Kato, Kurokawa, Saito - Number Theory I. Fermat's Dream s

Translations of

M&THEMATICAL MONOGRAPHS

Volume 186

Number Theory 1 Fermat’s Dream

Kazuya Kato Nobushige Kurokawa Takeshi Saito

Translated by Masato Kuwata

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American Mathematical Society

Providence, Rhode Ilsand

Contents

Preface

Preface to the English Edition

Objectives and Outline of these Books

ix

xi

. . . x111

Notation xv

Chapter 0. Introduction - Fermat and Number Theory ~

0.1. Before Fermat 0.2. Prime numbers and the sum of two squares 0.3. p = x2 + 2y2, p = x2 + 3y2,. . . 0.4. Pell’s equations 0.5. Triangular numbers, quadrangular numbers, pentagonal

numbers 0.6. Triangular numbers, squares, cubes 0.7. Right triangles and elliptic curves 0.8. Fermat’s Last Theorem

Exercises

8 10 11 12 14

Chapter 1. Rational Points on Elliptic Curves 1.1. Fermat and elliptic curves

1.2. Group structure of an elliptic curve 1.3. Mordell’s theorem

Summary Exercises

Chapter 2. Conies and padic Numbers 2.1. Conies 2.2. Congruence 2.3. Conies and quadratic residue symbols 2.4. p-adic number fields

vii

17 17 25 30 43 43

45 45 49 53 58

“Ill CONTENTS

2.5. Multiplicative structure of the p-adic number field 69

2.6. Rational points on tonics 74

Summary 78

Exercises 78

Chapter 3. < 81

3.1. Three wonders of the values of the < function 81

3.2. Values at positive integers 84

3.3. Values at negative integers 89

Summary 99

Exercises 100

Chapter 4. Algebraic Number Theory 4.1. Method of algebraic number theory 4.2. The heart of algebraic number theory 4.3. The class number formula

for imaginary quadratic fields 4.4. Fermat’s Last Theorem and Kummer

Summary Exercises

103

104 113

124 127 132 132

Appendix A. Rudiments on Dedekind domains 135

A.l. Definition of a Dedekind domain 135

A.2. Fractional ideal 136

Answers to Questions

Answers to Exercises

139

145

Index 153

Preface

This book was written in 1996, two hundred years after 1796, which was a very fruitful year for the great Gauss, who made many fundamental contributions to modern number theory. Gauss was in his late teens at the time. On March 30 he discovered a method of construction of a regular 17-gon. On April 8 he proved the quadratic reciprocity law (see $2.2 in this volume), which he himself called a gem. On May 31 he conjectured what would later be called “the prime number theorem” concerning the distribution of prime numbers. On July 10 he proved that any natural number can be expressed as a sum of at most three triangular numbers (see 50.5). On October 1 he obtained a result on the number of solutions for an equation with coefficients in a finite field, which had a great impact on mathematics in later eras. All these contributions are discussed in these volumes, Number Theory 1, 2, 3.

One, two, three, four.. . as naive as it is, the world of numbers encompasses many wonders that fascinated young Gauss. A discovery in one epoch induces a more profound discovery by the following generation. A hundred years later, in 1896, the prime number theorem was proved. After some 120 years, the quadratic reciprocity law had grown into the class field theory. After 150 years, Andre Weil, who had examined Gauss’s result of October 1, proposed the so-called Weil conjectures. These conjectures influenced a great deal of algebraic geometry in the twentieth century. The brilliance of the gems polished by Gauss has increased through the efforts of the mathematicians of following generations. It is said that there is no unexplored place on the earth any longer, but the world of numbers is still full of mysteries. That makes us think of the profoundness and richness of nature.

Wandering naively in the wonderland of numbers, we would like to describe in this book the intricate world of numbers that modern

x PREFACE

number theory has discoverd. We will be very happy if the reader discovers the wonders of numbers and the grandeur of nature.

Kazuya Kato, Nobushige Kurokawa, Takeshi Saito

Preface to the English Edition

The authors hope that the readers enjoy the wonderful world of modern number theory through the book.

Our special thanks are due to Dr. Masato Kuwata, who not only translated the Japanese edition into English but also suggested many improvements on the text so that the present English edition is more readable than the original Japanese edition.

xi

Objectives and Outline of these Books

In thses books, Number Theory 1, 2, 3, we introduce core theories in modern number theory, such as class field theory, Iwasawa theory, the theory of modular forms, etc. The structure of this book is as follows.

The starting point of number theory is astonishment at the wonders of numbers. The work of Fermat, who is considered to be a founding father of modern number theory, illustrates very well the wonder of numbers. We first discuss the work of Fermat on number theory in the introduction to Number Theory 1. The reader will learn how mathematicians of later eras little by little found a fascinating world behind each fact discovered by Fermat. In Number Theory 1 we study some important topics in modern number theory, such as elliptic curves (Chapter l), p-adic numbers (Chapter 2), the C-function (Chapter 3)) and number fields (Chapter 4). These chapters are more or less independent; the material in the earlier chapters is not necessary to understand each succeeding chapter. Chapters 2 and 3 may be easier to read than Chapter 1. The reader should not hesitate to skip parts that are difficult to understand.

Number Theory 2 is devoted to class field theory. We also study the <-function once again. In Number Theory 3 we explain Iwasawa theory and the theory of modular forms, before coming back to elliptic curves once again.

These books are part of the series Fundamentals of Modern Math- ematics, but we were not satisfied with the introduction of fundamentals. We tried to include today’s developments in number theory. For example, we included some important theories developed in recent years, such as the arithemetic theory of elliptic curves, which is part of arithmetic algebraic geometry, and Iwasawa theory, to which we did not find an introduction elsewhere. We hope that we convey the best of modern number theory.

Xl,,

xiv OBJECTIVES AND OUTLINE OF THESE BOOKS

We wanted to include more topics, but we had to omit many of them due to the limitation on the number of pages. We regret that we could not mention Diophantine approximations and transcendental number theory, both of which are seeing new developments in recent years.

Prerequisites to Number Theory 1 are the fundamentals of groups, rings and fields. In Number Theory 2 we recommend that the reader be familiar with Galois theory.

The reader is advised to write down simple and easy examples on scratch paper. Just as astronomical observations are indispensable to the study of astronomy, it is indispensable to observe the numbers in order to study number theory. The wonders are there to be discovered. Also, number theory has a long history, which teaches us interesting lessons. We advise you to take an interest in the history of mathematics.

Notation

Throughout the book we use the following symbols:

Z the set of all integers Q the set of all rational numbers lR the set of all real numbers C the set of all complex numbers

A ring is always assumed to have an identity element (written l), and a homomorphism of rings is assumed to send 1 to 1.

If A is a ring, AX denotes the group of invertible elements of A. In particular, if A is a field, AX is the multiplicative group consisting of all the nonzero elements of A.

CHAPTER 0

Introduction -- Fermat and Number Theory

In September 1994 Andrew Wiles proved Fermat’s Last Theorem, which states:

“For n greater than or equal to 3, there exist no natural numbers x, y, z satisfying the equation

xn + yn = Zn.”

Fermat’s Last Theorem had resisted a proof for more than 350 years. Fermat (1601-65) wrote his “Last Theorem” around 1630 in the

margin of a book he owned. Fermat also left a phrase (‘1 found a remarkable proof for this fact, but this margin is too narrow to write it down.” In spite of the efforts of many people, the proof has not been discovered.

In this chapter we focus on Fermat, who is considered to be a “founder of modern number theory”. We review his work on number

theory, and see how his work has been developed and extended in later eras. We introduce our treatment of Fermat’s work in this book from a modern viewpoint.

0.1. Before Fermat

Fermat wrote down his “Last Theorem” in the margin of his copy of Arithmetica by Diophantus, an ancient Greek mathematician. It was on the page where the positive integral solutions of the equation x2 + y2 = z2 were discussed. Fermat replaced the power of the equation by 3,4,5, . . . .

There are many positive integral solutions to x2 + y2 = ,z2, such as

32 + 42 = 52, 52 + 122 = 132, 82 + 152 = 172

2 0. INTRODUCTION

/ 5 4Lll

4 13

3 5’

\ 17 12 D 15

/

a

FIGURE 0.1. Pythagorean Theorem

(see 92.1). By the Pythagorean Theorem, such a solution corresponds to the three sides of a right triangle as we see in Figure 0.1. As a result, this equation has been studied since ancient times. In the middle of this century archaeologists succeeded in deciphering the writings on a plate found at an ancient Babylonian site of 4000 years ago. On it were inscribed many solutions to z2 + y2 = z2, such as

11g2 + 1202 = 16g2.

The author of this plate apparently knew how to find such x, y and Z.

In ancient Greece many superb mathematicians emerged, and Pythagoras (572-492 B.C.) is among them. The theorem is named for Pythagoras because he is considered to be the first to prove it. Some people consider Pythagoras to be the originator of number theory. He was fascinated by the mystery of numbers, and he said “Everything is a number.” Pythagoras found that two chords whose length have an integer ratio give a beautiful harmony, and he invented a musi- cal scale. He attached great importance to integer ratios, but he is considered to be the first one to find the existence of irrational numbers, namely numbers which cannot be expressed as the ratio of two integers.

Rational numbers, which can be expressed as the ratio of two integers, seem to be tightly packed in the line formed by real numbers, but there are numbers such as & which are not rational numbers. This fact cannot be seen by the naked eye. We are capable of seeing it through the method known as “proof”, which was invented by the ancient Greeks. Pythagoras was astonished by the proved existence of an irrational number. He proved it by himself, but he agonized over the interpretation of this fact. (Pythagoras thought that the existence

0.1. BEFORE FERMAT 3

of an irrational number was an error of the gods. So, he prohibited his disciples from telling this fact to anyone else. Legend has it that a disciple broke the ban, and then lost his life in a shipwreck because of the anger of the gods.)

Euclid’s Elements, which was written in the third century B.C., is a compilation of ancient Greek mathematics. It includes a proof of the existence of infinitely many prime numbers, and it discusses greatest common divisors and least common multiples (volumes 7 and 9 in the thirteen volumes of Elements). Knowing the existence of irrational numbers, Elements treats the question “How can we give a foundation for real numbers based on rational numbers?“, and it develops an excellent theory of real numbers (Elements, volume 5). Pythagoras agonized over this question, and Elements discusses it a great deal. It was only in the nineteenth century that a complete answer was given (see $2.4 in this volume).

However, the theory of real numbers developed in the nineteenth century did not put to rest the question posed by the ancient Greeks: “What are the numbers?” Around a hundred years ago, using a method similar to the construction of real numbers out of rational numbers, a world of numbers called “$-adic numbers” was established for each prime number p out of rational numbers. They form a quite different world from the world of real numbers, but they turn out to be as natural and as important as the world of real numbers.

{padic numbers} > {rational numbers} c {real numbers)

Diophantus was a mathematician of the third century, and he was a descendant of the ancient Greek school of mathematicians. He wrote the book Arithmetica, which discusses rational solutions to algebraic equations. After Diophantus, the development of number theory slowed down until Fermat. The Renaissance revived the free spirits of the ancient Greece, and Arithmetica was republished. Fermat was stimulated by Arithmetica and began to study number theory.

Fermat was a lawyer in Toulouse in France. He founded a method of describing a geometric figure by an equation (for example, express-

ing an ellipse by the equation $ + $ = 1) independent of Descartes. He obtained maxima and minima of a function using a method similar to calculus. Later this work served as a clue to the discovery of calculus. He also did some important work on number theory. He was the greatest mathematician of the first half of the seventeenth century.

4 0. INTRODUCTION

In the following sections we introduce some propositions Fermat claimed to have proved. Each of them surpassed the level of ancient mathematics, and they began the epoch of modern number theory. Fermat himself seldom wrote down a proof, but mathematicians of later eras made efforts to give a proof to each of these propositions. These propositions concern integral or rational solutions to algebraic equations. It appears as if they are just a compilation of bits of facts on different equations. Indeed, his contemporaries had a tendency to think that way.

However, we believe that Fermat, who had a deep affection for these propositions, understood intuitively that the study of integral or rational solutions to equations leads us to a profound part of mathematics. As it turned out, these theorems are the tip of the iceberg of deep mathematics.

0.2. Prime numbers and the sum of two squares

Fermat left forty-eight comments in the margin of his copy of Arithmetica about his work related to the text. These comments were published after the death of Fermat by his son. The so-called “Last Theorem” is the second among these comments. (See, for example, Number Theory by A. Weil.)

The seventh comment is related the following propositions obtained by Fermat.

PROPOSITION 0.1. Let p be a prime number congruent to 1 mod-

ulo 4 (e.g., 5,13,17). Then there exists a right triangle with integer sides such that the length of tile hypotenuse is p. Conversely, no such right triangle exists for any prime number congruent to 3 modulo 4 (e.g., 3,7,11).

Notice that in Figure 0.1 the prime numbers 5,13,17 are hy- potenuses of right triangles. It can be shown, however, that there is no right triangle having 21 (which is not a prime) as its hypotenuse, even though 21 is congruent to 1 modulo 4. As we mentioned earlier, right triangles whose sides are integers have been studied since ancient times. However, Fermat was the first to discover such relations between prime numbers and right triangles.

PROPOSITION 0.2. If p is a prime number congruent to 1 module 4, then there exist natural numbers x and y satisfying

p=x2+y2.

0.2. PRIME NUMBERS AND THE SUM OF TWO SQUARES 5

For example, we have

Conversely, for a prime number p congruent to 3 module 4 there do not exist rational numbers x and y satisfying p = x2 + y2.

Propositions 0.1 and 0.2 were “preludes” to class field theory, which is one of the greatest theories of twentieth century mathematics. We will discuss class field theory in Volume 2. Using the complex number i = a, we can interpret Proposition 0.2 as follows. A prime number p congruent to 1 modulo 4 loses its irreducibility as a prime number in the ring

Z[i]={a+bi/ a,bEZ} (Z is the ring of all integers)

and it factors into the product of two numbers, such as

5 = 22 + l2 = (2 + i)(2 - i),

13 = 32 + 22 = (3 + 2i)(3 - 2i),

17 = 42 + l2 = (4 + i)(4 - i).

The numbers such as 2 + i, 2 - i a.nd 3 + 2i that appear in the above factorizations are “prime elements” in Z[i] which correspond to prime numbers in Z. Just as any nonzero integer can be uniquely factored into the product of prime numbers up to a multiple of fl, any nonzero element of Z[i] can be factored into the product of prime elements up to a multiple of fl or fi. A prime number congruent to 1 modulo 4 is the product of two prime elements in Z[i], while a prime number congruent to 3 modulo 4 is a prime element in Z[i]. This is the idea behind Proposition 0.2.

We can also prove Proposition 0.1 using the idea of “prime factorization in Z[i]“, as we see

52 = (2 + i)2(2 - i)2 = (3 + 4i)(3 - 4i) = 32 + 42,

132 = (3 + 2i)2(3 - 2i)2 = (5 + 12i)(5 - 12i) = 52 + 122,

172 = (4 + i)2(4 - iy = (15 + 8i)(15 - 8i) = 152 + 82.

Therefore, Propositions 0.1 and 0.2 are reflections of the fact that as we extend the notion of numbers from Z to Zbi], the factorization of a prime number in Z[i] is determined by its residue modulo 4. One of the main themes of class field theory is the factorization of prime numbers when we extend the world of numbers, and Fermat’s

6 0. INTRODUCTION

Propositions 0.1 and 0.2 may be called the “prelude to class field theory”. We will come back to class field theory once again in $0.3.

0.3. p = x2 + 2y2, p = x2 + 3y2,. . .

Fermat also discovered the following fact.

PROPOSITION 0.3. If p is a prime number congruent to 1 or 3 modulo 8, then there exist natural numbers x and y satisfying

p=x2+2y?


3=12+2x12, 11=32+2x12, 17=32+2x22.

Conversely, for a prime number p congruent to 5 or 7 module 8 there do not exist rational numbers x and y satisfying p = x2 + 2y2.

PROPOSITION 0.4. If p is a prime number congruent to 1 modulo 3, then there exist natural numbers x and y satisfying

p=x2+3y?


7=22+3x12, 13=12+3x22, 19=42+3x12.

Conversely, for a prime number p congruent to 2 module 3 there do not exist rational numbers x and y satisfying p = x2 + 3y2.

PROPOSITION 0.5. If p is a prime number congruent to 1 or 7 module 8, then there exist natural numbers x and y satisfying

p=xa-2y2.


7 = 32 - 2 x 12, 17 = 52 - 2 x 22, 23 = 52 - 2 x 12.

Conversely, for a prime number p congruent to 3 or 5 modulo 8 there do not exist rational numbers x and y satisfying p = x2 - 2y2.

We will give a proof of these propositions in Chapter 4, together with a proof of Propositions 0.1 and 0.2. Through the eyes of modern mathematics, all these propositions may be regarded as preludes to class field theory. Consider the identities

3 = l2 + 2 x l2 = (1+ G)(l - J-2),

7 = 22 + 3 x l2 = (2 + Q)(2 - a),

7 = 32 - 2 x l2 = (3 + Jz)(3 - Jz).

0.4. PELL'S EQUATIONS 7

TABLE 0.1

~

primes congruent to 1 or 3 modulo 8

acm I primes congruent to 1 or 7 modulo 8

We see that Propositions 0.3, 0.4 and 0.5 are reflections of how prime numbers are factorized in Q(n) = {u + b&2 1 a, b E Q} (where Q is the set of all rational numbers), Q(a), and Q(d), respectively. Together with Proposition 0.2, we summarize the factorization of prime numbers in Table 0.1.

Class field theory tells us the correspondence between the extensions of the rational number field Q and the factorization of prime numbers. Furthermore, it tells us the correspondence between the extensions Q( J--r) and Q(a) and the factorization of prime elements of Q(&i) and Q(a). See Chapter 4 for details.

Class field theory is one of the summits attained by Teiji Takagi around 1920 after contributions by Fermat, Gauss, Kummer, Weber, Hilbert , and others.

Also, there is an interesting theory on the existence of rational solutions to equations of the type a~’ + by2 = c (a, b, c are rational numbers), such as x2 + y2 = 5, x2 + 2y2 = 7. We will discuss it in Chapter 2.

0.4. Pell’s equations

Fermat also declared that he proved the following.

PROPOSITION 0.6. Let N be a natural number which is not a square of another natural number. Then the equation

x2 - Ny’ = 1

has injinitely many natural number solutions.

For example, the equation x2 - 2y2 = 1 has infinitely many natural number solutions such as

32 - 2 x 22 = 1, 172 - 2 x 122 = 1, 9g2 - 2 x 702 = 1.

An equation cf the form x2 - Ny2 = 1 is called a Pell’s equation.

8 0. INTRODUCTION

Through the eyes of modern mathematics Proposition 0.6 may be regarded as a statement about the ring Z[&V] = {a+bfl ( a, b E Z}. If integers z and y satisfy 2’ - Ny2 = 1, then x + yfl is a unit of the ring Z[&V] ( an element that has an inverse in %[&I), since we have the relation (x + yv%)(z - yfi) = 1. For example, it can be seen that the set of units of Z[Jz] is the set {f(l + a)” j n E Z}, and the fact that iZ[fi] h as infinitely many units is the reason why the equation x2 - 2y2 = 1 has infinitely many solutions in natural numbers. The situation is significantly different with the ring Z[i], whose set of units is the finite set (51, rti}. We will study such sets of units in Chapter 4, where we introduce “Dirichlet’s unit theorem” (see 54.2; the proof will be given in 56.2). In $4.2 we will prove Proposition 0.6 using Dirichlet’s unit theorem.

0.5. Triangular numbers, quadrangular numbers, pentagonal numbers

The eighteenth comment of Fermat in the margin of Arithmetica is the following proposition.

PROPOSITION 0.7. If n > 3, any natural number can be expressed as the sum of less than or equal to n n-gonal numbers.

Here, an n-gonal number is the number of dots when you draw a regular n-gon in such a way as in Figure 0.2. Pythagoras and his disciples showed great interest in these numbers. For example, 1,3,6,10 ,... are triangular numbers, which can be expressed as iz(x + 1) with a natural number z. Quadrangular numbers are nothing but squares.

In the place where he wrote down Proposition 0.7, Fermat said that Proposition 0.7 was related to many profound mysteries in number theory and that he intended to write a book about them. Unfor- tunately, however, the book was never written.

If we extract the part about the quadrangular numbers from Proposition 0.7, we have the following.

PROPOSITION 0.8. Let n be a natural number. Then, there exist integers x, y, z and u satisfying

n=x2+y2+z2+u2.

0.5. TRIANGULAR, QUADRANGULAR.. PENTAGONAL NUMBERS 9

. OQ tsl FIGURE 0.2. n-gonal numbers


5=22+12+02+02, 7=22+12+12+12,

15 = 32 + 22 + 12 + 12.

Euler, the greatest mathematician of the eighteenth century, was quite impressed by Fermat’s Proposition 0.7, and was disappointed that Fermat had not written the proof. He became the successor to Fermat in number theory by giving proofs to many of the statements Fermat made. It is said that Euler struggled greatly when he attempted to prove Proposition 0.8. A proof of Proposition 0.8 was given in 1772 by Lagrange, who took over Euler’s effort.

In 1882 Jacobi gave a new proof of Proposition 0.8 using automorphic forms. We will present Jacobi’s proof in Chapter 9 on automorphic forms in Volume 3 (Theorem 9.22). Jacobi’s method of proof is so strong that it gives the number a(n) of quadruples (5, y, z, U) that

10 0. INTRODUCTION

satisfy

n = x2 + y2 + z2 + u2

for each integer n > 0. Jacobi’s method uses the fact that the series

-g a(n)Fnz n=O

is an automorphic form, and it is a typical example of applications of automorphic forms to the arithmetic of quadratic forms.

Propositons 0.1-0.8 solve some of the problems of representing integers or rational numbers by quadratic forms such as x2 + y2 and x2 + y2 + z2 + u2. The arithmetic of quadratic forms grew out of these questions.

0.6. Triangular numbers, squares, cubes

Until now, all the work of Fermat we introduced concerns squares of numbers. We now consider cubes of numbers. A natural number that is the cube of another natural number is called a cubic number. Fermat compared cubic numbers to triangular numbers, and cubic numbers to square numbers. He stated the following.

PROPOSITION 0.9. A triangular number dinerent from 1 is not a cubic number.

PROPOSITION 0.10. The only case where a square number added to 2 becomes a cubic number is 52 + 2 = 33.

PROPOSITION 0.11. The only cases where a square number added to 4 becomes a cubic number are 22 + 4 = 23 and 112 + 4 = 53.

Propositions 0.9, 0.10 and 0.11 concern natural number solutions to

iY(Y + 1) = x3, y2+2=x3, y2+4=23.

It is very difficult to prove these propositions (as well as Propo- sitions 0.1-0.8) by hand without using any significant tools. In at- tempting to prove these propositions we are naturally led to profound mathematics.

In $4.1 we will prove Propositions 0.10 and 0.11 by methods of algebraic number theory. Rewriting the equations y2 + 2 = x3 and y2 + 4 = x3 as

(y + d=)(y - J-2) = x3 and (y + 2a)(y - 2&i) = x3,

0.7. RIGHT TRIANGLES AND ELLIPTIC CURVES 11

FIGURE 0.3. The elliptic curve y2 = x3 - 2

respectively, we can prove Propositions 0.10 and 0.11 using the arithmetic of iZ[J-“i] and Z[&i], respectively.

We can view Propositions 0.9-0.11 as solving the equations of the form

(0.1) y2 = (polynomial of degree 3),

where the cubic polynomial on the right-hand side has no multiple root. (In Proposition 0.9 we can rewrite iy(y+ 1) = x3 as (2y+ 1)2 = (2~)~ + 1, and we obtain an equation of the form (0.1) by replacing

2~ + 1 by Y.) A curve defined by an equation of the form (0.1) is called an

elliptic curme (see Figure 0.3). An elliptic curve is not an ellipse; it is so named due to the fact that it is related to the length of the perimeter of an ellipse. From here on all the work of Fermat we discuss will be related to elliptic curves. Fermat studied elliptic curves a great deal, although he did not realize it consciously. Elliptic curves are rich mathematical objects. We will discuss elliptic curves in Chapter 1 and in Volume 3.

0.7. Right triangles and elliptic curves

Fermat’s twenty-third comment in the margin of Arithmetica is Proposition 0.12, and his forty-fifth comment is Proposition 0.13. He also mentions Proposition 0.14.

12 0. INTRODUCTION

PROPOSITION 0.12. Given a triangle whose sides have rational length, there exist infinitely many triangles with rational sides that have the same area as the given triangle.

For example, the area of the triangle whose sides are 3,4,5 is 6, and Fermat explained a method to obtain the triangle (&, y, w) that has the same area 6.

PROPOSITION 0.13. The area of a right triangle whose sides are integers is not a square.

PROPOSITION 0.14. The area of a right triangle whose sides are integers is not twice a square.

Propositions 0.13 and 0.14 say that there does not exist a triangle whose sides are rational numbers and whose area is 1 or 2, respectively. If such a triangle existed, we would be able to obtain, by multiplying all three sides by a suitable integer, a triangle whose sides are integers and whose area is a square or twice a square.

As we will show in §l. 1, finding a right triangle whose sides are rational numbers and whose area is a positive rational number d is essentially the same as finding a rational solution to the equation y2 = x3 - d2x other than (x, y) = (O,O), (fd, 0). Thus, Proposition 0.13 and 0.14 state that the equation y2 = x3 - d2x for d = 1,2 does not have a rational solution except for (x, y) = (O,O), (fd,O) (which we will show in the case d = 1 in §1.3), whereas Proposition 0.12 states that if y2 = x3 - d2x has a rational solution other than (O,O), (fd, 0), then it has infinitely many rational solutions.

A very important conjecture, called the Birch and Swinnerton- Dyer conjecture, has been proposed to provide a method of deter- mining whether or not an equation of an elliptic curve with rational coefficients has a rational solution (see $12.1(e) in Volume 3); this is currently an active field of research. Wiles, who proved Fermat’s Last Theorem, started his career by studying the Birch and Swinnerton- Dyer conjecture (3. Coats and A. Wiles, On the conjecture of Birch and Swinnerton-Dyer, Invent. Math. 39 (1977), 2233251).

0.8. Fermat’s Last Theorem

Statements made by Fermat have been proved by the efforts of mathematicians of later eras; however, Fermat’s Last Theorem re- mained unproved, and thus was called the “Last Theorem”.

It is known that Fermat had a complete proof for the case n = 4 (i.e., nonexistence of nontrivial solutions to the equation x4+y4 = z4).

0.8. FERMAT’S LAST THEOREM 13

Fermat seldom wrote a proof of his results, but he actually wrote down a proof of Proposition 0.13 in the margin of Arithmetica. The proof of Proposition 0.13 gives a proof of the Last Theorem for the case n = 4 as a by-product (see 31.1). Fermat told his acquaintances about the results mentioned in this chapter over and over again except for the Last Theorem. Later in life, he also mentioned the case n = 3 of the Last Theorem as his important discovery. Considering what he wrote about those results and the outline of the proofs in the letters, we guess that Fermat had a proof or something closer to a proof for those results. However, Fermat never discussed the Last Theorem in the case where n is greater than or equal to 5 except in t,he margin of Arithmetica. Considering how hard it was to prove the Last Theorem for the mathematicians of later eras, it is believed that Fermat thought wrongly that he had a proof for the Last Theorem.

Some attempts to prove Fermat’s Last Theorem by mathematicians of later eras brought advancements in mathematics. Among those are the work of Kummer and of Wiles. Kummer did the following. Fermat ‘s equation

xn + y” = zn

can be rewritten in the product form

Xn = (2 - Y)(Z - GLY) . ‘. (2 - c,“-‘YL

where cn is the n-th primitive root of unity cos(27r/n) + isin(2r/n). I f the ring

q&l = (a0 + a1Cn + . . + a,(~ 1 r 2 0, ~0,. . , a, E Z}

has the unique factorization property (i.e., the property that “any nonzero element can be factored uniquely into the product of prime elements” just as in Z), we can prove Fermat’s Last Theorem by factoring z and z - <ky (Ic = 0, 1, . . , n - 1). Unfortunately, for most n, Z[<J does not have a unique factorization property like Z or the ring Z[i] that appeared in 30.2.

Kummer discovered that in Z[&] there is a law called the unique factorization into prime ideals (see 34.2) which replaces the unique factorization into prime numbers. His discovery pioneered algebraic

number theory (the study of rings such as Z[&]), and he managed to prove Fermat’s Last Theorem for many n (94.4).

In the course of his work Kummer came close to discovering the notion of p-adic numbers, and he discovered a mysterious relation

14 0. INTRODUCTION

among three objects: the arithmetic of Z[&], p-adic numbers, and the < function

which was discovered by Euler in eighteenth century (see Chapter 3). Kummer’s work grew into Iwasawa theory in the twentieth century. We will discuss Iwasawa theory (see Chapter 10 in Volume 3). Wiles extended Iwasawa theory, used the theory of automorphic forms (see Chapter 9 in Volume 3), and studied the arithmetic of elliptic curves very deeply in order to prove Fermat’s Last Theorem.

Details of the proof given by Wiles will be discussed in the book Fermat’s Last Theorem in the Iwanami series The Development of Modern Mathematics. We will also explain the highlights of his proof in 512.2 in Volume 3.

We have seen the relation between the work of Fermat and modern mathematics. Fermat, who was the founder of modern number theory, noticed the depth of the world of numbers. Recently, a deeper part of number theory has been found to be tied up with a deeper part of theoretical physics as if it makes a harmony with the philoso- phy of Pythagoras that “everything is a number.” We think that the reason for the depth of the world of numbers fascinated Pythagoras, Fermat and many others is that it is a reflection of the depth of the universe. As number theory has been developed during the 350 years since Fermat’s era, we have discovered the enormous depth of the world of numbers.

Exercises

0.1. Show that the n-th root of 5 is an irrational number for n greater than 1.

0.2. Show that fi + fi is an irrational number.

0.3. Express 29, 37, 41, and 53 in the form x2 +y2 (x, y integers).

0.4. Diophantus states “65 = 5 x 13 is the product of 5 and 13, both of which can be the length of the hypotenuse of a right triangle with rational sides. Therefore, 65 can be the length of the hypotenuse of two different right triangles with rational sides as we have 652 = 632 + 162 = 562 + 332.” Explain this fact using prime factorization in Z[i] as in $0.2.

EXERCISES 15

0.5. If we form the fraction z/y from a natural number solution to x2 - 2y2 = 1, such as 172 - 2 x 122 = 1 and 9g2 - 2 x 702 = 1, we obtain a rational number very close to fi = 1.41421.. . as we have 17 = 1416

12 . 99 = 1.41428.. . . Explain why. “‘> 70

0.6. Show that there are infinitely many integers which are simultaneously both a triangular number and a square.

CHAPTER 1

Rational Points on Elliptic Curves

The aim of this chapter is to introduce elliptic curves and the main part of the proof of Mordell’s theorem, which plays an important role in the arithmetic of elliptic curves.

1.1. Fermat and elliptic curves

(a) x4 + y4 = z4 and elliptic curves. As we explained in $0.7, Fermat wrote down a proof of the fact that “there does not exist a right triangle whose sides are integers and whose area is a square”(Proposition 0.13) in the margin of his copy of Arithmetica. His proof implies the following proposition.

PROPOSITION 1.1. There is no solution (cc, y, z) to x4 + y4 = z4 satisfying xyz # 0.

In modern language, Fermat’s proof of Proposition 0.13 can be considered a study of the elliptic curve y2 = z3 - z. As we will see later in (c), Proposition 0.13 is equivalent to Proposition 1.2 below. Proposition 1.1 is also a consequence of Proposition 1.2.

PROPOSITION 1.2. The only rational solutions to y2 = x3 -x are

(z,y) = (0,O) and (kl,O).

We can see that Proposition 1.1 is a consequence of Proposi- tion 1.2 as follows. If there exist natural numbers 5, y and z satisfying

x*+y4=z4,

we see (by moving y4 to the other side and then multiplying by z2/y”) that they satisfy

(Ye),= ($)3Y;.

This implies that the equation y2 = x3 - x has a solution satisfying y # 0, which contradicts Proposition 1.2. Thus, we see that

17

18 1. RATIONAL POINTS ON ELLIPTIC CURVES

FIGURE 1.1. Elliptic curves

Proposition 1.1 follows from Proposition 1.2. We will give a proof of Proposition 1.2 in (d). Our proof is a translation of Fermat’s proof of Proposition 0.13 written in the margin of Arithmetica.

(b) Elliptic curves. In the Introduction we explained that Fer- mat’s statement “No triangular number different from 1 is the cube of a natural number” can be interpreted as a statement about the integer solutions to the equation y ’ = z3 + 1. We also said that Fer- mat stated that the only natural number solutions to y2 = x3 - 4 are (z, y) = (2,2) and (5,ll). The graphs of the elliptic curves

y2 = x3 - 5, y2 = x3 + 1, y2 = 23 - 4

are shown in Figure 1.1. An elliptic curve over Q is a curve given by an equation of the

following form:

(*I y2 = ax3 + bx2 + cx + d (a,b,c,d~Q!, a#O),

where the cubic polynomial of the right-hand side does not have a multiple root.

If K is a field of characteristic different from 2, then we define an elliptic curve over K by replacing a, b, c, d E Q by a, b, c, d E K in (*). In this section we consider only elliptic curves over Q, and we omit the definition of elliptic curves over a field of characteristic 2.

The curves defined by

y2 zz 53 and y2 = x2 (z + 1)

1.1. FERMAT AND ELLIPTIC CURVES 19

FIGURE 1.2. Curves that are not elliptic curves

are not elliptic curves since the cubic polynomials on the right-hand side have a multiple root. This can be seen in Figure 1.2 as they are graphically different from elliptic curves--each of them has a singular point at (0,O).

In Figure 1.1 the points indicated by . are integral points (points whose z- and y-coordinates are both integers) of each elliptic curve. A point whose x- and y-coordinates are rational numbers is called a rational point. Studying integral and rational points on a elliptic curve was Fermat’s favorite theme, and as we will explain in the book, it leads us to a profound part of mathematics.

The only integral points of elliptic curve in Figure 1.1 are the points marked by the dots . . (For y2 = x3 + 1, this statement contains Proposition 0.9. For y2 = x3 - 4, this statement corresponds to Proposition 0.11. A proof of Proposition 0.11 will be given in $4.1.)

In general, it is known that an elliptic curve over Q has only a finite number of integral points (Mordell, Siegel). Since y2 = x3 and y2 = x2(x+1) are not elliptic curves, they may have infinitely integral points. Indeed, (n3, n2) (n E Z) are integral points of y2 = z3, and (n2 - 1, n(n2 - 1)) (n E Z) are integral points of y2 = x2(x + 1). This suggests that the geometrical difference is related to the arithmetical difference.

On the other hand, an elliptic curve over Q may have a finite or infinite number of rational points. In Figure 1.1, all the rational points of y2 = x3 -x are the points indicated by the dots . (Proposition 1.2), and all the rational points of y2 = x3 + 1 are also the points indicated


by the dots . . However, there exit infinitely many rational points on y2 = x3 - 4, such as (7, 9). In 51.3 we will introduce Mordell’s theorem, which concerns rational points on elliptic curves. Studying rational points on an elliptic curve is still an active area of research where many studies are being done around the conjecture of Birch and Swinnerton-Dyer and other conjectures.

(c) Right triangles and elliptic curves. Fermat’s Proposi- tion 0.13 is equivalent to the statement “There is no triangle whose sides are rational numbers and whose area is 1.” This statement is equivalent to Proposition 1.2, which concerns the elliptic curve

Y 2 = x3 - z. This equivalence follows from the case d = 1 in Lemma 1.3.

LEMMA 1.3. Let d be a positive rational number. The following conditions (i) through (iii) are equivalent.

(i) There exists a triangle whose three sides are rational numbers and whose area is d.

(ii) There exist three squares of rational numbers that form an arithmetic progression of difference d.

(iii) There exists a rational solution to y2 = x3 - d2x other than (x, y) = (0,O) and (fd, 0).

For example, the area of the right triangle having sides 3, 4, 5

is 6. The sequence (f)‘, (f)‘, (g)” is an arithmetic progression of difference 6. The question “For which d does there exist a sequence of three squares of rational numbers that forms an arithmetic progression of difference d?” (which is equivalent to the question “Which numbers d can be the area of a right triangle whose sides are rational numbers” by Lemma 1.3) has drawn a great deal of attention for long time. In fact, we can find a reference in Arabian mathematics more than one thousand years ago. (Around that time ancient Greek mathematics was forgotten in Europe, but it was imported to the Arabic culture where it grew steadily. During the Renaissance, Europeans reintroduced Arabic mathematics.)

Lemma 1.3 follows from Lemma 1.4 below, since conditions (i), (ii) and (iii) in Lemma 1.3 imply that the sets Ad, Bd and cd in Lemma 1.4, respectively, are not empty when K = Q.


LEMMA 1.4. Let K be a field of characteristic different from 2.

For d E K define the sets Ad, Bd and Cd as ~0110~s:

Ad = {(x, y, z) E K x K x K ) x2 + y2 = z2, ;xy = d),

Bd={(u,u,w)EKxKxKIu2+d=u2, v2+d=w2},

Cd = {(x, y) E K x K ) y2 = x3 - d2x, y # O}.

Then there exist bijections between any two of Ad, Bd, and Cd.

Indeed, between Ad and Bd we have two maps

Ad-+Bd; (x,y,z)++ y,;,+

Bd -+ Ad; (u, u, w) - (w - u, w + 21, au),

and these maps are inverse to each other. For example, (3,4,5) E A6

corresponds to (i,$,g) E BG, and (f)“,(%)‘,(g)’ is an arithmetic progression with difference 6. (5,12,13) E A30 corresponds to

(S, y, 7) E B3o, and ($)’ , (y)’ , (q)” is an arithmetic progression with difference 30.

The fact that there is a one-to-one correspondence between Bd and Cd follows from the case a = d, b = 0, c = -d in Lemma 1.5.

LEMMA 1.5. Let K be a field of characteristic different from 2, and let a, b, and c be distinct elements in K. Define B, C and C by

B={(u,v,w)~KxKxK(u~+a=v~+b=w~+c},

C?= {(x,y) E K x K 1 y2 = (x - a)(x - b)(x - c)},

C = {(x,y) E K x K / y2 = (x - a)(x - b)(x - c), y # 0)

= 6’ - {(a, 01, (b, 01, (c, 0)).

Then

(1) There exist mutually inverse maps f : B --+ C and g : C + B given by

f(u, v, w) = (u2 + a + u21+ VW + wu, (u + v)(u + w)(w + u)),

dX,Yl) = $((x - a)2 - (b - a)(c- a)),

&((x - b)2 - (a - b)(c - b)), &(x - c)~ - (a - c)(b - c)))

(2) There is a map h : B --+ 6’ given by h(u, u, w) = (u2+a, uvw).


The proof of Lemma 1.5 is straightforward, and we leave it to the reader.

REMARK 1.6. The composition of two maps in Lemma 1.5, hog : C + C, is a map called the multiplication-by-2 map of the elliptic curve y2 = (x - a)(x - b)(x - c) (see 51.2). From the definition of h we see that the image of h o g (which coincides with the image of h since g is surjective) is

{(x,y) E K x K 1 y2 = (x - u)(x - b)(x - c),

x-a, x-b, 5 -c are squares in K}

We will use this fact later.

We now have seen that Proposition 0.13 and Proposition 1.2 are equivalent.

(d) Proof of Proposition 1.2. We will now prove that the only rational solutions to y2 = x3 - x are (0,O) and (kl,O).

Let a be a rational number and write a = z as a fraction in lowest terms. Define the height H(a) to be max(lm), In/), where max(a, b) indicates the greater of a and b. (If a = b, we define max(u, b) = a = b.) Also, min(u, b) i.s defined as the smaller of a and b, and if a = b, min(u, b) is defined as a (and thus b). For example, we have

H(-g) ~8, H(S) =7, H(O)=1 since O=i.

Suppose there is a rational solution to y2 = x3 - x other than

(0, (0, (fl, 0). Ch oose one of the solutions such that the height of the x-coordinate is the smallest possible, and denote it by (xo, ye). The strategy of the proof is to show that we can construct another rational solution to y 2 = x3 - x different from (0,O) and (fl, 0) such that the height of x-coordinate is smaller than that of xc. Fermat often used this method of construction of a “smaller solution” to the same equation. He called it the method of “infinite descent”.

The proof consists of the following three steps.

(i) Show that we may assume xc > 1. (ii) Letxe>l. Sincewehave(xc-l)xc(xe+l)=x$xe=yg,

(x0 - 1)X0(X0 + 1) is a square of a rational number. We show that each of xc - 1, xc and x0 + 1 is the square of a rational number.


(iii) Consider the case K = Q, a = 1, b = 0, c = -1 in Lemma 1.5 and consider the map in that lemma

h 0 g : c = {(XT, y) E Q x Q 1 y2 = x3 - 2, y # 0)

+ c = { (2, y) E Q x Q ) y2 = x3 - 22).

Since ze - 1, 20 and 50 + 1 are all squares, it follows from Remark 1.6 that there exists a point (zi,yl) E C such that h o g(zi, yi) = (ze, ye). We then show H(zi) < H(Q).

Let us show first that we may assume zo > 1. If (z, y) is a rational solution to y2 = x3 - II: different from (0, 0), then (-i, 3) is another solution, and we have H(z) = H (-i). Thus, we may assume x0 > 0. If ~0 > 0, then we have (20 - l)zo(ze + 1) = yi > 0, and thus 50 > 1.

Let us move on to the step (ii). Suppose ~0 > 1, and write 20 = z, m > n > 0, as a fraction in lowest terms. We first show that one of m and n is an even number. Suppose both m and n are odd numbers, and let

xb = * = cm + n)P 20 - 1 (m-n)/2

Then (zb, 2yo/(sc - 1)2) is another solution to y2 = x3 - 5. Since both y and “2” are positive integers, we have

m+n m-n < max(m,n) = H(Q).

This contradicts the minimality of H(Q). Thus one of m and n is even, and the other is odd since m and n are relatively prime. Since we have

(20 - l)so(zo + 1) = mn(m - n)(m + n)

n4

is the square of a rational number, it follows that mn(m - n) (m + n) is the square of an integer.

QUESTION 1. Here we used the fact, “If an integer a is the square of a rational number, a is the square of an integer.” Prove this fact.

Next we show that any two of m, n, m-n and m+n are relatively prime. The only thing we worry is that m - n and m + n may not be relatively prime. But a common factor of these two divides both 2m = (m - n) + (m + n) and 2n = (m + n) - (m - n), and thus it must be 2. Since m - n and m + n are both odd, 2 is not a common factor either.


It now follows from the case Ic = 2 in Lemma 1.7 below that all ofm,n,m-nandm+naresquares. Thus,ze=z,sc-l=y

and 20 + 1 = 9 are all squares of rational numbers.

LEMMA 1.7. Let k be a natural number and let al, . . , a, be pairwise relatively prime natural numbers such that the product al . . . a, is the k-th power of a natural number. Then ai is the k-th power of a natural number for each i = 1,. . . , r.

QUESTION 2. Prove Lemma 1.7. (Hint: Factor each a, into the product of

prime numbers).

Next we move on to step (iii). Let (si, yi) be the solution to y2 = 23 - x that is described in the outline of the proof. We show H(zi) < H(sc). By the definition of h o g we have

(XT + 1)”

x” = 4(X? - Xl).

Writing xi = i as a fraction in lowest terms, we have

(r2 + s2)2

xo = 4rs(r2 - s2) '

Here the greatest common divisor of the numerator and the denominator is at most 4. (Reason: It is easy to show that the common prime factor of the numerator and the denominator is at most 2, and thus the greatest common divisor is a power of 2. I f r2 + s2 is even, both r and s must be odd. Thus, both r2 and s2 are congruent to 1 modulo 4, and r2 + s2 is congruent to 2 modulo 4. This implies that (r” + s2)2 is not divisible by 8.) Hence, we have

23(x0) 2 a(r2 + s2)2 2: a max(lrl, 1~1)~ > m=4rl, IsI) = ff(zl).

Here the last > follows from the fact H(xi) 2 2 since xi # 0, &l. This completes the proof of Proposition 1.2.

This proof uses the group structure of an elliptic curve (which will be defined in $1.2) and the notion of “height”. In fact, as we see from Remark 1.6, we used the multiplication-by-2 map in step (iii). In steps (i) and (ii), given a point P(z, y) in y2 = x3 - 5, we

considered two points Q (- $, 3) and R 3, A). In terms of (

the group structure, they correspond to

Q=P+(O,O) and R=-P+(l,O).

1.2. GROUP STRUCTURE OF AN ELLIPTIC CURVE 25

1.2. Group structure of an elliptic curve

Given a rational point in an elliptic curve, there is a way to obtain another rational point. Consider the elliptic curve y2 = z3 - 4 in Figure 1.1. If we draw a tangent line to this elliptic curve at the rational point (2,2), we obtain the point (5,ll) as the other point of intersection between the elliptic curve and the tangent line. The third point of intersection between the elliptic curve and the line passing through (2,2) and (5, -11) is the rational point (y, - y).

This process is possible because an elliptic curve has a group structure. The theme of 51.2 is this group structure on an elliptic curve.

(a) Definition of the group structure on an elliptic curve. Let K be a field of characteristic different from 2. Consider the equation

y2 = ax3 + bz2 + cx + d

of an elliptic curve E over K. (Here, we assume a, b, c, d E K, a # 0, and the cubic polynomial of the right-hand side does not have a multiple root.) Let E(K) be the set of points in E defined over K together with a point 0, i.e.,

E(K) = {(x,y) E K x K / y2 = ax3 + bz2 + cz + d} u (0).

Note that 0 is not the point (0, 0), but it is an added point outside the plane. (The precise meaning of 0 will be discussed later.) We define a group structure on E(K) ( wri tt en additively) using the following principles (i)-(iii)

(i) 0 is the identity element. (ii) I f P,Q E E(K), P # 0, Q # 0, and R(z, y) is the third

point of intersection between the elliptic curve and the line passing through P and Q, then the point (2, -y) E E(K) is P + Q (see Figure 1.3).

(iii) I f P E E(K), P # 0, and the coordinates of P are (5, y), then the inverse element of P is (5, -y).

For example, consider K = Qp and the elliptic curve y2 = x3 - 4. I f P = (2,2), Q = (5,-ll), then P + Q = (y, 9). The above principle does not define P + Q when P and Q coincide. Let us define the sum of P + Q in E(K) more precisely.

IfP=O,thendefineO+Q=Q;ifQ=O,thendefineP+O= P. Suppose P # 0, Q # 0 and the coordinates of P are (xi, yi) and


R

P

u

<

p+Y

FIGURE 1.3

the coordinates of Q are (52, ~2). First we assume 51 # x2. Then the line passing through P and Q is given by the equation

(1.1) y = S(x - Xl) + y1.

In order to find the intersection points, substitute (1.1) in y2 = ax3 + bx2 + cx + d, and we have a cubic equation of the form

4x3 + TX2 + sx + t = 0 (4, r, s, t E K, 9 # 0).

Since x = x1 and x = x2 are solutions to this equation, qx3 + rx2 +

sx + t is divisible by (x - x1)(2 - x2) and it factors as

qx3 +rx2 + sx+t = q(x-x1)(x -x2)(x -x3) (x3 E w.

Substitute x = x3 in (1.1) and solve for y. Denote the solution by y4 and set ya = -y4. Then (xs,y4) is the third point of intersection, and (x3, ys) is P + Q. Explicitly, we have

(14 1 ~2 - YI 2 b

x3 = - ( >

~ ---x1-22, a x2 - Xl a

y3 = _ 92 - Yl -53 +

Y2Xl - 511x2

52 -x1 x2-51 . Next, consider the case xi = x2. If yi = -y2, define P + Q = 0. Suppose x1 = x2, and yi # -y/2. Then we have P = Q, and yi # 0. In this case the line joining P and Q in (ii) must be interpreted as the tangent line to the elliptic curve at P, which is given by

(1.3) Y= 3ax; + 2bxl + c

2Yl (x - Xl) + Yl.


In order to find the points of intersection, substitute (1.3) in y2 = ax3 + bx2 + cx + d, and we have a cubic equation of the form

4x3 + ?-x2 + sx + t = 0 (q,T,S,t E K, 4 # 0).

Since (1.3) is a tangent line, x = x1 is a double root of this equation, and thus the cubic factors to

9x3 + TX2 + sx + t = q(x - x1)2(x - 53) (23 E K).

Substitute x = x3 in (1.3) and solve for y. Denote the solution by ~4, and set y3 = -y4. We define P + Q(= P + P = 2P) as (x3, ~3). Explicitly, we have

(1.4) L(a2xf - 2acxf - 8adxl + c2 - 4bd), x3 =4ayf

4 y3 =8ayf a3xy + 2a2bxT + 5a2cxt

+ 20a2dxT + (20abd - 5ac2)xT

+ (8b2d - 2bc2 - 4acd)xl + (4bcd - 8ad2 - c”)).

For example, consider K = Q and the elliptic curve y2 = x3 - 4. If P = (2,2), then we have 2P = (5, -11).

We have defined P + Q. It is possible to prove that E(K) is an abelian group under this addition. (The associative law is difficult to prove. We can prove the associativity elegantly using algebraic geometry, but we do not discuss it here.)

QUESTION 3. Show that the set {P E E(K) 1 2P = 0) consists of 0 and

nonzero elements of E(K) whose y-coordinates are 0. Show that if K is an algebraically closed field, we have an isomorphism of groups

{P E E(K) 1 2P = 0) E Z/22 @ Z/22.

Let K be a field of characteristic different from 2, and a, b, c distinct elements in K. Consider the elliptic curve defined by

y2 = (x - a)(x - b)(x - c).

We have {P E E(K) 1 2P = 0) = (0, (a,O), (b,O), (c,O)} (see Ques- tion 3). The map in Lemma 1.5

h o g: C = E(K) - (0, (a, 0), (b, 0), (c, 0)) -+ c = E(K) - (0)

is nothing but the multiplication-by-2 map. This can be seen by comparing the definition of h o g and the formula (1.4), which gives the multiplication-by-2 map.

llllllllllllllllllllllllllllllllllllllllll 11111 lllll lllll l~llllll~llllllll F(,lDAN BOO12090492443 B km


(b) The meaning of 0. We now consider the meaning of 0. I f K is the field Iw of real numbers, then 0 is geometrically interpreted as the point at infinity. This can be seen as follows. If K = iw, then

{ (2, y) E R x lR 1 y2 = ax3 + bx2 + cx + d}

is the graph of the elliptic curve. 0 can be thought of as the limit point as we go higher and higher. It is also considered to be the limit as we go lower and lower.

This is consistent with the definition of P + Q. As an example, consider the elliptic curve y2 = x3 - 4 (see Figure 1.1). The sum of the points (2,2) and (2, -2) is 0 by definition. Let P be the point (2,2) and Q a point on the curve very close to but different from the point (2, -2). I f Q approaches to (2, -2) from below, the sum P + Q goes higher and higher to infinity. I f Q approaches to (2, -2) from above, then P + Q goes lower and lower. Therefore, it is natural to think that the limit to the upper direction and the limit to the lower direction should coincide, and the elliptic curve is connected at the point 0. Also, this interpretation is consistent with the fact P + 0 = P. When a point Q on the elliptic curve goes up or down to infinity, P + Q approaches P.

Let K be any field of characteristic different from 2. Let us consider the meaning of 0 in this case. Identify E(K) with the set

X = {ratio(x : y : Z) 1 z,y, z E K, (5, y, Z) # (O,O,O)

y2z = ax3 + bx2z + cxz2 + dz3}

as follows. Identify (2, y) E K x K that satisfies y2 = az”+bx2+cx+d with the ratio (CC : y : 1) E X, and identify 0 E E(K) with (0 : 1 : 0) E X. Here, we consider the ratio (x : y : 2) and the ratio (2’ : y’ : z’) to be the same if and only if there is a nonzero element c in K such that x’ = cx, y’ = cy, Z’ = cz. In X the point 0 acquires the same legitimacy as the points in E(K). (X is a subset of the projective plane consisting of all the ratios (x : y : z). For more detail on projective spaces, see, for example, J. H. Silverman and J. Tate, Rational Points on Elliptic Curves, Appendix A, and the references listed therein.)

I f K = Iw, we give a natural topology to X. When the point (x, y) on the elliptic curve goes higher and higher, or goes lower and lower,

the point (x, y) = ratio(x : y : 1) = ratio (f : 1 : $) converges to the

point 0 = (0 : 1 : 0).


FIGURE 1.4. y2 =x3 +1

(c) Examples. Let us see some examples of the group structure of E(Q) of an elliptic curve over Q.

EXAMPLE 1.8. If E is y2 = zr3 - 5, then each element of the set E(a) = (0, (O,O), (&l,O)} satisfies 2P = 0 (see Question 3). Thus, as a group we have

E(Q) = z/az a? z/az.

EXAMPLE 1.9. If E is y2 = x3 + 1, let P = (2,3) and we see that 2P = (0, I), 3P = (-l,O), 4P = (0, -l), 5P = (2, -3), 6P = 0 (see Figure 1.4). It can be proven that E(Q) consists of only these points, and thus

E(Q) g Z/6Z.

EXAMPLE 1.10. If E is y2 = 5s - 4, let P = (2,2) and we see that 2P = (5, -ll), 3P = (7, v). We do not prove it in this book, but it can be proved that we have

Z%E(Q);w-mP.

EXAMPLE 1.11. If E is y2 = IC’ - 2, let P = (3,5) and we have 2P = (#,-$$). W e d o not prove it in this book, but it can be proved that we have

zrE(Q);nHnP.


(d) Fermat’s method. As we mentioned in $0.7 (Proposi- tion 0.12), Fermat wrote that he had found a method to construct infinitely many right triangles whose sides are rational numbers and whose area is the same as that of a given right triangle with rational sides. He essentially found the fact that can be stated as follows using the notation in Lemma 1.4. Let d be a positive rational number. If (x, y, z) E Ad, then so is

2xy.z y2 - x2 z4 + 4xzyz

y2 -52’ 22 ’ 2(y2 - x2)z > E Ad.

The map Ad + Ad that sends (x, y, z) E Ad to this point (for example, it maps (3,4,5) to (y, &, +)) is nothing but the multiplication- by-2 map of y2 = x3 -d2x passing through the identification Ad g cd in Lemma 1.4.

As in the proof of Proposition 1.2 in $1.1, Fermat made the most out of the multiplication-by-2 map, even though he did not realize that an elliptic curve has a group structure.

The multiplication-by-2 map yielded very strong results for Fer- mat because the height (H(x) in $1.1) of the x-coordinate of 2P is usually much greater than that of P (see Example 1.11). For example, consider the point P = (5,ll) on the curve y2 = x3 - 4. The x-coordinate of 2P is $$, and its height is 785 since the numerator and the denominator are relatively prime. This phenomenon appeared in the proof of Proposition 1.2 at the end of 51.1, and it will be the key point to the proof of Mordell’s theorem in the next section. (The idea of the proof given by Mordell was probably influenced by Fermat . )

1.3. Mordell’s theorem

(a) Statement of Mordell’s theorem. Mordell proved the following theorem in 1922.

THEOREM 1.12 (Mordell’s theorem). Let E be an elliptic curve over Q. Then the group E(Q) is a finitely generated abelian group.

By the fundamental theorem on abelian groups, a finitely generated abelian group is isomorphic to

(1.5) Z@’ @ finite abelian group (r 2 01,

where Z@’ denotes the direct sum of r copies of Z. This number r is called the rank of the elliptic curve. For example, the rank of elliptic

1.3. MORDELL’S THEOREM 31

curves

y2 = x3 - 2, y2 = x3 + 1, y2 = 53 - 4, g = x3 _ 2

are, respectively, O,O, 1,1 (see Examples 1.8-1.11 in $1.2). It is generally believed that the rank of an elliptic curve over Q can be arbitrarily large, but this is an unsolved problem at present.

On the other hand, Mazur proved in 1977 that the finite abelian group part of (1.5), that is, the subgroup of E(Q) consisting of all the elements of finite order, must be one of the groups in the following list:

(1) Z/n& where 1 5 n 5 10 or n = 12; (2) Z/nZ @ Z/2& where n = 2,4,6,8.

(It is known that each of the groups in the above list occurs as the subgroup of all the elements of finite order of some elliptic curve over Q.)

In this section we give the main part of the proof of Mordell’s theorem. The rest of the proof will be given in Volume 3.

(b) Outline of the proof of Mordell’s theorem. Mordell’s theorem is proved using the following two facts.

(I) The weak Mordell theorem, which states that the quotient group E(Q)/2E(Q) is finite.

(II) The properties of heights of the rational points on E(Q).

We will explain (I) later. Here we discuss (II). In $1.1 we defined the height H(x) of x by max()ml, Inl) if we write z = E in lowest terms. For a rational point P on an elliptic curve E over Q we define the height H(P) as the height of the x-coordinate of P if P # 0, and we define H (0) = 1. We use the following two facts about the height.

(IIA) For any positive real number C the set

{P E E(Q) I H(P) I C>

is a finite set.

This follows from the trivial fact that for any real number C, the set {x E Q 1 H(x) 5 C} is finite.

(IIB) There exists a positive real number C satisfying the following two conditions:

(1) For any P E E(Q),

C. H(2P) > H(P)4;


(2) For any P, Q E E(Q),

C. H(P)H(Q) L min(H(P + Q), H(P - Q)).

(1) formulates the phenomenon we mentioned at the end of 51.2, namely, “H(2P) is much larger than H(P)".

Let us prove that Mordell’s theorem follows from (I), (IIA) and (IIB). More precisely, we prove

PROPOSITION 1.13. Let Q1,. . . , Qn E E(Q) be representatives of the elements of E(Q)/2E(Q). (That is, {Qz mod 2E(Q) 1 i = 1 . . > n} equals E(Q)/2E(U3)). Suppose a positive number C sat- iijies the properties (1) and (2) in (IIB). Let M be the largest of

H(Ql), . , H(Q,) and C. Th en E(a) is generated by the finite set

{P E E(Q) I H(P) 5 Ml.

PROOF. Suppose there exist elements of E(Q) outside the subgroup of E(a) generated by the set {P E E(Q) 1 H(P) 5 M}. Let PO be such an element whose height is the smallest. Clearly, we have H(Po) > M. The image of PO in E(Q)/2E(Q) coincides with Qi for some i. For this i, PO + Qi and PO - Qi belong to 2E(a). Let R be

the one of these whose height is smaller, and let PI E E(a) be an element satisfying R = 2Pl. By (1) of (IIB) we have

Hi < C.H(R) < M.H(R)

By (2) of (IIB) we have

H(R) I C. ff(Po)ff(Qi) 5 M2WPo).

Thus, we have

ffpq4 I M3w%).

Since H(Po) > M, we obtain Hi < H(Po)4. Thus, we have

ff(Pl) < H(h). Th e minimality of H(Po) implies that PI belongs to the subgroup generated by {P E E(Q) 1 H(P) < Ad}. Since PO equals either 2Pl + Qz or 2Pl - Qi, PO also belongs to the subgroup generated by {P E E(Q) 1 H(P) 5 hl}, which is a contradiction. This proves Proposition 1.13. 0

QUESTION 4. Let A be an abelian group. Show that A/2A is a finite group if A is finitely generated. On the other hand, show that A is not necessarily finitely

generated even if A/2A is a finite group. (Thus, Mordell’s theorem cannot be derived solely from the weak Mordell theorem, but we need the notion of height.)


(c) Main part of the proof of Mordell’s theorem. The remaining portion of this section is dedicated to the main part of the proof of the weak Mordell theorem for elliptic curves of the form

y/“=(x-a)(x-b)(x-c) (a, b, c are distinct rational numbers),

and the proof of part (IIB). Thus, for elliptic curves given by the above equation, the proof of Mordell’s theorem will be completed in this section. The general case will be treated later in Volume 3. The proof that follows is rather complicated; the first-time reader is advised to skip it and go directly to Chapter 2.

PROPOSITION 1.14. Let a, b, and c be distinct rational numbers. Consider the elliptic curve E defined by

y2 = (x-u)(x-b)(x-c).

I f P # 0, we denote the x-coordinate of P simply by x. Define the

map

8: E(Q) -+ Q”/(Qx)2 x Q”/W)” x Qx/Wx)2

by

d(P) =

I

- - - (x-a, x-b, x-c) ZfP # 0, (a, O>> (h 01, Cc, 01,

((u-b)(a-c), a-b, a-c) ifP=(a,O), ~-~__

(b - a, (b - a)(b - c), b - c) if P = (b,O),

(c-u, c-b, (c-a)(c-b)) ifP=(c,O),

(Ll, 1) ifP=O.

(Here - means mod(Qx)2.) Then we have

(1) The map i3 is a group homomorphism. (2) The kernel ofa is 2E(Q). (3) Let G be the subgroup of Qx/(Qx)” generated by the prime

factors of a - b, b - c, c - a and -1. Then the image of d is contained in G x G x G.

For those elliptic curves treated in Proposition 1.14, the weak Mordell theorem follows easily from Proposition 1.14. Indeed, Propo- sition 1.14 shows that E(Q)/2E(Q) is embedded in the finite group G x G x G by the homomorphism d.

Let us prove Proposition 1.14.


PROOF OF PROPOSITION 1.14( 1). We show that the first com- ponent of d is a homomorphism from E(a) to Qx/(a”)2. (The same argument holds for the second and third components.) Suppose

P,Q E E(Q) and P, Q, and P + Q are not 0 or (a, 0). (If one of P, Q, P + Q equals 0 or (a, 0), the proof is simpler and it is left to the reader.) Let (~1, yl) be the coordinates of P, (22,~~) those of Q, and (~3, ~3) those of P + Q. It suffices to show

(a-a)(22 - a)(23 -a) E (@y2.

(For this implies that 23 - a and (51 - U)(XZ - u) represent the same element in Cjx/(a”)2.) If y = AZ + /L is the equation of the line passing through P and Q, then

(x-u)(x-b)(x-c)-(Xx+p)2=0

is the equation for the x-coordinates of the points of intersection between the line and the elliptic curve. Thus, we have

(x - u)(x - b)(x - c) - (Xx +p)2 = (x-x1)(x-x2)(x - 23).

Letting x = a, we have

(x1 - u)(m - u)(x3 - u) = (Au + p)2 E (uyy2.

This completes the proof. 0

Proposition 1.14(2) follows from Remark 1.6 in 51.1. We need some preparation before proving Proposition 1.14(3).

DEFINITION 1.15. For a prime number p and a nonzero rational number t, we define the p-adic valuation of t, denoted by ord,(t), as the number m in the factorization t = pmu/v, m E Z, where u and v are not divisible by p. Then the following properties (i) and (ii) hold.

(i) ord,(st) = ord,(s) + ord,(t). (ii) For any nonzero rational numbers s and t

ord,(s - t) 2 min(ord,(s),ord,(t)).

If s and t satisfy ord,(s) # ord,(t), then

ord,(s - t) = min(ord,(s), ordp(t)).

PROOF OF PROPOSITION 1.14(3). Let p be a prime number that does not divide either the denominator or the numerator of any of a - b, b - c, and c - a. It suffices to show that for a rational solution (x,y) of y2 = (z-u)(x-b)(z-c) satisfying y # 0, each of ord,(x-a),


ord,(x - b) , and ord, (x - c) is an even number. It follows from

y2 = (x - u)(x - b)(x - c) and (i) that

(*I ord,(x - a) + ord,(x - 6) +- ordP(x - c) is even.

Suppose one of ord,(x - a), ord,(x - b), or ord,(x - c) is negative. Using property (ii), we see in this case that the fact that ord, of the difference of any two of x - a, x - b, and x - c is 0 implies that ord,(x - u) = ordP(x - b) = ord,(x - c). From this and (*) we see that ord,(x - a), ord,(x - b) and ord,(x - c) are all even. Suppose one of ord,(x - a), ord,(x - b) and ord,(x - c) is positive. In this case, the fact that ord, of the difference of any two of x - a, x - b and x - c is 0 implies that any two of ord, (x - u) , ord, (x - b) and ord, (x - c) are 0. From (*) we see that ord,(x - a), ord,(x - b), ord,(x - c) are all even. 0

Next we prove (IIB). Since the proof is complicated, we describe the outline first.

Let E be an elliptic curve over Q with equation

y2 = ax3 + bz2 + cz + d.

Outline of proof of (IIB) (1). We may omit P E E(Q) such that 2P = 0. That is, it suffices to find a positive real number C satisfying C.H(2P) 2 H(P)* for any P E E(Q) such that 2P # 0. For, if C’ is a number greater than both C and H(P)4 for all P E E(Q) satisfying 2P = 0 (there are at most 4 such P’s), then C’ . H(2P) 2 H(P)4 holds for any P E E(Q). Define polynomials f(T) and g(T) by

f(T) = uT3 + bT2 f CT + d,

g(T) = &(a2T4 - 2acT2 - 8adT + c2 - 4bd).

If (x, y) are the coordinates of P E E(Q) such that 2P # 0, it follows

from (1.4) that the x-coordinate of 2P is given by #. As we will see

later, f(T) and g(T) are relatively prime as polynomials (i.e., there is no polynomial of positive degree dividing both). Therefore, it suffices to show Lemma 1.16 below, which has nothing to do with elliptic curves.

LEMMA 1.16. Letf(T) and g(T) be relatively prime polynomials with Q coeficients. Let d be the greater of the degrees of f(T) and


g(T). Then there is a positive real number C such that

holds for all x satisfying f(x) # 0.

We will prove this lemma later. The outline of (IIB)(Z). It suffices to show that there is a positive

real number C such that H(P + Q) . H(P - Q) 5 C. H(P)‘H(Q)’ holds for P, Q in each of the following cases:

(i) P,QEE(Q), P=OorQ=O; (ii) P, Q E E(Q), P + Q = 0 or P - Q = 0;

(iii) P, Q E E(Q), P # 0, Q # 0, P + Q # 0, P - Q # 0.

Case (i) is clear. As for case (ii), we need to show that there exists a positive real

number C such that

H(2P) < c . H(P)4

for all P E E(Q). C onsidering the relation between the x-coordinate of P and that of 2P, it suffices to show Lemma 1.17 below, which has nothing to do with elliptic curves.

LEMMA 1.17. Let f(T) and g(T) be polynomials with Q coeficients. Suppose that the degree off(T) and that of g(T) are both no greater than a given natural number d. Then there is a positive real number C such that

j-g& ( > f(x)

5 c . Iqxy

holds for any x satisfying f(x) # 0.

Finally, consider case (iii). Suppose P, Q E E(Q) , P # 0, Q # 0, P + Q # 0, and P - Q # 0. Write the x-coordinate of P, Q, P + Q and P - Q as xl, x2, x+, and x-, respectively. Define s = xi + x2, t = x152, s’ = x+ +x-, and t’ = x+x-. Then we will later show that s’ and t’ can be expressed as

sI _ ds, t)

f (% t) ’

t’ = h(s,t)

f (% t) ’

where f (5 T), s(S, T) and h(S, T) are polynomials of two variables with Q coefficients whose total degree with respect to S and T is 2. For rational numbers u and u define the height H(u, V) of the pair


(u, u) as follows. Write u and ‘u as a fraction in lowest terms, respectively, and let n be the greatest common divisor of the denominators.

Write u = z and v = r$ and define

H(wv) = max(l4, Id, 14)

Then the question is reduced to Lemma 1.18 below, which has nothing to do with elliptic curves. For we have

H(z+)H(z-) 5 2H(s’, t’) (by Lemma 1.18(l))

I 2c. H(s, t)2 (by Lemma 1.18(2))

< 4c. H(x#H(Llg (by Lemma 1.18(l))

for the real number C appearing in Lemma 1.18(2). Thus, it suffices to replace C by 4C to prove the case (iii).

LEMMA 1.18. (1) F or any rational numbers u and v we have

;H(U)f(v) < H(u + v, UV) 5 2H(u)H(v).

(2) Let f(S, T), s(S, T) and h(S, T) be polynomials in two variables with Q coeficients. Suppose that the total degree with respect to S and T of each off (S, T), g(S, T) and h(S, T) is no greater than a given natural number d. Then there is a positive real number C such that

j-g

(

ds,t) h(s,t) < c. H(s qd

> f(s,t)’ f(s,t) - ’

holds for any rational numbers s and t satisfying f (s, t) # 0.

We will prove Lemmas 1.17 and 1.18 later. Now we discuss the details of the proof of (IIB). First, in the out-

line of the proof of (IIB)(l) the fact that f(T) and g(T) are relatively prime follows from the fact that

g(T) = if’(T)” - 2T + $ f(T) ( 1

(where f’(T) = 3aT2 + 2bT + c is a derivative of f(T)) and the fact that f(T) and f’(T) are relatively prime as polynomials, since f(T) does not have a multiple root. In the outline of proof of (IIB)(2)


case (iii) it suffices to define f(S, T), g(S, T) and h(S, T) as follows:

f(S,T) = S2 - 4T;

g(S, T) = ;(2aST + 2cS + 4bT + 4d);

h(S, T) = $(a2T2 - 2acT - 4adS + c2 - 4bd).

This can be seen from the addition formula (1.2) for the points on an elliptic curve.

In order to complete the proof of (IIB), it remains to prove Lem- mas 1.16, 1.17 and 1.18. We prove them in order of increasing difficulty. (The proof of Lemma 1.16 is the hardest, but the others are relatively easy.)

PROOF OF LEMMA 1.18(l). Writeuandvasu= zandw = 5 in lowest terms, respectively. We have

mn’ + m’n mm’ u+w=

nn’ ’ lLw=-----.

nn’

We show that the greatest common divisor of mn’ + m/n, mm’, and 7272’ is 1. Suppose 1 is a common prime factor of mn’ + m’n, mm’, and nn’. Then 1 divides mm’, and thus 1 divides either m or m’. If 1 divides m, then it divides m’n since it divides mn’ + m’n. Since m and n are relatively prime, 1 divides m’. On the other hand, 1 divides nn’, and thus it divides n’. This contradicts the fact that m’ and n’ are relatively prime. This shows that the greatest common divisor is 1. Consequently, we have

H(u + 21, UW) = max(lmn’ + m’nl, lmm’l, Inn’/)

by definition of the height. On the other hand, we have

H(u)H(v) = max()mm’l, lmn’l, Im’nl, (7272’1).

It follows easily from these that H(u + V,UV) 5 2H(u)H(v). To show @?(u)H(v) 5 H(u + U, UZI), it suffices to show that i lmn’l and ilrn’nl are less than or equal to max(lmn’ + m’nl, Imm’l, Inn’/). Consider ilrnn’l (the proof for ilrn’nl is similar). We may assume

mn’ # 0. Dividing by mn’, and setting y = 2 and y = 5, we need to show that

i I mdll + 4,bI, Ivl)

holds for all real numbers z and y.


This follows from the fact that the inequality 11 +xyl 2 1 - (i) 2 > i holds when 1x1 < $, and IyI < i. 0

PROOF OF LEMMA 1.17. By multiplying f(T) and g(T) by a common nonzero integer if necessary, we may assume that the coefficients of f(T) and g(T) are integers. Let C be d + 1 times the largest of the absolute value of all the coefficients of f(T) and g(T). If we define

f(T) = g-aiTi, g(T) = k biTi, i=o i=o

and write a rational number x satisfying f(x) # 0 as a fraction z in lowest terms, then we have

g(x) i$o bimi,d-i - = fCx) z$o aimind-i.

Therefore, we have

PROOF OF LEMMA 1.18(2). By multiplying f(S, T), g(S, T) and h(S, T) by a common nonzero integer if necessary, we may assume that the coefficients of these polynomials are integers. Let C be i (d + 1) (d + 2) times the largest of the absolute value of the coefficients of these polynomials. Define

f (S, T) = C aijSiT’, g(S, T) = c bi,S”TJ, Z>j id

h(S,T) = &SiT3, i>j

where (i, j) runs through all the pairs satisfying i > 0, j 2 0, i+j < d. For rational numbers s and t satisfying f(s, t) # 0, let n be the least common multiple of the denominators of s and t when we write them


in lowest terms, and let s = T and t = $. Then we have

d% t) & b,jmz(mf)jnd-i-.i c Cymym’)~nd-z-~

___ = h(s,t) .>’ f(S, t) fo = 5 u,.pn~(m')~?zd--i--J .

i.j ZJ

Hence we have

< max C aijrn’(n~‘)jn~-~-~ , i>j

I c . H(s, t)d.

PROOF OF LEMMA 1.16. By multiplying f(T) and g(T) by a common nonzero integer if necessary, we may assume that the coefficients of these polynomials are integers. We will show that there exist a nonzero integer R, a nonnegative integer e 2 0, and polynomials cI(T) (j = 1,2,3,4) with integer coefficients such that the degree of cl (T) is no greater than e for any j, and

(1.6) cl (VP(T) + c2V)dT) = R c3(T)f(T) + c4(T)g(T) = RTd+“.

Let C be 2(e + 1) times the largest of the absolute values of all the coefficients of c3 (T) (j = 1,2,3,4). For any rational number II: satis-

fying f(x) # 0, we show that Hi 5 C. H . Write x = E in

lowest terms. Set

d d e

f(T) = c aiTi, i=O

g(T) = c biTi, i=O

~1 (T) = C CijT’. a=0


Then

f(x)nd = -jy aimind-z, d

g(z)nd = c bimzndei, i=O z=o

cj (x)ne = 2 cijminee2 a=0

are all integers, and by (1.6) we have

(1.7) i

(cl(x)ne)(f(x)nd) + (cz(x)n”)(g(x)nd) = Rndfe

(c3(x)ne)(f(x)nd) + (c4(x)n”)(g(x)nd) = Rmdte.

From (1.7) we see that the greatest common divisor of f (x)nd and g(x)nd divides both Rndte and Rmdte, and thus it divides R since m and n are relatively prime. It follows from

(1.8)

that

(1.9) H # > R-‘max(lf(x)ndl, lg(x)ndl). ( >

(This is the key point of the proof; it shows that the denominator and the numerator of right-hand side of (1.8) will not cancel each other

very much, and thus H M ( >

stays large.)

On the other hand, from the expression of cJ (x)ne in lowest terms we have

lc3(x)nel < 2-I C. H(s)“.

Thus, by (1.7) we obtain the following inequality:

R. Hi+” = Rmax(lmld+e, Inld+e)

5 C. ~(xYm~(lf(xb4, ldx)ndl). In other words, we have

(1.10) H(xjd I CR-l max(lf(x)ndl, ldx)d)

From (1.9) and (1.10) we have


Finally, we show the existence of e, R and q(T) (j = 1,2,3,4) satisfying (1.6). Since f(T) and g(T) are relatively prime, there exist polynomials ~1 (7’) and IQ(T) with Q coefficients satisfying

Also from the fact that f(T) and g(T) are relatively prime we see easily that f ($) Td and g ($) Td are relatively prime polynomials with Q coefficients. Therefore, there exist polynomials VI(T) and 212(T) with Q coefficients satisfying

Let e be an integer greater than the degrees of ~1 (T), uz(T), q(T), and 212 (T) , and let R be a nonzero integer such that all of Rui (T) , Rwi(T) (i = 1,2) have integer coefficients. Define

cl(T) = Rw(T), ~~(5‘3 = Ru2(T),

Then cI)(T) (j = 1,2,3,4) are polynomials with integer coefficients of degree at least e, and they satisfy (1.6). 0

REMARK 1.19. For a point P in E(Q), it can be shown that log(H(2nP))/4n converges when n tends to infinity. So, we define

h(P) = Jim & log(H(2”P)).

For any P,Q E E(tJ), define

(P, Q) = ; (W + Q) - h(P) - h(Q)).

We have h(P) = (P, P), and we can show that the pairing ( , ) has properties of an “inner product”. Namely, for P, Q, R E E(Q) we have

(9 (P, Q) = (Q, P), (4 V’, Q + R) = P, Q) + (P, R),

(iii) (P, P) > 0, and (P, P) = 0 if and only if P is a point of finite order.

EXERCISES 43

Summary

1.1. An elliptic curve is a curve given by an equation of the form:

y2 = (polynomial of degree 3 in z without a multiple root).

1.2. The set of points definied over K of an elliptic curve over K, together with the point 0, forms an abelian group.

1.3. The set of rational points of an elliptic curve defined over Q, together with 0, forms a finitely generated abelian group (Mordell’s theorem).

1.4. In order to study rational points on an elliptic curve, it is important to use properties of the height of a rational point.

Exercises

1.1. Let E be the elliptic curve y2 = x3 + 1. Find the set

{P E E(C) I3P = 0).

1.2. If the z-coordinate of a rational point P of y2 = x3 - 4 is

given by E, the x-coordinate of 2P is given by $>~~$))~. Using this fact, show that

144. H(z-coordinate of 2P) 2 H(z-coordinate of P)4.

Using this fact, show that there exist infinitely many rational points in y 2=x3-4.

1.3. Let K be a field of characteristic different from 2 and 3. Take k E KX, and set

X = {(x, y) E K x K ) x3 + y3 = k},

Y = {(x,y) E K x K 1 y’ = 7x3 - f, z # 01.

Show that there is a map from X to Y given by

and that it is a bijection.

1. RATIONAL POINTS ON ELLIPTIC CURVES

1.4. Let K be a field of characteristic different from 2. Take kEKX,andset

X={(x,y)~KxKIy~=x~+k},

Y = {(x, y) E K x K 1 y2 = x3 - 4kx, (x, y) # (O,O)}.

Show that there is a map from X to Y given by

x -+ y; (X,Y) +x2 +y),4x(x2 -+Y)),

and that it is a bijection.

1.5. Let K be a field of characteristic different from 2. For k E KX, let E be the elliptic curve over K defined by y2 = x3 + kx. Let E’ be the elliptic curve over K defined by y2 = x3 - 4kx. Show that there are two maps f : E(K) + E’(K) and g : E’(K) + E(K) given by

f(P) = i

(x+;,Y(l-$)) if P = (2, Y) # (O,O), o

if P = (O,O), or P = 0.

g(P) = {

( 2 - 2,; (1+ $)) if P = (x, y) # (O,O),

0 if P = (O,O), or P = 0.

Show that g of : E(K) --f E(K) and f o g: E’(K) + E’(K) are the multiplication-by-2 maps. Show that the map

X --+ Y c E’(K) 3 E(K)

obtained by the composition with the map in Exercise 1.4 sends

C&Y) to (X24Y).

1.6. Using Exercises 1.4 and 1.5 and Proposition 1.2, find all the rational points on the following elliptic curves:

(i) y2 = x3 +4x, (ii) y2 = x4 - 1, (iii) y2 = x4 + 4

CHAPTER 2

Conies and p-adic Numbers

In the previous chapter we studied rational points on elliptic curves. In this chapter we study rational points on tonics, which are simpler objects than elliptic curves. The main goal of this chapter is to determine whether or not a given conic has a rational point, and if it does, to describe all the rational points. Even though they are “simpler” than elliptic curves, some interesting theories, such as

quadratic residues and p-adic numbers, arise in order to answer the question of the existence of a rational point on a conic. In addition, another goal of this chapter is to introduce p-adic numbers.

2.1. Conies

(a) Rational points on tonics. An integral solution of the equation

x2 + y2 = z2

with z # 0 determines a rational point on the circle x2 + y2 = 1, since

we have (z)’ + (y)” = 1. F or example, 3’ + 4’ = 5’ determines the

rational point (g, $) on t,he circle x2 + y2 = 1, and 5’ + 122 = 13’

determines the point (&, g).

Conversely, if a rational point on the circle x2 + y2 = 1 is given, we obtain an integer solution of x2 + y2 = .z2 satisfying z # 0 by clearing the denominators. Then, how many rational points does the circle x2 + y2 = 1 have? It turns out that it has infinitely many rational points, as we explain below.

Let us consider another circle x2 + y2 = 3. The fact is that this circle does not have any rational point at all. Can you tell by looking at Figures 2.1 and 2.2 that the right one does not have any rational points while the left one has infinitely many? I suspect not. Human vision cannot distinguish such a thing. In these figures rational numbers are hidden completely by real numbers, and under this

45

46 2. CONICS AND P-ADIC NUMBERS

FIGURE 2.1. FIGURE 2.2.

circumstance it is very difficult to tell something about rational numbers. Rational numbers must be seen under different lights, namely, under “the lights of prime numbers” (see Figure 2.3).

In this chapter we consider the conic

(2.1) ax2 + by2 = c

for nonzero rational numbers a, b and c. In s2.1 we prove that if the conic (2.1) has one rational point (as is the case for x2 + y2 = l), it has infinitely many of them. Moreover, we can write down all the rational points explicitly. On the other hand, it requires a deeper argument to determine whether or not the conic (2.1) has a rational point (see Theorem 2.3 in 52.3). Theorem 2.3 implies that the true feature about rational numbers emerges from obscurity if we see them under “the lights of prime numbers”, together with the light of real numbers.

It turns out that, for any prime number p, there exists “a world of p-adic numbers” analogous to the world of real numbers (see $2.4). In short, we can understand rational points on a conic if we consider it not only in the world of real numbers but also in the world of p-adic numbers for each prime number p.

For example, we know that x2 + y2 = -1 does not have a rational point since it does not have a solution in the world of real numbers. The fact that x2 + y2 = 3 does not have a rational point cannot be seen under the light of real numbers, but it can be seen by looking at it under the light of the prime number 2 or 3 since it has a solution

2.1. CONICS

A the lights of 2 T/prime numbers+

47

1” t . ,

, , ” n:‘” s the light &

5 thz/Zght $ + of 3 T-

2~ the light of fr (supplements 7, real numbers V each other)

-‘/ ,/ 2 the light 3 a the light 5 / of 5 3 of 7 C’. ’ ‘”

)..“3 3” .( 2 thelight 2 _.nj 4 of 11 =? ._.

9

FIGURE 2.3. The light of real numbers and the lights of prime numbers

neither in the world of 2-adic numbers nor in the world of 3-adic numbers. We will discuss this in $2.5.

(b) The case of x2 + y 2 = 1. Let us consider rational points on x2 + y2 = 1 (see Figure 2.4).

If (x, y) is a rational point on the circle x2 -t- y2 = 1 and if (x, y) # (-l,O), the slope of the line joining (x, y) and (-1,O) is the rational number $. Conversely, for a given rational number t, the points of intersection between the circle and the line of slope t passing through (-1,O) are (-1,O) and (&-$, &). The latter is of course a rational point.

FIGURE 2.4. Rational points of x2 + y” = 1


For example, if we replace t by i, i, i, i, $ successively, we obtain

If we let t = A, we obtain (#, z) Clearing the denominators of

(E)’ + (s)’ = 1, we obtain the identity 11g2 + 1202 = 16g2 of the ancient Babylonian plate mentioned in the introduction. To sum up, we have the following one-to-one correspondence:

rational points on x2 + y2 = 1 different from (- 1,O)

r {rational numbers},

(GY) - &?

1 - t2 2t

1+ t2 ’ 1+ t2 > - t.

(c) Conies that have a rational point. Let a, b and c be nonzero rational numbers. If the conic

ax2 + by2 = c

has at least one rational point, we can obtain all the rational points by the same method as above. If Q(xe, ye) is its rational point, we

have the correspondence

{(x,Y) I X,Y E Q, ax2 + by2 = cl

r Q U {co} - {at most 2 elements)

by associating a rational point P on ax2 + by2 = c to the slope of the line joining Q and P (called the line QP). When P = Q, we interpret the line QP as the tangent line to the conic at Q. Further, if the line QP is parallel to the y-axis, we interpret the slope as 00. The meaning of “at most 2 elements” is that we remove km from Q U {co} if -a/b is th e s q uare of a rational number, and we do not remove anything from Q U {oo} otherwise. When -a/b is the square of a rational number, the curve ax2 + by2 = c is a hyperbola, and km are the slopes of its asymptotes.

The reason for the existence of the one-to-one correspondence is the same as in the case x2 + y 2 = 1. If the slope of a line passing through Q is in (IJ U {co}, and it is different from *J-alb, the line intersects the conic in another point P, and P is a rational point. The problem of finding the points of intersection amounts to solving a quadratic equation in rational coefficients, and Q gives one of the

2.2. CONGRUENCE 49

two roots. Since it is a rational root, we see that the other root is also rational in view of the relations between the roots and the coefficients of the equation. That is why P is a rational point.

We can avoid the exceptions, i.e., the part “at most 2 elements”, in the above one-to-one correspondence in the following way. We put

X = { ratio (3~ : y : z) ( 5, y, z E Q,

(2, y, z) # (O,O, O), ax2 + by2 = cz2}.

As we did in $1.2 (b), we identify a solution (z, y) E Q x Q to ax2 + by2 = c with the ratio (z : y : 1) E X. Then the above one-to-one correspondence can be extended to the correspondence

If -u/b is the square of a rational number T, we associate r E Q to the element (1 : T : 0) in X.

The fact that we can describe all the rational points on a conic as soon as it has one point can be generalized to the case where the conic is defined over any field K of characteristic different from 2. Let a, b, c E KX and suppose there is an (z, y) E K x K satisfying ax2 + by2 = c. Then we obtain similarly the one-to-one correspondence

X = { ratio (x : y : 75) ( z,y, z E K, (5, y, z) # (O,O, 0),

ax2 + by2 = cz”}

‘;T’“Ub+

QUESTION 1. Find a rational point on x2 + y2 = 5 other than (3~1, k2), and (k2, ztl).

QUESTION 2. In the ancient Babylonian identity 119’ + 120’ = 169’, which

we mentioned in the Introduction, the ratio s of two sides of the corresponding right triangle is very close to 1. (The Babylonian who wrote the plate sorted the

solutions of x2 + y2 = *2 according to the ratio of I and y, and thus the above solution is found at the top of the list.) Find a solution whose ratio of z and y is closer to 1.

2.2. Congruence

If a conic ax2 + by2 = c with ra,tional coefficients has one rational point, we can find all the rational points, as we have seen in the previous section. On the contrary, it is a deeper question to determine whether or not a conic has a rational point. This question is related


to congruence equations and quadratic residues. In this section we explain congruence.

(a) Congruence and its fundamental properties. Let m be a natural number and a, b two integers. The notation

a-b modm

means that a - b is a multiple of m. (We say “a is congruent to b modulo m” .) For example,

28 ~3 mod 5, 35-O mod5.

We review here briefly the basic properties of the congruence. The quadratic reciprocity law will be introduced here and proved in Chapter 5 in Volume 2.

First, we immediately see the following:

(2.2) a E a mod m.

(2.3) a E b mod m implies b E a mod m.

(2.4) a E b mod m and b z c mod m imply a E c mod m.

(2.5) arbmodmandcrdmodmimplya+c~b+dmodm and ac E bd mod m.

In order to explain why congruences are useful when we study integral or rational solutions to an equation, we present a simple example. The equation x2 + y2 = a does not have an integral solution (z, y) if a is an integer satisfying a E 3 mod 4. Suppose there exist

such integers x and y. Then we have x2 + y2 E 3 mod 4. On the other hand, we have 0’ = 0, l2 E 1, 22 E 0, and 32 E 1 mod 4, and thus x2 + y2 = 3 mod 4 cannot be satisfied no matter how we choose x and y.

The properties (2.2), (2.3) and (2.4) show that the relation “ = mod m” is an equivalence relation. Taking (2.5) into account, we obtain a ring Z/mZ by identifying integers a and b satisfying a E b mod m. We assume that the reader is familiar with this fact. For a E Z we write a mod pm to indicate the class of a in Z/p-Z. Often, we abuse the notation to simply write a instead of a mod pm. For

example, Z/6Z consists of six elements 0, 1, 2, 3, 4 and 5, and it is a ring by the operations such as 3 + 4 = 7 = 1 and 2 x 3 = 6 = 0.

The proof of the following proposition will be left to the reader.

PROPOSITION 2.1. Let m be a natural number.

(1) Z/mZ is a field if and only if m is a prime number.

2.2. CONGRUENCE 51

(2) Let p be a prime number. (In this case we often use the notation F, instead of iZ/pZ.) The group IF: consisting of the nonzero elements of F, is a cyclic group of order p - 1.

(3) Let a be an integer. The image of a in Z/mZ is an invertible element in Z?/mZ if and only if a is relatively prime to m.

(4) (Chinese Remainder Theorem) Let m = pyl . . . p:r be the prime factorization of m. ( We assume pl, . . . ,p, are distinct prime numbers.) Then there is a natural isomorphism

Z/miZ-+Z/p~‘Z x . . . x Z/pFrZ.

(The map from left to right is given by regarding an integer mod m as an integer modpz” for each i.) In other words, if an integer ai is given for each i = 1,. . . , r, there exists an integer b satisfying

b E ai mod p&’ (i= l,...,r)

(the surjectivity of the map from left to right); and if b’ is another integer satisfying the same equations, we have b E b’ mod m (the injectivity of the map).

(b) Quadratic reciprocity law. The field iF5 has a square root of -1. Indeed, since we have 22 = 4 z -1 mod 5, 2 is a square root of -1 in Fs. By contrast, we can verify that iF7 does not have a square root of -1. In fact, if p is an odd prime number, F, has a square root of -1 if and only if p z 1 mod 4. For which prime numbers does there exist a square root of 5 in IF,? How about a square root of 3? The answers to these questions are given by the quadratic reciprocity law proved by Gauss in 1796. We introduce first the quadratic residue symbols.

Let p be an odd prime and a an integer prime to p. The quadratic residue symbol (% ) E {f 1) is defined as follows. If there exists a

square root of a in IF, (i.e., there exists an integer x satisfying z2 E a mod p), define (E) = 1, and if there is no such x, define (E) = -1.

For example, since we have O2 E 0, l2 = 42 E 1, 22 s 32 E 4 mod 5, we see that

(k) = (f!) = 1, (g) = (X) = -1.

From Proposition 2.1(2), the quotient group H,X/(F,X)2 is isomorphic to the multiplicative group {&l} of order 2. The symbol (E) E {fl} is nothing but the image of the class of a under the iso-

morphism of groups F,X/(!F,X)2 2 {fl}. Hence, for any integers a and


b prime to p, we have

THEOREM 2.2. Let p be an odd prime number.

(1) (Quadratic reciprocity law) If q is an odd prime number different from p, we have

(2) (First supplementary

-1

(-1 P = (-p

law)

1 Zfp=l mod4,

-1 ifpz3 mod4.

(3) (Second supplementary law)

The proof using a cyclotomic field will be given in Chapter 5 in Volume 2.

The law (2) tells us the existence or nonexistence of a square root of -1 inIF,. As an example of (l), let us consider a prime number p different from 2 and 5. Then we see from

(F) = (-l,+y) = (g)

that a square root of 5 exists in IF, if and only if p E 1 or 4 mod 5 (we have already determined (T)). I f p is a prime number different from 2 and 3, a square root of 3 exists in F, if and only if p G 1 or 11 mod 12. We can see this from

(;) = (-l)w+ (i) = (-$+ (5)

and the facts (i) = I, (2) = -1.

QUESTION 3. Let p be a prime number different from 2 and 3. Show that a

square root of -3 exists in FP if and only if p = 1 mod 3.

QUESTION 4. Let m be an integer and p a prime number that does not

divide 2m. Show that the existence of a square root of m in lFP can be determined

only by p mod 41ml (i.e., if p’ is a prime number which does not divide 2m and

which satisfies p z p’ mod 41ml, then we have the equivalence “there exists a

square root of m in FP H there exists a square root of m in F+“).

2.3. CONICS AND QUADRATIC RESIDUE SYMBOLS 53

2.3. Conies and quadratic residue symbols

(a) Existence of a rational point on a conic. In this section we state Theorem 2.3, which gives a criterion for the existence of a rational point on the conic ax2 + by2 = c (a, b, c E Q”). The proof of this theorem will be given in 52.6. Note first that it suffices to consider the case c = 1 since we can divide both sides of the equation by c.

Let a,b E Q”. We will define (a, b), E {fl} for each prime number p and (a, b)= E {kl}. The symbol (a,b)v (ZJ is a prime or co) is called the Hilbert symbol. (a, b), will be defined later using the

quadratic residue symbol (p). We define

(a,bb = if a > 0 or b > 0,

if a < 0 and b < 0.

We see immediately that

there exist real numbers II: and y such that ax2 + by2 = 1.

If there exist rational numbers x and y satisfying ax2 + by2 = 1, that means there exist real numbers satisfying ax2 + by2 = 1. The symbol (a, b)co tells us if this is the case. Of course, that is not sufficient to determine the existence of a rational solution. Not only “the light of reals” ( , )(x1 but also “the light of a prime number” ( , ), for every p is necessary to determine whether or not there exists a rational solution. To be precise, we will prove the following theorem after we finish defining ( , )P.

THEOREM 2.3. Let a, b E Q”. There exist rational numbers x and y satisfying ax2 + by2 = 1 if and only if we have (a, b)m = 1 and (a, b), = 1 for all prime numbers p.

(b) Definition and properties of the Hilbert symbol. Be- fore stating the definition of the Hilbert symbol (a, b), for a prime number p, we need some preliminaries. For a prime number p we define a subring Zc,) of Q by

z (PI = { f / a, b E Z, b is not divisible by p}


For n 2 1, the natural homomorphism Z --+ Z/pnZ (obtained by considering an integer modulo p”) is extended to the ring homomorphism

a a mod pn

b H bmodp” (a, b E Z, b is not divisible by p).

Here we used the fact that b mod pn is invertible in Z/pniZ. This homomorphism can also be understood in the following way. The natural homomorphism Z/p% + Z(,) /pnZc,) is an isomorphism, and the above homomorphism is nothing but the composition

%) ---) zb)lp”zb) 5 Z/p”Z.

For an element 2 in Zc,), its image in Z/pnZ will be written 2 mod pn.

The set of all the units in iZ(,), denoted by (77,~~)) x, is the set { % 1 a, b E Z, a, b are not divisible by p }. Any nonzero rational number can be written uniquely as pmzl (m E Z, u E (Zc,)) “).

For a prime number p and a, b E Q” , we define the Hilbert symbol (a, b)P as follows. Write

a = piu, b =pjv (i,j E z, %V E (q,,Y),

and put

r = (-l)i3a3b-” = (-1)‘ju3Ci E (ZC~))~.

If p # 2, we define

(a,b), = (T),

where the right-hand side is the quadratic residue symbol. If p = 2, we define

(a,b)2 = (-I)+ . (-l)q.w,

Here, the exponents of -1 in the right-hand side are elements of Z(z), but we regard them as elements of iZ/2iZ via the homomorphism ZC2) + z/22.

PROPOSITION 2.4. Let v be a prime number or 0~7. For a, b E Q” we have the following.

(1) (a, bL = (ha),. (2) (a,bc), = (a, b),(a,c),. (3) (a, -a), = 1. Ifa # 1, then (a, 1 - a), = 1.


(4) If p is an odd prime number and a, b E (Zt,))x, then we have the following.

(4-l) (a, b), = 1,

(4-2) (a,pb), = (T)

(5) If a, b E Z?y2,, then

(5-l) (a,b)2 =

{

i, ifa-lmod4orbElmod4,

ifaEb=-1mod4.

(5-2) (a, 2b)z

={

1 ifa~1mod8ora~1-2bmod8,

- 1 otherwise.

~ The proof of this proposition follows easily from the definition of the Hilbert symbols, and we leave it to the reader.

(c) Product formula for Hilbert symbols. The following theorem is a translation of the quadratic reciprocity law and the supplementary laws using Hilbert symbols.

THEOREM 2.5. Let a, b E Q”. Then (a, b)V is equal to 1 except for a finite number of ‘vu, and we have

where v runs through all the prime numbers and oo.

REMARK 2.6. By this theorem we only have to check the condition (a, b)V = 1 for all but one v in order to use Theorem 2.3, which requires that we verify the condition (a, b)V = 1 for all ZI.

PROOF OF THEOREM 2.5. The fact that (a, b)V is equal to 1 except for a finite number of v follows from the fact that a, b E (FE(,))’ for all but a finite number of primes p and Proposition 2.4(4-l). In order to show that the product for all the V’S is 1, it suffices to show it in the following cases (i)-(iii), since we have to prove it only for

each prime factor of a and b and for -1 thanks to Proposition 2.4(l), (2) and (3).

(i) a and b are two distinct odd prime numbers. (ii) a is an odd prime number, and b = -1 or 2.

(iii) a = -1, and b = -1 or 2.

2. CONICS AND P-ADIC NUMBERS

In case (i),

b 0 - a a

(a, b)v = 0 b

ifv=a,

if v = b,

(q+k$ ifw=2,

1 for other U.

Thus, the fact n,(a, b)v = 1 is in this case nothing but the quadratic reciprocity law (Theorem 2.2(l)).

In case (ii), it follows from Proposition 2.4 that

i

-1 (-1 ifv=a,

(a, -l)v = (-r)+ if u = 2,

for other v;

2

0 - if ‘u = a,

(a, 2), = (Q1)+ ifv=2,

1 for other v.

Thus, the fact n, (a, b)v = 1 is in this case nothing but the supplementary laws (Theorem 2.2(2) and (3)).

As for case (iii), a calculation shows that

(-1, -l)v = ,’ if z1 is 2 or 03,

otherwise;

(-1,2), = 1 for all v.

q

REMARK 2.7. Once we translate the quadratic reciprocity law into the form of Theorem 2.5 (which was done first by Hilbert), we realize that the quadratic reciprocity law expresses the harmony of “the light of real numbers” and “the lights of prime numbers”.


(d) Examples. Let us determine the existence of a rational point for some explicit examples using Theorem 2.3.

As a preliminary, we note the following. If a, b, c E Q”, the following conditions are equivalent.

(a) There exist 2, y E Q satisfying ax2 + by2 = c. (b) There exist 5, y, z E Q, (x, y, z) # (O,O, 0), satisfying ax2 +

by2 = cz2.

(a) + (b) is trivial. It suffices to put z = 1. Conversely, suppose

ax2 + by2 = cz2, X,Y,Z E Q, (X,Y,Z) # (O,O,O).

IfzfO, we have a (z)” + b (z)’ = c. If z = 0, then x # 0, and we

have a = c ($)’ - b(z)‘. Using th e results of 31.1, we see that the conic a = cu2 - bv2 has infinitely many rational points, and thus it has

a rational point satisfying 21 # 0. Hence we have a (i) 2 + b ($) 2 = c.

PROPOSITION 2.8. Let p be a prime number.

(1) There exist x, y E Q satisfying p = x2 +y2 if and only if p E 1 mod4 orp=2.

(2) There exist x, y E Q satisfying p = x2 + 5y2 if and only if pal or9mod20, orp=5.

(3) There exist x, y E Q satisfying p = x2 + 26y2 if and only if pal or3mod8, andp=1,3,4,9,10 or12mod13.

PROOF. Let a E Q”. Rewriting pz” = x2 + ay2 as x2 = pz2 - ay2 and using the equivalence of (a) and (b) above, we see that the existence of x, y E Q satisfying p = x2+ay2 is equivalent to (p, -a), = 1 for all prime numbers v = p and co. By Remark 2.6, we do not have to check the case v = p.

Proof of (1). A s we have already calculated in the proof of The-

orem 2.5, we have (p, -l)V = 1 if v # 2,p, and (p, -1)~ = (-l)q if p # 2. Then (1) follows from these facts.

Proof of (2). By Proposition 2.4(4-l), (p, -5), = 1 if v # 2,5,p.

We also have (p, -5)~ = (-l)q if p # 2, and (p, -5)s = (E) if p # 5. Now (2) follows from these.

Proof of (3). By Proposition 2.4(4-l) we have (p, -26), = 1 if v # 2,13,p. Also, we have (p, -26)~ = 1 if p E 1 or 3 mod 8, and (p, -26)2 = -1 if p E 5 or 7mod 8. If p # 13, we have (p, -13)13 = (6). c 1 1 t a cu a ing the square of each element of Z/132,

we see that (fi) = 1 if a s 1,3,4,9,10,12 mod 13 and (&) = -1 if a E 2,5,6,7,8,11 mod 13. Now (3) follows from these. cl


In Proposition 2.8 we looked for rational solutions to a quadratic equation. How about integral solutions? As Fermat says (see Chap- ter 0, §0.2), there exist 5, y E Z satisfying p = x2 + y2 if and only if p E 1 mod 4 or p = 2. This is the same as the condition for the existence of a rational solution. For the equation p = x2 + 5y2, the conditions for the existence of a rational solution and that of an integral solution are the same. As for p = x2 + 26y2, there exists a rational solution to 3 = x2 + 26y2 by Proposition 2.8(3). (For ex-

ample, 3 = ($) 2 + 26 (i)“.) Clearly, however, there is no integral

solution to 3 = x2 + 26~~. The difference between the existence of a rational solution and an integral solution is related to the class field theory, and we will discuss it in Chapter 5, §5.3(b) in Volume 2.

QUESTION 5. In Arithmetzca Diophantus says that the equation 15x2-36 =

y2 does not have a rational solution. Verify this using Theorem 2.3.

2.4. padic number fields

The meaning of the Hilbert symbol ( , )oo is that for a, b E Q” we have

(a, b)co = 1 ++ th ere exist x, y E Iw satisfying ax2 + by2 = 1.

For each prime number p we can interpret (a, b)P in the same manner. Namely, for each p there is an extension field Qp of Q, and for a, b E

Q” we have

(a, b)P = 1 _ th ere exist x, y E U& satisfying ax2 + by2 = 1.

Qp is called the p-udic number field, and its elements are called p-adic numbers. In this section we introduce the p-adic number fields, which are very important objects in number theory.

The p-adic numbers were originally introduced by Hensel around 1900. In the long history of mathematics a number meant a real number, and it is only relatively recently that we realized that there is a world of p-adic numbers. It is as if those who had seen the sky only during the day are marvelling at the night sky. The mathematical scenery is completely different. Q, emits “the light of prime number p” in the night sky as if it were a star that we could not see because of the sun, or the real number field R, which emits “the light of real numbers” during the day. Just as there are countless stars in the night sky, there is one U&, for each p. What each star is to the sun is what each Q, is to W. Just as we can see space objects better at

2.4. p-ADIC NUMBER FIELDS 59

1 26

a00 51 .

31

@(zj

FIGURE 2.5. Classification by mod 5”

night, we began to see the profound mathematical universe through the p-adic numbers.

We introduce the p-adic number fields in three different ways in

(b), Cc) and (d). W e would like you to get acquainted with it according to your taste.

(a) padic sense of distance. The sense of distance in the world of U&, is completely different from that of R. In Q,, p is close to 0 and the sequence p2 p” p4 . . approaches 0 rapidly. We explain here this “feeling” of d&&e.’ The distance in Q, comes from the congruence modulo p in the following sense.

For example, classifying the integers into the classes module 5 is analogous to putting them in five different rooms, one for the numbers congruent to 0 modulo 5, one for the numbers congruent to 1 modulo 5, and so on. We feel that the integers that enter the same room are close. We then divide the members of each room into the classes modulo 25; the room for the numbers congruent to 1 modulo 5, for example, is divided into five smaller rooms, one for the numbers congruent to 1 modulo 25, one for the numbers congruent to 6 modulo 25, one for the numbers congruent to 11 modulo 25, and so on. The numbers 1, 6 and 51 are all in the same room modulo 5. While 6 and 1 enter different small rooms, 51 and 1 still share the same small room. We thus think that 6 is closer to 1 than 4 is to 1, but 51 is even closer to 1 (see Figure 2.5).


Pushing this analogy further, we feel two integers a, b are very close to each other when we have a = b mod pn for a large number R. We call this sense of distance the p-adic sense of distance. If we push this to the limit, the p-adic number field emerges.

At present we know two different senses of distance in numbers: the sense coming from the real line and the sense coming from congruence. Both of them are compatible with addition and multiplication. In the case of congruence, the compatibility is nothing but the property (2.5). Among the distances coming from congruences, we consider only the congruence mod pn (p prime) for the following reason.

Let m be a natural number and m = py’ . . p:r (pi, . , p, distinct) be its prime factorization. For integers a, b, the congruence

a s b mod m is equivalent to a E b mod p:’ for all i = 1,. . . , r. This is a consequence of the Chinese Remainder Theorem (Proposi- tion 2.1(4)). Th us, the sense of distance “mod m” is a composition of the distances “mod pn”, and the sense of distance mod pn is fun-

damental. Let p be a prime number. For a rational number a we define the

p-adic valuation ord,(a) in the following way. As in Definition 1.15, for a # 0 we write

a=p “2 (m E Z, U, ‘u are not divisible by p), u

and we define ord,(a) = m. In other words, ord,(a) indicates exactly which power of p divides a. We also set ord,(O) = co. We have the following:

(2.6) ord,(ab) = ord,(a) + ordp(b);

(2.7) ord,(a + b) > min(ord,(a), ord,(b));

(2.8) ord,(a) # ord,(b) implies ord,(a+b) = min(ord,(a), ord,(b)).

Here we used the conventions oo+cc = co, 03 2 00, cx+n = n+cc = co, and 03 2 n for any integer n.

We generalize the p-adic distance to the rational numbers, and we consider two rational numbers a and b to be “p-adically close” if ord,(a - b) is large.

We say that a sequence of rational numbers (z~)~~I converges to a rational number a p-adically if we have

ord,(z, - u) + cc as n ---f 00.

61 2.4. p-ADIC NUMBER FIELDS

For example, if we let

2 n = 1 - 5 + 52 - 53 +. . . + (-5)“,

the sequence (x~)~>I diverges in the ordinary sense in the world of real numbers, but we can show that it converges 5-adically to i. In general, for a rational number a # 1, we have

an+l

1+a+a”+...+al’-A=-- l-a’

Replacing a by -5, we have

1

X71

(-l)n5n+l

6 . 6

Thus, as n + 00, we have

ordg(x,,-i) =ordi((-l)i’“+‘) =n+l+m.

As this example shows, convergence in the p-adic sense is quite different from convergence in ordinary sense. If we express the fact that (xn)+l converges to i by

(2.9) 2(-B)” = f (5-adically), i=o

it is as if we mistakenly put x = -5 in the ordinary formula

2x’=& if-l<x<l z=o

in the world of real numbers, but formula (2.9) is correct in the sense of 5-adic convergence.

QUESTION 6. Let p be a prime number, c a rational number, and ordp(c) 2

1. Show that

g c’ = & (padically);

i.e., if we put zrL = CFzo cz, the sequence (z,),> 1 converges padically to &

Formula (2.9) can be interpreted in the following way. For each n 2 1, it says that 1 - 5 + 52 - ... + (-5)“-l is the inverse of 6 in Z/5nZ. For example, in Z/252, 1 - 5 = -4 is the inverse of 6, while in Z/1252, 1 - 5 + 5’ = 21 is the inverse of 6. Indeed, we have 6 x 21 = 126 EE 1 mod 125.


QUESTION 7. Using (2.9), explain why 1 - 5 + 5’ - ‘.. + (-5)n-1 is the inverse of 6 in Z/5nZ.

QUESTION 8. Find the inverse of 4 in Z/34jz.

The p-adic convergence defined above can be considered to be convergence in a metric space. For a rational number a we define the p-adic absolute value /alp by

if a # 0 and 101, = 0. Thus lalp is the size of a in the p-adic sense. For example,

Ii,.=;> IP21p=$.

The p-adic absolute value expresses well that the sequence p, p2, p3, . . . converges to 0. (All the arguments in this section work well if we

replace the definition of Ialp by lalp = T”‘~P(~) for any T satisfying 0 < T < 1. However, it turns out that T = b is the most natural

choice, as we will explain at the end of subsection (c).) From properties (2.6) and (2.7) of ord, we see that

(2.10) I4 = MP . I%

(2.11) la+ bl, 5 max(lal,, lblP) (In particular, (a+ bl, i Ialp + IblP).

I f we define the p-adic metric d,(a, b) by

&(a, b) = la - bl,,

then d, satisfies

(2.12) d,(a, b) 2 0, d,(a, 6) = 0 if and only if a = b;

(2.13) d,(a, b) = d,(b, a);

(2.14) &,(a, 4 5 &(a, b) + d,(b, 4.

Thus Q is a metric space with respect to d, (see Introduction to Geom- etry 2 in the series Introduction to Modern Mathematics). A sequence (x~)Q~ of rational numbers converges p-adically to a if and only if

&(Zn,a) --+ 0 (n -+ oo), but this is the same as saying that (~~)+l converges to a with respect to the p-adic metric d,.


(b) QP as a completion of Q. In the world of real numbers a sequence of rational numbers may converge to a number which is not a rational number, as the following example shows:

1.4,1.41,1.414,1.4142,. + fi @ Q.

The world of rational numbers is an incomplete world where sequences such as the one above may not have a limit even if it “should converge”. From this point of view lR is an extension of Q where all the sequences that “should converge” in the ordinary sense do converge. (We will lat,er define the meaning of “should converge” precisely.)

With respect to padic convergence, the world of rational numbers is also incomplete, where some sequences that “should converge” may not have a limit. Q, is an extension of Q constructed so that all the sequences that “should converge” do converge with respect to p-adic convergence. In this regard both Iw and Q, are extensions introduced with the same motivation. We first review the precise definition of R, and then we introduce the definition of Q,.

As we stated in the Introduction, 50.1, ancient Greek mathematicians agonized over the problem “What are the real numbers with respect to the rational numbers?” (“How should we define the real numbers precisely based on the rational numbers?“), and it is only in the nineteenth century that this problem was finally solved. Here, we introduce the definition of the real numbers as the limits of sequences that “should converge”. This definition is due to Cantor at the end of the nineteenth century.

A sequence of rational numbers (xn)+i that “should converge” is defined to be a sequence satisfying condition (C) below. Such a sequence is called a Cauchy sequence.

(C) For any given rational number E, we can choose a natural number N such that

m,n 2 N implies IX, - 2,1 < E.

In the world of rational numbers, a sequence that converges to a rational number (in the ordinary sense) is a Cauchy sequence, but there are Cauchy sequences such as 1.4,1.41,1.414,1.4142,. . . that do not converge to a rational number. In the world of real numbers, however, a sequence is a Cauchy sequence if and only if it converges. Cantor’s idea is to reverse the direction and define a real number to be “the Cauchy sequences that converge to that number”. To be precise, let S be the set of all the Cauchy sequences of rational numbers,


and define an equivalence relation on S by saying that (x,),21 and (~~)~>i are equivalent if “for any rational number E, we can choose a natural number N such that

n>N implies Ix, - ynl < 2’.

We define R to be the quotient of S by this equivalence relation. (That two sequences are equivalent means that they converge to the same real number.) We can define addition and multiplication in lR

by

class of (x~)~z~ + class of (yn)n>i = class of (2, + yn)+i,

class of (z,),>i . class of (yn)+r = class of (z,y/,),>i,

and we can prove that lR is a field with respect to these operations. We now define Q,. We call a sequence of rational numbers

(x~)~z~ a p-adic Cauchy sequence (a sequence that “should converge” with respect to p-adic convergence) if it satisfies the following condition (C,):

(C,) For any given rational number E, we can choose a natural number N accordingly such that

m,n>N implies 15, - Ic& < E.

Let S, be the set of all p-adic Cauchy sequences, and define an equivalence relation on S, by saying that (x~)~z~ and (yn)+i are equivalent if “for any rational number E, we can choose a natural number N such that

n>N implies 15, - ylnlp < E”.

We define Q, as the quotient of S, by this equivalence relation. As in the case of R, we can define addition and multiplication in Q,, and QP becomes a field with respect to these operations.

The method of obtaining lR or QP from Q is known in general as completion of a metric space. IR is the completion of the metric space Q under the ordinary metric, and QP is the completion of Q under the p-adic metric.

We identify a rational number a with the element of Q, given by “the sequence identically equal to a” (which is a p-adic Cauchy sequence). This identification gives us an embedding of Q in QP.

We extend the p-adic valuation ordp, p-adic absolute value I IP, and p-adic metric dp to Q,. For an element a in QP we define ord,(a) E zU{m} in the following way. If a = 0, we put ord,(a) = co. Suppose a # 0. If we choose a p-adic Cauchy sequence of rational


numbers (z,),>r whose class is a, we can prove that ord,(z,) is constant for sufficiently large n using (2.6)-(2.8) (readers should check this). We define ord,(a) to be this constant. We can prove that ord,(a) defined in this way depends only on a and not on the choice

of the p-adic Cauchy sequence (x~)~>I. For an element a in U& we define /alp = 0 if a = 0, and lalp =

pP or’1p(a) if a # 0. We define &(a, b) = (a - bj, for a, b E QP. Then ordp, I lP, and d, defined this way in Q, satisfy (2.6)-(2.8), (2.10)- (2.14) for all a, b E Q,. We regard QP as a metric space with respect to d,.

A sequence (z,),>i of Q, converges if and only if (z,),>i satisfies condition (C,). Q is dense in Qip (i.e., each element of U&, is the limit of a certain sequence in Q). Indeed, if (x,),21 is a sequence of rational numbers and a is an element in Q,, then (%,),>I converges to a if and only if (zr,),>i is a Cauchy sequence and its class is a.

In QP the condition for the convergence of an infinite series is somewhat easier than in R.

LEMMA 2.9. Let alL E Q, (n 2 1). The series Cr=, a, converges in Qp (i.e., if we put s, = Cz=“=, ai, the sequence (s,),>l converges) if and only if lulLIp tends to 0 in Iw as n tends to 03 (i.e., ordr,(a,) tends to cc as n tends to CQ).

In R, x:=1 i does not converge even though 1; / --t 0 when n -+ 03, and thus the situation is more complicated. The difference comes from the fact that we have Iz+y[, 5 max(lzl,, IyIP) in QP, but we do not have Ix + y// 5 max(lzl, lyl) in R.

PROOF OF LEMMA 2.9. As we have already seen, (s~).,~>~ converges if and only if (s,,),>i satisfies the condition (C,). The latter can be seen equivalent to the condition (u~~(~ + 0 using the properties (2.10) and (2.11). I7

(c) Qp as an inverse limit. Define

Z, = {a E QP / ord,(a) > O}.

Z, is a subring of QP. (This follows from the fact that ord, : Qp + ZU {oo} satisfies (2.6) and (2.7).) An element of Z, is called a p-adic integer.

In this subsection (c) we explain that we can think of Z, as an “inverse limit” and that we can introduce Q, in a different manner.


DEFINITION 2.10. If a sequence of sets X, (n = 1,2,3,. . . ) and maps fn : Xn+i -+ X, (n = 1,2,3,. . . )

are given, the subset of n,,, X, defined by -

{(G)QI E &Xn I .fn(an+l) = a, for all n L 1) -

is called the inverse limit and is denoted l&,X,.

In Definition 2.10 we let X, = Z/pnZ and fn the natural pro- jection from Z/p n+lZ to Z/p”& and we consider the inverse limit @,Z/p”Z of the sequence

. . . -+ z/p4z + z/p3z 3 z/p% * z/pz.

An element (un)+i of l@,Z/p”Z has the following meaning. When we divide the set of all integers and put them into p rooms

following their values modulo p, al E Z/pZ is in one of the rooms. us is an element of Z/p2Z satisfying fi (us) = al. When we divide

the room of al into p small rooms mod p2, us corresponds to one of them.

us is an element of Z/p”Z satisfying f2 (~3) = ~2. When we divide the room of u2 into p tiny rooms mod p3, u3 corresponds to one of them.

To give an element of l@,Z/p”Z is to choose one of the small rooms in a room, then one of the tiny rooms in the small room, and so on.

As a matter of fact, l@,Z/p”Z is isomorphic to Z,. First we give the map l@,Z/p”Z + Z,. Let (a,),21 E l&,Z/p”Z. For each n > 1 we choose an integer x, such that the image of x, in Z/p”Z is a,. Then all x,, belong to the room al, they belong to u2 if n > 2, they belong to us if n > 3, and so on. This makes us feei that “(xn)n>i converges to something”. Indeed, we have x, = x, mod pN (i.e, /x, - x,1 < $) if m, n 2 A;. Thus (x~)~Z~ is a p-adic

Cauchy sequence, and it converges in Q,. Since ord,(x,) 2 0 for all n, the limit belongs to Z,.

We thus obtain a map l@,Z/p”Z + Z, by sending the element (un)~21 E l@,Z/pnZ to the limit of the p-adic Cauchy sequence

(GLQI in z,.


LEMMA 2.11. The map

defined as above is a bijection.

We will prove this lemma later. We now explain the definition of Q, using the inverse limit. We

first define Z, as l@ JZ/p”Z. In Definition 2.10, if all the X, (n > 1) are rings and all the fn are homomorphisms of rings, we can define a structure of ring on l&rnXn; we define addition and multiplication

of (GJ~~I and b&l by (a, + bn)+l and (anbnJnkl, respectively. We can prove that Z, so defined is an integral domain. We define QP as the quotient field of Z,.

In this definition we obtain Z, by letting n tend to infinity in Z/P’~Z. This definition is based on the idea that looking at an integer modulo pn for various n, we finally arrive at the world of Q,.

Before giving a proof of Lemma 2.11, we prove the following lemma. In the statement, Qp is the one defined in (b) as the completion of Q, and Z, is the subset {u E Q 1 ord,(a) 2 0) in Qp defined in (c).

LEMMA 2.12. (1) Z, is both open and closed in U&,. (2) If m is an integer, then we have

pmZp = {u E Qp 1 ord,(a) > m} .

(3) Z(,) c 27,. In U& we have Q n Z, = ?A(,). (4) For all integers m 2 0

z/pmz : Z&p?& -% z,/pmz$.

(5) Z, is the closure of Z&J in Qp. It is also the closure of Z

in Qp.

The image of a E Z, in Z/p”Z 2 Z,/pmZ, is written a mod

PrnZ,

PROOF. The proofs of (l), (2), (3) and the first isomorphism of (4) are easy, and we leave them to the reader.

Let us prove the second isomorphism of (4). It follows from (2) and (3) that YE(,) np “z, = p"z(,). Hence, ~(,)/P”~(,) --) WP”% is injective. Take a E Z,. There exists 5 E Q satisfying ord,(a: - a) > m since Q is dense in Qp. Since x - a E pm&, m 2 0 and a E Z,, weseethatsEQnZ,=Zc,). Thus,wehavea=z+(a-z)E

68 2. CONICS AND P-ADIC NURlBEKS

Z(,) + p”Z,, which shows that the map Z~,)/p’“Z&) --f Z,/p”Z, sends z to a. Thus this map is surjective.

To show (5), it suffices to show that Z and Zc,) are dense in Z,. But, this follows from (2) and (4). 0

PROOF OF LEMMA 2.11. For an element a in Z,, let a, be the image of a under the map

Z, + Z,/p”Z, 2 Z/p”Z (Lemma 2.12(4)).

Thus we have a map

Z, + l&Z/p”“Z; a H (a,),>~.

It is easy to see that this map is the inverse of the map in Lemma 2.11.

We not,e in passing that the definition of the p-adic absolute value / lp is a “natural” one. In the real field R, the scaling factor of the homothety R + R; 2 H az is the absolute value \a\. In other words, if 1 is an interval of length 1, the length of the interval al = {uz / 5 E I} is (a(. 1. On the other hand, in Q, the scaling factor of the homothety Q$ -+ Qp given by n: H us is the p-adic absolute value Ialp. For example, pZ, is a subgroup of iz, of index p, and we should think of the size of pZ, as $ the size of Z,. This means that the homothety of

scaling factor p reduces the size of Z, by $. In this way, the definition

of lplp = i has a natural meaning as the scaling factor of a homothety.

We will discuss this scaling factor, or “module”, in Volume 2.

(d) Definition of QP by padic expansion. In this section we explain that Qp may be defined by

Q, = 2 GLPn { 1

mEZ,c,E{0,1,..., p-l} 1L=VJ. 1

For example, we define an element of Qs to be something like

2~;+3~1+4~5+2~5”+4~5”+1x5”+~~~.

If m E Z and c, E Z (n = m, m+ 1, m+2,. ), the series c,“=,, c,pn converges in Qp as defined in subsection (b) (Lemma 2.9)) and thus the sum is an element of Q,. Conversely, we can prove that any element of Qp can be expressed in the form c,“=, c,pTL (m E Z, c,, E (0, 1, . . . ,p - 1)) in a unique way. (We call it the p-adic expansion of an element of Qp.)

2.5. MULTIPI,ICATIVE STRUCTURE OF THE p-ADIC NUMBER FIELD 69

Take an integer m satisfying ord,(a) > m. Then pPrna is an element of Z,, and there exists an integer c, E (0, 1, . ,p - l}

such that its image in Z,/pZ, coincides with that of p-nLa, since

the map Z/pZ 5 Z,/pZ, is an isomorphism. Since pema - c,,

belongs to pZ,, we have ord,(a -pmc,,) 2 m+ 1. The same argument shows that there exists an integer c,+r E (0, 1, . . . ,p - l} such that ord,(a - pmc,,, - pm+l~m+l) 2 m + 2. Repeating this process, we obtain the expansion

Examining the argument given above carefully, we see that the p-adic expansion is unique since each c, is uniquely determined.

REMARK 2.13. Let S be a subset of Z, such that the composition 5’ + Z, + i&/pZ, is a bijection. (The set (0, 1, ... ,p - 1) is an example of such a subset.) Then the same argument shows that any element of Q, can be written

2 cnpn (m E Z, c, E S) 71=?7l

in a unique manner.

QUESTION 9. A real number has a decimal expansion as we use it in every-

day life. Instead of 10, we can choose any natural number N > 2, and we can have an N-ary expansion of a real number. In particular, we can choose a prime number p. What is the difference between the pary expansion of a real number

in this sense and the padic expansion of a padic number?

2.5. Multiplicative structure of the p-adic number field

The real number field Iw has exponential and logarithmic functions, and they give an isomorphism between the additive group Iw and the multiplicative group formed by the positive real numbers

additive group Iw cz multiplicative group {t E R 1 t > 0},

x H ex, log(t) t--l t.

(Here, e is the base of the natural logarithm log.) Is there anything similar in Q,? In this section we introduce the exponential and logarithmic functions in Qp, and we determine the structure of the multiplicative group Q,” of nonzero p-adic numbers using these functions

(Propositions 2.16 and 2.17). An element a in Iwx is a square in IF?.’

70 2. CONIC3 AND P-ADIC NUMBERS

if and only if a > 0. Which elements in Q$ are squares? Propo-

sition 2.18 gives an answer to this question. For example, in Q)5” numbers such as 6 and 11, which are 5-adically close to the square 1, are squares, and -1, which is close to the square 4, is also a square. Just as in Rx, elements close to squares are also squares. (In this sense, the algebraic structure of lR or Qp is simpler than that of Q.)

(a) Exponential and logarithmic functions in Q,. In R or @ we have

(also written exp(z))

(where the right-hand side always converges), and when It - 11 < 1,

log(t) = c ‘-Y-l (t - 1)“. n=l

We consider an analog in Q,.

PROPOSITION 2.14. (1) Let IC E Q,. The series

2 5 (written exp(z)) n=O

converges if and only if x E pZ, in the case p # 2, and it converges if and only if x E 422 in the case p = 2. (That means the exponential function in Q, does not converge on all of Q,, as compared to the case o.f Iw or C.)

2 (-1Y-1 n (t - 1)” (written log(t)) n=l

converges if and only if t - 1 E pZ,. (3) If x1 and 22 are in the domain of convergence of exp(z), and

if tl and t2 are in the domain of convergence of log(t), then we have

exp(xl + x2) = exp(xl) ew(x2), log(tlt2) = log(tl) + log(t2).

(4) Weletm>lifp#2,andm>2ifp=2. Thenexpand log are isomorphisms, and they are inverse to each other:

additive group pmZ,

% multiplicative group 1 + p”Z, = { 1 + pma 1 a E Z,}.

2.5. MULTIPLICATIVE STRUCTURE OF THE p-ADIC NUMBER FIELD 71

In order to prove Proposition 2.14 we need to show the following lemma first.

LEMMA 2.15. (1) For any integer n > 0, we have

ord,(n!) = e F , is1 [ 1

where [x] is the “Gauss symbol” of z, which signifies the largest integer less than or equal to x.

(2) Let c be a real number. The condition nc - ord,(n!) --f co as n ----f co is equivalent to c > &. The condition nc -

ord,(n) --f cc as n -+ m is equivalent to c > 0.

(3) If c > &, then for any n > 1 we have

nc - ord,(n!) 2 c.

PROOF. We leave the proof of (1) to the reader. Let us prove (2). It follows from (1) that

nc - ord,(n!) 2 nc - 2 s 2 n

nc- -. i=l

P-l

The right-hand side tends to m as n -+ 00 if c > &. Also, if we put

n=pm, it follows from (1) that

1 nc - ord,(n!) = pmc - epm-z = pm (c - ~

>

1 + ~.

i=l P-l P-l

The right-hand side tends to 00 if and only if c > &.

If log,(n) is the logarithm of n with base p in the real number field, we have

nc - ord,(n) > nc - log,(n)

since ord,(n) 5 logp(n). The right-hand side tends to cc if c > 0. Letting n = pm, we have

nc - ord,(n) = pmc - m.

The right-hand side tends to 03 if and only if c > 0. Let us prove (3). Since ord,(n!) < C,“=, 5 = & and an integer

smaller than * is no greater than 2, we have ord,(n!) 5 3. Hence,

nc - ord,(n!) - c 2 (n - l,(c- &) 20. 0


PROOF OF PROPOSITION 2.14. In order to show (1) and (2), it suffices by Lemma 2.9 to find the conditions for the convergence of the following:

= nerd,(x) - ord,(n!),

(-1)-1 k$t) = nord,(t - 1) - ord,(n).

But they are given by Lemma 2.15(2). (Note that & < 1 if p # 2,

and 5 = 1 if p = 2.) The proof of (3) is similar to the case of IR or c.

Next we show (4). By Lemma 2.15(3), if x E pm&,, we have exp(x) E l+p’“Z,, since ord, (5) > m for n > 1; and if t E l+p”Z$,,

we have log(t) E pm&,, since ord, (

(-l)+‘v >

> m for 72 2 1. _

We can prove log(exp(x)) = x, t = exp(log(t)) for x and t in these domains of convergence in just the same way as the case of Iw or @. q

(b) Structure of Q,“.

PROPOSITION 2.16. (1) 1fp # 2, Q,” E Z @ Z/(p - 1)Z CE Z,.

(2) Ifp=2, Q,” “Z@Z/2Z@Z~.

This proposition follows from the following proposition and the fact F; g Z/(p - 1)Z (Proposition 2.1).

PROPOSITION 2.17. (1) Any element of Q$ can be written pnu, (n E Z, u E Z,“) in a unique manner. In other words,

zez; ‘Q;; (n,u) t-+pnu,

where ZF is the multiplicative group consisting of all the units in Z,.

(2) Let G = {x E Zp” ( xp-r = l}, and let Zp” -+ Fp” be the group homomorphism induced by the map Z, + Z,/pZ, = IF,. Then the composition G + Zc + IF; is a bijection, and

Zc is the direct product of G and 1 + pZ,.

(3) UP # 2, th e multiplicative group 1 +pZ, is isomorphic to Z,. I f p = 2, the multiplicative group 1-t 222 is the direct product of the subgroup {fl} and the subgroup 1 + 422. Moreover, we have 1+4& EC&.

PROOF. First, from the fact that Z?i = Ker(ord, : Q,” -+ Z) and ord,(p) = 1, (1) f 11 o ows easily. If p # 2, then (3) follows from the

2 5. MULI’IPLICATIVE STRUCTURE OF THE p-ADIC NUMBER FIELD 73

fact that 1 + pZ, E pZ, via exp and log, and the fact that the map Z, + pZ, given by a H pa is an isomorphism. If p = 2, then (3) follows from the fact that 1 + 4& ” 422 via exp and log, and the fact that iz2 ” 422.

Let us prove (2). Since the kernel of izz -+ IF: is 1 + pZ,, it

suffices to show that the composition map G -+ IF: is a bijection. For injectivity, it suffices to show that G n (1 + pZ,) = { 1). This is trivial if p = 2, since G = (1). I f p # 2, it follows from the fact that 1 + pZ, E Z, does not have any element of finite order except for the identity. We now prove that G + lFi is surjective. This is

trivial if p = 2, since IF,X = (1). I f p # 2, let a E “p” and let u E Z,

be an element whose image in IF, is a. Since up-’ = 1, we have up--l E 1 +pz,. Put

u = exp (

5 log(uP-l) >

and w = uu-I. Then we have w E G, because we have up-’ = exp(log(uP-I)) = up-l. Since v E 1 +pZ,, the image of w in ‘Fc is equal to a. cl

(c) Squares in Qp.

PROPOSITION 2.18. If we express an element a in Q: aspnu (n E

Z, u E Zc ) (Proposition 8.17( I)), a is a square in 0,” if and only if the following two conditions are satisfied.

(i) 72 is even.

(ii) I f p # 2, u mod p;Z, is a square in ‘Ft. I f p = 2, u E 1 mod 822.

PROOF. By Proposition 2.17, a is a square in Q,” if and only if

n is even and u is a square in Z:. I f p # 2, we have

1 + PZ, = exp(pZ,) = w$W,) = {exdp~p))2,

and thus an element of 1 + pZ, is a square in Et. Since we have

Zc/(l +pZ,) 2 “c, the case p # 2 is proved. If p = 2, we have

1 + 822 = exp(8Z2) = exp(2.42~) = {exp(4Z2)}2,

and thus an element of 1 + 822 is a square in Zc. Since we have Zc /(l + 8Z2) E (Z/8Z)x E Z/2Z C$ Z/22, the case p = 2 is also proved. 0


The following proposition follows from either Proposition 2.16 or Proposition 2.18.

PROPOSITION 2.19. (1) Ifp # 2, Q,“/(Q,“)2 S! Z/2Z CB Z/2Z.

(2) Q2” /(Q,“)” E z/az @? z/22 63 z/22.

QUESTION 10. Let a be an integer satisfying a = zkl mod 5. Show that there exists a square root of a in Qs

QUESTION 11. Show that there exists a square root of -1 in U& if and only ifp=l mod4.

QUESTION 12. Show that if p # 2, there exist exactly three quadratic ex-

tensions of U& Determine all three quadratic extensions of Q5.

2.6. Rational points on tonics

We begin this section by proving the statement at the beginning of 52.4:

If a, b E Q” and p is a prime number,

(a, b)p = 1 _ there exist II:, y E Qp such that ax2 + by2 = 1.

(This is contained in Proposition 2.20.) We then use it to prove Theorem 2.3.

(a) Conies defined over Qp. The Hilbert symbol ( , )p : Q” x

Q” -+ (51) can b e extended naturally to Qt x Q: + {fl}. Indeed,

for a, b E 0; we write

a = p’u, b =p-h (i,j E z, u,u E Z,x),

and we put

T = (-1)qgb-i = (-1)i&321-2 E q

If p # 2, define

(a,b), = (yq,

and if p = 2, define

(a,b)2 = (-I)+ (-l)++.

For this symbol ( , )p : Q: x QG + { z!~l}, Proposition 2.4 holds if

we replace (Zc,) ) ’ by Zc.

2.6. RATIONAL POINTS ON CONICS 75

PROPOSITION 2.20. For a, b E Q,“, the following two conditions are equivalent.

(i) (a, b)P = 1. (ii) There exist x, y E QP such that ax2 + by2 = 1.

PROOF. First we suppose that there exist z,y E Q, satisfying ax2 + by2 = 1, and we show (a, b), = 1. If z = 0, then b E (Q,X)“,

and if y = 0, then a E (Q,“)2. In both cases we have (a, b)P = 1. Suppose II: # 0, y # 0. Then (a, b)P = (ax2, by2)p = (ax’, 1 - UZ~)~, and we have (ax2, 1 - ax2 ), = 1, since Proposition 2.4(3) still holds for ( , ), : Q$ x Q,” 4 (51).

Next we suppose (a, b)P = 1 and show the existence of 5, y E Qp satisfying ax2 + by2 = 1. Conditions (i) and (ii) depend only on the image of a, b E 0,” in Q,“/(Q):)“. Thus, we may assume, by multiplying a and b by a suitable element in (Q,X)“, that a and b are both elements of Zc U pZc. If both a and b are in pZF, we may

replace a by -ab-‘; indeed, for (i) we have

(-abK1, b)P = (a, b)P . (-b, b)P = (a, b)P (Proposition 2.4(3)),

and for (ii) we have

32, y E Q, such that -ab-‘x2 + by2 = 1

M 3 z, y, z E Qp such that -ab-‘x2 + by2 = z2

and (x,Y,z) # (0,&O)

w 3 x, y, z E Q, such that (by)2 = ax2 + bz2

and (2, Y, z) # (0,&O)

u 3x, y E U& such that ax2 + by2 = 1.

Hence, it suffices to consider the case a E Zt, b E p. Z$ and the case a,bEZ,X.

(a) ThecaseaEZ,X, bEp.Z,X.

If P # 2, (a,b), = 1 means that a mod p E “c is a square. By Proposition 2.18, there exists t E Q,” such that t2 = a, and we have

a (i)’ + b. O2 = 1. If p = 2, (a, b)P = 1 means that “a E 1 mod 822 orazl-bmod822”. (This is because Proposition 2.4(5-2) holds for the Hilbert symbol extended to Q$ x Qc .) If a E 1 mod 8&, there exists t E Qc such that t2 = a (Proposition 2.18), and we have

a (i) 2 + b . O2 = 1. If a E 1 - b mod 822, there exists t E Q,” such

that t2 = e (Proposition 2.18), and we have at2 + b. l2 = 1.


(b) The case a, b E Zp”. Suppose p # 2. Then the condition (a,b), = 1 always holds,

and thus we must show that ax2 + by2 = 1 has a solution in U&.

We denote by a,& the images of a, b in IF,. Each of the two subsets

{au2 / u E IF,} and {I - bv2 ( v E IF,) has cardinality q, and thus t,heir intersection is nonempty. This implies that there exist 2, y E Z, such that ax2 z 1 - by2 mod pZ,. I f x $ 0 mod pZ,, there

exists t E Q,” such that t 2 - e by Proposition 2.18, and we have -

at2 + by2 = 1. If x = 0 mod pZ,, then 1 = by2 mod pZ,. Hence, there exists t E Q,” such that t2 = b by Proposition 2.18, and we have

a. O2 + b ($)” = 1. Now suppose p = 2. Since (a, b)2 = 1, we have a z 1 mod 422

or b = 1 mod 422. Suppose, say, a = 1 mod 422 (the case b G 1 mod 422 is similar). Then we have a = 1 mod 822 or a E 5 mod 822. If a = 1 mod 822, there exists t E Q,” such that t2 = a by Proposi-

tion 2.18, and we have a (f)” + b. O2 = 1. If a = 5 mod 822, then 4b G 4 mod 822 and thus we have a F 1 - 46 mod 822. Hence, there exists t E Qg such that t2 = e by Proposition 2.18, and we have at2+b+22=1. 0

(b) Proof of Theorem 2.3. By Proposition 2.20, Theorem 2.3 can be rewritten in the following form. Here, we write Qoc for R.

“Let a, b E Q”. The following conditions (i) and (ii) are equivalent.

(i) ux2 + by” = 1 has a solution in Q. (ii) ax2 + by 2 = 1 has a solution QV in for all primes u and

21 = m.”

Clearly, (i) implies (ii). So, all we need to prove is that (ii) implies (i).

Let u,b E QX, and suppose ax2 + by2 = 1 has a solution in Qv for all primes u and u = co. We need to prove that it has a solution in Q.

If we multiply a and b by the square of a rational number, it does not affect the existence of a solution in Q to ax2 + by2 = 1. Thus, we may assume that a and b are square-free integers. We prove the statement by induction on max(lal, lb\).

I f either a or b is 1, ux2 + by” = 1 clearly has a solution in Q.

2.6. EtA’I’IONAL POINTS ON CONICS 77

I f max(luj, lbl) = 1, we have a > 0 or b > 0, since we assumed that the equation has a solution in Iw. This means we have a = 1 or b = 1, and it has a solution in Q.

Suppose max([al, lb]) > 1. The statement is symmetric with respect to a, b, so we may assume Ial < lb/. Since b is square free, lb1 is a product of distinct prime numbers.

Let us prove that a mod b is a square in Z/bZ. If not, a mod p is not a square in F, for some prime factor p of b. (This follows from the Chinese Remainder Theorem.) Then p # 2, and we have

(u,b), = (;) = - 1. This implies that uz2 + by2 = 1 does not have a

solution in Qp, which is a contradiction. Hence, a mod b is a square in Z/biZ. We thus have an integer r such that r2 E a mod b. Since

any element of Z/bZ has a representative in --y 5 n 5 T, we may

assumeO<r< @J. Put - - 2

r‘J - a = bc, c E z.

I f c = 0, we have a = r2 and a (b)” + b. O2 = 1, which means that there is a solution in Q. Suppose c # 0. We have

(The last inequality is due to the fact that Ibl > 2.) By Lemma 2.21 below, all we need to consider is the case ax* + cy* = 1. If Ial < lbl, we can use the inductive hypothesis (since ICI < lbl). I f Ial = lbl, we can reduce to the case Ial < lbl, since /c( < Ibl. 0

LEMMA 2.21. Let K be a field; a, b, c E KX; r E K; and r2 -a = bc. Then there is a bijection between two sets

X = {(x, y, z) E K x K x K I ax2 + by* = z2, (2, y, z) # (O,O, 0)},

Y = {(x, y, z) E K x K x K I ax* + cy” = z2, (x, y, z) # (O,O, 0)).

PROOF. Define f : X -+ Y, g : Y -+ X by

f(x, y, z) = (rx + z, by, ax + rz),

dx, Y, ~1 = (rx - z, CY, --ax + rz),

and verify that g o f and f o g are the identities of X and Y, respectively. 0


Summary

2.1. If a conic defined over the rational number field has a rational point, it has infinitely many rational points, and we can describe them explicitly. (However, the main theme of the chapter is not this, but 2.2 and 2.3 below.)

2.2. For each prime number p there is an extension field of the rational number field called the p-adic number field. Each p-adic number field is considered to be as important as the real number field. The p-adic number field has a notion of convergence as does the real number field, but the properties of convergence are quite different from those in the real number field.

2.3. A conic defined over the rational number field has a rational point if and only if its equation has a solution in the real number field and in the Q, for all prime numbers p. The existence of a solution in Q, can be determined by the Hilbert symbol, which is related to the quadratic residue symbol.

Exercises

2.1. Find an example of a sequence of rational numbers which converges to 1 in Iw and which converges to 0 in Qz. Also find an example of a sequence of rational numbers which converges to 1 in Q3 and which converges to 0 in Qz.

2.2. Define

and define a ring structure on the set of all group homomorphisms from Z [l/p] /Z to itself, denoted by Horn (Z [l/p] /Z, Z [l/p] /Z), by defining the sum of f and g by (f + g)(z) = f(z) + g(z) for all z E ;Z [l/p] /Z, and the product off and g by the composition fog. Show that there is an isomorphism of rings

Z, “Horn (Z[i]/Z, z[~]/z) .

2.3. Find ords(4n - 1) (n E Z). (Hint: Use exp, log in the 3-adic number field to get 4” - 1 = exp(nlog(4)) - 1, then use Proposi- tion 2.14(4).

EXERCISES

2.4. Let p be a prime number. Show the following: (1) x2 = -2 has a solution in QP w p E 1,3 mod 8. (2) x2 +y2 = -2 has a solution in QP u p # 2. (3) x2 + y2 + 22 = -2 has a solution in QP for any p.

79

CHAPTER. 3

In this chapter we introduce an important !illlction called < (the

zeta function).

3.1. Three wonders of the values of the < function

The formula

(3.1)

was discovered by Euler around 1735. 1Ie had attempted to determine the infinite sum of the left-hand side for many years, and he was quit,e pleased to find the mysterious fact that, the sum is related t,o t,he number 7r.

The formula

is called Leibniz’s formula. He discovered it in 1673, and he felt that he found t,he mystery of Nature. It is said t,hat he decided t,o quit being a lawyer and diplomat, in order to pursue mat,hematics because of this discovery. Leibniz’s formula, however, had been discovered by Gregory shortly before Leibniz. and also by an Indian mat,hematician, Madhava, around 1400.

These formulas t,ogether wit,h Euler’s formula

(3.4) 7r3

1-$+&$+&&+... =E’

(3.5) 1-;++-;+ . ..= “iT 3&’

Xl

82 3. c

and Dirichlet’s formula

(3.6) l-~-;+~+L~-iL+~

+... (k signs repeat every 8 terms)

= 5 log(1 + v5)

are the formulas on the values of a class of functions called < functions. These formulas reveal their secrets as we study them more and more. In this section we introduce < functions and three interesting properties on the values of < functions. Define

This function c(s) is called the Riemann C function, named after Rie- mann who made important contributions to the study of this function in t,he 19th century. The formulas (3.1) and (3.3) may be expressed as

c(2) = $ and C(4) = $,

respectively, and thus they may be regarded as formulas for the values of the Riemann < function C(s). Let N be a natural number and (Z/NZ)x be the multiplicative group of units in the ring Z/NZ. A homomorphism from (Z/NZ) ’ to the multiplicative group of nonzero complex numbers Cx

x : (Z/NZ)X + Cx

is called a Dirichlet character (modulo N). We define

L(s, x) = 2 9 n=l

This is called the Dirichlet L function (with respect to x). Here, x(n) is defined as x(n mod N) if n and N are relatively prime, and 0 otherwise. The formulas (3.2) and (3.4) may be expressed respectively using the Dirichlet L functions as

L(l, xc) = : and L(3, x) = g,

I 3.1. THREE WONDERS OF THE VALUES OF THE C FUNCTION 83

where the character x is given by

~:(2/42)~ ={1mod4,3mod4}--+@X,

x(1 mod 4) = 1, x(3 mod 4) = -1.

The formula (3.5) may be regarded as a formula for the value of Dirichlet L function L(s, x)

L(l,x) = -!I- 3&i>

with the Dirichlet character x given by

x: (Z/SZ)’ ={1mod3,2mod3}+(GX,

x(1 mod 3) = 1, x(2 mod 3) = -1.

The formula (3.6) may be regarded as a formula for the value of Dirichlet L function L(s, x)

L(l,x) = +2 log(1 + J2)

with the Dirichlet character x given by

x: (Z/8Z)x = (1 mod8,3mod8,5mod8,7mod8}+~X,

x(1 mod 8) = x(7 mod 8) = 1,

x(3 mod 8) = x(5 mod 8) = -1.

These c(s) and L(s, x) are examples of the class of functions called < functions. c functions are so important in number theory that some people even claim that number theory is the study of < functions.

The first mystery of the values of C functions is that there exist unexpected formulas such as (3.1)-(3.6), where one side of the identities is quite different in nature from the other side. Many formulas of the following type have been known:

the value of a C function at s = integer

= (rational number)

x (the power of 7r or something similar to log(1 + A)).

For example, if T is a positive even integer, Euler proved the formula

c(r) = (rational number) x 7rr ($3.2, Corollary 3.9).

The second mystery of < functions is that their values at s = integers are related to the world of p-adic numbers in a quite unexpected way. For example, if T is a positive even integer, <(r)rPT is a rational number as mentioned above, and this rational number has

x.2 3. c

some p-adic properties. It was first studied by Kummer in the nineteenth century, and Kubota and Leopoldt clarified it around 1964. It seems as if the homeland where < functions originally come from is an unknown world which governs both the world of real numbers and the world of p-adic numbers.

The third mystery of < functions is that some values of < functions have subtle arithmetic meanings. For example, Leibniz’s formula (3.2) tells us that Z[i] is a principal ideal domain, as we will see in 54.3. This can be explained by the class number formula (see $4.3, and Chapter 7 on < function in Volume 2), which was discovered by Dirichlet in the nineteenth century. In the late twentieth century, an effort to understand the meanings of values of < functions more deeply than the class number formula turned into a theory called “Iwasawa theory”.

In $3.2 we discuss the first mystery about the values of c(s) and L(s,x) at s = positive integers, and we prove (3.1)-(3.5). (For the proof of (3.6), see Exercise 3.3.) In $3.3 we introduce the analytic continuation of these < functions to the entire complex plane, and we discuss the first mystery of the values of < functions at s = negative integers. We mention the second and third mysteries at the end of $3.3. We will further discuss these two mysteries in Chapter 10 in Volume 3.

We named this chapter “<” instead of “I functions”. We dropped the word “functions” because we feel more and more as we study < functions that C functions are something more than just functions.

3.2. Values at positive integers

(a) C(2). We first g’ ive one of the proofs Euler gave to the following theorem of Euler.

THEOREM 3.1.

C(2) = g .

PROOF. We use the product formula for the sine function

(3.7) sin(7rz) -=fi(l-$),

TX n=l

which was also discovered by Euler (see N. Bourbaki, Fonctions dine variable re’elle, Chapter VI $2, Theorem 2, or L. Ahlfors, Complex Analysis, Chapter 5, $2.3). We compare the Taylor expansion of

3.2. VALUES AT POSITIVE INTEGERS 85

both sides of (3.7) t a z = 0. By the Taylor expansion of sin(z),

3 - 27 X9

sin(x) = z - 5 + $ - T + sr - ... )

we have

T2 left-hand side of (3.7) = 1 - 31x2 + terms of degree 4 or higher.

On the other hand, the Taylor expansion of the right-hand side of (3.7) gives

right-hand side of (3.7)

Therefore, we have

&1

x2 + terms of degree 4 or higher

(b) Values at a general positive integer. Theorems 3.4 and 3.8 concern the values of c(s) and L(s, x) at positive integers.

DEFINITION 3.2. Define rational functions h,(t) (r = 1,2,3,. . . ) with rational coefficients by

hi(t) = JLiL 2(1 - t) ’

r-1

(hl (t)) (r >

For example

1).

(3.8) b(t) = (1 ” t)2 > b(t) = t + t”

b(t) = t + 49 + t”

(1 _ tj3> (1 - ty .

For any integer T greater than or equal to 1 we have

h7.(t) E Q t, L [ 1 1-t .

86 3. c

PROPOSITION 3.3. Let x E @, x $! Z and t = e2Tix.

(1)

w=-f&.(-&+-J-). nEz

(2) If r 2 2, then

h,(t) = (r - l)!. -& ( > T-c nEZ (x :c PROOF. Take $& log( ) of both sides of (3.7), and we have

(3.9) cot(7rx) = &c nEZ

-& + -J- . x-n >

Since cot(y) = a and we have

sin(y) = eYi - e-Yi

2i ’

we see that

cot(7rx) =

i(eXSi + e-~Zi)

eTxi _ e-m2

eYi + e-Y2

COS(Y) = 2 7

= -2ihl(t) (t = e2rzs),

which proves Proposition 3.3(l). Applying (t$)r-1 = (&)r-l to both sides of the above formula, we obtain Proposition 3.3(2). 0

Prom Proposition 3.3 we deduce Theorems 3.4 and 3.8.

THEOREM 3.4. Let N be a natural number greater than 1, x a Dirichlet character modulo N, and r a natural number. Suppose x(-l) = (-1)‘. I f we put CN = e2xi/N, then we have

Prom Theorem 3.4 we deduce the formulas (3.2), (3.4), and (3.5) in 53.1.

3.2. VALUES AT POSITIVE INTEGERS 87

EXAMPLE 3.5.

l-;+;+-L+...

27G =&. -4 ( > . ; . (hi(i) - h&.3))

+$().;.(3&2L) 2?ri 1 lr

zz 0 4 ‘?=4.

EXAMPLE 3.6.

1-;+;+-i+...

27ri 1 i (--I.-- n- = 3 2 A=- 3&’

EXAMPLE 3.7.

2Ti 3 1 =& -4 ( > . 5. (h&) - h3(i3))

=~.(-~)“+.(+-$) (from(3.8))

1 27ri 3 1 =-. 2 (-3 4

z. (-q = g.

THEOREM 3.8. Let r be a positive even integer. We have

c(r) = (r T l)! . 2’ \ 1 (a~ri)’ . ; . b-1).

COROLLARY 3.9. If r is a positive even integer, then rY<(r) is a rational number.

This is because h,(t) is a rational function with rational coefficients, and thus its value h,( -1) at -1 is a rational number.

We deduce the formulas (3.1) and (3.3) from Theorem 3.8.

72 6

EXAMPLE 3.11

C(4) = &y

irJ

90 .

REMARK 3.12

ii, (&4J. ; . -l’,.? - l (from (3.8))

C(5). C(7), 1‘(g), .:

Theorem 3.8 does not say anything about c(3), . Apkry proved that C(3) is an irrational number

in 1978. It is cori,jectured that c(5), c(7), c(9), are also irrational numbers, and if r is an odd integer at least 3, C(r) cannot be expressed as the product or sum of rational numbers and 7r unlike the case where T is even. But these conjectures have not been proved.

(from (3.8))

PROOF OF THEOREM 3.4. Using Proposition 3.3, we rewrite the sum C y(~)h,.((~~.) in th e right-hand side. If n 2 0, we have

at(Z/R;Z) x

If n < 0, we put n,’ = -n - 1 2 0 and we have

N’ c x(m) zzz m’.’

By Proposition 3.3 we obtain

PROOF OF THEOREM 3.8. In Theorem 3.4 take N = 2 and let x be the trivial homomorphism y : (Z/2Z)” --f Cx. Then. for a

3.3. V.L\LUES AT NEGATIVE INTEGERS

positive even integer r we have

27ri T 1 L(r,x) = (1: -2

( 1 . 2 h,.(-1).

I On the other hand, we have

Theorem 3.8 follows from this. 0

QUESTION 1. By letting z = i in Proposition 3.3, show the formula

2x + 1 ,27T - 1

QUESTION 2. Using Proposition 3.3(2) with r = 2, I = i and the formula in the previous question, show the formula

$e4T •t 2n2e2” - ;

(e 2n - 1)2

These formulas are not about thezvalues of ( functions, but they

belong to the same world as c(2) = $, and it has a flavor of <.

3.3. Values at negative integers

(a) Analytic continuation. I f we consider s to be a complex variable, C(s) and L(s,x) can be extended beyond the domain of convergence of the original infinite series, and we can consider their values at negative integers as we see in Proposition 3.15. In order to study the properties of the values at negative integers, it is convenient to introduce the partial Riemann < function and Hurwitz < function.

DEFINITION 3.13. For a natural number N and an integer a define

where t,he sum is taken over all natural numbers 72 satisfying n E a mod N. This function is called partial Riemann < function with respect to a mod N.

90 3. c


&(4)(S) = 1 + ,,: + ; + & + $ + ‘. .

DEFINITION 3.14. For a positive real number z define

C(s,x) =go (,:,,i

This is called the Hurwitz zeta function

We note that the notation &N)(S) is our own and it is not generally used.

From the definitions we have the following properties.

(3.10) La(l)W = I(s), as> 1) = C(s).

For x a Dirichlet character mod N

N (3.11)

a=1

(define x(a) = 0 for a that is not prime to N). For a natural number N and an integer a satisfying 1 <- a 5 N

(3.12) < (s, ;> = N” . L(N)(S).

PROPOSITION 3.15. (1) The defining series of C(s), L(s,x) (x is a Dirichlet character), C&~)(S) (N is a natural number and a is an integer) and C(s, x) (x is a positive real number) all converge absolutely for s satisfying Re(s) > 1, and they are holomorphic in this domain.

(2) The functions c(s), L(s, x), &(N)(S) and <(s, CC) have analytic continuation to the entire complex plane, and they are meromorphic functions. They are holomorphic in s # 1, and we have

liil (s - l)C(s) = 1, ;il (s - l)&(N)(S) = ;>

liil (s - l)C(s,x) = 1.

(3) I f the image of x : (Z/NZ)X -+ Cx is not {l}, the defining series of L(s, x) converge (the sum is taken in the order n = 1, 2, 3,. . . ) for s satisfying Re(s) > 0, and it is a holomorphic function in this domain. For such a x L(s, x) is holomorphic in the entire complex plane.

3.3. VALUES AT NEGATIVE INTEGERS 91

We give a proof of Proposition 3.15 at the end of this section.

(b) Values at negative integers and Bernoulli nwnbers and Bernoulli polynomials. Theorem 3.18 shows that the Rie- mann < function has rational values at nonpositive integers, and they can be expressed in terms of Bernoulli numbers and Bernoulli polynomials.

DEFINITION 3.16. The Bernoulli number B, (n = 0, 1, 2, 3,. . . ) is defined by the formula

Prom the formula 2 2 1 ~ =

= ez - 1 x+$+$+-. I+%+$+...

=I- ;+g+... ( > (

x x2 2

+ . . s+y+-* 1

-...,

we see that

(3.13) B. = 1, B1 = -;, B”=;, B4 = -&,

B8 = -$ BIO 5 691 7

= -, Blz = -- ’ B14 = 66 2730

-,.... 6

In particular, B,‘s are all rational numbers. Since & - 1 + 5 is an even function (i.e., invariant under J: H -x), we see that

(3.14) B, = 0 for n an integer greater than or equal to 3.

DEFINITION 3.17. The Bernoulli polynomial B,(z) (n = 0, 1, 2, 3,. . . ) is defined by

B,(z) = 2 (:) BiFi, i=o

where (1) = &.

From (3.13) we have

(3.15) B,,(z) = 1, Bl(s) =x- ;, B2(2) =x2 1

-x+-, 6

3 B3(x) = x3 - p2 + $x, By = x4 -2x3+x2-$,....

92 3. c

In particular, R,(z) is a polynomial in rational coefficients, and we have

B,(O) = B,.

THEOREM 3.18. (1) F or a natural number r and a positive real number x we have

<(l - 7-,x) = -r&.(x)

(2) For natural numbers r and N, and an integer a satisfying 1 5 a 5 N we have

COROLLARY 3.19. Let N be a natural number, a an integer, and m a nonpositive integer. Then, we have

CE ,(iv)(m) E Q.

In particular, we have C(m) E Q for any nonpositive integer m.

This can be seen from Theorem 3.18(2)

EXAMPLE 3.20. From Theorem 3.18 (2) and (3.15) we see

L a(N)(O) = -; + ;,

<@+l) = -& + ; - ;,

&q-2) = -& + ; - Jp.

COROLLARY 3.21. Let N be a natural number. If the image of a Dirichlet character x : (Z/NZ) ’ + Cx is diflerent from {l}, then we have

L(0, x) = -kc ax(a). a=1

This follows from Example 3.20 and the fact ~~=“=, x(a) = 0 (see Question 3).

QUESTION 3. Let G be a finite group and x: G ---) Cx a homomorphism whose image is different from (1). Show that CcAtc x(a) = 0.

3.3. VALIJES A’l’ NEGATIVE INTEGERS 93

Theorem 3.18(2) follows from Theorem 3.18(l) and the relation (3.12) between the Hurwitz < function and partial Riemann < function.

Before giving a proof of Theorem 3.18( 1) at the end of this section, we explain first why Theorem 3.18(l) . is a natural property in view of the nature of the Hurwitz < function and Bernoulli polynomials.

The Bernoulli polynomials first appeared in the formula for the sum of k-th power. For natural numbers T and 5, we have a formula

X-l

(3.16) c 72-l = g?(x) - 8.). n=o


1+2+3+.. . + (x - 1) = :(X2

1+ 2” + 32 + . + (x - 1)2 = ; (

x3 - Es2 + ix >

On the other hand, the Hurwitz < function satisfies by definition

C(%X + 1) - C(%X) = -;>

and therefore for any natural number x we have

(3.17)

X-1

c

1 - = -<(.5,x) + C(s). ns

n=l

In other words, when we consider the formula of the sum of I;-th powers when Ic is positive, Bernoulli polynomials appear, while we consider it when k is negative, the Hurwitz < function appears. This fact makes us feel that the formula ((1 - T, Z) = -f&(x) is very natural.

We now explain briefly why (3.16) holds. Let us consider the linear operator

D: Wd + @bl; f(x) H &XI

on the polynomial ring @[xl. W e see from the theory of Taylor ex-

pansion that the linear operator eD = c,“=,, $$ satisfies

eD(f(4) = f-(x + 1)

94 3. c

for all f(x) E @[xl . From the definition of B, we have

(3.18) D = (8 - 1) 2 $I-. n=O

Operate (3.18) on zT and use the formula

F $D’“(x’) = 2 (r,) BnsFn = BT(x), n=O n=O

and we obtain

(3.19) rx r-1 = B,.(z + 1) - BT(z)

The formula (3.16) follows easily from this.

QUESTION 4. If we tend s - 1 in (3.17), we have

for any natural number 2. Find the right-hand side when z = 5/2. (If we let r

2 = f in this formula, the left-hand side does not make sense, but it seems like

the sum of i from 1 to ?j.)

COROLLARY 3.22. (1) C(0) = -$. (2) If r is a natural number greater than or equal to 2, we have

<(l - r) = -fB,. (3) I f m is a negative even number, we have c(m) = 0.

PROOF. From Theorem 3.18(2) we have

((1 -r) = -iB,(l)

for a natural number r. Since B1 (x) = z - 5, we obtain c(O) = -$. If r 2 2, it follows from (3.19) that B,(l) = BT(0) = B,. From (3.14) we have B, = 0 for an odd integer r less than or equal to 3, and thus we have ((1 - r) = 0. 0

3.3. VALUES AT NEGATIVE INTEGERS 95

EXAMPLE 3.23. Prom 3.22 and (3.13)

((0) = -;, ((-1) = -A, ((-3) = l 23X3X5’

((-5) = - l 22 x 32 x 7’

((-7) = l 24X3X5)

<(-9)=- l 691

22 x 3 x 11’ <(-11) =

23X32X5X7X13’

C(-13) = -A,. “.

(c) Proof of Proposition 3.15 and Theorem 3.18.

PROOF OF PROPOSITION 3.15( 1). We prove this for the function L(s, x). (Similar proofs work for &N)(s) and <(s! z).) I f we let Re( s) = g > 1, we have

x(n) l-l ns

5;.

I f n > 2, we have

J n

12-l &dz,

and thus

~$,,.~=$dz=l+--&

n=l

This shows that the series C,“=, 9 converge absolutely, and, for any c > 1, they converge uniformly in the domain Re(s) > c. Since the limit of a uniformly convergent sequence of holomorphic functions is again holomorphic, Proposition 3.15(l) is proved. 0

PROOF OF PROPOSITIONS 3.15(2) AND 3.18(l). To prove Pro- position 3.15(2) it is sufficient to prove it for ((s, x). We prove it together with Proposition 3.18(l).

As a preliminary, we introduce the I function. For a complex number s satisfying Re(s) > 0 define

r(s) = IX e-“p f . I f s is a natural number, we hzve I’(s) = (s - l)!. l?(s) has an analytic continuation to a meromorphic function on the entire plane. We denote this extended function by I’(s) also. Then it is known

that I’(s) has the following properties. r(s) is holomorphic except for s = 0, -1. -2, -3, . , where it has a pole of order 1. I’(s) does not have a 0. For m > 0 we have

&Em(s + m)r(s) = (-1)“;.

Now, if Re(s) > 1, we have

x== =J’ c

e-(s+n)uUs * t We let u = ~

0 n=o u L7C+n’ >

=r

eCsu du (, Gus;.

In other words, we have

was, 4 = JX f(s, u)du, 0

where p(s,u) = &u S-l .

We divide the integral into two parts:

J’ ()_ f(s, u)du = /’ o f(s, u)du +

J’ lx f(s, u)du.

Since the function ePszL approaches 0 rapidly as u tends to infinity, the integral s;” f( , )d s u u converges for any complex number s, and

it is holomorphic on s. Consider J, f(s,u)du. By the definition of B, (z) we have

Therefore

x Kc(s) 12. uexu CTU =- 77=0

e” - 1’

3.3. VALUES AT NEGA'I'IVE INTEGERS 97

This has an analytic continuation to a meromorphic function s in the entire complex plane. It is holomorphic except at s = 1, 0, - 1, -2, -3,. . . ) where it has a pole of order 1.

Thus, r(s)(‘( ) s,x is extended to a meromorphic function on the whole complex plane, and it is holomorphic except at s = 1, 0, - 1, -2, -3,. . , where it has a pole of order 1. Therefore, C(s,z) has an analytic continuation to the whole complex plane, and it is holomorphic except at, s = 1, where it has a pole of order 1.

For an integer n > 0 we have

sJ~n(s + 71 - l)(r(s)((s, x)) = y . (-1)”

I f we let 72 = 0 and take the fact l?(l) = 1 into account, we have

li-i (s - l)<(s,z) = &(z) = 1.

If R > 1, we have lim,,l-,(s + 12 - l)r(s) = (-l)“-’ . A, and

thus we have

Bn (xl {(l- n,.,5) = -~ n

PROOF OF PROPOSITION 3.15(3). For s satisfying Re(s) > 0 and m 2 0, define

We have L( s, x) = fo (s) + Cz= 1 fr,, (s) . In the following we prove

(3.20) 5 I.fm(s)l I N. ISI . (1 + &) 77?=1 es

The inequality (3.20) shows that the series X:=1 fnL(s) converge uniformly in the domain {s E @ 1 IsI < C, Re( s) > C’}, for any real numbers C and C’, and thus the sum is holomorphic when Re(s) > 0.

Let us prove (3.20). Since the image of x : (Z/NE) ’ + @’ is not

{l}, we have C,“=, x(n) = 0 ( see Question 3). Hence, we have

98 3. c

We write mN+n s (@$+n)‘-&=- mN s -dx, xs+l

and thus, if we write 0 for Re(s), we have

Therefore, we have

and thus we have

ii: I.fm(s)l 5 N. ISI 2 -& I N. ISI. (1 + ;) . m=l m=l

(d) Functional equation. In Chapter 7, $7.2 in Volume 2 we will explain the fact that, when x : (i%/NZ) x + cx is a Dirichlet character and x-l : (Z/NZ)X --f cx is a Dirichlet character defined by x-‘(a) = ~(a)-i, th ere is a relation between L(s,x) and L(1 - s,x-‘) called th f t’ e uric zonal equation. It follows from the functional equation that we have the property that for an even number r no less

than 2, we have <(l - r) = 2 x (r - l)! x & for the Riemann <

function. For example, for r = 2 we have

<(-1) 7r2 1

= 2 x 1 x & x a4 = -2 x & X-=-- 6 12

(see Example 3.23).

(e) The second and third mysteries. We now discuss the second and third mysteries of the values of the < function.

The second mystery was proved by Kummer in the nineteenth century and part (2) of the following proposition is called “Kummer’s congruence”.

PROPOSITION 3.24. Let p be a prime number.

(1) If m is a nonpositive integer satisfying m $ 1 mod (p - l), then me have

SUMMARY 99

(2) If m and m’ are negative integers satisfying m c m’ q.k 1 mod (p - l), then we have

C(m) = <Cm’) mod PZ(,).

EXAMPLE 3.25. For a prime number p different from 2 and 3 we have - 1 $ 1 mod (p - 1). By Example 3.23 we have

<(-1) = -& E Z(,).

The above congruence relations mod p satisfied by the values of the Riemann < function at negative integers are generalized to congruence relations mod pn (n > l), and extended to the theory of p-adic L functions, which takes its values in p-adic numbers (theory of p-adic L functions of Kubota and Leopoldt).

QUESTION 5. Show by using Proposition 3.24 that if m is a nonpositive integer, any prime factor of the denominator of C(m) when we express C(m) as a quotient of relatively prime integers is no greater than 2 - m. (For example,

in Example 3.23 the prime factors of the denominator of c(-11) are 2,3,5,7,13, and they are no greater than 2 - (-11) = 13.

As for the third mystery, we will discuss in 54.4 that some arithmetic information of the field obtained by adjoining a 691st root of unity to the field Q is related to the fact that in Example 3.23 the numerator of <( -11) has the prime factor 691. In $4.3 we also discuss the arithmetic meaning of the values L(l,x) and L(0, x) for the Dirichlet character satisfying x( -1) = -1.

For the second and third mysteries, see Chapter 10 in Volume 2. The values of C functions appear in many areas of mathematics

in an unexpected way, and they do not cease to mystify us.

Summary

3.1. The value of the Riemann C function at a positive even integer r is of the form rational number x 7rT. The Riemann < function has an analytic continuation to the entire complex plane, and its value at a negative integer or 0 is a rational number.

3.2. The Dirichlet L function is a generalization of the Riemann < function. It has similar (but not exactly the same) properties to the Riemann C function.

100 3. c

3.3. The value at an integer of such a function called the < function has some mysterious arithmetical properties. (We will see later that it is related to the p-adic numbers and the ideal class group defined in Chapter 4.)

Exercises

3.1. Find the following sums: (1) (l+~-~-+)+(~+&-~-~)+....

(2) (l-+~++)+(i51;1-+-~+&+-..

3.2. (1) Show that if Re(s) > 1, we have (1 - 2l-“)<(s) = l-~+~-&t-~-&+....

(2) Using log(2) = 1 - i + f - i + i - i + . . , show

lili,(” - l)<(s) = 1.

Here lim,,r+e means s approaches 1 from the right on the

real line.

3.3. Let

For a = 1,3,5,7 define

S a= c 33 ET = -log(l - c‘“) 8 .

71=1 n

Prove formula (3.6) by calculation sr - ss - sg + ~7.

3.4. Let x,cr,. . . , ck be positive real numbers and define

(This is called the multiple Hurwitz < function.) Comparing to the proofs of Proposition 3.15(2) and Theorem 3.18, prove the following more general situation.

(1) The series c 1 7x1,” ,%>O (z+clnl+...+cknk)”

converges whenever

Re(s) > k. As a function on s, it has an analytic continuation to a meromorphic function on the entire complex plane, and it is holomorphic except at 1,2,. . , k.

EXERCISES 101

(2) Let m be a negative integer or 0. By multiplying by the product cl . . . ck, <(m, 2; cl,. . . , ck) becomes a polynomial in

Z,Cl,..., ck in ($ coefficients.

CHAPTER 4

Algebraic Number Theory

Algebraic number theory was founded by Kummer in the middle of the nineteenth century, and it was later developed by Dedekind and Kronecker.

Kummer had hoped that, by using his new theory, he could prove Fermat’s Last Theorem: If n is no less than 3, the equation x7% + yn = zn does not have an integral solution x, y, z satisfying xyz # 0. We can rewrite the equation x” + y* = zn as

(4.1) k=O

where & is the primitive n-th root of unity cos (%) + isin (s), and

(k is the Ic-th power of &. Both sides of (4.1) are in a product form, and we are tempted to apply the law of unique factorization to both sides of the equation. However, the formulas contain the number &, which is not a rational number, and thus Kummer was obliged to examine whether or not the law of unique factorization holds in the world of numbers containing such numbers as Cn.

A finite extension of the rational number field Q is called an algebraic number field. For example, Q(<,) is an algebraic number field. Algebraic number theory is a subject which studies how the unique factorization property for the natural numbers can be generalized (in a modified way, if necessary).

Even if a question is posed within the rational number field, it may not be answered within the rational number field, and it is often necessary to go to the world of algebraic number fields in order to answer the question. As a matter of fact, Kummer obtained a signif-

icant results to Fermat’s Last Theorem (see §4.4), which is originally a question within the rational number field.

In this chapter we will discuss the method and important results of algebraic number theory.

103

104 4. ALGEBRAIC NUMBER ‘THEORY

4.1. Method of algebraic number theory

In this section we will prove some of Fermat’s statements which we mentioned in the introduct,ion using the method of algebraic number theory, i.e., the method that enlarges the world of numbers. We also give a proof of Fermat’s Last Theorem in the case n = 3.

(a) Proof of Propositions 0.1-0.5, 0.10 and 0.11. We show

the propositions in the title by enlarging the world of numbers from Z to the rings such as Z[i] = {a+& 1 a, b E Z}, Z[J-2] = {a+bn j a,b~ Z},Z[&] = {u+b& 1 u,b~ Z},Z[fi] = {u+b&I u,b~ Z}. In the proofs below we will use the fact that these rings have the unique prime factorization property. Namely, if A is one of the following rings Z[i], Z[J--2], Z[&], Z[Jz], then A has the following property:

(*) I f an element a in A is nonzero and not a unit, then a can be factored into the form

a = a!1 “‘Q, (r 2 1, (~1, . . , (Y, are prime elements in A),

and this factorization is unique in the sense explained below.

The definition of a “prime element” is as follows. Rings such as Z[i] and Z[J--2] are integral domains (see Appendix A, §A. 1 for

the definition of an integral domain). An element (Y in an integral domain A is a prime element if the following conditions (i) and (ii) are satisfied:

(i) Q is nonzero and not a unit. (ii) I f a, b E A and ub E aA, then a E cuA or b E aA, where

aA = {OX 1 x E A}. (Condition (ii) says that if ub is divisible by CY, then a or b is divisible by CE.)

For example, a prime element in Z is a number of the form &prime number.

The meaning of the uniqueness is that if we have another factor- izationofu,u=~~...ab(s>l, c~~,...,~~~isaprimeelementofA), then we have T = s, and if we renumber a!;, . , (-Y: suitably, we have c(A = oiA (i.e., 0: = CQ x unit) for i = 1,. ,T.

An integral domain that has the property (*) is called a unique factorization domain. In 54.3 we will verify by using < functions that the rings Z[i], Z[J--2], and Z[&] are unique factorization domains.

PROOF OF PROPOSITION 0.2. The statement of Proposition 0.2 is the following:

4.1. METHOD OF ALGEBRAIC NUMBER THEORY 105

(1) If p is a prime number congruent to 1 modulo 4, then there exist 2, y E Z satisfying p = x2 + y2.

(2) If p is a prime number congruent to 3 modulo 4, then there do not exist x, y E Z satisfying p = x2 + y2.

We have already proved (2) as part of Proposition 2.8. Since we have a~ = x2 + y2 for an element o = II: + yi (x, y E Z) of Z[i] (where cu is the complex conjugate of a), the statement (1) follows from the following Proposition 4.1(l), which expresses the law of unique prime factorization.

PROPOSITION 4.1. (1) I f p is a prime number congruent to 1 modulo 4, then there is a prime element Q in Z[i] such that p = ~5 (5 is also a prime element ofZ[i]). Moreover, &[i] # nZ[i].

(2)UP’ P’ as a rzme number congruent to 3 modulo 4, then p is a prime element in Z[i].

(3) 2 = (1 + i)2 x (-i), and 1 + z are prime elments in Z[i], whereas -i is a unit in Z[i].

(4) Any prime element ofZ[i] is of the form (prime element up- peared in (1) (2) and (3)) x (unit).

(5) The set of units of Z[i] is {fl, &}.

PROOF. (1) Let p be a prime number congruent to 1 modulo 4.

Since we have 3 ( > P = 1 (Theorem 2.2(2)), there exists an integer a

such that a2 E -1 mod p. Since we have

(a + i)(a - i) = a2 + 1 E pZ[i], a + i $! pZ[i], a - i $J pZ[i],

we see that p is not a prime element in Z[i]. On the other hand, since p is not a unit in Z[i], it follows from the prime factorization of p in Z[i] that there exists a prime element Q in Z[i] dividing p. Write p = a!p, p E Z[i]. s ince p is not a prime element, p is not a unit. We have

- p2=a!yp.cYyp=cyG.pp,

and since both a~ and ,B,?? are natural numbers, cvcv must be one of the divisors of p2, namely 1, p, p2. If CVG = 1, (Y would be a unit, and it is a contradiction. If a6 = p2, then ,B would be a unit since /3p = 1. Thus, we have p = CUCU. Suppose &Z[i] = ZZ[i], and we deduce a contradiction. Take an integer a E Z such that (u + i)(a - i) E pZ[i]. Since a is a prime element, we have either a + i E (-YZ[i] or a - i E aZ[i]. Take the complex conjugates of these elements and use the fact

106 4. ALGEBRAIC NUMBER THEORY

oZ[i] = &Z[i], and we see that both a + i and a - i belong to crZ[i]. Since we have 2i = (a + i) - (u - i) E &Z[i], we have 2,p E aZ[i]. It follows that 1 is in &Z[i] and a is a unit, a contradiction.

(2) Let p be a prime number congruent to 3 modulo 4, and let (1~ be a prime element in Z[i] dividing p. We set p = a$ (,b E Z[i]). Then, as before, we have p2 = a~ ’ pp. Since we have a~ # 1 and we cannot write p = x2 + y2 (x, y E Z), we see p # as. Therefore, wehavec@%=p2, /3p=l, and p is a unit. Hence p = (Y/? is a prime element.

(3) We show that 1 + i is a prime element. Let cy be a prime element in Z[i] dividing 1 + i, and set 1 + i = ~$3. Then we have 2 = (1+ i)(l -i) = a3 . /3p. Since cucU # 1, we have a?% = 2, ,8p = 1, and therefore /3 is a unit. Hence 1 + i = cup is a prime element.

(4) Let cr be a prime element in Z[i]. Considering the prime factorization of the natural number a~, which is not equal to 1, we see that (Y divides a prime number.

(5) If 0 is a unit in Z[i], we set /3r = 1 (y E Z[i]). Then 1 = /3p. 77, and thus ,Dp = 1. If we write p = z + yi(z,y E Z), we have x2 + y2 = 1. The integer solutions of this equation are (x, y) = (fl,O), (0, *l). Hence, we have /3 E {fl, +A}. 0

PROOF OF PROPOSITION 0.1. Proposition 0.1 is about a prime number that can be the hypotenuse of a right triangle with rational sides.

Let p be a prime number congruent to 1 modulo 4. By Proposi- tion 4.1( 1) p can be written as p = a~, where o is a prime element in Z[i]. Ifweseto2=x+yi(x,yEZ),wehavep2=(r2~2=x2+y2. If we show x # 0, y # 0, we see that p is the length of the hypotenuse of the triangle whose three sides are 1x1, ]y], and p. If x = 0 or y = 0, the argument of cr is a multiple of 7r/4, and thus there exists an integer m such that

a=mp, wherep E (1, 1 + i, i, -1 + i}.

This contradicts the uniqueness of the prime factorization in Z[i]. Next, it is easy to see that the equation 22 = x2 + y2 does not

have a solution satisfying x # 0, y # 0. Let p be a prime number congruent to 3 modulo 4, and let p2 =

x2 + y2 (IL., y E Z). If we set a: = x + yyi, then we have p2 = a~%. By Proposition 4.1(2), p is a prime element in Z[i]. It follows from the uniqueness of factorization that Q = p x (kl, or * i ). This implies x=Oory=O. 0


PROOF OF PROPOSITION 0.11. Proposition 0.11 states that the only solutions to the equation y2 = x3 - 4 are (x, y) = (2,2), (5,ll). We rewrite the equation as

x3 = y2 + 4 = (y + 2i)(y - 2i).

Notice that the product of y + 2i and y - 2i is a cube. We will later show by using the unique factorization properties of Z[i] that each of y + 2i and y - 2i is a cube in Z[i]. Thus we have r

(4.2) y + 2i = (a + bi)3 (a, b E Z). I Expanding the right-hand side and comparing both side& we have ,,/

//’

(4.3)

\- _ /- 2 = 3a2b - b3 = (3a2 - b2)b.

Therefore, b is a divisor of 2, and thus b equals one of kl, 12. If we replace b = 1, -1,2, -2 in (4.3), we have 3a2 = 3, -1,5,3, respectively. Thus

(a, b) = (fl, 1) or (fl, -2).

It follows from this and (4.2) that y = 2 or 5, and we obtain 2 by substituting y in the equation y2 = x3 - 4.

We use the following Lemma 4.2 for the remaining part of the proof.

LEMMA 4.2. Let A be one of Z[i], Z[a], Z[<a], and Z[v!?]. Let cq,.. , (Y,, p be nonzero elements of A, k a natural number, and a1 “‘cl, = /3”. Furthermore, if i # j, suppose oi and a3 are prime to each other. Then for each i, cq can be written as cq = u& with an element ,0i in A and a unit ui.

This lemma can be proved by counting how many times each prime element appears in the prime factorization of 01, . . , Q,, /3, just as the proof of Lemma 1.7 in $1.1.

We now show that the equation x3 = (y + Zi)(y - 2i) (x, y E Z) implies that y + 2i, y - 2i are cubes in Z[i].

Let y be a prime element of Z[i] dividing both y + 2i and y - 2i. Since y divides (y + 2i) - (y - 2i) = 4i = -i( 1 + i)“, we have y = (1 + i) x (unit). Writing y + 2i = (1 + i)e~, e 2 1, (Y element of iZ[i] not dividing 1+ i, we have y - 2i = (1 - i)“??, 1 - i = (-i) x (1+ i), and thus y - 2i = (unit) x (1 + i)%. Hence,

Ql(Y2Q3 =X3,

(~1 = (unit) x (1-t i)2e, ~2 = Q, 03 = Cr.


Any two of al, (~2, and a3 do not have a common prime factor. By Lemma 4.2 all of al, CQ, and 0s are of the form (unit) x (cube in Z[i]). It follows from this that e is a multiple of 3. Thus, y + 2i = (1 + i)“o is of the form (unit) x (cube in Z[i]). But the units fl, fi in Z[i] are all cubes. Hence y + 2i is a cube in Z[i]. 0

PROOF OF PROPOSITION 0.3. Proposition 0.3 is about the qua- tion p = z2 + 2y2 (p is a prime number) and the residue of p dulo 8. If p f 5,7 mod 8, then (-2,~)s = -1. Thus, there do no exist

Proposition 2.8). L rational numbers 5 and y satisfying p = x2 + 2y2 (see the pr of of

Next, let p = 1,3 mod 8. We prove that there exist z,y E Z satisfying p = x2 + 2y2. For an element o = x + yfl (z, y E Z) in

Z[J--2] we have CG = x2 + 2y2. Therefore, it suffices to show the existence of Q E Z[J-- satisfying p = CUZ. This can be done by replacing Z[i] by Z[v’Z] in the proof of Proposition 0.2 and using

the fact 2 = 1. ( >

0

PROOF OF PROPOSITION 0.10. Proposition 0.10 states that the only solution in natural numbers to the equation y2 = x3 - 2 is (x, y) = (3,5). Rewriting the equation as

x3=(Y+J-2)(y-vq,

we can show that both y + &2 and y - n are cubes in Z[ J-- , as in the proof of Proposition 0.11. (Here we replace Z[i] by Z[J--. Instead of the prime element 1+ i in Z[i], the prime element fl in Z[J--2] appears in the proof. We also use the fact that the units of Z[J-- are fl instead of the fact that the units of Z[i] are ztl, zti.) We have

y+J--2=(a+bdq3 (a, b E Z).

Expanding both sides and comparing the imaginary parts, we have

1 = 3a2b - 2b3 = (3a2 - 2b2)b.

Therefore, b is a divisor of 1, and thus b = fl. It follows that

(a, b) = (fl, I),

and we obtain y = 55, x = 3. 0

PROOF OF PROPOSITION 0.4. Proposition 0.4 is about the equation p = x2 + 3y2 (p prime) and the residue of p modulo 3. If p s 2 mod 3, then we have (-3,~)s = -1, and thus there do not exist


rational numbers z, y satisfying p = z2 + 3y” (see the proof of Propo- sition 2.8).

Next, let p = 1 mod 3. We show the existence of 2, y E Z satisfying p = x2 + 3y2. For an element o = z + y&3 in Z[a] =

{a+bfl / a, b E Z}, we have 05 = x2 + 3y2. Therefore, it suffices to show the existence of (Y E Z[&?] satisfying p = cr~r. Using the fa

( > -3

P = 1 and replacing Z[i] by Z’[&] in the proof of Proposition 0.

?

,

we can show the existence of ,6 E ?!?[<a] satisfying p = pp. On the other hand, it is easy to show that any element of Z[&] belongs to Z[a] after multiplying one of 51, &<s, &<i (these are all of the units of Z[&?]). I f we set ck = up E Z[a], u E {*I, *<s, *@}, then we have p = pp = CUZ. 0

PROOF OF PROPOSITION 0.5. Proposition 0.5 is about the equation p = x2 - 2y2 (pprime number) and the residue of p congruent modulo 8. The proof of Proposition 0.5 is similar to that of Proposition 0.2. For an element o in Z[i] we considered the element

(Y. For an element (Y = x + yfi(x, y E Z) in Z[Jz] we consider the element o’ = x - yfi. I f p = 1,7 mod 8, then there exists cy = x + yfi E Z[Jz] (x, y E Z) such that

p = 3xm’ = *(x2 - 2y2).

I f p = --cyo’, we set /? = (1 + a)~: and obtain p = ,0@‘. 0

QUESTION 1. Show that (z, y) = (1,O) is the only integral solution to y2 = x3 - 1.

QUESTION 2. Using the fact that Z [W] is a unique factorization do-

main, show that the only integral solutions of y2 = x3 - 11 are (x, y) = (3, f4)

and (15, &58).

(b) x3 + y3 = z’. We give a proof of Fermat’s Last Theorem in the case 11 = 3. Our proof is essentially the same as the proof given by Euler. First, we give an outline without giving any detail in order to help understand the proof better. The strategy of the proof is to use the method we used to find integer solutions to the equation y2 = x3 - 4 and the method of “infinite descent”, which we used to prove Proposition 1.2 in $1.1.

PROOF. Suppose we have integers x, y, z satisfying x3 + y” = z3 and x # 0, y # 0, z # 0. Among those we choose x, y, z such


that max(]z], ]y], ]z]) is the smallest. We then induce a contradiction by showing that there is another solution x’, y’, z’ such that

madx’l, IY’I, Iz’l) < max(l4, Ivl, 14, 2’ # 0, Y’ # 0, z’ # 0. The outline of the proof is as follows.

(i) First we show that we may assume that y, z are odd without loss of generality.

(ii) Rewrite x3 + y3 = z3 as

x3 = (2 - Y)(Z - C3Y)(Z - 33Y).

(Notice that <,” = c3.) Use the same argument as the one we used to show that x3 = (y + 2i)(y - 2i) implies both y + 2i and y - 2i are cubes in Z[i], and we obtain the following. (ii-l) If x is not divisible by 3, then there exist c E Z and

Q E Z[&] such that (1) z - y = 9,

(2) z - 3Y = T3a3, (3) z - <3y = <37x3.

(ii-2) If z is a multiple of 3, then there exist c E Z and o E Z[&] such that

(1) z - y = 92,

(2) z - C3Y = (1 - C3)03> (3) z - c3y = (1 - T3)$.

(iii) Seta=a+b&(a,bEZ). (iii-l) If x is not divisible by 3, it follows from (ii-l)(2) that

y=u3-3ub2+b3, z = -u3 + 3a2b - b3.

Hence we have z-y = (a+b)(2u-b)(2b-a). Comparing this with (ii-l)(l), we have

c3 = (a + b)(2u - b)(2b - a).

We can show that any two of a + b, 2u - b, and 2b - a are relatively prime. Therefore, each of a + b, 2u - b, and 2b - a is the cube of an integer. Setting a + b = (z’)~, 2u - b = (x’)~, 2b - a = (Y’)~ (d, y’, z’ E Z), we have (x’)” + (Y’)~ = (z’)~, x’ # 0, y’ # 0, z’ #

0, max(l4, IY’I, VI) < max(l4, 1~1, 14). (iii-2) If x is a multiple of 3, it follows from (ii-2) (2) that we

have

~=a”-6u2b+3ub2+b3, z=u3+3u2b-6ub2+b3.


Thus, we have z-y = 9ab(a - b). Comparing this with (ii-2)(l), we have

c3 = ab(a - b).

We can show that any two of a, b, and a - b are relatively prime. Hence, each of a, b, and a - h is the cube of an integer. Setting a = (z’)“, b = ‘)3, a - b = (1~‘)~ (d, y’, z’ E Z), we obtain (z’)” + (Y’)~

(z’)3, 2’ # 0, Y’ # 0, z’ # 0, max(b’I, IY’I, 14) / “; max(l4, lyl, 14.

To complete the details of (i), (ii), (iii), we need the fo> mg / preliminaries (a) through (d).

(a) 1 - (3 is a prime element in Z[&]. (We can prove this by the same method as the proof of 1 + i being a prime element in Z[i].) 3 = (1 - <3)2 x (-T3).

(b) IC, y, z are pairwise relatively prime. In fact, if a prime 1 divides two of x, y, z, it divides the third because of the relation x3 + y3 = z3. Thus, (7, 7, f) is another integer solu-

tiontoz3+y3=z3, and it contradicts to the minimality of

max(lxl, IYI, 1~1). (c) If a prime element (Y of Z[&] divides two of z - y, z - <a y, and

z - T3y, then o = (1 - (3) x ( unit). In fact, if CY divides, say, z-y and z-&y, then o divides (z-y)-(z--&y) = (1-&)y, and thus cy divides y unless cy is of the form (1 - (3) x (unit). Since 01 also divides z - y, o divides both y and z, which contradicts the fact that y and z are relatively prime. The cases where o divides z-y and z - <sy, or z - <3y and z - c3y are similar.

(4 The ring ~[CY]/~Z[C 1 3 consists of four elements 0, 1, C;, and 1 + <a. Thus any nonzero element in this ring is a third root of unity. The set of units in Z[&] consists of 3~1, zt&, iC3. The image of Ifrl in Z[&]/2Z[<3] is the class of 1, the image of i<3 is the class of <3, and the image of ztc3 is the class of

1+<3.

About (i). It follows from (b) that only one of 2, y, z is even. Re- placing (x, y, z) by (y, 2,~) or (z, -y, z) if necessary, we may assume that both y and z are odd.

About (ii-l). By (a), (c) and the fact that z is not a multiple of 3 we see that any two of z - y, z - cay/, and z - c3y are not divisible bv a common mime element in Zlt,~l. It follows from Lemma 4.2 that


each of 2 - y, z - &y, and z - c3y is the product of a cube and a unit in Z[&]. Setting z-y = up3 (U is a unit in Z[&], p E Z[(s]), we have

(z - y)” = up”T$” = (L@)“.

Therefore (Z - y)” is the cube of an integer. By factoring into prime elements, we see that z - y is the cube of an integer. Next, we set

_____j\___ z - &y = ucy3 (U is a unit in Z[&,] o E Z[(;]). It suffices to show ‘u = +c3. We consider it modulo 2Z[&]. Since y E z E 1 mod 2, we have

wa3 E z - &y E 1 - (3 E <, mod 2Z[&].

1

From this and the fact that the cube of any nonzero eleme in z[&]/2z[<3] becomes 1 (by (d)) we see that u = c3 mo %I. Hence, we conclude v = *t3.

About (ii-2). Since 5 is divisible by 3, 2 is divisible by 1 - &. We

have x3 = (z - y) (z - &y) (z - <,y) and

z - y G z - &y E z - t3y mod (1 - &)iZ[&].

Therefore, each of z-y, z -&y, and z - c3y is divisible by 1 - (3. We have z - Cay $ 32[&] = (1 - &)‘Z[&], since if z - Cay E 32[&], both z and y are divisible by 3, contradiction to (b). I f we set orda(5) = m and ords(z - y) = R, the equation x3 = (z - y)(z - &y)(z - <,y) implies 6m = 2n + 1 + 1. Thus, we have n 2 2. Hence, we have

z - y = 9r, z - <3Y = (1 - C3)P, iJ -.3Y = (I- C3)P

(r E z, cp E W31). W e h ave (;)” = rcpq, and no two of r, cp, and (p are divisible by a common factor in Z[&]. It follows from these and Lemma 4.2 that each of r, cp, and $5 is the product of a unit and a cube in Z[<s]. Therefore, we have z - y = 9c3 (c E Z) and

z - &y = w(l - &)(Y3 ( v is a unit in Z[<3], Q E Z[(3]). It suffices to show that u = fl. By taking modulo 2Z[&], we have

Since the cube of any nonzero element in Z[C3]/2Z[&] equals 1, we have u E 1 mod 2iZ[c3], and thus w = &l.

About (iii-l). First, we have to show that a + b, 2a - b, and a - 2b are pairwise prime. If 1 divides two of these elements, 1 divides 3a = (u + b) + (au - b) and 3b = 2(u + b) - (2~ - b). But 1 divides z-y, which is the product of these three numbers. Therefore 1 divides x3, and thus x. Hence 1 # 3 by hypothesis. Thus, 1 divides a and b, and by the expression of y and z in terms of a and b, 1 divides

4.2. ‘I’HE HEART OF ALGERHAIC NUMl%ER ‘I’HEOHY 11:1

both y and Z, which contradicts (b). Next, it is easy to see that

2’ # 0, Y’ # 0, 2’ # 0 and max(l4, Iv’l, Iz’l) < max(l4, Ml 14). (iii-2) is similar to (iii-l) and we leave it to the reader. tl

This proof of Fermat’s Last Theorem in the case n = 3 is related to elliptic curves. By setting X = &, Y = z. the equation x3 +

y” = .z~ is written as

We denote this elliptic curve by E. Fermat’s Last Theorem in the case ,,,/ /’

n = 3 is equivalent to the fact E(Q) = (0, (O,fl)}, which means that E(Q) is a group of order 3; i.e., E(Q) z Z/32. Suppose that we have an elment Q in E(Q) such that Q # 0, Q # (0, 51). When we replaced (2, y, 2) by (y, 2, Z) or (z, -y, -CC) in the proof above, it corresponds to replacing Q by (0,l) - Q or (0, -1) - Q. Finding (z’, y’, z’) corresponds to finding P E E(Q) such that Q = 3P. The fact max(]5’], ly’], ]z’]) < max(]z], ]y], ]z]) is reflection of the fact that the height of 3P is far greater than that of P.

4.2. The heart of algebraic number theory

We will discuss the central facts in algebraic number theory, namely, about the ring of integers of an algebraic number field, prime factorization, the finiteness of class number and Dirichlet’s unit theorem. The last two are the two big theorems of algebraic number theory.

(a) The ring of integers of an algebraic number field. All the rings appeared in the previous section, ;Z[i], E:[n], Z[&], and

Z[fi] are examples of the ring of integers of an algebraic number field.

We explain the ring of integers of an algebraic number field. Just as the rational number field Q contains the ring of integers Z, each algebraic number field K contains a subring called the “ring of integers of K” (written OK). For example, if K = Q(&), then it is known

that

OK = z[<n] = { kai<:, 1 T 10, ~,...,a, E Z} z=o


TABLE 4.1

Algebraic number field Q Q( &) Q(a) Q(A)

Its ring of integers

The definition of OK is as follows: 0~ is the set of elements 0 in K such that Q: satisfies the equation

an + Clck n-1 + . . + c,, 1 0

for some n 2 1 and cl,...,c,, E Z. (The point here is that the coefficient of the highest degree (degree n) is 1.) An element in 0~ is called an integer of K, or an algebraic integer if we want to distinguish it from the usual integers. For example, <n satisfies (<7L)n - 1 = 0, and thus it is an integer of Q(&). I f we use the terminology of algebra, “integral closure”, 0~ is nothing but the “integral closure of Z in K”. See Appendix A, §A.1 for the generalities of “integral closure”.

When K is a quadratic field (degree 2 extension of Q), then 0~ is given as follows. We write K = Q(6), where m is a square free integer different from 1. Then we have

We remark that if m E 1 mod 4, v is a solution of 1~’ - z - v; i.e., it satisfies

(See Table 4.1; note that Q(o) is the field consisting of all numbers obtained from Q and cy by four fundamental operations, and Z[o] is the set of numbers cv written as a polynomial with Z coefficients.)

The ring of integers 0~ of K is isomorphic to iZ@” (n = [K : Q]) as additive group. In other words, there exist (3~1,. j Q~ E OK (n = [K: Q]) such that each element of 0~ can be writ,ten uniquely as

Clcyl Jr. ..+CT,N,(C1,... , c, E Z). This can be shown as follows using the general theory of integral closure. In general, if A is an integrally closed Noetherian domain (see Appendix A, §A.l), F is the quotient field of A, K is a finite separable extension of F, and B is the integral closure of A in K, then it is known that B is a finitely generated

4.2. THE HEART OF ALGEBRAIC NUMBER THEORY 115

A-module. Let A = Z (and thus F = Q). Then we have B = 0~. and 0~ is a finitely generated Z-module, that is, a finitely generated abelian group. Since 0~ does not have an element of finite order except for 0, the fundamental theorem of finitely generated abelian groups implies that there exists an integer n > 0 such that 0~ E Z@“. It is easy to see that this n is [K : Q].

QUESTION 3. Prove that the ring of integers of a quadratic field is as de-

scribed above.

(b) Failure of unique prime factorization. In the ring of integers of an algebraic integer field, the law of unique prime factorization (the condition (*) at the beginning of 54.1) may not hold, unlike Z, Z[J--T], Z[n], Z[&], Z[Jz]. For example, in the ring of

integers Z[J?Z?] = {u + bm / a, b E Z} of Q(m), there is no prime element dividing 3. In fact, we have

(4.4) 3” = (1-t J-26)(1 - J-26)

and even though the product of 1 + J-26 and 1 - &!% is in 3i%[q], neither is in 3Z[m]. Thus 3 is not a prime element in Z[J-26]. I f a: were a prime element of Z[a] dividing 3, the argument of Proposition 4.1 and the fact that 3 is not a prime element would impliy that 3 = ~3. If we write Q = IC + y&% (2, y E Z), we have 3 = x2 + 26y2, but it is easy to see that there are not such x and y. As this argument shows, the prime factorization does not work well with Z[&QG]. Thus, we cannot apply Lemma 4.2 to (4.4).

In fact, 1 + &??6 and 1 - m are not cubes in Z[m].

(c) Decomposition into prime ideals. As is explained in the previous paragraph, the prime factorization law may not hold for the ring of integers of an algebraic number field. However, the ring of intergers of algebraic number field has the “decomposition in prime ideals” instead. We now explain “ideals” and “prime ideals”.

DEFINITION 4.3. Let A be a commutative ring. A subset a of A is called an ideal in A if it satisfies conditions (i) and (ii).

(i) a is a subgroup of the additive group A (i.e., we have 0 E a and “a,bEa =+ a+b,a-bEa”).

(ii) a E A, b E a implies ab E a.

EXAMPLE 4.4. (1) For elements ~1,. , cy,, in A the set {uIcyl + . . + anan / ai, . . , a, E A} is an ideal of A. We call it the ideal of A generated by ~1,. , cy,, and we denote it by (~1,. . , a,,). In


particular, for an elment o in A we have (a) = cuA. An ideal of the form (0) is called a principal ideal.

We abbreviate (0) = (0) to 0 in what follows. (2) All the ideals of Z are principal ideals (n), n integer. In fact,

if a is a nonzero ideal of Z, it is easy to show that a = (n), where n

is a element of a whose absolute value is the smallest. An integral domain such that all of its ideals are principal ideals,

such as Z, is called a principal ideal domain. We often abbreviate principal ideal domain to PID. Z[i], Z[J-2], Z[&] and Z[Jz] are principal ideal domains (for Z[i], Z[G] and Z[&] see $4.3).

DEFINITION 4.5. Let A be a commutative ring. An ideal p in A is a prime ideal if (i) and (ii) hold.

(i) If a,b E A, ab E p, then a E p or b E p. (ii) 1 6 A (this is equivalent to p # A).

EXAMPLE 4.6. (1) If A is an integral domain and o is a nonzero element of A, then we have

(cy) is a prime ideal ++ a is a prime element

(2) All the prime ideals of Z are (p), where p is a prime number, and 0.

DEFINITION 4.7. For ideals a, b in a commutative ring A define the product ab as the set of elements of the form Cy=“=, azbi (n > 1, ui E a, b, E 6). ab is an ideal of A.

THEOREM 4.8. Let K be an algebraic number field and a an ideal of the ring of integers OK of K. Then a can be decomposed into the product of prime ideals of the form

a=pl...p, (r 2 0, pl,...,b are nonzero prime ideals in OK),

and this decomposition is unique in the following sense. I f a admits another decomposition

a=p\...pb (~20, pi,.. . , pb are nonzero prime ideals in OK),

then we have r = s, and by renumbering pi,. . . , pb suitably, we have p:=pz foralli=l,...,r.

The above decomposition of a is called the prime decomposition of a. Among pi,. . . ,pT, we regroup the same prime ideals and we

4.2. THE HEART OF ALGEBRAIC NIJMBEH THEORY 117

often write

a=pyl. .pp

(g > 0, pi are distinct nonzero prime ideals of 0~ eZ 2 1).

Thi theorem is a result of the fact that 0~ is a Dedekind domain i;

(see App dix A, sA.1). Though it is an important theorem, we L consider it to be~patimmof- the general theory of algebra, instead of number theory. Thus, we do not give a proof here. We give a summary of the theory of the Dedekind domain in Appendix A. We recommend that the reader look up any booki of ring theory for detail. The argument goes roughly as follows. The ring Z is a principal ideal domain (see Example 4.4), and a principal ideal domain is a Dedekind domain. Thus, Z is a Dedekind domain, and OK, the integral closure of Z in K, is also a Dedekind domain (see Appendix A, sA.1; the integral closure of a Dedekind domain in a finite extension of its field of fraction is once again a Dedekind domain).

In a Dedekind domain any nonzero ideal admits a unique decomposition into prime ideals (see Appendix A, 5A.2)

An ideal of a Dedekind domain can be written as (~1,. . . , a,) with a finite number of elements (~1,. . . , cy, (see Appendix A, sA.1). However, a Dedekind domain is not necessarily a principal ideal domain, and an ideal may not be written as (a).

EXAMPLE 4.9. Let K = Q(m). Consider an ideal

a = (3, 1+ J?%), b = (3,1- GZ)

in 01~ = Z[JZ%]. The ideals a, b are prime ideals which are not principal. We have

(3) = l-lb, (1 + J-26) = 113, (1- J-26) = 6”.

Both sides of (4.4) cannot be factored further, but as an ideal it decomposes as

(3”) = a”b” = ((1-t J-26)(1 - J-26)).

Theorem 4.8 can be generalized to Theorem 4.12 below, which is a theorem concerning “the decomposition of fractional ideals into prime ideals”.

- ‘See, for example, M. Atiyah and I. G. MacDonald, Introduction to Com-

mutatzwe Algebra, Addison-Wesley, Reading, Mass., or Chapter 7 of N. Bourbaki Algibre Commutatzve, Herman, Paris.


DEFINITION 4.10. Let K be an algebraic number field. A subset a of K is a fractional ideal of 0~ if one of the following two equivalent conditions holds.

an element c in 0~ such that ca is a nonzero

finitely generated OK-submodule of K.

For an elementninKX1Ye denote by (0) the fractional ideal @OK. A fractional ideal of the form (cr) (o E KX) is called a principal fractional ideal.

DEFINITION 4.11. Let K be an algebraic number field. For fractional ideals a, b in K we define the product ab as the set of all elements of the form Cy=“_, aibi (n 2 1, ai E a, b, E 6). ab is a fractional ideal of OK.

THEOREM 4.12. Let K be an algebraic numberfield, and a a fractional ideal of OK. Then a is written as

a= Jp, P

where p runs all nonzero prime ideals of OK, ep E Z and ep = 0 except for a finite number of p. The set of all fractional ideals of 0~ is a group under the multiplication defined in Definition 4.11; 0~ is its identity element and the inverse of a is given by

a -’ = {x E K / xa c OK}

This theorem, too, follows from the theory of the Dedekind domain since 0~ is a Dedekind domain (see Appendix A, ljA.2)

It was Kummer who first showed, around 1845, that the law of prime factorization fails for the ring Z[&] (n > l), but the law of prime decomposition of ideals holds. To be precise, Kummer introduced the notion of “ideal numbers” instead of ideals. It was Dedekind who, around 1863, defined the ring of integers of an algebraic number field and the notion of ideal, and proved the law of unique prime decomposition for the ideals. The notion of ideals, which was born in number theory, became quite important in all areas of mathematics such as algebraic geometry (see, for example, D. Eisenbud and J. Harris, Schemes-The Language of Modern Al- gebraic Geometry).

4.2.THE HEART OF ALGEBRAIC NUMBER THEORY 119

(d) Ideal class group and unit’ group. It is considered that the ideal class group is the most impbrtant group appearing in algebraic number theory, and the unit group is the second.

DEFINITION 4.13. Let K be an algebraic number field.

(1) The ideal class group of K is the quotient ot the group of fractional ideals of 0~ (Theorem 4.12) by the subgroup consisting of principal fractional ideals (Definition 4.10). We denote it by Cl(K) or C1(0~).

(2) The unit group of K is the group Og, the group that consists all the units in OK.

LEMMA 4.14. Let K be an algebraic number field. Then, conditions (i), (ii) and (iii) are equivalent.

(i) The ideal class group Cl(K) is reduced to the identity element.

(ii) The integer ring 0~ is a principal ideal domain. (iii) Every nonzero element in 0~ can be factored into the product

of prime elements, and the factorization is unique in the sense

of $4.1.

PROOF. The proof is left to the reader. 0

EXAMPLE 4.15. If K = Q, Cl(Q) consists of only one element, and the unit group of Q is Zx = {il}.

We discuss the meaning and importance of the ideal class group and the unit group.

We can say that the ideal class group and the unit group measure the difference between “numbers and ideals”. The ideal class group is the cokernel of the homomorphism from Kx to the group of fractional ideals given by cy H (a), while the unit group is its kernel. The sizes of the kernel and the cokernel indicate how this homomorphism differs from an isomorphism. Also, Lemma 4.14 shows that the ideal class group measures the “failure of the prime factorization law”. For example, consider the prime factorization of 7 in the ring Z[J2]. We have

(4.5) 7=(3+&?)(3-Jz)=(5+3Jz)(5-3Jz)

= (27 + 19&)(27 - 19fi) = ...

As we can see from the identity 3 + &’ = (5 - 3&)(1 + a)“, many different prime factorizations of 7 shown in (4.5) are obtained by


multiplying by a unit in a(&) such as (1 + a)“. Thus, the prime factorization of 7 is unique in the sense of §4.1(*). But, since there are

infinitely many units in Q(a), the situation of prime factorization in

Z[fi] seems somewhat different from that of Z, as we can see in (4.5).

Fermat did not arrive at the notion of the ring Z[Jz] or the units thereof, but he realized that “the equation 7 = x2 - 2y2 has infinitely many solutions suchas 7 = 32-2x12 = 52-2~32 = 27”-2x1g2 = ... because the equation 1 = x2 - 2y2 has many solutions”, and he began to study Pell’s equation in Proposition 0.6

As this shows, the ideal class group and the unit gr of K indicate how the law of prime factorization in 0~ is different r that of Z. If we understand how different the properties are, w

3 expect that we can understand K well (even if, for example, the 1 w of prime factorization, which played an essential role in $4.1, not hold). For this purpose it is important to know the ideal class group and the unit group. As we will see later, the ideal class group and the unit group play some mysterious roles, as they are related to zeta functions and class field theory.

(e) Two main theorems in algebraic number theory. Here we introduce two important theorems, Theorems 4.16 and 4.21, concerning the ideal class group and the unit group. Proofs of these theorems will be given in Chapter 6, 56.4 in Volume 2.

THEOREM 4.16. The ideal class group of an algebraic number field is a jinite group.

DEFINITION 4.17. The order of the ideal class group of an algebraic number field is called the class number of the algebraic number field.

EXAMPLE 4.18. Consider K = Q(m). We will see in 54.3 that the class number of K is 6. Setting a = (3,l + &%), c =

(2, v’?%?I), we have

a3 = (1+ pi%), c2 = (2),

and we have

Z/32 CE Z/2Z 3 Cl(Q(J-26));

(m,n) H (class of a)nL(class of c)”

In order to state Theorem 4.21 we define the notion of real places and imaginary places.

1.2. THE HEAItT OF ALGEBRAIC NUMBER ?'HEORY 121

DEFINITION 4.19. Let K be an algebraic number field.

(1) A real place of K is a homomorphism from K to IR. (2) An imaginary place of K is a homomorphism 0 from K to @

such that u does not satisfy a(K) c R. We regard 0 and its

conjugate a : K + @; 5 H g(z) as the same place.

PROPOSITION 4.20. Let r1 be the number of real places and r2 the number of imaginary places. Then, we have

[K: Q] =rl +27-z.

PROOF. We know from field theory that there are [K: Q,] homomorphisms from K to C. Among those there are r1 homomorphisms whose image is contained in IR, and 2~ whose image is not in If& 0

THEOREM 4.21 (Dirichlet’s unit theorem). The unit group of an algebraic number field K is a finitely genera.ted abelian group. More -,,’ precisely, if r1 is the number of real places, r2 is the number of imaginary places and r = r1 + r2 - 1, then we have

0; ” i%” @ (finite cyclic group).

The finite cyclic group above is the group formed by all the roots of unities in K.

EXAMPLE 4.22. For K = Q(a) we have ri = 2, r2 = 0, and

o;r{f(l+JZ)n 1 nEZ}“Z@z/22.

In general, for a real quadratic field K (a quadratic field Q(Jm) with positive m) we have ri = 2, r2 = 0, and K does not contain roots of unity except for *l. Thus, there is an element E in 0: such that

0; = {*En 1 n E 25).

Such a unit E is called a fundamental unit in K. For example 1 + fi is a fundamental unit in Q(a).

EXAMPLE 4.23. If K = Q(fi), we have ri = 1, rz = 1 and

0; ” {*(l - q-5)“; n E Z} rzC3z/2z.

EXAMPLE 4.24. We have ri + r2 - 1 = 0 only when K = Q or K is a quadratic imaginary field (i.e., K = Q(Jm), m a negative integer; in this case we have r1 = 0, r2 = 1). Therefore,

OK x is a finite group

M K = Q or K is an imaginary quadratic field.

122 4.ALGEBRAIC NUMBER THEORY

EXAMPLE 4.25. Let c7 be a primitive 7-th root of unity. If K = Q(<T), then ~1 = 0 and r2 = 3.

EXAMPLE 4.26. If K = Q(cT + C;l), then we have r1 = 3 and

We will explain below only Example 4.22. Examples 4.25 and 4.26 are parts of Iwasawa theory. They are also related to 54.4.

Let us use Dirichlet’s unit theorem to prove a statement related ’ ’ to Pell’s equation (§0.4), in particular Fermat’s Proposition 0.6.

PROPOSITION 4.27. Let N be a natural number that is not a

square. Define

PN = {(x, y) E Z x Z ) x2 - Ny2 = H},

p& = {(x,!/) E PN 1 32 > 1, !/ > 1).

(1) There is a bijection 0: PN + Z[m]‘; (x,y) H 2 + yfl

between the multiplicative group Z[fl] x consisting of all the

units in Z[JN] and the set PN.

(2) Let (XO,YO) b e an elem,ent of Pf, whose x-coordinate is the

smallest. Then, (x0, yo) is an element of PA whose y-coordinate is also the smallest, and we have

Z[Jlvl” = {&(x0 + yoJNy 1 n E Z},

Q(Ph) = ((x0 + y0JN)” 1 n 2 l}.

From Proposition 4.27(2) we see that Z[filx = {k(l + fi)lL 1 n E Z}, since both z- and y-coordinates of (1,l) E Pi are clearly minimal in Pk.

PROOF OF PROPOSITION 4.27. We show (1). The map

f:Z[v%]+Z; x+yv%-+(x+yv%)(x-ym)=x”-Ny”,

4.2. THE HEART OF ALGEBRAIC NUMBER THEORY 123

(z, y E Z) preserves multiplication. Thus, it maps the units in ;Z[fi] to the units in Z, namely, fl. Therefore, we have

x + yv% E I+‘%]’ M x2 - Ny2 = zkl

for any 2, y E Z. (1) follows from this.

To prove (2) we note that if u = ~7: + yfi E Z[filx (2, y E Z), we have

{&IL, &u-l} = {x + y&v, x-yyJN,-x+yJN,-x-y&v). \

Therefore, if u # 61, one and only one of fu and fu-’ belongs to

0(%). We prove (2) using Dirichlet’s unit theorem. First we show that

the group Z[&V] x is an infinite group. Let K = Q(m). We have

Z[&V] C OK, and it is easy to see that there exists a natural number

m such that mOK c Z[fi]. By Dirchlet’s unit theorem 0; has an

element u of infinite order. We prove that un E ZIJNlx for a certain n > 1. Since (O,/mO,)X is a group of finite order, there exists n > 1 such that the image of un in (O,/mOK)’ is 1. Thus, both

n - 1 and U-~ - 1 belong to mOK, and both un and uAn belong to

;,fi]. H ence we obtain un E Z[V%]’ . Therefore, Z[fi] ’ is an infinite subgroup of 02 E Z@iZ/2Z, and

it contains fl. Thus there exists E E Z[&Vlx such that Z[fi]’ = {ztzn / R E Z}. By replacing E by 3~ or fe-’ if necessary, we may

assume that E E d(Pf,i). Let E = XI+ ylv’% (21, y1 natural numbers). Then for n > 2, we have

(Xl + y1 JN)” = 2’ + y’fi, x’, y’ E z, 5’ > Xl, y’ > y1.

This means that (21, yi) is an element in PA whose z-coordiante and

y-coordinate are minimal. Hence we have @(PA) = ((51 + ylfi)” ) n 2 l}. 0

Finally, we prove Fermat’s Proposition 0.6.

PROOF OF PROPOSITION 0.6. Ifo E Z[v’8]‘, we have f(o”) = f(a)” = (11)2 = 1, where f is the map defined above. Therefore, elements (z> y) in the subset of P,TJ that corresponds to the infinite set

{cy” 1 a E Z[&V]x} through Proposition 4.27(l) satisfy the equation x2 - Ny2 = 1. Therfore, this equation admits an infinitely many soulutions. cl


4.3. The class number formula for imaginary quadratic fields

In order to study the arithmetic of an algebraic number field, it is important to know its class number. In this section we explain how the class number of an imaginary quadratic field is related to a value of a < function, and how it can be computed using this relation.

Let K be an imaginary quadratic field (a quadratic field that cannot be in R). We have K = Q(e), where m is a square free integer satisfying m < 0. Define

N= I;“‘, i

ifmcl mod4,

m if m E 2,3 mod 4.

By computing ( )

y using the quadratic reciprocity law, we see that

there is a unique Dirichlet character

x: (Z/NZ)X + {il} c Cx

such that for any prime number p not dividing m we have

(4.6) m

(-1 = x(p mod N)

P

(see Question 4 in Chapter 2). The character x can be expressed explicitly as follows. For an integer a relatively prime to N we have

x(u mod N) = (n (:)) O(U), I

where 1 runs all odd prime numbers dividing m, and e(a) is defined as follows.

(1) If m E 1 mod 4, then 19(a) = 1. (2) If m E 3 mod 4 and a E 1 mod 4, then 6(u) = 1; if m E 3

mod 4 and a E 3 mod 4, then e(u) = -1. (3) If m is even, then 19(u) = 1 for a E 1, 1 - m mod 8 and

0(u) = -1, otherwise.

The above definition of x may seem complicated. It is possible to define it in a simpler way using the material in Chapter 5, $5.2 of

Volume 2 as follows. Since K c Q(<N) (<N is a primitive N-th root of unity), we define x using Galois theory as follows:

(Z/NZ)X ” Gal(Q(C,v)/Q) ir) Gal(K/Q) ” {*l} c cx.

4.3. THE CLASS NUMBER FORMULA 125

(We will explain the first isomorphism in Chapter 5 in Volume 2. (*) is the restriction of an automorphism of Q(&) to K.)

THEOREM 4.28. Let K be an imaginary quadratic field and m, N, and x as above. Let hK be the class number of K, and wK the number of roots of unity contained in h’. Then we have

WKfi hK = yL(o,X) = FL&X).

Theorem 4.28 will be proved in Chapter 7, 57.5 in Volume 2.

QUESTION 4. Show that WK = 4 if K = (@(J-l), WK = 6 if K = Q(&3), and UJK = 2 for any other quadratic imaginary field K.

By Corollary 3.21 we have

COROLLARY 4.29. Let K, m, and N be as above. We have

hK = -$c ax(a). a=1

Theorem 4.28 or Corollary 4.29 is called the class number formula for the imaginary quadratic fields. Let us compute the class number of some examples using the class number formula.

EXAMPLE 4.30. K = Q(a). Then, we have WK = 4, N = 4, and x : (iZ/4Z)’ + Cx is given by x(1 mod 4) = 1, x(3 mod 4) = -1. By Corollary 4.29 we have

hK = -& eax(a) = -i(l - 3) = 1.

a=1

Note that using Theorem 4.28, we have

hK = wKV@ 4x2 FL&X) = 7 . L(l, x) = ; . L(l> x) .

Therefore, Leibniz’s formula

L(l,x)=l-~+:-:+~-lll+...--a

shows that hK = 1. It seems rather mysterious that Leibniz’s formula is related to the

fact that the class number of Q(n) is 1. This is the entrance to the “third mystery of the C function”.

126 4. ALGEBRAIC NUMBER 'THEORY

EXAMPLE 4.31. K = Q(a). Then, we have WK = 6, N = 3, and x : (Z/S@’ + cx is given by x(1 mod 3) = 1, x(2 mod 3) = -1. By Corollary 4.29 we have

hK=-- 2 z 3 -&zx(u) = -(l - 2) = 1. a=1

Using Theorem 4.28 once again, we have

h =sxa K ~. L(l,x) =

27r 7T

Thus, Euler’s formula L(l,x) = & expresses the fact that the

class number of Q(a) is 1. Note that even if we do not know that the exact value of L(1, x)

is &, we may be able to obtain hK = 1 from the formula hK =

@ . L(1, x). For, the formula 7r

L(l,x) = 1 - ; + ; - ; +. . . < 1

implies

hK = *. L(l,x) < !+ < 2, T 7r

and hK = 1 follows from this inequality since hK is a natural number. As we can see from this example, we may be able to compute hK from

the class number formula hK = eL(l,y), together with some

approximate value of L( 1, x) .

EXAMPLE 4.32. K = Q(m). We have WK = 2 and N = 4 x 26 = 104. In the proof of Proposition 2.8(3), we calculated (fi) for an integer a relatively prime to 104. Using this, we see that x(u) = 1 if and only if a mod 104 is one of the following:

1, 3, 5, 7, 9, 15, 17, 21, 25, 27, 31, 35, 37,

43, 45, 47, 49, 51, 63, 71, 75, 81, 85, 93.

It follows from this that Cp!i UX(U) = -624, and thus we have

hK = - & x (-624) = 6.

QUESTION 5. Using the class number formula for the imaginary quadratic

fields, find the class number of the following fields: Q( J??), Q(JT), Q(Jq)

and Q(J-10).

4.4. FERMAT’S LAST THEOREM AND KUMMER 127

Baker and Stark proved in 1967 that the only imaginary quadratic fields whose class number is 1 are the following nine fields:

Q(J-i), Q(J--2), Qw-1, Q(v-3, Q(J-il),

Q(m), Q(m), Q(d=@), Q(d=%).

Gauss conjectured that there are infinitely many real quadratic fields whose class number is 1, but this assertion has not been proved to this date.

4.4. Fermat’s Last Theorem and Kummer

In order to show Fermat’s Last Theorem:

If n 2 3 and (z, y, Z) is an integer solution to x7L + yTL = zn,

then zyz = 0,

it is sufficient to show this assertion when n = 4 and when n is an odd prime number. For, if Fermat’s Last Theorem holds for m and n = m . I-, the equation xn + yn = 2” implies xyz = 0 because we have (z~)~~ + (y”)” = (z’)~.

We proved the case n = 4 in Chapter 1, 51.1 and the case n = 3 in Chapter 4, $4.1. We consider the case where n is an odd prime number greater than or equal to 5. Following Kummer, we divide this into two cases: the case where none of x, y, z is divisible by p (the first case), and the case where one of 5, y; z is divisible by p (the second case). Kummer proved Fermat’s Last Theorem in the case n = p under the assumption that the class number of Q(&,) is not divisible by p. In the following we discuss Kummer’s proof in the first case.

(a) Proof of the first case.

PROPOSITION 4.33. Let p be a prime number greater than or equal to 5. Suppose that the class number of Q(&) is not divisible by p. If none of x, y, z E Z is divisible by p, then x, y, z do not satisfy

xp + yp = 2.

Unlike Q(&), which we considered in relation to the equation x3 + y3 = z3 in $4.1, the law of unique prime factorization in Q(&,) often fails. This is where the difficulty lies. (In fact, it is known that the class number of Q(6) is not 1 if p is greater than or equal to 23.) In the proof of Proposition 4.33 below we overcome this difficulty

128 4.ALGEBRAIC NUMBER THEORY

by studying the ideal class group and the unit group. The ideal class group appears in Proposition 4.33 under the form of the “class number”, and the unit group plays an important role in the proof of Proposition 4.33 (see Lemma 4.36 below). The following lemma replaces Lemma 4.2, which was proved in 54.1 using the unique prime factorization.

LEMMA 4.34. Let K be an algebraic number field, al,. , a, and b nonzero ideals of OK, k a natural n,umber and al . . . aT = 6”. Fur- thermore, if i # j, we suppose that a, and a3 are relatively prime (i.e., there is no prime ideal dividing a, and a3 simultaneously). Then for each i there is a nonzero ideal bi in 0~ such that ai = 6:.

This lemma can be proved by considering how many times each prime ideal of OK appears in the prime ideal decomposition of

al,..., a,. and 6. This is similar to the proofs of Lemmas 1.7 and 4.2, where we used the unique prime factorization of numbers.

We will prove the following Lemma 4.35 in Chapter 6, 56.3(e) in Volume 2.

LEMMA 4.35. Let p be a prime number. We denote cP by <, and Q(c) by A for simplicity. Then we have:

(1) A = WI. (2) [Q(c): Q] = p - 1 (left-h an sz d .d e is the degree of jield ezten-

sion). (3) The only roots of unity in Q(c) are of the form *(p-th root

of unity) . (4) The ideal (1 - <) zs a p rzme ideal of A, and (p) = (1 - 0P-l

is the prime ideal decomposition of the ideal (p) in A. (5) For 1 < i 2 p - 1, we have (1 - <) = (1 - <“).

PROOF OF PROPOSITION 4.33. We keep the notation < and A of Lemma 4.35. We suppose (z, y, 2) is an integer solution of zp+y” = 9 satisfying p { xyz, and we derive a contradiction. By dividing by t,he greatest common divisor of (x, y, z), we may assume that the greatest common divisor of (2, y, 2) is 1. Because of the equation XP + yp = 9, a prime factor of any two of x, y, z divides the third. Thus, x, y, z are pairwise relatively prime. Moving y” to right-hand side and factoring the equation in A = Z[<], we have

(4.7)

1.4. FERhlA’l”S L4S’I’ THEOIIE~l AN13 KU~IIVE:H 129

Using the fact that p does not divide the class number, we show that there exist a unit u in A and an element a in A such that

(4.8) z - <y = u. ap.

To do so, we first show that the ideals (Z -<‘y) (0 < i < p- l), which appear in the right-hand side of (4.7), are pairwise relatively prime.

Let 0 < i < j 5 p - 1 and let p be a nonzero prime ideal that divides both (Z - <“y) and (z - <J y). It follows from the fact z - <“y, z - CJy E p that (C’ - <J)y/, (<” - <J)z E p. Thus, we have

cc1 - PHY, 2) c P. s ince y and z are relat)ively prime: we have (y,z) = (1). By Lemma 4.35(5), we have (1 - <I-“) = (1 - <). Since (1 - <) is a prime ideal (Lemma 4.35(4)), we have (1 - <) = p. By (4.7) we have xP E p. and t)hus we have x E p. Since p n Z = (p), we have p 1 x, which contradicts the hypothesis p + xyz.

It follows from Lemma 4.34 that the ideal (z-<“y) (0 < i 5 p- 1) is the pth power of an ideal 6, in A.

If we set (Z - <y) = ap, the p-th power of the class of a in Cl(A) is the identity element. But, the order of Cl(A) does not divide p, and thus the only element whose p-th power is the identity is the identity itself. Therefore, a is a principal ideal. Let a be a generator of a. Then we have (Z - <y) . aep E AX, and (4.8) is proved.

Before going further, we show that we may assume y $ --z mod p. If y E --z mod p? then we use the substitution xi = -2, zi = -x. Then we have ~7 + y” = z:. It suffices to show that y $ -21 mod p. If not, we have IC E y E -Z mod p, and thus we see 2s” E -xv mod p by substituting in xp + yP = zP. S’ mce p # 3, we have x 3 0 mod p,

which is a contradiction. In order to derive a corkradiction from (4.8) we use the following

Lemmas 4.36 and 4.37.

LEMMA 4.36. Let p be an odd prime number, and let < und A be the same as in Lemma 4.35. Let r : Q(() + Q(c) be the complex conjugution, and let B = {a E A 1 T(Q) = a}. I f p,, = {CL 1 0 < i < p - l}, then we have

AX = pP x BX.

We will prove Lemma 4.36 later using Dirichlet’s unit theorem.

LEMMA 4.37. We denote also by r the automorphzsm of 2 = A/pA mduced by 7. Let 2 = {a E A ( ~(01) = 0~). Then we have the following:

(1) ,4 basis of 2 over IF, is given by {Cl 1 1 ( % 2 p - l}.


(2) A basis of B ower IF, is given by {<’ + CPi 1 1 < i < q}.

(3) xp = {a” E 2 ) c-u E A} is equal to IF,, and it is contained

in B.

Let us show that (4.8) induces a contradiction once we admit Lemmas 4.36 and 4.37. First, it follows from Lemma 4.36 that there exist <’ E pp and u E BX such that u = <‘u. We have u mod pA E B, and from Lemma 4.37 we have ap mod pA E B. Thus, we have

<‘-l (z - <y) mod pA = uap mod pA E B. We divide into cases according to C’. In the following we omit “mod PA” for simplicity.

(a) If <’ = 1, then z - <y E B. Since z E B, we have y< E B. It follows from Lemma 4.37(l) and (2) that y = 0 mod p, which contradicts the assumption.

(b) If C’ = <, then z . C-l - y E B. Since y E B, we have

z C-l E B. It follows from Lemma 4.37(l) and (2) that z E 0 mod p, which is a contradiction.

(c) If I’ # 1, <, then z. C’-’ - y<<‘-l E B. By Lemma 4.37(l)

and (2) we have C’ = <<‘-l and y z -z mod p. This also contradicts the assumption.

Therefore, if we admit Lemmas 4.36 and 4.37, we have finished the proof of Proposition 4.33.

We now prove Lemmas 4.36 and 4.37. We first prove Lemma 4.37.

PROOF OF LEMMA 4.37. From Lemma 4.35(l), (2) and the fact 1+ < + ‘. . + p-1 T 0, we can take {<” 1 1 5 i < p - l} as a basis of

A over Z. Thus, {<” 1 1 < i < p- l} is a basis for 2 = A/pA over IF,. This shows (1). Since 7 sends <” to <-“, (2) follows easily from (1).

Let us show (3). Take cy = CrI; aici E 3, a, E F,. In a ring,

such as A, in which p equals 0, the p-th power map preserves addition and multiplication. Thus, we have ap = CTil aL E F,. Therefore,

we have A” = F,. The other assertion is clear. 0

PROOF OF LEMMA 4.36. It suffices to show that the canonical map pup + AX/B x is an isomorphism. Consider the homomorphism of groups f : Ax + AX given by f(a) = Q/~(Q). The kernel of this homomorphism is given by (0 E AX I (u = T(D)} and it equals Bx Thus the image f(AX) is isomorphic t,o AX /BX. On the other hand, the restriction of f to pp is the square map pp + pp, and it, is an isomorphism onto pp. Thus, it suffices to show that the image f(AX)

equals fbp) = Pp.

4.4. FERMAT’S LAST THEOREM AND KUMMER 131

First we show that f(AX) is finite. It suffices to show that BX has finite index in AX; i.e., the ranks of AX and BX as finitely

generated abelian groups are the same. (An T such that ” 2ZBr @ (finite abelian group) is called the rank.) Let

K = {cl E a$(() I T(Q) = a} = Q!(( + <-I).

The ring B is the ring of integers of K. We compute the rank of AX and of BX using Dirichlet’s unit theorem. First Q(c) does not have a

real place and the number of the complex places is $ [Q(C) : Q] = q.

Thus, by Dirichlet’s unit theorem, the rank of AX is &$ - 1. Next, K does not have a complex place and the number of the real places is[K:Q]=%$. Th us, by Dirichlet’s unit theorem the rank of BX

is also &$ - 1. Thus the finiteness of f(AX) g Ax/B’ is proved.

Therefore, the image f(AX) consists of roots of unity in Q(&). It follows from Lemma 4.35(3) that the set of all the roots of unity in Q(c) equals {+$ 1 0 5 i 5 p - 1). We have already seen that pLp c f(AX), and thus it suffices to show -1 @ f(A”) in order to prove ,+ = f(AX). It suffices to derive a contradiction assuming cy E AX and T(Q) = --(Y. It follows from Lemma 4.35(5) that 7 preserves the ideal (< - 1). S’ mce A/(< - 1) g IF,, the action of 7 on A/(< - 1) is trivial. This contradicts the fact T(Q) E -a~ mod (< - 1). 0

(b) Kummer’s criterion. For which prime number p does Kummer’s assumption, “p divides the class number of a(&)“, hold? Kummer proved the following theorem, which relates this question to

the values of the < function. The following theorem is called Kum- mer’s criterion.

THEOREM 4.38. Let p be a prime number. Then the conditions (i), (ii), and (iii) are equivalent.

(i) The prime number p does not divide the class number of

Q,(G). (ii) For any negative odd number m the numerator of ((m) is not

divisible by p. (iii) For any negative odd number m satisfying Irnl < p - 4, the

denominator of C(m) is not divisible by p.

Thanks to the relation between c(m) and ((1 -m) shown in 53.3, condition (iii) is equivalent to the following condition (iii)‘.

(iii)’ For all positive even numbers f less than or equal to p - 3, the numerator of <(r)r?’ is not divisible by p.

132 4. AI,GEUI~AI( NUMBER THEORY

Using Example 3.23, we see (iii). The above theorem implies that all the prime numbers p less than or equal to 17 satisfy the condition, “p does not divide the class number of Q(&)“, but 691 divides the class number of Q(&i).

We discuss Kummer’s criterion at the beginning of Chat,per 10 in Volume 3.

Summary

4.1. A finite extension of the rational number field is called an algebraic number field. Algebraic number theory is a poweful theory that studies algebraic number fields.

4.2. Just as we have the ring of integers in the rational number field, an algebraic number field K contains a ring called the ring of

integers of K and it is denoted by 01~. In 0~ each element may not be factored uniquely into the product of prime elements, but any ideal may be factored uniquely into the product of prime ideals.

4.3. For an algebraic number field two important groups, the ideal class group and the unit group, are defined. These groups measure the difference between numbers and ideals. There are two important theorems: finiteness of the ideal class group and Dirch- let’s unit theorem.

4.4. These groups are related to { functions. In this chapter the relation between the ideal class groups of quadratic number fields and C functions (the class group formula for the quadratic number fields) is discussed.

Exercises

4.1. Let p be a prime number. Show that the following two properties (i) and (ii) are equivalent, using the fact, that the class number of Q(n) is 1.

(i) There exist integers z and y satisfying p = x2 + zy + 2y2. (ii) p E 1,2,4 mod 7 or p = 2,7.

4.2. Let n be a natural number. Show t,hat, the following t,wo properties (i) and (ii) are equivalent.

(i) There exist integers z and y satisfying n = 2’ + y2.

EXERC‘ISES 1x3

(ii) ord,(n) is even for any prime number p congruent to 3 modulo 4.

4.3. Let p be a prime number congruent t,o 1 modulo 4 and let n be a natural number. Show that there exists a unique triangle with integer sides such that the length of the hypotenuse is p” and the greatest of common divisor of the length of three sides is 1.

4.4. Show that the unit group of Q(d) is {~t(2+&)” 1 rl E Z}.

4.5. Let a and b be fractiona. ideals of a Dedekind domain A (see Appendix A. 5h.2). Suppose that

a= rJPQ> b = npb,

P P

are prime factorizations of a and 6. (Here p runs tjhrough all the nonzero prime ideals of A: and up and b, are integers that are nonzero only for a finitely many p’s) Define cP = max(aZp, bp) and d, = min(ap, hp). Show that the prime factorization of t)wo fractional ideals of A, a n b and a + b = (2: + y 1 z E a. y E 6) are given by

(see Appendix A, 9A.2):

P

4.6. Using t,he fact that the class number of Q(a) is 2. and thus it is not divisible by 3, show that (2, ;y) = (6,14) is the only

natural number solution of y2 = 2” - 20. (In 54.4 we used t,he fact that the class number Q(&) is not divisible by p to prove Fermat‘s Last Theorem in the first case. Use a similar method.)

APPENDIX A

Rudiments on Dedekind domains

In this appendix we give a summary on the fundamentals on Dedekind domains. In what follows a ring means a commutative ring.

A.l. Definition of a Dedekind domain

A ring A is a Dedekind domain if A satisfies the following conditions:

(1) A is a Noether ring. (2) A is an integrally closed domain. (3) Any nonzero prime ideal of A is maximal.

Let us explain the terminology that appears in the above definition. A ring A is a Noether ring if A satisfies the following condition:

(1) Any ideal of A is finitely generated.

This condition is equivalent to any of the following conditions:

(2) If ai c a2 c a3 c . . is an ascending chain of ideals of A, then there exists N such that aN = aN+i = a&7+2 = .

(3) If Q is a nonempty set of ideals of A, then there exists an ideal a in Q satisfying the condition: If b E 9 and b > a, then b = a.

(4) Any submodule of a finitely generated A-module is again finitely generated.

A ring A is a domain if A is different from (0) and if it satisfies the condition:

for a, b E A ab = 0 implies a = 0 or b = 0.

If A is a subring of B, an element z in B is said to be integral over A if x satisfies an equation with coefficients in A:

xn + UlX n-1 +. + (Jr2 = 0 (a, E A, n is a natural number ).

13 :,

The set {X E B 1 J: is integral over A} is a subring of B. This subring is called the inte.qml closure of A in B. If A is a domain, the integral closure of A in its field of fractions is called the integral closure of A. If the integral closure of A equals A itself, then A is said to be integrally closed.

An ideal a in A is a prime ideal if the quotient ring A/a is a domain. This is equivalent to (1) and (2) below:

(1) If ab E a, then a E a or b E a.

(2) 1$ a.

An ideal a in A is called a maximal ideal if the quotient ring .4/a is a field. This is equivalent to (1) and (2) below:

(1) An ideal of A containing a is eit,her A or a.

(2) I@ a.

A maximal ideal is a prime ideal, but the converse does not, hold in general. For example, the zero ideal of Z is a prime ideal but not a maximal ideal.

EXAMPLE A.1 (Dedekind domain). (1) A principal domain (see Example 4.4) is a Dedekind domain.

(2) Let A be a Dedekind domain, and K its field of fractions. If L is a finite extension of K and B is the integral closure of A in L, then B is a Dedekind domain.

A.2. Fractional ideal

Let, A be a domain. A fractional ideal of A is a nonzero finitely generated A-submodule in the field of fractions K of A. For a nonzero element a E KX , the set (a) = {ab 1 b E A} c K is a fractional ideal of A. Such a fractional ideal is called a principal fractional ideal.

For fractional ideals a and b of A we define the product a. b as the A-submodule of K generated by a. b (a E a, b E 6). I f for a fractional ideal a of A there exists a fractional ideal b satisfying a. b = A, a is said to be invertible. Since (a) . (a-‘) = A, any principal fractional ideal is invertible.

The set D(A) consisting of all invertible fractional ideals of A is an abelian group under the multiplication defined above. The ideal A is the identity element, and the inverse of a E D(A) is given by a-‘={bEK/bacA}. ThemapKX ----f D(A) given by a H (a) is a homomorphism of groups, and its kernel equals AX.

THEOREM A.2. Let A be a Dedekind domain and 5’~ the set of

all nonzero prime ideals of A. Then

R.2. FRACTIONAI> InEAL I37

(1) Any fractional ideal of A is in,vertible. (2) Let Z(‘.l) be the free abelian group generated by S.4. Then

the natural map

Z(““) + D(A); (eP)PESa H n p’p

PESA

is an isomorphism of groups.

(3) For a = n p’p an,{1 b = n p”; , a c b is equivalent to the fact

that for any p we have ep > eb.

For a Dedekind domain A the cokernel of the natural map KX + D(A) is called the ideal class group of A and written Cl(A). We have Cl(A) = {fractional ideals}/{p rincipal fractional ideals}. A is a principal ideal domain if and only if Cl(A) = 0.

Answers to Questions

In what follows we write ord,(a) to indicate which power of the prime

number p divides the integer a (see 51.3 and 52.4).

Chapter 1

1.1. Suppose that a is the square of a rational number T. For any

prime number p we have ord,(a) = 2ord,(r). Since ord,(a) 1 0, we have

ord,(r) 2 0. The number T is an integer since we have ord,(r) 2 0 for all

prime numbers p.

1.2. Suppose that p is a prime factor of a,. By hypothesis we have

ord,(a,) = 0 for all j different from i. Thus, we have ord,(ar . ..a.) =

ord,(a,). On the other hand, since ai a, is a k-th power, ord,(ur a,)

is a multiple of k. Thus, for any prime number p, ord,(a,) is a multiple of

k. This implies that a, is the product of integers of the form pkn’ (m is a

natural number), and thus u7. is a k-th power.

1.3. Let (~,y) be the coordinates of the nonzero element P in E(K).

Then, the coordinates of -P are (2, -y). The condition 2P = 0 is equiva-

lent to the condition P = -P. Thus, it is equivalent to y = -y, i.e., y = 0.

If K is an algebraically closed field, there are three nonzero elements P in E(K) whose y-coordinate is 0. Therefore, {P E E(K) 1 2P = 0} is a

group of order 4. Since twice of every element in the group E(K) is 0, we

see that E(K) is isomorphic to Z/22 G? Z/22.

1.4. The first part is easy. As for the second part, take A = Q. Then,

we have A/2A = {0}, but A is not finitely generated.

Chapter 2

2.1. For example, (y, g). Th’ is is the point of intersection between the

circle and the line with slope -3 passing through (2,l).

2.2. It suffices to find a rational point on the circle r2 + yi = 1 that

is very close to the rational point (h55). The slope of the line joining

C-1,0) and (&, &) is fi - 1 = 0.414 , while the slope of the line

joining (-1,0) and ($$. =) is & = 0.416’.., as we have seen in the text.

Thus, it suffices to take the line passing through (0, -1) whose slope is 0.415

and to find the other point of intersection with the circle. A calculation

shows tha.t the coordinate of the other point of intersection is given by (&cug), and we see that

33111’ + 33200’ = 46889’.

2.3. (@ = ($) (;) = (1’, (;)(-1,“” = (5).

2.4. Factor m as rn = 11 lk .T, where 11, , lk are odd prime numbers

and T E {&2” / n. > O}. If n, is odd, we have T E {fl}, and

(F) =(~)...($)(k)

= G).~4> x (number determined by p mod 4).

Similarly, if rrt is even, we have

(;) = C)...(%> x (number determined by p mod 8).

2.5. The fact that the circle gr’ - &y2 = 1 does not have a rational

point can be seen from the fact that (g, -$)P = (15, -l)P = -1 if p = 2

or p = 3.

2.6. ord,(C:l,,c’ - &) = ord,, (-K) > n + 1.

2.7. The equation (2.9) is equivalent to c,“=,, 6 x (-5)’ = 1 (5-addically).

The latter shows that we have Crl,, 6 x (-5)’ c 1 mod 5” when ?TL is

sufficiently large.

2.8. Wehave~=~=1-3+32-33"+3'-3~++3"-...=61-33;'+

3” - Therefore, 61 is the inverse of i.

2.9. If N is a natural number greater than 1, then the N-adic expansion

of a real number cy is to express cy as

cY= 2 a,3 ‘I, a,,~{O,l,..., N-l}.

Il=nL

On the other hand, the p-adic expansion of a p-adic number is of the form

c,TTV=,, a7&. The difference is that in the p-adic expansion of a real number,

the terms p7’ with negative n may appear infinitely many times and the

terms p” with positive n appear only finitely many times, whereas in the

p-adic expansion of a gadic number the terms p” with negative n may

ANSLVERY ‘I‘0 QUES’l’IONS 141

appear only finitely many times and the terms P’~ with positive n appear

infinitely many times.

2.10. The existence of a square root of a follows from Proposition 2.18

and the fact that &1 are squares in IF5

2.11. If p # 2, it follows from Proposition 2.18 that

Q, has a square root of -1 u Fp has a square root of -1

If p = 2, it follows from Proposition 2.18 and the fact -1 $ 1 mod 8 that

@ does not have a square root of -1.

2.12. It follows from field theory that any quadratic extension of a field

K of characteristic different from 2 is of the form K(A), (u E K, J71 @ K),

and

K( \/;;) = K(A) u ab5 ’ is a square in K

Thus, the correspondence that associates a mod (KX)” (u E K, fi 6 K)

to K( ,,6) is a one-to-one correspondence between the quadratic extensions

of K and the elements of KX/(KX)* different from the identity. If p # 2, then the order of Q,” /(Qc)” is 4 (Proposition 2.19(l)). Thus, the number of

quadratic extensions of Qj is 4 - 1 = 3. Furthermore, the group Qc /(at )”

consists of classes of 1, 2, 5 and 10, and thus Q5 (A), Q5 (A) and Q, ( fl)

are all the quadratic extensions of QR.

Chapter 3

3.1. By Proposition 3.3(l) we have

h,(i) = -1.1 CC 1 1

-fP 2 27~i ,,FE i + n 2-n > =&~A&

On the other hand, we have hr (i) = - & ,‘~~~~~~~~~~.

3.2. Use the formula

1 1 1 = - (n2 : 1)” -q&p 4(i ~ n)” + 4i i+n +&

>

3.3. The image of x is the set of all the n-th roots of unity {C:; 1 1 <

T < n} for some rt > 2. Let k be the order of the kernel of x. For each

‘r satisfying 1 < T < n, x takes the value <ii on I; different elements in G.

Thus, C,ec; x(a) = x;:-, k. C,‘; = 0.

142 ANSWERS 7’0 QUESTIONS

3.4. We have

Thus, we have

-c (%g) +c(s)) =;li~2-2”)c(s4+2+;

= !@I G(s - l)C(s) + 2 + g

= ; - 21og(2).

Here, we used lim,,r(s - l)<(s) = 1 (Proposition 3.15(2)) to prove the last

equality.

3.5. By Proposition 3.24(l), a prime factor of the denominator of C(m)

satisfiesmzlmodp-1. Sincep-ldividesl-m,wehavep-121-m.

Hence p 5 2 - m.

Chapter 4

4.1. Factor the equation as x3 = (y+i)(y-i), and use a similar argument

as in Proposition 0.11 to obtain

y+i=(a+bi)“, U,bEZ

Comparing the imaginary parts of both sides, we obtain 1 = 3a2b - b” =

(3~’ - b’)b. Thus, we have b = 51. The rest is easy.

4.2. Factor the equation as z3 = (y + &ii)(y - &ii), and use a

similar argument as in Proposition 0.10 to obtain

y+&ii= (a+b1+y)3, a,b E Z.

(Here, we used the fact that the only common prime factors of y + a

and y - &ii are *&ii and A2.) Comparing the imaginary parts of

both sides, we obtain 1 = 3 (a + $)’ $ - 11 (g)“. From this we.obtain

(3a2 + 3ab - 2b2)b = 2. Thus, we have b E {fl, ~t2}. The rest is easy.

ANSWERS TO QUESTIONS 143

4.3. Let m be an integer that is not divisible by any square except for 1.

LetK=Q(J\/m) ,o=z+y&(x,y~Q) andcr’=z-ye.

(i) First we show that LY E 0~ is equivalent to the fact that the rational

numbers 01 + o’ = 2s and cro = x2 - my2 both belong Z If o E OK.,

then by replacing (Y by o’ in the equation on + crcrnpl + ‘. + cn = 0

(n > 1, ~1,. , cn E Z), we see that 01’ E OK. Therefore, we have 01 + cy’,

eta’ E OK. Thus, these numbers belong to 0~ n Q = Z. Conversely, if we

have cx + CY’, cycv’ E Z, then LY satisfies the equation o2 + crcy + cs = 0 with

cl = -(o + o’) and c2 = oo’. This implies that cy belongs to OK.

(ii) By (i) it suffices to show the following: For x, y E Q

2x,x2-my2EZ I x,yEZ,

ifmE2,3mod4,and

2z,z2-my2EZ u 2x,2yEZandx-yEZ,

if m E 1 mod 4.

(iii) Show first that if x, y E Q satisfies 2x, x2 - my2 E Z, then we

have 2y E Z. If 1 is an odd prime number, it follows from ordl(x) 2 0

and x2 - my’ E Z that ordi(m) + 2ordl(y) > 0. Since ordl(m) < 1, we

have 2ordl(y) > -1. Thus we have ordl(y) > 0. Since ordz(x) 2 -1 and

x2 - my” E $ we have ords(m) + 2ords(y) 2 -2. Since ordn(m) < 1, we

have 2ord2(y) 2 -3. Thus, we have ords(y) > -1. Summing it all up, we

see that 2y E Z.

(iv) To show the equivalence in (ii), we may assume 2x, 2y E Z because

of (iii). Suppose 2x = u and 2y = v (u, u E Z). If m E 2,3 mod 4, it suffices

to show

2 u -mv2=0 mod4 u UEUEO mod2,

and if m E 1 mod 4, it suffices to show

2 u -mv2s0 mod4 I UEZI mod2

These are easy to show.

4.4. Similar to the proof or Proposition 4.1(5)

4.5. The answers are 1, 2, 2 and 2, respectively. As an example, we treat

the case Q(&2). We have WK = 2 and N = 8, and x : (Z/8Z)’ + Cx is

given by

x(1 mod 8) = x(3 mod 8) = 1, x(5 mod 8) = x(7 mod 8) = -1.

ByCorollary4.29wehavehK = -Y&C”,=, x(a)a = -$(1+3-5-7) = 1.

Answers to Exercises

Chapter 0

0.1. Suppose that the n-th root of 5 is a rational number and that it factors as fpP1 .. .psT (~1,. ,P,. distinct primes, e, integers satisfying e, # 0). Taking the n-th power, we have 5 = py” .‘.pF”“. This is a contradiction to the uniqueness of prime factorization since n 2 2.

0.2. If & + v’? is a rational number, so is 5 + 2&. Thus, & is a rational number. But we can show that I/% is an irrational number by a similar method as Exercise 0. 1.

0.3. 29 = 2’ + 5’, 37 = 1’ + 6’, 41 = 4’ + 52, 53 = 2’ + 72.

0.4. Combiningfactorizations 5 = (2+i)(2-i) and 13 = (3+2i)(3-2i), we have

652 = ((2 + i)(3 + 2i))2((2 - i)(3 - 2i))”

= (-33 + 56i)(-33 - 56i) = 33’ + 56”,

652 = ((2 + i)(3 - 2i))‘((2 - i)(3 + 2i))”

= (63 - 16i)(63 + 16i) = 63” + 162.

0.5. If x and y satisfy x2 - 2y2 = 1, then we have ($-a)(;+dq =

-$. Thus, we have 0 < E - ~‘2 < &. This shows that z becomes closer

to fi as y gets bigger.

0.6. It suffices to show that infinitely many pairs of natural numbers (2, y) satisfy iy(y + 1) = x2. Rewrite this equation as

(2y + 1)2 - 2(2x)2 = 1.

For n 2 1, define a, and b, by (1 + fi)” = an + b,&. We have

a: - 2b: = (an + b,h)(a, - b,h) = (1+ &)“(l - fi)” = (-1)‘“.

By expanding (1 + a)“, we see that a, = 1 + (even number) and b, =

n + (even number). Thus, if we take an even number as n, then we have

145

146 ANSWERS TO EXERCISES

2- 2bz = 1 with a, odd and b, even. If we set y = %$

zien we have, (2y + 1)2 - 2(2~)~ = 1.

andx= 3,

Chapter 1

1.1. Answer: The set in question consists of nine points 0, (0,&l),

(-@,*g), (-$‘%&,&a) and (-%C~,zt~), where <3 is aprim-

itive cube root of unity.

The method of finding these points: First, we see that 3P = 0 is

equivalent to 2P = -P. In general, if we denote by x(P) the x-coordinate

of P E E(C), P # 0, then we have

x(P) = x(Q) u Q = fP

for any P,Q E E(C). Thus, we have

3P= 0, P# 0 _ x(2P) =x(P) and Pf 0.

For any point P in E(C) satisfying 2P # 0 we have x(2P) = z~~‘c4pJfzc(l~)

(§1.2(1.4)). Therefore, x(2P) = x(P) if and only if x(P) = 0, -@, -NC3 or -@G.

1.2. Let m and n be relatively prime integers and let

A = l(m3 + 32n3)ml and B = 14(m3 - 4n3)nl.

Denote by D the greatest common divisor of A and B. In order to show

the inequality in question, it suffices to show that D is a divisor of 144.

For, if that is the case and if the x-coordinate of P is given by f (n # 0)

in lowest terms, then we have

H(x-coordinate of 2P) = H A

0 E = A max(A, B)

2 % max(m, n)4 = $H(x-coordinate of P)“.

Let p be a prime number. We have ord,(D) = min(ord,(A),ord,(B))

(since ord, indicates how many times p divides the number). If p is a prime

factor of D, then p does not divide n (since if it does, p does not divide m,

and thus p does not divide m3 + 32n3 and A). If p is a prime factor of D and p # 2, then p does not divide m (since if p # 2 and p divides m, then

p does not divide B). Thus, if p is a prime factor of D and p # 2, then we

have

ord,(D) = min(ord,(m3 + 32n3), ord,(m” - 4n3))

5 ord, ( (m3 + 32n3) - (m3 - 4n”))

= ord,(36n3) = ord,(36).

Hence, we have p = 3 and ordy (D) < 2.

ANSWERS TO EXERCISES 147

Next, we consider ordz(D). If m is odd, then ordz(A) = 0. If m is even, then ordz(m” - 47~‘) = 2 since n is even. Hence we have ordz(B) = 4. Therefore, D is a divisor of 24 ~3~ = 144, and thus the inequality in question is proved.

If T > 6, then we have the inequality &r4 > r. I f a rational point P on the elliptic curve in question satisfies H(z-coordinate of P) > 6, then we have H(z-coordinate of P) > H(z-coordinate of 2P). Then the height of the x-coordinate of P, 2P, 4P, 8P, 16P,. . are all different. Thus, these points are all distinct. This means there are infinitely many rational points on this elliptic curve. (To be more precise, we can show the following. If integers m and n satisfy m $ 0 mod 3 or n $ 0 mod 3, then m3 - 4n” $ 0 mod 9. This can be done by checking all the possibilities of 0 2 m 5 8, 0 5 n 5 8. Thus we see that D is a divisor of 24 x 3 = 48 and that

48 H(z-coordinate of 2P) 2 H(z-coordinate of P)”

If T > 4, then &r” > T. Thus, if P = (5, ll), then the z-coordinates of P, 2P, 4P, 8P, all have different heights. This implies that we see the existence of infinitely many rational points as soon as we find one rational point (5, ll).)

1.3. Since for (z,y) E X we have

x-y 2 (H

4k 1

x+Y +$+x2-xy+~2=-._

(x + Y)” 3 (x+y)3’

we have (A, z) E Y. This map is bijective since the map Y + X

given by (x, y) H (g, 2) t IS the inverse. (We omit the proof of the fact that we have (5, 2) E X for (x, y) E Y and that the compositions X + Y + X and Y + X + Y are both identities.)

1.4. The inverse is given by (x, y) H (&, $ + :).

1.5. We omit the proof of 1.5 since each verification is straightforward, as was the case with 1.3 and 1.4.

1.6. (i) Answer: (x, y) = (O,O), (2, f4). Reason: If (x, y) # (0,O) is a rational point on the curve y2 = x3 + 4x, then by considering the case k = -1 in Question 1.5, we see that g(z, y) = (2 - $, i (1 - 3)) is a rational point on the curve y2 = x3 - x. Prom Proposition 1.2 we know that this point is one of (O,O), (*l, 0). Therefore, we have i (1 - $2) = 0. Hence y = 0 or x = f2.

(ii) Answer: (x, y) = (&l,O). R eason: If (x, y) is a rational point on the curve y2 = x4 - 1, then by considering the case k = -1 in 1.4 and 1.5, the image of (x, y) by the map X + Y 4 E(K) given by (x, y) H (x2, xy) is a rational point on the curve y2 = x3 - x. Thus, we obtain xy = 0.


(iii) Answer: (z, y) = (0,12). R eason: Just as (ii), we see that if (z:, y)

is a rational point on the curve y2 = z4 + 4, then (x2, z:y) is a rational point

on the curve y2 = z3 + 42. Thus, it follows from (i) that (x2, zy) equals

one of (O,O), (2, *4).

Chapter 2

2.1. For example, $$ converges to 1 in 88, but it converges to 0 in QZ

The sequence $& converges to 1 in Qa (since 3” + 0), but it converges

to 1 in Ql.

2.2. Let f be an element of Horn (Z [t] /z, z [t] /z) For any n 2 1

we denote by fn the restriction of f to ( >

$Z /Z. The image of the map

fn : (*q/z + Z /Z is contained in the kernel [:I ( >

&Z /Zofthe

multiplication-by-p” map of Z 5 /Z ( [I

since every element of the subgroup

( > *Z /Z becomes 0 if it is multiplied by p”). Thus, fn is a homomorphism

from (&Z) /Z to ($Z) /Z, and it coincides with the multiplication-by-

czrL map of Z/p”Z for some element a,. Thus, we obtain a ring homomor-

phism

cp : Horn (z [i] P, z [i] P) + l&Z/p”Z; p(f) = (anJn>l. n

Conversely, we can find a ring homomorphism

1~ : lhr+Z/p”Z -+ Horn (Z [b] /z, z [ $1 /z)

as follows. Let (an)%>1 E l@,Z/p”Z. For 1c E Z b /Z there exists a pos- [I itive integer n such that z E

( > $Z /Z since Z [ ;] P = U,>l ($) /z.

We obtain a homomorphism f E Horn Z ( [t] /z, z [i] /z) by defining

f(z) = a,~. Define $((a,,),ri) = f. It is easy to check that $ o cp and

cp o + are the identities of Horn (Z [i] /z, z [t ] /z) and !im n~/pn~, re-

spectively. Hence, we have

n

2.3. For n # 0 define k = orda(n). From Proposition 2.14(4) we see

the following. Since 4 belongs to 1 + 3&, but not to 1 + 9&, log(4)

belongs to 323, but not to 9Za. Thus, nlog(4) belongs to 3kf1Zs, but not

to 3k+2Z:3. Thus, 4’” = exp(nlog(4)) belongs to 1 + 3k+1Z:s, but not to


1+ 3”+“&. Thus 4” - 1 belongs to 3k+1Zs, but not to 3”+“&. Hence, we

have ords(4” - 1) = lc + 1.

2.4. First, (1) follows from Proposition 2.18 and the fact that for an odd

prime number p we have

-2

(-> =l u pz1,3 mod8

P

Next, the equation x2 + y2 = -2 in (2) can be written as -ix’ - ay” =

1. The necessary and sufficient condition for the existence of z, y E U&

satisfying the equation is (-i, - $), = 1 by Proposition 2.20. But, if p # 2,

we have (-$,-f), = 1, (-i, -i), = -1. In order to show (3), it suffices

to show the existence of elements x, y and z in @ satisfying x2 + y2 + z2 =

-2 (since if p # 2, a solution of x2 + y2 = -2 satisfies x2 + y2 +02 = -2). In

Q2, 14 is very close to -2, and we have 12+22+32 = 14. Since $!j = -7 E 1

mod 8, it follows from Proposition 2.18 that there exists a E Q,” such that

a2 = 3;. Thus, we have -2 = 2 = (i)’ + (p)” + (z)‘.

Chapter 3

3.1. (1) Consider the Dirichlet character x : (Z/8Z) x + Cx given by

x(1 mod 8) = x(3 mod 8) = 1 and x(5 mod 8) = x(7 mod 8) = -1. Then

(1) is to find L(1, x). Since x(-l) = -1, it follows from Theorem 3.4 that

L(l,x) = -F ; (hl(C8) + hl(& - h(G) - hi(C)) = 5.

(2) Consider the Dirichlet character x : (Z/8Z)x + Cx given by

x(1 mod 8) = x(7 mod 8) = 1, x(3 mod 8) = x(5 mod 8) = -1. Then

(2) is to find L(2, x). Since x(-l) = 1, it follows from Theorem 3.4 that

L(2,x) = -F ( >

2 1 z (h2(<8) - h2(&) - WC3 + h2CC87)) = $x2.

3.2. (1) (1 - 2l.-“)<(s) = c;=, 5 - 2 c;& &s = 1 - +s + jk - $ +

‘-&+.... 5.3 (2) lim,++r+o(s-l)<(s) = lim,+l+s &.(l - & + & - & + ‘. ‘) =

&log2=1.

3.3. By calculating sr - ss - s:, + ~7, we have

( (l - G)(l - G8’)

-log (1-<;)(1-<,5) >

= (C8 - <; - <,j + <,7)q1,x),

where x is the same character as the one in Exercise 3.1(2). We have

-log ( (1 +1Jz)2) = 2JZULX).


Hence, L(1, x) = -& log(1 + a).

3.4. We omit the proof of absolute convergence, and we explain the

analytic continuation and the values at nonpositive integers. For simplicity,

we denote the sum over ni, , nk > 0 just by c. We have

r(s)(‘(s, 2; cl, , ck) = J-= 1

0

e-tt"$ . c (x + Clnl + + Cknk)’

= J-C e-(z+cllL1+...+cknk)uUSdZL 0 U

=JW o (1 - ,plU) ."T (1 - e-ck~)us~~ Let a > 0. Divide the integral sow = s,” + saw. Since eeZcu approaches

very rapidly as u goes to 00, the part s,” can be analytically continued

to the entire plane as a holomorphic function in s. Take a small enough

that 1 - eeCZ” (1 5 i 5 Ic) does not have a zero in 0 < IuI 5 a. Then, in

0 < u 5 a we have

epxu

c1 “‘ck (1 - e-clu). (1 - e-ck”) = u -k -g AnUn,

n=O

where A, is a polynomial in x, cl, , ck with Q coefficients. Thus, we have a w SC

s+n-k

Q...Ck. = A,. a 0 TX=0

s+n-k’

Therefore, [(s, xc; cl,. , ck) may be analytically continued to the entire

plane and we see that it is holomorphic outside 1,2, , k. If m is a negative

integer or 0, then we have

cl ck ((m; x, cl, . , ck)

= lim 1. A s-m l-(s) k-m z = z4k-m (-1)” Iml!.

Chapter 4

4.1. Sincex2+Zy+2y2= (x+yv) (x+yv),wehave

(i) u ThereexistsatZ[w] suchthat ~=a%.

Using a similar argument to the proofs of Propositions 0.2, 0.3 and 0.4 in

$4.1, we have

The above condition u

if p # 2,7.


4.2. If (ii) is satisfied, then we have n = m2 n,‘=, p,, where m is a natural number, r 2 0, and p, is a prime number congruent to 1 modulo 4 or p, = 2. We have p, = (Y~??~, crj E Z[&i]. Writing mn5=1 oj as /? and

puttingP=z+yi (x,y~Z), wehaven=pp=x2+y2. If (ii) is not satisfied, then there exists a prime number p = 3 mod 4

such that ord,(n) is odd. Since (-l,n), = (9) = -1, we see that there do not exist Z, y E Q such that n = x2 + y2.

4.3. Put p = crb (o is a prime element of Z[i]), and put cr2n = z + yi. Then we have pzn = cr2n~2n = x2 + y2. Because of the unique prime factorization property we have x # 0, y # 0. Thus, pn is the hypotenuse of the right triangle formed by x, y and p”. Since c?’ is not divisible p,

the greatest common divisor of the three sides is 1. We show that this is the only triangle up to congruence. Suppose that p2” = x2 + y2 with x, y natural numbers. We have p2” = (x + yi) (x - yi) . Consider the prime factorization of both sides, we see that x+yi = CY~&~P, x-yi = (Y%~F, T 2 0, s > 0, T + s = 2n, p E {fl, =ti}. I f T # 0, s # 0, x + yi is divisible by p, and thus x, y,p” are all divisible by p. If r = 0 or s = 0, we have x + yi = oyznp or x + yi = Eznp. Each gives a triangle equivalent to the one we obtained above.

4.4. Using the notation of Proposition 4.27, we see that (2,l) E 91 is the element in Pj whose y-coordinate is the smallest. The assertion now follows from Proposition 4.27.

4.5. By Theorem A.2 in Appendix A, n,, pep is the largest fractional ideal in A contained in both a and 6, and a n b has the same property. Similarly, flp pdp is the smallest fractional ideal in A containing both a and 6, and a + b has the same property.

4.6. We have < = (y + 2a)(y - 2&5). We show that y + 2&5 is a cube in Z[&5]. A prime ideal that contains both (y + 2J-5) and (y-2-) contains the element (y+2-)-(y-2-) = 40. Wecan show that the prime factorization of (2) is (2) = a’, where a = (2,1+-), and (fl) is a prime ideal. Thus, we have

(y + 2J-5) = a”(&%)“b, (y - 2&%) = am(G)“c, m 2 0, n 2 0,

where a, (v’?), b and c are pairwise relatively prime ideals. Prom (x)” = a2”(fl)2n bc we see that m and n are multiples of 3, and that b and c are cubes. Thus, (y + 2G) is the cube of an idea 0. This means that the cube of a is a principal ideal. But, since the class number is not divisible by 3, b itself is a principal ideal by a similar argument as the one used in s4.4. Put a = (o), cy E Z[J-51. Prom (y + a&%) = (03), we see that - y + 2v’3 = *a” = (I&)~. Hence, y + 2fl is a cube in Z[J-51; i.e.,

y + 2J-5 = (CL + h/T)“, a, b E Z[J-51.


Thus, we have y = a3 - 15ab2 and 2 = 3a2b - 56” = (3a2 - 5b’)b. The

latter shows that b = fl, 52. The rest is easy.

Index

algebraic number field, 103

analytic continuation, 90

arithmetic of quadratic form, 10

automorphic form, 9

Bernoulli number, 91

Bernoulli polynomial, 91

Chinese Remainder Theorem, 51

class field theory, 5

class number, 120

class number formula, 125

completion, 64

congruence, 50

conic, 46

cubic number, 10

Dedekind domain, 117

Dirichlet L function, 82

Dirichlet character, 82

Dirichlet’s unit theorem, 8, 121

elliptic curve, 11, 18

factorization in prime elements, 13

factorization into prime ideals, 13

Fermat’s Last Theorem, 14

First supplementary law, 52

fractional ideal, 118

functional equation, 98

fundamental theorem on

abelian groups, 30

fundamental unit, 121

P function, 95

group structure, 25

height, 22, 31

Hilbert symbol, 53

Hurwitz C function, 90

ideal, 115

ideal class group, 119

infinite descent, 22, 109

integral point, 19

inverse limit, 66

Iwasawa theory, 14

Kummer’s criterion, 131

metric space, 62

module, 68

Mordell’s theorem. 30

n-gonal number, 8

padic absolute value, 62

padic expansion, 68

padic integer, 65

padic L function, 99

padic metric, 62

padic number, 3, 58

padic number field, 58

padic valuation, 60

partial Riemann C function, 89

Pell’s equation, 7

point at infinity, 28

prime element, 5, 104

prime number, 5

principal fractional ideal, 118

principal ideal, 116

principal ideal domain, 116

Pythagorean Theorem, 2

quadratic reciprocity law, 50, 52

153

154 INDEX

rational number field, 7 rational point, 19, 45 Riemann < function, 82 ring homomorphism, 54 ring of integers, 113

Second supplementary law, 52 square numbers, 10

triangular number, 10

unique factorization domain, 104 unique prime factorization property,

104 unit, 8 unit group, 119

weak Mordell theorem, 31

C function, 82

Documents

Kato, Kurokawa, Saito - Number Theory I. Fermat's Dream s