Number Theory 1: Fermat's Dream

Translations

of

M&THEMATICAL MONOGRAPHS Volume

186

Number Fermat’s

Theory

1

Dream

Kazuya Kato Nobushige Kurokawa Takeshi Saito Translated by Masato Kuwata

lallalllllllllllllllllllllllllllllllllllllllllll FUDAN

BOO12090492443

American

Mathematical

Providence,

Rhode

B

Society

Ilsand

'!$

Contents Preface

ix

Preface to the English Edition

xi

Objectives

and Outline

. ..

of these Books

x111

xv

Notation 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 8 numbers 10 0.6. Triangular numbers, squares, cubes 11 0.7. Right triangles and elliptic curves 12 0.8. Fermat’s Last Theorem 14 Exercises Chapter 1.1. 1.2. 1.3.

1. Rational Points on Elliptic Curves Fermat and elliptic curves Group structure of an elliptic curve Mordell’s theorem Summary Exercises

17 17 25 30 43 43

Chapter 2.1. 2.2. 2.3. 2.4.

2. Conies and padic Numbers Conies Congruence Conies and quadratic residue symbols p-adic number fields

45 45 49 53 58

vii

CONTENTS

“Ill

2.5. 2.6.

Multiplicative structure of the p-adic Rational points on tonics Summary Exercises

Chapter 3.1. 3.2. 3.3.

3. < Three wonders of the values Values at positive integers Values at negative integers Summary Exercises

Chapter 4.1. 4.2. 4.3.

4. Algebraic Number Theory Method of algebraic number theory The heart of algebraic number theory The class number formula for imaginary quadratic fields Fermat’s Last Theorem and Kummer Summary Exercises

4.4.

number

of the < function

field

69 74 78 78 81 81 84 89 99 100 103 104 113 124 127 132 132

Appendix A.l. A.2.

A. Rudiments on Dedekind domains Definition of a Dedekind domain Fractional ideal

135 135 136

Answers

to Questions

139

Answers

to Exercises

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 {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, expressing 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. 0.1. Let p be a prime number congruent to 1 mod(e.g., 5,13,17). Then there exists a right triangle with integer such that the length of tile hypotenuse is p. Conversely, no such triangle exists for any prime number congruent to 3 modulo 4 3,7,11).

PROPOSITION

ulo 4 sides right (e.g.,

Notice that in Figure 0.1 the prime numbers 5,13,17 are hypotenuses 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

4, then there

0.2. If p is a prime number congruent exist natural numbers x and y satisfying p=x2+y2.

to 1 module

0.2.

PRIME

For example,

NUMBERS

AND

THE

SUM

OF

TWO

SQUARES

5

we have

Conversely, for a prime number p congruent not exist rational numbers x and y satisfying

to 3 module 4 there p = x2 + y2.

do

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/ and it factors

a,bEZ}

(Z is the ring of all integers)

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(2 132 = (3 + 2i)2(3 172 = (4 + i)2(4

- i)2 = (3 + 4i)(3

- 4i) = 32 + 42,

- 2i)2 = (5 + 12i)(5 - iy

= (15 + 8i)(15

- 12i) = 52 + 122, - 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? For example, we have 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 ulo 3, then there exist natural numbers x and y satisfying

to 1 mod-

p=x2+3y? For example, we have 7=22+3x12,

13=12+3x22,

Conversely, for a prime number p congruent not exist rational numbers x and y satisfying

19=42+3x12. to 2 module 3 there do 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. For example, we have 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

0.1

TABLE

~

acm

I

primes congruent

to 1 or 3 modulo

8

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. Fermat

equations

also declared that he proved the following.

PROPOSITION

square

Pell’s

of another

0.6. Let N be a natural number natural number. Then the equation x2 - Ny’

has injinitely

many natural number

which

is not

a

= 1 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 pentagonal numbers

The eighteenth comment is the following proposition.

of Fermat

numbers,

in the margin of Arithmetica

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. Unfortunately, however, the book was never written. If we extract the part about the quadrangular numbers from Proposition 0.7, we have the following. 0.8. Let n be a natural number. x, y, z and u satisfying

PROPOSITION

integers

n=x2+y2+z2+u2.

Then,

there

exist

0.5.

TRIANGULAR,

QUADRANGULAR..

PENTAGONAL

NUMBERS

9

. OQtsl FIGURE

For example,

0.2. n-gonal numbers

we have

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 cubic

dinerent

from

1 is not a

number.

PROPOSITION 0.10.

The only case where a cubic number is 52 + 2 = 33.

to 2 becomes

a square

number

added

PROPOSITION 0.11. The only caseswhere a square number added a cubic number are 22 + 4 = 23 and 112 + 4 = 53.

to 4 becomes

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 Propositions 0.1-0.8) by hand without using any significant tools. In attempting 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

FIGURE

TRIANGLES

AND

ELLIPTIC

CURVES

11

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 y2 = (polynomial

(0.1)

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.

0. INTRODUCTION

12

PROPOSITION

0.12. Given a triangle

length, there exist infinitely many triangles have the same area as the given triangle.

whose sides have rational with rational sides that

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 integers is not a square.

0.14. The area of a right

PROPOSITION

integers

is not twice

triangle whose sides are triangle

whose

sides are

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 sameas 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 SwinnertonDyer conjecture, has been proposed to provide a method of determining 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 SwinnertonDyer 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 remained 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 Xn

= (2 - Y)(Z - GLY) . ‘. (2 - c,“-‘YL

where cn is the n-th If the ring q&l

=

(a0

form

+

primitive

a1Cn

+

root

. .

+

a,(~

of unity

1r 2

cos(27r/n)

0,

~0,.

+ isin(2r/n).

. , 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 - 70 12 . 0.6. Show that there are infinitely many integers which multaneously both a triangular number and a square.

are si-

CHAPTER

Rational

Points

1

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 $0.7, Fermat wrote down a proof of the fact that exist a right triangle whose sides are integers and square”(Proposition 0.13) in the margin of his copy His proof implies the following proposition. PROPOSITION 1.1. There satisfying

is no solution

we explained in “there does not whose area is a of Arithmetica.

(cc, y, z) to x4 + y4 = z4

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

(z,y)

= (0,O)

and

solutions

to y2 = x3 -x

are

(kl,O).

We can see that Proposition 1.1 is a consequence of Proposition 1.2 as follows. If there exist natural numbers 5, y and z satisfying x*+y4=z4, we see (by moving that they satisfy

y4 to the other

(Ye),= This implies that ing y # 0, which

the equation contradicts

side and then multiplying

by z2/y”)

($)3Y;. y2 = x3 - x has a solution satisfyProposition 1.2. Thus, we see that

17

18

1. RATIONAL

POINTS

FIGURE

ON

ELLIPTIC

1.1. Elliptic

CURVES

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 Fermat’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 Fermat 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 following form:

of the

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

FIGURE

FERMAT

1.2. Curves

AND

ELLIPTIC

CURVES

that are not elliptic

19

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

20

1. RATIONAL

POINTS

ON

ELLIPTIC

CURVES

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

LEMMA 1.3. conditions (i) (ii) (iii)

Let d be a positive rational (i) through (iii) are equivalent.

number.

The following

There exists a triangle whose three sides are rational numbers and whose area is d. There exist three squares of rational numbers that form an arithmetic progression of difference d. 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.

1.1.

LEMMA

1.4.

For d E K define

FERMAT

AND

ELLIPTIC

21

CURVES

Let K be a field of characteristic different the sets Ad, Bd and Cd as ~0110~s:

Ad = {(x, y, z) E K x K x K ) x2 + y2 = z2, ;xy Bd={(u,u,w)EKxKxKIu2+d=u2,

there

exist

Indeed,

bijections

between

between

= d),

y # O}.

any two of Ad, Bd, and Cd.

Ad and Bd we have two maps

Ad-+Bd;

(x,y,z)++

Bd -+ Ad;

(u, u, w) -

and these maps are inverse corresponds to metic progression

2.

v2+d=w2},

Cd = {(x, y) E K x K ) y2 = x3 - d2x, Then

from

(i,$,g) with

y,;,+ (w - u, w + 21, au),

to each other.

For example,

(3,4,5)

E A6

E BG, and (f)“,(%)‘,(g)’ is an arithdifference 6. (5,12,13) E A30 corresponds to

, (q)” is an arithmetic progression (S, y, 7) E B3o, and ($)’ , (y)’ 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 given by

f(u,

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

dX,Yl)

=

&((x (2)

There

$((x

inverse

maps

- a)2 - (b - a)(c-

- b)2 - (a - b)(c - b)), &(x is a map h : B --+ 6’ given

f :B

--+ C and g : C + B

(u + v)(u + w)(w

+ u)),

a)), - c)~ - (a - c)(b - c))) by h(u, u, w) = (u2+a,

uvw).

1. RATIONAL

22

POINTS

ON

ELLIPTIC

The proof of Lemma 1.5 is straightforward, reader.

CURVES

and we leave it to the

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 equivalent.

0.13 and Proposition

1.2 are

(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 assumexc > 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.

1.1.

(iii)

FERMAT

AND

ELLIPTIC

23

CURVES

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 assumezo > 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 assumex0 > 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 mn(m - n)(m + n) (20 - l)so(zo + 1) = 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 rational number, a is the square of an integer.”

an integer Prove this

a is the fact.

square

of a

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.

24

1. RATIONAL

POINTS

ON

ELLIPTIC

CURVES

It now follows from the case Ic = 2 in Lemma 1.7 below ofm,n,m-nandm+naresquares. Thus,ze=z,sc-l=y and 20 + 1 = 9 are all squares of rational numbers.

that

all

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 prime

2.

Prove

Lemma

1.7.

(Hint:

Factor

each

a, into

the

product

of

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 (r2

xo

=

4rs(r2

terms,

we have

+ s2)2 -

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. If 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 the group

two

points

structure,

Q (- $, 3) they correspond

Q=P+(O,O)

and

and R to

(

3,

R=-P+(l,O).

A).

In terms

of

1.2.

1.2.

GROUP

STRUCTURE

Group

structure

OF

AN

ELLIPTIC

of an elliptic

CURVE

25

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

Let K be a field of characteristic tion

different

on an elliptic

from 2. Consider

curve. the equa-

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 using the following a group structure on E(K) (wri tt en additively) principles (i)-(iii) (i) 0 is the identity element. (ii) If 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) If 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. If 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

1. RATIONAL

26

POINTS

ON

ELLIPTIC

CURVES

R P u

p+Y

< 1.3

FIGURE

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, sx + t is divisible by (x - x1)(2 - x2) and it factors as qx3

+rx2

+

sx+t

=

q(x-x1)(x

-x2)(x

-x3)

qx3

(x3

+

rx2

+

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

x3 y3

= =

1 a

~2 ~ ( x2 -

_

92 - Yl

-53 52

YI Xl

2

b

---x1-22, a

> +

Y2Xl

-

511x2

x2-51 . -x1 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.

1.2.

GROUP

STRUCTURE

OF

AN

ELLIPTIC

27

CURVE

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 x3 =4ayf

- 2acxf - 8adxl + c2 - 4bd), a3xy + 2a2bxT + 5a2cxt

4 y3 =8ayf

+ 20a2dxT + (20abd - 5ac2)xT + (8b2d - 2bc2 - 4acd)xl

+ (4bcd - 8ad2 - c”)).

For example, consider K = Q and the If P = (2,2), then we have 2P = (5, -11). We have defined P + Q. It is possible abelian group under this addition. (The to prove. We can prove the associativity geometry, but we do not discuss it here.) QUESTION 3. Show that the set {P E E(K) nonzero elements of E(K) whose y-coordinates algebraically closed field, we have an isomorphism {P

E E(K)

1 2P = 0)

E Z/22

elliptic

curve y2 = x3 - 4.

to prove that E(K) is an associative law is difficult elegantly using algebraic

1 2P = 0) consists of 0 and are 0. Show that if K is an of groups @ 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)} tion 3). The map in Lemma 1.5 h o g: C = E(K)

- (0, (a, 0), (b, 0), (c, 0)) -+ c = E(K)

(see Ques- (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.

lllllllllllllllllllllllllllllllllllllll 11111 lllll lllll l~llllll~lllll F(,lDAN

BOO12090492443

B

km

1. RATIONAL

28

POINTS

ON

ELLIPTIC

CURVES

(b) The meaning of 0. We now consider the meaning of 0. If 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). If Q approaches to (2, -2) from below, the sum P + Q goes higher and higher to infinity. If 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.) If 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 point

(x, y) = ratio(x

0 = (0 : 1 : 0).

: y : 1) = ratio

(f

: 1 : $)

converges

to the

1.2. GROUP

STRUCTURE

FIGURE

(c) of E(Q)

OF

AN

ELLIPTIC

1.4. y2 =x3

CURVE

29

+1

Examples. Let us see some examples of the group structure 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.

30

1. RATIONAL

POINTS

ON

ELLIPTIC

CURVES

(d) Fermat’s method. As we mentioned in $0.7 (Proposition 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 -52’

y2 - x2 z4 + 4xzyz E Ad. 22 ’ 2(y2 - x2)z > The map Ad + Ad that sends (x, y, z) E Ad to this point (for example, is nothing but the multiplicationit maps (3,4,5) to (y, &, +)) 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 Fermat 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 (a) Statement of Mordell’s following theorem in 1922.

theorem theorem.

Mordell proved the

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& (2) Z/nZ

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

(It is known that each of the groups in the above list subgroup of all the elements of finite order of some over Q.) In this section we give the main part of the proof theorem. The rest of the proof will be given in Volume (b) Outline of the proof of Mordell’s theorem. theorem is proved using the following two facts.

occurs as the elliptic curve of Mordell’s 3. Mordell’s

(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 two conditions:

real number C satisfying

(1) For any P E E(Q), C. H(2P)

> H(P)4;

the following

32

1. RATIONAL

(2)

POINTS

ON

For any P, Q E E(Q), C. H(P)H(Q) L min(H(P

ELLIPTIC

+

CURVES

Q), H(P - Q)).

(1) formulates the phenomenon we mentioned namely, “H(2P) is much larger than H(P)". Let us prove that Mordell’s theorem follows (IIB). More precisely, we prove PROPOSITION 1.13. Let Q1,. of the elements of E(Q)/2E(Q). 1 . . > n} equals E(Q)/2E(U3)). iijies the properties (1) and (2) H(Ql), . , H(Q,) and C. Th en

{P E E(Q)

at the end of 51.2, from

(I),

(IIA)

and

. . , Qn E E(Q) be representatives (That is, {Qz mod 2E(Q) 1i = Suppose a positive number C satin (IIB). Let M be the largest of E(a) is generated by the finite set 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 By (2) of (IIB)

< M.H(R)

we have H(R)

Thus,

< C.H(R) I C.

ff(Po)ff(Qi)

5 M2WPo).

we have ffpq4

I M3w%).

Since H(Po) > M, we obtain Hi < H(Po)4. Thus, we have of H(Po) implies that PI belongs ff(Pl) < H(h). Th e minimality 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.)

1.3.

MORDELL’S

THEOREM

33

(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

and the proof above equation, in this section. The proof that advised to skip

numbers),

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

PROPOSITION Consider

rational

the elliptic

1.14. Let a, b, and c be distinct curve E defined by

rational numbers.

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

= (x-a,

x-b,

x-c)

((u-b)(a-c), a-b, a-c) ~-~__ (b - a, (b - a)(b - c), b - c) (c-u, I (Ll,

c-b,

(c-a)(c-b))

if P = (b,O), ifP=(c,O), ifP=O.

1)

(Here - means mod(Qx)2.)

ZfP # 0, (a, O>> (h 01, Cc, 01, ifP=(a,O),

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, Proposition 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.

34

1. RATIONAL

POINTS

ON

ELLIPTIC

CURVES

PROOF OF PROPOSITION 1.14( 1). We show that the first component 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 tween the line and the elliptic curve. Thus, we have (x

Letting

-

u)(x

-

b)(x

- c) -

(Xx

+p)2

=

(x-x1)(x-x2)(x

-

be-

23).

x = a, we have (x1 - u)(m

This completes

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

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 ord,(s If s and

-

s and t

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

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),

1.3.

MORDELL’S

THEOREM

ord,(x - b) , and ord, (x - c) is an even number. y2 = (x - u)(x - b)(x - c) and (i) that (*I

ord,(x

- a) + ord,(x

- 6) +- ordP(x - c)

35

It follows

from

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, the outline first. Let E be an elliptic curve over Q with equation

we describe

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

1. RATIONAL

36

g(T).

Then

holds for

POINTS

there is a positive

all x satisfying

f(x)

ON

ELLIPTIC

real number

CURVES

C such that

# 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). the relation between the x-coordinate C onsidering of P and that of 2P, it suffices to show Lemma 1.17 below, which has nothing to do with elliptic curves. 1.17. Let f(T) and g(T) be polynomials with Q coefiSuppose that the degree off(T) and that of g(T) are both no than a given natural number d. Then there is a positive real C such that

LEMMA

cients. greater number

j-g& 5 c . Iqxy (f(x) >

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) ’

of two variables where f (5 T), s(S, T) and h(S, T) are polynomials 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

1.3.

MORDELL’S

THEOREM

37

(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 to do with elliptic

is reduced to Lemma 1.18 below, curves. For we have

H(z+)H(z-)

which

has nothing

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 to replace C by 4C to prove the case (iii).

LEMMA 1.18. (1) F or any rational ;H(U)f(v)

1.18(2).

numbers

Thus,

it suffices

u and v we have

< 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)

( f(s,t)’

f(s,t)

holds for any rational




numbers

qd ’

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 outline 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)

38

1. RATIONAL

POINTS

ON

ELLIPTIC

CURVES

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 Lemmas 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.)

PROOFOF LEMMA 1.18(l). Writeuandvasu= in lowest terms, respectively. We have

zandw

= 5

mn’ + m’n mm’ lLw=-----. nn’ ’ 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 u+w=

H(u + 21,UW) = max(lmn’ by definition

of the height. H(u)H(v)

+ m’nl, lmm’l, Inn’/)

On the other hand, we have

= 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 holds for all real numbers

+ 4,bI,

z and y.

Ivl)

1.3.

MORDELL’S

THEOREM

This follows from the fact that the inequality i holds when 1x1 < $, and IyI < i.

39

11+xyl

2 1 - (i) 2 > 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, i=o

g(T) = k biTi, i=o

and write a rational number x satisfying lowest terms, then we have

f(x)

# 0 as a fraction

z in

g(x) i$obimi,d-i - = fCx) z$oaimind-i. Therefore,

we have

PROOFOF 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’, Z>j h(S,T)

g(S, T) = c bi,S”TJ, id = &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

40

1. RATIONAL

in lowest

POINTS

ON

terms, and let s = T and

ELLIPTIC

= $.

t

CURVES

Then we have

& b,jmz(mf)jnd-i-.i d% t) = ___ f(S,

c Cymym’)~nd-z-~ .>’

h(s,t)

t)

fo

=

ZJ

5 i.j

u,.pn~(m')~?zd--i--J

.

Hence we have

< max

C aijrn’(n~‘)jn~-~-~ i>j

,

I c . H(s, t)d.

PROOFOF 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: satisfying f(x) # 0, we show that Hi lowest terms. Set

5 C. H

d

f(T)

= c i=O

. Write x = E in

d

aiTi,

g(T) = c i=O

e

biTi,

~1

(T) = C CijT’. a=0

1.3.

MORDELL’S

THEOREM

41

Then d f(x)nd

= -jy aimind-z,

g(z)nd

i=O cj (x)ne

= c bimzndei, z=o

= 2 cijminee2 a=0

are all integers, and by (1.6) we have (1.7)

(cl(x)ne)(f(x)nd)

i (c3(x)ne)(f(x)nd)

+ (cz(x)n”)(g(x)nd)

= Rndfe

+ (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 # > R-‘max(lf(x)ndl, lg(x)ndl). ( > (This is the key point of the proof; it showsthat the denominator and the numerator of right-hand side of (1.8) will not cancel each other (1.9)

H

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,

From (1.9) and (1.10) we have

ldx)d)

1. RATIONAL

42

POINTS

ON

ELLIPTIC

CURVES

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

Then cI)(T) (j = 1,2,3,4) are polynomials degree at least e, and they satisfy (1.6). REMARK

log(H(2nP))/4n

=

Ru2(T),

with

integer coefficients

of 0

1.19. For a point P in E(Q), it can be shown that converges when n tends to infinity. So, we define h(P)

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

= Jim

& log(H(2”P)).

define

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.

43

EXERCISES

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

Using this fact, show that there exist infinitely in y 2=x3-4. 1.3. Let K be a field of characteristic Take k E KX, and set

of P)4.

many rational

different

from 2 and 3.

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.

points

1. RATIONAL

POINTS

ON

ELLIPTIC

CURVES

1.4. Let K be a field of characteristic kEKX,andset

different

from 2. Take

X={(x,y)~KxKIy~=x~+k}, Y = {(x, y) E K x K 1y2 = x3 - 4kx,

(x, y) # (O,O)}.

Show that there is a map from X to Y given by x -+ y; (X,Y) +x2 and that it is a bijection.

+y),4x(x2

-+Y)),

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 if P = (2, Y) # (O,O),

(x+;,Y(l-$))

f(P) = o i

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

2 - 2,; (1+ $)) if P ( if P { 0 Show that g of : E(K) --f E(K) and f the multiplication-by-2 maps. Show that g(P) =

X --+ Y c E’(K)

= (x, y) # (O,O), = (O,O), or P = 0. o g: E’(K) the map

+ E’(K)

are

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

Conies

and p-adic

2

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. (a) equation

Rational

points

on

Conies tonics.

An

integral

solution

of the

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

FIGURE

AND

P-ADIC

2.1.

NUMBERS

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 emergesfrom obscurity if we seethem 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

47

A the lights of 2 T/prime numbers+ 1”

t., ,,”

5 2~ the light of fr 7, real numbers V

(supplements each other)

n:‘”

thz/Zght

$

s +

the light of 3

& T-

-‘/

,/ 2 the light 3 of 5 a the light 5 / C’. ’ ‘” 3 of 7 3” .( )..“3 2 thelight 2 _.nj of 11 =? ._. 4 9

FIGURE 2.3. The light of real numbers of prime numbers neither in the world of 2-adic numbers numbers. We will discuss this in $2.5.

and the lights

nor in the world

of 3-adic

(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 $. Conversely, for a given rational number t, the points of intersection between the circle and the line of slope t passing through The latter is of course a rational (-1,O) are (-1,O) and (&-$, &). point.

FIGURE

2.4. Rational points of x2 + y” = 1

2. CONICS

48

For example,

If we let t = A,

AND

NUMBERS

t by i, i, i, i, $ successively,

if we replace

we obtain

P-ADIC

(#,

z)

Clearing

we obtain

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 different

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

r

{rational

(GY)

-

&?

1 - t2 2t 1+ t2 ’ 1+ t2 > (c) nonzero

Conies rational

-

that have a rational numbers. If the conic

numbers},

t. point.

Let

a, b and

c be

ax2 + by2 = c has at least one rational point, by the same method as above. have the correspondence {(x,Y)

I X,Y

we can obtain all the rational points If Q(xe, ye) is its rational point, we

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 sq 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.

49

CONGRUENCE

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+

(k2,

QUESTION ztl).

1.

Find

a rational

point

on x2 + y2 = 5 other

than

(3~1, k2),

and

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

2. CONICS

50

to congruence equations explain congruence.

(a) Congruence a natural

number

AND

P-ADIC

NUMBERS

and quadratic

residues.

and its fundamental

and a, b two integers. a-b

mod

section

properties.

The

we

Let m be

notation

modm

means that a - b is a multiple modulo m” .) For example, 28 ~3

In this

of m. 5,

(We say “a is congruent 35-O

mod5.

We review here briefly the basic properties The quadratic reciprocity law will be introduced 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

(2.4)

a E b mod

m and b z c mod

(2.5)

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

to b

of the congruence. here and proved in

b E a mod m. m imply

a E c 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 (1)

Z/mZ

is a field

number.

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 assumepl, . . . ,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 O2E 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 isomorphism of groups F,X/(!F,X)2 2 {fl}. Hence, for any integers a and

52

2. CONICS

b prime

AND

P-ADIC

NUMBERS

to p, we have

2.2.

THEOREM

Let p be an odd prime

(1)

(Quadratic reciprocity law) ferent from p, we have

(2)

(First

supplementary -1 (-1 P

(3)

(Second

number.

If q is an odd prime

number

dif-

law)

= (-p supplementary

1

Zfp=l

mod4,

-1

ifpz3

mod4.

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)). If 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+ and the facts QUESTION square

divide only which square

root

(i) 3.

of -3

= I, Let exists

(i) = (-$+

(5)

(2) = -1.

p be a prime in FP if and

number different from only if p = 1 mod 3.

2 and

3. Show

that

a

QUESTION 4. Let m be an 2m. Show that the existence by

p mod

41ml

(i.e.,

if p’

integer and p a prime number that does not of a square root of m in lFP can be determined is a prime number which does not divide 2m and

satisfies p z p’ mod 41ml, then we have the equivalence root of m in FP H there exists a square root of m in F+“).

“there

exists

a

2.3.

2.3.

CONICS

AND

Conies

QUADRATIC

and

RESIDUE

quadratic

residue

SYMBOLS

53

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 if a > 0 or b > 0,

(a,bb= We see immediately

there

exist

if a < 0 and b < 0.

that

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 fore stating the definition number p, we need some define a subring Zc,) of Q z

(PI

=

properties of the Hilbert symbol. Beof the Hilbert symbol (a, b), for a prime preliminaries. For a prime number p we by

{ f / a, b E Z, b is not divisible

by p}

54

2. CONICS

AND

P-ADIC

NUMBERS

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

a

b

(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 5

%) ---) zb)lp”zb)

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.

2.3. CONICS

(4)

AND

If p is an odd prime the following. (4-l) (a, b), = 1, (4-2)

(a,pb),

QUADRATIC

number

RESIDUE

55

SYMBOLS

and a, b E (Zt,))x,

then we have

= (T)

(5) If a, b E Z?y2,, then ifa-lmod4orbElmod4,

(5-l)

(a,b)2

=

i, {

ifaEb=-1mod4.

(5-2) (a, 2b)z ={ ~

1

ifa~1mod8ora~1-2bmod8,

- 1

otherwise.

The proof of this proposition follows easily from the Hilbert symbols, and we leave it to the reader. (c) Product formula for Hilbert theorem is a translation of the quadratic plementary laws using Hilbert symbols. THEOREM

for a finite

where

through

all the prime

of

symbols. The following reciprocity law and the sup-

2.5. Let a, b E Q”. Then number of ‘vu, and we have

v runs

the definition

numbers

(a, b)V is equal

to 1 except

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 since we have to prove it only for it in the following cases (i)-(iii), each prime factor of a and b and for -1 thanks to Proposition 2.4(l), (2) and (3).

(i) (ii) (iii)

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

2. CONICS

In case (i),

AND

P-ADIC

NUMBERS

0 -b a a

ifv=a, if v = b,

0b

(a, b)v =

(q+k$

ifw=2,

1

for other

U.

but the quadratic Thus, the fact n,(a, b)v = 1 is in this case nothing reciprocity law (Theorem 2.2(l)). In case (ii), it follows from Proposition 2.4 that

(a, -l)v

=

i

-1 (-1

ifv=a,

(-r)+

if u = 2, for other

2-

if ‘u = a,

0 (a,

2), =

(Q1)+

ifv=2,

1

for other

Thus, the fact n, (a, b)v = 1 is in this case nothing mentary laws (Theorem 2.2(2) and (3)). As for case (iii), a calculation shows that

= 1

v. but the supple-

if z1 is 2 or 03,

(-1, -l)v = ,’ (-1,2),

v;

otherwise; for all v.

q REMARK 2.7. Once we translate into the form of Theorem 2.5 (which realize that the quadratic reciprocity “the light of real numbers” and “the

the quadratic reciprocity law was done first by Hilbert), we law expresses the harmony of lights of prime numbers”.

2.3. CONICS

AND

QUADRATIC

RESIDUE

SYMBOLS

57

(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,

IfzfO,

X,Y,Z

E Q,

(X,Y,Z)

# (O,O,O).

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 Theorem 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 a1cu 1at 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 cl a E 2,5,6,7,8,11 mod 13. Now (3) follows from these.

58

2. CONICS

AND

P-ADIC

NUMBERS

In Proposition 2.8 we looked for rational solutions to a quadratic equation. How about integral solutions? As Fermat says (see Chapter 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 example, solution rational theory,

3 = ($) 2 + 26 (i)“.) Clearly, however, there to 3 = x2 + 26~~. The difference between the solution and an integral solution is related to and we will discuss it in Chapter 5, §5.3(b) in

QUESTION

5.

y2 does not have

In Arithmetzca Diophantus says that the equation a rational solution. Verify this using Theorem 2.3.

2.4. The we have

is no integral existence of a the class field Volume 2.

meaning

(a, b)co = 1 ++

padic

number

of the Hilbert

symbol

th ere exist

(a, b)P = 1 _

th ere exist

=

fields ( , )oo is that

x, y E Iw satisfying

For each prime number p we can interpret Namely, for each p there is an extension Q” we have

15x2-36

for a, b E Q”

ax2 + by2 = 1.

(a, b)P in the same manner. field Qp of Q, and for a, b E

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

59

FIELDS

1 26 51

a00

. 31

@(zj

FIGURE

2.5.

Classification

by mod

night, we began to see the profound mathematical the p-adic numbers. We introduce the p-adic number fields in three (b), Cc) and (d). W e would like you to get acquainted to your taste.

5” universe

through

different ways in with it according

(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).

2.

60

CONICS

AND

P-ADIC

NUMBERS

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, disFor integers a, b, the congruence tinct) be its prime factorization. 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 (Proposition 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 fundamental. 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

u

by p),

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)

(2.7)

ord,(a

(2.8)

ord,(a)

= ord,(a)

+ ordp(b);

+ b) > min(ord,(a), # ord,(b)

implies

ord,(b)); 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.

2.4.

For example,

p-ADIC

NUMBER

FIELDS

61

if we let 2

n = 1 - 5 + 52 - 53 +. . . + (-5)“,

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

of In

an+l 1+a+a”+...+al’-A=--

Replacing

Thus,

a by -5,

l-a’

we have

X71 as n + 00, we have ordg(x,,-i)

1

(-l)n5n+l

6

6

=ordi((-l)i’“+‘)

As this example shows, convergence ferent from convergence in ordinary (xn)+l converges to i by

(2.9)

2(-B)” i=o

it is as if we mistakenly

= f put x = -5

2x’=& z=o

6.

Let

p be

g i.e.,

if we put

zrL = CFzo

=n+l+m. in the p-adic sense is quite difsense. If we express the fact that

(5-adically), in the ordinary

formula

if-l<x

1 converges

padically

to

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

&

each of 6 of 6, have

62

inverse

2. CONICS QUESTION 7. Using of 6 in Z/5nZ.

QUESTION

8.

Find

(2.9),

AND explain

the inverse

P-ADIC why

NUMBERS 1 - 5 + 5’ - ‘..

+ (-5)n-1

is the

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 For example,

101, = 0. Thus

Ii,.=;> The p-adic absolute value converges to 0. (All the replace the definition of 0 < T < 1. However, it choice, as we will explain From properties (2.6)

(2.10) I4 (2.11)

sense.

IP21p=$.

expresses well that the sequence p, p2, p3, . . . arguments in this section work well if we Ialp by lalp = T”‘~P(~) for any T satisfying turns out that T = b is the most natural at the end of subsection (c).) and (2.7) of ord, we see that

= MP . I%

la+ bl, 5 max(lal,,

If we define

lalp is the size of a in the p-adic

the p-adic

lblP) (In particular,

metric

(a+ bl, i

Ialp + IblP).

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 Geometry 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,.

2.4.

p-ADIC

NUMBER

FIELDS

63

(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 (“How should we define the real respect to the rational numbers?” 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 N such that m,n 2 N

number

implies

E, we can choose a natural 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,

2.

64

CONICS

AND

P-ADIC

NUMBERS

and define an equivalence relation on S by saying (~~)~>i are equivalent if “for any rational number a natural number N such that n>N

implies

that (x,),21 and E, we can choose

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 (z,),>i

. class of (yn)+r

= class of (2, + yn)+i, = 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 N accordingly m,n>N

number E, we can choose such that

implies

15, - Ic&

a natural

< 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, For an element a in QP we define and p-adic metric dp to Q,. 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

2.4. p-ADIC NUMBER

FIELDS

65

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 n -+ 03, and thus the situation is more complicated. comes from the fact that we have Iz+y[, 5 max(lzl,, we do not have Ix + y// 5 max(lzl, lyl) in R.

1; / --t 0 when The difference IyIP) in QP, but

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) Z,

> O}.

is a subring of QP. (This follows from the {oo} satisfies (2.6) and (2.7).) An element integer. In this subsection (c) we explain that we “inverse limit” and that we can introduce Q,

ZU

fact that ord, : Qp + of Z, is called a p-adic can think of Z, as an in a different manner.

2. CONICS

66

DEFINITION

maps fn : Xn+i

AND

P-ADIC

NUMBERS

2.10. If a sequence of sets X, -+ X, (n = 1,2,3,. . . )

(n = 1,2,3,.

. . ) and

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, jection from Z/p n+lZ to Z/p”& @,Z/p”Z of the sequence . . . -+ z/p4z

+ z/p3z

= Z/pnZ and fn the natural proand we consider the inverse limit 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,.

2.4. p-ADIC LEMMA

defined

2.11.

NUMBER

FIELDS

67

The map

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)

2.12. (1) Z, is both open and closed in U&,. If m is an integer, then we have = {u E Qp 1ord,(a) > m} .

pmZp

(3) Z(,) c 27,. In U& we have Q (4) For all integers m 2 0 z/pmz

:

Z&p?&

(5) Z, is the closure in Qp.

of Z&J

n Z, = ?A(,). -%

in Qp.

The image of a E Z, in Z/p”Z

z,/pmz$. It is also the closure

2 Z,/pmZ,

of Z

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,. 0 But, this follows from (2) and (4). PROOF OF LEMMA 2.11. image of a under the map Z, + Z,/p”Z, Thus

For an element

2 Z/p”Z

a in Z,,

(Lemma

let a, be the

2.12(4)).

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 We will discuss this scaling factor, or “module”, (d) Definition of QP by padic explain that Qp may be defined by Q, = For example,

expansion.

factor of a homothety. in Volume 2. In this section

p-l} 2 GLPn mEZ,c,E{0,1,..., { 1L=VJ. 1 we define an element of Qs to be something

we

1 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

Take an integer element of Z,, and such that its image

STRUCTURE

OF

THE

p-ADIC

NUMBER

FIELD

69

m satisfying ord,(a) > m. Then pPrna is an there exists an integer c, E (0, 1, . ,p - l} in Z,/pZ, coincides with that of p-nLa, since

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

the argument given above carefully, we see that the p-adic 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 71=?7l in a unique

(m E Z, c, E S)

manner.

QUESTION 9. A real number has a decimal expansion as we use it in everyday 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 x H

ex,

log(t)

group

{t E R 1 t > 0},

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 logathe structure of the mulrithmic functions in Qp, and we determine tiplicative 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? Proposition 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 @ we have

and logarithmic

functions

(also written (where

the right-hand

side always

log(t) = c

In R or

exp(z))

converges), ‘-Y-l

in Q,.

and when It - 11 < 1,

(t - 1)“.

n=l

We consider

an analog in Q,.

PROPOSITION

2.14. (1) Let IC E Q,. 2 5 n=O

(written

The series exp(z))

converges

if and only converges if and only means the exponential

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

2 (-1Y-1

(t - 1)”

n

(written

log(t))

n=l

(3)

exp(xl

(4)

converges if and only if t - 1 E pZ,. If x1 and 22 are in the domain of convergence if tl and t2 are in the domain of convergence we have +

x2)

=

exp(xl)

ew(x2),

log(tlt2)

=

log(tl)

Weletm>lifp#2,andm>2ifp=2.

log are isomorphisms, additive

group

% multiplicative

and they are inverse

of exp(z), and of log(t), then

+

log(t2).

Thenexpand to each other:

pmZ, group

1 + p”Z,

= { 1 + pma 1 a E Z,}.

2.5.

MULTIPLICATIVE

STRUCTURE

In order to prove Proposition lemma first. LEMMA

OF

THE

p-ADIC

NUMBER

FIELD

71

2.14 we need to show the following

2.15. (1) For any integer

n > 0, we have

[1

ord,(n!) = e

F

is1

,

where [x] is the “Gauss symbol” of z, largest integer less than or equal to x. (2) Let c be a real number. The condition as n ----f co is equivalent to c > &. ord,(n) --f cc as n -+ m is equivalent to (3) If c > &, then for any n > 1 we have nc - ord,(n!)

which

signifies

nc - ord,(n!) The condition c > 0.

the

--f co nc -

2 c.

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

- ord,(n!)

s 2 nc-

2 nc - 2

-. P-l

i=l

The right-hand side tends to m as n -+ 00 if c > &. n=pm, it follows from (1) that nc - ord,(n!) = pmc - epm-z

n

= pm (c - ~

P-l

i=l

Also, if we put

1 >

+ ~.

1

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

72

2. CONICS

AND

P-ADIC

NUMBERS

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) But they are given and 5 = 1 if p = c. Next we show exp(x) E l+p’“Z,, we have log(t)

= nord,(t

- 1) - ord,(n).

by Lemma 2.15(2). (Note that & < 1 if p # 2, 2.) The proof of (3) is similar to the case of IR or (4). By Lemma 2.15(3), if x E pm&,, we have since ord, (5) > m for n > 1; and if t E l+p”Z$,,

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 CEZ,. (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 (n E Z, u E Z,“)

zez;

in a unique

‘Q;;

of Q$ can be written manner. In other words,

pnu,

(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,. group 1 +pZ, is isomorphic to Z,. (3) UP # 2, th e multiplicative If 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) fo11 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). If 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). If 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 exp(log(uP-I)) = up-l. equal to a. (c)

Squares

we have w E G, because we have up-’ = Since v E 1 +pZ,, the image of w in ‘Fc is cl

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) If p # 2, u mod p;Z, is a square If p = 2, u E 1 mod 822.

PROOF. By Proposition n is even and u is a square 1 + PZ,

= exp(pZ,)

in ‘Ft.

2.17, a is a square in Q,” if and only in Z:. If p # 2, we have = w$W,)

=

and thus an element of 1 + pZ, is a square Zc/(l +pZ,) 2 “c, the case p # 2 is proved. 1 + 822 = exp(8Z2)

= exp(2.42~)

and thus an element of 1 + 822 is a square Zc /(l + 8Z2) E (Z/8Z)x E Z/2Z C$ Z/22, proved.

if

{exdp~p))2,

in Et. Since we have If p = 2, we have = {exp(4Z2)}2, in Zc. Since we have the case p = 2 is also 0

74

2. CONICS

The following Proposition 2.18.

proposition

PROPOSITION

2.19.

(2) Q2”/(Q,“)”

E z/az

AND

P-ADIC

follows

(1)

Ifp

#

QUESTION 10. exists a square

Let a be an integer root of a in Qs

QUESTION mod4.

Show

ifp=l

QUESTION tensions of U&

that

12. Show Determine

there

this

If a, b E Q”

satisfying

a = zkl mod

a square

root

of -1

there

Show

that

in U& if and only

quadratic

ex-

the statement

at the beginning

number,

exist

in Proposition

II:, y E Qp such that 2.20.)

We then

(a) Conies defined over Qp. The Hilbert Q” -+ (51) can b e extended for a, b E 0; we write a = p’u,

5.

points on tonics

by proving

and p is a prime

(This is contained Theorem 2.3.

2.16 or

S! Z/2Z CBZ/2Z.

that if p # 2, there exist exactly three all three quadratic extensions of Q5.

section

(a, b)p = 1 _

Proposition

2, Q,“/(Q,“)2

exists

2.6. Rational We begin of 52.4:

from either

@? z/22 63z/22.

there

11.

NUMBERS

naturally

b =p-h

to Qt

ax2 + by2 = 1. use it to prove

symbol ( , )p : Q” x x Q: + {fl}. Indeed,

(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)+

For this symbol ( , )p : Q: we replace (Zc,) ) ’ by Zc.

x QG +

(-l)++. { z!~l},

Proposition

2.4 holds

if

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

w

3 x, y, z E Q, such that (by)2 = ax2 + bz2

u

3x, y E U& such that ax2 + by2 = 1.

+ by2 = z2 and (x,Y,z)

# (0,&O)

and (2, Y, z) # (0,&O)

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. means that a mod p E “c is a square. By If P # 2, (a,b), = 1 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.

76

and We {au2 thus 2, y

2. CONICS

AND

P-ADIC

NUMBERS

(b) The case a, b E Zp”. Suppose p # 2. Then the condition (a,b), = 1 always holds, thus we must show that ax2 + by2 = 1 has a solution in U&. denote by a,& the images of a, b in IF,. Each of the two subsets / u E IF,} and {I - bv2 ( v E IF,) has cardinality q, and t,heir intersection is nonempty. This implies that there exist E Z, such that ax2 z 1 - by2 mod pZ,. If 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 or b = 1 mod 422. Suppose, say, a = 1 mod 422 mod 422 is similar). Then we have a = 1 mod 822 or If a = 1 mod 822, there exists t E Q,” such that t2

a z (the a E = a

1 mod 422 case b G 1 5 mod 822. 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 can be rewritten “Let

of Theorem 2.3. By Proposition 2.20, Theorem 2.3 in the following form. Here, we write Qoc for R.

a, b E Q”.

The following

conditions

(i) and

(ii) are equiva-

lent. (i) ux2 + by” = 1 has a solution in Q. (ii) ax2 + by 2 = 1 has a solution QV in for all primes 21 = m.” Clearly,

(i) implies

(ii).

So, all we need to prove is that

u and

(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\). If either a or b is 1, ux2 + by” = 1 clearly has a solution in Q.

2.6.

EtA’I’IONAL

POINTS

ON

CONICS

77

If 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 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). If Ial = lbl, we can reduce to the case Ial < lbl, since /c( < Ibl. 0 LEMMA

bc. Then

2.21. Let K be a field; a, b, c E KX; there is a bijection between two sets

r E K;

and r2 -a

=

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

and verify that tively.

f : X -+ Y, g : Y -+ X by

f(x,

y, z) = (rx + z, by, ax + rz),

dx,

Y,

~1 = (rx - z,

CY,

--ax + rz),

g o f and f o g are the identities

of X and Y, respec0

2. CONICS

78

AND

P-ADIC

NUMBERS

Summary 2.1. If a conic defined over the rational tional point, it has infinitely many rational scribe them explicitly. (However, the main not this, but 2.2 and 2.3 below.)

number field has a rapoints, and we can detheme of the chapter is

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 from Z [l/p] /Z to itself, denoted by defining the sum of f and g by z E ;Z [l/p] /Z, and the product off Show that there is an isomorphism Z, “Horn

(Z[i]/Z,

set of all group homomorphisms by Horn (Z [l/p] /Z, Z [l/p] /Z), (f + g)(z) = f(z) + g(z) for all and g by the composition fog. of rings z[~]/z)

2.3. Find ords(4n - 1) (n E Z). (Hint: number field to get 4” - 1 = exp(nlog(4)) tion 2.14(4).

.

Use exp, log in the 3-adic - 1, then use Proposi-

EXERCISES

2.4. (1) (2) (3)

Let p be a prime number. Show the following: x2 = -2 has a solution in QP w p E 1,3 mod 8. x2 +y2 = -2 has a solution in QP u p # 2. x2 + y2 + 22 = -2 has a solution in QP for any p.

79

CHAPTER.

In this chapter zeta function).

3.1. The

Three

we introduce

wonders

3

an important

of the

values

!illlction

of the

called

< (the

< function

formula

(3.1) was discovered by Euler around the infinite sum of the left-hand pleased to find the mysterious number 7r. The formula

1735. 1Ie had attempted to determine side for many years, and he was quit,e fact that, the sum is related t,o t,he

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)

1-$+&$+&&+...

(3.5)

1-;++-;+

=E’ . ..= Xl

“iT 3&’

7r3

82

3. c

and Dirichlet’s (3.6)

formula l-~-;+~+L~-iL+~ +... = 5

(k signs repeat log(1

every 8 terms)

+ 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 Riemann 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 is called

a Dirichlet

character

+ Cx

(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

where the character

OF

THE

VALUES

OF

THE

C FUNCTION

83

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 Dirichlet L function L(s, x)

as a formula

for the value of

L(l,x) with the Dirichlet x:

= -!I3&i> x given by

character (Z/SZ)’

={1mod3,2mod3}+(GX,

x(1 mod 3) = 1,

x(2 mod 3) = -1.

The formula (3.6) may be regarded Dirichlet L function L(s, x) L(l,x) with the Dirichlet x: (Z/8Z)x

as a formula

for the value of

= +2 log(1 + J2)

character

x given by

= (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 = (rational

at s = integer

number)

x (the power of 7r or something For example, if T is a positive c(r)

= (rational

number)

similar to log(1 + A)).

even integer, Euler proved the formula 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, b(t)

For any integer

T greater

than

t + t”

= (1 _ tj3> or equal

b(t) =

to 1 we have

[ 1-t 1.

h7.(t) E Q t, L

t + 49 + t” (1 -

ty

.

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 cot(7rx) = &c

(3.9)

-&

+ -J-

Since cot(y) = a

.

x-n

nEZ

>

and we have eYi

-

sin(y) =

e-Yi

2i

eYi

’

COS(Y)

=

+

e-Y2 2

7

we see that i(eXSi cot(7rx)

=

eTxi

+

e-~Zi)

_

e-m2

= -2ihl(t)

(t = e2rzs),

which proves Proposition 3.3(l). Applying (t$)r-1 = (&)r-l both sides of the above formula, we obtain Proposition 3.3(2).

to 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)‘. If 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

3.5.

EXAMPLE

l-;+;+-L+... 27G

=&.

(-4 >

. ; . (hi(i)

- h&.3))

+$().;.(3&2L) 0 2?ri 4

zz

1 ‘?=4.

lr

3.6.

EXAMPLE

1-;+;+-i+...

=

(--I.-27ri 3

1

i

2

A=-

n3&’

3.7.

EXAMPLE

( > 2Ti

=&

3 1 . 5.

-4

(h&)

-

h3(i3))

=~.(-~)“+.(+-$) =-. 1

27ri (-3 4

2

3.8.

THEOREM

c(r) COROLLARY

a rational

(from(3.8)) 3 1

z. (-q

Let r be a positive

= (r T l)! 3.9.

= g.

. 2’ \

1

If r is a positive

even integer. (a~ri)’

We have

. ; . b-1).

even integer,

then rY 1, and they (2)

are holomorphic in this domain. The functions c(s), L(s, x), &(N)(S) and

= 1.

(3) If 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

We give a proof of Proposition

INTEGERS

91

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 Riemann < 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 is defined by the formula

number B, (n = 0, 1, 2, 3,. . . )

Prom the formula 2

~ = ez - 1 x+$+$+-. =I-

1

2

= I+%+$+... x

;+g+... (

.

>

.

+

(

2

x2

-..., s+y+-*

1

we see that (3.13)

B. = 1,

B”=;,

B1 = -;,

B4 = -&,

5 691 7 BIO = -, Blz = -’ B14 = -,.... 66 2730 6 B,‘s are all rational numbers. Since & - 1 + 5 is an (i.e., invariant under J: H -x), we see that

B8 = -$ In particular, even function (3.14)

B, = 0

for n an integer greater than or equal to 3.

DEFINITION 3.17. The Bernoulli 3,. . . ) is defined by

B,(z) where

= 2 (:) i=o

polynomial

B,(z)

(n = 0, 1, 2,

BiFi,

(1) = &.

From (3.13) we have (3.15)

B,,(z)

= 1,

Bl(s)

=x-

B3(x)

3 = x3 - p2 + $x,

;, By

B2(2)

=x2

-x+-,

= x4 -2x3+x2-$,....

1 6

92

3. c

In particular, have

R,(z)

is a polynomial B,(O)

THEOREM

3.18.

in rational

coefficients,

and we

= B,.

(1) F or a natural number r and a positive

real

number x we have and therefore

for any natural X-1

(3.17)

c n=l

1 ns -

number =

- 0 define

r(s)

=

f. IXe-“p

If 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 Now,

if Re(s)

+ m)r(s)

= (-1)“;.

> 1, we have

x==e-(s+n)uUs *u We =J’0 n=o c =r In other was,

words,

(,

eCsu Gus;.

t

let u = ~ L7C+n’ >

du

we have

4 = JX

f(s, u)du,

where

p(s,u)

= &u

lx

f(s, u)du.

S-l

.

0

We divide

the integral

J’

into two parts:

()_ f(s, u)du

= /’ o f(s, u)du +

Since the function ePszL approaches the integral s;” f( s , u )d u converges it is holomorphic B, (z) we have

on s. Consider

x Kc(s) CTU 77=0 Therefore

J’

0 rapidly as u tends to infinity, for any complex number s, and J, f(s,u)du. uexu 12. =- e” - 1’

By the definition

of

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. to a meromorphic function on the Thus, r(s)(‘( s,x ) is extended 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

If we let 72 = 0 and take the fact l?(l) li-i

(s - l) 1, we have lim,,l-,(s thus we have

PROOF

OF

= 1 into = &(z)

account,

we have

= 1.

+ 12 - l)r(s)

{(l-

. (-1)”

= (-l)“-’

. A,

and

n,.,5) = -~ Bn (xl n

3.15(3).

PROPOSITION

For

s satisfying

Re(s)

>

0

and m 2 0, define

We have L( s, x) = fo (s) + Cz= (3.20)

5 I.fm(s)l 77?=1

I N.

1 fr,, (s) . In the following ISI . (1 + &)

we prove

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

mN+n s (@$+n)‘-&=smN

We write

-dx, xs+l

and thus,

if we write

Therefore,

we have

0 for Re(s),

we have

and thus we have

ii: I.fm(s)l5 N. ISI2 -& 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 e f uric t’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