 D .1 C p

�5/.1 � p

�5/:

We claim that they are essentially different ways to factor 6 into primes. Let’s

see why.

Lemma 8.51. (i) The rational integers 2 and 3 are prime in R D ZŒ p

�5. (ii) 1C ˛ and 1 � ˛ are prime inR.

Proof. (i) If 2 D zw for non-units z and w, then

4 D N.zw/ D N.z/N.w/:

By the Fundamental Theorem in Z (and the fact that neither z nor w is a unit, so that neither has norm 1), N.z/ would be a proper factor of 4, that

is N.z/ D 2. But 2 can’t be written as a2 C 5b2. The proof for 3 uses exactly the same idea.

(ii) If 1C ˛ D zw for non-units z and w, then

6 D N.1 C ˛/ D N.z/N.w/:

By the Fundamental Theorem in Z (and the fact that neither z nor w is a

unit), N.z/ would be a proper factor of 6, say N.z/ D 2. But 2 can’t be written as a2 C 5b2. The proof for 1 � ˛ uses exactly the same idea.

So, we have two factorizations of 6 into primes in R. We’ve seen in other

rings that different-looking factorizations are really the same up to unit factors.

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366 Chapter 8 Cyclotomic Integers

But that doesn’t happen here because neither 2 nor 3 is associate to 1C ˛, for neither has norm 6 D N.1C ˛/. We have a problem!

Kummer was working on methods for factoring cyclotomic integers (not, as it turns out, towards a proof of Fermat’s Last Theorem, but towards another,

related question). He devised a way to think about our problem that actually

shows up in elementary school when children think that 14 � 15 and 10 � 21 are different factorizations of 210. The students are not going far enough in

their factorizations: if they write

14 D 2 � 7 and 15 D 3 � 5;

they see that the “other” factorization is just a rearrangement of the prime

factors of 14 � 15:

10 D 2 � 5 and 21 D 3 � 7:

Now, our problem is different in the sense that we already have prime factor-

izations. But Kummer’s idea was to imagine some “ghost factors” for each of 2, 3, 1C ˛, and 1 � ˛, sort of “super primes” behind the scenes, that could be rearranged to produce the different factorizations. Kummer called these ideal

numbers or divisors, and he imagined there was a further factorization into

ideal numbers J1; J2; J3; J4:

2 D J1J2 3 D J3J4

1C ˛ D J1J3 1 � ˛ D J2J4:

Kummer knew that no such Ji existed in R, but he was able to model these

ghost factors, not as elements of R but as “lists” of elements, each list contain- ing the non-associate divisors of 2, 3, 1 C ˛, and 1 � ˛. And he developed a theory extendingR to a new systemR0 in which there was unique factorization into ideal numbers. Later, Dedekind refined Kummer’s ideas, recasting idealActually, Kummer consid-

ered rings of cyclotomic

integers. We’re using R

here just for the sake of

example.

numbers into what we nowadays call ideals, a notion, as we’ve seen in this

book, that has utility far beyond investigations into Fermat’s Last Theorem.

We’ll use the contemporary notion of ideal to continue our story. The basic idea is that products of elements are replaced by products of ide-

Exercises 5.51 and 5.52

on page 220 define the

product of two ideals and

develop the properties of

the multiplication.

als. In a PID, nothing new is added, because there’s a bijection between ring

elements (up to associates) and principal ideals (Exercise 5.51(ii) on page 220).

But rings that are not UFDs are not PIDs (Theorem 6.50), so there’s a larger

stash of ideals that can enter into factorizations.

Example 8.52. We’ve seen, inR D ZŒ˛, where ˛ D p

�5, that

6 D 2 � 3 D .1 C ˛/.1 � ˛/:

The ghost factors that will resolve our problem are ideals in R generated by

two elements:

J1 D .2; 1C ˛/ D f2aC b.1C ˛/ W a; b 2 Rg J2 D .2; 1� ˛/ D f2aC b.1 � ˛/ W a; b 2 Rg J3 D .3; 1C ˛/ D f3aC b.1C ˛/ W a; b 2 Rg J4 D .3; 1 � ˛/ D f3aC b.1 � ˛/ W a; b 2 Rg:

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8.4 Approaches to the General Case 367

We claim that

Note that these equations

are equalities of ideals, not

numbers.

.2/ D J1J2

.3/ D J3J4 .1C ˛/ D J1J3 .1 � ˛/ D J2J4:

The verifications all use the same method, so we’ll carry it out for the first case

only, leaving the rest for you as Exercise 8.45 below.

Let’s show that .2/ D J1J2. Now the product of two ideals I and J is the set of all linear combinations of products rs where r 2 I and s 2 J (Exercise 5.51 on page 220). So, J1J2 D .2; 1C ˛/.2; 1 � ˛/ is the set of all linear combinations of the form (recall that .1 � ˛/.1C ˛/ D 6):

a.2 � 2/C b .2.1 � ˛//Cc .2.1C ˛//C d.1C ˛/.1 � ˛/ D 4aC 2b.1 � ˛/C 2c.1C ˛/C 6d;

where a; b; c; d 2 R. Well,

4aC 2b.1 � ˛/C 2c.1C ˛/C 6d D 2 Œ2aC b.1 � ˛/C 2c.1C ˛/C 3d  ;

so J1J2 � .2/. And, if .a; b; c; d / D .�1; 0; 0; 1/;

we have

4aC 2b.1 � ˛/C 2c.1C ˛/C 6d D 2; so that .2/ � J1J2. Hence

.2/ D J1J2 as claimed. The other verifications follow in the same way.

Ah, but there’s one glitch. What if one of the four ideals is .1/, the unit

ideal? If J1 D .1/ for example, we’d have .2/ D .1 � ˛/, and we’d still have the same problem. But we can show that none of the Ji is the unit ideal. Let’s

show that J1 ¤ .1/—the arguments for the others are the same (Exercise 8.46 below).

Suppose, on the contrary, that J1 D .2; 1 C ˛/ D .1/. Then there exist elements r C s˛ and t C u˛ in R, where r; s; t; u 2 Z, so that

1 D .r C s˛/ � 2C .t C u˛/.1C ˛/: Multiply this out, using the fact that ˛2 D �5, and write the result as x C y˛ to obtain

1 D .2r C t � 5u/C .2s C t C u/˛: It follows that

2r C t � 5u D 1 2s C t C u D 0:

Replace u by �2s � t in the first equation to obtain 2r � 4t C 10u D 1:

Since the left-hand side is even, this is impossible. N For rings of cyclotomic integers ZŒ�p , it turns

out that this new kind of

factorization into ideals is

unique.

Kummer introduced another brilliant idea. Call two ideals I and J equiva-

lent if there is a cyclotomic integer z so that

I D .z/J D fzb W b 2 J g:

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368 Chapter 8 Cyclotomic Integers

He was able to show that this gives an equivalence relation on nonzero ideals in

ZŒ�p  (for symmetry, the set of all ideals must be enlarged by adding in certain

subsets of Frac.ZŒ�p / D Q.�p/ called fractional ideals). Most importantly, Kummer showed that this relation has only finitely many equivalence classes,

and he called the number h.p/ of them the class number of ZŒ�p. If ZŒ�p 

has class number 1, then all ideals are principal, there is unique factorization,

and our proof of Fermat’s Last Theorem can be refined to produce a proof

for such exponents. In fact, Kummer generalized this, proving that if the class number h.p/ is not divisible byp, then there are no positive integer solutions to

apCbp D cp . This was a monumental achievement. Kummer called primes p such that p − h.p/ regular primes. For example, even though ZŒ�23 doesn’t have unique factorization, 23 is a regular prime—h.23/ D 3, and so Fermat’s Last Theorem holds for it. Alas, there are irregular primes. The smallest is 37, and the next two are 59 and 67. Unfortunately, it is known that there are

infinitely many irregular primes, and it’s unknown whether there are infinitely

many regular primes.

Let’s now say a bit more about Kummer’s ideal numbers (nowadays called

divisors), but we view his idea through the eyes of Dedekind. Take a cyclo-

tomic integer a 2 ZŒ�, and define its divisor

D.a/ D fz 2 ZŒ� W a is a divisor of zg:

Now D.a/ is closed under addition and multiplication by other cyclotomic integers; that is, if z; z0 2 D.a/, then zCz0 2 D.a/; if z 2 D.a/ and r 2 ZŒ�, then rz 2 D.a/. In other words, D.a/ is an ideal (in fact, a principal ideal) in precisely the sense we have been using the term in this book (and we seeIf a; b 2 R, where R

is a commutative ring,

then a j b if and only if D.a/ � D.b/; thus, if R is a domain, then

D.a/ D D.b/ if and only if a and b are associates.

how natural the idea is when viewed in this context). The definition of divisor

makes sense for any commutative ring R, not just for the rings ZŒ�. Now generalize the notion of divisor so that, instead of being a subset of

a commutative ring R of the form D.a/ for some a 2 R, it is a subset of R closed under addition and multiplication by elements ofR; that is, let’s replace

D.a/, which is a principal ideal, by any ideal. Thus, if a; b 2 R, then

D.a/ CD.b/ D fz C w W z 2 D.a/ and w 2 D.b/g

is a generalized divisor. If we denote D.a/ C D.b/ by D.c/, pretending that generalized divisors are just ordinary divisors, then we cannot declare that c is an element of R. Thus, c is a “ghost” element. Of course, if R is a PID, then

c is an element of R, but if R is not a PID, then c may be a creature of our

imagination.

Consider the ring R D ZŒ˛ in Example 8.52, where ˛ D p

�5. The fac- torizations of 6,

6 D 2 � 3 and 6 D .1C ˛/.1 � ˛/;

involve four elements of R, each of which gives a divisor. As in the example,

define

J1 D D.2/ CD.˛/ J2 D D.2/ CD.1 � ˛/ J3 D D.3/ CD.1 C ˛/ J4 D D.3/ CD.1 � ˛/:

We can pretend that there are ghosts ci so that Ji D D.ci / for i D 1; 2; 3; 4.

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8.4 Approaches to the General Case 369

To complete the story, we report that ghosts are primes: the ideals Ji can

be shown to be prime ideals, using the notion of the norm of ideals. Moreover,

one can prove that factorizations in terms of such ghosts are unique, using fractional ideals.

How to Think About It. One of the contributions of Fermat’s Last Theorem to algebra is that it attracted mathematicians of the first order and, as they

studied it, they enhanced the areas of mathematics impinging on it. For algebra

in particular, it brought the idea of commutative rings, factorization, and unique

factorization to the forefront. Kummer’s recognition that unique factorization

was not always present, and his restoration of it with his “ideal numbers,” led Dedekind to introduce ideals into the study of rings. Dedekind’s notion of ideal

was taken up by Hilbert and then later by Emmy Noether. It is today one of the

most fundamental ideas in modern algebra.

We have a confession to make. Our discussion in Chapter 6 explains par-

allels of the arithmetic of polynomials with coefficients in a field k with the

arithmetic of integers by saying that both kŒx and Z are PIDs. No doubt, our ancestors were aware of the analogy between these two systems, but viewing

them in terms of ideals is a modern viewpoint, after Dedekind, dating from

the 1920s. We wrote Chapter 6 using contemporary ideas because it unifies the

exposition.

Richard Dedekind was born in 1831 in Braunschweig (in what is now Ger-

many). He entered the University of Göttingen in 1850; it was a rather disap- pointing place to study mathematics at the time, for it had not yet become the

vigorous research center it turned into soon afterwards. Gauss taught courses

in mathematics, but mostly at an elementary level. Dedekind did his doctoral

work under Gauss’s supervision, receiving his doctorate in 1852; he was to be

the last pupil of Gauss. In 1854, both Riemann and Dedekind were awarded their habilitation de-

grees within a few weeks of each other. Dedekind was then qualified as a uni-

versity teacher, and he began teaching at Göttingen. Gauss died in 1855, and

Dirichlet was appointed to fill the vacant chair. This was an extremely impor-

tant event for Dedekind, who found working with Dirichlet extremely prof- itable. He attended courses by Dirichlet, and they soon became close friends;

the relationship was in many ways the making of Dedekind, whose mathemat-

ical interests took a new lease on life with their discussions. Around this time

Dedekind studied the work of Galois, and he was the first to lecture on Galois

theory when he taught a course on the topic at Göttingen.

In the spring of 1858, Dedekind was appointed to the Polytechnikum in Zurich. It was while he was thinking how to teach differential and integral

calculus that the idea of a Dedekind cut came to him. His idea was that every

real number r divides the rational numbers into two subsets, namely those

greater than r and those less than r . Dedekind’s brilliant idea was to represent

the real numbers by such divisions of the rationals. The Collegium Carolinum in Brunswick had been upgraded to the Brunswick

Polytechnikum by the 1860s, and Dedekind was appointed there in 1862. He

returned to his home town, remaining there for the rest of his life, retiring in

1894. Dedekind died in 1916.

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370 Chapter 8 Cyclotomic Integers

Dedekind made a number of highly significant contributions to mathematics

and his work would change the style of mathematics into what is familiar to us

today. One remarkable piece of work was his redefinition of irrational numbers in terms of Dedekind cuts, as we mentioned above. His work in number the-

ory, particularly in algebraic number fields, is of major importance. He edited

Dirichlet’s lectures, and it was in their third and fourth editions [8], published

in 1879 and 1894, that Dedekind wrote supplements in which he introduced

the notion of an ideal. Dedekind’s work was quickly accepted, partly because of the clarity with which he presented his ideas.

Dedekind’s brilliance consisted not only of the theorems and concepts that

he studied but, because of his ability to formulate and express his ideas so

clearly, his new style of mathematics has been a major influence on mathe-

maticians ever since. The full proof of Fermat’s Last Theorem had to wait for much more pow-

erful methods, developed in the latter half of the 20th century. More about this

in the next chapter.

Exercises

8.43 Let R D ZŒ p

�5 and let N W R ! Z be the norm map: N.z/ D z z. Show that (i) N.zw/ D N.z/N.w/ for all z; w 2 R.

(ii) u is a unit in R if and only if N.u/ D 1. (iii) If z 2 R, N.z/ D N.z/. (iv) If a 2 Z, N.a/ D a2.

8.44 Find all the units in R D ZŒ p

�5.

Figure 8.1. The front of a T-shirt.

8.45 Referring to Example 8.52, show that

.3/ D J3J4 .1 C ˛/ D J1J3 .1 � ˛/ D J2J4:

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8.5 Connections: Counting Sums of Squares 371

8.46 Referring to Example 8.52, show that none of J2; J3; J4 is the unit ideal in R.

8.47 Referring to Example 8.52,

(i) The ideal generated by the norms of elements in J1 is an ideal in Z, and hence

is principal. Find a generator for it.

(ii) Do the same for the other ideals Ji (i D 2; 3; 4).

8.48 Take It Further. Figure 8.1 is the front of a T-shirt that illustrates that

2 � 3 D .1 C p

�5/.1 C p

�5/:

Explain.

8.5 Connections: Counting Sums of Squares

This section investigates an extension of a question we asked in Section 8.2. You saw, in Corollary 8.20, that an odd rational prime can be written as a

sum of two squares if and only if it is congruent to 1 mod 4. What about

composite integers? For example, 15 can’t be written as a2 C b2, but 65 can: 65 D 82 C 12. In fact, 65 can be written as the sum of two squares in another way: 65 D 42 C 72. This leads to the following question:

In how many ways can a positive integer be written as a sum of two We should say “as a sum of two squares of

nonnegative integers.” squares?

The surprising answer to this question was first discovered by Fermat. Just as we used the arithmetic of Eisenstein integers to prove Fermat’s Last Theorem

for exponent 3, we’ll use the arithmetic of Gaussian integers to understand

Fermat’s discovery.

Before continuing, let’s first consider n D 5. Now 5 is a sum of two squares: 5 D 22 C 12. We recognize the norm of a Gaussian integer, for 5 D 22C 12 D N.2C i/. Is there another way to write 5 as a sum of two squares? Recall that 5 splits in ZŒi  as .2 C i/.2 � i/, which suggests writing 5 as N.2 � i/; that is, 5 D 22 C .�1/2. If we agree, when we write n D a2 C b2, that both a and b are nonnegative, then we can ignore the second equation 5 D N.2 � i/. Ah, but there’s another way to write it as N.a C bi/ with both a; b nonnegative. While 2 � i doesn’t have nonnegative real and imaginary parts, it is associate to 2C i , because

i.2 � i/ D 1C 2i I

and 2C i and 1C 2i are not associates (why?). So there are two bonafide non- associate Gaussian integers aCbi with nonnegative a and b and norm 5. Let’s agree, then, that 5 is a sum of two squares in two ways: 22 C 12 and 12 C 22. The following definition of a counting function makes sense.

Definition. The function r W N ! N is defined on nonnegative integers by

r.n/ D the number of non-associate Gaussian integers of norm n:

Since we are interested in equations n D a2 C b2 in which a and b are nonnegative, it is reasonable to restrict our attention to non-associate Gaussian

integers in the first quadrant. By Exercise 8.13 on page 343, every Gaussian

integer is associate to exactly one Gaussian integer in the first quadrant of

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372 Chapter 8 Cyclotomic Integers

the complex plane, We’ve been using the term “first quadrant” throughout the

book, often without properly defining it. We now insist that the positive x-axis

is in the first quadrant but that the positive y-axis is not. The reason is, viewing R2 as C, that we want to find a piece of the complex plane that contains one

Gaussian integer from each class of associates.

Definition. The first quadrant of the complex plane is

Q1 D fz D aC bi 2 C W a > 0; b � 0g:

The real number c is associate to ic (on the imaginary axis), and we don’t include ic in the first quadrant. Also, the origin 0 D 0C 0i does not lie in Q1.

In light of these remarks, we modify the definition of r.n/ for the purpose at

hand, counting only Gaussian integers in the first quadrant (as we wrote above,

Exercise 8.13 says that two such integers are necessarily not associate).

r.n/ D jfz 2 ZŒi  j N.z/ D n and z 2 Q1gj.

It’s worth calculating r.n/ to get a feel for what it is counting. High school

students should enjoy working out some of these numbers by hand (there is no need to mention machinery of ZŒi ). Here are some values for them to check.

You can also check that

r.15625/ D 7 and, if you are ambitious, that

r.815730721/ D 9.

n r.n/

1 1

2 1

3 0

4 1

5 2

6 0

7 0

8 1

9 1

10 2

n r.n/

11 0

12 0

13 2

14 0

15 0

16 1

17 2

18 1

19 0

20 2

n r.n/

21 0

22 0

23 0

24 0

25 3

26 2

27 0

28 0

29 2

30 0

n r.n/

31 0

32 1

33 0

34 2

35 0

36 1

37 2

38 0

39 0

40 2

n r.n/

41 2

42 0

43 0

44 0

45 2

46 0

47 0

48 0

49 1

50 3

Look for regularity in the table, make some conjectures, and try to prove them.

For example, can you see anything that the values of n for which r.n/ D 0 have in common?Or, pick a few primes p,

say 3; 5; 7; 11; 13 and

see what happens as

you calculate the values

of r.pk/. What about

products of two primes?

It’s likely that Fermat did exactly these kinds of investigations—lots of pur-

poseful numerical calculations—to arrive at an amazing result that we’ll prove in this section.

Theorem 8.53 (Fermat). The number of ways an integer n can be written as

a sum of two squares is the excess of the number of divisors of n of the form

4k C 1 over the number of divisors of n of the form 4k C 3I that is, if A.n/ D the number of divisors of n of form 4k C 1

and

B.n/ D the number of divisors of n of form 4k C 3; then

r.n/ D A.n/ � B.n/:

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8.5 Connections: Counting Sums of Squares 373

The proof that we’ll develop uses some new machinery as well as the law

of decomposition for Gaussian integers. First, a few examples that show some

of the delightful consequences of the theorem.

Example 8.54. (i) Consider n D 65 D 13 � 5. Its divisors are

1; 5; 13; 65:

There are four divisors congruent to 1 mod 4 and none congruent to

3 mod 4, so r.65/ D 4. Sure enough,

65 D 1C 64 D 64C 1 D 16C 49 D 49C 16:

(ii) Let n D 21. Its divisors are 1; 3; 7; 21:

There are two divisors that are 1 mod 4 and two that are 3 mod 4, so

r.21/ D 0. Thus, 21 is not a sum of two squares. (iii) Let n D 3m for some integer m. Odd powers of 3 are congruent to

3 mod 4, while even powers are 1 mod 4. The divisors of 3m are

1; 3; 32; 33; : : : ; 3m:

It follows that See Exercise 8.49 on page 377.

r .3m/ D ( 0 ifm is odd

1 ifm is even: N

Corollary 8.55. For any positive integer n, we have A.n/ � B.n/I that is, n has at least as many divisors of the form 4k C 1 as it has divisors of the form 4k C 3.

Proof. By Theorem 8.53, we have A.n/ � B.n/ D r.n/, and r.n/ � 0.

A Proof of Fermat’s Theorem on Divisors

Let’s now prove Theorem 8.53. Our proof requires a device that finds applica-

tions all over mathematics—a theory developed by Dirichlet that uses formal Once again, it’s time to pull out the pencil and paper.algebra to answer combinatorial questions in arithmetic.

Definition. A formal Dirichlet series is an expression of the form

1X

nD1

a.n/

ns D a.1/C a.2/

2s C a.3/

3s C � � � ;

where the a.n/ are complex numbers. (It will be useful to write a.n/ instead of the usual an.)

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374 Chapter 8 Cyclotomic Integers

The word “formal” is important here—we think of these series as book-

keeping devices keeping track of combinatorial or numerical data (as in Ex-

ample 2.31). So, we don’t worry about questions of convergence; we think of s simply as an indeterminate rather than as a variable that can be replaced by

a real or complex number. This misses many of the wonderful analytic appli-. . . [to] omit those parts of the subject, however,

is like listening to a

stereo broadcast of,

say Beethoven’s Ninth

Symphony, using only the

left audio channel. [39].

cations of such series, but it turns out that their formal algebraic properties are

all we need for this discussion.

Dirichlet series are added and multiplied formally. Addition is done term by term:

1X

nD1

a.n/

ns C

1X

nD1

b.n/

ns D

1X

nD1

a.n/ C b.n/ ns

:

Multiplication is also done term by term, but then one gathers up all terms with the same denominator. So, for example, if we’re looking for c.12/=12s inDirichlet series are not

formal power series, and

multiplication is not the

same as in CŒŒx.

1X

nD1

a.n/

ns

1X

nD1

b.n/

ns D

1X

nD1

c.n/

ns ;

then a denominator of 12s could come only from the products

a.1/

1s � b.12/ 12s

; a.2/

2s � b.6/ 6s

; a.3/

3s � b.4/ 4s

; a.4/

4s � b.3/ 3s

; a.6/

6s � b.2/ 2s

; a.12/

12s � b.1/ 1s

:

In general, the coefficient c.n/ in Eq. (8.5) is given by

c.n/ D X

d jn a.d/ � b

� n d

� ; (8.19)

where P d jn means that the sum is over the divisors of n.

The simplest Dirichlet series is the Riemann zeta function:

�.s/ D 1X

nD1

1

ns :

Eq. (8.19) implies that ifActually, the zeta function usually means the function

of a complex variable s that

analytically continues this

infinite series.

�.s/

1X

nD1

a.n/

ns D

1X

nD1

c.n/

ns ;

thenEq. (8.20) is, as we’ll see, extremely useful, and it’s

the reason for defining

Dirichlet series. c.n/ D

X

d jn a.d/: (8.20)

Let’s state this as a theorem.

Theorem 8.56. If

�.s/

1X

nD1

a.n/

ns D

1X

nD1

c.n/

ns ;

then c.n/ D P d jn a.d/:

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8.5 Connections: Counting Sums of Squares 375

Proof. Expand

1X

nD1

1

ns

! 1X

nD1

a.n/

ns

!

and gather terms with the same denominator.

Sometimes, the coefficients a.n/ have interesting properties. For example,

Definition. A function aW N ! C is strongly multiplicative if, for all nonneg- ative integersm; n,

a.mn/ D a.m/a.n/:

(A function aW N ! C is multiplicative if a.mn/ D a.m/a.n/ whenever gcd.m; n/ D 1.)

When a is strongly multiplicative, the Dirichlet series with coefficients a.n/

has an alternate form that shows its connection with arithmetic.

Theorem 8.57. If a is a strongly multiplicative function, then the Dirichlet

series

f .s/ D 1X

nD1

a.n/

ns

has a product expansion

f .s/ D Y

p

1

1 � a.p/ ps

! ;

where the product is over all prime numbers p.

Proof. Each factor on the right side is a geometric series: To be rigorous, we should put some restrictions on

the values of a.k/ to

ensure that the series

converges.

1

1 � a.p/ ps

D 1C � a.p/

ps

� C � a.p/

ps

�2 C � a.p/

ps

�3 C : : :

D 1C � a.p/

ps

� C � a.p2/

p2s

� C � a.p3/

p3s

� C : : : :

Multiply these together (one for each prime) and you get the sum of every

possible expression of the form

a.p e1 1 /a.p

e2 2 / : : : a.p

er r /

p e1s 1 p

e2s 2 : : : p

ers r

D a.p

e1 1 p

e2 2 : : : p

er r /�

p e1 1 p

e2 2 : : : p

er r

�s :

Since every n 2 Z can be written in one and only one way as a product of powers of primes (the fundamental theorem again), this is the same as the sum

1X

nD1

a.n/

ns :

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376 Chapter 8 Cyclotomic Integers

Example 8.58. (i) The constant function a.n/ D 1 is strongly multiplica- tive, so the Riemann zeta function has a product expansion

�.s/ D 1X

nD1

1

ns D Y

p

1

1 � 1 ps

:

(ii) Here’s a multiplicative function that’s connected to our work with Gaus-

sian integers:

�.n/ D

8 <̂

1 if n � 1 .mod 4/ �1 if n � 3 .mod 4/ 0 if n is even

:

You can check that � is strongly multiplicative, and so� is called a quadratic character.

1X

nD1

�.n/

ns D Y

p

1

1 � �.p/ ps

! : N

Now, by Theorem 8.56, if

�.s/

1X

nD1

�.n/

ns D

1X

nD1

a.n/

ns then a.n/ D

X

d jn �.d/: (8.21)

So, a.n/ is the excess of the number of divisors of n of the form 4kC1 over the number of divisors of n of the form 4kC 3. Bingo: this is exactly the function that is the at heart of Theorem 8.53. The idea, then, is to form the Dirichlet series with coefficients r.n/ and to show that

1X

nD1

r.n/

ns D �.s/

1X

nD1

�.n/

ns :

To do this, we’ll convert each of the sums to products. We already have done

this in Example 8.58 for the sums on the right-hand side; for the left-hand side,

we argue as follows.

Each term in the left-hand sum is a sum of unit fractions, and the number of such fractions is the number of Gaussian integers with given norm. For

example, 3=25s comes from

1

N.3C 4i/ C 1

N.4C 3i/ C 1

N.5C 0i/ :

Using this idea and the multiplicativity ofN , we get a product formula for the

left-hand side.

1X

nD1

r.n/

ns D

X

˛2Q1

1

.N.˛//s

D Y

p2Q1

1X

kD0

1� .N.p//k

�s (use the fundamental theorem in ZŒi /

D Y

p2Q1

1

1 � 1 N.p/s

(sum a geometric series) :

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8.5 Connections: Counting Sums of Squares 377

Here, the product is over all Gaussian primes in the first quadrant. This is

another example that is best understood by calculating a few coefficients by

hand. Now use Theorem 8.21 (the law of decomposition for ZŒi ). Every prime in

Q1 lies over one of these:

� the prime 2. There’s only one in the first quadrant: 1C i , and N.1C i/ D 2. � a prime p congruent to 1 mod 4. There are two for each such p—if

p D � �;

then both � and � have an associate inQ1 (and they are different), and each

has norm p.

� a primep congruent to 3 mod 4. There’s only one such prime inQ1, because such a p is inert and N.p/ D p2.

So,

1X

nD1

r.n/

ns D

Y

z2Q1

1

1 � 1

N.z/s

D 1

1 � 1

2s

� Y p�

1 mod 4

1

1 � 1

ps

�2� Y p�

3 mod 4

1

1 � 1 p2s

D 1

1 � 1

2s

� Y p�

1 mod 4

1

1 � 1 ps

�2� Y p�

3 mod 4

1

1 � 1 ps

�� Y p�

3 mod 4

1

1 C 1 ps

D 1

1 � 1 2s

�Y p odd

1

1 � 1 ps

�� Y p�1 mod 4

1

1 � 1 ps

�� Y p�3 mod 4

1

1 C 1 ps

D �.s/

� Y p�1 mod 4

1

1 � �.p/ ps

�� Y p�3 mod 4

1

1 � �.p/ ps

D �.s/

�Y p odd

1

1 � �.p/ ps

� D �.s/

1X

nD1

�.n/

ns

D 1X

nD1

a.n/

ns ;

where, by Eq. (8.21),

a.n/ D X

d jn �.d/:

It follows that r.n/ D a.n/, and we’ve proved Theorem 8.53.

Exercises

8.49 Suppose that m � 1 is an integer. Show that if p is a prime and p � 3 mod 4,

r � pm

� D (

0 if m is odd

1 if m is even:

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378 Chapter 8 Cyclotomic Integers

8.50 Suppose that m � 1 is an integer. Show that if p is a prime and p � 1 mod 4,

r � pm

� D m C 1:

8.51 Suppose that m � 1 is an integer. Show that

s � 2m �

D 1:

8.52 Show that if a is a strongly multiplicative function, so is b, where b is defined by

b.n/ D X

d jn a.d/:

8.53 A multiplicative function is a function a W N ! C so that a.mn/ D a.m/a.n/ whenever gcd.m; n/ D 1. (i) Give an example of a multiplicative function that is not strongly multiplica-

tive.

(ii) Show that if a is a multiplicative function, so is b, whereIf a is strongly multiplica- tive, is b?

b.n/ D X

d jn a.d/:

8.54 Show that, if gcd.m; n/ D 1, then

r.mn/ D r.m/r.n/:

8.55 Show that an integer can be written as a sum of two squares if and only if the

primes in its prime factorization that are congruent to 3 mod 4 show up with even

exponents.

8.56 Take It Further. Show that

�.s/

1X

nD1

�.n/

ns D �.s � 1/;

where � is the Euler �-function.

8.57 Take It Further. Tabulations of r show some erratic behavior with no apparent

pattern. When that happens with a function f , it’s often useful to look at theIt might be easier to see things if you use 4r

instead of r , allowing

Gaussian integers in all

four quadrants.

asymptotic behavior of its average value:

lim n!1

1

n

nX

kD1 f .k/:

Investigate the asymptotic behavior of the average value of r .

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9 Epilog Attempts to resolve Fermat’s Last Theorem have led to much modern alge-

bra. There were many other areas of mathematical research in the seventeenth,

eighteenth and nineteenth centuries, one of which was determining the roots of

polynomials. Informally, a polynomial is solvable by radicals if its roots can

be given by a formula generalizing the classical quadratic, cubic, and quar-

tic formulas. In 1824, Abel proved that there are quintic polynomials that are not solvable by radicals and in 1828 he found a class of polynomials, of any

degree, that are solvable by radicals. In 1830, Galois, the young wizard who

was killed before his 21st birthday, characterized all the polynomials which

are solvable by radicals, greatly generalizing Abel’s theorem. Galois’ brilliant

idea was to exploit symmetry through his invention of group theory. After a brief account of the lives of Abel and Galois, we will use ring theory

to make the notion of solvability by radicals precise. This will enable us to un-

derstand the work of Abel and Galois showing why there is no generalization

of the classical formulas to polynomials of higher degree. We will then intro-

duce some group theory, not only because groups were the basic new idea in

the study of polynomials, but because they are one of the essential ingredients in Wiles’ proof of Fermat’s Last Theorem in 1995. In fact, symmetry is an im-

portant fundamental idea arising throughout mathematics. In the last section,

we will say a bit about Andrew Wiles and his proof of Fermat’s Last Theorem.

9.1 Abel and Galois

Niels Abel was born in Frindöe, Norway, near Stavanger, in 1802. Norway was then suffering extreme poverty as a consequence of economic problems arising

from European involvement in the Napoleonic wars. Abel’s family was very

poor, although things improved a little in 1816 when his father, a Protestant

minister, became involved in politics (Norway, which had been part of Den-

mark, claimed independence, and then became a largely autonomous kingdom

in a union with Sweden). The next year, Abel was sent to a school in Christiana (present day Oslo), but he was an ordinary student there with poor teachers (the

best teachers having gone to the recently opened University of Christiana). But

two years later, a new mathematics teacher, B. Holmboë, joined the school and So, an inspiring teacher helped set the course of

modern algebra. inspired Abel to study mathematics. Holmboë was convinced Abel had great

talent, and he encouraged him to read the works of contemporary masters. In 1820, Abel’s father died; there was no money for Abel to complete his educa-

tion nor to enter the University. But Holmboë continued his support, helping

him to obtain a scholarship to enter the University in 1821; Abel graduated the

following year.

379

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380 Chapter 9 Epilog

In 1821, while in his final year at the University of Christiana, Abel thought

he had proved that quintic polynomials are solvable by radicals, and he submit-

ted a paper to the Danish mathematician Degen for publication by the Royal Society of Copenhagen. Degen asked Abel to give a numerical example of hisThe importance of nu-

merical examples can’t be

overestimated. method and, in trying to do this, Abel discovered a mistake in his paper. De-

gen had also advised Abel to study elliptic integrals, and Abel wrote several

important fundamental papers on the subject. In 1824, Abel returned to quin-

tic polynomials, proving that the general quintic polynomial is not solvable by radicals.

In 1825, having now done brilliant work in two areas of mathematics, the

Norwegian government gave Abel a scholarship to travel abroad. He went to

Germany and France, hoping to meet eminent mathematicians, but Gauss was

not interested in Abel’s work on the quintic, and the mathematicians in Paris did not yet appreciate his remarkable theorems on elliptic functions. By 1827,

Abel’s health deteriorated, he was heavily in debt, and he returned home to

Norway. In 1828, he briefly returned to polynomials, proving a theorem de-

scribing a class of polynomials (of any degree) that are solvable by radicals.

By this time, Abel’s fame had spread to all mathematical centers. Legendre

saw the new ideas in papers of Abel and of Jacobi, and he wrote

Through these works you two .Abel and Jacobi/ will be placed in the

class of the foremost analysts of our times.

Strong efforts were made to secure a suitable position for Abel by a group from the French Academy, who addressed King Bernadotte of Norway-Sweden;

Crelle also worked to secure a professorship for him in Berlin. But it was too

late. Abel died in 1829, at age 26.

An imprecise measure of Abel’s influence on modern mathematics is the

number of areas named after him: abelian groups, abelian varieties, abelian differentials, abelian integrals, abelian categories, abelian extensions, abelian

number fields, abelian functions. The Niels Henrik Abel Memorial Fund was

established in 2002, and the Norwegian Academy of Science and Letters awards

the Abel Prize for outstanding scientific work.

Évariste Galois was born in Bourg La Reine, near Paris, in 1811. France,

and especially Paris, was then in the throes of great political and social change

as a consequence of the French Revolution in 1789, the Napoleonic era 1799– 1815, the restoration of the French monarchy with King Louis XVIII in 1815,

his overthrow by King Charles X in 1824, and another revolution in 1830.

In April 1829, Galois’ first mathematics paper (on continued fractions) was

published; he was then 17 years old. In May and June, he submitted articles

on the algebraic solution of equations to Cauchy at the Academy of Science.

Cauchy advised him to rewrite his article, and Galois submitted On the condi- tion that an equation be solvable by radicals in February 1830. The paper was

sent to Fourier, the secretary of the Academy, to be considered for the Grand

Prize in mathematics. But Fourier died in April 1830, Galois’ paper was never

subsequently found, and so it was never considered for the prize. July 1830

saw another revolution. King Charles X fled France, and there was rioting in the streets of Paris. Later that year, Galois (now age 19) was arrested for mak-

ing threats against the king at a public dinner, but he was acquitted. Galois

was invited by Poisson to submit a third version of his memoir on equations to

the Academy, and he did so in January 1831. On July 14, Galois was arrested

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9.2 Solvability by Radicals 381

again. While in prison he received a rejection of his memoir. Poisson reported

that “His argument is neither sufficiently clear nor sufficiently developed to al-

low us to judge its rigor. . . . There is no good way of deciding whether a given polynomial . . . is solvable.” He did, however, encourage Galois to publish a

more complete account of his work.

In March 1832, a cholera epidemic swept Paris, and prisoners were trans-

ferred to boarding houses. The prisoner Galois was moved to a pension, where

he apparently fell in love with Stéphanie-Félice du Motel, the daughter of the resident physician. After he was released on April 29, Galois exchanged letters

with Stéphanie, and it is clear that she tried to distance herself from the affair.

Galois fought a duel on May 30, the reason for the duel not being clear but

certainly linked with Stéphanie. A note in the margin of the manuscript that

Galois wrote the night before the duel reads, “There is something to complete in this proof. I do not have the time.” It is this note which has led to the legend

that he spent his last night writing out all he knew about group theory (but this

story appears to have been exaggerated). Galois was mortally wounded in the

duel, and he died the next day, only 20 years old. His funeral was the focus of

a Republican rally, and riots lasting several days followed.

According to Galois’ wish, his friend Chevalier and Galois’ brother Alfred copied Galois’ mathematical papers and sent them to Gauss, Jacobi, and others.

No record exists of any comment these men may have made. Eventually, the

papers reached Liouville who, in September 1843, announced to the Academy

that he had found in Galois’ papers a concise solution “. . . as correct as it is

deep, of this lovely problem: given an irreducible equation of prime degree, decide whether or not it is solvable by radicals.” Liouville published these

papers of Galois in his Journal in 1846. What Galois outlined in these papers

is called Galois Theory today.

The following quotation is from the Epilog of Tignol’s book [33].

After the publication of Galois’ memoir by Liouville, its importance

dawned upon the mathematical world, and it was eventually realized that

Galois had discovered a mathematical gem much more valuable than

any hypothetical external characterization of solvable equations. After

all, the problem of solving equations by radicals was utterly artificial.

It had focused the efforts of several generations of brilliant mathemati-

cians because it displayed some strange, puzzling phenomena. It con-

tained something mysterious, profoundly appealing. Galois had taken

the pith out of the problem, by showing that the difficulty of an equation

was related to the ambiguity of its roots and pointing out how this ambi-

guity could be measured by means of a group. He had thus set the theory

of equations and, indeed, the whole subject of algebra, on a completely

different track.

We have chosen Fermat’s Last Theorem as an organizing principle of this

book, but an interesting abstract algebra text could be written centered on

group theory and roots of polynomials.

9.2 Solvability by Radicals

Informally, a polynomial is solvable by radicals if there is a formula for its

roots that generalizes the classical quadratic, cubic, and quartic formulas. Let

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382 Chapter 9 Epilog

us now examine the classical formulas to make this rather vague idea more

precise.

How to Think About It. Even though much of what we shall say applies

to polynomials over any field, the reader may assume all fields coming up are

subfields of the complex numbers C. We point out, however, that some famil- iar results may not be true for all fields. For example, the quadratic formula

doesn’t hold in kŒx when k has characteristic 2 (for 1 2

doesn’t make sense in

k); similarly, neither the cubic formula nor the quartic formula holds in kŒx

when k has characteristic either 2 or 3.

Definition. A field extension K=k is a pure extension if K D k.u/, where un 2 k for some n � 1.

In more detail, K D k.u/, where u is a root of xn � a for some a 2 k; that is, u D n

p a, and so we are adjoining an nth root of a to k. But there are

several nth roots of a in C, namely

n p a; � n

p a; : : : ; �n�1 n

p a;

where � D �n D cos.2�=n/ C i sin.2�=n/ is a primitive nth root of unity. To avoid having to decide which one to adjoin, let’s adjoin all of them by adjoining any one of them together with all the nth roots of unity. This is reasonable, for

we are seeking formulas for roots of polynomial equations that involve square

roots, cube roots, etc., and roots of numbers appear explicitly in the classical

formulas.

Let’s consider the classical formulas for polynomials of small degree, for we’ll see that they give rise to a sequence of pure extensions.

Quadratics

If f .x/ D x2 C bx C c, then the quadratic formula gives its roots as 1 2

� �b ˙

p b2 � 4c

� :

Let k D Q.b; c/. Define K1 D k.u/, where u D p b2 � 4c. Then K1 is a

pure extension, for u2 2 k. Moreover, the quadratic formula implies thatK1 is the splitting field of f .

Cubics

Let f .X/ D X3 C bX2 C cX C d , and let k D Q.b; c; d /. The change of variable X D x � 1

3 b yields a new polynomial ef .x/ D x3 C qx C r 2 kŒx

having the same splitting field E (for if u is a root of ef , then u � 1 3 b is a root

of f ); it follows that ef is solvable by radicals if and only if f is. The cubic formula gives the roots of ef as

g C h; !g C !2h; and !2g C !h;

where g3 D 1 2

� �r C

p R � , h D �q=3g, R D r2 C 4

27 q3, and ! is a primitive

cube root of unity. Because of the constraint gh D �1 3 q, each choice of g D

3

q 1 2 .�r C

p R/ has a “mate,” namely h D �q=.3g/, �q=.3!g/ D !2h, and

�q=.3!2g/ D !h.

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9.2 Solvability by Radicals 383

DefineK1 D k. p R/, and defineK2 D K1.g/, where g3 D 12 .�r C

p R/.

The cubic formula shows that K2 contains the root g C h of ef , where h D �q=3g. Finally, define K3 D K2.!/, where !3 D 1. The other roots of ef are !g C !2h and !2g C !h, both of which lie inK3, and so E � K3.

Thus, a sequence of pure extensions seems to capture the notion we are seeking. We now give the formal definition of solvability by radicals, after

which we will show that all polynomials of degree � 4 are solvable by radicals.

Definition. A radical extension of a field k is a field extensionK=k for which

there exists a tower of field extensions

k D K0 � K1 � � � � � Kt D K;

where Ki=Ki�1 is a pure extension for all i � 1. A polynomial f .x/ 2 kŒx is solvable by radicals if there is a splitting field

E=k and a radical extension K=k withE � K.

Quadratics are solvable by radicals, and the cubic formula shows that every

cubic f .x/ D x3 C qx C r 2 QŒx is solvable by radicals: a radical extension containing a splitting field of f is

Q.q; r/ D K0 � K1 D K0.!/ � K2 D K1. p R/ � K3 D K2.g/;

where we are using the notation in the cubic formula.

Why do we say in the definition of solvable by radicals, that E � K instead of E D K? That is, why don’t we say that some splitting field is a radical extension? The answer is that it isn’t. Consider the following theorem, due to

Hölder.

Theorem 9.1 (Casus Irreducibilis). Let f .x/ 2 QŒx be an irreducible cubic having three real roots. If E � C is the splitting field of f and K is a radical extension of Q containingE , then K ¨ R.

Proof. [25], p. 217.

If f .x/ 2 QŒx is an irreducible cubic all of whose roots are real (f .x/ D 3x3�3xC 1 is such a cubic, by Example 6.58), then its splitting field E � R. We have just seen that f is solvable by radicals, so there is a radical extension

K=Q with E � K. But the Casus Irreducibilis says that K ¨ R. Therefore, E ¤ K (because E � R); that is, the splitting field of f is not itself a radical extension.

Here is a more remarkable consequence of the Casus Irreducibilis. In down-

to-earth language, it says that any formula for the roots of an irreducible cubic

in QŒx having all roots real requires the presence of complex numbers! After

all, the formula involves a cube root of unity. In other words, it is impossible

to “simplify” the cubic formula to eliminate i ; we must use complex numbers to find real roots! How would this have played in Piazza San Marco?

We now show that quartic polynomials are solvable by radicals.

Proposition 9.2. Every polynomial f .X/ D X4 C bX3 C cX2 C dX C e 2 QŒx is solvable by radicals.

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384 Chapter 9 Epilog

Proof. Let k D Q.b; c; d; e/. The change of variable X D x � 1 4 b yields a

new polynomial ef .x/ D x4 C qx2 C rx C s 2 kŒx; moreover, the splitting field E of f is equal to the splitting field of ef , for if u is a root of ef , then u � 14b is a root of f . Factor ef in CŒx:

ef .x/ D x4 C qx2 C rx C s D .x2 C jx C `/.x2 � jx Cm/;

and determine j , `, and m. Now j 2 is a root of the resolvent cubic:

.j 2/3 C 2q.j 2/2 C .q2 � 4s/j 2 � r2:

The cubic formula gives j 2, from which we can determinem and `, and hence the roots of the quartic.

Define a radical extension

k D K0 � K1 � K2 � K3;

as in the cubic case, so that j 2 2 K3. DefineK4 D K3.j / (so that `; m 2 K4). Finally, defineK5 D K4

�p j 2 � 4`

� andK6 D K5

�p j 2 � 4m

� , giving roots

of the quadratic factors x2CjxC` and x2�jxCm ofef . The quartic formula gives E � K6. Therefore, f is solvable by radicals.

Example 9.3. f .x/ D x5 � 1 2 QŒx is solvable by radicals. We know that f .x/ D .x � 1/h.x/, where h is a quartic. But we have just seen that quartics are solvable by radicals. (Actually, Gauss proved that xn � 1 is solvable by radicals for all n � 1, and this led to his construction of the regular 17-gon by ruler and compass.) N

We have just seen that quadratics, cubics, and quartics in QŒx are solvable

by radicals. Conversely, let f .x/ 2 QŒx be a polynomial of any degree, and let E=Q be a splitting field. If f is solvable by radicals, we claim that there is

a formula that expresses its roots in terms of its coefficients. Suppose thatQ can be replaced by any field of characteristic 0.

Q D K0 � K1 � � � � � Kt

is a radical extension with E � Kt . Let z be a root of f . Now z 2 Kt D Kt�1.u/, where u is an mth root of some element ˛ 2 Kt�1; hence, z can be expressed in terms of u and Kt�1; that is, z can be expressed in terms of m

p ˛

and Kt�1. ButKt�1 D Kt�2.v/, where some power of v lies inKt�2. Hence, z can be expressed in terms of u, v, and Kt�2. Ultimately, z is expressed by a formula analogous to the classical formulas. Therefore, solvability by radicals

has now been translated into the language of fields.

9.3 Symmetry

Recognizing and exploiting symmetry is an important ingredient in geometry,

algebra, number theory, and, indeed, in all of mathematics. Here is the basic idea: an object is symmetric if, when you transform it in a

certain way, you get the same object back. For example, what do we mean

when we say that an isosceles triangle  is symmetric? Figure 9.1 shows

 D ABC with its base AB on the x-axis and with the y-axis being the

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9.3 Symmetry 385

A

C

B

Figure 9.1. Isosceles triangle ABC .

perpendicular-bisector of AB . Close your eyes; pretend that the y-axis is a Reflection in the y-axis is the function .x;y/ 7! .�x;y/. Reflection in the x-axis

is the function .x;y/ 7! .x;�y/.

mirror, and let  be reflected in the y-axis (so that the vertices A and B are

interchanged); open your eyes. You cannot tell that  has been reflected; that is,  is symmetric in the y-axis. On the other hand, if  were reflected in the

x-axis, then it would be obvious, once your eyes are reopened, that a reflection

had taken place; that is,  is not symmetric in the x-axis.

Here is a non-geometric example: the polynomial f .x; y/ D x3C y3 � xy is symmetric because, if you transform it by interchanging x and y, you get

the same polynomial back. Another example arises from g.x/ D x6 � x2 C 3. This polynomial is symmetric because g.�x/ D g.x/; this symmetry induces symmetry of the graph of g.x/ in the y-axis, for .�x; y/ lies on the graph if and only if .x; y/ does.

The transformations involved in defining symmetry are usually permuta-

tions.

Definition. A permutation of a set X is a bijection ˛WX ! X .

Here is a precise definition of symmetry in geometry.

Definition. An isometry of the plane is a function 'W R2 ! R2 that is distance preserving: for all points P D .a; b/ and Q D .c; d / in R2, Now you see how im-

portant the Pythagorean

Theorem really is; it allows

us to define distance. k'.P / � '.Q/k D kP �Qk;

where kP �Qk D p .a � c/2 C .b � d/2 is the distance from P toQ.

A symmetry of a subset  of the plane is an isometry � with �./ D  (by definition, �./ D f�.!/ W ! 2 g).

It is clear that every isometry is an injection, for ifP ¤ Q, then kP�Qk ¤ 0, hence k�.P / � �.Q/k ¤ 0, and �.P / ¤ �.Q/. It is also true (but harder to prove) that isometries are surjections ([26], p. 141). Thus, isometries are

bijections; that is, they are permutations of the plane.

Some figures have more symmetries than others. Consider the triangles in

Figure 9.2. The first, equilateral, triangle has six symmetries: rotations by 120ı, 240ı, 360ı D 0ı about its center, and reflections in each of the three angle bisectors. The second, isosceles, triangle has only two symmetries, the identity

isometry and the reflection in the angle bisector, while the scalene triangle

has only one symmetry, the identity isometry. A circle has infinitely many

symmetries (for example, all rotations about its center).

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386 Chapter 9 Epilog

P

QU

P

Q U

P

QU

Figure 9.2. Triangles.

We now introduce symmetry in an algebraic setting.

Definition. An automorphism of a commutative ring R is an isomorphism

� WR ! R. Given a field extension E=k, an automorphism � WE ! E fixes k if �.a/ D a for every a 2 k.

The following theorem should remind you of Theorem 3.12 (which is the

special case when E=k D C=R and � is complex conjugation). Of course, automorphisms are certain kinds of permutations.

Theorem 9.4. Let k be a field, let f .x/ 2 kŒx, and let E=k be a splitting field of f . If � WE ! E is an automorphism fixing k, then � permutes the set  of all the roots of f .

Proof. Let f .x/ D a0 C a1x C � � � C anxn, where ai 2 k for all i . If u 2 E is a root of f , then f .u/ D 0 and

0 D �.f .u// D �.a0 C a1uC � � � C anun/ D �.a0/C �.a1/�.u/ C � � � C �.an/�.un/ D a0 C a1�.u/ C � � � C an�.u/n

D f .�.u//:

Therefore, �.u/ is also a root of f , so that �./ � ; that is, im .� j/ �  and the restriction � j is a function ! . But � j is an injection, because � is, and the Pigeonhole Principle, Exercise A.11 on page 419, says that it is a

permutation.

The following definition, due to E. Artin around 1930, modernizes and sim-

plifies Galois’ original definition given 100 years earlier (it is equivalent to

Galois’ definition).

Definition. If k is a field, f .x/ 2 kŒx, and E=k is a splitting field of f , then the Galois group of f is

Gal.f / D fautomorphisms � WE ! E fixing kg: Just as some triangles are more symmetric than others, some polynomials

are more symmetric than others. For example, consider f .x/ D x2 � 2 and g.x/ D x2 � 9, where we consider both polynomials as lying in QŒx. The splitting field of f is E D Q.

p 2/, and there is an automorphism � WE ! E

that interchanges the roots p 2 and �

p 2, namely � W aC b

p 2 7! a�b

p 2. On

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9.3 Symmetry 387

the other hand, the splitting field of g is Q, for both 3 and �3 lie in Q, and so Gal.g/ consists only of the identity permutation.

The astute reader may have noticed that Gal.f / really depends only on the fields k and E; two polynomials in kŒx having the same splitting field have

the same Galois group. For this reason, we usually write

Gal.f / D Gal.E=k/:

Example 9.5. We show that not every permutation of the roots of a polyno- mial f is the restriction of some automorphism � 2 Gal.f /. We saw in Ex- ample 3.7 that the roots of

f .x/ D x4 � 10x2 C 1 2 QŒx are

˛ D p 2C

p 3; ˇ D

p 2 �

p 3; D �

p 2C

p 3; ı D �

p 2 �

p 3:

Let E=Q be a field extension containing these four roots, and let � be the

permutation that interchanges ˇ and and fixes the other two roots:

�.˛/ D ˛; �.ˇ/ D ; �. / D ˇ; �.ı/ D ı:

In E , we have ˛ � ˇ D 2 p 3. Suppose there is an automorphism � of E with

� jf˛; ˇ; ; ıg D � . Then �.˛ � ˇ/ D �.2 p 3/, and

�.˛ � ˇ/ D �.˛/ � �.ˇ/ D �.˛/ � �.ˇ/ D ˛ � D 2 p 2:

Hence, �.2 p 3/ D 2

p 2. Square both sides:

�.2 p 3/2 D .2

p 2/2 D 8:

The left-hand side is �.2 p 3/2 D �

� .2

p 3/2 �

D �.12/ D 12, and this is a contradiction. Therefore, � … Gal.f /. N

An important class of symmetric polynomials are the elementary symmet- ric polynomials in n variables ˛1; : : : ; ˛n. For two variables, the elementary

symmetric polynomials are ˛1 C ˛2 and ˛1˛3. For three variables, they are

s1 D ˛1 C ˛2 C ˛3; s2 D ˛1˛2 C ˛1˛3 C ˛2˛3; s3 D ˛1˛2˛3

and, for n variables, si is the sum of all products of the ˛i , taken i at a time.

We’ve met these before in the context of roots of polynomials. The coefficients

of a monic polynomial are the elementary symmetric polynomials of its roots,

with alternating sign, so that, for example,

.x � ˛1/.x � ˛2/.x � ˛3/ D x3 � .˛1 C ˛2 C ˛3/x2 C .˛1˛2 C ˛1˛3 C ˛2˛3/x � ˛1˛2˛3:

It turns out that every symmetric polynomial in n variables can be expressed

as a polynomial in the elementary symmetric polynomials si (see [2], Chap-

ter IIG). For example,

˛21 C ˛22 C ˛23 D .˛1 C ˛2 C ˛3/ 2 � 2.˛1˛2 C ˛1˛3 C ˛1˛3/

D s21 � 2s2:

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388 Chapter 9 Epilog

Example 9.6. The elementary symmetric polynomials can be used to give al-

ternate derivations of the quadratic and cubic formulas (Exercises 9.1 and 9.3

below). Let’s sketch a derivation of the cubic formula along these lines. Assume our cubic has been reduced and is, as before, of the form

x3 C qx C r: Suppose further that we let its roots be ˛1, ˛2 and ˛3. Then we know that

˛1 C ˛2 C ˛3 D 0; ˛1˛2 C ˛1˛3 C ˛2˛3 D q;

˛1˛2˛3 D �r:

Form the two expressions s and u:As usual, ! D 1 2

�1C i p

3 �

. s D ˛1 C ˛2! C ˛3!2

u D ˛1 C ˛2!2 C ˛3!:

So, we have three expressions in the roots

0 D ˛1 C ˛2 C ˛3 s D ˛1 C ˛2! C ˛3!2

u D ˛1 C ˛2!2 C ˛3!:

Adding the equations, we see that s C u D 3˛1. Hence, if s and u can be expressed in terms of q and r , then ˛1 can be so expressed (and, by symmetry,

the other roots can be expressed in terms of q and r ). Experimenting with a

CAS or by hand, we find that

su D ˛21 C ˛22 C ˛23 � ˛1˛2 � ˛1˛3 � ˛2˛3 D .˛1 C ˛2 C ˛3/2 � 3 .˛1˛2 C ˛1˛3 C ˛2˛3/ D 0 � 3q D �3q:

Expanding s3 C u3 and factoring the result, we getA CAS is a great help here (See Appendix A.6).

s3Cu3

D �.˛1 C ˛2 � 2˛3/.˛1 C ˛3 � 2˛2/.˛2 C ˛3 � 2˛1/ D �.˛1 C ˛2 C ˛3 � 3˛3/.˛1 C ˛2 C ˛3 � 3˛2/.˛1 C ˛2 C ˛3 � 3˛1/ D �.�3˛3/.�3˛2/.�3˛1/ D 27˛1˛2˛3 D �27r:

From su D �3q, we get s3u3 D �27q3. Coupled with s3 C u3 D �27r , we see that s3 and u3 are roots of the quadratic polynomial

x2 C 27rx � 27q3: We can solve this for s3 and u3, take cube roots, and recover ˛1, leading to Cardano’s formula (Exercise 9.3 below). N

Exercises

9.1 * Suppose the roots of x2Cbx Cc are ˛ and ˇ. Find, without using the quadratic formula, an expression for and ˛ � ˇ in terms of ˛ C ˇ and ˛ˇ . Use it and the

Exercise 9.1 shows how

to derive the quadratic

formula without completing

the square. fact that ˛ C ˇ D �b to find ˛ in terms of b and c.

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9.4 Groups 389

9.2 Find the roots of the cubic from page 83,

x3 � 18x � 35;

using the method of Example 9.6.

9.3 * Finish the derivation of the cubic formula outlined in Example 9.6.

9.4 Groups

Galois invented groups to exploit symmetry. Our purpose here is only to dis-

play Galois’ ideas in enough detail so that Theorem 9.16 below is plausible;

we wish to dispel some of the mystery that would arise if we merely cited the

ultimate result (you can follow the proofs in [26], Chapter 5).

Commutative rings are sets with two binary operations; a group is a set having only one binary operation. Permutations, as any functions from a set

X to itself, can be composed and, as we show in Appendix A.1, composition

equips the family of all permutations of X with a a binary operation. This

viewpoint begets a kind of algebra, called group theory.

Definition. A group is a nonempty set G with a binary operation

�WG �G ! G;

where �W .a; b/ 7! a � b, satisfying the following properties:

(i) .a � b/ � c D a � .b � c/ for all a; b; c 2 G, (ii) there is e 2 G with e � a D a D a � e for all a 2 G,

(iii) for all a 2 G, there is a0 2 G with a0 � a D e D a � a0. The element e is called the identity of G, and the element a0 is called the inverse of a (the inverse of a is usually denoted by a�1).

It is not difficult to prove, for groups as for commutative rings, that the

identity element is unique (if e0 � a D a D a � e0 for all a 2 G, then e0 D e), and the inverse of every element is unique (if a00 � a D e D a � a00, then a00 D a0).

Example 9.7. Theorem A.12 in Appendix A.1 shows that SX , the family of

all the permutations of a nonempty set X , is a group with composition as its

binary operation. In the special case when X D f1; 2; : : : ; ng, denote SX by

Sn;

and call it the symmetric group on n letters. N

Example 9.8. Just because we call a Galois group a group doesn’t make it so.

Recall that the Galois group Gal.E=k/ of a field extensionE=k consists of all

the automorphisms � ofE that fix k. We now show that Gal.E=k/with binary operation composition is a group.

If �; � 2 Gal.E=k/, then their composite �� is an automorphism of E fixing k; that is, �� 2 Gal.E=k/, so that composition is a binary operation on Gal.E=k/. Proposition A.5 says that composition of functions is always

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390 Chapter 9 Epilog

associative. The identity 1E WE ! E is an automorphism fixing k, and it is a routine calculation to show that if � 2 Gal.E=k/, then its inverse ��1 also lies in Gal.E=k/. In particular, if f .x/ 2 kŒx and E=k is a splitting field of f , then Gal.f / D Gal.E=k/ is a group. We have assigned a group to every polynomial. N

Example 9.9. The set G D GL.2;R/ of all nonsingular 2 � 2 matrices with real entries is a group with binary operation matrix multiplication�. First, we

have �WG � G ! G because the product of two nonsingular matrices is also nonsingular. Matrix multiplication is associative, and the identity matrix

� 1 0 0 1

is the identity element. Finally, nonsingular matrices have inverses (this is the

definition of nonsingular!), and so G is a group. N

How to Think About It. We warn the reader that new terms are going to

be introduced at a furious pace. You need not digest everything; if a new idea seems only a little reasonable, continue reading nevertheless. One way to keep

your head above water is to see that definitions and constructions for groups

(subgroups, homomorphisms, kernels, normal subgroups, quotient groups) are

parallel to what we have already done for commutative rings (subrings, ho-

momorphisms, kernels, ideals, quotient rings). Your reward will be a better appreciation of the beautiful results of Abel and Galois.

Definition. A subgroup of a group G is a nonempty subset S � G such that s; t 2 S implies s � t 2 S and s 2 S implies s�1 2 S .

Subgroups S � G are themselves groups, for they satisfy the axioms in the definition. In particular, since S is not empty, there is some s 2 S ; by definition, its inverse s�1 also lies in S , and so e D s � s�1 lies in S .

Food for Thought. If  is a set, we may view any subgroup of S, the

group of all permutations of , as symmetries of it. The notion of symme-

try depends on the permutation group: isometries of the plane are one kind

of symmetry; another kind arises from the group of all homeomorphisms of the plane; yet another arises from the group of all nonsingular linear transfor-

mations. This observation is the basis of Klein’s Erlanger Programm, which

classifies different types of geometries according to which geometric proper-

ties of figures are left invariant.

Multiplication in a commutative ring R is, by definition, commutative: if

a; b 2 R, then ab D ba. But multiplication in a group need not be commuta- tive: a � b and b � a may be different. For example, composition in the sym- metric group S3 is not commutative: define �; � 2 S3 by �.1/ D 2, �.2/ D 1, �.3/ D 3, and �.1/ D 1, �.2/ D 3, �.3/ D 2. It is easy to see that � ı� ¤ � ı� (for ��.1/ D �.1/ D 2 and ��.1/ D �.2/ D 3).

You also know that the product of two matrices depends on which is written

first: AB ¤ BA is possible. Hence, the group GL.2;R/ is not commutative.

Definition. A groupG is abelian if a � b D b � a for all a; b 2 G.

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9.4 Groups 391

Abel’s 1828 theorem says, in modern language, that a polynomial with an

abelian Galois group is solvable by radicals; this is why abelian groups are so-

called. From now on, we shall simplify notation by writing the product of two group elements as ab instead of by a � b and the identity as 1 instead of by e.

Proposition 9.10. If k is a field, a 2 k, and k contains all the nth roots of unity, �; �2; : : : ; �n D 1, then the Galois group of f .x/ D xn � a is abelian. There is always a primitive

nth root of unity; that

is, an element � with

�n D 1 such that every nth root of unity is a

power of �. In particular,

a primitive complex nth

root of unity is e2�i=n D cos.2�=n/C i sin.2�=n/.

Sketch of proof. Since k contains all the nth roots of unity,

E D k � !iˇ W ˇn D a and 1 � i � n

is a splitting field of f . Any automorphism � of E must permute the roots, and so �.!iˇ/ D !jˇ for some j depending on i . Similarly, if � is another automorphism, then �.!jˇ/ D !`ˇ. It follows that both �� and �� send !iˇ to !jC`ˇ D !`Cjˇ; from this fact it is not hard to see that Gal.E=k/ is an abelian group. �

Galois was able to translate a polynomial being solvable by radicals into a

certain property of its Galois group by constructing analogs for groups of the

constructions we have done in earlier chapters for commutative rings. Of course, abstract algebra did not exist in Galois’ time.

In particular, rings, fields,

and homomorphisms were

not in anyone’s vocabulary;

nor were groups.

Definition. If G and H are groups, then a homomorphism is a function

'WG ! H such that, for all a; b 2 G,

'.ab/ D '.a/'.b/:

An isomorphism is a homomorphism that is a bijection. If there is an isomor-

phism 'WG ! H , then we say that G and H are isomorphic and we write G Š H .

We can be more precise. If groups are denoted by .G; �/ and .H; ı/, where � and ı are binary operations, then a homomorphism 'WG ! H is a function for which

'.a � b/ D '.a/ ı '.b/:

It is easy to see that if ' is a homomorphism, then '.1/ D e, where 1 is the identity of G and e is the identity of H ; moreover, for each a 2 G, we have '.a�1/ D '.a/�1 (the latter being the inverse of the element '.a/ in H ). If a; b 2 G commute, then ab D ba. Hence, if 'WG ! H is a homomorphism, then

'.a/'.b/ D '.ab/ D '.ba/ D '.b/'.a/:

It follows that if G is abelian and ' is an isomorphism, then H is abelian.

Every polynomial determines a group of symmetries of its roots.

Theorem 9.11. If a polynomial f .x/ 2 kŒx has n roots, then its Galois group If X and Y are sets of n elements, then

SX Š Sn Š SY . Thus, groups don’t care if you

are permuting n numbers,

n roots, or n monkeys.

Gal.E=k/ is isomorphic to a subgroup of the symmetric group Sn.

Proof. By Theorem 9.4, elements of Gal.E=k/ permute the roots of f . Now

see [26], p. 454.

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392 Chapter 9 Epilog

Definition. The kernel of a homomorphism 'WG ! H of groups is

ker ' D fa 2 G W '.a/ D 1g;

where 1 is the identity element of H . The image of ' is

im' D fh 2 H W h D '.g/ for some g 2 Gg:

If 'WG ! H is a homomorphism, then ker' is a subgroup of G and im ' is a subgroup of H .

Just as the kernel of a ring homomorphism has special properties—it is an

ideal—so, too, is the kernel of a group homomorphism special. If a 2 ker' and b 2 G, then

'.bab�1/ D '.b/'.a/'.b�1/ D '.b/1'.b/�1

D '.b/'.b/�1 D 1:

Definition. A subgroup N of a group G is a normal subgroup if, for each

a 2 N , we have bab�1 2 N for every b 2 G.

Thus, kernels of homomorphisms are always normal subgroups. In an abelian group, every subgroup is normal but, in general, there are subgroups that are

not normal. For example, if � is the permutation of X D f1; 2; 3g that inter- changes 1 and 2 and fixes 3, then S D fe; �g is a subgroup of the symmetric group S3. But S is not a normal subgroup of S3, for if � is the permutation that

fixes 1 and interchanges 2 and 3, then ����1.1/ D ��.1/ D �.2/ D 3. Hence, ����1 ¤ e and ����1 ¤ � ; that is, � 2 S but ����1 … S .

Just as we used ideals to construct quotient rings, we can use normal sub-

groups to construct quotient groups. If N is a subgroup of a group G, define

certain subsets, called cosets of N inG, as follows: if a 2 G, then

aN D fas W s 2 N g � G:

The family of all cosets of N is denoted by

G=N D fall cosets aN W a 2 Gg:

When N is a normal subgroup, then G=N is a group if we define a binary

operation by

aN � bN D abN

(normality of N is needed to prove that this multiplication is well-defined: if

a0N D aN and b0N D bN , then a0b0N D abN ). The group G=N is called the quotient group.

There is an isomorphism theorem for groups analogous to the isomorphism

theorem for commutative rings.

Theorem 9.12 (First Isomorphism Theorem). If 'WG ! H is a group ho- momorphism, then im ' is a subgroup of H , N D ker' is a normal subgroup of G, and there is an isomorphismˆWG=N ! im ' given by ˆW aN 7! '.a/.

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9.4 Groups 393

Sketch of proof. Adapt the proof of the First Isomorphism Theorem for com-

mutative rings. �

How to Think About It. Without a doubt, this section contains too much

new material; there’s too much to digest. Fortunately, you have seen analogs of these definitions for commutative rings so, at least, they sound familiar. You

can now sympathize with the members of the Academy in Paris in 1830 as

they struggled, without benefit of ever having seen any abstract algebra at all,

to read such things in the paper Galois submitted to them!

Let us now see why normal subgroups are important for polynomials. If k � B � E , then Gal.E=B/ is a subset of Gal.E=k/:

Gal.E=B/ D f� 2 Gal.E=k/ W �.B/ D Bg:

It is easy to check that Gal.E=B/ is a subgroup of Gal.E=k/.

Theorem 9.13. Let k � B � E be a tower of fields, where E=k is a splitting field of some f .x/ 2 kŒx. If B is the splitting field of some g.x/ 2 kŒx, then Gal.E=B/ is a normal subgroup of Gal.E=k/, and

Gal.E=k/=Gal.E=B/ Š Gal.B=k/:

Sketch of proof. Define 'W Gal.E=k/ ! Gal.E=B/ by 'W � 7! � jB . The restriction � jB does send B to itself, since B is a splitting field (so automor- phisms permute the roots of g, by Theorem 9.4), and so � jB 2 Gal.B=k/. By Theorem 9.12, the First Isomorphism Theorem, it suffices to find im ' and ker'. It is obvious that Gal.E=B/ � ker', and a short calculation gives equal- ity. We can prove that ' is surjective and hence im ' D Gal.B=k/ (the proof of surjectivity ([26] p. 455) is not straightforward). �

The converse of Theorem 9.13 is true: the Fundamental Theorem of Galois Theory says, in part, that if N is a normal subgroup of Gal.E=k/, then there

is a subfield

B D f˛ 2 E W �.˛/ D ˛ for all � 2 N g � E

that is a splitting field of some polynomial in kŒx.

How to Think About It.

The subgroups get smaller as the field extensions get bigger. If K � L � E , then Gal.E=L/ � Gal.E=K/: if � is an automorphism of E that fixes everything in L, then surely � fixes everything inK � L.

Lemma 9.14. LetK � L � E be a tower of fields, whereK contains all roots of unity. If L=K is a pure extension, then Gal.E=L/ is a normal subgroup of It is not true that that if

A is a normal subgroup

of B and B is a normal

subgroup of C , then A is a

normal subgroup of C .

Gal.E=K/ and the quotient group Gal.E=K/=Gal.E=L/ is abelian.

Sketch of proof. The field extension L=K is a splitting field (because the

subfield K contains all needed roots of unity), and so Theorem 9.12 gives

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394 Chapter 9 Epilog

Gal.E=L/ a normal subgroup of Gal.E=K/. By Proposition 9.10, the quotient

group is abelian. �

We conclude the story by applying Lemma 9.14 to each pure extension Ki=Ki�1 in a tower of a radical extension.

Lemma 9.15. Let k be a field containing all roots of unity. If f .x/ 2 kŒx is solvable by radicals, then there is a chain of subgroups

G0 D Gal.Kt=k/ � G1 � G2 � � � � � Gt D f1g;

where eachGiC1 is a normal subgroup ofGi and each quotient groupGi=GiC1 is abelian.

This lemma suggests the following definition.

Definition. A groupG is solvable if there is a chain of subgroups

G D G0 � G1 � G2 � � � � � Gt D f1g

where eachGiC1 is a normal subgroup ofGi and each quotient groupGi=GiC1 is abelian.

Clearly, every abelian group is solvable—take G1 D f1g. It is shown in [26], p. 466, that the symmetric group S4 and all its subgroups are solvable

groups, but that S5 is not solvable.

Using these ideas, Galois proved the following beautiful theorem.

Theorem 9.16 (Galois). Let k be a field and f .x/ 2 kŒx. If f is solvable by radicals, then its Galois group is a solvable group. If k has characteristic 0,

then the converse is true.

Proof. [25], p. 189 and p. 208.

Galois’ Theorem explains why the classical theorems hold for polynomials of degree � 4.

Corollary 9.17. If k is a field of characteristic 0 and f .x/ 2 kŒx has degree � 4, then f is solvable by radicals.

Proof. Since deg.f / � 4, Theorem 9.11 says that Gal.f / is (isomorphic to) a subgroup of S4 and, hence, it is a solvable group. Theorem 9.16 now says that

f is solvable by radicals.

Finally, the next theorem explains why degree 5 was so troublesome.

Corollary 9.18 (Abel–Ruffini). The general polynomial of degree 5 is not Ruffini’s name occurs here

because he published

a proof of this result in

1799. Although his ideas

were correct, there were

gaps in his proof, and it

was not accepted by his

contemporaries.

solvable by radicals.

Proof. The Galois group of the general quintic f .x/ 2 QŒx is S5 ([26], p. 468), which is not a solvable group, and so Galois’ Theorem says that f

is not solvable by radicals.

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9.4 Groups 395

Here is an explicit numerical example. The quintic

f .x/ D x5 � 4x C 2 2 QŒx

(see Figure 9.3) is not solvable by radicals because its Galois group is S5 ([26],

p. 469).

−2 −1 1 2

−6

−4

−2

2

4

6

Figure 9.3. f .x/ D x5 � 4x C 2.

Corollary 9.18 is often misquoted. It says the general quintic is not solvable

by radicals: there is no formula involving only addition, subtraction, multipli- cation, division, and extraction of roots that expresses the roots of the general

quintic polynomial in terms of its coefficients. But it doesn’t say that roots of

quintics cannot be found. There are other kinds of formulas; for example, New-

ton’s method gives the roots as limn!1 xn, where xnC1 D xn�f .xn/=f 0.xn/. Thus, it is not accurate to say that there is no formula finding the roots of a quintic polynomial.

Exercises

9.4 Prove that every subgroup of an abelian group is abelian.

9.5 Let f .x/; g.x/ 2 QŒx be solvable by radicals. (i) Show that f .x/g.x/ is also solvable by radicals.

(ii) Give an example showing that f .x/ C g.x/ need not be solvable by radicals.

9.6 Assuming that xn � 1 is solvable by radicals, prove that xn � a is solvable by radicals, where a 2 Q.

9.7 Prove that S3 is a solvable group and that it is not abelian.

9.8 Recall Exercise 1.56 on page 35: if m � 2 is an integer, gcd.k; m/ D 1, and gcd.k0; m/ D 1, then gcd.kk0; m/ D 1.

Prove that

Um D fŒk 2 Zm W gcd.k; m/ D 1g

is a group under multiplication.

9.9 If k is a field, prove that k� D fa 2 k W a ¤ 0g is a group under multiplication.

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396 Chapter 9 Epilog

9.10 If R is a commutative ring, prove that R is an abelian group under addition. (Note

that 0 is the identity element and that �a is the (additive) inverse of a.)

9.11 Let k be a field.

(i) Prove that k� is an abelian group under multiplication, where k� denotes the set of nonzero elements of k.

(ii) Prove that GL2.k/, the set of all 2 � 2 nonsingular matrices with entries in k, is a group under matrix multiplication.

(iii) Prove that the determinant function,GL2.k/ is called the General Linear group, and

SL2.k/ is called the Special

Linear group.

detW GL2.k/ ! k�;

is a surjective homomorphism of groups.

(iv) Prove that ker.det/ D SL2.k/, the set of all 2 � 2 matrices over k having determinant 1.

(v) Prove that GL2.k/=SL2.k/ Š k�.

9.12 (i) Prove that R is an abelian group with addition as binary operation.

(ii) Prove that Q is an abelian group with addition as binary operation; indeed, it

is a subgroup of R.

(iii) Let R> be the group of positive real numbers. Show that R> is a group with

addition as as binary operation.

9.13 (i) Prove that expW R ! R>, defined by a 7! ea , is a group homomorphism.The “laws of exponents” from high school algebra

preview the results of

Exercise 9.13.

(ii) Prove that logW R> ! R, defined by b 7! log b, is a group homomorphism. (iii) Prove that exp is an isomorphism by showing that its inverse is log.

9.14 (i) Prove that R>, the set of all positive real numbers, is an abelian group with

multiplication as binary operation, and prove that Q>, the set of all positive

rational numbers, is a subgroup of R>.

(ii) Prove that ZŒx is an abelian group under addition.

(iii) Use the Fundamental Theorem of Arithmetic to prove that the additive group

ZŒx is isomorphic to the multiplicative group Q> of all positive rational

numbers.

Hint: Define 'W ZŒx ! Q> by

'W e0 C e1x C � � � C enxn 7! pe00 p e1 1 � � � p

en n ;

where p0 D 2, p1 D 3, p2 D 5; : : : is the list of all primes.

9.5 Wiles and Fermat’s Last Theorem

Andrew Wiles proved Fermat’s Last Theorem in 1995: Modular elliptic curves

and Fermat’s last theorem, Ann. Math. (2) 141 (1995), pp. 443-551. He has

said,

I was a ten year old and one day I happened to be looking in my local

public library and I found a book on maths and it told a bit about the

history of this problem and I, a ten year old, could understand it. From

that moment I tried to solve it myselfI it was such a challenge, such a beautiful problem. This problem was Fermat’s Last Theorem.

and

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9.5 Wiles and Fermat’s Last Theorem 397

There’s no other problem that will mean the same to me. I had this very

rare privilege of being able to pursue in my adult life what had been my

childhood dream. I know it’s a rare privilege, but I know if one can do

this it’s more rewarding than anything one can imagine.

Andrew Wiles was born in Cambridge, England in 1953. He was awarded

a doctorate in 1980 from the University of Cambridge, then spent a year in

Bonn before joining the Institute for Advanced Study in Princeton. In 1982,

he was appointed Professor at Princeton. Around 1985, Wiles learned that the

Shimura–Taniyama–Weil conjecture about elliptic curves, if true, would im- ply Fermat’s Last Theorem (we will say more about this below). Wiles was

able to prove a special case of this conjecture (the full conjecture was proved

in 2001) which was strong enough to give Fermat’s Last Theorem.

In 1994, Wiles was appointed Eugene Higgins Professor of Mathematics at

Princeton. Wiles received many honors for his outstanding work. For exam- ple, he was awarded the Schock Prize in Mathematics from the Royal Swedish

Academy of Sciences, the Prix Fermat from Université Paul Sabatier, the Wolf

Prize in Mathematics from the Wolf Foundation in Israel, and the Cole Prize

from the American Mathematical Society. He was elected a member of the

National Academy of Sciences of the United States, receiving its mathematics

prize, and Andrew Wiles became “Sir Andrew Wiles” when he was knighted by the Queen of England. In 1998, not being eligible for a Fields medal (the

mathematics prize equivalent to a Nobel prize) because he was over forty

years of age, the International Mathematical Union presented him with a silver

plaque at the International Congress of Mathematicians.

Elliptic Integrals and Elliptic Functions

The context of Wiles’ proof of Fermat’s Last Theorem is that of elliptic curves,

an area with an interesting history. Leibniz, one of the founders of calcu- We refer the reader to the books of Siegel [30],

Silverman–Tate [31], and

Stillwell [32], as well

as to the article by M.

Rosen, Abel’s Theorem

on the Lemniscate, Amer.

Math. Monthly 88 (1981),

pp. 387–395, for further

details of this discussion.

lus, posed the problem of determining which integrals could be expressed in

“closed form;” that is, as linear combinations of familiar functions such as ra-

tional functions, exponentials, logarithms, trigonometric functions, and their

inverse functions. One of the first integrals that could not be so expressed (al- though the proof of this fact, by Liouville, waited until 1833) is the arclength

of an ellipse. If f .x; y/ D 0 is the equation of a curve in the plane, then its arclength is given in terms of the indefinite integral

Z p 1C .dy=dx/2 dx:

Consider the ellipse with equation

x2

a2 C y

2

b2 D 1;

where a > b > 0. We have y D b p 1 � .x2=a2/, so that

dy

dx D �bx a2 p 1 � .x2=a2/

;

and the arclength integral is

1

a

Z r a4 � .a2 � b2/x2

a2 � x2 dx:

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398 Chapter 9 Epilog

The eccentricity of the ellipse is

E D p 1 � .b=a/2:

Make the substitution x D a sin � , so that cos � D 1 a

p a2 � x2 and dx D

a cos � d� , to obtain

1

a

Z r a4 � .a2 � b2/x2

a2 � x2 dx D a Z s

1 � � a2 � b2 a2

� sin2 � d�

D a Z p

1 �E2 sin2 � d�:

Finally, we rewrite the last integral using the tangent half-angle formula t D tan.�=2/ in Chapter 1 (so that d� D 2dt=.1 C t2/ and sin � D 2t=.1 C t2/). We obtain

a

Z p 1 �E2 sin2 � d� D 2a

Z p g.t/

.1 C t2/2 dt;

where g.t/ D t4 C .2 � 4E2/t2 C 1. Thus, if R.x; y/ is the rational function in two variables,

R.x; y/ D y .1C x2/2

;

then the arclength of an ellipse has the form

2a

Z R � t; p g.t/

� dt;

where g.t/ is a quartic polynomial. A similar integral arises from the arclength

of the hyperbola x2=a2 � y2=b2 D 1.

Definition. An elliptic integral is an indefinite integral of the form

Z R � t; p g.t/

� dt;

where R.x; y/ is a rational function and g.t/ is either a cubic or a quartic

polynomial having no repeated roots.The substitution t D 1=u transforms the cubic

integrand

dtp .t � a/.t � b/.t � c/

into the quartic

�dup u.1�ua/.1�ub/.i�uc/

:

These integrals are so called because, as we have just seen, the arclength

of an ellipse was one of the first examples of them. Another example of an

elliptic integral, studied by Jacob Bernoulli in 1679, arises from computing the arclength of a spiral. In 1694, James Bernoulli examined the shape an elas-

tic rod takes if its ends are compressed; he found the resulting curve to be

the lemniscate r2 D cos 2� ; see Figure 9.4 (there are eight mathematicians in the Bernoulli family, in the seventeenth and eighteenth centuries, listed in

the MacTutor History of Mathematics Archive). Recall that the arclength of a

curve r D f .�/ in polar coordinates is Z p

1C r2.d�=dr/2 dr:

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9.5 Wiles and Fermat’s Last Theorem 399

Figure 9.4. Lemniscate r2 D cos 2� .

If r2 D cos 2� , then

1C r2.d�=dr/2 D 1C r4

sin2 2�

D 1C r 4

1 � cos2 2�

D 1C r 4

1 � r4

D 1 1 � r4 :

Therefore, the arclength of the lemniscate is

Z drp 1 � r4

.

Yet another example, Z

dtp .1 � t2/.1 � at2/

;

arises when calculating the period of oscillation of a simple pendulum. Com- puting the electrical capacity of an ellipsoid with equation x2=a2 C y2=b2 C z2=c2 D 1 involves the integral

Z dtp

.a2 C t/.b2 C t/.c2 C t/ ;

an elliptic integral involving a cubic.

Since there are interesting elliptic integrals and they are difficult to evalu-

ate, they were the subject of much investigation. In 1718, Fagnano proved a

Duplication Formula:

2

Z u

0

dtp 1 � t4

D Z Q.u/

0

dtp 1 � t4

;

whereQ.u/ D 2u p 1 � u4=.1Cu4/. In proving this, Fagnano inverted I.x/ DR x

0 1=

p 1 � t4 dt , getting the inverse function I�1.x/ D

p 2x=

p 1C x4. In

1751, Euler generalized the duplication formula of Fagnano, obtaining an

Addition Theorem: Z u

0

dtp 1 � t4

C Z v

0

dtp 1 � t4

D Z P.u;v/

0

dtp 1 � t4

;

where P.u; v/ D � u

p 1 � v4 C v

p 1 � u4

� =.1 C u2v2/ (so that P.u; u/ D

Q.u/). Euler further generalized this by replacing the integrand by 1= p p.t/,

where p.t/ is any quartic polynomial.

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400 Chapter 9 Epilog

In 1797, Gauss considered the elliptic integrals R u 0

dtp 1�t3

and R u 0

dtp 1�t4

.

He saw an analogy (as, most likely, did Fagnano and Euler) with

sin�1 u D Z u

0

dtp 1 � t2

;

and he inverted many elliptic integrals; after all, sinx is the inverse function of sin�1 x. Nowadays, inverse functions of elliptic integrals are called elliptic functions. Just as sinx is periodic, that is, sin.x C 2�/ D sin x for all x, so, too, are elliptic functions f ; there is some number p with f .x C p/ D f .x/ for all x. Gauss then studied complex elliptic integrals I.z/ D

R z 0

d�p g.�/

; their

inverse functions f .z/ D I�1.z/ are called elliptic functions of a complex variable. Gauss saw that complex elliptic functions are doubly periodic: there

are (noncollinear) complex numbers p and q with

f .z Cmp C nq/ D f .z/

for all complex z and all m; n 2 Z. This fact has important geometric conse- quences, both for elliptic functions and for complex variables in general. Alas,

Gauss never published these ideas, and they became known only later.

In 1823, Abel investigated elliptic functions, rediscovered many of Gauss’s

theorems, and proved new beautiful results about them. For example, just as

Gauss had found all n for which one can divide the circle into n equal arcs using ruler and compass (n D 2mp1 � � �pk , where m � 0 and the pi are dis- tinct primes of the form 22

t C 1), Abel obtained the same result (for exactly the same n) for the lemniscate. At the same time, Jacobi began his investiga-

tions of elliptic functions, further explaining and generalizing work of Euler

by introducing theta functions and modular curves.

Congruent Numbers Revisited

The search for congruent numbers can be viewed as the search for “general-

ized Pythagorean triples”—right triangles with rational side-lengths and inte-

ger area. Recall from Chapter 1 that a congruent number is a positive integer n

that arises from asking which integers are areas of right triangles having ratio-

nal side-lengths; that is, there are positive rational numbers a, b, and c such that

a2 C b2 D c2 and 1 2 ab D n:

Let’s loosen the constraints a bit and allow a, b, and c to be negative rationalSince ab D 2n > 0, we have a ¤ 0 and b ¤ 0. It follows that c ¤ 0, too.

numbers as well. We’d like to replace the two equations in four unknowns with

a simpler set of constraints. We’ll see that the solution can be realized as the

search for rational points on a polynomial curve.

We now turn the pair of defining equations into a single equation in twoIn Theorem 1.9, we reduced the defining pair

(for n D 2) to a degree 4 equation in three variables.

We’ll do a little better here.

variables. The equation a2 C b2 D c2 can be written as

b2 D c2 � a2 D .c � a/.c C a/:

Let k D c � a, so that we have

b2 D k.c C a/:

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9.5 Wiles and Fermat’s Last Theorem 401

Since c D k C a, this is the same as

b2 D k.k C 2a/ D k2 C 2ak; or

2ak D b2 � k2:

Since a D 2b=n, this is equivalent to

4nk

b D b2 � k2;

or

4nk D b3 � k2b:

This is beginning to look like a cubic in b. To homogenize it, multiply both

sides by � n k

�3 to get

4n4

k2 D � bn

k

�3 � n2

� bn

k

or � 2n2

k

�2 D � bn

k

�3 � n2

� bn

k

� I

remembering that k D c � a, we have a single cubic equation satisfied by a,

We’re assuming that

c � a ¤ 0, or else b D0.

b, c, and n:

� 2n2

c � a

�2 D � bn

c � a

�3 � n2

� bn

c � a

� :

This shows that if a2 C b2 D c2 and ab D 2n, then � bn

c � a ; 2n2

c � a

is a rational point on the graph of y2 D x3�n2x (the graph of y2 D x3�25x is shown in Figure 9.5).

Except for sign changes and points on the x-axis, the correspondence goes

both ways.

–5 5

Figure 9.5. y2 D x3 � 25x

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402 Chapter 9 Epilog

Theorem 9.19. Let n be a positive integer. There’s a bijection between triples

of rational numbers .a; b; c/ satisfying

a2 C b2 D c2 and ab D 2n

and rational points on the graph of y2 D x3 � n2x with y ¤ 0.

Proof. The calculation preceding the statement of the theorem shows that a triple produces such a point on the graph. Going the other way, if .x; y/ is a

point on the graph with y ¤ 0, we can solve the system

x D bn c � a

y D 2n2

c � a c2 D a2 C b2

for a, b, and c, either by hand or CAS (see Figure 9.6) to find

a D x 2 � n2 y

b D 2nx y

c D x 2 C n2 y

:

It is easily checked that this produces a triple of rational numbers of the desired

type.

Figure 9.6. Solving for a, b, and c.

The first part of what

the CAS returns (for

.x2 � cy � n2/=y ¤ 0) combined with the value

given for c, just say that

n¤ 0. Let’s state the conversion formulas explicitly.

Corollary 9.20. The bijection guaranteed by Theorem 9.19 is given explicitly

by

.a; b; c/ 7! � bn

c � a ; 2n2

c � a

� and .x; y/ 7!

� x2 � n2 y

; 2nx

y ; x2 C n2 y

Example 9.21. The correspondence between rational right triangles with in-

teger area and cubic curves allows us to generate infinitely many congruent

triangles with the same area from a given such triangle.

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9.5 Wiles and Fermat’s Last Theorem 403

For example, on page 18, we saw that there are two rational right triangles

with area 5. One comes from a scaled copy of 4.9; 40; 41/whose area is 5 �62. To find the second one, it would take a very long time (even with a computer) to find 4.2420640; 2307361; 3344161/whose area is 5 � 7473482. But we can use an idea related to the “sweeping lines” method of Diophantus. The rational

right triangle with side-lengths � 3 2 ; 20 3 ; 41 6

� corresponds, via the formulas in

Corollary 9.20, to the point P D � 25 4 ; 75 8

� on the curve C defined by

y2 D x3 � 25x:

The idea is to take a line tangent to C through P ; it intersects C in a second

point P 0, which is also rational because P is (see Exercise 9.16 below). From A CAS (or at least a calculator) is a very useful

tool for these calculations. this new rational point, we can build a new right triangle.

From y2 D x3 � 25x, we have, using implicit differentiation that

dy

dx D 3x

2 � 25 2y

:

Using this, we find that the slope of the tangent to C at P is 59=12, and hence

the tangent line to C at P has equation

y D 59 12

� x � 25

4

� C 75

8 :

Solving the system (see Figure 9.7)

y2 D x3 � 25x

y D 59 12

� x � 25

4

� C 75

8 ;

we get P and

P 0 D � 1681

144 ; 62279

1728

� :

Figure 9.7. P and P 0

Finally, using Corollary 9.20 again, we recover .a; b; c/ from P 0:

a D 1519 492

; b D 4920 1519

; c D 334416 747348

:

These are the side-lengths of the triangle on page 18. N

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404 Chapter 9 Epilog

Exercises

9.15 Show that there are no rational points .x; y/ with y ¤ 0 on the graph of (i) y2 D x3 � x

(ii) y2 D x3 � 4x

9.16 Show that a cubic equation with rational coefficients and two rational roots has,

in fact, three rational roots.

9.17 Find a third rational right triangle with area 5, different from the two we found in

Example 9.21.

Elliptic Curves

The curves defined by the equation in Theorem 9.19,

y2 D a cubic polynomial in x

show up all across mathematics. We just saw how they can be used to generate

congruent numbers.

Before that, we saw that the integral defining arcsine, R dt=

p 1 � t2, sug-

gested studying elliptic functions, the inverse functions of elliptic integrals.

Just as the unit circle is parametrized by sine and cosine (it consists of the

points .sin �; cos �/), Gauss, Abel, and Jacobi considered curves parametrized by elliptic functions; that is, curves consisting of the points .f .u/; f 0.u//, where f is an elliptic function (cosine is the derivative of sine). What sort of

curves are these? Expand the integrand of an elliptic integral as a power series

(since it has a denominator, the series begins with a negative power), and then

integrate term by term. There results a differential equation involving x D f and y D f 0, which turns out to be a cubic in two variables (see [9], pp. 17– 19). After some manipulations, one obtains a Weierstrass normal form for the

points .x; y/ on the curve y2 D ax3Cbx2CcxCd (there is another, simpler, Weierstrass normal form, y2 D 4x3 � g2x � g3, where g2; g3 are constants).

Definition. An elliptic curve over a field k is a curve C � k2 with equationThis definition is not quite accurate, for an

elliptic curve is really an

equivalence class of such

curves.

y2 D g.x/;

where g.x/ D ax3 C bx2 C cx C d 2 kŒx has no repeated roots.

The most interesting elliptic curves are over C (for complex variables) orCurves over C are two- dimensional surfaces when

viewed over R. over Q (for number theory), while elliptic curves over finite fields Fq give rise to public access codes that are more secure than the RSA codes we discussed

in Chapter 4.

Elliptic functions and elliptic curves, whose humble origins are in arclength

problems, occur in analysis, geometry, and complex variables. In the previous

subsection, we saw that congruent numbers lead to rational points on elliptic

curves. More generally, let’s now see the connection with number theory and with Fermat’s Last Theorem in particular.

Definition. A Diophantine equation is an equation f .x1; : : : ; xm/ D 0, where f .x1; : : : ; xm/ 2 QŒx1; : : : ; xm is a polynomial in several variables having ra- tional coefficients.

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9.5 Wiles and Fermat’s Last Theorem 405

A solution to a Diophantine equation f .x1; : : : ; xm/ D 0 is an m-tuple .q1; : : : ; qm/ 2 Qm for which f .q1; : : : ; qm/ D 0I an integer solution is a solution in Zm.

For example, xnCyn�1 D 0 is a Diophantine equation. Rational solutions .u; v/ D .a=c; b=c/ give rise, by clearing denominators, to integer solutions of the Diophantine equation xn C yn D zn. Of course, this example arises from Fermat’s Last Theorem.

A curve in the plane is the locus of solutions to an equation f .x; y/ D 0. Let’s focus on polynomials f .x; y/ 2 QŒx; y; that is, on Diophantine equa- tions of two variables. A rational point .u; v/ on the curve f .x; y/ D 0 is a geometric way of viewing a solution to the Diophantine equation f .x; y/ D 0. The method of Diophantus classifies Pythagorean triples by intersecting the

unit circle x2 C y2 D 1 with lines through the rational point .�1; 0/, thereby parametrizing the circle with rational functions, and then finding the rational points on the circle that correspond to Pythagorean triples. We saw, in Chap-

ter 1, that it is worthwhile to generalize the method by replacing the unit circle

by conic sections.

Now pass from conic sections, curves in the plane corresponding to quadratic

polynomials f .x; y/ 2 RŒx; y, to cubic polynomials of two variables. For ex- ample, rational points on x3 C y3 D 1 correspond to integer solutions of a3 C b3 D c3, so that the truth of Fermat’s Last Theorem for n D 3 says that the curve x3 C y3 D 1 has no rational points.

Diophantus also studied cubic curves. There was no analytic geometry in his

day, and geometry was not explicit in his results. However, both Fermat and,

later, Newton believed that geometry explains Diophantus’s method of finding solutions of cubic Diophantine equations f .x; y/ D 0; indeed, Newton called it the chord–tangent construction. Just as lines usually intersect a conic section

in two points, lines usually intersect cubic curves in three points. If y D mxCh is the equation of a line L, then L meets the curve in points .x; mx C h/ for which f .x;mxCh/ D 0. But f .x;mxCh/ is a cubic, and so it has three roots (if we admit complex numbers). In particular, given rational points P;Q on a

cubic curve C , say P D .a; b/ andQ D .c; d /, then the slope .d �b/=.c�a/ of the line L they determine is rational. If C has equation f .x; y/ D 0, where f .x; y/ 2 QŒx; y, then L intersects C in a third point, which is also a rational point. Thus, the chord joining rational pointsP andQ on C determines a third rational point on C (see Figure 9.8); denote the third point by

P �Q:

If we think of the tangent line T to C at P as a limit of chords through P , then it is natural to consider where T meets C . The slope of T is also rational:

if

A.x/y3 C B.x/y2 C C.x/y CD.x/ D 0;

where A;B; C;D 2 QŒx have degrees, respectively, 0; 1; 2; 3, then implicit differentiation gives

y0.x; y/ D �B 0y2 C C 0y CD0

3Ay2 C 2By C C I

since the coefficients of A;B; C;D are rational and P D .a; b/ is a rational point, the slope y0.a; b/ of T is rational. It follows easily that if T meets C ,

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406 Chapter 9 Epilog

P

Q

y

x

P * Q

P + Q

Figure 9.8. Adding points on a cubic.

then the point of intersection is another rational point; denote such a point by

P � P

(thus, the tangent line T intersects C in another rational point, say Q, and the

two points P;Q determine a third rational point).

If we are considering cubic curves C in the plane, then it is possible that a line meets C in only one point, not three (a cubic in RŒx always has a real root,

but its other roots may not be real). To make all work smoothly, we enlarge theThis is a good reason to consider cubics over the

complex numbers. plane to the Riemann sphere bC D R2 [ f1g, where we regard 1 as a point “at infinity.” We agree that lines through 1 are precisely the vertical lines; we declare that 1 lies on every cubicC and that 1 is a rational point onC . Given two pointsP;Q on the curve, the line they determine meets the curve in a third

rational pointP �Q. Define PCQ to be the intersection ofC with the vertical line V through P � Q; that is, V is the line determined by the two rational points 1 and P � Q (see Figure 9.8). The wonderful discovery is that this allows us to “add” pointsP;Q on elliptic curves (indeed, the set of all rational

points on an elliptic curve is an abelian group under this binary operation). In

particular, if C is the elliptic curve arising from the lemniscate (or any of the

elliptic functions considered by Euler), then the limits of integration in Euler’s Addition Theorem are given by the chord–tangent construction: for example,

Z P

0

dtp 1 � t4

C Z Q

0

dtp 1 � t4

D Z PCQ

0

dtp 1 � t4

:

As we have seen, congruent numbers n arising from rational side-lengths .a; b; c/ of a right triangle correspond to rational points on the elliptic curve

y2 D x3 � n2x. The binary operation shows how to construct new congruent numbers from given ones. The importance of this operation is illustrated by

Theorem 9.19.

What has this discussion to do with Fermat’s Last Theorem? The abelian group of rational points on elliptic curves, an example of complex multipli-

cation, is only the beginning of deep connections between Diophantine equa-

tions and elliptic curves. The following account by the number theorist Andrew

Granville summarizes the recent history.

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9.5 Wiles and Fermat’s Last Theorem 407

It all began in 1955, with a question posed by the Japanese mathe- We refer the reader to [31] for more about elliptic

curves and Diophantine

equations. We also rec-

ommend the expository

article of Cox, Introduction

to Fermat’s Last Theorem,

Amer. Math. Monthly 101

(1994), pp. 3–14, for more

details.

matician Yutaka TaniyamaW Could one explain the properties of ellip- tic curves, equations of the form y2 D x3 C ax C b with a and b given whole numbers, in terms of a few well-chosen curves? That is,

is there some very special class of equations that in some way encap-

sulate everything there is to know about our elliptic curves‹ Taniyama

was fairly specific about these very special curves .the so-called modu-

lar curves) and, in 1968, André Weil, one of the leading mathematicians of the twentieth century, made explicit which modular curve should de-

scribe which elliptic curve. In 1971, the first significant proven evidence

in favor of this abstract understanding of equations was given by Goro

Shimura at Princeton University, who showed that it works for a very

special class of equations. This somewhat esoteric proposed approach to

understanding elliptic curves is now known as the Shimura–Taniyama–

Weil conjecture. There the matter stood until 1986, when Gerhard Frey

made the most surprising and innovative link between this very abstract

conjecture and Fermat’s Last Theorem. What he realized was that if

cn D an C bn, then it seemed unlikely that one could understand the equation y2 D x.x � an/.x C bn/ in the way proposed by Taniyama. It took deep and difficult reasoning by Jean-Pierre Serre and Ken Ribet

to strengthen Frey’s original concept to the point that a counterexam-

ple to Fermat’s Last Theorem would directly contradict the Shimura–

Taniyama–Weil conjecture.

This is the point where Wiles enters the picture. Wiles drew together a

vast array of techniques to attack this question. Motivated by extraor-

dinary new methods of Victor Kolyvagin and Barry Mazur, Wiles estab-

lished the Shimura–Taniyama–Weil conjecture for an important class of

examples, including those relevant to proving Fermat’s Last Theorem.

His work can be viewed as a blend of arithmetic and geometry, and

has its origins way back in Diophantus’s Arithmetic. However he em-

ploys the latest ideas from a score of different fields, from the theories

of L-functions, group schemes, crystalline cohomology, Galois repre-

sentations, modular forms, deformation theory, Gorenstein rings, Euler

systems and many others. He uses, in an essential way, concepts due to

many mathematicians from around the world who were thinking about

very different questions.

The work of Wiles is a tour de force, and will stand as one of the sci-

entific achievements of the century. His work is not to be seen in iso-

lation, but rather as the culmination of much recent thinking in many

directions. Wiles’ proof, starting from scratch, would surely be over a

thousand pages long.

The story of this important discovery is a tribute to the deeper and more

abstruse levels of abstract understanding that mathematicians have long

claimed is essential. Many of us, while hailing Wiles’ magnificent achieve-

ment, yearn for Fermat to have been correct, and for the truly marvel-

lous, and presumably comparatively straightforward, proof to be recov-

ered.

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A Appendices A.1 Functions

Pick up any calculus book; somewhere near the beginning is a definition of

function which reads something like this: a function f is a rule that assigns to

each element a in a set A exactly one element, denoted by f .a/, in a set B . Ac-

tually, this isn’t too bad. The spirit is right: f is dynamic; it is like a machine, whose input consists of elements of A and whose output consists of certain

elements of B . The sets A and B may be made up of numbers, but they don’t

have to be.

There is a slight notational surprise. We are used to writing a function, not

as f , but as f .x/. For example, integrals are written Z f .x/ dx:

Logically, one notation for a function, say f , and another for its value at a point a in A, say f .a/, does make sense. However, some notation is grand-

fathered in. For example, we will continue to write polynomials as f .x/ D anx

n C an�1xn�1 C � � � C a0, trigonometric functions as sinx and cos x, and the exponential as ex (but some authors denote the exponential function by

exp). Still, the simpler notation f is usually a good idea. One problem we have with the calculus definition of function involves the

word rule. To see why this causes problems, we ask when two functions are

equal. If f is the function f .x/ D x2 C 2x C 1 and g is the function g.x/ D .x C 1/2, is f D g? We usually think of a rule as a recipe, a set of directions. With this understanding, f and g are surely different: f .5/ D 25C10C1 and g.5/ D 62. These are different recipes, but note that both cook the same dish: for example, f .5/ D 36 D g.5/.

A second problem with the calculus definition is just what is allowed to be

a rule. Must a rule be a formula? If so, then f .x/, defined by

f .x/ D ( 1 if x is rational

0 if x is irrational,

is not a function. Or is it? The simplest way to deal with these problems is to

avoid the imprecise word rule. If A is a set, then we write

a 2 A;

which abbreviates “a belongs to A” or “a is an element of A.”

409

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410 Appendix A Appendices

Definition. If A1; A2; : : : ; An are sets, their cartesian product is

A1 � A2 � � � � � An D ˚ .a1; a2; : : : ; an/ W ai 2 Ai for all i

:

We can define ordered

pair from scratch: see

Exercise A.2 on page 418.

Cartesian product could

then be defined by induc-

tion on n.

An element .a1; a2/ 2 A1 � A2 is called an ordered pair, and .a1; a2/ D .a01; a

0 2/ if and only if a1 D a01 and a2 D a02. More generally, two n-tuples

.a1; a2; : : : ; an/ and .a 0 1; a

0 2; : : : ; a

0 n/ are equal if ai D a0i for all subscripts i .

We now review subsets and equality, for functions f WA ! B are subsets of A � B . We say that U is a subset of V or U is contained in V , denoted by

U � V;

if, for all u 2 U , we have u 2 V . Formally: .8u/.u 2 U ) u 2 V /. Two subsets U and V of a set X are equal, that is,

U D V;

if they are comprised of exactly the same elements: thus, U D V if and only if U � V and V � U . This obvious remark is important because many proofs of equality of subsets break into two parts, each part showing that one subset

is contained in the other. For example, let

U D fx 2 R W x � 0g and V D fx 2 R W there exists y 2 R with x D y2g:

Now U � V because x D . p x/2 2 V , while V � U because y2 � 0 for

every real number y (if y < 0, then y D �a for a > 0 and y2 D a2). Hence, U D V .

If U is a proper subset of V ; that is, if U � V but U ¤ V , then we write

U ¨ V:

An empty set, denoted by ¿, is a set with no elements. Given a set X , it is always true that ¿ � X . To see this, observe that the negation

.9u/.u 2 ¿ ) u … V /

is false, for there is no u 2 ¿. The empty set is unique: if ¿0 is also an empty set, then ¿ � ¿0 and ¿0 � ¿, so that ¿ D ¿0. There is only one empty set.

Informally, a function is what we usually call its graph.

Definition. Let A and B be sets. A function f WA ! B is a subset f � A�B such that, for each a 2 A, there is a unique b 2 B with .a; b/ 2 f . The set A is called its domain, and the set B is called its target.

If f is a function and .a; b/ 2 f , then we write f .a/ D b and we call b the value of f at a. Define the image (or range) of f , denoted by im f , to be

the subset of B consisting of all the values of f .

When we say that A is the domain of a function f WA ! B , we mean that f .a/ is defined for every a 2 A. Thus, the reciprocal f .x/ D 1=.x � 1/ is not a function R ! R, but it is a function R0 ! R, where R0 denotes the set of all real numbers not equal to 1.

The second problem above—is f W R ! R a function, where f .x/ D 1 if x is rational and f .x/ D 0 if x is irrational—can now be resolved; yes, f is a function:

f D f.x; 1/ W x is rationalg [ f.x; 0/ W x is irrationalg � R � R:

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A.1 Functions 411

Let’s look at more examples before resolving the first problem arising from

the imprecise term rule.

Example A.1. (i) The squaring function f W R ! R, given by f .x/ D x2, is the parabola consisting of all points in the plane R � R of the form .a; a2/. It satisfies the definition, and so f is a function (if f wasn’t a

bona fide function, we would change the definition!).

(ii) If A and B are sets and b0 2 B , then the constant function at b0 is the function f WA ! B defined by f .a/ D b0 for all a 2 A (when A D R D B , then the graph of a constant function is a horizontal line).

(iii) For a set A, the identity function 1AWA ! A is the function consisting of all .a; a/ (the diagonal of A �A), and 1A.a/ D a for all a 2 A.

(iv) The usual functions appearing in calculus are also functions according to

the definition just given. For example, the domain of sinx is R, its target

is usually R, and its image is the closed interval Œ�1; 1. N

How to Think About It. A function f WA ! B is the subset of A � B consisting of all the ordered pairs .a; f .a// (this subset is the function but,

informally, it is usually called its graph). In order to maintain the spirit of a function being dynamic, we often use the notation

f W a 7! b

instead of f .a/ D b. For example, we may write the squaring function as f W a 7! a2 instead of f .a/ D a2. We often say that f sends a to f .a/.

Let’s return to our first complaint about rules; when are two functions equal?

Proposition A.2. Let f WA ! B and gWA ! B be functions. Then f D g if and only if f .a/ D g.a/ for every a 2 A.

Proof. Assume that f D g. Functions are subsets of A � B , and so f D g means that each of f and g is a subset of the other. If a 2 A, then .a; f .a// 2 f ; since f D g, we have .a; f .a// 2 g. But there is only one ordered pair in g with first coordinate a, namely, .a; g.a//, because the definition of function

says that g gives a unique value to a. Therefore, .a; f .a// D .a; g.a//, and equality of ordered pairs gives f .a/ D g.a/, as desired.

Conversely, assume that f .a/ D g.a/ for every a 2 A. To see that f D g, it suffices to show that f � g and g � f . Each element of f has the form .a; f .a//. Since f .a/ D g.a/, we have .a; f .a// D .a; g.a// and hence .a; f .a// 2 g. Therefore, f � g. The reverse inclusion g � f is proved similarly. Therefore, f D g.

Proposition A.2 resolves the first problem arising from the term rule: if

f; gW R ! R are given by f .x/ D x2 C 2x C 1 and g.x/ D .x C 1/2, then f D g because f .a/ D g.a/ for every number a.

Let us clarify another point. Can functions f WA ! B and gWA0 ! B 0 be equal? Here is the commonly accepted usage.

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412 Appendix A Appendices

Definition. Functions f WA ! B and gWA0 ! B 0 are equal if A D A0, B D B 0, and f .a/ D g.a/ for all a 2 A.

Thus, a function f WA ! B has three ingredients—its domain A, its tar- get B , and its graph—and we are saying that functions are equal if and only if

they have the same domains, the same targets, and the same graphs.

It is plain that the domain and the graph are essential parts of a function;

why should we care about the target of a function when its image is more

important? As a practical matter, when first defining a function, one usually

doesn’t know its image. For example, what’s the image of f W R ! R, defined by

f .x/ D log

1C jxje�x 5 p x2 C cos2 x

! � Z x

0

dt 7 p 1C t6

We must analyze f to find its image, and this is no small task. But if targets

have to be images, then we couldn’t even write down f W R ! R without having first found the image of f . Thus, targets are convenient to use.See the discussion on

page 437 for a more

sophisticated reason why

targets are important.

If A is a subset of a set B , the inclusion i WA ! B is the function given by i.a/ D a for all a 2 A; that is, i is the subset of A � B consisting of all .a; a/ with a 2 A. If S is a proper subset of a set A, then the inclusion i WS ! A is not the identity function 1S because its target is A, not S ; it is not the identity function 1A because its domain is S , not A.

Instead of saying that the values of a function f are unique, we sometimes

say that f is single-valued or that it is well-defined. For example, if R� de- notes the set of nonnegative reals, then

p W R� ! R� is a function because

we agree that p a � 0 for every nonnegative number a. On the other hand,

g.a/ D ˙ p a is not single-valued, and hence it is not a function. The sim-

plest way to verify whether an alleged function f is single-valued is to phrase

uniqueness of values as an implication:

if a D a0; then f .a/ D f .a0/:

For example, consider the addition function ˛W R � R ! R. To say that ˛ is well-defined is to say that if .u; v/ D .u0; v0/ in R � R, then ˛.u; v/ D ˛.u0; v0/; that is, if u D u0 and v D v0, then uC v D u0 C v0. This is usually called the Law of Substitution.

Another example is addition of fractions. We define

a

b C c

d D ad C bc bd

:

But fractions have many names. If a=b D a0=b0 and c=d D c0=d 0, is .ad C bc/=bd D .a0d 0 C b0c0/=b0d 0? We verified that this formula does not depend on the choices of names of the fractions on page 193. On the other hand, the

operation

a

b � c d

D aC c bd

is not well-defined: 1 2

D 2 4

, but 1 2

� 3 4

D 4 8

, while 2 4

� 3 4

D 5 16

¤ 4 8

.

There is a name for functions whose image is equal to the whole target.

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A.1 Functions 413

Definition. A function f WA ! B is surjective (or onto or a surjection) if

im f D B:

Thus, f is surjective if, for each b 2 B , there is some a 2 A (depending on b) with b D f .a/.

Example A.3. (i) The identity function 1AWA ! A is a surjection. (ii) The sine function R ! R is not surjective, for its image is Œ�1; 1, a

proper subset of its target R. The function sW R ! Œ�1; 1, defined by s.x/ D sinx, is surjective.

(iii) The functions x2W R ! R and exW R ! R have target R. Now im x2 consists of the nonnegative reals and im ex consists of the positive reals,

so that neither x2 nor ex is surjective.

(iv) Let f W R ! R be defined by

f .a/ D 6aC 4:

To see whether f is a surjection, we ask whether every b 2 R has the form b D f .a/ for some a; that is, given b, can we find a so that

6aC 4 D b‹

Since a D 1 6 .b � 4/, this equation can always be solved for a, and so f

is a surjection.

(v) Let f W R � ˚ 3 2

! R be defined by

f .a/ D 6aC 4 2a � 3

:

To see whether f is a surjection, we seek, given b, a solution a: can we

solve

b D f .a/ D 6aC 4 2a � 3 ‹

This leads to the equation a.6 � 2b/ D �3b � 4, which can be solved for a if 6 � 2b ¤ 0 (note that .�3b � 4/=.6 � 2b/ ¤ 3=2). On the other hand, it suggests that there is no solution when b D 3 and, indeed, there is not: if .6a C 4/=.2a � 3/ D 3, cross multiplying gives the false equation 6a C 4 D 6a � 9. Thus, 3 … im f , and f is not a surjection (in fact, im f D R � f3g). N

The following definition gives another important property a function may

have.

Definition. A function f WA ! B is injective (or one-to-one or an injection) if, whenever a and a0 are distinct elements of A, then f .a/ ¤ f .a0/. Equiva- lently, the contrapositive states that f is injective if, for every pair a, a0 2 A, we have

f .a/ D f .a0/ implies a D a0:

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414 Appendix A Appendices

Being injective is the converse of being single-valued: f is single-valued if

a D a0 implies f .a/ D f .a0/; f is injective if f .a/ D f .a0/ implies a D a0. Most functions are neither injective nor surjective. For example, the squar-

ing function f W R ! R, defined by f .x/ D x2, is neither.

Example A.4. (i) The identity function 1AWA ! A is injective. (ii) Let f W R �

˚ 3 2

! R be defined by

f .a/ D 6a C 4 2a � 3

:

To check whether f is injective, suppose that f .a/ D f .b/: 6aC 4 2a � 3 D

6bC 4 2b � 3 :

Cross multiplying yields

12abC 8b � 18a � 12 D 12ab C 8a � 18b � 12;

which simplifies to 26a D 26b and hence a D b. We conclude that f is injective. (We saw, in Example A.3, that f is not surjective.)

(iii) Consider f W R ! R, given by f .x/ D x2 � 2x � 3. If we try to check whether f is an injection by looking at the consequences of f .a/ D f .b/, as in part (ii), we arrive at the equation a2 � 2a D b2 � 2b; it is not instantly clear whether this forces a D b. Instead, we seek the roots of f , which are 3 and �1. It follows that f is not injective, for f .3/ D 0 D f .�1/; that is, there are two distinct numbers having the same value. N

Sometimes there is a way of combining two functions to form another func-

tion, their composite.

Definition. If f WA ! B and gWB ! C are functions (the target of f is the domain of g), then their composite, denoted by g ı f , is the function A ! C given by

g ı f W a 7! g.f .a//I

that is, first evaluate f on a, and then evaluate g on f .a/.

We usually abbreviate the notation for composites in the text, writing gf

instead of g ı f , but we shall always write g ı f in this Appendix. Composition is thus a two-step process: a 7! f .a/ 7! g.f .a//. For exam-

ple, the function hW R ! R, defined by h.x/ D ecosx , is the composite g ı f , where f .x/ D cos x and g.x/ D ex. This factorization is plain as soon as one tries to evaluate, say h.�/; one must first evaluate f .�/ D cos� D �1 and then evaluate

h.�/ D g.f .�// D g.�1/ D e�1:

The chain rule in calculus is a formula that computes the derivative .g ıf /0 in terms of g0 and f 0:

.g ı f /0.x/ D .g0 ı f /.x/ � f 0.x/ D g0.f .x// � f 0.x/:

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A.1 Functions 415

If f WA ! B is a function, and if S is a subset of A, then the restriction of f to S is the function f jS WS ! B , defined by .f jS/.s/ D f .s/ for all s 2 S . It is easy to see that if i WS ! A is the inclusion, then f jS D f ı i ; that is, the functions f jS and f ı i have the same domain, the same target, and the same graph (see Exercise A.4 on page 419).

If f W N ! N and gW N ! R are functions, then g ı f W N ! R is defined, but f ı g is not defined (for target.g/ D R ¤ N D domain.f /). Even when f WA ! B and gWB ! A, so that both composites gıf and f ıg are defined, they need not be equal. For example, define f , gW N ! N by f Wn 7! n2 and gWn 7! 3n; then g ı f W 2 7! g.4/ D 12 and f ı gW 2 7! f .6/ D 36. Hence, g ı f ¤ f ı g.

Given a set A, let

AA D fall functionsA ! Ag:

The composite of two functions in AA is always defined, and it is, again, a

function in AA. As we have just seen, composition is not commutative; that is,

f ıg and gıf need not be equal. Let us now show that composition is always associative.

Proposition A.5. Composition of functions is associativeW given functions f W A ! B , gWB ! C , and hWC ! D, then

h ı .g ı f / D .h ı g/ ı f:

Proof. We show that the value of either composite on an element a 2 A is just d D h.g.f .a///. If a 2 A, then

h ı .g ı f /W a 7! .g ı f /.a/ D g.f .a// 7! h.g.f .a/// D d;

and

.h ı g/ ı f W a 7! f .a/ 7! .h ı g/.f .a// D h.g.f .a/// D d:

Since both are functions A ! D, it follows from Proposition A.2 that the composites are equal.

In light of this proposition, we need not write parentheses: the notation

h ı g ı f is unambiguous. Suppose that f WA ! B and gWC ! D are functions. If B � C , then

some authors define the composite hWA ! D by h.a/ D g.f .a//. We do not allow composition if B ¤ C . However, we can define h as the composite h D g ı i ı f , where i WB ! C is the inclusion.

The next result implies that the identity function 1A behaves for composition

in AA just as the number one does for multiplication of numbers.

Proposition A.6. If f WA ! B , then 1B ı f D f D f ı 1A.

Proof. If a 2 A, then

1B ı f W a 7! f .a/ 7! f .a/

and

f ı 1AW a 7! a 7! f .a/:

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416 Appendix A Appendices

Are there “reciprocals” in AA; that is, are there any functions f WA ! A for which there is g 2 AA with f ı g D 1A and g ı f D 1A? The following discussion will allow us to answer this question.

Definition. A function f WA ! B is bijective (or a one-one correspondence or a bijection) if it is both injective and surjective.

Example A.7. (i) Identity functions are always bijections.

(ii) Let X D f1; 2; 3g and define f WX ! X by

f .1/ D 2; f .2/ D 3; f .3/ D 1:

It is easy to see that f is a bijection. N

We can draw a picture of a function f WX ! Y in the special case when X and Y are finite sets (see Figure A.1). Let X D f1; 2; 3; 4; 5g, let Y D fa; b; c; d; eg, and define f WX ! Y by

f .1/ D b f .2/ D e f .3/ D a f .4/ D b f .5/ D c:

Now f is not injective, because f .1/ D b D f .4/, and f is not surjective,

1

2

3

4

5

a

b

c

d

e

X Y

Figure A.1. Picture of a function.

because there is no x 2 X with f .x/ D d . Can we reverse the arrows to get a function gWY ! X? There are two reasons why we can’t. First, there is no arrow going to d , and so g.d/ is not defined. Second, what is g.b/? Is it 1 or

is it 4? The first problem is that the domain of g is not all of Y , and it arises

because f is not surjective; the second problem is that g is not single-valued, and it arises because f is not injective (this reflects the fact that being single-

valued is the converse of being injective). Neither problem arises when f is a

bijection.

Definition. A function f WX ! Y is invertible if there exists a function gWY ! X , called its inverse, with both composites g ı f and f ı g being identity functions.

We do not say that every function f is invertible; on the contrary, we have

just given two reasons why a function may not have an inverse. Notice that if

an inverse function g does exist, then it “reverses the arrows” in Figure A.1. If f .a/ D y, then there is an arrow from a to y. Now g ı f being the identity says that a D .g ı f /.a/ D g.f .a// D g.y/; therefore gWy 7! a, and so the picture of g is obtained from the picture of f by reversing arrows. If f twists

something, then its inverse g untwists it.

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A.1 Functions 417

Lemma A.8. If f WX ! Y and gWY ! X are functions such that gıf D 1X , then f is injective and g is surjective.

Proof. Suppose that f .a/ D f .a0/. Apply g to obtain g.f .a// D g.f .a0//; that is, a D a0 (because g.f .a// D a), and so f is injective. If x 2 X , then x D g.f .x//, so that x 2 im g; hence g is surjective.

Proposition A.9. A function f WX ! Y has an inverse gWY ! X if and only if it is a bijection.

Proof. If f has an inverse g, then Lemma A.8 shows that f is injective and surjective, for both composites g ı f and f ı g are identities.

Assume that f is a bijection. Let y 2 Y . Since f is surjective, there is some a 2 X with f .a/ D y; since f is injective, a is unique. Defining g.y/ D a thus gives a (single-valued) function whose domain is Y (g merely “reverses

arrows;” since f .a/ D y, there is an arrow from a to y, and the reversed arrow goes from y to a). It is plain that g is the inverse of f ; that is, f .g.y// D f .a/ D y for all y 2 Y and g.f .a// D g.y/ D a for all x 2 X .

Example A.10. If a is a real number, then multiplication by a is the function

�aW R ! R, defined by r 7! ar for all r 2 R. If a ¤ 0, then �a is a bijection; its inverse function is division by a, namely, ıaW R ! R, defined by r 7! 1a r ; of course, ıa D �1=a. If r 2 R, then

�a ı ıaW r 7! 1a r 7! a 1 a r D r I

hence, �a ı ıa D 1R. Similarly, ıa ı �a D 1R. If a D 0, however, then �a D �0 is the constant function�0W r 7! 0 for all

r 2 R, which has no inverse function. N

Etymology. The inverse of a bijection f is denoted by f �1 (Exercise A.6 on page 419 says that a function cannot have two inverses). This is the same

notation used for inverse trigonometric functions in calculus; for example,

sin�1 x D arcsin x satisfies sin.arcsin.x// D x and arcsin.sin.x// D x. Of What are the domain and image of arcsin?course, sin�1 does not denote the reciprocal 1= sinx, which is csc x.

Example A.11. Here is an example of two functions f; gW N ! N with one composite the identity, but with the other composite not the identity; thus, f

and g are not inverse functions.

Define f , gW N ! N as follows:

f .n/ D nC 1I

g.n/ D (

0 if n D 0 n � 1 if n � 1.

The composite gıf D 1N , for g.f .n// D g.nC1/ D n (because nC1 � 1). On the other hand, f ı g ¤ 1N because f .g.0// D f .0/ D 1 ¤ 0. N

Two strategies are now available to determine whether a given function is

a bijection: use the definitions of injective and surjective, or find an inverse

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418 Appendix A Appendices

function. For example, if R> denotes the positive real numbers, let us showWhat are the domains and targets of log and exp? that the exponential function f W R ! R>, defined by f .x/ D ex D

P xn=nŠ,

is a bijection. A direct proof that f is injective would require showing that if ea D eb, then a D b; a direct proof showing that f is surjective would involve showing that every positive real number c has the form ea for some a. It is

simplest to prove these statements using the (natural) logarithm g.y/ D logy. The usual formulas elogy D y and log ex D x show that both composites f ıg and g ı f are identities, and so f and g are inverse functions. Therefore, f is a bijection, for it has an inverse.

The next theorem summarizes some results of this section. IfX is a nonempty

set, define the symmetric group on X :

SX D fbijections � WX ! Xg:

Theorem A.12. If X is a nonempty set, then composition .f; g/ 7! g ı f is a function SX � SX ! SX satisfying the following propertiesW

(i) .f ı g/ ı h D f ı .g ı h/ for all f; g; h 2 SX I (ii) there is 1X 2 SX with 1X ı f D f D f ı 1X for all f 2 SX I

(iii) for all f 2 SX , there is f 0 2 SX with f 0 ı f D 1X D f ı f 0.

Proof. Exercise A.12(iii) on page 420 says that the composite of two bijec-

tions is itself a bijection, and so composition has target SX . Part (i) is Proposi- tion A.5, part (ii) is Proposition A.6, and part (iii) is Proposition A.9.

Exercises

A.1 True or false, with reasons.

(i) If S � T and T � X , then S � X . (ii) Any two functions f W X ! Y and gW Y ! Z have a composite f ı gW X ! Z.

(iii) Any two functions f W X ! Y and gW Y ! Z have a composite g ı f W X ! Z. (iv) For every set X , we have X � ¿ D ¿. (v) If f W X ! Y and j W im f ! Y is the inclusion, then there is a surjection

gW X ! im f with f D j ı g. (vi) If f W X ! Y is a function for which there is a function gW Y ! X with

f ı g D 1Y , then f is a bijection. (vii) The formula f .a=b/ D .a C b/.a � b/ is a well-defined function Q ! Z.

(viii) If f W N ! N is given by f .n/ D n C 1 and gW N ! N is given by g.n/ D n2, then the composite g ı f is n 7! n2.n C 1/.

(ix) Complex conjugation z D a C ib 7! z D a � ib is a bijection C ! C.

Hint: (i) True. (ii) False. (iii) True. (iv) True. (v) True. (vi) False. (vii) False.

(viii) False. (ix) True.

A.2 * Let A and B be sets, and let a 2 A and b 2 B . Define their ordered pair as follows:

.a; b/ D fa; fa; bgg:

If a0 2 A and b0 2 B , prove that .a0; b0/ D .a; b/ if and only if a0 D a and b0 D b.

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A.1 Functions 419

Hint: In any formal treatment, one is obliged to define new terms carefully. In

particular, in set theory, one must discuss the membership relation 2. Does x 2 x make sense? If it does, is it ever true? One of the axioms constraining 2 is that the statement a 2 x 2 a is always false.

A.3 Let L D f.x; x/ W x 2 Rg; thus, L is the line in the plane that passes through the origin and makes an angle of 45ı with the x-axis.

(i) If P D .a; b/ is a point in the plane with a ¤ b, prove that L is the perpendicular-bisector of the segment PP 0 having endpoints P D .a; b/ and P 0 D .b; a/.

Hint: You may use Lemma 3.16 and the fact that .12 .a C c/; 1 2 .b C d// is

the midpoint of the line segment having endpoints .a; b/ and .c; d/.

(ii) If f W R ! R is a bijection whose graph consists of certain points .a; b/ (of course, b D f .a/), prove that the graph of f �1 is

f.b; a/ W .a; b/ 2 f g:

A.4 * Let X and Y be sets, and let f W X ! Y be a function. (i) If S is a subset of X , prove that the restriction f jS is equal to the composite

f ı i , where i W S ! X is the inclusion map.

Hint: Use the definition of equality of functions on page 412.

(ii) If im f D A � Y , prove that there exists a surjection f 0W X ! A with f D j ı f 0, where j WA ! Y is the inclusion.

A.5 If f W X ! Y has an inverse g, show that g is a bijection.

Hint: Does g have an inverse?

A.6 * Show that if f W X ! Y is a bijection, then it has exactly one inverse.

A.7 Show that f W R ! R, defined by f .x/ D 3x C 5, is injective and surjective, and find its inverse.

A.8 Determine whether f W Q � Q ! Q, given by

f .a=b;c=d/ D .a C c/=.b C d/ is a function.

Hint: It isn’t.

A.9 Let X D fx1; : : : ; xmg and Y D fy1; : : : ; yng be finite sets, where the xi are distinct and the yj are distinct. Show that there is a bijection f W X ! Y if and only if jX j D jY j; that is, m D n.

Hint: If f is a bijection, there are m distinct elements f .x1/; : : : , f .xm/ in Y ,

and so m � n; using the bijection f �1 in place of f gives the reverse inequality n � m.

A.10 Suppose there are 11 pigeons, each sitting in some pigeonhole. If there are only

10 pigeonholes, prove that there is a hole containing more than one pigeon.

A.11 � .Pigeonhole Principle/: If X and Y are finite sets with the same number of elements, show that the following conditions are equivalent for a function

f W X ! Y : (i) f is injective (ii) f is bijective (iii) f is surjective.

Hint: If A � X and jAj D n D jX j, then A D X ; after all, how many elements are in X but not in A?

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420 Appendix A Appendices

A.12 * Let f W X ! Y and gW Y ! Z be functions. (i) If both f and g are injective, prove that g ı f is injective.

(ii) If both f and g are surjective, prove that g ı f is surjective. (iii) If both f and g are bijective, prove that g ı f is bijective. (iv) If g ı f is a bijection, prove that f is an injection and g is a surjection.

A.13 (i) If f W .��=2; �=2/ ! R is defined by a 7! tan a, then f has an inverse function g; indeed, g D arctan. Hint: Compute composites.

(ii) Show that each of arcsin x and arccos x is an inverse function (of sin x and

cos x, respectively) as defined in this section. (Domains and targets must be

chosen with care.)

A.2 Equivalence Relations

When fractions are first discussed in grammar school, students are told that 1 3

D 2 6

because 1�6 D 3�2; cross-multiplying makes it so! Don’t believe your eyes that 1 ¤ 2 and 3 ¤ 6. Doesn’t everyone see that 1 � 6 D 6 D 3 � 2? Of course, a good teacher wouldn’t just say this. Further explanation is required,

and here it is. We begin with the general notion of relation.

Definition. Let X and Y be sets. A relation from X to Y is a subset R of

X � Y (if X D Y , then we say that R is a relation on X). We usually write xRy instead of .x; y/ 2 R.

Here is a concrete example. Certainly � should be a relation on R; to see that it is, define the subset

R D f.x; y/ 2 R � R W .x; y/ lies on or above the line y D xg:

You should check that .x; y/ 2 R if the second coordinate is bigger than the first. Thus, xRy here coincides with the usual meaning x � y.

Example A.13. (i) Every function f WX ! Y is a relation from X to Y . (ii) Equality is a relation on any set X ; it is the diagonal

f.x; x/ W x 2 Xg � X �X:

(iii) For every natural number m, congruence mod m is a relation on Z. Can

you describe the subset of Z � Z? (iv) If X D f.a; b/ 2 Z � Z W b ¤ 0g, then cross multiplication defines a

relation � on X by

.a; b/ � .c; d / if ad D bc: N

Definition. A relation x � y on a set X is (i) reflexive if x � x for all x 2 X ,

(ii) symmetric if x � y implies y � x for all x; y 2 X , (iii) transitive if x � y and y � z imply x � z for all x; y; z 2 X . An equivalence relation on a set X is a relation on X that has all three prop-

erties: reflexivity, symmetry, and transitivity.

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A.2 Equivalence Relations 421

Example A.14. (i) Ordinary equality is an equivalence relation on any set.

(ii) If m � 0, then Proposition 4.3 says that x � y mod m is an equivalence relation on Z.

(iii) If I is an ideal in a commutative ring R, then Proposition 7.1 shows that

congruence mod I is an equivalence relation on R.

(iv) We claim that cross multiplication is an equivalence relation on X D f.a; b/ 2 Z � Z W b ¤ 0g. Verification of reflexivity and symmetry is easy. For transitivity, assume that .a; b/ � .c; d / and .c; d / � .e; f /. Now ad D bc gives adf D bcf , and cf D de gives bcf D bde; thus, adf D bde. We may cancel the nonzero integer d to get af D be; that is, .a; b/ � .e; f /.

(v) In calculus, equivalence relations are implicit in the discussion of vectors.

An arrow from a pointP to a pointQ can be denoted by the ordered pair

.P;Q/; call P its foot and Q its head. An equivalence relation on arrows

can be defined by saying that .P;Q/ � .P 0; Q0/ if the arrows have the same length and the same direction. More precisely, .P;Q/ � .P 0; Q0/ if the quadrilateral obtained by joining P to P 0 and Q to Q0 is a paral- lelogram (this definition is incomplete, for one must also relate collinear

arrows as well as “degenerate” arrows .P; P /). The direction of an arrow

from P to Q is important; if P ¤ Q, then .P;Q/ 6� .Q; P /. N

An equivalence relation on a set X yields a family of subsets of X .

Definition. Let � be an equivalence relation on a nonempty set X . If a 2 X , the equivalence class of a, denoted by Œa, is defined by

Œa D fx 2 X W x � ag � X:

We now display the equivalence classes arising from the equivalence rela- tions in Example A.14.

Example A.15. (i) Let � be equality on a set X . If a 2 X , then Œa D fag, the subset having only one element, namely, a. After all, if x D a, then x and a are equal!

(ii) Consider the relation of congruence mod m on Z, and let a 2 Z. The congruence class of a, defined by

fx 2 Z W x D aC km where k 2 Zg;

is equal to the equivalence class of a, namely

Œa D fx 2 Z W x � a mod mg:

(iii) If I is an ideal in a commutative ringR, then the equivalence class of an

element a 2 R is the coset aC I . (iv) The equivalence class of .a; b/ under cross multiplication, where a; b 2 Z

and b ¤ 0, is Œ.a; b/ D f.c; d / W ad D bcg:

If we denote Œ.a; b/ by a=b, then the equivalence class is precisely the

fraction usually denoted by a=b. After all, it is plain that .1; 3/ ¤ .2; 6/, but Œ.1; 3/ D Œ.2; 6/ because 1 � 6 D 3 � 2; that is, 1=3 D 2=6.

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422 Appendix A Appendices

(v) An equivalence class Œ.P;Q/ of arrows, as in Example A.14, is called a

vector; we denote it by Œ.P;Q/ D ��!PQ. N

It is instructive to compare rational numbers and vectors, for both are de-

fined as equivalence classes. Every rational a=b has a “favorite” name—its ex- pression in lowest terms; every vector has a favorite name—an arrow .O;Q/

with its foot at the origin O . Although it is good to have familiar favorites,

working with fractions in lowest terms is not always convenient; for example,

even if both a=b and c=d are in lowest terms, their sum .ad C bc/=bd and product ac=bd may not be. Similarly, it is not always best to think of vectors as arrows with foot at the origin. Vector addition is defined by the parallelo-

gram law (see Figure A.2): ��! OP C ��!OQ D ��!OR, where O , P , Q, and R are the

vertices of a parallelogram. But ��! OQ D �!PR, because .O;Q/ � .P; R/, and it

is more natural to write ��! OP C ��!OQ D ��!OP C �!PR D ��!OR.

O

P a,b= ( )

R a+c,b+d= ( )

Q c,d= ( )

Figure A.2. Parallelogram Law.

The next lemma says that we can replace equivalence by honest equality at

the cost of replacing elements by their equivalence classes.

Lemma A.16. If � is an equivalence relation on a set X , then x � y if and only if Œx D Œy.

Proof. Assume that x � y. If z 2 Œx, then z � x, and so transitivity gives z � y; hence Œx � Œy. By symmetry, y � x, and this gives the reverse inclusion Œy � Œx. Thus, Œx D Œy.

Conversely, if Œx D Œy, then x 2 Œx, by reflexivity, and so x 2 Œx D Œy. Therefore, x � y.

Here is a set-theoretic idea, partitions, that we’ll see is intimately involved

with equivalence relations.

Definition. Subsets A and B of a set X are disjoint if A \ B D ¿; that is, no x 2 X lies in both A and B . A family P of subsets of a set X is called pairwise disjoint if, for all A;B 2 P, either A D B or A \ B D ¿.

A partition of a set X is a family P of nonempty pairwise disjoint subsets,

called blocks, whose union is X .

We are now going to prove that equivalence relations and partitions are

merely different ways of viewing the same thing.

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A.2 Equivalence Relations 423

Proposition A.17. If � is an equivalence relation on a nonempty set X , then the equivalence classes form a partition of X . Conversely, given a partition P

of X , there is an equivalence relation on X whose equivalence classes are the

blocks in P.

Proof. Assume that an equivalence relation � on X is given. Each x 2 X lies in the equivalence class Œx because � is reflexive; it follows that the equiva- lence classes are nonempty subsets whose union is X . To prove pairwise dis-

jointness, assume that a 2 Œx \ Œy, so that a � x and a � y. By symmetry, x � a, and so transitivity gives x � y. Therefore, Œx D Œy, by Lemma A.16, and so the equivalence classes form a partition of X .

Conversely, let P be a partition of X . If x; y 2 X , define x � y if there is A 2 P with x 2 A and y 2 A. It is plain that � is reflexive and symmetric. To see that � is transitive, assume that x � y and y � z; that is, there are A;B 2 P with x, y 2 A and y, z 2 B . Since y 2 A\B , pairwise disjointness gives A D B and so x, z 2 A; that is, x � z. We have shown that � is an equivalence relation.

It remains to show that the equivalence classes are the blocks in P. If x 2 X , then x 2 A for some A 2 P. By the definition of �, if y 2 A, then y � x and y 2 Œx; hence, A � Œx. For the reverse inclusion, let z 2 Œx, so that z � x. There is some B with x 2 B and z 2 B; thus, x 2 A\B . By pairwise disjointness,A D B , so that z 2 A, and Œx � A. Hence, Œx D A.

Corollary A.18. If � is an equivalence relation on a set X and a; b 2 X , then Œa \ Œb ¤ ¿ implies Œa D Œb.

Example A.19. (i) If � is the identity relation on a set X , then the blocks are the one-point subsets of X .

(ii) Let X D Œ0; 2�, and define the partition of X whose blocks are f0; 2�g and the singletons fxg, where 0 < x < 2� . This partition identifies the endpoints of the interval (and nothing else), and so we may regard this as

a construction of a circle. N

Exercises

A.14 Let X D frock; paper; scissorsg. Recall the game whose rules are: paper domi- nates rock, rock dominates scissors, and scissors dominates paper. Draw a subset

of X � X showing that domination is a relation on X .

A.15 Which of the following relations are equivalence relations? State your reasons.

(i) The relation � on R. (ii) The relation R on Z given by m R n if m � n is odd.

(iii) The relation R on Z given by m R n if m � n is even. (iv) The relation on a group of people of having a common friend.

A.16 Let f W X ! Y be a function. Define a relation on X by x � x0 if f .x/ D f .x0/. Prove that � is an equivalence relation. If x 2 X and f .x/ D y, the equivalence class Œx is denoted by f �1.y/; it is called the fiber over y.

A.17 (i) Find the error in the following argument that claims to prove that a symmet-

ric and transitive relation R on a set X must be reflexive; that is, R is an

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424 Appendix A Appendices

equivalence relation on X . If x 2 X and xRy, then symmetry gives yRx and transitivity gives xRx.

Hint: What is y?

(ii) Give an example of a symmetric and transitive relation on the closed unit

interval X D Œ0; 1 that is not reflexive.

A.3 Vector Spaces

Linear algebra is the study of vector spaces and their homomorphisms (namely,

linear transformations), with applications to systems of linear equations. We

assume that most readers have had a course involving matrices with real en-

tries. Such courses deal mainly with computational aspects of the subject, such as Gaussian elimination, finding inverses, and determinants, but here we do not

emphasize this important aspect of linear algebra. Instead, we focus on vector

spaces with only a few words about linear transformations.

Introductory linear algebra courses begin with vector spaces whose scalars

are real numbers but, toward the end of the course, scalars are allowed to be complex numbers. The instructor usually says that the results about vec-

tor spaces over R hold, more generally, for vector spaces over C. This hand-

waving bothers most students. We are now going to generalize the definition

of vector space so that scalars may belong to any field k, and we will prove

that the usual theorems about vector spaces over R do, in fact, hold not only

for vector spaces over C but for vector spaces over k. In particular, they hold for vector spaces over Q or over Fq .

The first definitions do not change when we allow more general scalars.

Definition. If k is a field, then a vector space over k is a set V equipped with

addition V � V ! V , denoted by .u; v/ 7! uC v, that satisfies (i) .uC v/C w D uC .v C w/ for all u; v; w 2 V ,

(ii) there is 0 2 V with 0C v D v for all v 2 V , (iii) for each v 2 V , there is �v 2 V with �v C v D 0, (iv) uC v D v C u for all u; v 2 V ; and scalar multiplication k � V ! V , denoted by .a; v/ 7! av, that satisfies, for all a; b; 1 2 k and all u; v 2 V ,

(i) a.uC v/ D auC av, (ii) .a C b/v D av C bv,

(iii) .ab/v D a.bv/, (iv) 1v D v.

The elements of V are called vectors and the elements of k are called

scalars. It is not difficult to prove that the vector �v in the third axiom of addition is equal to the scalar product .�1/v.

Etymology. The word vector comes from the Latin word meaning “to carry;”

vectors in Euclidean space carry the data of length and direction. The word

scalar comes from regarding v 7! av as a change of scale. The terms scale

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A.3 Vector Spaces 425

and scalar come from the Latin word meaning “ladder,” for the rungs of a

ladder are evenly spaced.

Example A.20. (i) Euclidean space V D Rn is a vector space over R. Vec- tors are n-tuples .a1; : : : ; an/, where ai 2 R for all i . Picture a vector v as an arrow from the origin to the point having coordinates .a1; : : : ; an/.

Addition is given by

.a1; : : : ; an/C .b1; : : : ; bn/ D .a1 C b1; : : : ; an C bn/I

geometrically, the sum of two vectors is described by the parallelogram

law (see Figure A.2 on page 422).

Scalar multiplication is given by

av D a.a1; : : : ; an/ D .aa1; : : : ; aan/:

Scalar multiplication v 7! av “stretches” v by a factor jaj, reversing its direction when a is negative (we put quotes around stretches because av

is shorter than v when jaj < 1). (ii) The example in part (i) can be generalized. If k is any field, define V D

kn, the set of all n � 1 column vectors v D .a1; : : : ; an/, where ai 2 k for all i . Addition is given by

.a1; : : : ; an/C .b1; : : : ; bn/ D .a1 C b1; : : : ; an C bn/;

and scalar multiplication is given by

av D a.a1; : : : ; an/ D .aa1; : : : ; aan/:

(iii) The polynomial ring R D kŒx, where k is a field, is another example of a vector space over k. Vectors are polynomials f , scalars are elements

a 2 k, and scalar multiplication gives the polynomial af ; that is, if

f D bnxn C � � � C b1x C b0; then

af D abnxn C � � � C ab1x C ab0:

Thus, the polynomial ring kŒx is a vector space over k.

(iv) Let R be a commutative ring and let k be a subring; if k is a field, then

R can be viewed as a vector space over k. Regard the elements of R as

vectors and the elements of k as scalars; define scalar multiplication av,

where a 2 k and v 2 R, to be the given product of two elements in R. The axioms in the definition of vector space are just particular cases of axioms holding in the commutative ring R. For example, if a field k is a

subfield of a larger field E , then E is a vector space over k; in particular,

C is a vector space over R, and it is also a vector space over Q.

(v) The set CŒ0; 1 of all continuous real-valued functions on the closed in- terval Œ0; 1 is a vector space over R with the usual operations: if f; g 2 CŒ0; 1 and c 2 R, then

f C gW a 7! f .a/C g.a/ cf W a 7! cf .a/: N

Informally, a subspace of a vector space V is a nonempty subset of V that

is closed under addition and scalar multiplication in V .

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426 Appendix A Appendices

Definition. If V is a vector space over a field k, then a subspace of V is a

subset U of V such that

(i) 0 2 U , (ii) u; u0 2 U imply uC u0 2 U ,

(iii) u 2 U and a 2 k imply au 2 U .

Every subspace U of a vector space V is itself a vector space. For example,

since u1 C .u2 C u3/ D .u1 C u2/C u3 holds for all vectors u1; u2; u3 2 V , it holds, in particular, for all vectors u1; u2; u3 2 U .

Example A.21. (i) The extreme cases U D V and U D f0g (where f0g de- notes the subset consisting of the zero vector alone) are always subspaces

of a vector space. A subspace U � V withU ¤ V is called a proper sub- space of V ; we may write U ¨ V to denote U being a proper subspace of V .

(ii) If v D .a1; : : : ; an/ is a nonzero vector in Rn, then the line through the origin,

` D fav W a 2 Rg;

is a subspace of Rn.

Similarly, a plane through the origin consists of all vectors of the form

av1 C bv2, where v1; v2 is a fixed pair of noncollinear vectors, and a; b vary over R. It is easy to check that planes through the origin are sub-

spaces of Rn.

(iii) If k is a field, then a homogeneous linear system over k of m equations

in n unknowns is a set of equations

a11x1 C � � � C a1nxn D 0 a21x1 C � � � C a2nxn D 0

::: :::

am1x1 C � � � C amnxn D 0;

where aj i 2 k. A solution of this system is an n � 1 column vector c D .c1; : : : ; cn/ 2 kn, where

P i aj ici D 0 for all j ; a solution .c1; : : : ; cn/ is

nontrivial if some ci ¤ 0. The set of all solutions forms a subspace of kn, called the solution space (or nullspace) of the system. Using matrices, we

can say this more succinctly: if A D Œaij  is them � n coefficient matrix, then the linear system isAx D 0 and a solution is an n�1 column vector c for which Ac D 0.

In particular, we can solve systems of linear equations over Fp, where

p is a prime. This says that we can treat a not necessarily homogeneous

system of congruences mod p just as one treats an ordinary system of

equations.

For example, the system of congruences

3x � 2y C z � 1 mod 7 x C y � 2z � 0 mod 7

�x C 2y C z � 4 mod 7

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A.3 Vector Spaces 427

can be regarded as a system of equations over the field F7. The system

can be solved just as in high school, for inverses mod 7 are now known:

Œ2Œ4 D Œ1; Œ3Œ5 D Œ1; Œ6Œ6 D Œ1. The solution is

.x; y; z/ D .Œ5; Œ4; Œ1/: N

Bases and Dimension

The key observation in getting the “right” definition of dimension is to under- stand why R3 is 3-dimensional. Every vector .x; y; z/ is a linear combination

of the three vectors e1 D .1; 0; 0/, e2 D .0; 1; 0/, and e3 D .0; 0; 1/; that is,

.x; y; z/ D xe1 C ye2 C ze3:

It is not so important that every vector is a linear combination of these specific

vectors; what is important is that there are three of them, for it turns out that three is the smallest number of vectors with this property; that is, one cannot

find two vectors u D .a; b; c/ and u0 D .a0; b0; c0/ with every vector a linear combination of u and u0.

Definition. A list in a vector space V is an ordered set X D v1; : : : ; vn of vectors in V .

More precisely, we are saying that there is some n � 1 and a function

'W f1; 2; : : : ; ng ! V;

with '.i/ D vi for all i . Thus, the subset im ' is ordered in the sense that there is a first vector v1, a second vector v2, and so forth. A vector may appear

several times on a list; that is, ' need not be injective.

Definition. Let V be a vector space over a field k. A k-linear combination of We often write linear combination instead of

k-linear combination if it

is clear where the scalar

coefficients live.

a list v1; : : : ; vn in V is a vector v of the form

v D a1v1 C � � � C anvn;

where ai 2 k for all i .

Definition. If X D v1; : : : ; vm is a list in a vector space V , then

Span hXi D hv1; : : : ; vmi ;

the set of all the k-linear combinations of v1; : : : ; vm, is called the subspace

spanned by X. We also say that v1; : : : ; vm spans Spanhv1; : : : ; vmi.

It is easy to check that Span hv1; : : : ; vmi is, indeed, a subspace.

Lemma A.22. Let V be a vector space over a field k.

(i) Every intersection of subspaces of V is itself a subspace.

(ii) If X D v1; : : : ; vm is a list in V , then the intersection of all the sub- spaces of V containing X is Span hv1; : : : ; vmi, the subspace spanned by v1; : : : ; vm, and so Span hv1; : : : ; vmi is the smallest subspace of V containingX .

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428 Appendix A Appendices

Proof. Part (i) is routine. For part (ii), let X D fv1; : : : ; vmg and let S denote the family of all the subspaces of V containingX ; we claim that

\

S2S S D Span hv1; : : : ; vmi :

The inclusion � is clear, because S D Span hv1; : : : ; vmi 2 S. For the reverse inclusion, note that if S 2 S, then S contains v1; : : : ; vm, and so it contains the set of all k-linear combination of v1; : : : ; vm, namely, Span hv1; : : : ; vmi.

The next observation is important.

Corollary A.23. The subspace spanned by a list X D v1; : : : ; vm does not depend on the ordering of the vectors, but only on the set of vectors themselves.

Proof. This follows from part (ii) of Lemma A.22.

See Exercise A.24 on page 440 to see other properties of a list that do not depend on the ordering of its vectors.

If X D ¿, then Span hXi D T S2S S , where S is the family of all the

subspaces of V containing X . Now f0g � Span h¿i D T S2S S , for f0g is

contained in every subspace S of V . For the reverse inclusion, one of the sub-

spaces S of V occurring in the intersection is f0g itself, and so Span h¿i DT S�V S � f0g. Therefore, Span h¿i D f0g.

Example A.24. (i) Let V D R2, let e1 D .1; 0/, and let e2 D .0; 1/. Now V D Span he1; e2i, for if v D .a; b/ 2 V , then

v D .a; 0/C .0; b/ D a.1; 0/C b.0; 1/ D ae1 C be2 2 Span he1; e2i :

(ii) If k is a field and V D kn, define ei as the n�1 column vector having 1 in the i th coordinate and 0s elsewhere. The reader may adapt the argument

in part (i) to show that e1; : : : ; en spans k n.

(iii) A vector space V need not be spanned by a finite list. For example, let

V D kŒx, and suppose that X D f1.x/; : : : ; fm.x/ is a finite list in V . If d is the largest degree of any of the fi , then every (nonzero) k-linear

combination of f1; : : : ; fm has degree at most d . Thus, x dC1 is not a

k-linear combination of vectors inX , and so X does not span kŒx. N

The following definition makes sense even though we have not yet defined

dimension.

Definition. A vector space V is called finite-dimensional if it is spanned by a

finite list; otherwise, V is called infinite-dimensional.

Part (ii) of Example A.24 shows that kn is finite-dimensional, while part (iii)

shows that kŒx is infinite-dimensional. Now C is a vector space over R, and it

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A.3 Vector Spaces 429

is finite-dimensional (C is spanned by 1; i ); by Example A.20(iii), both R and

C are vector spaces over Q (each can be shown to be infinite-dimensional).

If a subspace U of a vector space V is finite-dimensional, then there is a

list v1; v2; : : : ; vm that spans U . But there are many such lists: if u is a vector

in U , then the extended list v1; v2; : : : ; vm; u also spans U . Let us, therefore,

seek a shortest list that spans U .

Notation. If v1; : : : ; vm is a list, then v1; : : : ; bvi : : : ; vm is the shorter list with vi deleted.

Proposition A.25. If V is a vector space over a field k, then the following

conditions on a list v1; : : : ; vm spanning V are equivalentW

(i) v1; : : : ; vm is not a shortest spanning list

(ii) some vi is in the subspace spanned by the others

(iii) there are scalars a1; : : : ; am, not all zero, with

mX

`D1 a`v` D 0:

Proof. (i) ) (ii). If v1; : : : ; vm is not a shortest spanning list, then one of the vectors, say vi , can be thrown out, and the shorter list still spans. Hence,

vi 2 Span hv1; : : : ; bvi ; : : : ; vmi. (ii) ) (iii). If vi D

P j¤i cj vj , define ai D �1 ¤ 0 and aj D cj for all

j ¤ i .

(iii) ) (i). The given equation implies that one term, say aivi , is nonzero. Since k is a field, a�1i exists, and

vi D �

� a�1i �X

j¤i aj vj : (A.1)

Deleting vi gives a shorter list that still spans V : write any v 2 V as a lin- ear combination of all the vj (including vi ); now substitute the expression Eq. (A.1) for vi and collect terms.

We now give a name to lists described in Proposition A.25.

Definition. A list X D v1; : : : ; vm in a vector space V is linearly dependent if there are scalars a1; : : : ; am, not all zero, with

Pm `D1 a`v` D 0; otherwise,

X is called linearly independent.

The empty set ¿ is defined to be linearly independent (we interpret ¿ as a list of length 0).

Example A.26. (i) A list X D v1; : : : ; vm containing the zero vector is lin- early dependent: if vj D 0, then

P i aivi D 0, where aj D 1 and ai D 0

for i ¤ j . (ii) A list v1 of length 1 is linearly dependent if and only if v1 D 0; hence, a

list v1 of length 1 is linearly independent if and only if v1 ¤ 0.

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430 Appendix A Appendices

(iii) A list v1; v2 is linearly dependent if and only if one of the vectors is

a scalar multiple of the other: if a1v1 C a2v2 D 0 and a1 ¤ 0, then v1 D �.a2=a1/v2. Conversely, if v2 D cv1, then cv1 � v2 D 0 and the list v1; v2 is linearly dependent (for the coefficient �1 of v2 is nonzero).

(iv) If there is a repetition in the list v1; : : : ; vm (that is, if vi D vj for some i ¤ j ), then v1; : : : ; vm is linearly dependent: define ci D 1, cj D �1, and all other c D 0. Therefore, if v1; : : : ; vm is linearly independent, then all the vectors vi are distinct. N

Linear independence has been defined indirectly, as not being linearly de-

pendent. Because of the importance of linear independence, let us define it directly.

Definition. A list v1; : : : ; vm is linearly independent if, whenever a k-linear

combination Pm `D1 a`v` D 0, then every ai D 0.

It follows that every sublist of a linearly independent list is itself linearly

independent (this is one reason for decreeing that ¿ be linearly independent).

Corollary A.27. If X D v1; : : : ; vm is a list spanning a vector space V , then X is a shortest spanning list if and only ifX is linearly independent.

Proof. These are just the contrapositives of (i) ) (iii) and (iii) ) (i) in Propo- sition A.25.

We have arrived at the notion we have been seeking.

Definition. A basis of a vector space V is a linearly independent list that

spans V .

Thus, bases are shortest spanning lists. Of course, all the vectors in a linearly

independent list v1; : : : ; vn are distinct, by Example A.26(iv).

Example A.28. In Example A.24(ii), we saw that X D e1; : : : ; en spans kn, where ei is the n � 1 column vector having 1 in the i th coordinate and 0s elsewhere. We now show that X is linearly independent. If 0 D

P i ciei , then

c1e1 D .c1; 0; 0; : : : ; 0/ C c2e2 D C .0; c2; 0; : : : ; 0/

::: :::

C cnen D C .0; 0; 0; : : : ; cn/ 0 D .c1; c2; : : : ; cn/:

Hence, ci D 0 for all i , X is linearly independent, and X is a basis; it is called the standard basis of kn. N

Proposition A.29. A list X D v1; : : : ; vn in a vector space V over a field k is a basis of V if and only if each v 2 V has a unique expression as a k-linear combination of the vectors in X .

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A.3 Vector Spaces 431

Proof. Since X is a basis, it spans V , and so each vector v 2 V is a k-linear combination: v D

P aivi . If also v D

P bivi , then

P .ai � bi/vi D 0, and

linear independence gives ai D bi for all i ; that is, the expression is unique. Conversely, existence of an expression shows that the listX spans V . More-

over, if 0 D P civi with not all ci D 0, then the vector 0 does not have a

unique expression as a linear combination of the vi : a second expression is

0 D P aivi with all ai D 0.

Definition. If X D v1; : : : ; vn is a basis of a vector space V and v 2 V , then Proposition A.29 says that there are unique scalars a1; : : : ; an with

v D nX

iD1 aivi :

The n-tuple .a1; : : : ; an/ is called the coordinate list of a vector v 2 V relative to the basis X .

If v1; : : : ; vn is the standard basis of V D kn, then this coordinate list coincides with the usual coordinate list.

How to Think About It. If v1; : : : ; vn is a basis of a vector space V over a field k, then each vector v 2 V has a unique expression

v D a1v1 C a2v2 C � � � C anvn;

where ai 2 k for all i . Since there is a first vector v1, a second vector v2, and so forth, the coefficients in this k-linear combination determine a unique

n-tuple .a1; a2; : : : ; an/. Were a basis merely a subset of V and not a list (i.e.,

an ordered subset), then there would be nŠ coordinate lists for every vector. But see Exercise A.24(iv) on page 440.

We are going to define the dimension of a vector space V to be the number of

vectors in a basis. Two questions arise at once.

(i) Does every vector space have a basis?

(ii) Do all bases of a vector space have the same number of elements?

The first question is easy to answer; the second needs some thought.

Theorem A.30. Every finite-dimensional vector space V has a basis.

Proof. A finite spanning list X exists, since V is finite-dimensional. If X is

linearly independent, it is a basis; if not, Proposition A.25 says that we can

throw out some element from X , leaving a shorter spanning list, say X 0. If X 0 is linearly independent, it is a basis; if not, we can throw out an element from X 0 leaving a shorter spanning sublist. Eventually, we arrive at a shortest spanning list, which is linearly independent, by Corollary A.27 and hence it is

a basis.

The definitions of spanning and linear independence can be extended to

infinite lists in a vector space, and we can then prove that infinite-dimensional

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432 Appendix A Appendices

vector spaces also have bases. For example, it turns out that a basis of kŒx is

1; x; x2; : : : ; xn; : : : .

We can now prove invariance of dimension, perhaps the most important result about vector spaces.

Lemma A.31. Let u1; : : : ; un; v1; : : : ; vm be elements in a vector space V

with v1; : : : ; vm 2 Spanhu1; : : : ; uni. If m > n, then v1; : : : ; vm is linearly dependent.

Proof. The proof is by induction on n � 1. Base Step. If n D 1, there are at least two vectors v1; v2, and v1 D a1u1 and v2 D a2u1. If u1 D 0, then v1 D 0 and the list of v’s is linearly dependent (by Example A.26(i)). Suppose u1 ¤ 0. We may assume that v1 ¤ 0, or we are done; hence, a1 ¤ 0. Therefore, v1; v2 is linearly dependent, for 1 � v2 � a2a

�1 1 v1 D 0 and hence the larger list v1; : : : ; vm is linearly dependent.

Inductive Step. There are equations

vi D ai1u1 C � � � C ainun

for i D 1; : : : ; m. We may assume that some ai1 ¤ 0, otherwise v1; : : : ; vm 2 hu2; : : : ; uni, and the inductive hypothesis applies. Changing notation if nec- essary (that is, by re-ordering the v’s), we may assume that a11 ¤ 0. For eachWrite out the proof in the

special case m D 3 and nD 2.

i � 2, define

v0i D vi � ai1a�111 v1 2 Spanhu2; : : : ; uni:

Each v0i is a linear combination of the u’s, and the coefficient of u1 is ai1 � .ai1a

�1 11 /a11 D 0. Sincem� 1 > n� 1, the inductive hypothesis gives scalars

b2; : : : ; bm, not all 0, with

b2v 0 2 C � � � C bmv0m D 0:

Rewrite this equation using the definition of v0i :

� � X

i�2 biai1a

�1 11

� v1 C b2v2 C � � � C bmvm D 0:

Not all the coefficients are 0, and so v1; : : : ; vm is linearly dependent.

The following familiar fact illustrates the intimate relation between linear

algebra and systems of linear equations.

Corollary A.32. If a homogeneous system of linear equations over a field k

has more unknowns than equations, then it has a non-trivial solution.

Proof. Recall that an n-tuple .ˇ1; : : : ; ˇn/ is a solution of a system

˛11x1 C � � � C ˛1nxn D 0 :::

::: :::

˛m1x1 C � � � C ˛mnxn D 0

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A.3 Vector Spaces 433

if ˛i1ˇ1 C � � � C ˛inˇn D 0 for all i . In other words, if c1; : : : ; cn are the columns of them � n coefficient matrix A D Œ˛ij  (note that ci 2 km), then

ˇ1c1 C � � � C ˇncn D 0:

Now km can be spanned by m vectors (the standard basis, for example).

Since n > m, by hypothesis, Lemma A.31 shows that the list c1; : : : ; cn is

linearly dependent; there are scalars 1; : : : ; n, not all zero, with 1c1 C � � �C ncn D 0. Therefore, . 1; : : : ; n/ is a nontrivial solution of the system.

Theorem A.33 (Invariance of Dimension). If X D x1; : : : ; xn and Y D y1; : : : , ym are bases of a vector space V , then m D n.

Proof. If m ¤ n, then either n < m or m < n. In the first case, y1; : : : ; ym 2 Spanhx1; : : : ; xni, because X spans V , and Lemma A.31 gives Y linearly de- pendent, a contradiction. A similar contradiction arises if m < n, and so we

must have m D n.

It is now permissible to make the following definition, for all bases of a

vector space have the same size.

Definition. If V is a finite-dimensional vector space over a field k, then its

dimension, denoted by dimk.V / or by dim.V /, is the number of elements in

a basis of V .

Corollary A.34. Let k be a finite field withq elements. If V is an n-dimensional

vector space over k, then jV j D qn.

Proof. If v1; : : : ; vn is a basis of V , then every v 2 V has a unique expression

v D c1v1 C � � � C cnvn;

where ci 2 k for all i . There are q choices for each ci , and so there are qn vectors in V .

Example A.35. (i) Example A.28 shows that kn has dimension n, which

agrees with our intuition when k D R: the plane R � R is 2-dimensional, and R3 is 3-dimensional!

(ii) If V D f0g, then dim.V / D 0, for there are no elements in its basis ¿. (This is another good reason for defining ¿ to be linearly independent.)

(iii) Let I be a finite set with n elements. Define

kI D ffunctions f W I ! kg:

Now kI is a vector space if we define addition f C f 0 to be

f C f 0W i 7! f .i/C f 0.i/

and scalar multiplication af , for a 2 k and f W I ! k, by

af W i 7! af .i/

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434 Appendix A Appendices

(see Exercise A.18(i) on page 439). It is easy to check that the set of n

functions of the form fi , where i 2 I , defined by

fi .j / D ( 1 if j D i 0 if j ¤ i

form a basis, and so dim.kI / D n D jI j. This is not a new example: an n-tuple .a1; : : : ; an/ is really a func-

tion f W f1; : : : ; ng ! k with f .i/ D ai for all i . Thus, the functions fi comprise the standard basis. N

Definition. A longest (or maximal) linearly independent listX D u1; : : : ; um is a linearly independent list for which there is no vector v 2 V such that u1; : : : ; um; v is linearly independent.

Lemma A.36. If V is a finite-dimensional vector space, then a longest linearly

independent list X D v1; : : : ; vn is a basis of V .

Proof. If v1; : : : ; vn is not a basis, then it does not span V , for this list is lin-

early independent. Thus, there is w 2 V with w … Span hv1; : : : ; vni. But the longer list v1; : : : ; vn; w is linearly independent, by Proposition A.25, contra- dictingX being a longest linearly independent list.

The converse of Lemma A.36 is true; bases are longest linearly independent

lists. This follows from the next proposition, which is quite useful in its own

right.

Proposition A.37. Let V be an n-dimensional vector space. IfZ D u1; : : : ; um is a linearly independent list in V , where m < n, then Z can be extended to

a basisI that is, there are vectors vmC1; : : : ; vn 2 V such that u1; : : : ; um, vmC1; : : : ; vn is a basis of V .

Proof. If the linearly independent list Z does not span V , there is vmC1 2 V with vmC1 … Span hu1; : : : ; umi, and the longer list u1; : : : ; um; vmC1 is lin- early independent, by Proposition A.25. If u1; : : : ; um; vmC1 does not span V , there is vmC2 2 V with vmC2 … Span hu1; : : : ; um; vmC1i. Since dim.V / D n, Lemma A.31 says that the length of these lists can never exceed n, and so this

process of adjoining elements vmC1; vmC2; : : : must stop. But the only reason a list stops is that it spans V ; hence, it is a basis.

Corollary A.38. Let V be an n-dimensional vector space. Then a list in V is

a basis if and only if it is a longest linearly independent list.

Proof. Lemma A.36 shows that longest linearly independent lists are bases.

Conversely, ifX is a basis, it must be a longest linearly independent list: other-

wise, Proposition A.37 says we could lengthenX to obtain a basis of V which is too long.

We now paraphrase Lemma A.31.

Corollary A.39. If dim.V / D n, then a list of nC1 or more vectors is linearly dependent.

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A.3 Vector Spaces 435

Proof. Otherwise, the list could be extended to a basis having too many ele-

ments.

Corollary A.40. Let V be a vector space with dim.V / D n. (i) A list of n vectors that spans V must be linearly independent.

(ii) Any linearly independent list of n vectors must span V .

In either case, the list is a basis of V .

Proof. (i) Were the list linearly dependent, then it could be shortened to give

a basis, and this basis is too small.

(ii) If the list does not span, then it could be lengthened to give a basis, and

this basis is too large.

Corollary A.41. Let U be a subspace of a vector space V of dimension n.

(i) U is finite-dimensional.

(ii) dim.U / � dim.V /. (iii) If dim.U / D dim.V /, then U D V .

Proof. (i) Take u1 2 U . If U D Span hu1i, then U is finite-dimensional. If U ¤ Span hu1i, there is u2 … Span hu1i. By Proposition A.25, u1; u2 is linearly independent. If U D Span hu1; u2i, we are done; if not, there is u3 … Span hu1; u2i, and the list u1; u2; u3 is linearly independent. This process cannot be repeated nC 1 times, for then u1; : : : ; unC1 would be a linearly independent list in U � V , contradicting Corollary A.39.

(ii) A basis of U is linearly independent, and so it can be extended to a basis

of V .

(iii) If dim.U / D dim.V /, then a basis of U is already a basis of V (otherwise it could be extended to a basis of V that would be too large).

Linear Transformations

Linear transformations are homomorphisms of vector spaces; they are really

much more important than vector spaces, but vector spaces are needed in order to define them, and bases of vector spaces are needed to describe them by

matrices. (You are surely familiar with the next definition, at least for k D R.)

Definition. Let V and W be vector spaces over a field k. A linear transfor-

mation is a function T WV ! W such that, for all vectors v; v0 2 V and scalars a 2 k, we have

(i) T .v C v0/ D T .v/C T .v0/ (ii) T .av/ D aT .v/.

It follows by induction on n � 1 that linear transformations preserve linear combinations:

T .a1v1 C � � � C anvn/ D a1T .v1/C � � � C anT .vn/:

You’ve certainly seen many examples of linear transformations in a linear

algebra course. Here are a few more.

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436 Appendix A Appendices

Example A.42. (i) If A is an m � n matrix, then x 7! Ax, where x is an n � 1 column vector, is a linear transformation kn ! km.

(ii) If we regard the complex numbers C as a 2-dimensional vector space

over R, then complex conjugation T W z 7! z is a linear transformation. (iii) Let V D kŒx, where k is a field. If a 2 k, then evaluation eaW kŒx ! k

is a linear transformation (we can view k as a 1-dimensional vector space

over itself).

(iv) Integration f 7! R 1 0 f .x/ dx is a linear transformationCŒ0; 1 ! R (see

Example A.20(v)). N

We now associate matrices to linear transformations.

Theorem A.43. Let V andW be vector spaces over a field k. If v1; : : : ; vn is a

basis of V andw1; : : : ; wn is a list of elements inW .possibly with repetitions/,

then there exists a unique linear transformation T WV ! W with T .vj / D wj for all j .

Proof. By Proposition A.29, every vector v 2 V has a unique expression as a linear combination of basis vectors:

v D a1v1 C � � � C anvn:

Therefore, there is a well-defined function T WV ! W with T .vj / D wj for all j , namely

T .a1v1 C � � � C anvn/ D a1w1 C � � � C anwn:

It is routine to check that T is a linear transformation. If v0 D a01v1C� � �Ca0nvn, then

v C v0 D .a1 C a01/v1 C � � � C .an C a0n/vn; and

T .v C v0/ D .a1 C a01/w1 C � � � C .an C a0n/wn D � a1w1 C � � � C anwn

� C � a01w1 C � � � C a0nwn

D T .v/C T .v0/:

Similarly,

T .av/ D T � a.a1v1 C � � � C anvn/

D T .aa1v1 C � � � C aanvn/ D aa1w1 C � � � C aanwn D a.a1w1 C � � � C anwn/ D aT .v/:

To prove uniqueness, suppose that S WV ! W is a linear transformation with S.vj / D wj for all j . Since S preserves linear combinations,

S.a1v1 C � � � C anvn/ D a1S.v1/C � � � C anS.vn/ D a1w1 C � � � C anwn D T .a1v1 C � � � C anvn/;

and so S D T .

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A.3 Vector Spaces 437

Definition. Let T WV ! W be a linear transformation. Given bases v1; : : : ; vn and w1; : : : ; wm of V and W , respectively, each T .vj / is a linear combination

of the w’s:

T .vj / D a1jw1 C � � � C amjwm:

Them�n matrix A D Œaij  whose j th column is a1j ; : : : ; amj , the coordinate list of T .vj / with respect to the w’s, is called the matrix associated to T .

Example A.44. As in Example A.42(ii), view C as R2. A basis is 1; i ; that

is, .1; 0/; .0; 1/. Since 1 D 1 and i D �i D .0;�1/, the matrix of complex conjugation relative to the basis is

A D " 1 0

0 �1

# : N

Of course, the matrix A associated to a linear transformation T WV ! W depends on the choices of bases of V and of W .

The next theorem shows why the notation a1jw1 C � � � C amjwm is chosen instead of aj1w1 C � � � C ajmwm.

Theorem A.45. If T W kn ! km is a linear transformation, then

T .v/ D Av;

where A is the matrix associated to T from the standard bases of kn and km,

v is an n � 1 column vector, and Av is matrix multiplication.

Proof. If A is an m � n matrix and vj is the n � 1 column vector whose j th entry is 1 and whose other entries are 0, thenAvj is the j th column of A. Thus,

Avj D T .vj / for all j , and so Av D T .v/ for all v 2 kn, by Exercise A.29 on page 441.

In Appendix A.1, we defined functions to be equal if they have the same

domain, same target, and same graph. It is natural to require the same domain

and the same graph, but why should we care about the target? The coming

discussion gives a (persuasive) reason why targets are important.

Definition. Let V be a vector space over k, and regard k as a 1-dimensional

vector space over itself. A functional on V is a linear transformation

f WV !k. The dual space V � of a vector space V is the set of all functionals on V .

It is shown in Example A.42 that evaluation and integration give rise to

functionals.

Proposition A.46. Let T WV ! W be a linear transformation,where V andW are vector spaces over a field k.

(i) V � is a vector space over k.

(ii) If f 2 W �, then the composite f ı T is in V �:

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438 Appendix A Appendices

(iii) The function T �WW � ! V �, given by T �Wf 7! f ı T , is a linear transformation.

Proof. (i) This is Exercise A.30 on page 441.

(ii) Since T WV ! W and f WW ! k are linear transformations, so is their composite f ı T WV ! k; that is, f ı T is a functional on V .

(iii) That f 7! f ıT is a linear transformationW � ! V � follows easily from the formulas .f C g/ ı T D f ı T C g ı T and .cf / ı T D c.f ı T /,Use Proposition A.2 to

prove .f C g/ ı T D f ı T C g ı T : just evaluate both sides on

v 2 V .

where f; g 2 W � and c 2 k (note that cf is a functional on W , for W � is a vector space).

Proposition A.47. If v1; : : : ; vn is a basis of a vector space V over a field k,

then there are functionals v�j WV ! k, for each j , with

v�j .vi / D ( 1 if i D j 0 if i ¤ j;

and v�1 ; : : : ; v � n is a basis of V

� .it is called the dual basis/.

Proof. By Theorem A.43, it suffices to prescribe the values of v�j on a basis of V .

Linear Independence: If P j cj v

� j D 0, then

P j cj v

� j .v/ D 0 for all v 2

V . But P j cjv

� j .vj / D cj , so that all the coefficients cj are 0 and, hence,

v�1 ; : : : ; v � n is linearly independent.

Spanning: If g 2 V �, then g.vj / D dj 2 k for all j . But g D P j dj v

� j , for

both sides send each vj to dj . Thus, g is a linear combination of v � 1 ; : : : ; v

� n .

Corollary A.48. If V is an n-dimensional vector space, then

dim.V �/ D n D dim.V /:

Proof. A basis of V and its dual basis have the same number of elements.

If T WV ! W is a linear transformation, what is a matrix associated to the linear transformation T �WW � ! V �?

Lemma A.49. Let v1; : : : ; vn be a basis of a vector space V over k. If g 2 V �, then g D d1v�1 C � � � C dnv�n , where dj D g.vj / for all j . Therefore, the coordinate list of g relative to the dual basis v�1 ; : : : ; v

� n of V

� is d1; : : : ; dn.

Proof. We saw this in the proof of Proposition A.47, when showing that the

dual basis spans V �.

The next result shows that dual spaces are intimately related to transposing matrices. If A D Œaij  is an m � n matrix, then its transpose A> is the n �m matrix Œaj i  whose ij entry is aj i . In words, for each i , the i th row ai1; : : : ; ain of A is the i th column of A> (and, necessarily, each j th column of A is the j th row of A>).

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A.3 Vector Spaces 439

Proposition A.50. If T WV ! W is a linear transformation andA is the matrix of T arising from bases v1; : : : ; vn of V andw1; : : : ; wm ofW , then the matrix

of T �WW � ! V � arising from the dual bases is the transpose A> of A.

Proof. Let B be the matrix associated to T �WW � ! V �. The recipe for con- structing B says that if w�i is a basis element, then the i th column of B is the coordinate list of T �.w�i /. Let’s unwind this. First, T

�.w�i / D w�i ıT . Second, the coordinate list of w�i ı T is obtained by writing it as a linear combination of v�1 ; : : : ; v

� n : Lemma A.49 does this by computing

� w�i ı T

� .vj / for all j .

Now

� w�i ı T

� .vj / D w�i .T .vj // D w�i

� a1jw1 C � � � C amjwm

� D aij :

Thus, the i th column of B is ai1; : : : ; aim; that is, the i th column of B is the

i th row of A. In other words, B D A>.

If T WV ! W is a linear transformation, then the domain of T � is W �, which depends on the target of T . Suppose that W is a subspace of a vector

space U ; let i WW ! U be the inclusion. Now S D i ı T WV ! U is also a linear transformation. The transformations T and S have the same domain,

namely V , and the same graph (for T .v/ D S.v/ for all v 2 V ); they differ only in their targets. Now T �WW � ! V �, while S�WU � ! V �. Since T � and S� have different domains, they are certainly different functions, for we have agreed that the domain of a function is a necessary ingredient of its definition.

We conclude that S and T are distinct; that is, if you like transposes of matri-

ces, then you must admit that targets are essential ingredients of functions.

Exercises

A.18 (i) * If k is a field, c 2 k, and f W k ! k is a function, define a new function cf W k ! k by a 7! cf .a/. With this definition of scalar multiplication, prove that the commutative ring kk of all functions on k is a vector space

over k (see Example A.35(iii)).

(ii) Prove that Poly.k/, the set of all polynomial functions k ! k, is a subspace of kk .

A.19 If the only subspaces of a vector space V are f0g and V itself, prove that dim.V / � 1.

A.20 Prove, in the presence of all the other axioms in the definition of vector space,

that the commutative law for vector addition is redundant; that is, if V satisfies

all the other axioms, then u C v D v C u for all u; v 2 V .

Hint: If u; v 2 V , evaluate �Œ.�v/ C .�u/ in two ways.

A.21 If V is a vector space over F2 and if v1 ¤ v2 are nonzero vectors in V , prove that v1; v2 is linearly independent. Is this true for vector spaces over any other

field?

A.22 Prove that the columns of an m�n matrix A over a field k are linearly dependent in km if and only if the homogeneous system Ax D 0 has a nontrivial solution.

A.23 Prove that the list of polynomials 1; x; x2; x3; : : : ; x100 is a linearly independent

list in kŒx, where k is a field.

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440 Appendix A Appendices

A.24 * Let X D v1; : : : ; vn be a list in a vector space V , and let Y D y1; : : : ; yn be a permutation of v1; : : : ; vn.

(i) Prove that X spans V if and only if Y spans V .

(ii) Prove that X is linearly independent if and only if Y is linearly independent.

(iii) Prove that X is a basis of V if and only if Y is a basis of V .

(iv) Conclude that spanning, being linearly independent, or being a basis are

properties of a subset of vectors, not merely of a list of vectors. See Corol-

lary A.23.

A.25 It is shown in analytic geometry that if `1 and `2 are lines with slopes m1 and

m2 , respectively, then `1 and `2 are perpendicular if and only if m1m2 D �1. If

`i D f˛vi C ui W ˛ 2 Rg;

for i D 1; 2; prove that m1m2 D �1 if and only if the dot product v1 � v2 D 0. (Since both lines have slopes, neither is vertical.) See Lemma 3.16.

A.26 (i) In calculus, a line in space passing through a point u is defined as

fu C ˛w W ˛ 2 Rg � R3;

where w is a fixed nonzero vector. Show that every line through the origin is

a one-dimensional subspace of R3.

(ii) In calculus, a plane in space passing through a point u is defined as the subset

fv 2 R3 W .v � u/ � n D 0g � R3;

where n ¤ 0 is a fixed normal vector. Prove that a plane through the origin is a two-dimensional subspace of R3.

If the origin .0; 0; 0/ lies on a plane H , then u D 0 and

H D fv D .x; y; z/ 2 R3 W v � n D 0g;

where n D .˛; ˇ; / is a (nonzero) normal vector; that is, H is the set of all vectors orthogonal to n.

A.27 If U and W are subspaces of a vector space V , define

U C W D fu C w W u 2 U and w 2 W g:

(i) Prove that U C W is a subspace of V . (ii) If U and U 0 are subspaces of a finite-dimensional vector space V , prove that

dim.U / C dim.U 0/ D dim.U \ U 0/ C dim.U C U 0/:

Hint: Extend a basis of U \ U 0 to a basis of U and to a basis of U 0.

A.28 If U and W are vector spaces over a field k, define their direct sum to be the set

of all ordered pairs,

U ˚ W D f.u; w/ W u 2 U and w 2 W g;

with addition

.u; w/ C .u0; w0/ D .u C u0; w C w0/

and scalar multiplication

˛.u; w/ D .˛u; ˛w/:

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A.4 Inequalities 441

(i) Show that U ˚ W is a vector space. (ii) If U and W are finite-dimensional vector spaces over a field k, prove that

dim.U ˚ W / D dim.U / C dim.W /:

A.29 * Let S; T W V ! W be linear transformations, where V and W are vector spaces over a field k. Prove that if there is a basis v1; : : : ; vn of V for which S.vj / D T .vj / for all j , then S D T .

A.30 * Prove that the dual space V � of a vector space V over a field k is a vector space over k.

A.4 Inequalities

Many properties of inequality follow from a few basic properties. Denote the Recall that N is the set of all nonnegative integers,

so that N D P [ f0g. set of all positive real numbers by P (we do not regard 0 as positive). We assume the set P satisfies

(i) a; b 2 P implies a C b 2 P (ii) a; b 2 P implies ab 2 P

(iii) Trichotomy W If a is a number, then exactly one of the following is true:

a 2 P; a D 0; �a 2 P:

The first two properties say that P is closed under addition and multiplica- tion. We now define inequality.

Definition. Given real numbers a and b, we say that a is less than b, written Other notation: if a < b, we may write b > a and,

if b � a, we may write a � b.

a < b, if b � a 2 P ; we say that a is less than or equal to b, written a � b, if b � a 2 N; that is, a < b or a D b.

Thus, a is positive if 0 < a (that is, a 2 P ), and a is negative if a < 0 (that Just to complete the

picture, a > b means

b < a (and a � b means b � a).

is, �a 2 P ).

Here are some standard properties of inequality.

Proposition A.51. Let a; b; B be real numbers with b < B .

(i) If a > 0, then ab < aBI if a < 0, then ab > aB . (ii) If a > 0 and b < 0, then ab < 0.

(iii) If b > 0, then b�1 > 0; if b < 0, then b�1 < 0.

(iv) aC b < a CB and a � b > a � B . (v) If c; d are positive, then d < c if and only if d�1 > c�1.

Proof. We prove the first three parts; the last two proofs are similar and appear

in Exercise A.31 below.

(i) By definition, b < B means that B � b 2 P . � Suppose that a > 0; that is, a 2 P . To show that ab < aB , we must

show that aB �ab D a.B �b/ 2 P , and this follows from Property (i) of P .

� If a < 0, then �a 2 P . Therefore, .�a/.B � b/ 2 P , and so

.�a/.B � b/ D .�1/a.B � b/ D a.b � B/ 2 P:

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442 Appendix A Appendices

(ii) The first part says that if b < B and both sides are multiplied by a positive

number, then the sense of the inequality stays the same. So, if B < 0, then

aB < a � 0 D 0. (iii) Suppose that b > 0. If b�1 < 0, then

1 D b � b�1 < b � 0 D 0;

a contradiction. If b�1 D 0 then

1 D bb�1 D b � 0 D 0;

another contradiction. Hence, Trichotomy gives b�1 > 0.

Exercises

A.31 * Prove parts (iv) and (v) of Proposition A.51.

A.32 Prove, or disprove and salvage if possible. Suppose a, b, c, and d are real num-

bers.“Disprove” here means “give a concrete counterex-

ample.” “Salvage” means

“add a hypotheis to make it

true.”

(i) If a < b, then a2 < b2.

(ii) If a2 < b2, then a < b.

(iii) If a < b and c < d , then ac < bd .

(iv) If a3 > 0, then a > 0.

A.33 Does C have a subset P 0 like P ; that is, P 0 is closed under addition and multi- plication, and it satisfies Trichotomy?

A.5 Generalized Associativity

Recall that a set with a binary operation is a set G equipped with a function

G�G ! G; we denote the value of the function by .a; b/ 7! a�b. ExamplesWe do not assume that � is commutative; that is, a � b ¤ b � a is allowed.

of such sets are the real numbers and the complex numbers, each of which

is usually viewed as having two binary operations: addition .a; b/ 7! a C b and multiplication .a; b/ 7! ab. More generally, every commutative ring has binary operations addition and multiplication. Another example is given by

G D XX , the family of all functions from a set X to itself: composition of functions is a binary operation on G.

The adjective binary means two: two elements a; b 2 G are combined to produce the element a�b 2 G. But it is often necessary to combine more than two elements: for example, we may have to multiply several numbers. The

binary operations in the examples cited above are associative; we can combine

three elements unambiguously. If a; b; c 2 G, then

a � .b � c/ D .a � b/ � c:

Since we are told only how to combine two elements, there is a choice when

confronted with three elements: first combine b and c, obtaining b � c, and then combine this new element with a to get a � .b � c/, or first get a � b and then combine it with c to get .a � b/ � c. Associativity says that either choice yields the same element of G. Thus, there is no confusion in writing

a � b � c without parentheses. In contrast, subtraction is not associative, for it

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A.5 Generalized Associativity 443

is not clear whether a� b � c means .a� b/� c or a� .b � c/, and these may be different: 9 � .5 � 3/ D 7 while .9 � 5/� 3 D 1.

Suppose we want to combine more than three elements; must we assume more complicated identities? Consider powers of real numbers, for example.

Is it obvious that a3a2 D � aŒaa2

� a? The remarkable fact is: assuming we

don’t need parentheses for three factors, we don’t need parentheses for more

than three factors. To make all concrete, we now call a binary operation mul-

tiplication, and we simplify notation by omitting � and writing ab instead of a � b.

Definition. Let G be a set with a binary operation; an expression in G is an

n-tuple .a1; a2; : : : ; an/ 2 G � � � � �G that is rewritten as a1a2 � � �an; we call the ai factors of the expression.

An expression yields many elements of G by the following procedure. Choose two adjacent a’s, multiply them, and obtain an expression with n � 1 factors: the new product just formed and n � 2 original factors. In the shorter new expression, choose two adjacent factors (either an original pair or an orig-

inal one together with the new product from the first step) and multiply them.

Repeat this procedure until there is a penultimate expression having only two

factors; multiply them and obtain an element of G that we call an ultimate product. For example, consider the expression abcd . We may first multiply

ab, obtaining .ab/cd , an expression with three factors, namely, ab, c, d . We

may now choose either the pair c, d or the pair ab, c; in either case, multi-

ply them to obtain expressions having two factors: ab; cd , or .ab/c, d . The

two factors in the last expressions can now be multiplied to give two ultimate products from abcd , namely .ab/.cd/ and ..ab/c/d . Other ultimate products

derived from the expression abcd arise from multiplying bc or cd as the first

step. It is not obvious whether the ultimate products from a given expression

are equal.

Definition. Let G be a set with a binary operation. An expression a1a2 � � �an inG needs no parentheses if all its ultimate products are equal elements ofG.

Theorem A.52 (Generalized Associativity). IfG is a set with an associative

binary operation, then every expression a1a2 � � �an inG needs no parentheses.

Proof. The proof is by induction on n � 3. The base step holds because the operation is associative. For the inductive step, consider two ultimate prod-

ucts U and V obtained from a given expression a1a2 � � �an after two series of choices:

U D .a1 � � �ai /.aiC1 � � �an/ and V D .a1 � � �aj /.ajC1 � � �an/I

the parentheses indicate the penultimate products displaying the last two fac-

tors that multiply to give U and V , respectively; there are many parentheses

inside each of the shorter expressions. We may assume that i � j . Since each of the four expressions in parentheses has fewer than n factors, the inductive

hypothesis says that each of them needs no parentheses. It follows thatU D V if i D j . If i < j , then the inductive hypothesis allows the first expression to be rewritten as

U D .a1 � � �ai / � ŒaiC1 � � �aj ŒajC1 � � �an

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444 Appendix A Appendices

and the second to be rewritten as

V D � Œa1 � � �ai ŒaiC1 � � �aj 

� .ajC1 � � �an/;

where each of the expressions a1 � � �ai ; aiC1 � � �aj , and ajC1 � � �an needs no parentheses. Thus, the three expressions yield unique elements A, B , and C in

G, respectively. The first expression gives U D A.BC/ inG, the second gives V D .AB/C in G, and so U D V in G, by associativity.

Corollary A.53. If G is a set with a binary operation, a 2 G, and m, n � 1, then

amCn D aman and .am/n D amn:

Proof. In the first case, both elements arise from the expression havingmC n factors each equal to a; in the second case, both elements arise from the ex-

pression havingmn factors each equal to a.

A.6 A Cyclotomic Integer Calculator

Several times in the previous chapters, we’ve advised you to use a CAS. Some uses are simply to reduce the computational overhead of algebraic calculations,

such as the expansion in Lagrange interpolation on page 272. For this, you can

use the CAS “right out of the box:” all the functionality you need is built in

with commands like expand or simplify.

Other applications require programming that uses specific syntax for the CAS in use. A good example is the formula in Example 6.61 on page 265.

The recursive formula for ˆn can be implemented in almost any CAS, but the

details for how to get a product over the divisors of an integer (especially if

the product is in the denominator of an expression) can either be trivial, if theComputer Algebra is used regularly in many

high school classrooms,

implemented either on

handheld devices or with

tablet apps.

functionality is built-in, or extremely tricky to implement if it is not. There are

many CAS environments, so it would be of little use to include actual programs here.

What we can do in this Appendix is point out how to use Proposition 7.20

on page 290 to model Q.˛/, where ˛ is algebraic over Q and its minimal

polynomial p D irr.˛;Q/ is known. The essential piece of that Proposition is that

Q.˛/ Š QŒx=.p.x//I

so, as long as your CAS can find the remainder when one polynomial is divided

by another, you can use it to perform “modular arithmetic” with polynomials

in Q.˛/. If p is a prime in Z and � D cos.2�=p/C i sin.2�=p/, we know that

irr.�;Q/ D 1C x C x2 C � � � C xp�1:

This is easily implemented in a CAS with something like

Phi(x,p):= sum(xˆk,k,0,p-1)

Suppose that your CAS command for polynomial remainder is pmod. For ex-

ample,

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A.6 A Cyclotomic Integer Calculator 445

pmod(xˆ3+4xˆ2-3x+1,xˆ2+1)

returns �4x� 3, the remainder when x3 C 4x2� 3xC 1 is divided by x2C 1.

The two functions, Phi and pmod, allow us to calculate in Q.�/. Let’s look

at some examples.

Eisenstein Integers

Arithmetic with complex numbers is built into most CAS environments, so that

you can do calculations with Gaussian integers right away. Arithmetic with Eisenstein integers isn’t usually built in, but you can build a model of ZŒ! by

thinking of an Eisenstein integer as a congruence class mod x2 C x C 1:

cl(f) := pmod (f,phi(x,3))

or even

cl(f) := pmod (f, xˆ2+x+1).

Addition and multiplication of classes are defined as in Chapter 7:

add(f,g) = cl(f+g)

mult(f,g) = cl(fg).

So, now we can compute: to find, for example, 3!5 � !2 C 1, you want the class of 3x5 � x2 C 1 mod x2 C x C 1

cl (3xˆ5-xˆ2+1)

> -2x-1

And, sure enough,

3!5 � !2 C 1 D �1 � 2!:

Your model can do generic calculations, giving the rules for addition and mul-

tiplication in ZŒ!:

add(a+b*x,c+d*x)

> a+c+(b+d)*x

mult(a+b*x,c+d*x)

> a*c-b*d + (a*d+b*c-b*d)*x

You can generate Eisenstein triples by squaring Eisenstein integers:

mult(3+2*x,3+2*x)

> 5+8*x

mult(5+x,5+x)

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446 Appendix A Appendices

> 24+9*x

mult(4+3*x,4+3*x)

> 7+15*x

Symmetric Polynomials

In Example 9.6 on page 388, we derived the cubic formula via symmetric

polynomials. There, we defined

s D ˛1 C ˛2! C ˛3!2

u D ˛1 C ˛2!2 C ˛3!;

and we saw that s C u D 3˛. We also claimed that

s3 C u3 D 27˛1˛2˛3:

Our CAS Eisenstein calculator can help. Replacing ˛; ˇ; by a; b; c, we have

add((a+b*x+c*xˆ2)ˆ3,(a+b*xˆ2+c*x)ˆ3)

> 2*aˆ3-3*aˆ2*b+aˆ2*c-3*a*bˆ2-12*a**b*c-3*a*cˆ2+2* bˆ3-3*bˆ2*c-3*b*cˆ2+2*cˆ3

factor(2*aˆ3-3*aˆ2*b+aˆ2*c-3*a*bˆ2-12*a**b*c-3*a* cˆ2+2*bˆ3-3*bˆ2*c-3*b*cˆ2+2*cˆ3)

> (a+b-2*c)*(a-2*b+c)*(2*a-b-c)

This can be written (with an eye to symmetric polynomials) as

.aC b C c � 3c/.a C b C c � 3b/.3a � .a C b C c//:

Since aC b C c D 0, this is exactly what we claimed in Example 9.6.

Algebra with Periods

One last example: in Section 7.3, we outlined Gauss’s construction of the reg-

ular 17-gon with ruler and compass. Central to that is the specification of “pe-

riods” of various lengths, listed on page 323. They are constructed according to the formula on page 325: if ef D 16, the periods of length f are given by

�e;k D f�1X

jD0 �3

kCje ; 0 � k < e:

A CAS model follows the syntax pretty closely:mod is the CAS built-in “mod” function.

n(e,k):=sum(xˆ(mod(3ˆ(k+e*j),17)),j,0,(16/e)-1).

Now change cl so that it gives the congruence class mod ˆ17.x/:

cl(f) := pmod (f, phi(x,17)),

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A.6 A Cyclotomic Integer Calculator 447

and we can calculate with the classes of the periods:

n(2,0)

> xˆ16+xˆ15+xˆ13+xˆ9+xˆ8+xˆ)+xˆ2+x

n(2,1)

> xˆ14+xˆ12+xˆ11+xˆ10+xˆ7+xˆ6+xˆ5+xˆ3

cl( add (n(2,0), n(2, 1)) )

>-1

cl( mult (n(2,0), n(2, 1)) )

> -4

So, as we claimed on page 323, �2;0 and �2;1 are roots of

x2 C x � 4:

Exercises

A.34 Find a polynomial in QŒx that has roots �0;k ; �1;k ; �3;k ; �4;k :

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References

[1] Apostol, Tom M., Introduction to Analytic Number Theory, Under-

graduate Texts in Mathematics, Springer-Verlag, New York-Heidelberg,

1976.

[2] Artin, E., Galois Theory, Edwards Brothers, Ann Arbor, 1948.

[3] Baker, A., Transcendental Number Theory, 2d ed., Cambridge Univer- sity Press, Cambridge, 1990.

[4] Barbeau, E. J., Polynomials, Springer-Verlag, New York, 1989.

[5] Borevich, Z. I., and Shafarevich, I. R., Number Theory, Academic Press, New York, 1966.

[6] Cajori, F., A History of Mathematical Notation, Open Court, 1928;

Dover reprint, Mineola NY, 1993.

[7] Cuoco, A., Mathematical Connections, Classroom Resource Materials,

MAA, Washington, 2005.

[8] Dirichlet, P. G. L., Lectures on Number Theory, Supplements by R.

Dedekind. Translated from the 1863 German original and with an intro-

duction by John Stillwell. History of Mathematics, 16. American Math-

ematical Society, Providence; London Mathematical Society, London, 1999.

[9] Du Val, P., Elliptic Functions and Elliptic Curves, London Mathemati-

cal Society Lecture Note Series 9, Cambridge University Press, London,

1973.

[10] EDC, The CME Project, Pearson, Boston, 2013

[11] Edwards, H. M., Fermat’s Last Theorem. A Genetic Introduction to

Number Theory, Graduate Texts in Mathematics vol. 50, Springer-

Verlag, New York, 1977.

[12] Euler, L., Elements of Algebra, Translated from the German by John

Hewlett. Reprint of the 1840 edition. With an introduction by C. Trues- dell, Springer-Verlag, New York, 1984.

[13] Flannery, S., In Code. A Mathematical Journey, Reprint of the 2000 original, Workman Publishing, New York, 2001.

[14] Gauss, C. F., Disquisitiones Arithmeticae, Arthur A. Clarke (trans), Yale

University Press, New Haven, 1966.

[15] Hadlock, C. R., Field Theory and Its Classical Problems, Carus Mathe-

matical Monographs 19, Mathematical Association of America, Wash-

ington, 1978.

449

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450 References

[16] Heath, T. L., The Thirteen Books of Euclid’s Elements, Cambridge Uni-

versity Press, 1926; Dover reprint, Mineola NY, 1956.

[17] Ireland, K. and Rosen, M., A Classical Introduction to Modern Number

Theory, Springer-Verlag, New York, 1982.

[18] Koblitz, N., A Course in Number Theory and Cryptography, Springer-

Verlag, New York, 1987.

[19] ——, Introduction to Elliptic Curves and Modular Forms, Springer Ver-

lag, New York, 1993.

[20] Loomis, E. S., The Pythagorean Proposition, Edwards Brothers., Ann

Arbor, 1940.

[21] Montgomery, S., and Ralston, E. W., Selected Papers in Algebra, Ray-

mond W. Brink Selected Mathematical Papers, Vol. 3, Mathematical As-

sociation of America, Washington, 1977.

[22] Needham, T., Visual Complex Analysis, Clarendon Press, Oxford, 1997.

[23] Ribenboim, P., Thirteen Lectures on Fermat’s Last Theorem, Springer–

Verlag, New York, 1979.

[24] Rosen, K. H., Elementary Number Theory and Its Applications, 4th ed., Addison-Wesley, Reading, MA, 2000.

[25] Rotman, J. J., Advanced Modern Algebra, 2d ed., Graduate Studies in Mathematics vol. 114, American Mathematical Society, Providence,

2010.

[26] ——, A First Course in Abstract Algebra, 3d ed., Prentice Hall, Upper Saddle River NJ, 2006.

[27] ——, Galois Theory, 2d ed., Springer-Verlag, New York, 1998.

[28] ——, Journey into Mathematics, Prentice Hall, Upper Saddle River NJ,

1998; Dover reprint, Mineola NY, 2007.

[29] Samuel, P., Algebraic Theory of Numbers, Houghton-Mifflin, Boston,

1992.

[30] Siegel, C. L., Topics in Complex Function Theory Vol. I, Elliptic Func-

tions and Uniformization Theory, Wiley, New York, 1969.

[31] Silverman, J. H., and Tate, J., Rational Points on Elliptic Curves,

Springer–Verlag, New York, 1992.

[32] Stillwell, J., Mathematics and Its History, 3d ed., Springer, New York,

2010.

[33] Tignol, J.-P., Galois’ Theory of Equations, World Scientific Publishing,

Singapore, 2001.

[34] van der Waerden, B. L., Science Awakening, John Wiley, New York,

1963.

[35] ——, Geometry and Algebra in Ancient Civilizations, Springer–Verlag, New York, 1983.

[36] Washington, L.C., Introduction to Cyclotomic Fields, Springer, New York, 1982.

[37] Weil, A., Number Theory: An Approach Through History. From Ham-

murapi to Legendre, Birkhäuser, Boston, 1984.

[38] Weyl, H., Algebraic Theory of Numbers, 6th printing, Princeton Univer-

sity Press, Princeton, 1971; Princeton Landmarks in Mathematics and Physics Series, 1998.

[39] Wilf, H., Generatingfunctionology, Academic Press, New York, 1994.

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Index

Abel, Niels H., 379, 391, 394, 400

abelian group, 390

Addition Theorem, 399

adjoining to field, 294

Adleman, Leonard M., 150

al-Khwarizmi, 240

algebraic

element, 293

extension field, 293

algebraic closure, 254

algebraically closed, 301

anagram, 70

antanairesis, 31

Archimedes, 321

Aristarchus, 3

Arithmetic–Geometric Mean Inequal-

ity, 6

Artin, Emil, 386

associate, 234

associated polynomial function, 204

associativity, 37

generalized, 443

atom, 227

automorphism, 386

b-adic digits, 139

Babylonian method, 5

base b, 139

basis

standard, 430

vector space, 430

Bernoulli, Jacob, 398

Bernoulli, James, 398

Bhaskara I, 149

bijective, bijection, 416

binary operation, 155

Binomial Theorem

in Z, 66

in commutative ring, 160

Boolean ring, 167, 223

Braunfeld, Peter, 53

calendar

formula

Conway, John H., 175

Gregorian, 173

Julian, 169

Mayan, 144

cancellation law, 192

Cardano, Girolamo, 81

cartesian product, 410

CAS = Computer Algebra System, 142

casting out 9s, 137

castle problem, 90

Casus Irreducibilis, 383

century year, 170

characteristic function, 209

characteristic of field, 288

Ch’in Chiu-shao = Qin Jiushao, 90, 146

Chinese Remainder Theorem, 142

chord–tangent construction, 405

closed under operation, 155

coconuts, 148

code (see RSA code), 150

coefficients, 197

commensurable, 28

common divisor, 24

in Z, 24

polynomials, 243

common multiple

in Z, 55

polynomials, 253

common year, 169

commutative, 37

commutative ring, 156

formal power series, 200

polynomial ring

one variable, 200

several variables, 205

451

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452 Index

compass, 309

complement, 168

completing the square, 2, 23

complex conjugate, 96

complex exponential, 108

complex number, 92

absolute value, 99

argument, 100

conjugate, 96

exponential form, 109

imaginary part, 92

modulus, 99

norm, 116

polar form, 101

real part, 92

composite

functions, 414

number, 22

congruence class, 154, 421

congruent mod I , 278

congruent mod m, 132, 270

congruent numbers, 400

constant function, 411

constant polynomial, 201

constant term, 201

constructible, 311

number, 312

point, 311

subfield, 313

contrapositive, 26

convex, 48

Conway, John H., 175

coordinate list, 431

coset

ideal, 279

subgroup, 392

Cramer’s Rule, 164

Crelle, August L., 380

cross multiplication, 193

cubic formula, 84

cubic polynomial, 382

cyclotomic integers, 157

cyclotomic polynomial, 265

day, 169

de Moivre’s Theorem, 107

exponential form, 109

de Moivre, Abraham, 107

De Morgan, Augustus, 169

decimal expansion, 177

Dedekind, Richard, 218, 366, 369

degree, 198

extension field, 291

derivative, 202

Descartes, René, 82

diagonal, 420

dimension, 433

Diophantine equation, 404

Diophantus, 5, 8, 12, 27, 405

direct product, 221

direct sum of vector spaces, 440

Dirichlet, J. P. G. Lejeune, 153, 358, 360

discriminant, 91

disjoint, 422

distance preserving, 385

distributive, 37

divides, 21

Division Algorithm

integers, 23

polynomials, 237

divisor, 21

commutative ring, 233

proper, 257

divisors, 368

domain, 192

Euclidean domain, 333

of function, 410

doomsday, 174

double angle formula, 107

dual basis, 438

dual space, 437

Eisenstein integers, 120

Eisenstein Criterion, 267

Eisenstein triple, 121

Eisenstein, F. Gotthold M., 120, 267

Elements, 2, 20

elliptic curve, 404

elliptic function, 400

elliptic integral, 398

empty set, 410

equality of functions, 412

equality of sets, 410

equivalence class, 421

equivalence relation, 420

Erlanger Programm, 390

etymology

abelian, 391

algebra, 240

algorithm, 240

arithmetic, xv

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Index 453

binomial, 63

calculus, 197

calendar, 173

casting out 9s, 137

coefficients, 198

corollary, 11

cubic, 202

degree

field extension, 291

polynomial, 198

geometry, xv

golden ratio, 76

homomorphism, 207

hypotenuse, 4

ideal, 218, 366

isomorphism, 207

lemma, 11

linear, 202

mathematics, xv

modulo, 132

monomial, 63

power, 51

proof, 11 proposition, 10

quadratic, 202

radical, 240

ring, 157

root, 239

scalar, 425

September, 171

theorem, 11

vector, 424

Euclid, 2, 20, 23

Elements, 20

Euclid’s Lemma

integers, 25

PID, 256

polynomials, 248

Euclidean Algorithm I

integers, 32

polynomials, 249

Euclidean Algorithm II

integers, 32

polynomials, 250

Eudoxus, 28, 308

Euler �-function, 111, 232

Euler, Leonhard, 108, 111, 131, 326,

330, 399

evaluation homomorphism, 215

extension field, 291

algebraic, 293

degree, 291

finite, 291

pure, 382

radical, 383

Factor Theorem, 240

factorial, 51

Fagnano, Giulio, 399

Fermat prime, 325

Fermat’s n D 4 Theorem, 15 Fermat’s Last Theorem, 14

Fermat’s Little Theorem, 136

Fermat’s Theorem on Divisors, 372

Fermat’s Two-Square Theorem, 342

Fermat, Pierre de, 14

Ferrari, Ludovico, 87

Fibonacci sequence, 75

Fibonacci, Leonardo, 19, 75

field, 163

algebraically closed, 301

extension, 291

finite, 305

prime, 293

rational functions, 205

splitting, 301

finite extension field, 291

finite-dimensional, 428

first day of month, 173

first form of induction, 57

First Isomorphism Theorem

commutative rings, 282

groups, 392

first quadrant, 372

fixes, 386

formal power series, 197

fraction field, 194

Frey, Gerhard, 407

function, 410

bijective, 416

constant, 411

identity, 411

inclusion, 412

injective, 413

restriction, 415

surjective, 413

functional, 437

Fundamental Theorem

Algebra, 105

Arithmetic, 54

Galois Theory, 393

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454 Index

Galois field, 306

Galois group, 386

Galois, Évariste, 305, 306, 380, 394

Gauss’s Lemma, 260

Gauss, Carl F., 50, 258, 268, 321, 380,

384, 400

Gauss-Wantzel theorem, 325

Gaussian integers, 7, 119, 157

gcd, 24, 243, 255

Generalized Associativity, 443

Germain, Marie-Sophie, 358

golden ratio, 76

googol, 140

Granville, Andrew J., 406

graph of function, 411

greatest common divisor, 24, 243, 255

greatest integer function, 29

Gregorian calendar, 170

group, 389

abelian, 390

quotient, 392

solvable, 394

symmetric, 389

Heath, Thomas L., 2, 20

Helikon, 308

Heron triangle, 14

Hilbert, David, 157, 369

hockey stick, 68

Hölder, Otto L., 383

Holmes, Sherlock, 176

homomorphism, 207

group, 391

Hume, James, 82

Hungerbühler, Norbert, 320

hyperbolic cosine, 6

ideal, 218

generated by X , 168

maximal, 286

prime, 286

principal, 218

product, 220

proper, 218

sum, 220

identity element

commutative ring, 156

group, 389

identity function, 411

image

function, 410

homomorphism, 217

inclusion, 412

Inclusion–Exclusion, 230

independent list

longest, 434

indeterminate, 200

indirect proof, 27

induction

base step, 47

first form, 47

inductive hypothesis, 47

inductive step, 47, 57

strong, 57

inductive reasoning, 45

inequalities, 441

inert, 342

infinite descent, 12

infinite-dimensional, 428

injective, injection, 413

integers modm, 154

integers Z, 21

rational, 339

integers, cyclotomic, 359

integration, 42

invariance of dimension, 433

inverse element

additive, 37

group, 389

multiplicative, 37, 156

inverse function, 416

irr.z; k/, 296

irreducible in commutative ring, 234

isometry, 385

isomorphism, 207

group, 391

Jacobi, Carl G. J., 360, 380, 400

Joachimsthal, Ferdinand, 359

Julian calendar, 169

k-linear combination, 427

Kaplansky, Irving, 236

kernel, 217

group, 392

Klein, Felix C., 390

Kolyvagin, Victor, 407

Kronecker, Leopold, 28, 300, 359

Kummer, Eduard, 218, 358, 359

Lagrange Interpolation, 272

Lagrange, Joseph-Louis, 272

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Index 455

Lamé, Gabriel, 32, 358

lattice point, 13, 35

Law of Decomposition

Eisenstein integes, 346

Gaussian integers, 343

Law of Substitution, 37, 155, 412

Laws of Exponents, 52

leading coefficient, 198

leap year, 169

least common multiple

in Z, 55

polynomials, 253

Least Integer Axiom, 21

Legendre, Adrien-Marie, 358

Leibniz, Gottfried W., 68, 397

Levi ben Gershon, 52

lies above, 340

Lindemann, C. L. Ferdinand von, 293

linear combination, 218

in Z, 24

vector space, 427

linear transformation, 435

linearly dependent, 429

linearly independent, 429, 430

Liouville, Joseph, 381, 397

list, 427

longest independent list, 434

lowest terms

in Q, 26

rational functions, 248

Mascheroni, Lorenzo, 320

matrix associated to linear transforma-

tion, 437

Maurolico, Francesco, 52

maximal ideal, 286

Mayan calendar, 144

Mazur, Barry, 407

minimal polynomial, 296

Möbius, August F., 263

modulus, 132

Mohr, Georg, 320

monic polynomial, 201

Moore, Eliakim H., 306

Motzkin, Theodore S., 335

multiple, 21

multiplicative function, 378

strongly multiplicative function,

375

multiplicity, 254

natural map, 282

natural numbers N, 21

n choose r , 63

negative, 37, 156

Newton, Isaac, 197, 405

Nine Chapters on the Mathematical Art, 3, 6

Noether, Emmy, 369

norm, 116

normal subgroup, 392

one-one correspondence

see bijective, 416

one-to-one

see injective, 413

onto (function)

see surjective, 413

order of power series, 203

order of unit, 165

ordered pair, 410

Oresme, Nicole, 45

origin, 310

Oughtred, William, 82

pairwise disjoint, 422

parallelogram law, 93, 425

parity, 131

partition, 422

Pascal’s triangle, 63

Pascal, Blaise, 52, 64

period, 179

permutation, 385

PID, 255

Pigeonhole Principle, 419

Plato, 308

Plimpton 322, 4

Pogrebishte, 176

pointwise addition, 157

pointwise multiplication, 157

polar form, 101

polynomial

n variables, 205

monic, 201

one variable, 197

reduced, 83

splits, 301

zero, 198

polynomial function, 204

polynomial ring

one variable, 200

several variables, 205

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456 Index

power series, 197

powers, 51

commutative ring, 160

predecessor, 57

prime, 22

inert, 342

ramifies, 342

rational, 339

splits, 340

prime factorization, 54, 252

prime field, 293

prime ideal, 286

primes

irregular, 368

regular, 368

primitive element, 325

primitive Pythagorean triple, 27

primitive root of unity, 111, 264

principal ideal, 218

principal ideal domain, 255

private key, 150

Proclus, 20

product expansion, 375

proof

by contradiction, 27

indirect, 27

proper

ideal, 218

subset, 410

subspace, 426

public key, 150

pure extension, 382

Pythagoras, 2

Pythagorean

point, 10

Theorem, 2

converse, 7

triple, 4

primitive, 27

similar, 9

Qin Jiushao, 90, 146

quadratic formula, 2

quartic formula, 87

quartic polynomial, 383

quotient, 23, 38

quotient group, 392

quotient ring, 281

radical extension, 383

Rahn, Johann H., 82

rational

block, 179

function, 205

integer, 339

line, 13

period, 179

point, 10

prime, 339

repeats, 179

terminates, 179

Rational Root Theorem, 260

Recorde, Robert, 82

recurrence, 74

reduced polynomial, 83

reduction mod p, 260

reflexive relation, 420

relation, 420

relatively prime

integers, 26

polynomials, 248

remainder, 23

Remainder Theorem, 240

repeated roots, 263

resolvent cubic, 87, 384

restriction, 415

Ribet, Kenneth A., 407

Riemann zeta function, 374

Riemann, G. F. Bernhard, 369

ring, commutative, 156

Rivest, Ronald L., 150

root

multiplicity, 254

polynomial, 81, 239

root of unity, 111

commutative ring, 165

cube root, 82

primitive, 111, 264

RSA code, 150

Ruffini, Paolo, 394

ruler, 309

scalar, 424

multiplication, 424

Schönemann, Theodor, 267

Scherk, Heinrich F., 359

Scipione, 81

Serre, Jean-Pierre, 407

Shamir, Adi, 150

Shimura, Goro, 397, 407

similar, 9

Singer, Richard, 267

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Index 457

single-valued, 412

size function, 333

smallest subspace, 427

solution of linear system, 426

solution space, 426

solvable by radicals, 383

solvable group, 394

spans, 427

splits

polynomial, 301

prime, 340

splitting field, 301

squarefree integer, 34

standard basis, 430

straightedge (see ruler), 309

strong induction, 57

subfield, 167

generated by X , 293

subgroup, 390

normal, 392

subring, 166

generated by X , 168

subset, 410

subspace, 426

proper, 426

smallest, 427

spanned by X , 427

subtraction, 38

surjective, 413

surjective, surjection, 413

symmetric difference, 166

symmetric group, 389, 418

symmetric relation, 420

symmetry, 385

tangent half-angle formula, 43

Taniyama, Yutaka, 397, 407

target, 410

Tartaglia = Nicolo Fontana, 81

transcendental element, 293

transitive relation, 420

transpose, 438

triangular number, 68

trichotomy, 441

trigonometric identities, 42

triple angle formula, 110

Trotsky, Leon, 170

tzolkin calendar, 144

UFD, 258

Uncle Ben, 176

Uncle Charles, 174

unique, 23

unique factorization in Z, 54

polynomials, 252

unique factorization domain, 258

unit, 162

unit n-gon, 110 unit circle, 8

unit vector, 99

valuation, 350 value of function, 410

van der Waerden, Bartel L., 21, 28

Vandermonde’s Identity, 73

Vandermonde, Alexandre, 73

vector space, 424 vectors, 424

Venn diagram, 222

Venn, John, 222

Viète, Francois, 82, 88

Wantzel, Pierre L., 318

Weierstrass, Karl, 404

Weil, André, 407

well-defined, 412 Well-Ordering Axiom, 21

Wessel, Caspar, 92

Widman, Johannes, 82

Wiles, Andrew J., 14, 396, 407

Williams, Kenneth S., 335 Wilson’s Theorem, 263

Wilson, Jack C., 335

Wilson, John, 263

year, 169

century year, 170

common, 169

leap year, 169

zero polynomial, 198

zero ring, 159

zero divisor, 192

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About the Authors

Al Cuoco is Distinguished Scholar and Director of the Center for Mathematics

Education at Education Development Center. He is lead author for The CME

Project, a four-year NSF-funded high school curriculum, published by Pear- son. He also co-directs Focus on Mathematics, a mathematics-science part-

nership that has established a mathematical community of mathematicians,

teachers, and mathematics educators. The partnership evolved from his 25-year

collaboration with Glenn Stevens (BU) on Boston University’s PROMYS for

Teachers, a professional development program for teachers based on the Ross program (an immersion experience in mathematics). Al taught high school

mathematics to a wide range of students in the Woburn, Massachusetts public

schools from 1969 until 1993. A student of Ralph Greenberg, Cuoco holds a

Ph.D. from Brandeis, with a thesis and research in Iwasawa theory. In addition

to this book, MAA published his Mathematical Connections: a Companion for

Teachers and Others. But his favorite publication is a 1991 paper in the Amer- ican Mathematical Monthly, described by his wife as an attempt to explain a

number system that no one understands with a picture that no one can see.

Joseph Rotman was born in Chicago on May 26, 1934. He studied at the

University of Chicago, receiving the degrees BA, MA, and Ph.D. there in 1954, 1956, and 1959, respectively; his thesis director was Irving Kaplansky.

Rotman has been on the faculty of the mathematics department of the Uni-

versity of Illinois at Urbana-Champaign since 1959, with the following ranks:

Research Associate 1959–1961; Assistant Professor 1961–1963; Associate Pro-

fessor 1963–1968; Professor 1968–2004; Professor Emeritus 2004–present. He has held the following visiting appointments: Queen Mary College, Lon-

don, England 1965, 1985; Aarhus University, Denmark, Summer 1970; He-

brew University, Jerusalem, Israel 1970; University of Padua, Italy, 1972; Tech-

nion, Israel Institute of Technology and Hebrew University, Jerusalem (Lady

Davis Professor), 1977–78; Tel Aviv University, Israel, 1982; Bar Ilan Univer-

sity, Israel, Summer 1982; Annual visiting lecture, South African Mathemati- cal Society, 1985; Oxford University, England, 1990.

Professor Rotman was an editor of Proceedings of American Mathematical

Society, 1970, 1971; managing editor, 1972, 1973.

Aside from writing research articles, mostly in algebra, he has written the

following textbooks: Group Theory 1965, 1973, 1984, 1995; Homological Al- gebra 1970, 1979, 2009; Algebraic Topology 1988; Galois Theory 1990, 1998;

Journey into Mathematics 1998, 2007; First Course in Abstract Algebra 1996,

2000, 2006; Advanced Modern Algebra 2002.

459

  • Cover
  • Copyright
  • Title
  • Contents
  • Preface
    • Some Features of This Book
    • A Note to Students
    • A Note to Instructors
  • Notation
  • 1. Early Number Theory
    • 1.1 Ancient Mathematics
    • 1.2 Diophantus
      • Geometry and Pythagorean Triples
      • The Method of Diophantus
      • Fermat’s Last Theorem
      • Connections: Congruent Numbers
    • 1.3 Euclid
      • Greek Number Theory
      • Division and Remainders
      • Linear Combinations and Euclid’s Lemma
      • Euclidean Algorithm
    • 1.4 Nine Fundamental Properties
    • 1.5 Connections
      • Trigonometry
      • Integration
  • 2. Induction
    • 2.1 Induction and Applications
      • Unique Factorization
      • Strong Induction
      • Differential Equations
    • 2.2 Binomial Theorem
      • Combinatorics
    • 2.3 Connections
      • An Approach to Induction
      • Fibonacci Sequence
  • 3. Renaissance
    • 3.1 Classical Formulas
      • Cubics
      • Quartics
    • 3.2 Complex Numbers
      • The Complex Plane
      • Algebraic Operations
      • Absolute Value and Direction
      • The Geometry Behind Multiplication
    • 3.3 Roots and Powers
      • Roots of Unity
    • 3.4 Connections: Designing Good Problems
      • Norms
      • Pippins and Cheese
      • Gaussian Integers: Pythagorean Triples Revisited
      • Eisenstein Integers
      • Eisenstein Triples and Diophantus
      • Nice Boxes
      • Nice Functions for Calculus Problems
      • Lattice Point Triangles
  • 4. Modular Arithmetic
    • 4.1 Congruence
    • 4.2 Public Key Codes
    • 4.3 Commutative Rings
      • Properties of Commutative Rings
      • Units and Fields
      • Subrings and Subfields
    • 4.4 Connections: Julius and Gregory
    • 4.5 Connections: Patterns in Decimal Expansions
      • Real Numbers
      • Decimal Expansions of Rationals
      • Periods and Blocks
  • 5. Abstract Algebra
    • 5.1 Domains and Fraction Fields
      • Fraction Fields
    • 5.2 Polynomials
      • Polynomial Functions
    • 5.3 Homomorphisms
      • Extensions of Homomorphisms
      • Kernel, Image, and Ideals
    • 5.4 Connections: Boolean Things
      • Inclusion-Exclusion
  • 6. Arithmetic of Polynomials
    • 6.1 Parallels to Z
      • Divisibility
      • Roots
      • Greatest Common Divisors
      • Unique Factorization
      • Principal Ideal Domains
    • 6.2 Irreducibility
      • Roots of Unity
    • 6.3 Connections: Lagrange Interpolation
  • 7. Quotients, Fields, and Classical Problems
    • 7.1 Quotient Rings
    • 7.2 Field Theory
      • Characteristics
      • Extension Fields
      • Algebraic Extensions
      • Splitting Fields
      • Classification of Finite Fields
    • 7.3 Connections: Ruler-Compass Constructions
      • Constructing Regular n-gons
      • Gauss’s construction of the 17-gon
  • 8. Cyclotomic Integers
    • 8.1 Arithmetic in Gaussian and Eisenstein Integers
      • Euclidean Domains
    • 8.2 Primes Upstairs and Primes Downstairs
      • Laws of Decomposition
    • 8.3 Fermat’s Last Theorem for Exponent 3
      • Preliminaries
      • The First Case
      • Gauss’s Proof of the Second Case
    • 8.4 Approaches to the General Case
      • Cyclotomic integers
      • Kummer, Ideal Numbers, and Dedekind
    • 8.5 Connections: Counting Sums of Squares
      • A Proof of Fermat’s Theorem on Divisors
  • 9. Epilog
    • 9.1 Abel and Galois
    • 9.2 Solvability by Radicals
    • 9.3 Symmetry
    • 9.4 Groups
    • 9.5 Wiles and Fermat’s Last Theorem
      • Elliptic Integrals and Elliptic Functions
      • Congruent Numbers Revisited
      • Elliptic Curves
  • A. Appendices
    • A.1 Functions
    • A.2 Equivalence Relations
    • A.3 Vector Spaces
      • Bases and Dimension
      • Linear Transformations
    • A.4 Inequalities
    • A.5 Generalized Associativity
    • A.6 A Cyclotomic Integer Calculator
      • Eisenstein Integers
      • Symmetric Polynomials
      • Algebra with Periods
  • References
  • Index
  • About the Authors

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