Number theory, Volume 20 (Pure and Applied Mathematics)

399884

Number Theory by

Z. I. BOREVICH and I. R. SHAFAREVICH Translated by NEWCOMB GREENLEAF for SCRIPTA TECHNICA, Translators New York, New York

ACADEMIC PRESS

New York San Francisco London

A Subsidiary of Harcourt Brace Jovanovich, Publishers

COPYRIGHT © 1966, BY ACADEMIC PRESS INC. ALL RIGHTS RESERVED. NO PART OF THIS BOOK MAY BE REPRODUCED IN ANY FORM, BY PHOTOSTAT, MICROFILM, OR ANY OTHER MEANS, WITHOUT WRITTEN PERMISSION FROM THE PUBLISHERS.

ACADEMIC PRESS INC.

III Fifth Avenue, New York, New York 10003

United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD. 24/28 Oval Road. London NWI

LIBRARY OF CONGRESS CATALOG CARD NUMBER:

65-28624

PRINTED IN THE UNITED STATES OF AMERICA

Original Russian Edition: Teoriya C;sel, Moscow, 1964

Translafor's Preface This book was written as a text for the learning of number theory, not as a referente work, and we have attempted to preserve the informal, slowplaced style of the original. The emphasis of the book is on number theory as a living branch of modern mathematics, rather than as a collection of miscellaneous results. The book should prove accessible to any advanced undergraduate in mathematics, or to any graduate student. The reader should be familiar with the basic concepts of abstract algebra, and should have followed analysis through a standard advanced calculus course. While some results from elementary number theory are occasionally used, a previous course in number theory is certainly not necessary, though the reader without such a course may have a few occasions for consulting a more elementary text. Almost all of the notation and terminology is standard. The only difficulty arose in the terminology for valuations. Since there does not seem to be any universally adopted terminology in English, we have, after some hesitation, followed that of the authors, which has the advantage of being clear and simple in this context. Thus we reserve the term "valuation" for the case when the value group is the integers. Mappings into the positive reals are called "metrics." We would like to mention some additional references. The theory of quadratic forms receives a systematic development in the book Introduction to Quadratic Furms, by O. T. O'Meara (New York, 1963). In particular, O'Meare presents a proof of the Hasse-Minkowski theorem which does not use the Dirichlet theorem on primes in arithmetic progressions. In Chapter 7 of Commutative Algebre (Paris, 1965), Bourbaki gives a complete exposition of the theory of divisors: from. a somewhat more abstract standpoint than that found in Chapter 3. NEWCOMB GREENLEAF

Rochester, New York

v

Foreword This book is written for the student in mathematics. Its goal is to give a view of the theory of numbers, of the problems with which this theory deals, and of the methods that are used. We have avoided that style which gives a systematic development of the apparatus and have used instead a freer style, in which the problems and the methods of solution are closely interwoven. We start from concrete problems in number theory. General theories arise as tools for solving these problems. As a rule, these theories are developed sufficiently far so that the reader can see for himself their strength and beauty, and so that he learns to apply them. Most of the questions that are examined in this book are connected with the theory of diophantine equations - that is, with the theory of the solutions in integers of equations in several variables. However, we also consider questions of other types; for example, we derive the theorem of Dirichlet on prime numbers in arithmetic progressions and investigate the growth of the number of solutions of congruences. The methods that we use are primarily algebraic. More precisely, we work with finite field extensions and with metrics on them. However, analytic methods have a considerable place. Chapter 5 is devoted to them, and p-adic analytic functions are used in Chapter 4. Geometric concepts playa considerable role in several spots. The book does not presuppose a great deal of knowledge on the part of the reader. For reading most of it, two university courses would be completely satisfactory. Some facts on analytic functions are used in the last two chapters. The necessary prerequisites of an algebraic nature are given in the "Algebraic Supplement" at the end of the book. There the reader will find definitions, results, and some proofs that are used in the book but might not appear in a university course in higher algebra. This book grew out of a course taught by one of the authors at Moscow University. We would like to thank A. G. Postnikov, who allowed us the use of his notes from this course. vii

viii

PREFACE

We are also extremely grateful to Dmitri Constantine Faddeev, who made many contributions to this book. He should receive credit for some of the proofs that appear in this book, for example, the new p-adic proof of the theorem of Kummer on the second factor in the number of divisor classes of a cyclotomic field. Moscow

THE AUTHORS

Contents

Translator's Preface

v

vii

Foreword

Chapter 1.

Congruences I. 2. 3. 4. 5. 6. 7.

1

Congruences with Prime Modulus Trigonometric Sums p-Adic Numbers An Axiomatic Characterization of the Field of p-adic Numbers Congruences and p-adic Integers Quadratic Forms with p-adic Coefficients Rational Quadratic Forms

3

9 18 32

40 47 61

Chapter 2.

Representation of Numbers by Decomposable Forms I. 2. 3. 4.

5. 6. 7.

Decomposable Forms Full Modules and Their Rings of Coefficients Geometric Methods The Groups of Units The Solution of the Problem of the Representation of Rational Numbers by FulI Decomposable Forms Classes of Modules Representation of Numbers by Binary Quadratic Forms

ix

75 77 83

94 107 116

123 129

CONTENTS

x

Chapter 3. The Theory of Divisibility

155

I. 2. 3. 4. 5. §. 7. 8.

156 164 170 180

Some Special Cases of Fermat's Theorem Decomposition into Factors Divisors Valuations Theories of Divisors for Finite Extensions Dedekind Rings Divisors in Algehraic Number Fields Quadratic Fields

193 207 216 234

Chapter 4. Local Methods

251

I. 2. 3.

253 267

4. 5. 6. 7.

Fields Complete with Respect to a Valuation Finite Extensions of Fields with Valuations Factorization of Polynomials in a Field Complete with Respect to a Valuation Metrics on Algebraic Number Fields Analytic Functions in Complete Fields Skolem's Method Local Analytic Manifolds

272

277 282 290 302

Chapter 5. 309

Analytic Methods 1. 2. 3. 4. 5. 6. 7. 8.

Analytic Formulas for the Number of Divisor Classes The Number of Divisor Classes of Cyclotomic Fields Dirichlet's Theorem on Prime Numbers in Arithmetic Progressions The Number of Divisor Classes of Quadratic Fields The Number of Divisor Classes of Prime Cyclotomic Fields A Criterion for Regularity The Second Case of Fermat's Theorem for Regular Exponents Bernoulli Numbers

Quadratic Forms over Arbitrary Fields of Characteristic Algebraic Extensions Finite Fields Some Results on Commutative Rings Characters

325 338 342 355 367 378 382

390

Algebraic Supplement 1. 2. 3. 4. 5.

309

t:-

2

390 396

405 410 415

Tables

422

Sub;ect Index

433

CHAPTER ..

Congruences

This chapter is devoted to the theory of congruences and to its application to equations in several variables. The connection between congruences and equations is based on the simple remark that if the equation (0.1) where F is a pOlynomial with integral coefficients, has a solution in integers, then the congruence (0.2) is solvable for any value of the modulus m. Since the question of the solvability of a congruence can always be decided (if only by trial and error, as there are only finitely many residue classes), we have a sequence of necessary conditions for the solvability of (0.1) in integers. The question of the sufficiency of these conditions is much more difficult. The assertion that" an equation is solvable if and only if it is solvable as a congruence modulo any integer" is in general false (see, for example, Problem 4), but it is true for certain special classes of equations. In this chapter we shall prove it in the case where F is a quadratic form which satisfies the additional condition, clearly necessary, that (0.1) be solvable in real numbers. (Note that if F is a form, then by the solvability of the equation F = 0 we shall understand the existence of a nonzero solution.) The p-adic numbers, which we shall study and apply to the theory of congruences and equations, will be our basic tool. We now indicate their role. From the elementary theory of numbers it is known that if the congruences F(x 1 ,

... ,

x n)

== 0 (mod

p/i)

2

CONGRUENCES

[Chap. 1

are solvable for i = I, ... , r, where Pl' ... ,P, are distinct primes, then the congruence (0.2) is solvable modulo m, where m = p/' ... p,k,. Thus the solvability of the congruence (0.2) for all m is equivalent to its solvability modulo all powers of primes. We fix a prime p and ask whether the congruence F(x l , ... , x n) == 0 (mod pk)

(0.3)

is solvable for all natural numbers k. It was in connection with this problem that Hensel constructed, for each prime p, a new kind of number, calling it p-adic. He showed that the solvability of (0.3) for all k was equivalent to the solvability of (0.1) in p-adic numbers. Hence we may say that the solvability of the congruence (0.2) for all m is equivalent to the solvability of (0.1) in p-adic numbers for all prime numbers p. Using p-adic numbers, our theorem on quadratic fQrms then receives the following formulation (its proof will appear in Section 7): If F(x1> ... , xn ) is a quadratic form with integral coefficients, then (0.1) is solvable in integers if and only if it is solvable in p-adic numbers for all p and also in real numbers. In the formulation of this theorem, called the Hasse-Minkowski theorem, and in many other instances, the p-adic numbers occur on equal terms with the real numbers. If the real numbers are necessary for the study of rational numbers from the standpoint of their size, the p-adic numbers playa completely analogous role in question connected with divisibility by powers of the prime number p. The analogy between real and p-adic numbers can be developed in other ways. It will be shown that the p-adic numbers can be constructed starting from the rational numbers, in exactly the same way that the real numbers are constructed-by adjoining the limits of Cauchy sequences. We shall arrive at different types of numbers by giving different meanings to the notion of convergence. We make one further remark. If F is a form, then the solvability of (0.1) in integers is equivalent to its solvability in rational numbers. Thus one may speak of rational solvability instead of integral solvability in the HasseMinkowski theorem. This obvious remark becomes important when one considers an arbitrary quadratic polynomial F, since the analogous theorem then only holds when one speaks of rational solvability. Hence when we study equations of the second degree, we shall consider not just integral, but also rational solutions.

PROBLEMS

1. Show that the equation 15x 2 - 7 y 2 = 9 has no integral solution. 2. Show that the equation 5x 3 + l1 y 3 + 13z 3 = 0 has no integral solution other than

x=y= z=o.

Sec. 1]

CONGRUENCES WITH PRIME MODULUS

3

3. Show that an integer of the form 8n + 7 cannot be represented as the sum of three squares. 4. Using the properties of the Legendre symbol, show that the congruence (x 2

-

13)(x 2

-

17)(x 2

-

221) == 0 (mod m)

is solvable for all m. It is clear that the equation (x 2 - 13)(x 2 - 17)(x 2 - 221) = 0 has no integral solutions. 5. Show that the equation alXl + '" + anx. = b, where a.. ••. , an, b are integers, is solvable in integers if and only if the corresponding congruence is solvable for all values of the modulus m. 6. Prove the analogous assertion for systems of linear equations.

1. Congruences with Prime Modulus

1.1. Equivalence of Polynomials

We first consider congruences with prime modulus p. The residue classes modulo p form a finite field with p elements, and a congruence with modulus p can be considered as an equation in this field. We shall denote the field of residue classes modulo p by Zp. There exist finite fields other than the various Zp. All considerations of the next two sections carryover word for word in the general case of any finite field. To do this it is necessary only to replace the number p by the number q = pm of elements of this field. But we shall confine our attention to the field Zp and shall use the notation of congruences rather than equations. Only in the construction of the example following Theorem 3 will We need to employ other finite fields. The field of residue classes modulo a prime (and more generally any finite field) has several properties which distinguish it from the familiar fields of elementary algebra, the fields of rational, real, and complex numbers. Most important in our considerations is the fact that the well-known theorem that polynomials which take equal values for all values of the variables must have equal coefficients is no longer true for this field. For example, by the small Fermat theorem, the polnomials x P and x take equal values in the field Zp for all values of the variable x, but their coefficients are unequal. [The following holds for any finite field: If lXI' ... , IXq are the elements of the field, then the polynomial (x - IXI) ••• (x - IXq), which has nonzero coefficients, has value zero for every value for x in the field.] We write F(xI'"'' x n)

E

G(x1 ,

...

,xn) (modp)

and call the polynomials F and G congruent, if the coefficients of corresponding terms on the right and left sides are congruent modulo p. If for any set of values cl , ... , Cn we have

CONGRUENCES

4

F(c l , '" , cn)

== G(c1 ,

... ,

[Chap. 1

cn) (modp),

then we write F", G and call F and G equivalent. It is clear that if F == G, then F'" G, but the example of the polynomials x P and x shows that the converse is, in general, false. Since, if F '" G, the congruences F == 0 (mod p) and G == 0 (mod p) have the same solutions, it is natural that in the theory of congruences one needs to be able to replace a polynomial F by a polynomial which is equivalent to it but is possibly in a simpler form. We now return to this problem. If any variable Xi occurs in the polynomial F to a power not less than p, then using the equivalence Xi P '" Xi' which follows from the small Fermat theorem, we may replace xt in F by Xi' Since equivalence is preserved under addition and multiplication, we shall obtain a polynomial which is equivalent to F but which contains Xi to a lower degree. This process can be continued until we arrive at an equivalent polynomial which is of degree less than p in each variable Xi' Such a polynomial we call reduced. It is clear that when xt is replaced by Xi' the total degree of F (in all its variables) is not increased. Hence we obtain the following result. Theorem 1. Every polynomial F is equivalent to a reduced polynomial F*, whose total degree is not greater than that of F.

We now show that the reduced polynomial equivalent to a given polynomial is uniquely determined. Theorem 2. If two reduced polynomials are equivalent, then they are

congruent. This theorem is proved in precisely the same way as is the theorem mentioned above on the identity of polynomials, namely, by induction on the number of variables. It clearly suffices to show that if the polynomial F is reduced and F", 0, then F == 0 (mod p). We consider first the case n = 1. If the degree of F(x) is less than p and F(c) == 0 (mod p) for all c, then F has more roots than its degree, and this is possible only if all coefficients of F are divisible by p, that is, F == 0 (mod p). For arbitrary n ~ 2, we write F in the form F(xl> ... , x n) = AO(XI' ... , Xn-l) + AI(XI, ... , I +Ap-I(XI, ... , xn_I)X/- .

Take an arbitrary set of values cn - l )

= ao , '"

Xl

= CI ,

... ,

xn -

l

=

Xn-l)Xn

Cn-l'

+ ...

and set Ao(cl> ... ,

, Ap-I(CI' ... , cn - l ) = ap-l' Then F(c l , ... ,

Cn - l ,

x n)

=

ao + alXn

+ '" + ap- l

We have obtained a polynomial in one variable

Xn ,

x/- l

which is equivalent to

Sec. 1)

CONGRUENCES WITH PRIME MODULUS

5

zero, since F", O. But for polynomials in one variable the theorem has been proved, and therefore the above polynomial must be congruent to zero. Thus AO(c l , ... , Cn-l)

== 0 (mod p),

Ap-I(C I ,,,,, Cn-l)

== 0 (mod p),

that is, A o '" 0, ... ,A p - l '" 0 (since C b ... ,Cn-l were arbitrary). Since the polynomials A j are clearly reduced and depend on n - I variables (and for such polynomials the theorem is true by the induction hypothesis), then A o == 0 (modp), ... , A p -

I

== 0 (modp),

from which it follows that F", 0 (mod p). 1.2. Theorems on the Number of Solutions of Congruences

From Theorems I and 2 we can deduce some corollaries on the number of solutions of congruences. Theorem 3. If the congruence F(XI' .,. , x n ) == 0 (mod p) has at least one solution, and the total degree of the polynomial F is less than the number of variables, then the congruence has at least two solutions. Proof. Assume that the polynomial F(x l , ... , x n) with total degree r is such that the congruence F(x l , ... , x n) == 0 (mod p) has the unique solution Xl

==

al

(mod p), ... ,

Xn

==

an

(mod p).

Set H(x l , ... , x n) = I - F(x l , ... , xny-l. By the small Fermat theorem and the assumptions on F we have for

Xl

==

G I , .... X n

==

an

(mod p),

otherwise.

Denote by H* the reduced polynomial equivalent to H, by Theorem I. H* takes the same values as H. But, on the other hand, we can explicitly construct a reduced polynomial taking the same values as H, namely, the polynomial n

TI (I -

(Xi - aiy-l).

i= I

By Theorem 2 we have n

H*

==

TI (1 i= I

(x j

-

aiy-l) (mod p).

(1.1)

[Chap. 1

CONGRUENCES

6

From Theorem 1 it follows that the degree of H* is not greater than the degree of H, that is, not greater than rep - 1). Thus the degree of the left side of (Ll) is not more than rep - 1) and the degree of the right side equals n(p - 1). Hence n(p - 1) ~ rep - 1), and this proves the assertion for the case r < n. Corollary (Chevalley's Theorem). If F(xl> '" ,xn ) is a form of degree

less than n, then the congruence F(Xl> ... , x n ) =:

°

(mod p)

has a nonzero solution. The existence of such a solution follows from Theorem 3, since one solution, namely, zero, always exists in this case. To complete the picture, we shall show that the inequality r < n cannot be weakened if Chevalley's theorem is to remain valid. We shall construct for every n a form F(x l , .. , , x n ) of degree n, such that the congruence (1.2)

F(Xl> ... , x n ) =: 0 (mod p)

has only the zero solution. We use the fact that for any n ~ I there is a finite field ~ with pn elements, which contains Zp as a subfield (see the Supplement, Section 3, Theorem 2). Let W l , ... , W n be a basis for the field rover Zp. Consider the linear form X1Wl + .,. + XnW n , in which Xl' ... 'X n may take arbitrary values in Zp. Its norm Nr./Zp(X1W l + ... + XnW n) = '" , x n) = 0 has only the zero solution in the field Zp. Now replace each coefficient of the form ... , with " + ... + < n. and the system of congruences

'm

F1(xl> ... , x n) Fm(x" .... x n )

=0 (mod p), =0 (modp).

'm

(1.7)

has at least one solution, then it has at least two solutions. 6. Show that if the conditions of Problem 5 are fulfilled, then the number of solutions of the system (1.7) is divisible by p. 7. Show that if [is a quadratic form of rank;;' 2 over the field Zl' and a;t= 0 (mod p) then the congruence f= a (modp) has a solution. 8. Using Theorems 2 and 3 of Supplement, Section 1, prove that two nonsingular quadratic forms of the same rank over the field Zp (p '# 2) are equivalent if and only if the product of their discriminants is a square. 9. Determine the Witt group of classes of quadratic forms over the field Zl' (p '# 2) (see Problem 5 of Section I of the Supplement). 10. Show that the number of nonzero solutions of the congruence [(x, y) 0 (mod p). where [(x, y) is a quadratic form with discriminant d;t= 0 (mod p), is equal to (p - 1) (1 + ( -dip). 11. Using Theorem 7 of Supplement, Section 1, show that if [(Xl> .... x.) is a quadratic form with discriminant d;t= 0 (mod p) and p '# 2. then the number of nonzero solutions of the congruence [(x" .... x.) 0 (mod p) is equal to

=

=

1)./2d) pl./2)-1 p. -, - 1 + (p - 1) --p-( p.-' - 1

for n even, for n odd.

12. Under the assumptions of Problem 11. find the number of solutions to the congruence [(x" .... x n ) a (mod p).

=

2. Trigonometric Sums

2.1. Congruences and Trigonometric Sums

In this section (as in the preceding one) we shall consider congruences modulo a prime p, but from a somewhat different point of view. In the theorems of Section 1 we drew conclusions about the number of solutions of congruences, depending on the degrees and the number of variables of the polynomials involved. Here the principal role will be played by the value of the prime modulus p. We first note that for the equation F(x 1 , ... , x n) = 0 to have a solution, it is necessary that for all m the congruence F == 0 (mod m) have a solution. Even if we limit our considerations to prime values of m, we still have an

10

CONGRUENCES

[Chap. 1

infinite number of necessary conditions. Clearly, these conditions can be used only if we have a finite method (a method involving a finite number of operations) for verifying them. It can be shown that for a very important class of polynomials such a method (moreover, a very simple one) exists. Namely, for a given polynomial F with integer coefficients from this class the congruence F == 0 (mod p) has solutions for all values of p larger than some bound. The polynomials of which we speak are described by the following definition. Definition. A polynomial F(Xl' ... , x.) with rational coefficients is called absolutely irreducible if it cannot be factored in a nontrivial manner in any extension of the field of rational numbers.

The following fundamental theorem holds. Theorem A. If F(x 1 , ... ,x.) is an absolutely irreducible polynomial with integer coefficients, then the congruence F(x 1 ,

'" ,

x.)

== 0 (mod p)

(2.1)

is solvable for all prime numbers p, larger than some bound which depends only on the polynomial F. An analogous result holds for nonzero solutions when F is homogeneous, and (when the definition of absolute irreducibility is suitably generalized) the result generalizes to systems of congruences. For n ;; 1 Theorem A is trivial (any polynomial of degree greater than 1 factors in the field of complex numbers and for polynomials of degree I the assertion is obvious). But already in the case n = 2 its proof requires the use of deep results from algebraic geometry. The first proof of Theorem A for n = 2 was given by WeiI (A. Weil, Sur les courbes algebriques et les varietes qui s'en deduisent, Act. Sci. Ind. 1041, Paris, Hermann, 1948). The best versions of the proof which have appeared are contained in S. Lang, "Abelian Varieties," Wiley (lnterscience), New York, 1959; and A. Mattuck and J. Tate, On an inequality of Castelnuevo-Severi, Abhandl. Math. Sem. Hamburg 22, 195-199 (1959). The transition from n = 2 to the general case proved to be much easier. This was done in L. B. Nisnevich, On the number of points of an algebraic variety in a finite prime field, Dokl. Akad. Nauk SSSR 99, No.1, 17-20 (1954), and in S. Lang and A. Weil, Number of points of varieties in finite fields, Am. J. Math. 76, No.4, 819-827 (1954). These last two papers actually contain a result considerably stronger than the assertion of Theorem A. Namely, they show that if the form Fis fixed and the prime modulus p varies, then the number N of solutions to the congruence (2.1) becomes arbitrarily large as p increases, and they even give an estimate for the rate of increase of N. Their result can be formulated precisely as follows.

Sec. 2]

TRIGONOMETRIC SUMS

11

Theorem B. The number N(F,p) of solutions of the congruence (2.1) satisfies the inequality IN(F, p) - pn-ll < C(F)pn-l-(1f2),

where the constant C(F) depends only on the polynomial F and not on p. All known proofs of Theorem A go by way of Theorem B. But the proof of Theorem B delpands an algebraic apparatus much more complex than any which we shall use in this book. Therefore we shall not give the proofs of Theorems A and B but instead shall describe a method which can be used to prove these theorems in certain cases, and we shall work out one of these cases. All our work will be based on the fact that the number of solutions to (2.1) can be given in an explicit formula, or more precisely, can be represented as a sum of certain pth roots of unity. Sums of this type are called trigonometric. We set up the following notations. If f(x) or f(x l , .. , , x n) are complexvalued functions whose value depends only on the residue class of the integers x, Xl' ... , X n modulo p, then by Lf(x)

and

x

Xt,.·.,X n

we denote the sums where the values of X and of Xl' full system of residues modulo p, and by

... , X n

are taken from a

L'f(x) x

the sum where X takes all values from a reduced system of residues. Let' be some fixed primitive pth root of 1. Then it is clear that for for

y == 0 (mod p), y"l- 0 (mod p).

(2.2)

It is this equation which makes it possible to find an explicit formula for the number of solutions of the congruence (2.1). Consider the sum

If the values of Xl'

'"

, Xn

give a solution of (2.1), then, by (2.2),

L

(XF(xl •...•x n )

= p.

X

The sum of all such terms entering into S is therefore equal to Np, where N is the number of solutions to the congruence (2.1). If F(Xl' ... , x n) ¢. 0 (mod p), then again, by (2.2),

[Chap. 1

CONGRUENCES

12

I

(xF(Xt, ... ,X n )

= O.

x

The sum of all such terms in the formula for S is then zero and we have that S = Np. We have thus proved Theorem 1. The number N of solutions to the congruence (2.1) is given by the formula

(2.3) All terms in which x == 0 (mod p) enter into the sum (2.3). Since each such term is equal to I, and they are p' in number (each of the variables Xl> ... , x. taking on p different values independently), then N = p.-1

+ -1 I'

I

P

Xl.···.X n

X

cF(Xt, ... ,X n ).

(2.4)

In this form of the formula for N we see a suggestion of Theorem B. The term p.-1 is already singled out. We must only show (but this is where all the difficulties lie!) that as p increases the sum of all remaining terms increases in ab-

solute value more slowly than does the principal term p' -1. 2.2. Sums of Powers

We now apply the general method of the preceding section to the case when the polynomial F is equal to a sum of powers of the variables, i.e., ai

'I- 0 (mod p).

We shall assume that n ~ 3, since for n = 1 and n = 2 the number of solutions of the congruence F == 0 (mod p) can be found by an elementary method. By formula (2.4) the number N of solutions to the congruence a1xt" + .,. +a.x.'n == 0 (mod p) is given by the expression

which can be written in the form (2.5)

Hence we must investigate sums of the form

I y

(ayr(a

'I- 0 (mod

p».

Sec. 2]

TRIGONOMETRIC SUMS

Clearly,

I

,a y' = I m(x)(ax,

13

(2.6)

.
3 the inequality (2.16) is much stronger than that of Theorem B. Remark. For the proof of Theorem 3 it would suffice, by (2.5), to find a bound for the absolute value of the sum ,ax'. Such a bound can be found, moreover, by a shorter route, without the use of Gaussian sums (see Problems 9 to 12, for which the authors thank N. M. Korobov). We have chosen the proof involving Gaussian sums because Gaussian sums have many other uses in number theory.

Ix

2.3. The Absolute Value of Gaussian Sums Consider the set lj of all complex functions f(x), defined for rational integers x, and satisfying the condition: f(x) = f(y) if x == y (modp). Since each functionf(x) E lj is determined by its values on a full system of residues modulo p, lj is a p-dimensional linear space over the field of complex numbers. We introduce a Hermitian inner product on lj by setting

(f, g)

1

=-

p

If(x)g(x) x

(f,

g E lj).

16

CONGRUENCES

[Chap. 1

It is easily checked that with respect to this inner product the p functions (a a residue (modp))

(2.17)

form an orthonormal basis for (Y. Indeed, by (2.2), (fa ,fa') =

l~

c(a) satisfies all conditions of being a metric and hence turns I 1, for which cp(a) > I, or else cp(n) ~ 1 for all natural numbers n. Consider the first case. Since

cp(n) = cp(1

+ ... + 1) ~ cp(I) + ... + cp(1) = n,

(4.3)

[Chap. 1

CONGRUENCES

38

we may set (4.4) where IY. is real and satisfies 0 < IY. < I. Taking an arbitrary natural number N, we decompose it in powers of a N

where 0 equality

~ Xi ~

= Xo + xla + '" + xk-lif-t,

a - I (0

~ i ~

k - I),

Xk-I

~

1. Hence N satisfies the in-

if-I~N CIW,

where the constant C I does not depend on N. Let m again denote an arbitrary

Sec. 4]

AXIOMATIC CHARACTERIZATION OF p-ADIC NUMBERS

39

natural number. If N is replaced by N in the preceding mequality, then nl

from which and as m -+

00

this yields cp(N)

~

W.

(4.6)

Comparing (4.5) and (4.6), we see that cp(N) = N a for any natural number N. Now let x = ±Nt!N2 be an arbitrary rational number, different from zero (N! and N 2 are natural numbers). Then a

N 1) cp(Nd N 1 cp(x) = cp ( = - - = -a = N2 cp(N 2 ) N 2

Ixl a .

We have shown that if cp(a) > 1 for at least one natural number a, then the metric cp is of the form (4.2). We now turn to the case where cp(n)

~

(4.7)

1

for all natural n. If for every prime p, we had cp(p) = 1, then by condition (3) we would also have cp(n) = 1 for all natural n, and thus also cp(x) = 1 for all rational x # O. But this would contradict the assumption that cp is nontrivial. Thus for some prime p we have cp(p) < 1. Assume that for some other prime q # p we also had cp(q) < 1. We take exponents k and I so that cp(p)k
... ,xn) == 0 (modp)

(5.7)

satisfies the inequality (5.8) in which the constant L is equal to the total degree of F. We prove the lemma by induction on n. For n = 1 it follows from the fact that the number of roots of a nonzero polynomial in the field Zp cannot exceed its degree. If n > I, consider F(x l , ... , x n) as a polynomial in XI' ... , Xn- I ' the coefficients of which are polynomials in Xn • Denote by f(x n ) the greatest common divisor of these coefficients modulo p. Then

where for any a the polynomial FI(x l , ... , Xn- I , a) is not identically congruent to zero modulo p. Let I and L I be the degrees off and F I , respectively. It is clear thatfand FI can be chosen so that 1+ L I ~ L. We can now bound the number of solutions (c I , ... , cn) to the congruence (5.7) by considering the different values for X n in these solutions. Consider first those solutions for which f(c n) == 0 (modp). (5.9) If (5.9) is fulfilled, then for any choice of CI , ... , Cn - I we obtain a solution of (5.7). Since the numbers of values of Cn , modulo p, is at most I, then the number of solutions of (5.7), for which (5.9) holds, is at most lpn-I. Consider now solutions for which f(c n ) =t= 0 (mod p). All such solutions clearly satisfy the congruence FI(x l , ... ,xn) == 0 (modp). Since F(XI' ... ,Xn- I , cn) is not identically congruent to zero modulo p, then by the induction hypothesis the number N(p, cn) of solutions of the congruence F(x l , ... , Xn-l> cn) == 0 (mod p) satisfies the inequality N(p, cn) ~ L lp n-2. Since Cn takes not more than p values, the total numbe~ of solutions of this type does not exceed L1pn-l. Thus the total number of solutions of (5.7) does not exceed Ipn-l + L1pn-1 ~ Lpn-I, which is what was to be proved. Proof of TheoremC. We may, of course, assume that the polynomial F

CONGRUENCES

40

[Chap. 1

actually depencis on the variable X n • Consider F as a polynomial in X n with coefficients which are polynomials in XI' ... ,Xn-I' Since F is absolutely irreducible, it follows that the discrimin:mt Drn(xI' ... , x n - I ) of the polynomial F, considered as a polynomi,'ll in X n , is a polvnomial in Xl' ... , Xn-I which is not identically zero, since otherwise Fwould be divisible by the square of some polynomial. Consider a prime number p, which does not divide all coefficients of Drn(x l , ... , x n - I ), and let NI(fJ) be the number of sohltions of (5.6). If (CI' ... , cn) is a solution of (5.6), then Cn is a common root, modulo p, of the polynomials F(c l , ... ,C n - I , x) and F'rn(cI' ... , Cn - t , x n ) and therefore

By the lemma, the number of solutions to this con!?;ruence does not exceed K lpn-2, where K I is some constant which depends only on the polynomial F. For given CI , ... , Cn-I the values of Cn are determined by the congruence F(c l ,

... , Cn-I'

x n)

== 0 (mod p)

and therefore there are at most m of them, where m is the degree of the polynomial Fin X n . Thus the number NI(p) of solutions to system (5.6) does not exceed Kpn - 2, where K = mKI . We now show that for sufficiently large p, the number N(p) solutions to the congruence (5.7) is larger than the number NI(p) of solutions to system (5.6). Indeed, by Theorem B, N(p)

> pn-I _ Cpn -

I -(1/2l,

and we have just shown that N I (p) < Kpn - 2. Thus N(p) - NI(p)

> pn-I _ C p n-I-(l/2) _ K p n-2 = pn-2(p _ C p l/2 - K),

which means that N(p) > NI(p) for sufficiently large p. Thus, for sufficiently large p, the congruence F == 0 (mod p) has a solution (YI' ... , r'n) for which

aF

-a (YI"'" Yn) i= 0 (mod p). Xn

By the corollary of Theorem 3, this proves that the equation F tion in the ring Op for all p, larger than some given constant.

= 0 has a solu-

PROBLEMS

1. Show that if m and p are relatively prime, then any p-adic unit E, satisfying the congruence E =; 1 (mod p), is an mth power in R p • 2. Let m = lma, (ma, p) = I, and let E ==: 1 (mod p2 6+1). Show that the p-adic unit E is an mth power in R p •

Sec. 5]

QUADRATIC FORMS WITH p-ADIC COEFFICIENTS

47

3. If p # 2 and the p-adic integers <X and f3 are not divisible by p, show that the solvability of the congruence <Xx' ~ f3 (mod p2) implies the solvability of the equation <Xx 2 = f3 in the field R•. 4. Assume that the coefficients E, in the form G = E,X,' + ... + EnX n' are p-adic units (p #2). Show that if the congruence G =0 (l (mod p") has a solution in which at least one of the variables is not divisible by p, then the equation G = 0 has a nonzero solution in the field R•. 5. Let all coefficients of the form G = <X,x,' -+- ... + <xnxn' be p-adic Integers which are divisible at most by the (p - I)th power of p. If the congruence G ~ 0 (mod p'+ 2) has a solution in which not all variables are divisible by p, show that the equation G = 0 has a nonzero solution in the field R•. [If p # 2, it suffices to have a solution to the congruence G =0 0 (mod p'+').] 6. Let the quadratic form F= <X,X,2 + ... + <xnx/ have coefficients which are p-adic integers (p # 2) not divisible by p. Show that if the congruence F == 0 (mod pO) has a solution in which not all values of the variables are divisible by p, then the equation F = 0 has a nonzero solution in the field R•. 7. If the form F= <X,x,m + ... + <xnxnm has coefficients which are nonzero p-adic integers, set r = v.(m), S = max(v p «(",(,), ... , vp(<X n» and N = 2(r + s) + 1. Show that the equation F = 0 has a nonzero solution in the field R. if and only if the congruence F =0 0 (mod p/'l) has a solution in whicn not all values of the variables are divisible by p. 8. Show that the form 3x' + 4 y 3 + 5z' represents zero in the field R. for all p (see Problem D, Section 2). 9. Let the polynomial F(x" ... , x n ) have coefficients in 0. and, denote by C m (m ;:> 0) the number ()f solutions to the congruence F(x" ... , x n ) ~ 0 (mod pm). Consider the series cp(t) = L.:.'=o Cmf m. It has been conjectured that the series cpU), called the Poincare series of the polynomial F, represems a rational function of f. Find the Poincare series for the polynomial F = E,X,2 + ... + EnX/, where E, is a p-aaic unit, and check that the function cp(f) is rational. 10. Find the Poincare series for a pOlynomial F(x" ... , x n ) withp-adic integral coefficients, which satisfies the condition that for any solution of the congruence F ",,0 (mod p), of/ox, "'" 0 (mod p) for some i = 1, ... , n. 11. Compute the Poincare series for the polynomial F(x, y) = x 2 _ y3.

6. Quadratic Forms with p-Adic Coefficients

In this and the next sectIOn we shall apply tne theory of p-adic numbers which we'have developed to the investigation of the simplest types of equations. We shall consider the problem of the representation of p-adic and rational numbers by quadratic forms. The algebraic preliminaries that we shall need on the properties of quadratic forms over arbitrary fields are given in the Supplement, Section 1. 6.1. Squares in the Field of p- Adic Numbers

For the study of quadratic forms over a given field it is important to know which of the elements of the field are squares. Therefore we first turn to the

[Chap. 1

CONGRUENCES

48

study of squares in the field R p of p-adic numbers. We know (Section 3, Theorem 4) that every nonzero p-adic number rx can be represented uniquely in the form rx = pm e, where e is a p-adic unit (that is, e is a unit in the ring Op of p-adic integers). If rx is the square of the p-adic number y = pk eo , then m = 2k and e = e/. To determine alI squares of the field R p , we must thus determine which units of Op are squares. Theorem 1. Let p # 2. In order that the p-adic unit

(0 ~ ci < p,

Co

# 0)

(6.1)

be a square, it is necessary and sufficient that the integer Co be a quadratic residue modulo p. Proof. If e = 1]2 and I] == b (modp) (b a rational integer), then Co == b2 (mod p). Conversely, if Co == b 2 (mod p), let F(x) = x 2 - e. We have F(b) == 0 (modp) and F(b) = 2b ¢ 0 (modp). By the corollary of Theorem 3 of Section 5 there is a I] E Op such that F(I]) = 0 and I] == b (mod p). Thus e = 1]2, and the theorem is proved. Corollary 1. If p # 2, any p-adic unit which is congruent to I modulo p is a square in R p • Corollary 2. If p # 2, the index (R p * : R p *2) of the subgroup of squares R p *2 in the multiplicative group of the field R p is equal to 4.

For if e is not a square, then the quotient of any pair of numbers from I, e, p, pe is not a square in R p • But any nonzero p-adic number can be represented as the product of one of the numbers 1, e, p, pe with some square. If p # 2 and the unit e is given by (6.1), set if e is a square in R, otherwise. By Theorem 1 we have

where (co/p) is the Legendre symbol. If e is a rational integer relatively prime to p, then the symbol (e/p) which we have defined clearly coincides with the Legendre symbol. It is easily seen that for p-adic units e and I] we have

(e;) = (~)(~). We turn to the case p = 2.

Sec. 6]

QUADRATIC FORMS WITH p-ADIC COEFFICIENTS

49

Theorem 2. In order that the 2-adic unit e be a square (in the field R 2 ), it is necessary and sufficient that e == 1 (mod 8). Proof. The necessity follows from the fact that the square of an odd integer is always congruent to 1 modulo 8. To prove sufficiency, set F(x) = x 2 - e and apply Theorem 3 of Section 5, taking b = 1 and y = 1. Since F(l) == 0 (mod 8) and P(l) = 2 ¢ 0 (mod 4), the theorem implies that there is an ,., == 1 (mod 4), such that F(,.,) = 0; that is, e = ,.,2.

Corollary (R 2 * : R 2 *2) = 8, where R 2 *2 is the subgroup of squares of the multiplicative group R 2 * of the field of 2-adic numbers.

By the above theorem the reduced system of residues modulo 8, namely, I, 3, 5, 7, forms a system of coset representatives for the subgroup of squares in the group of all 2-adic units. If we also take the products 2· 1, 2· 3, 2· 5, 2· 7, then we obtain a full system of coset representatives for the subgroup R 2 *2 of the group R 2 *.

6.2. Representation of Zero by p-Adic Quadratic Forms As is the case in any field, a nonsingular quadratic form over the field R p can be put in the form (a i =1= 0) after a linear change of variables (see the Supplement, Section 1.1). If a i = p2k'ei or ai = p2k i +l ei (ei a unit in Op), then after the substitution pk'X i = Yi we obtain a form in which all coefficients are p-adic integers which are divisible at most by the first power of p. Thus any nonsingular quadratic form over the field R p is equivalent to a form

F

= F o + pF I = e\x/ + ... + erx/ + p(er+lxr+/ + '" + enx/),

(6.2)

where the ei are p-adic units. While considering the question of the representation of zero, we may assume that r ~ n - r. The form pF is clearly equivalent to the form F I + pFo . Since F and pF simultaneously represent zero, we may take the form F1 + pFo instead of Fo + pFI' We first consider the case p =1= 2. Theorem 3. Let p =1= 2 and 0 < r < n. The form (6.2) represents zero in the field R p if and only if at least one of the forms F o or F I represents zero. Proof. Let the form (6.2) represent zero:

el~/

+ ... + er~/ + p(er+J~r+/ + ... + en~/) = O.

(6.3)

CONGRUENCES

50

[Chap. 1

We may assume that all ~ i are integers and that at least one of them is not divisible by p. If not all ~1' ... , ~r are divisible by p, say ~1 ¢ 0 (mod p), then, considering (6.3) modulo p, we have FO(~I' ... , ~r) =:

0 (mod p),

oFo (~1' ... '~r) = 2Bl~1 ¢ 0 (mod p).

-

OXI

By the corollary of Theorem 3 of Section 5, the form F o represents zero. Assume now that ~1' .,. , ~r are all divisible by p, so that B1 ~/ + .,. + Br~/ =: (mod p2). We consider (6.3) modulo p2. Dividing this congruence by p, we obtain

where at least one of ~r+ l ' '" , ~n is not divisible by p. Again appiying the corollary of Theorem 3 of Section 5, we conclude that in this case the form F 1 represents zero. Since the sufficiency of the condition is obvious, Theorem 3 is proved. "i'he foilowing corollaries are immediate. are p-aaic umts and p =I 2, then the form represents zero in R p if and only if the congruence =: 0 (mOd p) has a nomnviai solution in Op.

Corollary '1. If /

k

B1

/(x 1 ,

B1 , ••• ,B r

x/ + ... + BrX/

•• , ,

xr )

Corollary l. if we also assume that r ~ 3, then the form /(Xb '" , x r ) always represents zero in R p •

For by Theorem 5 of Section 1, the congruencej(x1 , ... ,xr ) =: 0 (modp) has a nomrivlal solution. In the proof of Theorem 3 the equality (6.3) was not actually used; we used only the congruences F =: 0 (mod p) and F =: 0 (mod p2). Thus the solvability of the second of these congruences already implies that one of the forms F o or F 1 , and hence F, represents zero. Hence we have Corollary 3. If p =I 2 the form (6.2) represents zero if and only if the congruence F == 0 (mod p2) has a solution in which not all variables are divisible by p.

We now cOIJsioer quaoratlc forms over the field of 2-adic nurnoers. In this case Theorem 3 and all its coroliaries are false. For example, if / = x/ + x/ + x/ + x/, then tne equationj= 0 has no nontrivial solution (since the congruence/=: 0 (mod 8) has no solution in Integers, at least one of which is odd). But we shall see that the form / + 2x/ does represent zero in R 2 (Theorem 5).

Sec. 6]

QUADRATIC FORMS WITH p-ADIC COEFFICIENTS

51

Theorem 4. The form (6.2) (with p = 2) represents zero in the field of 2-adic numbers if and only if the congruence F == 0 (mod 16) has a solution in which at least one of the variables is odd.

Proof Let F(~I' ... , ~n) == 0 (mod 16), where not all of the 2-adic integers ~; are divisible by 2. We fin~t assume that ~ i ¥= 0 (mod 2) for some i ~ r, say ~l ¥= 0 (mod 2). Since F(~l' ... , ~n) == 0 (mod 8) and (aF/aXl) (~1l '" , ~n) =2el~1 ¥= 0 (mod 4), by Theorem 3 of Section 5, the form F represents zero. If ~1l ... ,~r are all divisible by 2, set ~; = 21'/; (l ~ i ~ r), where 1'/; is a 2-adic integer. Divide the congruence r

4

L

i=1

n

10;1'//

+2

L

i=r+1

e;~/ == 0 (mod 16)

by 2 to obtain n

r

L e,e + 2 ;;1 L 10;1'/;2 == 0 (mod 8), ;;r+l where at least one of ~r+l' ... , ~n is not divisible by 2. As above it follows from this congruence that the form Fl + 2Fo represents zero. Since the forms F and 2F represent zero simultaneously, the sufficiency of the condition is proved. The converse is obvious. In the course of the proof we have obtained the following result. Corollary. If the congruence F == 0 (mod 8), where F is given by (2) with p = 2, has a solution in which at least one of the variables Xl' ••• , X r takes an odd value, then this form re9resents zero in the field R 2 . Theorem 5. Any quadratic form over the field R p of p-adic numbers in five or more variables always represend zero. ,

~ n - r. Since n ~ 5, then r ~ 3. Corollary 2 of Theorem 3 then implies that the form F o , and hence also the form F, represents zero. The theorem is proved if p =I- 2. Let p = 2. If n - r> 0, consider the" partial" form 1= elx12 + e2x22 + e3x/ + 2e nx/. We claim that such a form always represents zero in R 2 . Since 101 + 102 = 2a (ct a 2-adic integer), then 101 + 10 2 + 2e na 2 == 2a + 2a 2 = 2a(l + a) == 0 (mod 4), that is, 10 1 + 102 + 2en IX 2 = 4f3, where f3 is a 2-adic integer. Setting Xl = X2 = 1, X3 = 2f3, Xn = IX, we have

Proof. We may assume that our form is (6.2) with r

By the corollary of Theorem 4 the form I represents zero. But then F also represents zero. In the case n = r, we take as a partial form I = el x 1 2 + f: 2 x/

52

CONGRUENCES

[Chap. 1

+ E3X32 + E4X/ + EsX/. If 1'1 + 1'2 == 1'3 + 1'4 == 2 (mod 4) then set Xl = x 2 = = x 4 = 1, and if, say, el + e 2 == 0 (mod 4), set Xl = X 2 = 1, X3 = X 4 = O. In general we get 1'1 X1 2 + E2 X/ + E3X/ + E4X/ = 4y, where y is a 2-adic integer. Set X 2 = 2y, and then X3

f== 4y + 4y2 == 0 (mod 8). We complete the proof by applying the corollary of Theorem 4. Theorem 5 is completely proved. By Theorem 6 of Section 1 of the Supplement, Theorem 5 implies the following corollary. Corollary 1. Any nonsingular quadratic form in four or more variables over the field R p represents all p-adic numbers. Corollary 2. Let F(x l , ... ,x.) be a nonsingular quadratic form whose coefficients are rational integers. If n ~ 5, then for any m the congruence F(xl> ... , x.) == I{ (mod m) has a nontrivial solution.

Indeed, since the form F represents zero in R p , then for any s ~ 1, the congruence F == 0 (mod pS) has a solution in which at least one variable is not divisible by p.

6.3. Binary Forms

Binary quadratic forms form an important special case of the general theory. We consider the question of the representation of numbers of the field R p by the quadratic form tl =I

0,

(6.4)

(Any nonsingular binary form can be put in this form by a change of variables and by multiplying the form by some p-adic number.) Let H a denote the set of all nonzero p-adic numbers represented by the form (6.4). This set has the surprising property of being a group under multiplication. Indeed, if p = x2 - tly2, Pl = x/ - tly/, then a simple computation shows that

Sec. 5]

53

QUADRATIC FORMS WITH p-ADIC COEFFICIENTS

Another proof of this fact can be given, using the quadratic extension Rp(-Ja) of the field R p (assuming that (1. is not a square in R p )' The equation P= x 2 - (1.y2 simply says that P is the norm of the number ~ = x + y-Ja of the field R p ( -Ja). But if P = N(~) and PI = N(~I)' then PPI = N(~~I) and p-l = N(C 1 ). If (1. is a square in R, then the form (6.4) represents zero, and hence represents all numbers of R p • Hence in this case HI1. coincides with the entire multiplicative group R p * of the field R p • Since the form (6.4) represents all squares of the field R p (set y = 0), then R p*2 C HI1.' By the corollaries to Theorems I and 2 the index (R p* : R p*2) is finite, so that the group HI1. has finite index in R/. Theorem 6. If the number

(1.

E

R p * is not a square, then (R p * : H a)

= 2.

Proof. First note that the form (6.4) represents the p-adic number and only if the form

P if (6.5)

represents zero (Theorem 6 of the Supplement, Section I). The represeotability of zero will not be changed if (1. and P are multiplied by squares. Hence we may assume that (1. and Pare taken from some fixed system of coset representatives of R p*2 inR p*. First, let p # 2. We claim that HI1. # R p *2. This is clear if - (1. is not a square (since -(1. E HI1.)' If -(1. is a square, then the form x 2 - (1.y2 is equivalent to the form x 2 + y2, which represents all p-adic units (Corollary 2 of Theorem 3), so that HI1. does oot coincide with R p *2. Further, we claim that HI1. does not coincide with R p * (assuming, of course, that (1. ¢ R p *2). If e is a nonsquare p-adic unit, then we may assume that (1. is e, p, or pe. But by Theorem 3 (and Theorem 10 of the Supplement, Section I) the form (6.5) does not represent zero if (1. = e, P = p, or when (1. = p or pe and P = e. Thus HI1. # R/. Since 2 R/ 2 c HI1. c R/, the index (R/ : HI1.) must divide the index (R/: R/ ) = 4 (by Corollary 2 of Theorem I). But we have shown that the index is neither 4 nor I, so that (R/ : HI1.) = 2 and Theorem 6 is proved in the case p #2. Now let p = 2. In this case we have (R 2 * : R 2 *2) = 8, and as coset representatives we may take the numbers I, 3, 5,7,2'1,2'3,2'5,2'7. We shall therefore assume that (1. and P, in the form (6.5), are taken from this set. We thus need to check which of these forms represents zero in R. The answer is given in the following table, in which a "+" sign denotes that for the corresponding (1. and P the form (6.5) represents zero in R,and an empty square denotes that the form does not represent zero.

CONGRUENCES

54

ZI

~ ex

f3

I

I

1

I

3

: I 5

I I

7

[Chap. 1

I I 2·1

I II

I

I

2·3

I[

2·5

2·7

I-+-I-+-:-+-I-+-!-+-I-+-\-+-!-+3 1_+_1=1 + I=!=I + :=! +5 1+1+[+'+1 I -------I---------'------~-----7 ,+1 1+1 1+1 1+1 '----'----'--:--,--1-2·1 I +! i ; + , I+ ,I + I 1

I

1

I

I

______1

I-+-:-+-i--I--l-~I--,-+-'-~

2.3

_------_1

_ _ _2_.5_ _ 1

i

2·7

!

I

1

'

I

I

+ I=!=I ~ 1=1 + + 1= + +! +I +I i I

[Since the form (6.5) is symmetric in ex and fJ, the table is symmetric about its main diagonal.] We see that in each row except the first one the" +" occurs in four columns. This means that for each nonsquare ex E R 2 *, the form (6.4) represents precisely four cosets of the subgroup R 2 *2. Thus (Ha : R 2 *2) = 4 and since (R 2 * : R 2 *2) = 8 (corollary of Theorem 2), then (R 2 * : H a ) = 2. We use the results of Section 6.2 to verify the table. Let ex = 21', fJ = 2f/, where I' and f/ are 2-adic units, and let 2ex 2 + 2f/y 2 - Z2

= O.

(6.6)

We may assume that x, Y and z are integers, not all divisible by 2. It is clear that z == 0 (mod 2), and that neither x nor yare divisible by 2 [otherwise the left side of (6.6) would not be divisible by 4]. Setting z = 2t, we put (6.6) in the form 2 2 2 ex + f/y - 2t = O. This equation, by the corollary of Theorem 4, is equivalent to the corresponding congruence modulo 8 (with x andy odd). Sincex 2 == y 2 == 1(mod 8), and either 2t 2 == 2 (mod 8) or 2t 2 == 0 (mod 8), then (6.6) is solvable if and only if one of the following holds:

+ f/ == 2 (mod 8); I' + f/ == 0 (mod 8). Let now a = 21', fJ = 'I. In the equation 2ex 2 + '1y2 - Z2 = 0 (with x, y and z I'

2-adic integers not all divisible by 2) we obtain, by similar reasoning, y

=1=

0

Sec. 6]

QUADRATIC FORMS WITH p-ADIC COEFFICIENTS

55

(mod 2) and z '1= 0 (mod 2). Hence (again by the corollary of Theorem 4), this equation can be satisfied if and only if we have one of the congruences: 2E

+ 11 ==

1 (mod 8);

11 == 1 (mod 8);

(6.7)

which correspond to the cases 2 .t x and 21 x. Only the case 11. = E, fl = 11 remains. If in the equation EX 2 + l1y 2 - Z2 = 0 the p-adic integers x, y, and z are not all divisible by 2, then precisely one of them is divisible by 2 and the other two are not. If z == 0 (mod 2) then EX 2 + l1y2 == E + 11 == 0 (mod 4), so that either E == 1 (mod 4) or 11 == 1 (mod 4). If z '1= 0 (mod 2), then EX 2 + l1y2 == 1 (mod 4), and since precisely one of the numbers x and y is divisible by 2, we again find that at least one of the congruences (6.8) E == 1 (mod 4), 11 == 1 (mod 4) holds. Conversely, assume, say, that E == 1 (mod 4). Then the congruence 2 + l1y2 - Z2 == 0 (mod 8) is satisfied by x = 1, y = 0, z = 1 if E == 1 (mod 8), and by x = 1, y = 2, z = 1 if E == 5 (mod 8), and this means that the form 2 EX + l1y2 - z2 represents zero. This ends the verification of the table and hence the proof of Theorem 6. From Theorem 6 it follows that if ex is a nonzero p-adic number which is not a square, then the factor group R/IH~ is a cyclic group of order 2. We can thus establish an isomorphism between this factor group and the group {I, -l} of square roots of I. The unique isomorphism between R/ (H~ and {I, - I} sends the subgroup H 7 to the number + 1, and the coset flH~, distinct from H~, to the number - 1. It will be easier for us to deal with the homomorphism of the group R p * onto the group {I, -I} with kernel H~, since then we will have a function on R p * (and not on the factor group EX

R/IH~).

Definition. For any pair 11. =f. 0, fl =f. 0 of p-adic numbers, we define the symbol (:x, fl) to be equal to + 1 or to -1, depending on whether the form exx 2 + f3y 2 - Z2 represents zero in the field R p * or not. The symbol (C(,f3) is called the Hilbert symbol. It follows immediately from the definition that if ex is a square, then (C(, 13) = 1 for all 13. If :x ¢; R p *2, then (C(,f3) = 1 if and only if 13 E H~. Thus for any ex =f. 0, the mappingf3 -> (:x, p) is a homomorphism of the group R p * to the group {I, -I} with kernel H,. In other words,

= (:x, f3,)(:x, fJ2)' (6.9) Further, the definition of the symbol (ct, 13) depends on the solvability of (6.5), which is symmetric in :x and 13, so that (6.10) (13,11.) = (ct, 13), (:x, fllf32)

56

CONGRUENCES

[Chap. 1

from which, by (6.9), (6.11) We note that

(ct, -ct) = I

(6.12)

for any ct E R p * (since the equation ctX 2 - cty 2 x = y = 1, Z = 0), and thus, by (6.9),

Z

= 0 has the solution

(ct, ct) = (ct, -I).

(6.13)

Using (6.9) to (6.13) the computation of(ct, fJ) in the general case is reduced to the computation of(p, B) and (B, 11), where Band 11 are p-adic units. Indeed, if ct = pkB, fJ = pll], then from these formulas we obtain

(lB, p l 11) = (p, ptl(B, p)l(p, 11)k(B, 1]) = (p, BI11k( - l t l)(B, 11). We now compute (p, B) and (S,11). If p =f: 2, then by Theorem 3 the form px 2 + By2 - Z2 represents zero if and only if By2 - Z2 represents zero, that is, if and only if the unit Bis a square. Thus (p, B) = (Bjp) for p =f. 2 (see Section 6.1). By Corollary 2 of Theorem 3, the form BX 2 + 11y2 - Z2 always represents zero, and thus (B, 11) = + 1 for any p-adic units Band 11 (p =f. 2). If p = 2, the values of the symbols (2, B) and (B, 11) have already essentially been found in the proof of Theorem 6. For by (6.7), with B = 1, the form 2x 2 + 11y2 - Z2 represents zero if and only if 11 == ± 1 (mod 8). Hence (2, 11) = ( _l)(~2 -1)/8. Further, the form BX 2 + 11y2 - Z2 represents zero if and only if one of the congruences of (6.8) is fulfilled. Thus (B, 11) = ( -1 )[('-1)/2J[(~-l)/21. Summing up, we have Theorem 7. The values of the iHilbert symbols (p, B) and (B, 11) for p-adic units Band 11 are given by the forrt1Ulas

G), (f' 11) = 1 I

(p, s)

=

(2, B) = (_1)(,2_ 1 )/8,

(B,

for

p =f: 2,

1J~ = (_1)[('-1)/2][(~-1)/2]

for p = 2.

I

6.4. Equivalence of Binary Forms The Hilbert symbol allows us to give explicit conditions for the equivalence of two binary quadratic forms over the field R p • Let f(x, y) and g(x, y) be two binary nonsingular quadratic forms over R p with determinants b(f) and b(g). For f and g to be equivalent, it is necessary that b(f) and b(g) differ by a factor which lies in R p *2 (Theorem 1 of the Supplement, Section 1). To formulate another necessary condition for equivalence, which, along with the above one will be sufficient, we need the following fact.

Sec. 6]

QUADRATIC FORMS WITH p-ADIC COEFFICIENTS

57

Theorem 8. Let the binary form f have determinant b l' O. Then the Hilbert symbol (a, -b) takes the same value for all nonzero p-adic numbers a represented by f

Proof. Let a and a' be two nonzero p-adic numbers represented by the form By Theorem 2 of Section 1 of the Supplement the formf is equivalent to a form fl of the type ax 2 + fJ y 2. Since ct' is also represented by f, then a' = axo2 + fJYo 2, so that act' - afJYo 7 - (ax o)2 = O. Hence the form aa'x 2 afJy 2 - Z2 represents zero, so that (aa', - afJ) = 1. But afJ differs from b by a square factor, so that (aa', -b) = 1, and thus by (a, -b) = (a', -b), which

f

proves the theorem. Theorem 8 implies that the binary form f has a new invariant, and we set e(f) = (a, -b(f)), where a is any nonzero p-adic number which is represented by f

Theorem 9. Letf and g be two nonsingular binary quadratic forms over the field R p • f and g are equivalent if and only if both of the following conditions hold: (I) b(f) = b(g)y2, (2) e(f) = e(g).

Proof The necessity of both conditions is clear. To prove sufficiency we first show that the two forms represent the same p-adic numbers. Let the number y E R p * be represented by the form g. Lettingf = ax 2 + fJ y 2, we have (a, -a/l) = e(f) = e(g) = (y, -b(g)) = (y, -afJ), by which

(ya- 1 , -afJ)

= 1.

By the definition of the Hilbert symbol this means that we can solve the equation ya- 1 x 2 - afJy 2 - Z2 = 0 in nonzero x, y, and z. But then

(ay)2 , (2)2 +/l-;

y=a~

that is, y is represented by the form f The equivalence off and g now follows from Theorem 11 of Section 1 of the Supplement.

6.5. Remarks on Forms of Higher Degree Theorem 5 on quadratic forms over the field R p is one of a class of theorems in number theory which run &s follows: "All is well as long as the number of

58

CONGRUENCES

[Chap. 1

variables is sufficiently large." In this case "well" means that the quadratic form represents zero over the field of p-adic numbers, and "sufficiently large" means that the number of variables is at least five. It would be most interesting to observe this phenomenon also in the case of forms of higher degree over the field of p-adic numbers. The precise formulation is this. For any natural number r there is a number N(r) such that any form of degree r over the field R p represents zero, provided that the number of variables exceeds N (r). We note that there is no reason to believe, a priori, that any such number N(r) exists, but Brauer showed that it does. However, his bound is rather large [R. Brauer, "A note on systems of homogeneous algebraic equations," Bull. Am. Math. Soc. 51 (1945) pp. 749-755]. For r = 2, Theorem 5 shows that N(r) = r2 . For r = 3, Demyanov and Lewis showed that N(r) = r2 also; that is, any cubic form over the field of p-adic numbers in at least 10 variables represents zero [V. B. Demyanov, "On cubic forms over discrete normed fields," Dokl. Akad. Nauk SSSR 74 (1950) pp. 889-891; D. J. Lewis, "Cubic homogeneous polynomials over p-adic fields," Ann. Math. 56 (1952) pp. 473478]. It was believed for some time that N(r) = r2 is true in general, but recently, G. Terjanian found a counterexample. One may also consider systems of equations

(6.14)

in which F 1 , ... , F k are forms with p-adic coefficients of degrees r1 , ... , rk' In the case of two quadratic forms~ with m :::::" 9, the solvability of the system I was shown by Demyanov [a simple proof of this result of Demyanov was given in B. J. Birch, D. J. Lewis, a~d T. G. Murphy, "Simultaneous quadratic forms," Am. J. Math. 84 (1962) p. 110-115]. A general method is known which shows how to get solutions for systems (6.14) when m is sufficiently large compared to r1 , ... , rk provided one knows the function N(r) mentioned in the preceding paragraph [see, for instance, S. Lang, "On quasialgebraic closure," Ann. Math. 55 (1952) pp. 373-390]. Finally, it is easily shown that the hypothesized value for N(r) is the best possible, that is, for any r there is a form of degree r in r 2 variables which does not represent zero over the field of p-adic numbers. We give an example of such a form. Recall that in Section 2.1 we constructed a form F(x 1 , ... , x n )

Sec. 6]

QUADRATIC FORMS WITH p-ADIC COEFFICIENTS

59

of degree n in n variables such that the congruence F(x 1, ... , x n) == 0 (mod p)

had only the zero solution: Xl == 0 (modp), ... , Xn == 0 (modp).

(6.15)

Set (X 1, ... ,

X n 2)

= F(x 1, ... , x n )

+ pF(Xn+1'"'' x 2n ) + ... + pn- 1F(x n

L

n+1"'" Xnl)

We shall show that the form does not represent zero in the field of p-adic numbers. Assume the contrary, that is, assume that the equation (X 1, ... ,

X n 2)

=0

(6.16)

has a nonzero solution. Since is homogeneous we may assume that all variables are integers and that at least one of them is not divisible by p. Considering as a congruence modulo p, we obtain that F(x 1, ... , x n) == 0 (modp), from which it follows by (6.15) that Xl = PX1', ... , Xn = px n'. Equation (6.16) then takes the form pnF(x 1', ... , x n') + pF(x n+1, ... , X2n) + ... + pn- 1F(x n2_ n+l' ... , Xn2) = 0 or, after dividing by p, F(x n+l'

... , X2n)

+ ... + pn- 1 F(x n2_ n+l'

... ,

Xn2)

+ pn-1 F(X1',

As in the previous step, we obtain here that x n+ l> Repeating this process n times, we obtain that xl> which is a contradiction.

... , Xn') = O.

, x2n are divisible by p. , X n2 are divisible by p,

PROBLEMS

1. Verify the following properties of the Hilbert symbol: (1)

(a,l-a)=+l,a#l;

(2)

(a, f3) = (y, - a{3), y = ag> + f37Jl # 0;

(3)

(ay, f3y)

2. Let f = alxl + .., + a"x. (a, symbol by the formula 1

l

E

=

(a, f3)(y, - af3).

R.*) be a quadratic form, and define the Hasse

c p(f)=(-l,-l)

n

(a"a)

l~i~j~n

60

CONGRUENCES

[Chap. 1

Show that c.(o:x 2

-+- I) = c.(f)(o:, -8),

c.(o:x 2 -+- {3y 2 -+- f)

=

c.U)(o:{3, -8)(0:, {3)

(8 is the determinant of the form f). 3. Let the form 1= 0:1 X 12 + '" + rx.x/ with p-adic coefficients represent the number y # 0 of R•. Show that there is a representation y = O:tgt 2 + ... + 0:.g.2 (g, E R.) such that all .. partial sums" y. = O:tg1 2 + ... + 0:.g.2(1 .;; k .;; n) are nonzero. (Use Theorems 5 and 8 of Section 1 of the Supplement.) 4. Using the same notation, show that the form I is equivalent to a diagonal form 9 = YYt 2 + {32Y2 2 + ... + {3.y/, for which c.(g) = c.U). (First show that the form o:x 2 + {3y 2 is transformed by the substitution x = f-LX - v{3Y, Y = vX + f-LO:Y(0:f-L2 + {3v 2 = y # 0) into yX 2 + rx{3yy 2, and (0:,{3) = (y, o:f3y).) 5. Show, by induction on the number of variables, that equivalent diagonal nonsingular quadratic forms over the field R. have the same Hasse symbol (use Theorem 4 of Section I of the Supplement). The Hasse symbol can thus be defined for arbitrary nonsingular quadratic forms: If the form lis equivalent to the diagonal form 10, set c.U) = c.Uo). 6. Let/t and 12 be two quadratic forms over the field R. with determinants 8 1 # 0 and 82 # O. Show that

IX

7. Let I be a nonsingular quadratic form over the field R., with 8 its discriminant and a nonzero number from R• . Show that c.( 0:/) =

c.U)(o:, (_1)(0+1)/2 ( c.U)( 0:, ( -1)./28),

if n is odd, if n is even.

8. Show that a nonsingular quadratic form in three variables over the field R. represents zero if and only if c.U) = + 1. 9. Let I be a nonsingular quadratic form in four variables over the field R. with determinant 8. Show that I does not represent zero in R if and only if 8 is a square in R. and c.U) = -1. 10. Let I be a nonsingular quadratic form in n variables over R. with determinant 8. Show that I represents the nonzero p-adic number 0: if and only if one of the following holds: (a) n = 1 and 0:{3 is a square in R•. (b) n = 2 and c.(f) =( -0:, -8). (c) n = 3, -0:8 is a square in R. and c.U) = 1. (d) n = 3 and -0:8 is not a square in R•. (e) n;;. 4.

11. Give the conditions under which a nonsingular quadratic form over the field R p does not represent zero (nontrivially), but otherwise does represent all p-adic numbers. 12. In which p-adic fields does the form 2x 2 - 15y2 + 14z 2 fail to represent zero? 13. Which 5-adic numbers are represented by the form 2x 2 + 5y 2? 14. Let I and f' be nonsingular quadratic forms in n variables over the field R. with determinants 8 and 8'. Show that I and f' are equivalent if and only if c.U) = c.U') and

8=

8'0: 2 (0: E R.).

RATIONAL QUADRATIC FORMS

Sec. 7]

61

7. Rational Quadratic Forms

7.1. The Hasse-Minkowski Theorem In this section we shall give the proof of one of the most important results of number theory-the so-called Hasse-Minkowski theorem-of which we have already spoken at the beginning of this chapter. Theorem 1 (Hasse-Minkowski). A quadratic form with rational coefficients represents zero in the field of rational numbers if and only if it represents zero in the field of real numbers and in all fields of p-adic numbers (for all primes p).

The proof of this theorem depends essentially on the number n of variables of the quadratic form. For n = 1 the assertion of the theorem is trivial. In the case n = 2 the proof is very simple. If the binary rational quadratic form f with discriminant d "# 0 represents zero in the field of real numbers, then - d > 0 (see Theorem 10 of Section 1 of the Supplement); hence - d = Pl k, ••• Psk " where the Pi are distinct primes. If now frepresents zero in the field R p , then since - d is a square in R p " the exponent k i must be even (i = 1, .. ". , s). But in this case - d is a square in the field of rational numbers and hence f represents zero in R. The proof of the theorem for n ~ 3 is rather difficult. The various cases which occur will be analyzed in the following paragraphs. But first we make some preliminary remarks. We may assume that the coefficients of the quadratic form f(x 1 , .,. ,xn ) are rational integers (if not, multiply it by the least common multiple of the denominators of the coefficients). It is clear that the equation f(x 1 ,

.••

,xn ) = 0

(7.1)

can be solved in the field of rational numbers R (or in the field of p-adic numbers R p ) if and only if it can be solved in the ring Z of integers (respectively, in the ring Op of p-adic integers). Further, (7.1) is solvable in real numbers if and only if the form f is indefinite. Hence, by Theorem 2 of Section 5, we may formulate the Hasse-Minkowski theorem as follows: Equation (7.1) is solvable in rational integers if and only if the formfis indefinite and for any modulus p the congruence I

f(x" ... , x n) == 0 (mod pm) has a solution in which at least one of the variables is not divisible by p. By Theorem 5 of Section 6 any form in five or more variables represents zero in the field of p-adic numbers. Hence, for such forms the Hasse-Minkowski theorem reads: In order that a nonsingular rational quadratic form in

62

CONGRUENCES

[Chap. 1

n ~ 5 variables represent zero in the field of rational numbers, it is necessary and sufficient that it be indefinite. Thus the conditions for solvability in p-adic fields actually need only be verified for n = 3 and n = 4. For these values of n the Hasse-Minkowski theorem gives us an effective criterion for the solvability of (7.1). Indeed, if the form f is given by f = Ia;x/, then Corollary 2 of Theorem 3, Section 6, implies that, for any odd prime p which does not provide any of the a;, the form f with n ~ 3 always represents zero in R. Thus only a finite number of primes p actually need be considered. For each of these p the theorems of Section 6 decide the question of the representation of zero by fin R p • By Theorem 6, Section I, of the Supplement, Theorem I implies the following. Corollary. A nonsingular quadratic form with rational coefficients represents the rational number a if and only if it represents a in the field of real numbers and in the field of p-adic numbers for all primes p.

7.2. Forms with Three Variables We now turn to the proof of the Hasse-Minkowski theorem, treating the case n = 3 in this section. For forms in three variables Theorel)1 I was proved (in somewhat different terminology) by Legendre. The formulation of Legendre is given in Problem I. Let the form be given as a 1 x 2 + azy2 + a 3z 2. Since the form is indefinite, not all the coefficients a1 , a z , a3 have the same sign. Multiplying (if necessary) the form by - I, we may assume that two coefficients are positive and one negative. We may assume that a1> a z , a3 are integers, square-free and relatively prime (they can be divided by their greatest common divisor). Further, if, say, a 1 and a z have a common prime factor p, then, multiplying the form by p and taking px and pyas new variables, we obtain a form with coefficients a 1 !p, az!p, pa 3 . Repeating this process as necessary, we arrive at a form (7.2) whose coefficients are positive integers a, b, and c which are pairwise relatively prime (and square-free). Let p be some odd prime divisor of the number c. Since by assumption the form (7.2) represents zero in R p , by Theorem 3 of Section 6 and Corollary I of that theorem, the congruence ax z + byz == 0 (mod p) has a nontrivial solution, say (x o ,Yo). Then the form ax z + byz factors, modulo p, into linear factors:

63

RATIONAL QUADRATIC FORMS

Sec. 7]

The same also holds for the form (7.2), so that we have ax z + by 2 - ez z == L(p)(x, y, z)M(P)(x, y, z) (mod p),

(7.3)

where L(p) and M(p) are integral linear forms. An analogous congruence also holds for the odd prime divisors of the coefficients a and b, and also for the prime 2, since ax 2 + byz - ez 2 == (ax + by - ez)2 (mod 2). We find linear forms L(x, y, z) and M(x, y, z) such that

L(x, y, z) == L(p)(x, y, z) (modp), M(x, y, z) == M(p)(x, y, z) (mod p) for all prime divisors p of the coefficients a, b, and e. The congruence (7.3) shows that ax 2 + byZ - ez z == L(x, Y, z)M(x, y, z) (mod abc). (7.4) We shall give integer values to the variables x, y, and z, satisfying the inequalities o ~ x < Jbe, 0 ~ Y < J ae, 0 ~ z < abo (7.5)

J

If we exclude from consideration the case a = b = e = I (the assertion of the theorem is obvious for the form X Z + yZ - zZ, since it represents zero in any field), then, since a, b, e are pairwise relatively prime, the numbers ae, and ab will not all be integers. Hence the number of triples (x, y, z) satisfying (7.5) will be strictly greater than Jab· Jbe·Jae = abc. Since the number of triples (x, y, z) is greater than the number of residues modulo abc, there are two distinct triples (Xl' Y1' Zl) and (x z , Yz, zz) such that

J7Jc,

J

J

L(x 1, Y1, Zl) == L(xz, Yz, zz) (mod abc). The linearity of L implies that

L(xo , Yo, zo) == 0 (mod abc), where

XO=X1- X2,

YO=Y1-Y2,

ZO=Zl- Z2'

From (7.4) it follows that

axoz

+ byo Z -

ezoz == 0 (mod abc).

(7.6)

Since the triples (Xl' Yl, Zl) and (X2' Yz, zz) satisfy (7.5),

IXol < Jbe, IYot < J ae, IZol < Jab, so that (7.7)

64

[Chap. 1

CONGRUENCES

The inequality (7.7) and the congruence (7.6) combine to give either

ax 02 + bY0 2 - ez 02 = 0,

(7.8)

or (7.9) In the first case we have a nontrivial representation of zero by the form (7.2), which is what was required. In the second case we rely on the following lemma. Lemma 1. If the form (7.2) represents abc, then it also represents zero.

Let xo, Yo, zo satisfy (7.9). It is easily seen that then

a(xozo

+ bYO)2 + b(yozo -

axO)2 - e(z0 2 + ab)2 = O.

(7.10)

If z02 + ab i= 0, then this equality proves the lemma. If -ab = Z02, then the form ax 2 + by 2 represents zero (Theorem 10 of Section I of the Supplement). But then (7.2) also represents zero, so that the lemma is proved. This proof is very short, but is based on the computation involved in (7.10). We shall give another proof which uses more general methods. If be is a square, then the form by 2 - ez 2 , and hence also (7.2), represents zero. Assume that be is not a square. It will be shown that in this case the representability of zero by (7.2) is equivalent to ae being the norm of some element from the field R( -jbe). Indeed, from (7.8) (where we may assume that x i= 0) it follows that

ac = (ez o)2 _ bc(Yo)2 = N(ez o + ~o Jbe). Xo Xo x o Xo Conversely, if ae

= N(u + v-jIX),

then

ae 2 + b(ev)2 - eu 2 = O. Assume now that (7.9) holds. Multiplying it bye, we obtain either

ae(x0 2 - be)

= (ez o)2 -

bey0 2

or

aeN(rx) where rx = x o + -jbe,

= N(fl),

fJ = ezo + Yo -jbe. But then fl

ac = N(y), Y = -

rx

-

E

R(Jbc),

and this, as we have seen, means that (7.2) represents zero in R. We now note the following fact. In the proof of Theorem I for the case of three variables we have never used the fact that the form (7.2) represents zero over the field of2-adic numbers. Hence, from the solvability of(7.2) in the field

Sec. 7]

RATIONAL QUADRATIC FORMS

65

of real numbers and also in the field of p-adic numbers for all odd p it follows that (7.2) is solvable in the field R 2 • It will be shown that an analogous result holds also for any other field R q • Namely, if the rational quadratic form in three variables represents zero in the field of real numbers and also in all fields R p , with the possible exception of the field R q , then it represents zero in the field Rq (and hence, by what has been proved, also in the field R). We shall try to explain the cause of this phenomenon. CO'lsider the conditions for the representability of zero by the form (7.11) in all fields R p and in the field of real numbers (here a and b are arbitrary nonzero rational numbers; hence any nonsingular rational quadratic form in three variables can, after change of variables and multiplication by some rational number, be put into this form). By Section 3.6, the condition for the representability of zero in the field of p-adic numbers can be expressed as

(a/) =

I,

(7.12)

where (a, blp) is the Hilbert symbol in the field R p • For rational a and b we use the notation (a, blp) for the Hilbert symbol (a, b) to denote the field in which it is being considered. This change in notation is necessary because we will now be considering the Hilbert symbol simultaneously in different fields. As for the real numbers, the form (7.11) clearly represents zero if and only if at least one of the numbers a, b is positive. To write this condition in the form (7.12), we carryover the results of Section 6.3 to the field of real numbers. We first agree on the following notation. All p-adic fields R p and the field of real numbers together comprise all completions of the field R of rational numbers (Section 4.2). The fields Rp are in one-to-one correspondence with the rational primes p. To extend this corresoondence to the field of real numbers, we introduce the symbol CD, which we call the infinite prime, and we say that the real numbers are the completion of the field R with respect to the infinite prime. An ordinary prime p, in contrast, is called a finite prime. By analogy with the notation Rf] for the p-adic fields, we denote the field of real numbers by Roo . For any (X from the multiplicative group Roo * of the field Rc1J , we consider the form (7.13)

and by H~ we denote the set of all fJ E Roc *, represented by this form. If 0, that is, r:J. E Roo *2, then the form (7.13) represents all real numbers, and thus H, = Roo *. If r:J. < 0, that is, r:J. is not a square, then the form (7.13) represents only positive numbers, and therefore as in Theorem 6 of Section 6,

(X>

66

CONGRUENCES

[Chap. 1

we have (7.14) For rx, fl E Roo * we set (rx, fJ) equal to + 1 or -1 depending on whether the form represents fl or not, and it follows from the above that the symbol (rx, fl) will have all the properties (6.9) to (6.13). In analogy with Theorem 7 of Section 6 by which the Hilbert symbol was computed for the p-adic fields, we have here the much simpler relations

(rx, fJ) =

+ 1,

if rx > 0 or fJ > 0, (7.15)

(rx, fJ) = -1,

if rx < 0 and fl < O.

For rational a and b we denote the value of the Hilbert symbol in the field

Roo by (a, bloo). Using the Hilbert symbols (a, blp) we can now reformulate Theorem I for forms in three variables as follows. The form ax 2 + by 2 - Z2 with nonzero rational coefficients a and b represents zero in the field of rational numbers if and only if for all p (including p = 00)

(a/) = 1.

(7.16)

For any nonzero rational numbers a and b the symbol (a, blp) differs from many values of p. Indeed, if p is not equal to 2 or 00 and if p does not enter into the factorizations of a and b into prime powers (which means that a and bare p-adic units), then, by Corollary 2 of Theorem 3 of Section 6, the form (7.11) represents zero in R p and thus for all such p the symbol (a, blp) equals + 1. Besides this, it will now be shown that the value of the symbol (a, blp) for fixed a and b is subject to one further condition. Namely, the number of p (includingp = ro) for which (a, blp) = -1 is always even. Another way of expressing this fact is to say that

+ 1 only for finitely

I) C/) = 1,

(7.17)

where p runs through all prime numbers and the symbol roo For the formal infinite product on the left contains only a finite number of terms different from + 1, so the product will be I if and only if the number of p for which (a, blp) = -1 is even. We now prove (7.17). Factoring a and b into prime powers and using formulas (6.9) to (6.13) (also valid, as mentioned, for p = 00), we easily reduce the proof of the general formula (7.17) to the proof of the following special cases:

Sec. 7] (1) a (2) a (3) a

67

RATIONAL QUADRATIC FORMS

= -I, b = -1. = q, b = -I (q a prime). = q, b = q' (q and q' distinct primes).

By Theorem 7 of Section 6 and (7.15), we have

(-I' -1)(-1 '=(-1)'(-1)=1, -I) 11 ( -I ' -I) = p

2

P

2,-1) - = 11(p

CD

(2,-1)(2,-1) - - - - =1'1=1

p

11 (-q, p-I) _- (q,- q-I)(q,- 2-I) =

2

CD

'

(-I) q~I.-12-1 (-I) =1,

p

q

I} e~q) = (2~q)(\q) = G)(-lt;1 = 1, I} (q~q') = (q'qq')(q~;')(q,/) = (~')(:,)< _1)q·~I.q~1 = 1. These computations, in which q and q' denote distinct odd primes, prove the relation (7.17). Note that in the proof of (7.17) we have used the quadratic reciprocity law of Gauss. On the other hand, knowing the explicit formulas for the Hilbert symbol (Theorem 7, Section 6), we can deduce all parts of the law of quadratic reciprocity from the formula (7.17). Thus (7.17) is equivalent to Gauss' reciprocity law. Assume now that the form (7.11) represents zero in all fields R p , except perhaps for R q • From the equality (7.17), along with the fact that (a, blp) = I for all p =f. q, we deduce that (a, blq) = I. In other words we have the following assertion. Lemma 2. If a rational quadratic form in three variables represents zero in all fields R p (p running through all prime numbers and the symbol 00), except possibly for R q , then it also represents zero in R q •

7.3. Forms in Four Variables We shall assume that our form is given by (7.18)

where all a i are square-free integers. Since the form is indefinite we may assume

68

CONGRUENCES

[Chap. 1

that al > 0 and a4 < O. Along with the form (7.18) we consider the forms and The idea of the proof of the Hasse-Minkowski theorem is as follows. Using the fact that the form (7.18) represents zero in the fields R p , we shall show that there is a rational integer a i= 0 which is simultaneously rationally represented by the forms 9 and h. This immediately gives us a rational representation of zero by the form (7.18). Let PI> ... ,Ps be all distinct odd primes dividing the coefficients at> a 2 , a3 , a4' For each of the preceding primes and also for P = 2, choose a representation of zero, a1~12 + a2~/ + a3~/ + a4~/ = 0, in the field R p for which all plement) and set

b p = a1~12

~i

i= 0 (see Theorem 8, Section I, of the Sup-

+ a2~22 =

-a3~/ - a4~42.

Our representation can be chosen so that each b p is a nonzero p-adic integer divisible at most by the first power of P (if b p = 0, then the forms f and 9 represent zero in R p and hence by Theorem 5 of Section 1 of the Supplement they represent all numbers of R p ). Consider the system of congruences a

== b 2 (mod 16),

a

== b p , (mod p/),

a

== b p, (mod p/).

(7.19)

A rational number a, satisfying these congruences, is uniquely determined modulo m = 16p/ ... p/. Since bPi is divisible by at most the first power of Pi' bpp-l is ap-adic unit, and

b pi a- 1 == 1 (mod p;). By Corollary 1, Theorem I, of Section 6 the quantity bpp-l is a square in the field R p • Analogously, since b 2 is not divisible by any higher power of 2 than the first, b 2 a- 1 == 1 (mod 8), and therefore (Theorem 2 of Section 6) b 2 a -1 is a squ 0 and

Sec. 7]

RATIONAL QUADRATIC FORMS

69

-a4 > 0 the forms (7.20) represent zero in the field of real numbers. Finally, if P is different from 2, PI' ... ,Ps and does not divide a; that is, if P is odd and does not divide the coefficients of the form (7.20), then, by Corollary 2 of Theorem 3 of Section 6, these forms represent zero in R p • If, in addition to 2, P, ... ,P, there were at most one more prime q dividing the integer a, th~n we could apply Lemma 2 and conclude (using the Hasse-Minkowski theorem for forms in three variables) that the forms (7.20) represent zero in the field of rational numbers. In such a case we would have the representations

with rational c;, from which

a 1c l 2

+ azc z z + a3c3z + a4c4z = 0,

and the Hasse-Minkowski theorem would be proved for forms in four variables. We will now show that a number a, satisfying the congruence (7.19) and possessing the desired additional property, always can be found. To do this we shall have to apply the theorem of Dirichlet on prime numbers in arithmetic progressions, which we shall prove in Chapter 5, Section 3.2.t Dirichlet's theorem asserts that if the increment and first term of an infinite arithmetic progression are relatively prime, then the progression contains an infinite number of primes. Let a* > 0 be any number satisfying (7.19). Let d denote the greatest common divisor of a* and m. Since a*ld and mid are relatively prime, Dirichlet's theorem implies that there is an integer k ~ 0 such that a*ld + kmld = q is prime. As a we may then take

a = a*

+ km

= dq.

Since all the divisors of d are among 2, PI' .. , ,Ps' this choice of a allows us to finish the proof of Theorem I for forms in four variables.

7.4. Forms in Five and More Variables Let an indefinite rational quadratic form in five variables be given by (7.21) where all a i are square-free integers. We can assume that at > 0 and a 5 < O. Set Reasoning precisely as in the case n = 4, we use Dirichlet's theorem to find a rational integer a> 0 which is represented by the forms 9 and h in the field

[Chap. 1

CONGRUENCES

70

of real numbers and in all the fields R p , with the possible exception of R q , where q is some prime number which does not divide the coefficients a i • We claim that the forms 9 and h represent a in the field R q • For the form 9 this is established exactly as before, using Lemma 2. The form h represents zero in R (Corollary 2 of Theorem 3, Section 6), and thus represents all numbers in R q (Theorem 5 of Section I of the Supplement). By the Corollary to the Hasse-Minkowski theorem (see the end of Section 7.1), which has already been proved for forms in two and three variables, we find that the forms 9 and h represent a in the field of rational numbers. As before it easily follows that the form (7.21) admits a rational representation of zero. For the proof of Theorem I in the case n> 5, we simply note that any indefinite quadratic form f, after being diagonalized, is easily represented as f = fa f1' where fa is an indefinite form in five variables. We have proved that fa represents zero in the field of rational numbers, and hence so does f The Hasse-Minkowski theorem is completely proved.

+

7.5. Rational Equivalence The Hasse-Minkowski theorem allows us to solve another important question on rational quadratic forms, the question of their equivalence. Theorem 2. In order that two nonsingular quadratic forms with rational coefficients be equivalent over the field of rational numbers, it is necessary and sufficient that they be equivalent over the field of real numbers and over all the p-adic fields R p •

Proof. The necessity of the condition is clear. The proof of sufficiency is carried out by induction on the number of variables. Let n = 1. The forms ax 2 and bx 2 are equivalent over any field provided only that alb is a square in that field. But, if alb is a square in the real field and also in all p-adic fields, then as we saw in Section 7.1, alb is a square in the field R of rational numbers. Hence for n = I Theorem 2 holds. Now let n> 1. Let a =I- 0 be a rational number represented by the form f (in the field R). Since equivalent forms represent the same numbers, the form 9 represents a in the real field and also in all fields R p • By the corollary to the Hasse-Minkowski theorem, the form 9 represents a in the field R. Applying Theorem 2 of the Supplement, Section I, we obtain 2 f - ax +f1'

2

9 - ax +gl,

where f and 9 are quadratic forms in n - I variables over the field R (the sign - denotes equivalence over R). Since the forms f and 9 are equivalent in the fields R p , it follows (Supplement, Section I, Theorem 4) that the forms!l and

Sec. 7]

RATIONAL QUADRATIC FORMS

71

gl are also equivalent in all these fields. By the induction hypothesis,f1 and 91

are equivalent in the rational field R. Then f and g are also equivalent in R, and Theorem 2 is proved. As an example we consider the question of the equivalence of binary quadratic forms. The determinant d(f) of a nonsingular rational quadratic form has a unique representation as where do(f) is a square-free integer. When we pass to an equivalent form the value of do(f) does not change (Supplement, Section 1, Theorem 1) and thus it is an invariant of the equivalence class of rationally equivalent forms. Let a be any nonzero rational number represented by the nonsingular binary formj. For each prime p (including p = 00) set

By Theorem 8 of Section 6 (which clearly also holds for the real field R p ), the value of ep(f) does not depend on the choice of a. It is consequently also an invariant off under rational equivalence. Combining Theorem 2 with Theorem 9 of Section 6 (which also holds for R",), we obtain the following criterion for rational equivalence of binary quadratic forms. Theorem 3. Two binary quadratic formsf and g are rationally equivalent if and only if

for all p. Note that while, formally, an infinite number of invariants appear, their number is actually finite, since ep(f) = + 1 for all but a finite number of p.

7.6. Remarks on Forms of Higher Degree As was done for forms with p-adic coefficients in relation to Theorem 5, Section 6, it would be interesting to include the Hasse-Minkowski theorem, or its corollary for n ~ 5, in a system of more general results, or at least hypotheses, concerning forms of higher degree. It is natural to ask first if the analog of the Hasse-Minkowski theorem for forms of higher degree is true; that is, if a form represents zero in all p-adic fields and in the real field, does it represent zero in the rationals? It is easy to construct examples which disprove this hypothesis. For instance, if q, I, q', I' are distinct primes such that (ljq) = -1 and (/'jq') = -1 and the form

CONGRUENCES

72

x2

+ qy 2

-

[Chap. 1

Iz 2 represents zero in the field R 2 , then the form in four variables

(7.22) represents zero in all fields R p and in the field of real numbers, but fails to represent zero in the field of rational numbers. Indeed, in the field R 2 the first factor represents zero by hypothesis. If p is odd and different from q and I, then the first factor represents zero in R p by Corollary 2, Theorem 3, Section 6. As for q and I, the second factor represents zero in R q and R, for the same reason. However, neither factor represents zero in R, since the first factor fails to represent zero in R q and the second in R q • [since (llq) = -1 and (1'Iq') = - I}. As a numerical example of (7.22) consider (x 2

+ 3y 2

_

l7z 2 )(x 2

+ 5y 2

_

7z 2 ).

This example may perhaps appear somewhat artificial, since the form (7.22) is reducible, and the cause of this phenomenon may lie in its reducibility. Selmer gave a simple example free from this deficiency [E. S. Selmer, The diophantine equation ax 3 + + cz 3 = 0, Acta. Math. 85, 203-362 (l951)}. 3 He showed that the form 3x + 4y 3 + 5z 3 represents zero in every p-adic field R p and in the real field, but does not represent zero in the field of rational numbers. The fact that this form represents zero in all fields R p is ea~ily shown (see Problem 8, Section 5). But the nonrepresentability of zero over the rational numbers is a more delicate question (see Problem 23, Section 7, Chapter 3). The analog of the Hasse-Minkowski theorem for forms of higher degree is not even true when the number of variables is large. For example, the form

hi

(x/

+ ... + X/)2

_ 2(Y1 2 +

... + y/)2

with n ~ 5 represents zero in all p-adic fields and in the real field, but does not represent zero in the field of rational numbers for any n. The same also holds for the form 3(x/ + .. , + X/)3 + 4(y/ + ... + y/)3 - 5(Z1 2 + '" + Z/)3, which. unlike the previous example. is absolutely irreducible. In the preceding examples both forms had even degree. Analogous examples for forms of odd degree have never been found. Hence it is possible that the analog of the Hasse-Minkowski theorem holds for forms of odd degree in sufficiently many variables. Since Brauer's theorem says that forms in sufficiently many variables represent zero in all p-adic fields (see Section 6.5), we are led to the following hypothesis: A rational form of odd degree in sufficiently many variables represents zero rationally. This hypothesis was proved by Birch [B. 1. Birch, "Homogeneous forms of odd degree in a large number of variables," Mathematika 4 (1957) pp.

Sec. 7]

RATIONAL QUADRATIC FORMS

73

102-105], who showed that forms of odd degree represent zero in the field of rational numbers provided that the number of variables is sufficiently large compared to the degree.

PROBLEMS

1. Prove the following theorem of Legendre: If a, b, and c are rational integers, pairwise relatively prime, square-free, and not all of the same sign, then the equation ax 2 + by 2 + cz 2 = 0

is solvable in rational numbers if and only if the congruences x 2 == - be (mod a),

x 2 == -ca (mod b), x 2 == - ab (mod c), are all solvable. 2. Do either of the forms 3x 2 + 5y 2 -7z 2, 3x 2 - 5y 2 -7z 2, represent zero in the field of rational numbers? 3. Which prime integers are represented by the forms x 2 + y2, x 2 + 5y 2, x 2 - 5y 2? 4. Give a description of the set of all rational numbers represented by the form 2x 2 - 5y 2. 5. Which rational numbers are represented by the form 2x 2 - 6y 2 + 15z 2 ? 6. Let f be a nonsingular quadratic form over the field of rational numbers, with the number of variables not equal to 4. Show thatfrepresents zero if and only if it represents all rational numbers. 7. For which rational mtegers a does the form x 2 + 2y 2 - az 2 represent zero rationally? 8. Find all solutions of the equation x 2 + y2 - 2z 2 = 0 in rational numbers. 9. Which of the forms x 2 _ 2y 2 + 5z 2, x 2 _ y2

+ lO z 2,

3x2 _ y2 + 30z2

are equivalent over the field of rational numbers? 10. Let a and b be square-free rational integers with \al > Ibl. Iftheform ax 2 + by 2 - Z2 represents zero in all p-adic fields, show that there are rational integers a and c, such that

aa, = c2 -

b,

la,l < lal.

74

CONGRUENCES

[Chap. 1

(The equation aal + b - c 2 = 0 shows that the form aalx 2 + by 2 - Z2 represents zero rationally.) 11. By consideration of forms ax 2 + by 2 - Z2, where a and b are square-free integers, prove the Hasse-Minkowski theorem for forms in three variables by induction on m = max(lal, Ibl) (use Problem 10 and Problem 3, Section 1, the Supplement).

CHAPTER

2

Representation of

Numbers by Decomposable Forms In Chapter 1 we considered questions dealing with the existence and determination of rational solutions to equations. This chapter deals with the same questions, but only with respect to integral solutions. We consider a simple example. The problem consists in finding all integral solutions to the equation (0.1)

We may assume that x> 0, Y > 0 (the remaining solutions are obtained by change of sign). This equation has the solutions (3, I) and (5, 3). From these two solutions we can obtain an infinite number of others by the foIlowing method: if (x, y) is a solution of (0.1), then (3x + 4y, 2x + 3y) is also a solution, as is shown by substitution. Starting fro~ the solution (x o , Yo) = (3, 1), we thus obtain an infinite sequence of solutions (x n , Yn), determined by the recursion formula x n + 1 = 3xn + 4Yn,

Yn+l = 2x n + 3Yn.

(0.2)

Starting from the solution (x o', Yo') = (5, 3) we use the same formula to obtain another infinite sequence of solutions (x o', Yo'). It can be shown that these two sequences exhaust all solutions to (0.1) with x > 0 and Y > O. This completely elementary solution of (0.1) was obtained by computation. We can connect it with some general concepts and lay the groundwork for future generalizations. 75

76

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

We note that the form x 2 - 2y 2 is irreducible over the field R of rational numbers, but in the extension field R( J2) it can be factored into linear factors (x + y J2)(x - y J2). If we use the concept of the norm for the extension R( J2)! R (Supplement, Section 2.2), then (0.1) can be written in the form N(~) = N(x

+ yJ2) = 7.

(0.3)

The problem then is to find in the field R( J2) all numbers ~ = x + Y J2, where x and yare rational integers, whose norms are equal to 7. If the norm of the number e = u + vJ2(u and v rational integers) equals I, then by the multiplicativity of the norm, if ~ is a solution of (0.3) so are all numbers of the form ~en. Since N(3 + 2 J2) = 1, we may take e to be 3 + 2 J2. The passage from ~ to ~e corresponds to that from (x, y) to (3x + 4y, 2x + 3y). The two infinite sequences given by (0.2) now take the form X

n+ YnJ2 = (3 + J2)(3 + 2J2t

x n'

n

~

o.

+ y/J2 = (5 + 3J2)(3 + 2J2)n

The possibility of Obtaining an infinite number of solutions of (0.1) from one solution thus depends on the existence of a number e = u + v J2 with integral u and v for which N(e) = 1. In turn, the question of the existence of such numbers is connected to the basic concepts of the theory of algebraic numbers. Consider the set of all numbers of the form x + y ,/i~ where x and yare rational integers. It is easily checked that this set, which we denote by .0, forms a ring. In the arithmetic of this ring a major role is naturally played by the units, that is, those numbers IX E .0 such that IX -1 E .0 also. It is easily shown that a is a unit in .0 if and only if N(a) = ± I. This indicates the deeper significance of those numbers e E .0 whose norm is I; along with the numbers of norm -1, they form all units of the ring .0. In this chapter we consider the general theory, of which (0.1) is one of the simplest examples. Our succeSS with the equation (0.1) was based on the fact that the form x 2 - 2y 2 is irreducible over the rational numbers and factors into linear factors over the field R( Jf), allowing the equation to be written in the form (0.3). Our general theory will deal with forms which factor, in some extension of the field of rational numbers, into a product of linear forms. Although our goal is the investigation of equations in which the coefficients and values of the variables are integers, we shall find it necessary to consider the more general case of forms with rational coefficients. The values of the variables will always be assumed to be integers.

DECOMPOSABLE FORMS

Sec. 1]

77

1. Decomposable Forms

1.1. Integral Equivalence of Forms Definition. Two forms F(x 1 , .,. , x m) and G(Yb ... ,y,) of the same degree with rational coefficients are called integrally equivalent if each can be obtained from the other by a linear change of variables with rational integer coefficients. For example, the forms x 2 + 7y 2 + Z2 - 6xy - 2xz + 6yz and 2u 2 - v2 are equivalent, since the linear substitutions

= 3v, = u + v, z = -u + v,

x

y

u = -x + 2y + z, v=x-y-z

take one into the other. In the case of forms which depend on the same number of variables, this is equivalent to saying that one of the forms can be transformed into the other by a linear change of variables with unimodular matrix (that is, an integral square matrix with determinant equal to ± I). If the forms F and G are equivalent, then, knowing all integral solutions of the equation F = a, we can obtain all integral solutions of the equation G = a, and conversely. Hence if we are interested in integral solutions of an equation of the form F = a, we may take instead of the form F any form which is equivalent to it. Lemma 1. Any form of degree n is equivalent to a form in which the nth power of one of the variables occurs with nonzero coefficient.

Let F(x 1 , .,. ,xm ) be a form of degree n. We shall show that there exist rational integers a2' '" , am, so that

The proof goes by induction on m. If m = I, the form F is given by Ax 1 n , where A -# 0, so that F(l) -# 0. Assume that the lemma has already been proved for any form in m - 1 variables (m ~ 2). Write F as

k ~ n) is either zero or is a form of degree k in the variables (we say that a form i5 of degree zero if it is a nonzero constant). All the Gk are not zero, since F, as a form of degree n, has at least one nonzero coefficient. By the induction assumption there exist integers a2, ... , a m-1 such that Gk(l, a 2 , ... , am -1) -# for at least one k. Since the polynomial F(I, a2' ... , am-I' x m) in the single variable X m is not identically zero, we may

where Gk (0

~

XI' ... ,X m - 1

°

78

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

choose the value of am distinct from its roots and thus can obtain F(I,a2"'"

an.) 01-0.

We now make the following change of variables:

After this transformation the form F becomes

Since the matrix of the transformation is integral and has determinant I, the forms F and G are equivalent, and the coefficient of Yin is G(1,O, ... ,0)

= F(1, a 2 ,

... ,

am),

which is nonzero. Lemma I is proved. 1.2. Construction of Decomposable Forms Definition. The form F(x l , .•. , x m ) with coefficients in the field of rational numbers is called decomposable if it factors into linear factors in some extension o.jR. An example of a decomposable form is the form F(x, y)

= aoxn + a1xn-1y + ... + anyn

in two variables (ao 01- 0). Indeed, if 0. is a splitting field for the polynomial F(x, I) and al> ... , Iln are its roots, then in 0. we have the factorization

Among the nonsingular quadratic forms considered in Chapter I, the only decomposable forms are those in one or two variables (Problem I). It is clear that if F is decomposable, then so are all forms equivalent to F. In the definition of decomposable forms no mention was made of the nature of the field 0., in which the form factors into linear terms. We shall now show that 0. may always be taken to be a finite extension of R. The basic tools here are the results from the theory of finite extensions of fields. The results which we shall need are collectl+i in the Supplement, Section 2. Definition. A finite extension field of the field of rational numbers is called an algebraic numberfield, and its elements are called algebraic numbers.

79

DECOMPOSABLE FORMS

Sec. 1 ]

Theorem 1. Any rational decomposable form factors into linear terms in some algebraic number field. Proof. By Lemma 1 we may assume that we are given

in which the coefficient of XI n is nonzero. Since in this case the coefficients (1 ~ i ~ n) are all nonzero, we may set

IXiI

F

= A(x i + PI2X2 + ... + PlmXm)'"

(XI

+ Pn2 X 2 + ... + Pnmxm),

(1.1)

where A = IX II ' " <Xnl and Pij = IXiPiI -I. The number A is rational because it is the ccefficient of Xl n • For some fixed} (2 ~) ~ n) we set x j = 1, and we set all remaining variables, except Xb equal to zero. Then F(x l , 0, .. , , 1, ... ,0) = A(xi

+ Plj)'"

(XI

+ Pn).

Since on the left there is a polynomial (of degree n) with rational coefficients, it follows that the Pij are algebraic numbers. Let L denote the subfield of Q generated over R by all Pij. The extension LjR is clearly finite (Supplement, Section 2.1); that is, L is an algebraic number field. From now on we shall consider only forms which are irreducible over the field of rational numbers, since for such forms the question of integral representation of rational numbers is of greatest interest. We now give a method for constructing irreducible decomposable forms. Let K be any algebraic number field of degree n, and let (J be a primitive element for Kover R, so that K = R(O) (Supplement, Section 2.3). The minimum polynomial qJ(t) of the number () over the field R has degree n. Construct an extension Lover K in which qJ(t) factors completely, qJ(t) =(t - O(l) .. ·(t _

o(n»

(O(l) = 0)

we may assume that L = R(O(1), ... ,o(n». For any number [J(t) a polynomial with rational coefficients] we set IX(i) =

f( (J(i»

E

IX

= f(O)

E

K

R( (J(i» c L.

Then the norm N(IX) = NK/R(IX) satisfies N(IX)

= IX O JIX (2) ... IX(n)

(Supplement, Section 2.3). Now let /-I b ... , Jim be any set of nonzero elements of K. These numbers determine a form (1.2) Since Jik (i) = fk(O(i) [1 ~ k ~ m,fk(t) a polynomial with rational coefficients],

NUMBERS BY DECOMPOSABLE FORMS

80

[Chap. 2

the coefficients of the form (1.2) are symmetric functions in 8(1), .•. , 8(n), which means that they are rational functions of the coefficients of the polynomial ((J(t). Hence the form (1.2) has rational coefficients. If we substitute arbitrary rational numbers for the variables Xl' '" , X m , then, since Xlf.1.I(i)

+ ... + X mf.1.m(i)

=

+ ... + X mf.1.m)(il, number X l f.1.l + ... + X m f.1.m

(X l f.1.1

the product (1.2) will be the norm of the (with respect to the extension KjR). Hence (1.2) can more simply be written F(x l ,

... ,

+ ... + X mf.1.m).

x m) = N(X l f.1.l

(1.3)

The form (1.2) will not always be irreducible. For example, if in the field we take til = )2, /12 = )3, then the corresponding form will be (2X 1 2 - 3X/)2. However, we have the following theorem.

R(,)2, )3)

Theorem 2. If the numbers ,11m), then the form

f.1.2' .,. , f.1.m

generate the field K, I.e., K =

R(f.1.2' ...

F(x l ,

... ,

x m) = N(x I

+ X2f.1.2 + ... + X mf.1.m)

(1.4)

is irreducible (over the field of rational numbers). Conversely, every irreducible decomposable form is equivalent to some constant multiple of a form of the type (1.4). Proof. Assume that

F=GH, where the forms G and H have rational coefficients. Since factorization in polynomial rings is unique (up to constant factors), each of the linear forms

must divide either G or H. Let L I = that is,

Xl

+ X 2P2 + ... + XmPm

divide G;

G = LIMI .

In this last equation, replace all coefficients by their images under the isomorphism a --. a(i) of the field K = R(8) onto the field R(8(i»). Since the coefficients of the form G are rational, we obtain

G=LiM i , which means that L i divides G for all i = 1, ... , n [n = (K : R)]. Note that the isomorphism a -> ali), a E R(P2' ... ,Pm) is completely determined by the images f.1.2(i), .. , , f.1.m(i) of the numbers f.1.2' ... , f.1.m' From this it follows that the sets of numbers f.1.2(i), ... , /1m(i) (1 ~ i ~ n) are pairwise-distinct (since the isomorphisms a -> a(i) are pairwise-distinct), which means that the forms L I , ... ,L n are pairwise-distinct. Since the coefficient of Xl in each form L i

Sec. 1 ]

DECOMPOSABLE FORMS

81

is equal to 1, these forms are pairwise-nonproportional. Using again the uniqueness of factorization, we conclude that G is divisible by the product L I ... L n ; that is, G is divisible by F. Hence H is a constant and the first assertion of the theorem is proved. We prove now the second assertion. Let F*(x\l .. , , x m ) be any irreducible decomposable form of degree n. By Lemma I we may assume that the coefficient of Xl n is nonzero, so that F* will have a factorization ofthetype(1.1), where Pij are some algebraic numbers. Set PI} = Ilj (2 ~j ~ m) and consider the field K = R(1l2 , ... ,11m), whose degree we denote by r. By what has been proved, the form F

= N(x I + X21l2 + ... + Xmll m)

is irreducible, and one of its linear factors, L I = XI + X 21l2 + ... + Xmll m , is a divisor of the form F*. Replacing all coefficients in the equation F* = LIMI by their images under the isomorphism a -> a(i) (a E K, 1 ~ i ~ r), we obtain F* = LiMi . We have already seen that the forms L\l ... ,Lr are pairwise-nonproportional, so that F* is divisible by their product L I ... L" which coincides with F. Since F is irreducible, F = AF, where A is a constant, and Theorem 2 is proved. (In the process we have also proved that r = n.) 1.3. Modules It is clear that the question of integral solutions to the equation F(x\l ... , x m ) = a, where F is given by (1.3), reduces to the determination of all numbers ~ in the field K which can be represented in the form

(1.5) with XI' ... , X m rational integers, and for which N(~) = a. It is thus natural to study the set of all numbers of the form (1.5). Definition. Let K be an algebraic number field, and let Ill' ... ,11m be an arbitrary finite set of elements of K. The set M of all linear combinations CIIlI

+ ... + cmll m

with rational integer coefficients C i (1 ~ i ~ m) is called a module in K. The numbers Ill' ... ,11m are called generators for the module M. A given module M can be generated by many different sets. If Ill, ... ,11m is a set of generators for the module M, we write M = {Ill, ... ,11m}' We consider how the form (1.3) changes if, instead of Ill' ... ,11m we take another set of numbers PI' ... , PI' generating the same module M. We have m

Pj

= k= I ICjkll k

(1 ~ j ~ I)

82

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

with rational integers c jk . Let G(YI' ... , Yt)

= N(YIPI + '" + YtPt)·

Since I

I

)

= k!;l j!;ICjkYj Jik' m

j!;/jPj

(

then the linear substitution t

Xk

=

L cjkYj

(l~k~m)

j=l

takes the form F into the form G. As the sets of generators Jik and Pj of the module M playa symmetric role, there is also an integral linear change of variables which takes G into F. This means that different systems of generators for the same module M correspond to equivalent forms; that is, with each module M ofthe field K is associated a uniquely determined class of equivalent decomposable forms. For each module M = {Jil' ... 'Jlm} and each number a E K, we denote by aM the set of all products a~, where ~ is any element of M. It is clear that aM coincides with the set of all integral linear combinations of the numbers aJlI' ... , aJlm; that is, M = {aJlI' ... , aJim}' Definition. Two modules M and M I in the algebraic number field K are called similar if M I = aM for some a#-O in K.

The forms associated to similar modules M and aM differ only by a constant multiple, equal to N(a). Hence if we are considering forms only up to constant multiples, we may replace the module M by any module similar to it, and in particular we may assume that one of the generators ofthe module, say Jll' equals I. We may now formulate our problem on the representation of numbers by irreducible decomposable form as follows. If the form F is given by F(x b

.. , ,

x m)

= AN(XIJlI + .. , + XmJlm)

(for suitable choice of the field K), then the finding of all integral solutions to the equation F(x b ... , x m ) = a is equivalent to the finding in the module M = {Jib ... , Jim} of all numbers a, such that N(a) equals the rational number a/A. Hence in the future we shall start from the problem of finding in a given module all numbers with given norm. We have seen that this is equivalent to finding all numbers in the similar module jiM with norm N(Ji)a/A. Hence we may replace the given module by any similar module whenever such replacement is helpful. If the degree of the algebraic number field K equals n, then any module M of the field K contains at most n linearly independent numbers (over R).

Sec. 2]

FULL MODULES AND THEIR COEFFICIENT RINGS

83

Definition. Let K be an algebraic number field of degree n; Let M be a module in K. If M contains n linearly independent elements (over the field of rational numbers), then it is calledfull, otherwise nonfull. The forms connected with the module are correspondingly called full or nonfull. For example, if the rational integer d is not a cube, then the numbers 1, form a basis for the field R(~d) over R, and thus the form

Vd. VP

N(x

+ yVd + zV d 1 )

=

x + dy 3

3

+ d2z 3

-

3dxyz

is full. As an example of a nonfull form, take N(x

+ yVd)

=

x3

+ dy 3.

If {I, 112 , '" , Jim} is a full module of the field K, then K = R(Ji2 , '" , 11m). By Theorem 1 it follows that any full form is irreducible. The problem of the representation of numbers by nonfull irreducible forms is very difficult, and at this time there is little satisfactory general theory. A particular case wil\ be considered in Chapter 4. The problem of the representation of rational numbers by full forms is much easier and is essentially solved. We shall deal with it in this chapter. This problem, as we have noted, is equivalent to the problem of finding in a fixed full module of an algebraic number field K all numbers with given norm.

PROBLEMS

1. Show that a rational quadratic form is decomposable if and only if its rank is ,;; 2. 2. Show that a form connected with an arbitrary module of an algebraic number field K is a constant multiple of a power of an irreducible form. 3. Show that in the field of rational numbers R any module has the form aZ, where a E R (2 is the ring of rational integers).

2. Full Modules and Their Rings of Coefficients 2.1. Bases of Modules

Definition. A system (Xl' ... , am of generators of the module M is called a basis for M if it is linearly independent over the ring of integers, that is, if the equation (a i

occurs only when all a i are zero.

E

Z),

84

[Chap. 2

NUMBERS BY DECOMPOSABLE FORMS

It is clear that if lXI' ••• , rxm is a basis for the module M, then any rx a unique representation in the form (C; E

E

M has (2.1 )

Z).

We now show that any module has a basis. The proof of this does not depend on the fact that the module consists of numbers from some algebraic number field, but only on the fact that the module is a finitely generated Abelian group under addition, containing no elements of finite order. Therefore we shall prove the result we need in the theory of Abelian groups. We use the following terminology. A system of elements IXl> .,. , IXm of an Abelian group M (whose operation is written additively) is called a system of generators if every C( E M can be represented in the form (2.1). In this case we write: M = {lXI' ... , IX m}. If this system satisfies the above definition, it is called a basis for M. Theorem 1. If an Abelian group without elements Jf finite order possesses a finite system of generators, then it possesses a basis. Proof. Let

IXI> ... , IXs be an arbitrary set of generators of the group M, First, note that if any integral multiple of one generator is added to another. the resulting system is also a system of generators. Let, for instance, lXI' = IX I + kIXz. Then for any IX E M we have IX

= CIIX I + CzIXz + '" + CsIXs =

CIIX I '

+ (c z -

kCI)IX Z +

... + CSIX.,

where all· coefficients are integers, which means that M = {lXI', IXz, ••• IX.}. If the elements IX I , ... , IXs are linearly independent, they form a basis of M. Assume that they are linearly dependent, that is, that (2.2) for some set of integers c, not all zero. Choose among the nonzero coefficients the smallest one in absolute value. Let it be C I . Assume that not all coefficients C i are divisible by C I , say, Cz = clq + c', where 0 < c' < Iell. Ifwe pass to the new set of generators C

then the relation (2.2) takes the form

and in this relation the coefficient c' > 0 appears, which is less than c. Thus, if for the generators IXl> ... , IXs we have a nontrivial relation (2.2), in which the nonzero coefficient of smallest absolute value does not divide all remaining coefficients, then we can construct another system of generators for which we also have a nontrivial relation with integer coefficients in which the nonzero

Sec. 2]

FULL MODULES AND THEIR COEFFICIENT RINGS

85

coefficient of smallest absolute value is smaller (in absolute value) than the analogous quantity in the first system. Hence after a finite number of such transformations, we arrive at a new system of generators PI> ... , P.. for which we have the dependence (2.3) with integer coefficients k i, where one of the coefficients, say, k j , is a divisor of all the others. Dividing the relation (2.3) by k l (this can be done since we have assumed that M contains no elements of finite order), we obtain (2.4) with integers 12 , .. , , Is. From (2.4) it follows that PI can be dropped from the system of generators; that is, M = {P2' .. , ,Ps}. We have shown that if some system of generators of M is linearly dependent, then we can construct a new system with fewer generators. After carrying out this procedure several times, we must arrive at a system of generators which is linearly independent, that is, a basis for the group M. Corollary. Any module in an algebraic number field K has a basis.

The number of elements m in any basis of the module M is equal to the maximal number of linearly independent (over R) elements in M. Hence this number will be the same for all bases. It is called the rank of the module M. The rank of the module consisting only of zero is set equal to zero. Let WI' .. , , W m and WI" ... , w m ' be any two bases of a module M of rank m. It is clear that the matrix of transition C from the first to the second basis is integral. By symmetry the transition matrix from the second basis to the first, that is, C- I , is also integral. Consequently, det C = ± 1. We thus obtain that any transition matrix from one basis of a module of rank m to another is unimodular of rank m. If the degree of the field Kover R is equal to n, then the rank of any module of K does not exceed n. It is clear that the rank of a module is equal to n if and only if it is a full module. Nonfull modules are thus characterized by having rank less than n, the degree of the the field. Any system of generators of a module of rank m contains not less than m elements. It follows that among the forms associated with a given module there are forms in m variables and there are no forms in less than m variables. A full form of degree n could thus be characterized as an irreducible decomposable form which is not equivalent to a form in less than n variables. Theorem 2. Let M be a finitely generated Abelian group without elements of finite order and let N be a subgroup. Then N has a finite set of generators

86

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

and hence a basis. For any basis WI' .. " , W m of the group M (for some ordering of this basis) there is a basis for N of the form

where the c ij are integers with

C ii

> 0, k

~

m.

Proof. The theorem will be proved by induction on the rank m of the group

°

M, that is, on the number of elements of the basis of M. The case m = is trivial. Let m ;;, 1. If N consists only of zero, then k = and the theorem is valid. If IX E N, IX #- 0, then

°

(2.5) where at least one of the coefficients C i is not zero. By reordering the basis we may assume that Cl i= O. If Cl < then the coefficient of W l in -IX will be positive. Among all elements of the subgroup N choose that element

°

°

in which the coefficient Cll > of Wl is smallest. We now claim that for any N the coefficient Cll will be divisible by Cll' Indeed, if C l = cllq + c', o < C' < C (q an integer), then for the element IX - qtl, we have IX E

so by the minimality of Cll it follows that c' = O. Consider now in M the subgroup M o = {W2' ... , w m }. Since the intersection N n M o is a subgroup of the group M o , by the induction hypothesis N n M o has a basis of the type

where Cij are integers, Cii > 0, k - 1 ~ m - 1 (for suitable ordering of the basis elements W2' ... , wn ). We assert that N consists of all integral linear combinations of the elements til' tl2, ... , tlk' Let IX be an arbitrary element of N. If we write IX in the form (2.5), then since we have shown that C l = cllql, with ql an integer, which lies in the intersection M o n N. By the induction assumption we have IX -qltll

=q2t12

+ ...

+qktlk,

Sec.2]

FULL MODULES AND THEIR COEFFICIENT RINGS

87

where the qi are integers, so that ':1. = ql'11 + ... + qk'1k' We have thus shown that N = {'1l, '12 , '" , '1d· The generators '11' ... , '1k, as is easily seen, are linearly independent over Z, which means that they form a basis for N of the required type. The proof of Theorem 2 essentially reproduces Gauss' method for eliminating variables in systems of linear equations. The only difference is that in our case the coefficients lie not in a field but in the ring of integers. Corollary. Any subgroup N of a module M in an algebraic number field K is also a module (a submodule of the module M).

2.2. Coefficient Ring8 Definition. A number ':1. of the algebraic number field K is called a coefficient of the full module M of the field K jf ':1.M c M, that is, if for any ~ E M the product a~ also belongs to M. The set .oM of all coefficients of a module M forms a ring. For if a and {J belong to .oM' then for any ~ E M we have (a - (J)~ = a~ - {J~ E M and (':J.{J)~ = ':1.({JO E M; that is, a - (J E.o M and ':1.{J E.oM' The ring .oM is called the ring ofcoefficients of the full module M. Since I E.oM,.o M is a ring with unit. To ascertain whether a given number a E K lies in the ring .oM, it is not necessary to check for all ~ E M whether the product Ci~ lies in M or not. It suffices to check this only for any basis ill' '" , I1n of the module M. Indeed, if ':1.il i EM for all i = I, ... ,n, then for ~ = clill + '" + c.il. E M we have 11~ = CI (IlI1I)

+ ... + c.(l1il.) E M.

We now show that the coefficient ring .oM is a full module in K. Let l' be an arbitrary nonzero element of M. Since 111' EM for any a E D M , then 1'D M c M. The set of all numbers 1'.0 M is clearly a group under addition and thus by the Corollary of Theorem 2, 1'D M is a module. But then .oM = 1'-1 (y.oM) is also a module. We now need to show that this module is full. Let 11 be any nonzero element of K and denote by c a common denominator for all rational numbers au, determined by

':1.l1i =

•

I aiji1j j=1

(l~i~Il).

(2.6)

Since the products ca ij are integers, C':1.l1i EM and thus C':1. E.o M . If we now take an arbitrary basis ':1. 1, '" , ':1. n for the field K, then by what has just been proved for some rational integers c I , ... ,C., the products C I I1 I , ... ,C.Il. will all belong to D M . We thus see that D M contains n linearly independent numbers, and this means that .oM is a full module.

[Chap. 2

NUMBERS BY DECOMPOSABLE FORMS

88

Definition. A full module in the field of algebraic numbers K which contains the number 1 and is a ring is called an order of the field K.

Using this definition we can formulate our result as follows. Theorem 3. The coefficient ring for any full module of the algebraic number field K is an order of this field.

The converse also holds: Any order .0 of the field K is the coefficient ring for some full module, for example, for itself (since 1 ED, aD c .0, if and only if a ED). For any number)' i' 0 of K the condition a~ E M is equivalent to the Condition a(/,O E )'M (here ~ EM). It follows that the similar modules M and )'M have the same coefficient rings; that is,

D YM

.oM'

=

Let Jil' '" ,/In be a basis for the module M, and WI' coefficient ring .oM' For each i = 1, '" ,n, we have

'"

'W n

a basis for its

n

Jii

=

L bijw

j'

j= 1

where the bij are rational numbers. If b is a common denominator for all the coefficients bij' then the number bJii will be an integral linear combination of the basis elements of .oM; that is, bJii will lie in .oM' The module bM thus satisfies bM c .oM' We summarize these results. Lemma 1. The coefficient rings of similar full modules coincide. Every full module is similar to a module contained in its coefficient ring.

2.3. Units Consider the problem of integral representation of rational numbers by full decomposable forms. In Section 1.3 we saw that this problem reduces to the determination in a full module M of all numbers Ji, for which N(/l)

= a.

(2.7)

For any W of the coefficient ring .0 = .oM' the product wJilies in M and by the multiplicativity of the norm, N(w/l) = N(w)a.

If N(w) = 1, then (2.7) still holds, with Ji replaced by W/l. Thus the coefficients

Sec. 2]

FULL MODULES AND THEIR COEFFICIENT RINGS

89

w of norm I allow us to obtain a whole class of new solutions of (2.7) from one solution. This fact is the foundation of the method of solving (2.7) which we are going to describe. We shall show that the coefficients wED, with N(w) = I, are contained in the set of elements e of the ring D for which e - I also belongs to D. Such numbers e are called the units of the ring D (Supplement, Section 4.1). Since the inclusions eM c M and e - I Me M are equivalent to the equality eM = M, the units of the ring D M can be characterized as being those elements a E K for which aM = M. Lemma 2. If the number a belongs to the order D, then its characteristic and minimum polynomials have integer coefficients. In particular, the norm N(a) = NK/R(a) and trace Sp(a) = SPK/R(a) are rational integers.

Proof. Let the order D be the coefficient ring of the module M = {Jib'" , Jin} (for example, we may take M = D). If a E D, then in (2.6) the coefficients au are integers, from which it follows that the characteristic polynomial of the number a (with respect to the extension K/R) has integer coefficients. The remaining assertions of the lemma are now obvious. Theorem 4. Let D be an arbitrary order of the algebraic number field K. In order that the number e E D be a unit of the ring D, it is necessary and sufficient that N(e) = ± 1.

Proof. We first show that for any a -# 0 of D the norm N(a) is divisible (in the ring D) by a. By Lemma 2 the characteristic polynomial cp(t) = t n + n I clt + ... + Cn of the number a has integer coefficients. Since cp(a) = 0, then N(a)/a lies in D, which means that N(a) is divisible by a. Now if N(a) = ± I, then I is divisible by a; that is, a is a unit of the ring D. Conversely, if e is a unit of the ring D, so that ee' = I for some e' E D, then since N(e) and N(e') are integers, the equation N(e)N(e') = I implies that N(e) = ± I. Theorem 4 is proved. To find all coefficients weD with N(w) = I, we thus must determine all units of the ring D, and then isolate those units with norm + 1. Two numbers /ll and /l2 of the full module M are called associates if their quotient /l1/Ji2 = e is a unit of the coefficient ring D = D M • It is clear that if M = D, then this concept coincides with the usual notion of associates in commutative rings with unit (Supplement, Section 4.1). This relation induces an equivalence relation on the set of all solutions to (2.7), and therefore the set of all solutions to (2.7) is divided into equivalence classes of associate solutions. If Jil and Ji2 are two associate solutions, then Jil = /l2e, where e is a unit of the ring D with N(e) = I. Conversely, if e is any unit of norm I and /l is a solution of (2.7), then /le is also a solution of (2.7) and /l and Jie are associates. Thus all

90

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

solutions from a given class of associate solutions are obtained by multiplying one solution by all units with norm 1. We now show that the number of such classes of solutions is finite. Theorem 5. An order .0 contains only a finite number of nonassociate elements of given norm. WI' .•. , W n be a basis of the order.o and let e > I be an arbitrary natural number. Using the general definition of the Supplement, Section 4.1, we say that two numbers a and fJ of .0 are congruent modulo e if their difference a - fJ is divisible by e (in the ring .0). It is clear that any a E .0 is congruent to a unique number of the form

Proof. Let

Hence .0 contains en congruence classes modulo e. Let the numbers a and fJ belong to the same congruence class and satisfy IN{a)1 = IN{~)I = e. The equation a - fJ = ey, 'I ED, implies that alfJ = 1 ± [N{fJ)lfJ]y ED [since N{fJ)lfJ ED; see the start of the proof of Theorem 4], and analogously fJla = I ± [N{a)la]y ED. Thus the numbers a and fJ divide one another, which means that they are associates in the ring .0. This proves that .0 can contain only a finite number (not greater than en) of pairwise-nonassociate elements whose norm in absolute value is equal to e. Corollary. A full module M of the field K contains only a finite number of pairwise-nonassociate elements with given norm.

Indeed, if.o is the coefficient ring of the module M, then for some natural number b the module bM is contained in .0. If 'II, .. , , "Yk are pairwise-nonassociate elements of M with norm e, then the numbers bYI, ... , bYk of.o have norm bne and are pairwise-nonassociate in .0. Thus the number k cannot be arbitrarily large. Remark. The proof of Theorem 5 shows that in the ring.o (and also in the module M) there is a finite set of numbers with given norm e such that any number of .0 (or of M) with the same norm e is associate with one of these. However the proof is noneffective, that is, it does not allow us to find these numbers, although it does give an effective bound on their number.

Our basic problem of finding all solutions to (2.7) thus splits into the following two problems: (l) Find all units e in the coefficient ring.o M with norm N{e) = 1. (2) Find numbers Jil' ... , Jik in M with norm a such that they are pairwise-

Sec. 2]

FULL MODULES AND THEIR COEFFICIENT RINGS

91

nonassociate and such that any 11 E M with norm a is associative to one of them, that is, Ji = Jiie, where 1 ~ i ~ k and e is a unit of the coefficient ring .oM' If these two problems are solved, then we will have solved the problem of integral representation of rational numbers by full decomposable forms.

2.4. Maximal Orders The concept of an order leads naturally to the question of the relationship between different orders in a given algebraic number field K. In this section we show that among the various orders of the field K there is one maximal one which contains all other orders. Lemma 2 shows that the minimum polynomial of any number in any order has integer coefficients. We shall see below that the maximal order of an algebraic number £eld K coincides with the set .0 of all numbers of K whose minimum polynomial has integer coefficients. We first prove the following lemmas. Lemma 3. If a E .0, that is, if the minimum polynomial t m + c1t m - 1 + '" +c M of the number a has integer coefficients, then the module M = {I, a, :.. , am-I} is a ring. Proof. It is clearly sufficient to show that any power ak (k ~ 0) of the number a lies in M. For k ~ m - 1 this is true by the definition of M. Further, am = -Clam-I - ... - CM with integers Ci , so that am EM. Let k > m, and

assume that it is already proved that ak -) EM; that is, ak -) = a l am -) + am with integers ai' Then

ak = aak -

I

+ ...

= alex'" + a 2 a m - 1 + ... + ama.

Since all terms on the right lie in M, ak also belongs to M. Lemma 3 is proved. Lemma 4. If .0 is any order of the field K and a E £5, then the ring .o[a], consisting of all polynomials in ex with coefficients from .0, also is an order of the field K. Proof. Since .0 c .o[a], the ring D[a] contains n = (K: R) linearly inde·

pendent numbers over R. We thus need only show that .o[a] is a module (that is, that it is finitely generated). Let WI' ... , W n be a basis for the order .o. By Lemma 3, any power ak (k ~ 0) can be represented in the form ao + a l a + ... + am -I am - I with integers ai' where m is the degree of the minimum polynomial of the number a. From this it easily follows that any number of .o[a] can be represented as an integral linear combination of the products Wiai (l ~ i ~ n, 0 ~j ~ m - 1), and this means that .o[a] is a module. Repeated application of Lemma 4 gives us the following.

92

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

Corollary. If .0 is an order and IX l , '" , IXp are numbers of .0, then the ring D[IX b ... , IX p ] of all polynomials in a l , .. , , IX p with coefficients in .0 is also an order. Theorem 6. The set of all numbers of the algebraic number field K whose minimum polynomial has integer coefficients is the maximal order of the field K.

Proof. Let .0 be any order of the field K and let IX and fJ be arbitrary numbers of .0. By the CoroIlary of Lemma 4 the ring D[IX, fJ] is an order, and hence it is contained in .0 (Lemma 2). But then the difference IX - {J and the product IXfJ are also contained in .0. This proves that .0 is a ring. Since .0 cD, is contains n linearly independent numbers. We thus need only show that D is a module. Let W l , .,. , W o be any basis of the order .0, and let W l *, ... , W.* be the dual basis to it in the field K (Supplement, Section 2.3). We shaIl show that the ring D is contained in the module .0* = {w l *, ... , w.*}. Let IX be any element of the ring D. Represent it in the form IX =

with rational

Ci'

C1W l

Multiplying by Ci

Wi

* + ... + cow.* and taking the trace, we obtain

= Sp IXW i

(l ~ i ~ n)

(we are using here the fact that Sp WjW j * = I and Sp wjw/ = 0 for i #- j). All products aW j are contained in the order D[IX], and therefore by Lemma 2 all numbers c j are integers, and this means that IX E .0*. Thus.o c .0*. By the Corollary of Theorem 2 we now conclude that .0 is a module, and Theorem 6 is proved. The proof which we have given that .0 is a ring is of very general character; that is, it remains valid (with insignificant changes) also in the general theory of commutative rings without zero divisors. The corresponding concepts in the general case are given in Section 4 of the Supplement. Using the termin~ ology introduced there we can say that the maximal order of an algebraic number field K is the integral closure of the ring Z of rational integers in the field K. Here the maximal order .0 wiIl frequently be called the ring of integers of K, and any number in D win be ~alled an integer of K. The units of the maximal order .0 are also caIled the units of the algebraic number field K. 2.5. The Discriminant of a Full Module Let /110 .. , , /10 and /11" ... , /10' be two bases for the full module M of the algebraic number field K. We have seen (Section 2.1) that the transition

Sec.2]

FULL MODULES AND THEIR COEFFICIENT RINGS

93

matrix from one basis to the other is unimodular (that is, it is an integral matrix with determinant ± I). It follows that the discriminants D(Jil' .,. , Jin) and D(Jil', ... ,Ji.') are equal [Supplement, Eq. (2.12)]. All bases of the module thus have the same discriminant. This common value is clearly a rational number, and it is called the discriminant of the module M. Every order of the field K is a full module in K. Hence we may speak of the discriminant of an order. Since the trace of any number in an order is an integer, the discriminant of an order will always be a rational integer (the same holds for any full module contained in .6). A basis of the maximal order .6 of the algebraic number field K is frequently called a fundamental basis of K, and its discriminant is called the discriminant of the field K. The discriminant of an algebraic number field is a very important arithme'tic invariant and will playa key role in many questions.

PROBLEMS 1. Let.w 1, W2, W3 be linearly independent numbers of the algebraic number field K. Show that the set of numbers of the form aWl + bW2 + CW3, where the rational integers a, b, C satisfy 2a + 3b + 5c = 0, forms a module in K, and find its basis.

2. Find the coefficient ring of the module {2, v2/2} in the field R( vi). Show that the module {I, v2} is the maximal order of the field R(v2). 3. Show that the field of rational numbers contains only one order, the ring of rational integers. 4. Show that in the order {I, ~2, f /4} of the field R(~i) every number with norm 2 is an associate of ~2. 5. Show that the intersection of two full modules is again a full module. 6. Show that any module of an algebraic number field which is a ring is contained in the maximal order. 7. Let M = {(Xl> ... , (X.} and N = {{31> ... , {3.} be two full modules of the field K. The module generated by the products (X1{3J (l ~ i,j ~ n) does not depend on the choice of the bases (X/ and {3J. It is called the product of the modules M and N and is denoted by MN. Show that the coefficient rings of the modules M and N are contained in the coefficient ring of their product MN. 8. Let M be a full module contained in the maximal order D of the algebraic number field K. Show that if the discriminant of the module M is not divisible by the square of any integer other than I, then M coincides with D. 9. Let 8 bea primitive element of the algebraic number field K ofdegree n, with 8 contained in the maximal order. Show that if the discriminant of the minimum polynomial of the number 8 is not divisible by any square, then the numbers 1,8, ... ,8· -1 form a fundamental basis of the field K. 10. Find a fundamental basis and the discriminant of the field R( ~i). 11. Find a fundamental basis and the discriminant of the field R(p), where p is a root of the equation x 3 - x -I =0.

94

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

12. Let M be a full module of the algebraic number field K. Show that the set M * of all ~ E K, for which Sp IX~ E Z for all IX E M, is also a full module of the field K. The module M * is called the dual of the module M. Show that if fll' ... , fl. is a basis of M, then the dual basis fl' *, .,. , fL. * of the field K (with respect to R) is a basis of M *. 13. Show that (M*)* = M; that is, the dual of the module M* coincides with M. 14. Show that the dual modules M and M* have the same coefficient ring. 15. Show that for full modules M, and M, the inclusions M I C M, and M I * =oM,* are equivalent. 16. Let 8 be a primitive element of the algebraic number field K of degree n, with 8 contained in the maximal order ~, and let f(t) be the minimum polynomial of 8 over R. Show that for the module M = {I, 8, ... ,8· -I) (which is clearly an order), the dual module M* coincides with (I/f'(8»M. 17. Let M be a full module in K with coefficient ring D. Show that the product MM* (see Problem 7) coincides with D*. 18. Let M = (4,8,8') be a module in the field R(8), where 8 3 = 2. Show that the coefficient ring of M is the order {I, 28, 28'}, while that of the module M' = (2, 28, 8'} is the maximal order {I, 8, 8 2 }. 19. The polynomialt· + alt" - I + ... + a. with rational integer coefficients is called an Eisenstein polynomial with respect to the prime number p if all the coefficients a" ... , a. are divisible by p, and the constant term a., while divisible by p, is not divisible by p'. Show that if a primitive element 8 of an algebraic number field K of degree n is a root of an Eisenstein polynomial with respect to p, then N(co+c I 8+

...

+c._18·-I)~co·(modp)

for any rational integers Co, C" ... , C._I' 20. If 8 is a primitive element of the algebraic number field K and 8 lies in is, then the index of the order {I, 8, ... ,8· -'} in the maximal order is called the index of the number 8. Show that if (J is a root of an Eisenstein polynomial with respect to the prime p, then p does not divide the index of 8. 21. Show that each of the three cubic fields KI

=

K,

=

K3

=

R(8), R(8), R(8),

83 83 83

-

-

188 6 = 0, 368 - 78 = 0, 548 - 150 = 0,

have as a fundamental basis 1,8, (J'. Verify that all three fields have the same discriminant, namely, 22356 = 23·2'·3'. (It follows from Problem 14, Section 7, Chapter 3, that the three fields K" K" K 3 are distinct.) 22. Show that a fundamental basis for the field R(8), 8 3 - 8 - 4 = 0, is given by I, (J, (8 + ( 2 )/2. 23. Let a and b be relatively prime natural numbers which are square-free. Set k = ab if a' - b' ==: 0 (mod 9), and k = 3ab if a' - b' "# 0 (mod 9). Show that the discriminant of the field R( ~ ;;p) is D = - 3k'. 24. Show that the numbers I, ~!6, (v 6)' form a fundamental basis for the field R( V6).

3. Geometric Methods

Two problems were formulated at the end of Section 2.3 (to which we were led by the question of the representation of numbers by full decomposable

Sec. 3]

95

GEOMETRIC METHODS

forms) whose sol ution requires the introduction of new concepts of a geometric character. At the base of these concepts is a method of representing algebraic numbers as points in n-dimensional space, analogous to the wel1-known planar representation of the complex numbers.

3.1. Geometric Representation of Algebraic Numbers If the algebraic number field K is of degree n over the rational numbers R, then there are precisely n distinct isomorphisms of this field into the field C of al1 complex numbers. Definition. If the image of the field K under the isomorphism a: K -. C is contained in the real numbers, then the isomorphism a is cal1ed real, and, if this does not hold, it is cal1ed complex. Thus for the cubic field K = R(O), where 03 = 2, the isomorphism R(O) -. R(~/i), for which () -. ~2, is real (by ~2 we understand here the real root). The two other isomorphisms R(O) -. R(e~i) and R(O) -. R(e 2 Vi) [e = cos(2n:/3) + i sin(2n:/3)] are complex. If d is a nonsquare rational number, then for the field R(O), 0 2 = d, both isomorphisms are real if d > 0, and both are complex if d < O. In general, if 0 is a primitive element of the arbitrary algebraic number field K, which is a root of the irreducible polynomial rp(t) over R, and if 01 , .,. , On are the roots of cp(t) in the field C, then the isomorphism

K

=

R(O) -. R(O;) c C,

(0 -.

OJ

(3.1)

wil1 be real if the root 0 i is real, and complex otherwise. If)' = x + yi is any complex number (x and y real) we denote by }' the complex conjugate x - yi. Let a : K -. C be a complex isomorphism. The mapping (j : K -. C, defined by (ct

E

K),

is also a complex isomorphism of K into C. This isomorphism (j is cal1ed conjugate to a. Since (j -# a and (j = a, the set of all complex isomorphisms of K into C is divided into pairs of conjugate isomorphisms. In particular, the number of complex isomorphisms is always even. Two complex isomorphisms of the form (3.1) are conjugate if and only if the corresponding roots 0i and ()j are complex-conjugate numbers. Assume that among the isomorphisms of K into C there are s real ones al> ... , as and 2t complex ones, so that s + 2t = n = (K: R). From each pair of complex-conjugate isomorphisms, choose one. Denote this set of

96

isomorphisms by takes the form

[Chap. 2

NUMBERS BY DECOMPOSABLE FORMS

O"s+1' .. , , O"S+I'

The set of all isomorphisms of K into C then

In the future we shall always assume the isomorphisms to be enumerated in this manner. Clearly there exist fields with no real isomorphisms (s = 0) or no complex isomorphisms (t = 0). Consider the set £5,1 of all rows of the form (3.2) in which the first s components, Xl' ... , X S ' are real, and the remaining ones, X 5 + 1, ... , XS+1' are complex numbers. Addition and multiplication of these rows, as well as scalar multiplication by a real number, are defined componentwise. Under these operations £5,1 becomes a commutative ring with unit (1, ... , 1) and at the same time a real linear space. The rows (3.2) will be called vectors or points of the space £5,1. As a basis of £5,1 (over the field of real numbers) we may clearly take the vectors

~1~

'?~~, :".' O)]s,

(0,

, 1; 0, ." , 0)

(0, .. , ,0; 1.'

,0))

(0, (0,

,1) , i)

(3.3)

(0, ... ,0; I, ,0) . . . . . . . . . .. 21. ,0;0, , 0; 0,

Thus the dimension of the space £5,1 over R equals n = s

(j

= 1,

+ 2t.

If we set

... , t),

then the vector (3.2) will have coordinates (3.4) with respect to the basis (3.3). In cases where £5,1 is being considered as an n-dimensional real linear space, we shall also denote it by 9l n • Fix some point X in £5,1. The transformation X' ~ xx' (x' E £5,1), that is, multiplication of an arbitrary point of £5,1 by x, is clearly a linear transformation of the real space £5,1 = 9l n • In terms of the basis (3.3), the matrix of this transformation is seen to be

GEOMETRIC METHODS

Sec. 3]

97

y, - z, z, y, where all other entries are zero. The determinant of this matrix is Xl ...

xs(y/

+ z/) ... (y/ + z/) = Xl ... Xs lxs +1 12 · .. lx s +,1 2 .

This suggests the following definition. By the norm N(x) of any point X = (Xl' ... , x s+,) E £s", we shall understand the expression N(x)

= Xl'" Xs lx s +1 12 ···lxs +,1 2 •

Thus we have just shown that the norm N(x) of a point X can be defined as the determinant of the matrix of the linear transformation x' ..... xx'. The norm is clearly multiplicative: N(xx') = N(x)N(x').

We now turn to the representation of numbers of the field K by points of the space £s". Each number et. of K will be made to correspond to the point (3.5) of £s.'. This point is the geometric representation of the number et.. If et. and fJ are different numbers of K, then for any k = 1, ... , s + t, the numbers O"k(et.) and O"k(fJ) are distinct, and therefore x(et.) #- x(fJ). Thus the embedding et. ..... x(et.) (et. E K) is one-to-one. (Of course it is not a mapping" onto"; that is, not every point of £s,1 is the image of some number of the field K.) Since O"iet. + fJ) = O"k(et.) + O"k(fJ) and O"k(et.fJ) = O"k(et.)O"k(fJ),

x(et.

+ fJ) = X(IX) + x(fJ), X(lXfJ) = x(et.)x(fJ);

(3.6) (3.7)

that is, if numbers of K are added or multiplied, the corresponding points are also added or multiplied. Further, if a is a rational number, then O"k(alX) = O"k(a)O"k(et.) = aO"k(IX), so that x(aet.) = ax(et.). (3.8)

98

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

Thus by Section 2.3 of the Supplement we have N(rx)

= N K/R(rx) =

0" l(rx)

O".(rx)O"s+ 1 (rx)O',,+ 1 (rx) ... 0"" + lrx)O's + I(rx)

=

O"l(rx)

O".(rx)!O"S+l(rxW ... 100s+l(rxW,

so that the norm N(x(rx)) of the point x(rx) coincides with the norm N(rx) of the number (1.: (rx E K). N(x(rx)) = N(rx), We consider two simple examples. If d is a positive rational number which is not a square, then for the real quadratic field R(O), 02 = d, the geometric representation of the number rx = a + bO (a and b rational) will be the point x(rx) = (a + b~d, a - b~d). In the case of the imaginary quadratic field R(rJ), rJ2 = - d, the representation of the number rx = a + brJ will be the point in the complex plane with coordinates (a, b ~d) [the basis (3.3) in this case consists of the numbers I and i]. We shall show that if rx1' ... , rxn is any basis of the field K (over R), then the corresponding vectors X(rx1), .. , ,x(rxn) of £5.1 = 91 n are linearly independent (over the reals). For this set O"k(rx,) = Xk (I) O"s+/rx,) = Y/')

(1

+ iz/')

~

k

~

s),

(l ~ j ~ t).

Since the vectors

with the basis (3.3) have the coordinates

then to prove our assertion we need only show that the determinant X1(1) ... xP)

d= Xl (n) ...

x/ n )

Y1(1) Y1

(n)

zP)···Y?)

z (1)

Y?)

z (n)

Zl (n) •••

1

1

is nonzero. Consider instead the determinant X (1) ... X (l) 1 s d* = Xl (n) ...

x/ n)

Y1(1)

+ iz/ 1 )

Y1(n)+iz 1(n)

which can also be written in the form

y/ 1) _

izP) ...

Y1(n) - iz 1 (n) •••

Sec. 3]

GEOMETRIC METHODS

d* =

99

al(cx l ) .,. a'(cxd

.

a I(cx n)

0",( cx n)

In the determinant d add to the column number s + I the succeeding column and then take the 2 outside the determinant sign. Then subtract this [Jt:W column from the succeeding one, and take the - i outside the determinant. Performing this operation on eHch succeeding pair of columns, we W1l1d up with the equation d* = (-2ir d. (3.9) In Section 2.3 of the Supplement it is shown that (3.10) where D = D(cx l , '" ,cx n) is the discriminant of the basis CX I , ... 'CX n (with respect to the extension KjR). Since D -# 0, it follows from (3.9) and (3.10) that the determinant d also is nonzero. Assume now that CX I , ... , CX n is a basis for the full module M of the field K. From (3.6) and (3.8) it follows that if cx = alcx l + ... + ancx n is in M (ai, ... ,tin rational integers), then the geometric representation of cx in £,,1 will be the vector x(cx) = alx(cxl) + .. , + anx(cx n). We thus have the following result. Theorem 1. Let K be a number field of degree n = s + 2t, with M = , cx n } a full module in K. Under the geometric representation ofnurnbers of K by points of the space 91 n , the numbers of M are represented by the set of all integral linear combinations of the n linearly independent (in the space 9t n) vectors X(CXI), ... , x(cx n).

{CXI' '"

3.2. Lattices The geometric study of full modules is based on the fact established in Theorem I. We therefore consider sets of vectors in 9t n of the above type, without necessarily assuming that they represent the numbers of some fuJi module. Definition. Let e!, ... , em' m ~ n, be a linearly independent set of vectors in 9tn • The set ~R of all vectors of the form

where the a i independently take on all rational integral values, is called an m-dimensionallattice in 9t n , and the vectors el> .. , , em are called a basis of this lattice. If m = n, the lattice is called full; otherwise it is called nonfull.

100

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

Theorem 1 thus states that the geometric representation of the numbers of a full module is some full lattice. It is easily seen that two linearly independent sets e l , '" , em andfl' ... ,fm determine the same lattice if and only if they are connected by a unimodular transformation, that is, if m

Ii = j=l L cijej

(l ~ i ~ m),

where (c) is an integral matrix with determinant ± 1. The more detailed study of lattices is based on the metric properties of the space mn • We introduce an inner product on ,2s.t = mn by taking the vectors (3.3) to be an orthonormal basis. If the vectors x and x' have the coordinates (Xl> ... , x n ) and (Xl" ... , x n') with respect to this basis, then the inner product of x and x', (x, x') is given by the formula The length of a vector x will be denoted by Ilxll. Let r be a positive real number. The set of all points x with coordinates (Xl' ... , x n) [with respect to the basis (3.3)], for which

Ilxll =

JX 1

2

+ ... + x n 2 < r,

will be denoted by U(r). The set U(r) is called the (open) ball of radius r with center at the origin. A set of points in 9\n is called bounded if it is contained in some ball U(r). A set of points is called discrete if for every r > there are only a finite number of points of the set in the ball U(r).

°

Lemma 1. The set of points of any lattice Wl in

m is discrete. n

Proof. Since any nonfull lattice can be embedded in a full lattice (in many ways), it suffices only to consider a full lattice Wl. Let e l , ... , en be any basis for IDl. The conditions (x, e 2) = 0, ... , (x, en) =

°

give us a system of n - 1 homogeneous linear equations in n unknowns. Since such a system has a nonzero solution, there is a nonzero vector x which is orthogonal to the vectors e 2 , ... , en' If we also had (x, e l ) = 0, then the vector x would be orthogonal to all vectors of the space mn , which is impossible. Hence (x, e l ) -# 0. The vector fl = [l!(x, e l )] x will also be orthogonal to all the vectors e2' ... , en, and (fl' el) = 1. In this manner, for every i, 1 ~ i ~ n, we can choose a vector fi' for which if j = i, if j =I- i.

Sec. 3]

GEOMETRIC METHODS

101

Assume now that the vector z = aIel + ... + anen ofIDl (a j rational integers) lies in the ball U(r); that is, Ilzll < r. Since ak = (Z,fk)' by the CauchySchwartz inequality we have

where rllfkll does not depend on z. Thus there are only a finite number of possibilities for the integers ak> so that the set of all z E IDl for which Ilzll < r is finite. Lemma 1 is proved. Let X be some set of points of the space 91 n and z a point of 91 n• The set of all points of the form x + z, where x is in X, is called the translate of the set X by the vector z and is denoted by X + z. Definition. Let el , points of the form

.. , ,

em be any basis for the lattice IDl. The set T of

where CX I , ••• 'CX m independently take on all real values satisfying 0 ~ CX j < 1, is called a fundamental parallelepiped of the lattice IDl. A fundamental parallelepiped is not uniquely determined by the lattice; it depends on the choice of basis. Lemma 2. If T is a fundamental parallelepiped of the full lattice IDl, then the sets

T z = T+ z, where z runs through all points of the lattice IDl, are pairwise-disjoint and fill the entire space 91n • Proof. Let el , ... , en be the basis of IDl used to construct the parallelepiped

T. We must show that every point x = Xlel + ... + xnen of 91n lies in one and only one set Tz • For each j write the real number Xi in the form Xi = k j + CX i , where k j is a rational integer and CX i satisfies the condition 0 ~ CXi < 1. Setting z = kle l + ... + knen and u = CXlel + ... +cxnen, we have X=u+z

(u E T, z E IDl),

which means that x E T z • Now if x E T z • , that is, x = u' + z' (u' E T, z' E IDl), then by comparing the coefficients of e j in the equation u + z = u' + z' we easily obtain z = z'. Lemma 2 is proved. Lemma 3. For any real number r > 0 there are only a finite number of sets T z (see the notation of Lemma 2) which intersect the ball U(r). Proof. Let el , ... , en be the basis of the lattice used to construct the

102

NUMBERS BY DECOMPOSABLE FORMS

parallelepiped T. If we set d = llelll + '" + cxnen E T, we will have

[Chap. 2

+ ... + llenl!, then for any vector

u

=

CX I el

Assume that the set T z (z E 9Jl) intersects U(r). This means that for some vector x = u + z, where u E T, Z E 9Jl, we have IIxll < r. Since Z = x - u,

+ II-ull < r + d, that is, the point z is contained in the ball U(r + d). By Lemma 1 there are Ilzll ~ IIxll I

only finitely many such points, and Lemma 3 is proved. It is clear that the vectors of a lattice form a group under the operation of vector addition. In other words, every lattice is a subgroup of the additive group mn. Lemma 1 shows that all subgroups are not lattices. We now show that the property of lattices which was established in that lemma characterizes lattices among all subgroups of the group mn. Lemma 4. A subgroup 9Jl of the group is a lattice.

mn, the points of which are discrete,

Proof. Denote by G> the smallest linear subspace of the space mn which contains the set 9Jl, and by m the dimension of G>. We can then choose m vectors el' '" ,em in 9Jl which form a basis for the subspace G>. Denote by 9Jl o the lattice with basis el' .. , , em' Clearly 9Jl o c 9Jl. We shall show that the index (9Jl : 9Jl o) is finite. Indeed, we may represent any vector x of 9Jl (even any

vector of G» in the form x= u

+ z,

(3.11)

where Z E 9Jl o and u lies in the fundamental parallelepiped T of the lattice 9Jl o , constructed with the basis el , ••• , em' Since x E 9Jl and Z E 9Jl o C 9Jl, and since 9Jl is a group, u E 9Jl. But T is a bounded set, and since 9Jl is discrete, T can contain only a finite number of vectors of 9Jl. This shows that the number of vectors u which occur in (3.11) for all x E 9Jl is finite, which means that the index (9Jl: 9Jl o) is finite. Let (9Jl: 9Jl o) = j. Since the order of any element of the factor group 9Jl/9Jl o is a divisor of j, then jx E 9Jl o for all x E 9Jl, which means that x is a linear combination, with integer coefficients, of (l1J)el' ... , (llJ)em. The group 9Jl is thus contained in the lattice 9Jl* with basis (l/j)el' ... ,(llJ)em • Applying Theorem 2 of Section 2, we see that the subgroup 9Jl of the group 9Jl* must possess a basis of I ~ m vectors!l, ... ,j;. To show that 9Jl is a lattice, we need only verify that the vectors!l' ... ,j; are linearly independent over the real numbers. But this follows from the fact that the m linearly independent (over the reals) vectors el' ... ,em are linear combinations of the!i (since 9Jl o £: 9Jl). Lemma 4 is proved.

103

GEOMETRIC METHODS

Sec. 3]

3.3. The Logarithmic Space Along with the above geometric representation of the numbers of the field K, in which the operation of multiplication of numbers was represented by the operation of multiplication of vectors in ~W, we must consider another geometric representation, in which the operation of multiplication also has a simple interpretation. Let there be s real and 2t complex isomorphisms of the algebraic number field K into the field C of complex numbers. We shall assume that these isomorphisms are indexed as in Section 3.1. Consider the real linear space 9ls + 1 of dimension s + t, consisting of rows (A'I' ... , As + 1) with real components. If x E £S,I is of the form (3.2), with all components different from zero, set lk(x) l,+)x)

= Inlx k \ = Inlxs+jl2

for k = I, ... ,s, for j = 1, .. , ,t.

(3.12)

We associate to each such point x of f!S,1 the vector (3.13) of the space 9ls + l . If xand x' are any points of f!s,1 with nonzero components, then (1

~

k

~

s

+ t),

so that l(xx') = l(x)

+ l(x').

(3.14)

The collection of all points x E £S,I of the form (3.2) with nonzero components [that is, for which N(x) "# 0] form a group under componentwise multiplication. Equation (3.14) shows that the mapping x --+ l(x) is a homomorphism of this multiplicative group onto the additive group of the vector space 9ls + l . Comparing (3.12) with the definition of the norm N(x) of a point x E £S,I, we easily see that s+1

I

lk(x)

= InIN(x)l·

(3.15)

k=1

If Ct. is a nonzero number from the field K, set 1(Ct.)

= l(x(Ct.»,

where x(Ct.) is the representation of the number Ct. in the space .1!S,1 descnbed in Section 3.1. From (3.5), (3.12), and (3.13) we see that the vector 1(Ct.) has the form l(rx) = (Inla 1 (:x)I, ... , Inla"(:x)I, In\a s + 1 (Ct.W, ... ,lnias+,(Ct.W).

104

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

We call the vector I(rx) E 9p+tthe logarithmic representation of the nonzero number rx E K, and call the space 9l s + t the logarithmic space of the field K. From (3.7) and (3.14) it follows that l(rxfJ)

= I(rx) + l(fJ)

(rx =! 0, fJ =! 0).

(3.16)

The mapping rx -+ I(rx) is thus a homomorphism of the multiplicative group of thefieldKinto the group of vectors of the space 9l s + t • In particular it follows that (rx =! 0). The sum of the components (1 ::0;; k ::0;; s

+ t),

of the vector I(rx) is given by the formula s+t I lk(rx) = InIN(rx)l· k=l

(3.17)

Indeed, the sum on the left is the logarithm of the absolute value of the product O'l(rx) .. · O',(rx)O's+ 1 (rx)O's + l(rx) ... O's+t(rx)O's+t(rx),

and this product (Supplement, Section 2.3) equals the norm N(rt.) (with respect to the extension KjR). This proof of (3.17) [which does not rely on (3.15)] shows why the definition (3.12) of the components Ik(x) of the vector I(x) distinguished between real and complex isomorphisms. The component Is + ix) corresponds not to one, but to the two complex-conjugate isomorphisms O's + j and ii s + j • 3.4. Geometric Representation of Units Let .0 be some fixed order of the field K. In the logarithmic space 9l s + t consider the set of all vectors I(e), where e is a unit in the ring.o. The mapping e -+ I(e) is not one-to-one. For if the unit tI E .0 is a root of}, that is, if tI'" = 1 for some natural number m, then IO'k(tI) I = I for all k = 1, ... , s + t, so that 1(,,) is the zero vector. Thus all roots of I (and the order .0 contains at least two: + 1 and - 1) are mapped to the zero vector. In order to use the mapping e -+ I(e) to clarify the structure of the group of units, we must answer the following two questions: (1) Which units e E .0 are mapped to the zero vector? (2) What is the form of the set of all vectors I(e)? We start with the first question. Denote by W the set of all numbers rx ED for which I(rx) = O. By (3.16) the product of any two numbers of Wagain lies in W. Since the condition I(rx) = 0 is equivalent to

GEOMETRIC METHODS

Sec. 3]

o'iex) 1 = 1

1

105

(1 ~ k ~ s + t),

n

the set of all points x(ex) E 9l = £s,t for all ex E W is bounded, that is, it is contained in some ball U(r). Applying Lemma 1 we find that the set W is finite. For an arbitrary number ex E W consider its powers 1, ex, ... , exl , ...• Since these powers are contained in W, there is some equality rl = ex', / > k. Setting m = /- k, we obtain exm = 1. Thus all numbers of Ware roots of 1, and this means that W is a finite group, contained in the group of units of the ring ,o. Since the group W contains a subgroup of order 2 (consisting of + 1 and -1), then it has even order. Further, all finite subgroups of the multiplicative group of a field are cyclic (Supplement, Section 3), and therefore W is cyclic. We therefore have the following answer to the first question. Theorem 2. The units of the order ,0, for which /(e) is the zero vector, form a finite cyclic group of even order. This group consists of all roots of 1 contained in ,o.

We therefore turn to the second question, that is, we shall seek to clarify the structure of the set (f in 9l s + t which consists of all vectors /(e), where e is a unit of the ring ,o. By Theorem 4 of Section 2 the norm of any unit e of ,0 equals ± 1, and therefore InIN(e)1 = O. By (3.17) we therefore have s+t

L lk(e) = O.

(3.18)

k=I

This means that all points /(e) belong to the subspace £ c 9l s + t , consisting of all points (AI' ... ,As +t ) E 9l s +t , for which Al + ... + As +t = O. The dimension of the subspace £ clearly equals s + t - 1. We shall show that (f is a lattice. Since (f is a subgroup of the additive group of the vector space 9l s + t , by Lemma 4 it suffices to prove that the set (f is discrete. (As an orthonormal basis in 9l s + t we take those vectors with one component equal to 1 and the rest equal to 0.) Let r be any positive real number and let 11/(e)11 < r. Since lk(e) ~ 11k(e)1 ~

11/(e)ll, then

Ik(e) < r

(l ~ k ~ s + t),

which means that !uk(e)1 < e'

(k = 1, .. , , s),

lu.,+/eW < e'

(j = 1, ." , t).

From this it follows that the set of points x(e) in 9ln, where e runs through all units of ,0 for which II/(e)II < r, is bounded. But since the vectors x(ex) E 9ln for all ex E ,0 form a lattice (Theorem 1), it follows from Lemma 1 that the

106

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

number of such e is finite. Thus the number of vectors I(e) such that II l(e) II < r is also fimte and this means that the seT G: is discrete. Since the Ianice G: is contained in the subspace .2, its dirm:nsion does not exceed s + t - 1. We have thus proved the following fact. Theorem 3. The set of all points I(e), where e is a unit ot'the order .0, forms a lattice G: in the logarithmic space £ of dimension r ~ s + t - 1.

3.5. Partial Results on the Group of Units Theorems 2 and 3, which were derived from very simple geometric considerations, contain much important information on the structure ot' the group of units of the order .0. From these theorems it follows that in .0 there exist units el' ... , e" r ~ s + t - 1, such that every unit e is uniquely representable in the form (3.19) where aI' ... , a, are rational integers and ( is some root ot' 1 contained in D. In other words, the group of units of the order .0 is the product of a finite group and r infimte cyc~ic groups. To prove this assertion we take any basis for the lattice If, say, I(el)' ... , I(e,), and will show that the units el' ... , e, satisfy the desired properties. Let e be an arbitrary unit of the ring .0. Since l(e) E G:, then

I(e) = a 1 /(e l ) +

.. , + o,l(e,),

where a l are rational integers. Consider the unit: By formula (3.16) we have 1(0 = I(e) - ol/(el) - ... - a,l(e,), and then by Theorem 2 ( is a root of 1. Hence e has a representation (3.19). We now prove that this representation is unique. Assume that we also have e = ('e/' ... e/r • Since the vectors I(ed, .. , , I(e,) are linearly independent, it foHows from I(e) = b1 /(el) + ... + b,/(e,) that a1 = b1 , ... , Q, = b,. But then also ( = (', and our assertion IS compleTely proved. Therc remains the open question of the precise value of t:le number r, since we have proved only that it does not exceed s + t - 1. Using the meThods we have relied on so far, we cannot even guarantee that r> U (when s + t 1 > 0). In Section 4 we shall show that actually r = s + t - 1. But this will be an existence theorem; it will establish the existence of s + t - 1 llldependent units. It is therefore not surprising that its proof will require some new concepts.

Sec. 4]

THE GROUP OF UNITS

107

By Theorem 3 the remaining assertion is equivalent to the fact that the dimension of the lattice G:, which represents the units of the order D in the logarithmic soace. is of dimension equal to s + t - 1.

PROBI.EMS

1. Show th:H the set of :111 images x( ex) E )l{" of numbers ex of the algebraic number field K of degree n is an everywhere-etense slJbset of the soace ffi". 2. Assume that s l' 0, th:1t is, that there is an isomorohisrn of the field K into the field of real numbers. Show that the grouD of roots of 1 contained in K consists only of two numbers: + 1 and -1. (This condition always holds when the cte!!ree of K is odd.) 3. Determine all roots of 1 which call be contained in an algebraic number field of degree 4. 4. Find all units of the field R( Y3). 5. Show that in the field R(~2) any unit has the form ±(l + ,,3/ 2)". 6. Let the algebraic number field K contain a complex root of 1. Show that then the norm of any c£ l' 0 of K is positive.

4. The Group of Units 4.1. A Criterion for the Fullness of a Lattice

In this section we finish our investigation of the structure of the group of units of an order in an algebraic number field. The basic problem, which we are going to solve, has already been considered at the end of the preceding section. It is to prove that the lattice G:, the vectors of which represent the units of the order D in the logarithmic representation, has dimension s + t - I (we preserve all notations from Section 3). The lattice G: lies in the space 9{s+ r and is contained in the subspace .2, which consists of all points (A'l'"'' A.,+,) for which Ai + ... + As +r = O. Since the dimension of .2 is equal to s + t - I, we need to show that G: is a full lattice in the space .2. This will be proved in Section 4.3, using the follow-. ing criterion for the fullness of a lattice. Theorem 1. A lattice min a linear space .2 is full if and only if there exists a bounded set U in .2, sue!'. that the translates of U by all vectors of!m occupy the whole space .2 (they may intersect).

m is full, then as U we may take any of its fundamental parallelepipeds. Lemma 2 of Section 3 now implies that the translates of a fundamental parallelepiped by all points of the lattice fill the entire space (it is clear that a fundamental parallelepiped is bounded). Assume now that the Proof. If the lattice

108

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

lattice 9R is not full, and let U be an arbitrary bounded set in £. We shall show that in this case the translates of U by all vectors of 9R cannot fill the entire space £. Since U is bounded, there exists a real number r > 0 such that Ilull < r for all u E U. Let £' denote the subspace generated by the vectors of IDl. Since the lattice 9R is not full, £' is a proper subspace of £, and therefore there exists in £ vectors of arbitrary length which are orthogonal to the subspace £' (and hence to all vectors in 9R). We claim that any such vector y, for which Ilyll ~ r, cannot lie within a translate of Uby a vector of9R. For suppose that such a y (orthogonal to £') is contained in some translate, say, y = u + z, where u E U, Z E IDl. By the Cauchy-Schwarz inequality

IIYl12 = (y, y) = (y, u)

~

Ilyll Ilull < rllyll,

so that lIyll < r. Theorem I is proved. (The geometric meaning of the proof is that all translates of U by vectors of the nonfull lattice 9R lie in the strip consisting of all points at distance less than r from the subspace £'). Remark. In topological terms, the fullness of the lattice 9R in the space £ is equivalent to the compactness of the factor group £/IDl (£ being considered as a topological group under addition).

4.2. Minkowski's Lemma Our proof of the existence of s + t - I independent units will be based on a simple geometric fact which has many applications in number theory. The formulation and proof of this assertion (Theorem 3) use the concept of volume in n-dimensional space and some of its properties. The volume v(X) of the set X in the n-dimensional space 9\n can be defined as the multiple integral v(X) =

J... JdX

1

dX2'" dx n ,

(Xl

carried out over the set X. [Here we sometimes deviate from (3.4) and denote the coordinates of the point x in 9t n by (Xl' '" , x n).] We shall not enter into the question of conditions under which the volume exists. In the cases which we shall consider, the set X will be given by some inequalities of a very simple type, and the question of the existence of the volume can be decided by elementary considerations. We list some simple properties of volume, easily verified from properties of the integral.(We assume that all volumes considered exist.) (1) If X is contained in X', then

v(X)

~

v(X').

Sec. 4]

THE GROUP OF UNITS

109

(2) If the sets X and X' do not intersect, then

veX u X') = veX)

+ veX').

(3) Any translate of a set has the same volume; that is,

veX + z)

= veX).

(4) Let Ct. be a positive real number. Let Ct.X denote the set of all points of the form Ct.x, where x runs through all points of X. (The set Ct.X is called the expansion of X by Ct..) Then

v(Ct.X)

= Ct.·v(X).

We now compute the volume of a fundamental parallelepiped T of the full lattice 9Jl in 91', which is constructed from some basis e l , ... , e•. Let (I ~j~n).

ej = (ali' ... , a.j) We shall show that then

(4.1) In the integral

v(T)

=

f.. ·f dxl,,·dx. (T)

we change variables by the formula

•

Xi

=

LI aijx/

(1

~

i ~ n).

j=

The Jacobian of this transformation is det(aij)' which is nonzero since the vectors e l , ... ,e. are linearly independent. Under this transformation the set T is taken to the set To, consisting of all points (Xl" ... , X.'), for which o ~ x;' < 1 (i = 1, ... ,n), so that

f.. ·f Idet(aij)1 dx l' .. · dx.' = Idet(aij)! f··· f dx dx.' = Idet(aij)l,

veT) =

(Tol

I

I

o

0

l ' .. •

and (4.1) is proved. Let the mapping X -+ x' give a nonsingular linear transformation of the space 91' into itself. The lattice 9Jl is taken by this transformation into some lattice 9Jl' (clearly full), and its fundamental paralIelepiped T is taken to a fundamental parallelepiped T' of the lattice 9Jl'. It is clear that the parallelepiped T' will be constructed from the images el', ... , e.' of the vectors of basis el, ... , e•. If e/ = (blj' ... ,b.) (I ~j ~ n), then the volume veT') = Idet(bij)l. Let C = (c i) denote the matrix of the linear transformation x -+ x'

110

NUMBERS BY DECOMPOSABLE FORMS

with respect to the basis el ,

... ,

[Chap. 2

e", so that

e/ = i=I" I cijej

(l~j~n).

I;

It is easily seen that b ij = = 1 ai,c,; ; that is, the matrix (bi;) is the product of (a j ;) and (c ij), which m-:ans that v(T')

= v(T) ·Idet q.

(4.2)

Now assume that el' ... ,e" and el" ... ,e"' are two bases of the same lattice 9Jl. Since these bases are obtained from one another by unimodular transformations (that is, C has integer entries and det C = ± 1), it follows from (4.2) that v(T') = v(T). This shows that the volume of a fundamental parallelepiped of a lattice depends only on the lattice itself and not On the choice of basis. Combining (4.1) with (3.9) and (3.10) we obtain the following strengthening of Theorem 1, Section 3. Theorem 2. Let K be a number field of degree n = s + 21 and let M be a full module with discriminant D. Under the geometric representation of numbers of K by points of £5.1 = ~W, the points which represent the numbers of M form a full lattice with the volume of a fundamental parallelepiped 1 equal to 2- JIDI. To formulate the basic proposition of this section we need two more geometric concepts. A set X c 91" is called centrally symmetric if whenever x E X, then also -XEX.

A set X is called convex if for any two points x E X and x' E X, all points of the form ax + (1 - a)x', where a is a real number satisfying 0 ~ a ~ 1, lie in X. In other words, X is convex if the entire line segment connecting any two points of X lies in X. Theorem 3 (Minkowski's Lemma on Convex Bodies). Let 9Jl be a full lattice in the n-dimensional space 91", with the volume of a fundamental parallelepiped of 9Jl given by ~, and let X be a bounded, centrally symmetric, convex set with volume v(X). If v(X) > 2"~, then the set X contains at least one nonzero point of the lattice 9Jl.

Proof. We base our proof on the following (intuitively obvious) assertion: If a bounded set Y c 91" has the property that its translates Y z = Y + z by vectors Z E 9Jl are pairwise-nonintersecting, then v( Y) ~ ~. To prove this assertion we consider some fundamental parallelepiped T of the lattice 9Jl and look at the intersections YilT_ z of the set Y with all translates T _ z = T - z

THE GROUP OF UNITS

Sec. 4]

111

of the parallelepiped T. It is clear that v(Y) =

L v(Y II L

z ).

ZEm

(Although this sum is formally infinite, it contains only a finite number of nonzero terms, since the set Y is bounded and thus intersects only a finite number of the T _= ; see Lemma 3 of Section 3.) The translate of the set YilT _. by the vector Z is clearly equal to Y. II T, and therefore v( YilT ~.) = v( Y. II T), so that v(Y) =

L v(Yz II T). ZEm

If the translates Y z are pairwise-nonintersecting, then the intersections Y. II T are also pairwise-nonintersecting, and since they are all contained in T, the sum on the right of the above equation cannot be more than v(T). Hence 1'(Y) ~ l'(T), and our assertion is proved. Consider now the set tx (obtained from X by contracting by a factor of 2). From the assumptions of the theorem it follows that v(tX) = (I/2")v(X) > ~. If all translates -!- X + Z by vectors Z E 9Jl were pairwise-nonintersecting, then we would necessarily have l{tX) ~ ~, which is not the case. Hence for two distinct vectors ZI and Z2 of 9Jl, the sets tx + Zt and tx + Z2 have a common point: -!-x'

+ ZI = ,±x" + Z2

(x', x" EX).

We write this in the form

Since the set X is centrally symmetric, -x' tx" - tx'

E

X, and since it is convex,

= tx" +!( -x') EX.

Thus the nonzero point ZI - Zz of WI lies in the set X and the theorem is proved. From the first part of the proof we may draw the following obvious corollary (which will be used in Section 5). Lemma 1. If the union of all translates of the set Y by vectors of the lattice 9Jl completely fills the space m", then 1'( Y) ~ ~. For in this case the intersections Y= II T completely fill the fundamental paral1elepiped T (possibly overlapping) and therefore v(Y) =

L v(Yz II T) ~ v(T) = ~. ZEm

To investigate the group of units we shal1 apply Minkowski's lemma to

112

[Chap. 2

NUMBERS BY DECOMPOSABLE FORMS

lattices in the space £S.I, and as the set Xwe shall take all points x of the form (3.2) for which

Ixll < C1 ,

... ,

IXsl
,C S + 1 satisfy TIt:: Ci > (4/n)'~, then there is a nonZero vector x = (XI> , X S + 1) in the lattice 9Jl for which

4.3. The Structure of the Group of Units

We can now completely solve the question of the structure of the group of units of an arbitrary order. Theorem 5 (Dirichlet's Theorem). Let .0 be any order of the algebraic number field K of degree n = s + 2t. Then there exist units el , ... , en r = s + t - I, such that every unit e E .0 has a unique representation in the form G=

'el Q, '"

Gr

Or

,

where ai' '" , a, are rational integers and' is some root of I contained in D. Proof. We have already remarked that we need only prove that that lattice

G:, which is the image of the units of .0, is full in the space £ (which is of dimension s + t - I). From Theorem I we see that it will suffice to show that there is a bounded set U in £ such that the translates of U by all Vectors of G: cover the entire space £. We rephrase this last assertion in terms of the space £S.I.

Sec. 4]

113

THE GROUP OF UNITS

It is clear that any point (.1. 1, ... , As+ 1) of £ (and also of ~R'+I) is the image of some point x of £8,1 under the mapping x ~ l(x). It follows immediately from (3.15) that the image of a point x E £8.1 (with nonzero components) under the logarithmic mapping lies in the subspace £ of 91 8 + 1 if and only if IN(x) I = 1. Denote by S the set of all x E £8.1 for which IN(x) I = 1. If X o is an arbitrary bounded subset of S, then its image I(Xo) is also bounded. For if the point x == (Xl' .. , , X S + 1) has norm ± I and satisfies (l

~

k

then lk(x) < In C for all k = 1, ... , s lk(x) = -

(l~j~t);

~ s),

+t

L li(x) >

ii'k

and thus

-(s

+t

-

l)C,

so that I(Xo) is bounded. Since the norm is multiplicative, if XES and Xo c S, then the product Xox is also contained in S. In particular, for any unit e of the order.o, we have Xox(e) c S [since N(x(e)) = N(e) = ± 1]. If the products Xox(e) for all units e completely cover S, then the translates I(Xo) + l(e) clearly cover the entire space £. We have thus shown that to prove Theorem 5 it suffices to find in S a bounded subset X o whose "multiplicative translates" Xox(e) completely cover S. Let y be any point of S and let 9Jl be the lattice of points in £S.I corresponding to the numbers of the order .0. We map the space £S.I into itself by the linear transformation x ~yx (x E £8.1). In Section 3.1 we saw that the determinant of the matrix of this transformation was equal to N(y), that is, to ± 1. Hence, by (4.2), the volumes of fundamental parallelepipeds for the lattices 9Jl and yIDl are the same. Denote this volume by ~. Choose positive real numbers Cl , ... , cs + 1 so that

Q=

C l '" C8 + 1

>

(~r~,

and denote by X the set of all points x E £s.1 for which the inequality (4.3) holds. Theorem 4 implies that there is a nonzero point x = yx(ex) (ex ED, ex # 0) contained in X. Since N(x) = N(y)N(ex) = ±N(ex) and IN(x) I < Cl ... Cs + 1 = Q, IN(Cl) I < Q. By Theorem 5 of Section 2 there are only finitely many pairwise-nonassociate numbers in the order .0 whose norms have absolute value less than Q. Fix some set ex l , ,., , ex", of nonzero numbers of .0 such that any nonzero number of .0, whose norm has absolute value less than Q, is associate with one of these. Then for some i (l ~ i ~ N) we have Ill:: = Ili' where e is a unit in .0. Then y can be represented as (4.4)

114

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

Set

o

X = S

n

C~/X(ai-I)).

(4.5)

Since X is bounded, each of the sets Xx(a;-I) is also bounded and thus X o is bounded. Further, the choice of the numbers c l , ... , Cs + I which determine the set X and of the numbers ai' ... , aN did not depend on the point y, and therefore the set (4.5) is completely determined by the order D. Since y and X(E) lie in S, then by (4.4) the point xx(a; - I) also lies in S, which means that it belongs to X o . Equation (4.4) thus shows that the point y of S, which we chose arbitrarily, is contained in the set XoX(E). Thus S is completely covered by alI these sets (for all units e), and this proves Theorem 5. As we remarked in Section 3.5, Dirichlet's theorem implies that the group of units in any order of an algebraic number field of degree n = s + 2t is the product of a finite group with s + t - 1 infinite cyclic groups. If s + t = 1 (and this is the case only for the field of rational numbers and for imaginary quadratic fields), then r = O. In this case the lattice G: consists only of the zero vector and the group of units of the order D is just the finite group of roots of 1. The units Eb ... , En whose existence :s established by Di.ichlet's theorem, are calIed fundamental units of the order D. From Section 3.5 it is clear that a set of units 8 1, .,. , E, is fundamental if and only if the vectors I(EI), .. , , leE,) form a basis for the lattice l.i:. From this it easily follows that the units

E/

= (iE 1UiJ

...

E/"

(I :( i :( r)

(where C is an arbitrary root or 1 contained in D) will be a fundamental set of units if and only if the matrix (au) is unimodular. Remark. The proof of Dirichlet's theorem is not effective in that it does not give an algorithm for finding some set of fundamental units for the order .o. This was caused by our use of the fulI system of nonassociate numbers ai' ... , aN whose norm is iess than Q. The existence of such a system was established in a noneffective manner. We return to the question of effectiveness in Section 5. Dirichlet's theorem also holds (as does Theorem 2 of Section 3) for the maximal order D of the field K. Fundamental units for the maximal order D are also calIed fundamental units for the algebraic number field K.

4.4. The Regulator By the construction of Sections 3.3 and 3.4, there is associated to each order .0 of an algebraic number field K of degree r = s + 2t a lattice G: of

THE GROUP OF UNITS

Sec.4]

115

m

dimension r = s + t - 1 in the subspace ~ c s + 1. The volume v of a fundamental parallelepiped of this lattice does not depend on the choice of basis and thus is completely determined by the lattice G:. We now compute this volume. Let To be a fundamental parallelepiped of the lattice G: constructed from the basis I(e,), ... ,/(er) (here e" ... ,er is a system of fundamental units of the order D). The vector 10 = I( ~s + t (I, ... , I) in 91 S + 1 is clearly orthogonal to the subspace ~ and has unit length. Then the r-dimensional volume v = c(To) equals the (s + t)-dimwsional volume of the parallelepiped T determined by the vectors 10 , I(E,), .. ' ,/(er ). By (4.1) the volume v is equal to the absolute value of the determinant of the matrix whose rows are the components of these vectors. If we now add all other columns to the ith column and use (3.18), we may expand the determinant by the ith column and obtain v = ~;+tR, where R is the absolute value of any of the minors of order r of the matrix

( ~'~E~).": [~+'/~.')).

(4.6)

It(E r ) "'[S+I(e r )

It thus follows that all rth-order minors of (4.6) have the same absolute value, which is, moreover, independent of the choice of the system of fundamental units El' ... , Er • The number R (as well as v) thus depends only on the order .o. It is called the regulator of the order .o. The regulator of the maximal order D is called the regulator of the algebraic number field K. (For the field of rational numbers and for imaginary quadratic fields the regulator is by definition equal to 1.) PROBLEMS

1. Show that the inequality veX) > 2°t>. in Minkowski's lemma cannot be weakened. To do this construct a convex bounded centrally symmetric set X with volume veX) = 2°t>. which does not contain any nonzero point of the laltice. 2. Let a be a positive real number. Show that the volume of the set X c ~'.l, consisting of all points x for which IXII

+ ...

-j-

!x,l-r 2 \'y~i-+ Zl2 + ... + 2 \

y/

+ z/ < a

I

[in the coordinates (3.4)], equals veX)

=

2' TT)' - - I an.

(2

n!

Check that the set X is bounded, centrally symmetric, and convex. 3. Let a and b be natural numbers which are not squares. Show that fundamental units for the order {I, V~} of the field Rh/~) are also fundamental units for the order {I, V;, V=b, -b} in the field R(V a~ V =b)~

v av

116

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

4. Show that the group of units of an arbitrary order is a subgroup of finite index in the group of units of the maximal order D. 5. Let the units TJ " .,. , TJ, (r = s + I - 1) of the order D be such that the vectors [(TJ'), ... , [(TJ') are linearly independent. Show that the group of all units of the form TJ' c, '" TJn c, is a subgroup of finite index of the group of all units of the order D. 6. Let c" ... , Cn be positive real numbers and let (a/j) be a nonsingular real matrix of order n. Show that if c, ... Cn >d = Idet (a,}I, then there exist rational integers x" ... , x n , not all zero, such that n

I ~ aijxJI
0 and hence (5.1) either has no solution or has infinitely many.

Remark. Theorem 1 shows us the structure of the set of all solutions of (5.1), but it does not give an effective means for finding these solutions. For the practical solution of (5.1) we must find an effective method for finding a system of fundamental units for the order .0, and a method for finding a full set of pairwise-nonassociate numbers PI' .. , , Pk in the module M with given norm. in the following parts of this section we shall show that both of these problems can actually be solved in a finite number of steps. It must be said, however, that the general effective methods to be described for finding fundamental units and numbers with given nOrm in a module are very ill-suited to actual computation, in view of the very large amount of unnecessary computation. Our goal is only to show that in principle it is possible to carry out these constructions in a finite number of steps. In any given case, by using other considerations and examining the particular behavior of the special

PROBLEM OF RATIONAL-NUMBER REPRESENTATION

Sec. 5]

119

case, it is usually possible to find a much shorter route. In Section 5.3, by way of example, we shall give a simple method for solving our problems in the case of quadratic fields.

5.3. The Effective Construction of a System of Fundamental Units Let a1> ... , a. denote all isomorphisms of the algebraic number field K into the field of complex numbers. Lemma 1. Let cl , ... , c. be arbitrary positive real numbers. In any full module M of the field K there exist only finitely many numbers a for which

c.,

!at(a)1 < c t , ... , la.(a)1
M::> M 2 ). ... , ~" s = (MI : M 2 ), be any system of coset representatives of M 2 in MI. If a1> ... ,an is a basis of M 2 , then every element 0 E M 1 has a unique representation 0 = ~k + C I a l + ... + cnan, where ~k is one of these representatives, and C I , ... , Cn are rational integers. Let 01 , ••• , On be a basis of the intermediate module M. Then each OJ has a representation OJ = ~kj + CIPI + ... + cnpn with integral cij. Therefore,

Proof. Let ~I'

M

=

{Ol' ... , On}

=

{Ol' ... , On, a1> ... , an}

= {~kl' .•. , ~k"'

a1 ,

••• ,

Since there are only finitely many possibilities for the set ~kl ' •.. , are only finitely many possibilities for the intermediate module M.

an}·

~k"

there

Corollary. If M o is any full module in the field K and r is any natural number, there are only finitely many full modules in K which contain M o such that (M: M o) = r.

For by the finiteness of the factor group MjMo we have rM c M o , and hence (ljr)Mo ::> M::> Mo. Lemma 3. Let K be an algebraic number field of degree n = s + 2t and let M be a full module in K with discriminant D. Then there exists a nonzero number a in M whose norm satisfies

(6.5) Proof. We take positive real numbers C 1 ... Cs + t

C1 , •.. ,Cs + t

so that

2)t _

= (-; JIDI + t;,

(6.6)

where t; is an arbitrary positive real number. From Theorems 2 and 4 of Section 4 it follows that there exists a number a 0 in M satisfying

*

luk(a)1
D,

(2)' JIDol.

(M':D) ~ ~

(6.7)

By the Corollary of Lemma 2 there are only finitely many such modules M' satisfying (6.7). Hence the number of classes of modules with D as coefficient ring is finite, and Theorem 3 is proved. Remark. If M 1 and M 2 are any two full modules of the algebraic number field K, we may effectively determine whether or not they are similar. To do this we first determine their coefficient rings. If these are different, then M I and M 2 are not similar. Suppose that M 1 and M 2 have the same coefficient ring D. Replacing, if necessary, one of our modules by a module similar to it, we may assume that M 1 :::> M 2 • We compute the index (M 1 : M 2 ) = a. If IXM I = M 2 , then Ct E D and jN(Ct) \ = a. Therefore we find a full set of pairwisenonassociate numbers Ct 1 , ... , Ct k in the ring D whose ,norms are equal in absolute value to a (by Section 5.4 such a system can be effectively computed). If Ct is any number of the ring D for which IN(Ct) I = a, then IX is associate with some lXi' and CtM1 = Ct i M 1 • We therefore compare the modules M 2 and IX i M 1 (1 ~ i ~ k). The modules M 1 and M 2 will be similar if and only if the module M 2 coincides with one of the CtiMI •

Sec. 7]

REPRESENTATION OF NUMBERS BY QUADRATIC FORMS

129

PROBLEMS

1. Show that any algebraic number field other than the field of rational numbers contains an infinite number of orders. (Hence the number of equivalence classes of similar modules corresponding to all possible orders is infinite.) 2. Use Problem 2 of Section 4 to show that in a full module M with discriminant D there is a number ct # 0 for which n! IN(ct)l,,; ->VIDI 7r nn

4)t (

+ 2t being the degree of the algebraic number field). 3. Use Stirling's formula

(n = s

n!

=

vi 27Tn

(;)"e8

112n

(0

< 8 < 1)

and Problem 2 to show that the discriminant Do of an algebraic number field K with degree n = s + 2t satisfies

IDol>

t

(::.)2 _1_ 4 27rn

e2n-('/6n).

Thus as n increases, the discriminants of algebraic number fields of degree n converge to infinity. 4. Show that the discriminant of any algebraic number field of degree n > I is not equal to ± I (Minkowski's theorem). 5. Show that there exist only finitely many algebraic number fields with given discriminant (Hermite's theorem). Remark. By Problem 3 it suffices to show that there exist only finitely many fields K with fixed degree n = S + 2t whose di~criminant is a given value Do. In the space jRn [consisting of all points (x" ... , x s ' y" z" ... , y" z,)] consider the set X defined in the case s > 0 by (2 ,,; k ,,; s),

y/ + z/ and in the case s

=

< I

0 by

Applying Minkowski's lemma on convex bodies to the set X and to the lattice representing the numbers of the maximal order D, deduce that K Gontains a primitive element (J E :0 whose minimum polynomial has bounded coefficients.

7. Representation of Numbers by Binary Quadratic Forms In this section we make a more detailed study of the questions of this chapter in the case of binary quadratic forms. Since any irreducible rational form ax 2 + bxy + cy2 decomposes into linear factors in some quadratic field, our

130

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

problem is connected with the study of full modules and their coefficient rings in quadratic fields. 7.1. Quadratic Fields We call any extension of the rational field of degree 2 a quadratic field. We first describe this simplest class of algebraic number fields. Let d 1 be a square-free rational integer (positive or negative). Since the polynomial t 2 - d is irreducible over the rationals, the field R(O), obtained from R by adjoining a root 0 of this polynomial, is of degree 2 over R; R(O) is a quadratic field. We shall denote it R( Jd). It is easily seen that any quadratic field K is of this type. We prove this. If IX lies in K and is not rational, then clearly K = R(a). The minimum polynomial for IX over R will have degree 2, so for some rational p and q we have a 2 + pa + q = 0. Set fJ = a + (P12); then fJ2 = (p2/4) - q. The rational number (p 2/4) - q can be represented in the form c 2 d, where d is a square-free integer. It is clear that d 1, since otherwise fJ and also IX would be rational. If 0 = fJlc, then 02 = d and K = R(O); that is, K = R(Jd). We now show that for distinct d (not equal to 1 and square-free), the fields R(Jd) are distinct. For if R(Jd') = R(Jd), then

*

*

Jd'

= x + yJd

for some rational x and y, so that d'

= x 2 + dy 2 + 2xyJd

and consequently 2xy = 0. If y = 0, then d' = x 2 , which is impossible. If x = 0, then d' = dy 2, which means that d' = d. We have shown that there is a one-to-one correspondence between quadratic fields and square-free rational integers d 1.

*

7.2. Orders in Quadratic Fields Any number of the field R( Jd) has the form (J,

= x + yJd,

where x and yare rational. Since the characteristic polynomial of (J, equals t 2 _ 2xt + x 2 _ dy 2, then

(J,

will lie in the maximal order

.0

of the field R( Jd) if and only if

Sec. 7]

REPRESENTATION OF NUMBERS BY QUADRATIC FORMS

131

and x 2 - d/ = N(ct) are rational integers. Set 2x = m. Since (m /4) - dy 2 will be an integer and d is square-free, the denominator of the rational number y (in reduced form) must be 2; that is, y = nl2 with integral n. Clearly, N(ct) = (m 2 /4) - d(n 2 /4) will be integral only if 2x

= Sp(ct)

2

m2

dn 2

-

=0 (mod 4).

(7.1)

The solvability of this congruence depends on the residue class of d modulo 4. Since d is square-free, d i:- 0 (mod 4), and we have the three possibilities d= I (mod 4),

=

d

=2 (mod 4),

d

= 3 (mod 4).

=

If d I (mod 4), then the congruence (7.1) takes the form m 2 n 2 (mod 4), which is equivalent to m n (mod 2); that is, m = n + 2, and we obtain

=

m n ct=-+-

2

2

J-d=l+n-----"-1 + Jd 2

with integral I and n. Hence in this case, as a basis for the maximal order D (that is, as a fundamental basis for the field R( ,jd), see the end of Section 2), we may take the numbers I and w = (1 + ,jd)/2. Now let d 2 (mod 4) or d 3 (mod 4). If the congruence (7.1) had a solution with n odd, then from d m 2 (mod 4) it would follow that d 0 (mod 4) for m even and d I (mod 4) for m odd. But this contradicts our assumptions. If n is even, then from the congruence m 2 0 (mod 4) we find that m is even. Thus in these cases the number x + y,jJ belongs to the maximal order ~ of the field R( ,jd) only when x = ml2 and y = nl2 are integers. As a basis for the order ~ we may thus take the numbers I and w = ,jd. In the future, when we speak of a basis for the maximal order of the field R( ,jJ), we shall always have in mind the basis I, w, where w = (1 + ,jd)/2 for d I (mod 4) and w = ,jd for d 2, 3 (mod 4). Now consider an arbitrary order .0 of the field R( .jd). Since D is contained in the maximal order f5 (see Section 2.4), then all numbers of .0 have the form x + yw with integral x and y. We choose from these numbers one for which y takes its smallest positive value, say, a + fw. Since a, being a rational integer, is contained in .0, thenfw E D. It is then clear that for any x + yw E .0, the coefficient y is divisible by f, and hence D = {I, f w}. Conversely, by Lemma 3 of Section 2, for any natural number fthe module {I,fw} is a ring and hence is an order in the field R( ,jd). Since for distinct natural numbersfthe orders {l,fw} are distinct, we obtain the following fact: The set of all orders of a quadratic field is in one-to-one correspondence with the set of all natural numbers. In the future we shall denote the order {I,f w} by .of' It is easily seen that

=

=

=

= =

=

=

=

132

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

the number f equals the index of the order 'of in the maximal order.o = ,01 = {I, w}. Thus an order of a quadratic field is completely determined by its index in the maximal order. We now turn to the computation of the discriminant D f of the order n f . We first assume that d == 1 (mod 4). Since Sp-Jd = 0, Sp w

Sp w 2

= Sp ( 1 +2'

/d) = 1,

= Sp(d + 1 + .Jd) = d + 1 4

2

2

These formulas for D f show that each order of a quadratic field is uniquely determined by its discriminant. The results of this section are summarized in the following theorem. Theorem 1. Let d be a square-free rational integer, d #- 1. As a basis for the maximal order D of the quadratic field R( /d) we may take the numbers land w, where w = (l + -Jd)/2 when d == 1 (mod 4) and w = -Jd for d == 2, 3 (mod 4). The discriminant D 1 of the maximal order .0 [that is, the discriminant of the field R( Jd)] equals d in the first case and 4d in the second case. Any order .0 of the field R( -Jd) is of the form .of = {l,fw}, where f is the index (D: D). The discriminant of the order 'of equals Dd 2 •

7.3. Units Since any number of the order ,0 f is represented in the form x + y f w with x and y rational integers, then by Theorem 4 of Section 2 we shall find all units in tl f if we solve the equation

N(x+yfw)=

± 1,

(7.2)

Sec. 7]

REPRESENTATION OF NUMBERS BY QUADRATIC FORMS

133

that is, the equation x 2 + fx;

+ f2

1- d

-4- y2 =

±1

(7.3)

for d == 1 (mod 4) and the equation

x2

_

dj2 y 2 = ±

1

(7.4)

for d == 2, 3 (mod 4). For an imaginary quadratic field s = 0, t = 1, r = s + t - 1 = 0, So that the group of units of any order of such a field is finite and consists of roots of 1. This fact also follows from a direct examination of (7.3) and (7.4), which only have a finite number of integral solutions when d < O. When d = -1, f = 1, (7.4) has the four solutions: x = ± 1, y = 0; x = 0, y = ± 1, which correspond to the numbers ± 1, ± i, which are the fourth roots of 1. When d = - 3, f= 1,(7.3)hassixsolutions:x= ±1,y=O;x=O,y= ±1;x= l,y= -1; x = -1, y = 1, which correspond to the numbers ± 1, ± t ± (i ..}3/2), which are the sixth roots of 1. For all remaining orders of imaginary quadratic number fields (7.3) or (7.4) have only two solutions: x = ± 1, y = 0; that is, ± 1 are the only units. The case of real quadratic fields is more complicated. For the field R( ..}d) with d> 0 we have s = 2, t = 0, and hence r = 1, so all units of the order D f have the form ± en, where e is a fundamental unit of the order D f' Our problem is thus to determine the fundamental unit e. Along with e the numbers lie, - e, and -lie will also be fundamental units. We may thus assume that e > 1. It is clear that the condition e > 1 determines the fundamental unit e uniquely. Let '1 > 1 be any unit of n J' We shall show that in the representation '1 = x + yfw the coefficients x and yare positive (for d = 5,/ = 1, it is possible that x = 0). For any (J, E F( ..}d) we denote by (J,' its conjugate, that is, the image of IX under the automorphism ..) d --+ - ..}d of the field R( ..}d). It is easy to check that w - w' > O. Since N('1) = '1'1' = ± 1, then '1' equals either 1/'1 or - 1/'1; and in both cases 1/ - '1' > 0; that is, yf(w - w') > 0, and hence y > O. Further, since 1/(1 = Ix + yfw'l < 1 and fw' < -1 with the exception of the cased=5,f=l, we have x>O [ifd=5,f=l, then -1 x and Yl > y. Hence to find the fundamental unit e > 1, we must find the integral solution of(7.2) with smallest positive values for x and y. Using the results of Section 5.3 we may find an upper bound C for the desired values of x and y and then find them after a finite number of steps.

134

[Chap. 2

NUMBERS BY DECOMPOSABLE FORMS

We now show that the number of steps in the location of a fundamental unit can be significantly reduced if a basic result from the theory of continued fractions is employed. We speak of the theorem which asserts that if for the real number > 0 as well as the relatively prime natural numbers x and y we have

e

1

~_el 0, then the polynomial at 2 + bt + c is uniquely determined. We denote it by ({)y(t). If y' is the conjugate of y, then we have ({)y,(t) = ({)/t), and if (()yJt) = cp/t), then either Yi = Y or Yi = y'. Lemma 1. Ify is an irrational number of R( .jd)with ({)y(t) = at 2 + bt + c, then the coefficient ring of the module M = {I, y} is the order {l, ay} with discriminant D = b 2 - 4ac. Proof. Consider the number a = x + yy with rational x and y. Since the inclusion aM c: M is equivalent to the assertions that al = x + yy E M and a' y

cY + ( x = - -;

bY) y E M, - -;

then a belongs to the coefficient ring D if and only if the rational numbt:rs

cy by x,y,-,a

a

are all integers. Since (a, b, c) = 1, this will occur only when x and yare integers and y is divisible by a. This shows that D = {I, ay}. To finish the proof of the lemma we compute the discriminant of the order D:

Sec. 7]

REPRESENTATION OF NUMBERS BY QUADRATIC FORMS

D = Sp I Sp ay 1 = 2 I Sp ay Sp a y2

1

2 -

- b b

b2

-

I

137

= b 2 - 4ac.

2ac

Corollary. Under the same notations as above, the norm of the module {I, y} is equal to lla. Indeed, the matrix of transition from the basis {l, ay} to the basis {l, "I} is

Lemma 2. In order that the modules {I, yd and {I, "I} be similar, it is necessary and sufficient that the numbers "I and "11 be connected by a relation of the form ky + 1 (7.5) "11 = my + n'

where k, I, m, n are rational integers such that (7.6) Proof. Since different bases of the same module are connected by unimodular transformations (see Section 2.1), then from the equation {a, ayd = {l, "I}

it follows that aY1

a

= ky + I,

= my + n,

where the rational integers k, I, m, n satisfy (7.6). Dividing the first equation by the second, we obtain (7.5). Conversely, let "11 andy be connected by relation (7.5). Then {I,yd=

I

my

+n

{my+n,ky+I}=

I

my

+n

{I, "I}

[{my + n, ky + I} = {I, "I} in view of (7.6)]. The proof of the lemma is complete. Consider the set of all modules in the field R( lJ) which belong to some fixed order D (that is, for which D is the coefficient ring). By Theorem 3 of Section 6, all such modules are divided into finitely many equivalence classes of similar modules. We now introduce the operation of multiplication

138

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

of classes and show that under this operation the set of classes belonging to a given order becomes an Abelian group. If M = {et, f3} and M l = {etl> f31}, then MM l denotes the module {etet l , etf31' f3et l , f3f3d (see Problem 7 of Section 2). It is clear that if A =f. 0 and /l =f. 0, then (AM)(/lM l ) = A/l(MM1)·

(7.7)

If M is any module, we denote by [M] the class of similar modules which contains M. From (7.7) it follows that the class [MMd depends only on the classes [M] and [Mil The class [MM1] is called the product of the classes [M] and [Md. Hence to multiply two classes we choose arbitrary representatives of each class and multiply them. The class which contains this product will be the product of the classes. If M is any module, we denote by M' the module consisting of all numbers et', where et is any number of M. If M is a full module, then M' is also a full module. It is easily checked that if .0 is any order, then the conjugate module .0' coincides with D. From this it easily follows that conjugate modules have the same coefficient rings. We shall prove the formula MM' = N(M)D,

(7.8)

where .0 denotes the coefficient ring of M and N(M) the norm of M. We first assume that the module M has the form {l, y}. In this case, using the notation of Lemma I, MM' = {I, y}{I, y'} = {l, y, 1", yy'}

=

{I, 1', -

Y-

~, -

;}

b C} 1 = { I, 1', - -, - - = - {a, b, c, ay}. a a a

Since a, b, and c are relatively prime, every rational integer is a linear combination of a, b, c with integer coefficients and hence MM'

1

= a

{I, ay}

1

= a

.0 = N(M)D

(by the corollary of Lemma I). If M is now an arbitrary module, it can be represented in the form M = etMl' where M l has the form {l, y}. By Theorem 2 of Section 6 we have MM' = etet'M1M l' = N(et)N(Ml)D

= IN(et)IN(Ml)D = N(M)D, and formula (7.8) is proved in general.

Sec.7]

REPRESENTATION OF NUMBERS BY QUADRATIC FORMS

139

Now let M and M I be two modules belonging to the same order D. If .0 is the coefficient ring of the product MMI , then by formula (7.8), MMI(MMI )'

= N(MMdD.

On the other hand, since multiplication of modules is clearly commutative and associative, by use of MM' = N(M)D and MIMI' = N(MI)D, we obtain MM'

= N(M)D

and

MM\(MM I )'

MIM{

= N(MI)D,

= N(M)N(MI)D.

Comparing this equation with the previous one and recalling that two distinct orders cannot be similar, we obtain that .0 =.0. Incidentally, since the equality aD = bD for positive rational a and b is possible only when a = b, we obtain Thus if the modules M and M I belong to the order .0, then their product MMI also belongs to D. Since for any module M with coefficient ring .0 we have both MD = M and M[(ljN(M))M'] =.0, then we obtain the following result. Theorem 2. The set of all modules of a quadratic field which belong to a fixed order becomes an Abelian group under the operation of multiplication of modules.

From this theorem and Theorem 3 of Section 6, we easily obtain Theorem 3. The set of classes of similar modules in a quadratic field with given coefficient ring forms a finite Abelian group. Note that Theorems 2 and 3 hold only for quadratic fields and cease to be true for modules which belong to nonmaximal orders in arbitrary algebraic number fields (see Problem 18 of Section 2).

7.5. The Correspondence between Modules and Forms

As shown in Section 1.3, each basis rt, {3 of the full module M c R( ~d) corresponds uniquely to the binary quadratic form N(rtx + {3y) with rational coefficients. Since for different bases of M the corresponding forms are equivalent, the module M corresponds to a class of equivalent forms. If we replace M by the similar module yM, then each corresponding form is multiplied by the constant factor N(y). Hence, considering forms only up to a constant multiple, we may say that any class of similar modules corresponds to a class of equivalent forms. But this correspondence is not one-to-one.

·140

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

Indeed, conjugate modules M and M' are, in general, not similar, but their corresponding forms coincide. An analogous phenomenon clearly also holds for decomposable forms of any degree. In general, there is no natural way to rectify this lack of correspondence between modules and forms. But for quadratic fields we shall see that it is possible to establish? one-to-one correspondence by slightly changing the definitions of equivalence of forms and similarity of modules.

+ Bxy + Cy 2 with rational integer coefficients is called primitive if the greatest common divisor of the coefficients is I. The integer B 2 - 4AC is called the discriminant of the primitive formr Definition. The binary quadratic form f(x, y) = Ax 2

The discriminant of a primitive form hence differs from its determinant AC - (B 2 j4) by a factor of -4. It is easily seen that any form equivalent to a primitive form is also primitive. Under a linear change of variables with matrix C the determinant of a quadratic form is multiplied by (det C)2, and hence does not change if det C = ± 1. Hence equivalent primitive forms have the same discriminant. Definition. Two primitive forms are called properly equivalent if one can be obtained from the other by a linear change of variables with determinant + 1.

The collection of all primitive binary quadratic forms is broken up into classes of properly equivalent forms. For the rest of this section, when we speak of equivalent forms, we shall always mean properly equivalent forms. It will frequently happen that two forms which are nonproperly equivalent (that is, carried into each other by linear substitutions with determinant -I) will also be properly equivalent. We now give a new definition for similarity of modules. Definition. Two modules M and M 1 in a quadratic field are called strictly similar if M 1 = rtM for some rt with positive norm.

Since in imaginary quadratic fields the norm of any nonzero rt is positive, in such fields the concept of strict similarity does not differ from the usual concept. The situation will be the same in real quadratic fields when the coefficient ring .0 of the module M contains a unit e with N(e) = - I. Indeed, if M 1 = rtM and N(rt) < 0, then, since eM = M, we have M 1 = (rte)M, with N(rte) > O. Conversely, suppose that the two concepts of similarity coincide. Then if M 1 = rtM, N(rt) < 0, there exists a number 13 for which N(f3) > 0 and M 1 = 13M. Setting e = rtf3-1, we have eM = M, and this means that e is a unit in the coefficient ring .0 with N(e) = -1.

Sec. 7]

REPRESENTATION OF NUMBERS BY QUADRATIC FORMS

141

Hence the concept of strict similarity differs from the usual concept of similarity precisely for those modules in a real quadratic field whose coefficient rings contain only units with norm + 1. It is clear that in this case every class of modules, similar in the usual sense, breaks up into two classes of strictly similar modules. We now describe a correspondence between classes of modules and classes of forms. In each module M of the field R( JJ) we shall only consider those bases IX, 13 for which the determinant 11

=

IX IX'

I

PI 13'

(7.9)

satisfies 11>0

for d > 0,

-11 > 0

for d < O.

(7.10)

1 j

As previously, IX' and 13' here denote the conjugates of IX and 13 in R( J(i). [A basis in M which satisfies (7.10) can always be found; if any basis lXI' IX z does not work, interchange IX I and IX z .] We set each basis lX,f3 of the module M which satisfies (7.10) in correspondence with the form f(x, y)

=

Ax Z + Bxy N(lXx

+ f3y)

+ cyz (IXX

+ f3y)(IX'x + f3'y)

N(M)

N(M)

(7.11)

[N(M) is the norm of the module M]. If for the number y = - f3llX we consider ({)it) = at Z + bt + c (see Section 4), then we shall clearly have N(lXx

N(IX)

+ f3y) = -

a

(ax z + bxy

+ cyz).

On the other hand, by the Corollary of Lemma 1 and by Theorem 2 of Section 6 the module M = 1X{1, y} has norm IN(IX)lla. Hence the coefficients A, B, C differ from a, b, c at most in sign. The form (7.11) is primitive and its discriminant B Z - 4AC coincides with the discriminant b Z - 4ac of the coefficient ring of the module M (Lemma 1). Thus we have the mapping {IX, f3} ~ f(x, Y),

(7.12)

Jd)

which associates to each basis IX, 13 of the field R( which satisfies condition (7.10) the primitive formf(x, y) (if the field is real, the coefficient A may

142

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

be negative). It is clear that if the field is imaginary quadratic, then the form (7.11) will always be positive-definite, so negative-definite forms are not included in the correspondence (7.12). Theorem 4. Let rol be the set of all classes of strictly similar modules similar in the narrow sense of the quadratic field R( Jd). When d > 0, let lY be the set of all classes of properly equivalent binary quadratic forms which split into linear factors in R( Jd). When d < 0, we consider only positivedefinite forms. Then the mapping (7.12) establishes a one-to-one correspondence between rol and lY. If some .class of modules has a coefficient ring with discriminant D, then the corresponding forms also have discriminant D.

Let IX, 13 and 1X 1, 131 be two bases of the field R( /J) for which the determinant (7.9) satisfies (7.10) and let these bases correspond to the formsfand fl' To prove Theorem 4 we must show that the forms f and fl are properly equivalent if and only if the modules {IX, f3} and {1X 1, f3d are strictly similar. Further, we must show that for any irreducible primitive form g(x, y) [which splits up into linear factors in R( /d), and is positive-definite in the case d < 0] there is a basis IY., 13 satisfying (7.10) for which the form (7.11) coincides with g(x, y). We leave the simple details of this verification to the reader. In Section 7.4 we defined the product of two classes of similar modules. In precisely the same way one can define the product of two classes of strictly similar modules. Since the mapping 9Jl -+ lY is one-to-one the multiplication of classes of modules induces a multiplication on the set of classes of forms. The operation of multiplication in lY is called composition of classes offorms (a term introduced by Gauss, who first studied this operation). Since the set of all classes of modules with some fixed coefficient ring is a group, the set of classes of primitive forms with fixed discriminant D (only positive-definite forms for D < 0) also forms a group.

7.6. The Representation ofNumbers by Binary Forms and Similarity ofModules In this section we show that the problem of finding representations of integers by binary quadratic forms can be reduced to the problem of similarity of modules in a quadratic field. Let f(x, y) be a primitive binary quadratic form with discriminant D l' 0, which splits into linear factors in the field R(Jd) and let m be a natural number. In the case D < we further assume thatfis positive-definite. Our problem is to find all integral solutions of the equation

°

f(x,y) = m. (We only consider positive values for m, since

(7.13) In

the case m < 0, D> 0.

Sec. 7]

REPRESENTATION OF NUMBERS BY QUADRATIC FORMS

143

we can replace f by the form -f) By Theorem 4 we can represent f in the form N(ax + fJy) (7.14) f(x, y) = N(M) , where the basis a, fJ of the module M satisfies (7.10). The mapping (x, Y) -+ ~ = ax + fJy establishes a one-to-one correspondence between solutions to (7.13) and numbers ~ E M with norm N(~) = mN(M). Two solutions of (13) are called associate if the corresponding numbers of M are associate. It is easily verified that the concept of associate solutions does not depend on the choice of the representation (7.14). We denote the coefficient ring of the module M by .0 and denote the class of strictly similar modules which contains M by C. By Theorem 4 C is uniquely determined by f Assume that we have a number ~ E M with norm mN(M). Consider the module A = ~M-I. Since AM = ~M-I M = ~.o c M, the module A is contained in .o. Its norm is N(~)N(M)-I = m. Also it is clear that A is contained in the class C -I, the inverse of the class of M. Conversely, assume that in the class C- I there is a module A which is contained in the ring .0 and has norm m. Then for some ~ with positive norm we have A = ~M-I, so that ~ E MA c: M and N(O = m. If Al is any other module of the class C- I which is contained in .0 and has norm m, and if Al = ~IM-I with N(~I) > 0, then Al = ~ICIA. Hence A coincides with Al if and only if ~ I is associate with ~. We have thus proved the following theorem. Theorem 5. Let the form f(x, y) correspond to the class C of strictly similar modules with coefficient ring .o. The set of classes of associate solutions of (7.13) is in one-to-one correspondence with the set of modules A which are in the class C- I , are contained in the coefficient ring .0, and have norm m. The solutions (x, y) which correspond to the module A are given by the numbers ~ for which A = ~M-I, N(~) > 0, where M is a module of the class C.

For any natural number m we can easily find the set of all modules A with coefficient ring .0 which are contained in .0 and have norm m. Let A be such a module, and let k be the smallest natural number contained in A. The module A can then be written in the form

A

=

{k, ky}

=

k{l, y}.

The number y is determined except for sign and addition or subtraction of integers. We may therefore choose y so that Imy > 0

for d < 0,

Irr y > 0

for d > 0

(7.15)

NUMBERS BY DECOMPOSABLE FORMS

144

[Chap. 2

(irr y denotes the irrational part of the number y), and also so that the rational part of y is contained in the interval (-1. H In the notation of Lemma 1 we may write y=

-b+JD 2a

(7.16)

'

where -a,,;;b 2y 2 > x 2+ y2

=

IYI2,

which proves our assertion. Now let y and YI be two reduced numbers. If the modules {I, y} and {I, yd are similar, then by Theorem 8 the bases {I, y}

Sec.7]

REPRESENTATION OF NUMBERS BY QUADRATIC FORMS

149

and {l, yd are similar. But this is possible if and only if Y = Yl' Theorem 9 is completely proved. To solve the question of the similarity of modules in an imaginary quadratic field we must have an algorithm for finding the reduced module similar to a given module. Such an algorithm is formulated in Problem 24. Two given modules M 1 and M 2 are similar if and only if their reduced modules coincide. Remark. In the proof of Theorem 9 we never actually used the fact that the module under consideration was contained in some imaginary quadratic field. The assertion of the theorem hence is true for any lattice in the plane: any lattice in the complex plane is similar to one and only one lattice of the form {l, y}, where Y is some number of the domain r, which is described in Figure 6. By Lemma 2, which is applicable to arbitrary lattices in the plane without any changes, two lattices {l, y} and {I, A} are similar if and only if A and yare connected by

A = ky m"

+I + 11'

kll-ml=±I,

with rational integers k, I, m, n. Two such nonreal complex numbers are called modularlyequimlent. Hence we have shown that every nonreal complex number is modularly equivalent to one and only one number of the region r. The region r itself is called the modular domain. Its points are in one-to-one correspondence with the set of classes of similar lattices in the plane. The question of similarity of planar lattices is connected with many important questions in the theory of elliptic functions. A field of elliptic functions is given by its period lattice, and two such fields are isomorphic if and only if their corresponding period lattices are similar (see, for example, C. Chevalley, "Introduction to the Theory of Algebraic Functions of One Variable," 1951). Hence the points of the modular domain r are in one-to-one correspondence with isomorphism classes of fields of elliptic functions. Consider now the classes of similar modules which belong to some fixed order D with discriminant D < 0. Let the module {l, y}, y E r, belong to the order D. If we use the notations of Lemma 1 and write y in the form

y

=

JIDI

- b+ i

2a

'

then conditions (7.23) and (7.24) yield - a

~

b < a, ~

c?a

for

b

c>a

for

b > 0.

0,

(7.25)

150

NUMBERS BY DECOMPOSABLE FORMS

[Chap. 2

Hence to find a full system of reduced modules of an imaginary quadratic field which belong to the order with discriminant D, we need only find all triples of integers a > 0, b, c which satisfy (7.25) and also the condition

D = b2

4ac,

-

(a, b, c) = 1.

(7.26)

By Theorem 3 of Section 6 the number of such triples is finite, a fact that can be directly verified from the inequalities

jDI = 4ac

- b 2 ~ 4a 2

Ibl~a
1 not be a perfect square. Show that in the continued fraction expansion of the sequence of entries has the form

vim

(hereql+1 =q,_I, i= 0, .__ , s - I). 5. Under the same assumptions, show that if P,I Q, is the convergent (s again denoting the period), then P, -+- Q, is a fundamental unit of the order {I, V;;;) (in the field

vm

R(v-;;;».

6. Let the modules M 1 and M 2 of a quadratic field have for coefficient rings the orders e'l and 1:'2 (using the notation of Section 7.2). Show that the product M I M 2 belongs to the order C" where I is the greatest common divisor of II and 12. 7. For any natural number f let \l, denote the group of modules in a given quadratic field which belong to the order C, (see Section 7.4). Show that if d is a divisor of f, then the mapping M -+ M Cd (M E 1\,) is a homomorphism from 11, to the group 11 d •

e

8. Let be a number of the maximal order I: = (I, w) of a quadratic field which is relatively prime to f. Show that the coefficient ring of the module M = {f,fw, 0 is 1:, and that M I: = C. Further, show the converse, that is, that any moduli: M which belongs to the and satisfies the property = 6 is of the form M = if, Iw, for some E D order which is relatively prime to f.

C,

Me

e)

e

el and e2 be two numbers of ~ which are relatively prime to f Show that el) ~ if. Iw, e2) if and only if Sel e2 (mod!) for some rational integer s.

9. Let {f, fw,

==c

10. Show that if M I and M 2 are any two full modules of a quadratic field (not necessarily belonging to the same order), then

11. Let h denote the number of classes of similar modules belonging to the maximal

Sec. 7]

REPRESENTATION OF NUMBERS BY QUADRATIC FORMS

153

order 0 of a quadratic field, and let h, denote the number of classes of similar modules belonging to the order 0, (of index f). Show that

where rtJ(f) is the number of residue classes in 0 modulo f which consist of numbers relatively prime to f(rtJ is analogous to the Euler ep-function), and is ~he index of the group of units of the order 0, in the group of units of the maximal order D. 12. A number I' of a real quadratic field is called reduced if it satisfies 0 < I' < 1 and its conjugate satisfies 1" < -I. If I' is reduced the module {I, y} is also called reduced. Using the notation of Lemma 1, show that the number I' is reduced if and only if

e,

-b+ YD < 2a < b+YD.

O 2. It is clear that if Fermat's theorem is proved for some exponent n, then it is also automatically proved for all exponents which are multiples of n. Since any integer n > 2 is either divisible by 4 or by some odd prime, we may limit our consideration to the cases where n = 4 or n is an odd prime. For n = 4, an elementary proof was given by Euler. We thus consider only

x'

+ y' = Zl,

(1.1)

where the exponent 1 is an odd prime number. We may clearly assume that the numbers x, y, z in (1.1) are relatively prime. For those values of 1for which a proof of Fermat's theorem has been found, the proof is usually divided into two parts: first, it is shown that (1.1) has no solution in integers x, y, z, which are not divisible by I, and second, that (1.1) has no solution in integers x, y, z precisely one of which is divisible by I. These two are called the first and second cases of Fermat's theorem. From the extant proofs of various cases of Fermat's theorem we can deduce that the principal difficulties in the first and second cases of Fermat's theorem are roughly the same, although the techniques used in the first case are more simple. Here we consider only the first case of Fermat's theorem. The connection between Fermat's theorem and the problem of the decomposition of algebraic numbers into prime factors is given in the following simple observation. If' denotes a primitive lth root of 1, then (1.1) may be written 1-1

TI (x + 'k y ) = Zl.

(1.2)

k=O

If a product of pairwise relatively prime rational integers is an lth power, then each of the factors is an lth power. The factors on the left side of (1.2) belong to the algebraic field R(O of degree 1 - lover R. (It is easily seen that the polynomial t ' - 1 + t ' - 2 + ... + t + 1, 1 a prime, is irreducible over the field of rational numbers; see Problem 6 or Theorem 1 of Section 2, Chapter 5.) Consider in the field R(,) the order .0 = {I, " ... , ,'- 2} [by Theorem 1 of Section 5, Chapter 5, .0 is the maximal order of the field R(Ol. Assume that in the ring.o factorization into primes is unique. Then for any IX ED, IX # 0, we have a factorization

Sec. 1 ]

SOME SPECIAL CASES OF FERMAT'S THEOREM

157

where e is a unit of the ring .0, the prime numbers 7tl' '" , 7t r are pairwisenonassociate, and the exponents aI' ... , a r are uniquely determined (see Section 2.2). Then every prime 7t which occurs in the factorization of Zl occurs with an exponent divisible by I. But we shall show below that when we are dealing with the first case of Fermat's theorem, the numbers x + y (k = 0, 1, ... , I - 1) are pairwise relatively prime. Hence if we represent x + y as a product of prime factors, each prime will occur with an exponent divisible by I. This means that, up to a unit factor, x + y is an lth power and, in particular, (1.3) x + = ea l ,

,k

'k

'k

'y

where e is a unit of the ring .0 and ex E .0. Since I is odd, we may also write (1.1) in the form Xl

+ (_Z)l = (_y)l,

and analogously we obtain (1.3')

Equations (1.3) and (1.3') lead, in a fairly easy manner, to a contradiction. When this is done, we shall prove that (1.1) has no solution in integers x,)I, z not divisible by I (under the hypothesis made on the ring .0). After this introduction we now establish some auxiliary facts concerning the ring .0.

1.2. The Ring Z[,]. Lemma 1. In the ring .0

= Z[,] the number I - , is prime, and I has the

factorization 1= e*(1 -

0 /- 1

(1.4)

where e* is a unit in .0. Proof. In the decomposition tl-

1

+ t l - 2 + ... + t + I

= (t - O(t - ,2) ... (t _

,1-1),

set t equal to I. Then (1.5)

If ex = r(O is any number of the field R(O [here r(t) is a polynomial with rational coefficients], then the numbers (1

~

k

~

I - I)

(1.6)

are the images of ex under the isomorphisms of the field R(O into the field of all complex numbers. In the terminology of Section 2.3 of the Supplement,

I 158

I

[Chap. 3

the numbers (1.6) are the conjugates of a, and thus N(a) = particular, for s ¥= 0 (mod I), we have I-I

N(1 - 'S)

=

TI (1 -

k=1

TIL: ~ r(,k).

In

I-I

,kS)

=

TI (1 -

,k) = I.

k=1

,2, ... ,

From this it follows that 1 - " 1 1 - , I - I are primes in the ring .0. Indeed, if 1 = afJ, then N(a)N(fJ) = I, and then either N(a) = 1 or N(fJ) = 1; that is, one of the factors is a unit (Theorem 4 of Section 2, Chapter 2). Taking norms in the equation

'S

'S,

we obtain N(E s ) = 1, and thus Es is a unit in .0. Hence the numbers 1 _ for s ¥= 0 (mod I), are associate with 1 - ,. The decomposition (1.4) now follows from (1.5) and (1.7). Lemma 2. If the rational integer a is divisible by 1 - , (in the ring .0), then it is also divisible by I. Proof. Let a = (1 - ')a, where a ED. Taking the norm of both sides, we obtain d- I = IN(a), where N(a) is a rational integer. Since 1 is prime, then a is divisible by I.

,I ,

i

Lemma 3. The only roots of 1 contained in the field R(,) are those whose degree divides 2/.

i

j

I

~

I

II ~ :

,

1

I

III.

I

THEORY OF DIVISIBILITY

!

:

r/I' I:

i~ j

I

Proof. Any root of 1 in R(,) clearly lies in the maximal order. By Theorem 2 of Section 3, Chapter 2, the set of all roots of I in R(O forms a finite cyclic group. Let m denote the order of this group and let 1'/ be any primitive mth root of 1. Since - , belongs to R(O and is a root of degree 21 of 1, m is divisible by 2/. In Section 2 of Chapter 5 (corollary of Theorem 1), it is proved that the degree of the field R(I'/) over R equals cp(m), where cp(m) is Euler's function. Set m = l'mo, (mo, I) = 1 (r ~ 1, mo ~ 2).

Since R(I'/) is contained in R(O, and the latter field has degree 1 - 1, then

From this inequality it follows that r = 1 and cp(m o) = 1. Since cp(mo) mo ~ 2 only when mo = 2, m = 2/, and Lemma 3 is proved.

= 1 for

Lemma 4 (Kummer's Lemma). Any unit of the ring.o is a product ofa power of' with a real unit.

Sec. 1 ]

SOME SPECIAL CASES OF FERMAT'S THEOREM

159

Proof. Let

e=a O+a l '+···+a l _ 2,1-2=rm

(aiEZ)

be any unit of .0. It is clear that the complex conjugate e = r(C l) = r(,I-I) is also a unit of .0. Consider the unit 11 = eje E .0. By (1.6) any conjugate of 11 has the form

Since r(,k) and r(C k ) are complex conjugate, then Iak (ll) I = 1 (k = 1, ... ,1- 1). By Theorem 2 of Section 3, Chapter 2, 11 is a root of 1, and then by Lemma 3, 11

= ±,o.

We shall now show that the plus sign always occurs on the right. For otherwise we would have ·Consider this equation as a congruence in the ring .0 modulo A = I - ,. Since' == 1 (mod A), all powers of' are congruent to 1 modulo ).. and we have e ==

e == a o + al + ...

+a l - 2 = M (mod A),

which means that M == -M (mod A), or 2M == 0 (mod A). By Lemma 2 2M == 0 (mod I),

M== 0 (mod I),

M == 0 (mod A),

so that e == 0 (mod A),

which contradicts the fact that e is a unit of the ring .0. Thus e=

'''e.

We now take an integer s so that 2s == a (mod I). Then e = e can be written in the form

,2s

'0 = ,25 and the equation

This shows that the unit '1 = ens is real. Hence we have represented e as the product of and the real unit '1, and the lemma is proved.

'5

Lemma 5. Let x, y, m, n be rational integers, m "I- n (mod I). Then x + ,my and x + y are relatively prime if and only if x and yare relatively prime and x + y is not divisible by I.

,n

160

THEORY OF DIVISIBILITY

[Chap 3

Proof. If x and y have a common divisor d> I, then x + ,my and x are both divisible by d. If x + y is divisible by I, then x + ,my and x have a common divisor I - , (which is not a unit). Indeed,

x

+

,my =

X

+Y +

+ ,ny + ,ny

(,m - I)y

= (x + y) - (l - Oemy == 0 (mod I - O. Thus the necessity of both conditions is proved. To prove their sufficiency we shall show that there exist numbers ¢o and '10 in .0 such that

Consider the set A of all numbers of the form

. "

,

:

where ¢ and '1 independently run through all numbers of .0. It is clear that if a and fJ belong to A, then any linear combination a¢' + fJ'1' with coefficients ¢', '1' ED also belongs to A. We need to show that I belongs to A. From

!

(x (x

I

+ ,my) _

+ ,myKn -

(x

(x

+ ,ny ) = ,m(l

+ ,nyKm =

- C-m)y = ,men_m(l - Oy,

_,m(l - ,n-m)x

= _,men_m(l - ')x,

we conclude that (l - Oy and (I - Ox belong to A (since ,men_m is a unit in the ring .0). Since x and yare relatively prime, there exist rational integers a and b such that ax + by = I, and therefore

j' I

I, i!

Further,

'- !

l'

IfI: .

!

I

i

I:I, I,I

I I I i!

I I,

I

and thus x + YEA. Since I is divisible by I - " then lEA. But we are also assuming that x + y and I are relatively prime. Hence for some rational integers u and v we have (x + y)u + Iv = I, so that lEA. Lemma 5 is proved. 1.3. Fermat's Theorem in the Case of Unique Factorization Theorem 1. Let I be a prime integer and let' be a primitive Ith root of I. If decomposition into prime factors is unique in the order .0 = Z[,] = {I, " ... , ,1- 2} of the field R(O, then the equation

has no solution in integers x, y, z not divisible by I.

,I

Sec. 1 ]

SOME SPECIAL CASES OF FERMAT'S THEOREM

161

Proof. The prime 3 will play a special role in our proof, so we consider the case f = 3 separately. We shall show that not only the equation x 3 + y3 = Z3, but also the congruence

has no solution in integers not divisible by 3. For assume that there is such a solution of this congruence. But from the congruence x 3 + y3 == Z3 (mod 3) it easily follows (from the little Fermat theorem) that x + y == z (mod 3); that is, Z = x + Y + 3u, and hence x

3

+ y3 == (x + Y + 3U)3 ==

x 3 + y3

+ 3x 2Y + 3xy2 (mod 9),

and 0== x 2y

+ xy2 =

xy(x

+ y) ==

xyz (mod 3).

Thus one of the numbers x, y, z is divisible by 3, and our assertion is proved. Now let f ~ 5. We prove the theorem by contradiction, assuming that for some rational integers, x, y, z, pairwise relatively prime and not divisible by f, we have x' + l' = Zl, which we also write in the form (1.2). Since x + y == x' + l' = Zl "I- (mod f), and x and yare relatively prime, then by Lemma 5, all the numbers x + (ky (k = 0, 1, ... ,f- 1) are pairwise relatively prime. Then, as has already been shown in Section 1.1, from the unique factorization of the numbers appearing in (1.2) it follows that

°

x

+ (y = eal,

(1.3)

x - (z = ela l ' ,

(1.3')

where e and e l are units in the ring .0. We have already remarked that (1.3) and (1.3') lead to a contradiction. We now show that this contradiction even arises from the corresponding congruences modulo f in the ring .0. Let a = a o + al ( + ... + a ,_ 2(1-2 with ao , ... , a l - 2 rational integers. Then al == ao l + all(l

+ ... + al_ 2(1(l-2) == M

(mod I),

where M = ao + al + ... + a l -2' By Kummer's lemma the unit e can be represented in the form e = (5'1, where '1 is a real unit. Hence from (1.3) we obtain the congruence

with the real number ¢ ED. We may also write this congruence in the form

C

5

(x

+ (y) == ¢ (mod l).

(1.8)

We now note that for any a ED the complex conjugate fi. also belongs to .0. If we have the congruence a == f3 (mod f), then a - f3 = fy, so that fi. - /3 = fy

THEORY OF DIVISIBILITY

162

[Chap. 3

and hence Ci. == /3 (mod I). Passing now from the congruence (1.8) to its complex conjugate, we obtain

'S(x

i

.1

:I

But ~

= ¢,

+ ely) == ~ (mod I).

(1.9)

and therefore from (1.8) and (1.9) it follows that

eS(x

+ 'y) == 'S(x+

ely) (mod I),

or (1.10) It is clear that a number of .0, represented in the canonical form ao + al ' + .. , + a l _2,'-2, is divisible by I if and only if all coefficients ao , ... , a, -2 are divisible by I. If the exponents

s - 1,

s,

) I.

I'

I ~

ii

-s,

I-s

(1.11)

are pairwise-noncongruent modulo I and also noncongruent to 1- 1, then the number on the left side of the congruence (1.10) is in canonical form and hence all its coefficients are divisible by I. Thus in this case x == 0 (mod I) and y == 0 (mod I), which is impossible, since x and yare relatively prime (and also not divisible by I). Consider the case when the left side of (1.10) is not in canonical form, that is, when one of the integers (1.11) is congruent to 1- 1 modulo I, or two of them are congruent modulo I. One of the exponents (1.11) will be congruent to I - 1 modulo I only in the following cases:

j

.! :

J

f!

Ii

;

~"

s

I s-1

-s

1- s

---- -- - ---

2

Ii

/- 1

/- 2

1

0

/ -1

0

1

I

1

0

/- 1

0

2

1

/-2

/- 1

, I:

i'I

I

J

I

!i

We see that in each of these cases only one of the exponents is congruent to 1- 1 (since I ~ 5). To write the left side of (1.10) in canonical form, we must use the equation

I'

i

!i

If

Substituting this expression in the left side of (1.10), we replace the term with exponent I - 1 by a sum of the monomials 1, " ... , ,'- 2 each with coefficient ± x or ±y. Since the number of these terms is equal to I - 1 ~ 4 (since I ~ 5), after we combine terms in which the exponent of' is the same, there will be at least one term in which the coefficient is ±x or ±y. But this again would

Sec.1]

SOME SPECIAL CASES OF FERMAT'S THEOREM

163

imply that x == 0 (mod I) or y == 0 (mod I), which is impossible', since we have assumed that x and yare not divisible by I. We now need only consider the case when two of the exponents in (1.11) are congruent modulo I. The congruences s == s - 1 (mod I) and -s == 1 - s (mod I) are clearly impossible. If s == -s (mod I) or s - 1 == 1 - s (mod I), then we have s == 0 (mod I) or s == 1. (mod I), and we have again the cases considered above where s - 1 == 1- 1 (mod I) or -s == 1- 1 (mod I). In the remaining (equivalent) possibilities s == 1 - s (mod I) and s - 1 == -s (mod I), we have s == (l + 1)/2 (mod I). In this case the congruence (1.10) takes the form (x - y)(l+ 1)/2

+ (y

_

X)((I-l)/2

== 0 (mod I).

Since the left side of this congruence is in the canonical form [the exponents (l + 1)/2 and (l - 1)/2 are neither congruent to each other nor to I - 1], it follows that x == y (mod I). Analogously, we deduce from (1.3') that

x == -z (mod I). Then from the congruences x + y == x' + y' = Zl == z (mod I) it follows that 2x == -x (mod I) or 3x == 0 (mod I). Since 1# 3, x == 0 (mod I) and we again have a contradiction. This completes the proof of Theorem 1. By using more subtle arguments involving the integers of the field R«(), Kummer showed that if the prime I satisfies the conditions of Theorem 1, then the second case of Fermat's theorem also holds for the prime I. We shall generalize Theorem 1 to a wider class of exponents in Section 7.3. For this wider class of exponents we shall prove the second case of Fermat's theorem in Section 7.1 of Chapter 5. We make some remarks about Theorem 1. Remark 1 . The main part of the proof of the theorem is the verification of the impossibility of certain congruences modulo I. Of course it does not follow from this that the congruence x' + y' == Zl (mod I) is impossible, since this congruence is equivalent to x + Y == z (mod I), which always has solutions in integers not divisible by I. Moreover, it can be shown that, for example, when 1= 7, the equation x' + y' = Zl, when considered as a congruence, has, for any modulus, solutions not divisible by 7. Thus the proof of the unsolvability of (1.1) is achieved first by using unique factorization in the ring Z[(] to obtain Equations (1.3) and (1.3'), and then by applying the theory of congruences to these latter equations.

164

THEORY OF DIVISIBILITY

[Chap. 3

Remark 2. It is clear that the methods which we have applied in this section to the solution of Fermat's theorem can also be applied to analogous problems, by using other algebraic number fields instead of the field R«() (Problem 2). Remark 3. If we wish to apply the theorem to some particular prime I, we discover that this cannot be done, since we have no means for determining whether factorization into primes is unique for the field R(O.

Hence we come to the following two basic problems of number theory: (1) In which algebraic number fields K is decomposition into prime factors unique? (2) What are the arithmetic properties of those fields K in which decomposition into prime factors is not unique?

c

It

r

r

z e

PROBLEMS

1. Show that the congruence x 3 + y3 == Z3 (mod 52) has no solution in rational integers x, y, z not divisible by 5. Let w be a primitive cube root of I. Assume it known that decomposition into prime factors is unique in the field R(w). Show that the equation x 3 + y3 = 5z 3 has no solution in rational integers x, y, z not divisible by 3. 3. Let I be a prime number, ~ a primitive Ith root of I, x and y rational integers, and d the greatest common divisor of x and y. If x + y ~ 0 (mod I) set I) = d, and if x + y == 0 (mod I) set I) = d(l - ~). Show that I) is a common divisor of the numbers x + ~my and x + ~.y which is divisible by all other common divisors of these numbers. 4. Show that in the order {l,~, ... , ~1-2 of the field R(~) a product 1Xf3 is divisible by I - ~ if and only if IX or f3 is divisible by I - ~. 5. Using the concept of congruence of integral polynomials (Section I.l of Chapter I), show that 1'-1+ ... +1+ I ==(1_1)1-1 (modI).

6. Show that the polynomial 11- 1 + ... + 1 + I is irreducible over the field of rational numbers by considering congruence of integral polynomials modulo 12.

2. Decomposition into Factors

2.1. Prime Factors In Section 1 we saw how a problem of number theory can reduce to a question of decomposition into prime factors in some order of an algebraic number field. We shall see other such examples later. We now consider the general problem of decomposition into prime factors. In order to speak of decomposition into primes, we must be dealing with a

I

v

e f tl v

n a

E r

o 2

p

fI

Sec. 2]

DECOMPOSITION INTO FACTORS

165

fixed ring .0, the elements of which we are decomposing into factors. We formulate our problem in the general case where .0 is any commutative ring without divisors of zero and possessing a unit element. In the future these conditions will be assumed without special mention. Definition. An element n: of the ring.o, nonzero and not a unit, is called prime if it cannot be decomposed into factors n: = exfJ, neither of which is a unit in .0.

Thus an element is prime if it is divisible only by units and associates. In some rings there are no prime elements and hence not every element of a ring can be represented as a product of primes. For example, let .0 be the ring of all algebraic integers. Any ex # 0, which is not a unit, has the factorization ex = -J~ -J~, in which both factors lie in .0 and are nonunits. Thus every nonunit of.o has nontrivial factorizations and there are no prime elements in .0. For examples of rings in which decomposition into prime factors is always possible, consider orders in algebraic number fields (it is these rings which will interest us most). We shall call prime elements in orders prime numbers. Theorem 1. In any order .0 of an algebraic number field K, every nonzero element which is not a unit can be represented as a product of prime numbers. Proof. By Theorem 4 of Section 2 of Chapter 2, the units of .0 are characterized by having norm ± 1. We prove the theorem by induction on the absolute value IN(ex) I of the number ex ED. If the number ex is itself prime, there is nothing to prove. Otherwise ex = fJy, where fJ and yare numbers of.o which are not units, so that

I < IN(fJ) I < IN(ex)I,

1 < IN(y)1

< IN(ex)l.

By the induction assumption, fJ and yare products of prime numbers of the ring.o. But then since ex = fJy, the number ex is also a product of prime numbers of the ring.o. Hence Theorem 1 is proved.

2.2. Uniqueness of Factorization We assume now that in the ring .0 decomposition into prime factors is possible, and we turn to the question of the uniqueness of such factorizations. Definition. We shall say that decomposition into prime factors in the ring .0 is unique if for any two decompositions

I I

166

THEORY OF DIVISIBILITY

[Chap. 3

the number of factors is always the same (r = s) and for suitable indexing of the factors, the prime elements 7t; and 7t;' are associate (i = I, ... , r). In the decomposition a = 7t 1 ••• 7t" associate prime elements can be made equal by multiplying by a suitable unit. We may then group equal factors into powers and obtain a factorization a

= 1>7t/'

... 7t m

km

,

in which the prime elements 7t 1 , ... , 7t m are pairwise-nonassociate and I> is a unit of the ring D. In case factorization is unique, the prime elements 7t 1 , ••• , 7t m are determined up to associates and the exponents k l , ... , k m are uniquely determined. The classical example of a ring with unique factorization is the ring of rational integers. It is far from true that decomposition into prime factors is unique in all rings. Thus the result of Problem I shows that among orders in algebraic number fields, unique factorization can occur only for maximal orders. Unique factorization for the ring Z of rational integers follows from the theorem on division with remainder, which asserts that for any a and b # 0 of Z there exist integers q and r, such that a = bq + rand r < Ibl. If in a ring.c there is an analog of division with remainder, then we can prove uniqueness of factorization in .0 just as in Z. 1 I'

Definition. We say that the ring .0 has the division with remainder property if there is a function Iiall on nonzero elements a E .0 which takes nonnegative integral values and is such that the following conditions hold:

I I

I

iI

(I) If IX # 0 is divisible by fl, then lIall ~ IIflll· (2) For any elements IX and fl # 0 of .0, there exist I' and p such that

l

! I

a = fly + p, where either p = 0, or Euclidean.

I

\, !

I:

Consider the proof of unique factorization in the ring of rational integers.

I ,

Ilpll < IIflll. The ring .0 itself is then called

j

It uses, in addition to the general properties of rings, only the theorem on

division with remainder. Therefore, by translating this proof, we obtain the following result. Theorem 2. In every Euclidean ring, factorization into primes is

unique. Con sider as an example the maximal order .0 of the quadratic field R( -J~). We shall show that .0 has the division with remainder property with lIall = N(IX). Let a and fl # 0 be arbitrary numbers of .0. Then

~=u+vFI,

Sec. 2]

DECOMPOSITION INTO FACTORS

167

where u and v are rational numbers and we choose rational integers x ann y so that

lu - xl If we now set y

1-.

~

Iv - yl

~!-.

= x + y -J-::}, p = a - fJ y, then from N

a)y = (u (P-

x)

2

+ (v

2

- y) ~

4:1 + 4:1
11{311. 3. Let \JR be a lattice in the complex plane, the points of which represent the numbers of the maximal order 0 of an imaginary quadratic field. Show that we obtain an algorithm for division with remainder in 0 by using the norm N(a) if and only if the translates of the unit disc (without boundary) by all vectors of the lattice \JR completely cover the plane. 4. Show that in the maximal order of the imaginary quadratic field R( Vd), an algorithm for division with remainder is obtained by using the norm if and only if d is one of the values -I, -2, -3, -7, -II. 5. Letd < 0 be square-free and not equal to -I, -2, -3, -7, - I I. Show that the norm of any integer of R( v'd), except 0 and ± I, is greater than 3. 6. Show that, except for the five fields indicated in Problem 4, the maximal order of an imaginary quadratic number field is never a Euclidean ring. (Hint: Carry out the proof by contradiction. Assume that there is a function Iiall on the elements of the maximal order 0 which satisfies the conditions given in Section 2.2. Among the numbers of 0 which are not units, choose y so that Ily [1 is as small as possible. Then any a E 0 will be congruent modulo y to one of the numbers 0, I, - I.) 7. Show that there exists an algorithm for division with remainder in the maximal order of the field R( v2t

v-

8. Show that in the maximal order of the field R( I) every odd rational prime p of the form 4k + 3 remains prime, while every odd rational prime p of the form 4k + I factors into p = 7T7T', where7T and 7T' are nonassociate primes. Find the decomposition of the number 2 into prime factors. 9. Let 0 be a ring with unique factorization. Show that for any two numbers a and {3 of 0 (not both equal to zero), there is a common divisor 0 which is divisible by all common divisors of a and {3 (0 is called the greatest common divisor of a and (3). 10. Show that in the maximal order of the field R(v -6) the following are essentially different prime factorizations:

55 = 5· II = (7 + v - 6)(7 - v - 6), 6=2'3= -(v-6)2. 11. Show that in the maximal order of the field R(V -23) the following are essentially different prime factorizations: .

6=2'3=

l+v-231-v-23 2 2'

27 = 3·3·3 = (2 + v - 23X2 - v - 23). Find all possible factorizations of the number 8 in this ring.

170

I:

[Chap. 3

3. Divisors

3.1. An Axiomatic Description of Divisors We consider an arbitrary commutative ring .0 (with unit element and without divisors of zero), and we shall try to clarify the idea mentioned in Section 2.3 of mapping the nonzero elements of the ring .0 into some new domain, in which decomposition into prime factors is unique. Our theory must clearly consist of two parts: the construction of some set :?fl of new objects in which decomposition into prime factors in unique, and the determination of the mapping of the nonzero elements of the ring.o into the set [0. We start with the first part. In order to be able to speak of decomposition into prime factors in [0, we must have an operation of multiplication defined in:?fl; that is, we must associate to each pair of elements of [0 a third element, their product. We shall require that this operation be associative and commutative. A set with such an operation is called a commutative semigroup. We shall further require that the set [0 contain a unit element, that is, an element e such that ea = a for all a E [0. In a commutative semigroup [j} with unit e we may speak of divisibility of elements; an element a E:?fl is divisible by b E [0 if there exists aCE [0 such that a = be (we also say that b divides a). An element P E [0, distinct from e, is called prime if it is divisible only by itself and by the unit e. We further say that the semigroup [0 has unique factorization into prime elements if every element a E [0 can be represented as a product of prime elements

11

I

THEORY OF DIVISIBILITY

" I,'

I

I, ,I!:

a = PI ... Pr

I'

(r

~

0),

! :

I

I

I

and this decomposition is unique up to the order of the factors (for r = 0 this product is set equal to e). Thus uniqueness of factorization implies that e is the only invertible element (divisor of e) in the semigroup:?fl. It is clear that a semigroup with unique factorization is completely determined by its set of prime elements (essentially by the cardinality of this set). As a simple example of a semigroup with unique factorization we may take the set of all natural numbers under the operation of multiplication. In a semigroup with unique factorization, any two elements have a greatest common divisor (a common divisor which IS divisible by all common divisors of the two elements), and also a least common multiple. Two elements of [0 are called relatively prime if their greatest common divisor is equal to e. We note some elementary properties of divisibility in [0: If a product ab is divisible by e and a is relatively prime to e, then b is divisible bye; if e is divisible by the relatively prime elements a and b, then e is divisible by their product ab;

·t

Sec. 3]

DIVISORS

171

if a product ab is divisible by a prime element p, then at least one of the factors is divisible by p. We now pass to the second part of the theory, the conditions which must be satisfied by the mapping from the ring .0 to the semigroup f:0. Let .0* denote the set of all nonzero elements of the ring.o. Since we have assumed that .0 does not have divisors of zero, the set .0* is a semigroup under the operation of multiplication. Suppose that we have a mapping of the semigroup.o* into the semigroup :?fl which has unique factorization. We denote the image of an element ex E .0* by (ex). It is clear that we can use the semigroup f:0 to study the multiplicative structure of the ring .0 only if under the mapping ex ~ (ex), the product of two elements in .0* is mapped onto the product of their images in f:0, that is, only if (ex{3) = (ex)({3) for all ex and {3 in .0*. Hence we must assume that the mapping ex ~ (ex) is a homomorphism of the semigroup .0* into the semigroup f:0. If ex is divisible by fJ in the ring .0, it will then follow that (ex) is divisible by ({3) in the semigroup f:0. In order that divisibility in .0 should closely correspond to divisibility in :?fl, we shall also demand the converse: If (ex) is divisible by ({3) in f:0, then ex is divisible by {3 in .0. We shall also say that the element ex # 0 of.o is divisible by the element a E f:0, and shall write alex, if (ex) is divisible by a in the semigroup f:0. We shall suppose 0 to be divisible by all elements of f:0. If ex E .0*, the set of all elements of.o which are divisible by ex is closed under addition and subtraction. It is natural to assume that this property is preserved for divisors a of the semigroup f:0. Our last requirement is that f:0 not contain any "unnecessary" elements. By this we shall mean that distinct elements of f:0 must not divide precisely the same elements of .0*. We thus give the following definition. Definition. By a theory of divisors for the ring.o we shall mean the giving of some semigroup f:0 with unique factorization, along with a homomorphism ex ~ (ex) of the semigroup .0* into f:0, satisfying the following conditions:

(1) An element ex E.o* is divisible by {3 E.o* in the ring .0 if and only if (ex) is divisible by ({3) in the semigroup f:0. (2) If ex and {3 of.o are divisible by a E f:0, then ex ± {3 are also divisible by a. (3) If a and b are two elements of:?fl and the set of all elements ex E .0 which are divisible by a coincides with the set of all elements {3 E .0 which are divisible

by b, then a = b. The elements of the semigroup :?fl are called divisors of the ring .0, and divisors of the form (ex), ex E .0*, are called principal divisors. The unit element e of the semigroup is called the unit divisor.

THEORY OF DIVISIBILITY

172

[Chap. 3

Condition (1) in the definition of a theory of divisors clearly implies the following assertion: The equality (a) = ({3) holds if and only if a and {3 are associate in the ring .0. In particular, units e of the ring .0 are characterized by (e) = e. We shall denote a theory of divisors for the ring .0 by .0* --+:?fi. Our definition of a theory of divisors only fixed what we shall mean by such a theory. It does not at all guarantee the existence or uniqueness of the homomorphism .0 --+ [0. In the next section we consider the question of the uniqueness of a theory of divisors, assuming that one exists, and in Section 3.3 we indicate an important necessary (but not sufficient) condition for existence. The existence of a theory of divisors for maximal orders in algebraic number fields will be proved in Section 5 (Theorem 3 implies that such a theory does not exist for nonmaximal orders). 3.2. Uniqueness Theorem 1. If a ring.o has a theory of divisors, then it has only one. More precisely, if we have two homomorphisms .0* --+ [0 and .0* --+ [0', satisfying all requirements of the definition, then there is an isomorphism [0 ~ :?fi' under which the principal divisors in :?fi and :?fi' which correspond to a given element a E .0* are identified. Proof Let.o* --+ [0 and .0* --+ [0' be two theories of divisors for the ring .0. Let P E [0 and P' E:?fi' be prime divisors. Denote by Pand p' the sets ofelements

of the ring which are divisible by P and by P' (with respect to the theory .0* --+ [0 for P, and with respect to .0* --+ [0' for p'). We now show that for any prime divisor p' E [0' there is a prime divisor p E [0 such that V c V'. Assume that this is not the case, that is, that p ¢ p' for all prime divisors p E [0. From condition (3) it easily follows that any divisor must divide a nonzero element of the ring.o. Choose in D an element {3 ¥- 0 which is divisible by p', and decompose the divisor ({3) E [0 into prime factors: ({3) =

p/l ... p/r

(Pi' ... , Pr are prime divisors of the semigroup [0). Since we have assumed that Pi ¢ p', then for each i = 1, ... , r there is an element Yi ED which is divisible by Pi but not divisible by p'. The product Y = Yl kl ••. Yr kr is divisible

by Plk 1 ••• p/ and this means, by condition (1), that Y is divisible by {3 in the ring.o. But then Y must be divisible by p'. Thus we have a contradiction, since the product Yl kl ••• y/ r cannot be divisible by p', since p' is prime and does not divide any of the Yi. Hence for any prime divisor p' E [0' there is a prime divisor p E [0 such that r

,

Sec. 3]

DIVISORS

173

pcp'. By symmetry, there is a prime divisor q' E [0' for which i1' c p . We shall show that q' = p', and hence i1' = P = p'. Indeed, by condition (3) there is an element ¢ in .0 which is divisible by q' and not divisible by q'p'. If we assume that q' ¥- p', then the element ¢ will not be divisible by p', which is impossible since q' c p'. Since (for given p' E [0') there is one and only one prime divisor p E :!fi such that p = p' [condition (3)], we obtain a one-to-one correspondence p ...... p' between prime divisors of [0 and prime divisors of [0'. This correspondence can clearly be extended (in. a unique manner) to an isomorphism :!fi ~ :!fi'. Namely, if PI ...... PI" ... , p, ...... p/, then

We now need only show that under this isomorphism, the divisors (a) E [0 and (a)' E [0' (for given a E .0*) correspond tD one another. Let p E:!fi and p' E [0' be corresponding prime divisors, and assume that they occur in the factorizations of (a) and (a)' with exponents k and I, respectively. From condition (3) it follows that there is an element nED which is divisible by p and not divisible by p2. Since p = p', the element n is also divisible by p'. The principal divisor (n) hence has the form (n) = pb, where b is not divisible by p. Now choose in .0 an element w which is divisible by bk and not divisible by bkp. Since p does not divide b\ then w is not divisible by p or by p'. Consider the product aw. Since a is divisible by p\ and w is divisible by b\ then aw is divisible by pkb k = (n k), and by condition (I) we have aw = nk'1, '1 ED. But p'ln, and hence aw is divisible by p'\ and since p' ,r w, then p'kl a . This means that in the factorization of the divisor (a)' E [0', the prime divisor p' occurs with exponent not less than k; that is, I ~ k. But by symmetry also k ~ I, and thus k = l. We have thus shown that if (a)=PIk1 ... p/r and PI ...... PI" ... 'P, ...... p/, then (a)' = PI,k p/k r , and this means that under the above isomorphism :!fi ~ :!fi', the principal divisors (a) E:!fi and (a)' E:!fi' correspond to each other. If the ring .0 has unique factorization, then we can easily construct a theory of divisors .0* -. [0, and in this theory all divisors will be principal. Indeed, break up the set of all nonzero elements of.o into classes of associate elements, and consider the set [0 of all such classes. For a E .0*, denote by (a) the class of elements associate with a. It is easily seen that under the operation of multiplication (a)(fJ) = (afJ), the set [0 becomes a semigroup with unique factorization, and that the mapping a -. (a), a E .0*, defines a theory of divisors for the ring .0. [The prime divisors in this theory are just the divisors of the form (n), where n is a prime element of .0.] By Theorem I any theory of divisors for this ring coincides with the one just constructed. Assume now the converse, that we have for some ring.o a theory of divisors .0* -. [0, in which all divisors of:!fi are principal. We now show that an element l

...

THEORY OF DIVISIBILITY

174

[Chap. 3

n :1= 0 of the ring .0 will be prime if and only if the corresponding divisor (n) is prime. Indeed, if (n) = P is a prime divisor and y divides n in the ring .0, then the divisor (y) must divide P (in the semigroup ~) and then, since P is prime, either (y) is equal to P or to the unit divisor c. In the first case y is associate with n, and in the second case y is a unit in .0, and this means that n is a prime element of the ring .0. Now let (a) be neither prime nor the unit divisor. Then (a) is divisible by some prime divisor P = (n), and a is divisible by the prime element n and is not associate with it. Hence a cannot be prime. We have shown that if every divisor is principal, then the element n is prime if and only if the divisor (n) is prime. Let a be any element of .0*. Ifwe have the factorization

(3.1)

(a) = P1 ... Pr

in Pr

!!)

(the prime divisors Pi are not necessarily distinct), and if PI then in the ring .0 we have the factorization

= (11: r ),

= (n 1 ),

... ,

(3.2) where e is a unit of the ring.o. Since any factorization of the form (3.2) induces a factorization of the form (3.1), we must have unique factorization in the ring .0. We have obtained the following result. Theorem 2. In order that the ring .0 have unique factorization, it is necessary and sufficient that .0 have a theory of divisors .0* -+ ~ in which every divisor of ~ is principal.

3.3. Divisors and Integrally Closed Rings We have already noted that not every ring has a theory of divisors. The existence of a homomorphism a -+ (a) which satisfies the requirements of a theory of divisors imposes strong restrictions on a ring. One such restriction is given in the following theorem. Theorem 3. If the ring .0 has a theory of divisors, then .0 is integrally closed in its quotient field K. Proof. Assume that the element

e of K satisfies an equation (a 1

, ... ,

an ED),

e

Ii I

I I ,'

1'1

but does not belong to .0. We represent it in the form = alP, where a E .0 and PE .0, and decompose the principal divisors (a) and (P) into prime factors. Since a is not divisible by Pin the ring .0 (we have assumed that rt .0), (a) is

e

Sec. 3]

DIVISORS

175

not divisible by (fl) [by condition (1)]. This means that some prime divisor p occurs in (fl) with greater exponent than in (a). Let p occur in (a) with exponent k ~ O. Since (fl) is divisible by pH I, we deduce by condition (2) that the right side of a"= -alfla"-I- ... -a"fl" is divisible by pk"+ I. But p occurs in (a") = (a)" with exponent kn, and thus a" is not divisible by pk"+I. This contradiction shows that ~ E ,0, and Theorem 3 is proved. Another necessary condition for the existence of a theory of divisors is given in Problem 1. Since the only orders in algebraic number fields which are integrally closed are the maximal orders, only maximal orders can possibly have a theory of divisors.

3.4. The Theory of Divisors and Valuations We now turn to the question of the practical construction of theories of divisors. We first assume that a theory of divisors ,0* -+ £0 exists for the ring ,0, and then proceed to clarify how this theory could be constructed. Taking an arbitrary prime divisor p, we can construct with it a function vp(a), which is similar to the p-adic valuation with respect to a prime p which was constructed in Chapter 1. Namely, for any a#-O of ,0, by vp(a) we denote the power to which p enters in the factorization of the principal divisor (a) into prime factors. Clearly, vp(a) is characterized by p.p(a)la

and

p.p(a) + l,r a.

Since zero is divisible by arbitrarily large powers of p, it is natural to set vp(O) = 00. It easily follows from the definition that vp(afl) = vp(a) vp(a

+ fl)

~

+ vp(fl),

min (vp(a), vp(fl))

(3.3) (3.4)

[for the proof of (3.4) we must use condition (2)]. The function vp(a) can be extended to the quotient field K of the ring ,0 in such a way that (3.3) and (3.4) still hold. For any ~ = a/fl E K (a, fl E D) set vp(~)

= vp(a) - vp(fl).

The value of vp ( ( ) dearly does not depend on the choice of the representation of ~ in the form ~ = a/fl. It is now easily verified that (3.3) and (3.4) still hold for the extended function vp • We shall now see what values the function vp(a) takes as a ranges through K.

176

I,

[Chap. 3

TH EORY OF DIVISI BI L1TY

Since the divisors P and p 2 are distinct, by condition (3) there is an element y E D which is divisible by p but not by p2. For this element we have vp(y) = 1. But then vp(/) = k for any integer k. Hence the function vp(C() takes on all rational integral values. Definition. Let K be any field. A function v(C(), defined for C( E K, is called a valuation of the field K, if it satisfies the following conditions: (1) v(C() takes on all rational integral values as C( ranges through the nonzero elements of K; v(O) = 00; (2) v( C(fJ) = v(C() + v(fJ); (3) v(C( + fJ) ~ min (v(C(), v(fJ)). We can now say that every prime divisor p of the ring D determines a valuation vp(C() of the quotient field K. It is easily seen that distinct prime divisors determine different valuations. For if p and q are distinct prime divisors, then by condition (3) the ring D contains an element y divisible by p and not divisible by q. But then vp(y) ~ 1 and vq(y) = 0, and hence vp i= vq . All valuations of the field K of the form vp clearly satisfy

vp(C()

~

0

for all

C(

E

D.

(3.5)

In terms of valuations we can give a simple expression for the factorization of the principal divisor (C(), C( E D*. The prime divisors which enter into this decomposition are characterized by vplC() > O. Then we have (C()

=

TI

Pivpi(a),

(3.6)

i

where Pi runs through all prime divisors for which vpi(C() > O. We thus see that the semigroup f0 of divisors and the homomorphism D -+ g are completely determined by the set of all valuations vp of the field K which correspond to prime divisors p. For the set of all divisors and the operation of multiplication are determined as soon as the set of all prime divisors is known (each divisor is a product of prime divisors with nonnegative exponents, and when divisors are multiplied the corresponding exponents are added). But the prime divisors are in one-to-one correspondence with the valuations vp ' Finally, the homomorphism D* -+ f0 is determined from (3.6). This means that the concept of a valuation can be used as a foundation for the construction of a theory of divisors. We shall proceed to develop this idea. We must first answer the following important question: How can we characterize the set 91 of valuations of the field K which must be taken to construct a theory of divisors for the ring D? The product (3.6) can contain only a finite number of factors. Hence, for any fixed C( E D*, the condition vp(C() = 0 must hold for almost all valuations of the set 91 (by" almost all" is meant" for all but a finite number ").

Sec. 3]

DIVISORS

177

From (3.5) we see that for all v E 91 we must have v(C() ~ 0 if C( E ,o. Conversely, assume that for some ~ ¥- 0 of K we have v(~) ~ 0 for all v E 91. If we represent ~ in the form ~ = C((fJ (C(, fJ E D), then we have v(C() ~ v(fJ) for all v E 91. But this means that the principal divisor (C() is divisible by the principal divisor (fJ)· Condition (l) now implies that C( is divisible by fJ in the ring ,0; that is, ~ E ,o. We hence have a second necessary condition: The set ofvaluations 91 must be such that v(C() ~ 0 for all v E 91 if and only if C( is an element of the ring ,o. We now give another necessary condition for 91. Take any finite set of valuations v1 , ••• , Vm of 91 which correspond to the prime divisors P1' ... , Pm' If k 1 , ••• ,km are fixed nonnegative integers, we consider the divisor 0 = p/ ' ... Pm km • From condition (3) it follows that the ring ,0 contains an element C(j which is divisible by OJ = 0P1 '" Pi-1 Pi+1 ... Pm and not divisible by 0iPi (l ~ i ~ m). Consider the sum

Using condition (2), we easily find that C( is divisible by p/' and not divisible by p/'+l. Hence the set 91 must satisfy the following condition: For any valuations Vb •.• , Vm of 91 and for any nonnegative integers k b ••• , kit. there exists an element C( in the ring ,0 for which vj(C() = k j (l ~ i ~ m). The necessary conditions which we have found. on 91 will now be shown sufficient to construct a theory of divisors for the ring ,o. To prove this, take a semigroup q,& with unique factorization, with the prime elements in one-to-one correspondence with the valuations of the set 91. The valuation v E 91, which corresponds to the prime element P E q,&, will again be denoted by vp ' By the first and second conditions, for any C( E ,0* the product (3.6) will make sense [the exponents vplC() are nonnegative and almost all of them are zero]. Since v(C(fJ) = v(C() + v(fJ) the mapping C( -+ (C() will be a homomorphism from ,0* to q,&. From the second condition it easily follows that C( is divisible by fJ in the ring ,0 if and only if v(C() ~ v(fJ) for all v E 91. Hence condition (l) is satisfied. Condition (2) is implied by the inequality v(C( ± fJ) ~ min (v(C(), v(fJ)). If 0 and b are two different elements of q,&, then some prime element P occurs in their factorizations with different exponents, say, k and I. Let k < I. By the third of the above conditions there is in ,0 an element C( which is divisible by 0 and for which vp(C() = k. But then C( is not divisible by b. This shows that condition (3) is also satisfied. Hence the homomorphism ,0* -+ q,& gives us a theory of divisors for the ring ,o. We formulate the results which we have obtained. Theorem 4. Let ,0 be a ring with quotient field K, and let 91 be a set of valuations of K. In order that the valuations of 91 induce a theory of divisors on ,0 it is necessary and sufficient that the following conditions hold:

I

'I'

I

II, ,

178

THEORY OF DIVISIBILITY

[Chap. 3

(1) For any rx #- 0 of ,0, v(rx) = 0 for almost all valuations v E 91. (2) An element rx of K belongs to ,0 if and only if v(rx) ~ 0 for all v E 91. (3) For any finite set of distinct valuations Vl' ... , Vm of 91 and for any set of nonnegative integers k l , ... , k m , there is an element rx E ,0 for which

I'

Henc.e the construction of a theory of divisors for the ring ,0 is reduced to the construction of the corresponding set 91 of valuations of its quotient field K. We shall not enter here into the determination of those integrally closed rings for which a theory of divisors can be constructed (see, for example, the book "Modern Algebra" by van der Waerden, Section 105, Ungar, New York, 1950). In Section 4 we show that if 0 is a ring with quotient field k and ,0 is the integral closure of 0 in some finite extension field K of k, then if 0 has a theory of divisors, so does D. Since the ring Z has a theory of divisors (being a ring with unique factorization), then we will have proved that there is a theory of divisors for the maximal order of any algebraic number field. The set of valuations 91 of the field K which must be taken to construct a theory of divisors depends essentially on the ring ,0, and, in general, this set will not consist of all valuations of the field K (Problem 6). It can even happen (Problem 7) that condition (1) of Theorem 4 will not hold for the set of all valuations of the fieldK. We now show, however, that in the case of the ringZ of rational integers, we must take all valuations of the field R of rational numbers (we shall see in the future that this is also true for maximal orders of algebraic number fields). To each prime number p E Z (that is, the prime divisor of the ring Z) there corresponds the valuation vp of the field R, the value of which is given for the nonzero rational number x

= pm_a b

(3.7)

(a and b integers not divisible by p) by

vix) = m.

(3.8)

This valuation vp is called the p-adic valuation of the field R [it is clear that the valuation (3.8) coincides with the p-adic valuation of the field R of p-adic numbers; see Section 3.2 of Chapter 1]. Theorem 5. Every valuation of the field of rational numbers is of the form vp for some prime p.

Sec. 3]

DIVISORS

179

Proof. Let v be any valuation of the field R. Since

v(l

+ ... + 1) ~

min (v(l), ... , v(l))

= 0,

then v(n) ~ 0 for all natural numbers n. If v(p) = 0 for all primes p, then we would also have v(a) = 0 for all a¥-O of R, which is impossible by condition (l) of the definition of a valuation. Hence for some prime p we must have v(p) = e > O. Suppose that for the prime q ¥- p we also had v(q) > O. Then from the equation pu + qu = 1 (u and u rational integers) we obtain 0= v(pu

+ qu) ~

min (v(pu), v(qu)) ~ min (v(p), v(q)) > O.

This contradiction shows that v(q) = 0 for all primes q except p. Hence v(a) = 0 for all integers a not divisible by p. For the rational number (3.7) we thus have v(x)

=

mv(p)

+ v(a) -

v(b)

=

me

=

evix).

Since the valuation v must take on all integral values, e = 1 and hence v = v ' p Theorem 5 is proved. Note that Theorem 5 could easily have been deduced from Theorem 3 of Section 4, Chapter 1, the second part of the proof of which we have essentially repeated above. We conclude this section by considering another special case. Assume that for some ring ,0 we have a theory of divisors ,0* -+ f0 with only finitely many prime divisors Pl' ... , Pm' Denote by v1 , ••. , Vm the corresponding valuations of the quotient field K. By condition (3) of Theorem 4 for any divisor a = Pl k, .,. Pm k m (k i ~ 0), there is an element rt. E ,0 for which V1 (rt.) = k b '" , vm(rt.) = k m. But this means that the divisor a coincides with the principal divisor (rt.). Thus all divisors of f0 are principal, and the ring ,0 has unique factorization (Theorem 2). If Pl = (n 1 ), •.. , Pm = (n m ), then the elements n 1 , '" , n m constitute a complete set of pairwise-nonassociate prime elements of the ring ,0 and every element rt. E ,0* has a unique representation in the form

where c; is a unit of the ring ,o. The prime elements n 1 , characterized by

•.. ,

n m are clearly

forj ¥- i. We have obtained the following result. Theorem 6. If for some ring ,0 we have a theory of divisors with only a finite number of prime divisors, then ,0 has unique factorization into primes.

180

THEORY OF DIVISIBILITY

[Chap. 3

PROBLEMS

1. If the ring D has a theory of divisors, show that every element of D has only a finite number of (nonassociate) factors. 2. Show that in any theory of divisors any divisor is the greatest common divisor of two principal divisors. 3. Let K = k(x) be the field of rational functions over a field k and let cp be some irreducible polynomial in k[x]. Every nomero rational function u of K can be written in the formu = cp'(f/g), wherefandg are polynomials in k[x] which are not divisible by cpo Show that the function V,p, given by v",(u) = k, is a valuation of the field K. 4. If f and g are nonzero polynomials of k[x] of degrees nand m, and u = f/g set v*(u) = m - n. Show that the function v* is a valuation of the field K = k(x).

E

k(x),

5. Let v be a valuation of the field k(x) such that v(a) = 0 for all nonzero a in k. Show that v is either of the form v", (for some irreducible polynomial cp E k[x]) or else v = v* (see Problems 3 and 4). 6. If we set D = k [x], determine the set SR of valuations of the fieldK = k(x) which satisfies the conditions of Theorem 4. Further, determine the set SR for the ring D' = k[l/x]. 7. Let K = k(x, y) be a field of rational functions in two variables over the field k. For any natural number n set x. = x/y"o A nonzero rational function u = u(x, y) E K can be represented in the form

J(x., y) u = u(x.y", y ) = y - ( - - ) , g x., y where the polynomialsfandg are not divisible by y. If we set v.(u) = k, show that thefunction v. is a valuation of the field K. Further, show that the valuations v. (n ;;. l) are all distinct, and that for all of them v.(x) > O. 8. Formulate and prove an .. Eisenstein irreducibility criterion" for polynomials over any ring 1) with a theory of divisors. 9. Show that if a ring D has a theory of divisors, then its quotient field K has algebraic extensions of all degrees. 10. Letfbe a nonzero polynomial in the ring D = k[x, y] of polynomials in two variables over the field k. Denote by (ii f) the smallest degree of a monomial which appears in fwith nonzero coefficient. Show that the function ii can be extended to a valuation of the field of rational functions k(x, y). Denote by ~I the set of all valuations of the field k(x, y) which correspond to irreducible polynomials of the ring D, and let SRI be obtained from SR by adjoining ii. Which of the conditions of Theorem 4 are not fulfilled for the ring D and the set SRI of valuations?

4. Valuations

Theorem 4 of Section 3 reduces the problem of constructing a theory of divisors for an integrally closed ring D to the determination of a set of valuations of the quotient field K which satisfy the conditions of the theorem. We turn to a systematic study of valuations.

Sec. 4]

VALUATIONS

181

4.1. Simple Properties of Valuations

From the definition of a valuation of a field K (Section 3.4) we immediately obtain v(±l)=O v( -C()

= v(C()

v(~) = v(C() v(C(n) v(C(,

= nv(C()

+ ... + C(n) ~

(f1 # 0),

v(f1), nEZ,

min (v(C(,), ... , v(C(n».

Now assume that v(C() # v(f1). If v(C() > v(f1), then v(C( + f1) ~ v(f1). On the other hand, since f1 = (C( + f1) - C(, v(f1) ~ min (v(C( + f1), v(C(», so that v(f1) ~ v(C( + f1). Hence v(C(

+ f1) = min (v(C(),

v(f1»

if v(C() # v(f1).

(4.1)

By induction we obtain v(C(,

+ ... + C(n) = min (v(C(l),

... , v(C(n»,

provided that the minimum value of V(C(l), '" , v(C(n) occurs only once. Definition. Let v be a valuation of the field K. The subring D. of the field K consisting of all elements C( E K for which v(C() ~ 0 is called the ring of the valuation v. The elements of D. are called integral with respect to the valuation v.

It is clear that all three conditions of Theorem 4 of Section 3 are fulfilled for the ring D and the set 91 consisting of the single valuation v. Hence the ring D. has a theory of divisors with a single prime divisor. From Theorems 3 and 6 of Section 3 we obtain the following results. y

Theorem 1. The ring D y of the valuation v of the field K is integrally

closed in K. Theorem 2. The ring D has (up to associates) a single prime element n, and any element C( # 0 of D. has a unique (for fixed n) representation in the form C( ~ E:n m , where E: is a unit in D y (m ~ 0). y

The prime element n is clearly characterized by v(n) = 1. In the ring D y , as in any ring, we can consider congruences with respect to the elements of D. (see the Supplement, Section 4.1). Since congruences

I

THEORY OF DIVISIBILITY

182

[Chap. 3

modulo associate elements are equivalent, the ring of residue classes modulo the prime element 1t does not depend on the choice of 1t but is completely determined by the ring D •. We denote this ring of residue classes by L. and will now show that it is a field. For if rx E D. and rx ;f. 0 (mod 1t), then v(rx) = 0 and this means that rx is a unit in D •. Then rx has an inverse ~ and rx~ == I (mod 1t), since rx~ = 1. The field L. is called the residue class field of the valuation v. 4.2. Independence of Valuations

Let the ring ,0 have a theory of divisors ,0* -+ g&, and let PI' ... ,Pm be distinct prime divisors of g&. By Theorem 4 of Section 3 there correspond to these prime divisors valuations VI' ... , Vm of the quotient field K, and these valuations are independent in the sense that there exist elements in K on which they take on any given set of values k l , ... , k m • For if we set k / = max (Q, k j ), i = 1, ... , m, and k;" = min (0, k;), then by condition (3) of Theorem 4 of Section 3 we can find elements rx and fJ in ,0 fur which vj(rx) = k/ and vj(fJ) = -k;", and then for the quotient ~ = rxlfJ we will have vl~) = k j (1 ~ i ~ m). We now show that this property of independence does not depend on the fact that the valuations V j corresponded to prime divisors in some theory of divisors, but is true for any finite set of valuations. Theorem 3. If VI' rational integers k l ,

are distinct valuations of the field K, then for any k m there exists an element ~ E K for which

, Vm ,

vIG)

= kl ,

... ,

vm(~)

= k m.

Let ,01'"'' ,om denote the rings of the valuations VI' ... 'V m and set = n;':.1 ,oj' Conditions (1) and (2) of Theorem 4 of Section 3 are clearly fulfilled for the ring ,0 and the set 91 consisting of the valuation SV I , ... , V m • From the formulation of Theorem 3 we see that condition (3) also holds, and hence the ring ,0 has a theory of divisors with a finite number of prime diyisors. Thus Theorem 3 implies that for any finite set of valuations VI' ... , V m of the field Kwe have a theory of divisors for the ring ,0 = n~= I D. From Theorem 6 of Section 3 we then derive the following result.

,0

Corollary. If

are distinct valuations of the field K with rings 0 ,0 i is a ring with unique factorization. Further, each nonzero element of ,0 has a unique representation in the form rx = e1tlk, .. , 1t mkm , where e is a unit in ,0, and 1t 1, ... , 1t m are fixed prime elements of ,0 characterized by ,01' .. , ,

VI' ... , V m

,om' then the intersection ,0 =

Vj(1tj)

= 1,

nr=

v/1t j)

=0

(j ¥- i).

Sec. 4]

VALUATIONS

183

= I, the assertion of the theorem is contained in the definition of a valuation. Assume that m ~ 2 and that the case of m - I valuations has already been proved. We show that then there do not exist rational integers C I , ... , Cm , not all zero, for which Proof of Theorem 3. For m

(4.2) for all nonzero ~ E K. Assume the contrary, that is, that (4.2) does hold. Among the coefficients at least two must be nonzero and have the same sign [otherwise there would be only two nonzero coefficients, say, C1 and C1 , with CI > and C1 < 0, and then from CI V1 (~) + C1v1(O = we would obtain vI(O = eVl(~) with positive e, and this is possible only for e = I and VI = v1 ]. Changing the numeration, if necessary, we can write (4.2) in the form

°

°

(4.3) where at least one of the rational coefficients OJ is negative. By the induction hypothesis there exist elements P and P' in the field K such that V

j(P)

vj(P)

for all i

= 0,

v;(P')

=I

= I, vj(P') =

°

if

OJ

~

if

OJ

< 0,

0,

= 2, .. , , m. Then (4.4)

Consider the sum P + p'. Since one of the numbers v j(P) and vj(P') (i = 2, .. , , m) equals and the other equals I, then vj(P + P') = min (vj(P), vj(P')) = 0. From the relation (4.3) we therefore obtain v1(P + P') = 0. On the other hand, from (4.4) we obtain v j(P + P') = min (v j(P), VI (P')) < 0.

°

This contradiction proves that (4.2) is impossible. Let ,0 now denote the intersection of the rings of the valuations V 1 , '" , Vm , and let E denote the group of units of this ring. Let n 1 , ... , n m denote prime elements of ,0, numbered so that v;(n;) = I (i = 2, ... , m) (recall that for the case of m - I valuations, Theorem 3, and hence also its corollary, are assumed). We now show that the valuation VI cannot be identically zero on the group E. Any element ~ E K* can be written in the form

where c; E E, k j = v;(~) (2 ~ i ~ m). If v1(c;) we would obtain Vl(~) = k1v1(nl) + '" which can also be written in the form

=

°for all

+ kmv1(n m),

(4.5) c;

E

E, then from (4.5)

!

184

THEORf OF DIVISIBILITY

[Chap. 3

where the rational integers ai = vlni) do not depend on ~, and this contradicts the fact that a relation of the form (4.2) is impossible. Hence the group E contains elements on which the valuation Vi is nonzero. Choose an element y in the group E on which the valuation Vi takes its smallest positive value I. It is clear that all values of Vi on E are divisible by I. We shall show that 1= 1. If all the values a 2 = Vi (n 2 ), •• , , am = Vi (n m) were divisible by I, then we would deduce from (4.5) that all values Vl(~) of the valuation Vi are divisible by I, which is possible only for I = 1. Consider the case where not all a j are divisible by I, say, a 2 is not divisible by I. Consider the element

where s is an integer chosen so that the number

satisfies the inequality 0 < Ii < I. It is clear that v1 (ct) i = 2, ... , m. Set e = y + ct. Since vle) = min (v;(y), vj(ct)) same time,

= Ii

and Vj(ct) > 0 for

= 0 for all i = 2, ... , m, then e E E. But at the

which contradicts the choice of y. This shows that the case when not all ai are divisible by I is impossible, and hence 1= 1. We may now assume that the prime elements n j (2 ~ i ~ m) of the ring.o are chosen so that vlni) = a j = O. For each ni can be replaced by n;' = niy-a" for which v1(n/) = a i - ajvi(y) = O. Setting nl = y, we have obtained a system of elements nl' n2' ... , n m , for which vj(ni) = I and Vini) = 0 for j #- i. If now k 1 , ... , k m are any integers, for the element ~ = nl k , ... n mk ", we have

Theorem 3 is proved. From Theorem 3 we easily deduce the following stronger result. Theorem 4 (Approximation Theorem). If Vi' ... , Vm are distinct valuations of the field K, then for any elements ~1' ... , ~m of K and any integer N, there exists an element ~ E K for which l

185

VALUATIONS

Sec. 4]

Proof. Choose in K elements ct 1, ... , (j #- i) and set ct k ~ = __ 1 _ ~1 1 ct/

+

ct m

such that Vj(ctj) = -1, v/ct i ) = 1 ct k

+ ... + _ m _k 1 + ct m

~m'

Since Vj(ctn #- 0 = vj(l) for all natural numbers k, then by (4.1) the value of vp + ct/) equals 0 for i #- j and equals -k for i = j, so that ct.k ) Vj ( 1 +' ct k = k

for

i #-j

and

Vj(~) = k. 1 + ct j

i

Hence V/~ -~) ~

It is clear that

~

min (k i

+ V/~i))'

will satisfy the theorem provided k ~ N - min v/

O.

i,j

4,3, Extension of Valuations

Let k be a field and K a finite extension of k. If v is some valuation of the field K, then by restricting v to the field k we obtain a function which clearly satisfies conditions (2) and (3) in the definition of a valuation (Section 3.4). The first condition may not be satisfied; that is, the values ofv on the elements of k may not exhaust the group Z. But v cannot be identically zero on k. If this were the case, then the field k would be contained in the ring of the valuation v, and since that ring is integrally closed (Theorem 1), then K would also be contained in it, and this is impossible. Thus v(a), a E k*, takes on both negative and positive values [if v(a) < 0, then v(a- 1 ) > 0], Let p denote any element of k on which v takes its least positive value v(P) = e. Then for any a E k* the value v(a) = m is divisible bye. For if m = es + r, 0 ~ r < e, then v(ap-S) = m - se = r, so by the minimality of e we have r = 0, Now setting (a

E

k*),

vo(O) =

00,

(4.6)

we obtain on k a fUQction vo, which takes all integral values and which consequently is a valuation of the field k. Definition. Let K be a finite extension of the field k. If the valuation Vo of the field k is related to the valuation v of the field K by (4.6), then we say that Vo is induced on k by the valuation v, and v is an extension of Vo to the

I

I I,

;I

186

THEORY OF DIVISIBILITY

[Chap. 3

field K. The uniquely determined integer e which appears in (4.6) is called the ramification index of v with respect to V o (or with respect to the subfield k). Note that in this defl.nition when e > 1, the term extension of a valuation does not correspond to the usual concept of the extension of functions to larger domains of definition. It follows from the above that every valuation v of K is induced by a unique valuation Vo of k. The converse assertion is also valid, that is, for any valuation V o of k there exists an extension to K (which is, in general, not unique). The proof of this fact is fairly difficult, and we shall give it in the next section. First, we consider some properties of extensions of a given vo , assumin'g that such extensions exist. Let k eKe K' be a tower of finite extensions, and let vo , v, v' be valuations of the fields k, K, K'. It is clear that if v is an extension of V o with ramification index e, and v' an extension of v with ramification index e', then v' is an extension of V o to the field K', and the ramification index of v' with respect to V o is equal to ee'. It is also easy to see that if v and Vo are induced by the valuation v', then v is an extension of V o ' Lemma 1. If K is a finite extension of the field k of degree n, then any valuation V o of the field k has at most n extensions to the field K. Vm be distinct extensions of V o to the field K. By Theorem 3 we can find elements ~l' ... , ~m for which v;(~i) = and viO = 1 for j # i. We shall show that these elements are linearly independent over k. Consider the linear combination

Proof. Let v:, ... ,

°

with coefficients aj of k, not all zero. Set k = min (vo(ad, ... , vo(a m)), and let i o be such that VO(aio) = k. Denoting by e the ramification index of Vio with respect to k, we have

and therefore

vIO(Y) = min (vio(a 1 ~ d, ... , vio(am~m)) = ek, so that Y # 0, which proves our assertion. From the linear independence of the elements ~l' ... , ~m over the field k, it follows that m ~ (K: k), and this means that the number of extensions Vi is not greater than n. Lemma 1 is proved.

Sec.4]

VALUATIONS

187

Assume now that Vb •.• , vm are all extensions of a fixed valuation V o of a field k to a finite extension K. Let 0 denote the ring of the valuation Vo and ,0 its integral closure in the field K, and let Db ... , ,om denote the rings of the valuations VI, ... , V m • Since 0 c Db and the ring ,oj is integrally closed in K, then ,0 c ,oj for i = I, ... , m, and hence

Later we shall see that equality actually occurs here. If this is so, then by the corollary of Theorem 3, ,0 has unique factorization with a finite number of nonassociate prime elements. Since the nonassociate prime elements 1tl' ... , 1t m of the ring ,0 are in one-to-one correspondence with the valuations VI' •.• , V m , we obtain a method for constructing valuations of K which are extensions of the valuation V o ' So assume that we know that the ring ,0, the integral closure of the ring of the valuation Vo in the field K, has unique factorization with a finite number of nonassociate prime elements. From Theorem 6 of Section 3, this assumption is equivalent to the existence in ,0 of a theory of divisors with a finite set of prime divisors PI' ... , Pm' We shall show that then the valuation Vo has precisely m extensions to the field K, these being the valuations VI' .•. , Vm of the field K which correspond to the prime divisors PI' ... , Pm . Let p be any prime element of the ring 0 of the valuation Vo [that is, any element of k such that vo(p) = 1], and let 1t 1, .. , , 1tm be a complete set of nonassociate prime elements of the ring ,0 [numbered so that v j (1tJ = 1]. Since oeD, the element p has a factorization p

= e1t I e,

'"

1tm e~

(4.7)

in the ring ,0, with nonnegative exponents ej (e a unit in D). Now if a is any element of k* and vo(a) = s, that is, a = pSu, where u is a unit in 0, then in ,0 we have

(4.8) If e j = 0, then Vj would be identically zero on k*, and we saw at the start of this section that this is impossible. Hence ej > O. Formula (4.8) now implies that each of the valuations V j (i = 1, ... , m) is an extension of V o to the field K. We also obtain that ej, the ramification index of V j with respect to vo , is given by (4.7) Assume now that V is an extension of the valuation Vo to the field K. Since 0 is contained in the ring of the valuation v, so is its integral closure ,0, that is, v(a) ~ 0 for all rx E ,0, and this means that v(e) = 0 for all units e E D. If V were distinct from VI' •.. , V m , then by Theorem 3 there would be a unit e of the ring ,0 such that v(e) i= O. Hence V must be one of the V j •

,

I'

1:

188

1

1
(a) is a unit in the ring o. We now compute lIy - ail· The characteristic polynomial of y - a equals f(t + a), and therefore

°

N K/k(Y - a) = ( - l)"f(a) = ( - l)"q>(a)p-',

so that

Ily-all =2-'< 1, Ilet-aPil < IIPII. Hence (4.10) is satisfied if we set ~ = a, p = et - ap. We have shown that ,0 is a Euclidean ring, and thus by Theorem 2 of Section 2 it has unique factorization. Let n be any prime element of the ring ,o. Since for any et E ,0*, its norm NK/k(et) is always divisible by et, then NK/k(n) = plu is divisible by n (u is a unit of 0, f ~ 1). But since n is prime and factorization into primes is unique, the element p is also divisible by n. Hence, if p has the factorization

in the ring ,0 (e a unit in D), then the prime elements n l , plete set of nonassociate primes of ,o. The proof of Lemma 2 is complete.

... ,

n m form a com-

We now turn to the proof of the basic results of this section. Theorem 5. Any valuation V o of the field k can be extended to any finite extension K of k. Theorem 6. Let 0 be the ring of the valuation Vo , and let ,0 be its integral closure in the field K. If Vi> ... , Vm are all extensions of the valuation Vo to the field K, and ,01' ... , ,om are their rings, then m

Theorem 7. Under the same notations, the ring ,0 has unique factorization, and the set of valuations of K which correspond to prime elements of ,0 is precisely the set of all extensions VI' ••• , Vm of the valuation Vo to the field K.

I,

,1

: ,I

190

[Chap. 3

THEORY OF DIVISIBILITY

If the prime elements 1r1> ... , 1r m of the ring ,0 are ordered so that v;(1rj) = I, and if the prime element p of the ring 0 has the factorization (c; a unit in D),

then ej is the ramification index of the valuation

Vi

with respect to

Vo'

Proof. If we assume that Theorems 5 and 6 have already been proved, then

i

,I

, I

'

by the corollary of Theorem 3 the ring has unique factorization (with a finite number of nonassociate prime elements), and hence all results obtained in the second half of Section 4.3 are valid. But these results are precisely the contents of Theorem 7. Theorems 5 and 6 will be proved by induction on the degree n of the extension Kjk. For n = I, there is nothing to prove. Let n> I and assume that Theorems 5 and 6 have already been proved for all extensions of degree O. Since the powers 1,0, ... ,on-2 form a basis for k' over k, the element y can be represented in the form y

= l(c o + cIO + .. , + Cn_2 0n - 2 ) = pk rx,

where all coefficients C j belong to 0 and at least one of them is a unit in o. We saw above that 0 E 0' and the classes 1,0, ... , on-2 of L' are linearly independent over LO' Hence the residue class Ii

= Co + CIl} + ... + Cn_20n-2

is nonzero (since at least one of the coefficients C j is nonzero). This means that rx is not divisible by p' (in the ring 0'); that is, vo'(rx) = O. Analogously we obtain vo"(rx) = O. Comparing the conditions vo'(y) = 0 and vo'(rx) = 0 with y = pk rx, we see that k = 0, and hence vo"(y) = vo"(rx) = O. But this contradicts the choice of y. Thus the valuation Vo has only one extension to the field k'. Since Theorem 6 is assumed valid by induction for the extension k'lk, the ring 0' of the valuation vo' coincides with the integral closure of the ring 0 in the field k'. Let ,0' denote the integral closure of the ring 0 in the field K'.

THEORY OF DIVISIBILITY

192

[Chap. 3

Since 0' c ,0' and the ring ,0' is integrally closed in K' (Supplement, Section 4.3), then ,0' is also the integral closure of the ring 0' in the field K'. Let VI" ••• , v: be all extensions of the valuation vo' to the field K', and let ,01" .. " ,0: be their rings. Since Theorem 6 holds for the extension K'lk' (by Lemma 2), then r

D'=nD/j;1

(4.11)

The set of valuations V/ is also the set of all extensions of the valuation Vo to the field K'. Equation (4.11) thus can be considered a proof of Theorem 6 for the extension K'ik and the valuation vo . Let VI, .,. , Vm denote all valuations of the field K which are induced by one of the valuations v/' and let ,01' ... , ,om denote their rings. If v/ is an extension of V; , then clearly ,0/ n K = D i • Noting that the intersection ,0' n K coincides with the integral closureD of the ring 0 in the field K, we have m

,0

= ,0' n

K

=n j~

(,0/ I

n K)

=n i; I

Di •

(4.12)

Suppose now that there is an extension V of the valuation V o to K different from VI' ••• , Vm • Then by Theorem 3 there would be an element rx in K for which v1(rx) ~ 0, ... , vm(rx) ~ (and hence rx E D) and v(rx) < 0. But this would contradict the fact that ,0 must be contained in the ring D. of the valuation v. Thus VI' ..• , Vm are the only extensions of the valuation Vo to the field K. Formula (4.12) coincides with the assertion of Theorem 6.

°

PROBLEMS

t~ " I 1. Show that an algebraically closed field has no valuations. 2. Let K = k(x) be a field of rational functions over the field k and let v be the valuation of K corresponding to the polynomial x - a (a E k). Show that the residue class field ~v of the valuation v is isomorphic to k. Show further that two elements f(x) and g(x) of the ring lie in the same residue class if and only if f(a) = g(a). 3. Let K = k(x) be a field of rational functions over the field k of real numbers, and let v be the valuation of K corresponding to the irreducible polynomial x 2 + I. Find the residue classfield ~v of the valuation v. 4. Let (I, and (12 be the rings of the valuations v, and V2 of some field K. Show that if (I, c (12. then v, = V2' 5. Find the integral closure of the ring of 3-integral rational numbers in the field R(V - 5) and determine all extensions of the 3-adic valuation v, to this field. 6. For all prime numbers p find all extensions of the p-adic valuation V p to the field R(V=!) and determine the corresponding ramification indices.

Sec. 5]

THEORIES OF DIVISORS FOR FINITE EXTENSIONS

7. Let KI k be a normal extension and Vo a valuation of the field k. Show that if extension of Vo to the field K, then all extensions have the form v'( IX) = v(a( IX»

193 v

is any

(IX E K),

where a runs through all automorphisms of Klk. 8. Let k be a field of characteristic p. Let Klk be a purely inseparable extension. Show that a valuation Vo of the field k has only one extension to the field K. [The extension Klk is called purely inseparable if every element of K is a root of degree p' (s ~ 0) of some element of k.]

9. Let k = ko(x, y) be a field of rational functions in x and y over some field k o . In the field of formal power series k o {t} (see Problem 7 of Section 4, Chapter 1, or Section 1.5 of Chapter 4) choose a series g(t) = L:=o cot O (co E ko) which is transcendental over the field of rational functions ko(t) [the existence of such series follows from the fact that the field k o {t} has higher cardinality than the field ko(t), and hence higher cardinality than the set of elements of ko{t} which are algebraic over ko(t)]. For a nonzero polynomial f = f(x, Y)E ko[x, y], the seriesf(t, g(t» will be nonzero by choice of g.1f tOis the smallest power of t which OCcurs in this series with nonzero coefficient, set vo(f) = n. Show that the function Vo (after suitable extension) is a valuation of the field k and that the residue class field of the valuation is isomorphic to the field k o .

5. Theories of Divisors for Finite Extensions

5.1.

Existence

Theorem 1. Let the ring 0 with quotient field k have a theory of divisors f0 which is determined by the set 91 0 of valuations. If K is a finite extension of the field k, then the set 91 of all valuations of K which are extensions of 0* ---+

valuations of 91 0 determines a theory of divisors for the integral closure D of the ring 0 in the field K. Proof. By Theorem 4 of Section 3 we need only verify that the set 91 satisfies

all conditions of that theorem. We first verify the second condition. For any valuation v E 91 and any a EO we clearly have v(a) ~ O. This means that 0 is contained in the ring of the valuation v. But then by Theorem I of Section 4 the integral closure of the ring 0 in the field K is also contained in the ring of the valuation v. In other words, v(C() ~ 0 for all C( E D. Conversely, let C( E K be an element such that v(C() ~ 0 for all v E 91. Let t r + a 1 t r - 1 + ... + ar denote the minimal polynomial of C( with respect to k. Let Vo be any valuation of k belonging to the set 91 0 , and let VI' ... , V m be its extensions to the field K. Since V1 (C() ~ 0, ... , vm(C() ~ 0, then by Theorem 6 of Section 4 the element C( lies in the integral closure in K of the ring of the valuation V o . But in this case all the coefficients ai' ... , a r must lie in the ring of the valuation Vo (see the Supplement, Section 4.3); that is, vo(a l ) ~ 0, "', VO(a r ) ~ O. Since this holds for all V o E 91 0 , the coefficients a1 , '" , ar belong to 0, and hence C( ED.

194

THEORY OF DIVISIBILTY

[Chap. 3

We now tum to the first condition. Let rx ED, rx #- 0, iL1.d let a, be determined as above. Then for all but a finite number of valuations vo of 91 0 we have vo(a,) = 0. Hence for all but a finite number of valuations v of 91 we have v(rx-l)=a,-l(rx'-l+.··+a'_l))~O, and this means that V(rx) =0. Hence v(rx) = for almost all v E 91. Only the third condition now remains to be verified. Let Vl' .,. , Vmbe distinct valuations of 91 and kl> ... , k m be nonnegative integers. Let VOl' ... , VOm be the corresponding valuations of 91 0 (the VOi are not necessarily all distinct). Expand the original set of valuations to the set Vl' ... , Vm, Vm+l' '" , V s ' consisting of all extensions of the valuations VOi to the field K. By Theorem 3 of Section 4 there is an element y in the field K for which

°

, I

vl(y) = kl> ... , vm(y) = k m,

Vm+l(}') = 0, ... , v.(y) = 0.

If this element y belongs to the ring ,0, just set rx = y. Assume that y does not belong to D. In this case denote by Vl', ... , v: the valuations of 91 which take negative values on y: and by VOl" ... , vo: the corresponding valuations of 91 0 (various vo/ also may be equal). Since each of the valuations vo/ is different from each of the valuations VOi, in 0 there is an element a such that vOi(a) =

°

(1 ~ i ~ m),

vo/(a) = / (1 ~j ~ r),

where / is taken equal to max (/1' ... , I,). Set rx = ya. Since v/(rx) = v/(y) + v/(a) ~ -/j + vo/(a) = -/j + / ~ 0,

then rx E D. Thus in any case we have in the ring ,0 an element rx such that vl(rx) = kl> ... , vm(rx) = k m , so that condition (3) of Theorem 4 of Section 3 also holds for the set 91 of valuations. The proof of Theorem I is complete. We apply Theorem I to the case of algebraic number fields. The maximal order ,0 of the algebraic number field K is, as we have seen, the integral closure in K of the ring Z of rational integers. Since Z has a theory of divisors (since it has unique factorization), then by Theorem I ,0 also has a theory of divisors. By Theorem 5 of Section 3 the theory of divisors for Z is induced by the set of all valuations of the field R of rational numbers, and since every valuation of the field K is an extension of some valuation of the field R we find that the theory of divisors for the ring ,0 is induced by the set of all valuations of the field K. We hence have the following theorem. Theorem 2. If ,0 is the maximal order of the algebraic number field K, there exists a theory of divisors ,0* ---+ g&, and this theory is induced by the set of all valuations of the field K.

Sec. 5]

THEORIES OF DIVISORS FOR FINITE EXTENSIONS

195

5.2. Norms of Divisors Let 0 be a ring with theory of divisors 0* --+!!J o and with quotient field k; let K be a finite extension of k, with .0 the integral closure of 0 in K, and let .0* --+!!J be a theory of divisors for the ring D. In this paragraph we establish a connection between the semigroups of divisors !!J o and !!J. Since 0 cD, elements of 0* correspond to principal divisors both in !!J o and in !!J. To distinguish these principal divisors, if a E 0*, we denote the corresponding principal divisor in !!J o by (a)k' and for any CI. E .0* we denote the corresponding principal divisor by (Cl.h. We have an inclusion isomorphism of semigroups 0* --+ .0*. Since any unit of the ring .0 which is contained in 0 is a unit of the ring 0, this inclusion induces an isomorphism (a)k --+ (a)K' for a E 0*, of the semigroup of principal divisors of the ring D. We now show that this isomorphism can be extended to an isomorphism !!J 0 --+!!J (which will not be onto). Theorem 3. There is an isomorphism of the semigroup !!J o into the semigroup !!J, which on principal divisors coincides with the isomorphism (ah --+ (a)K' a E 0*.

The isomorphism !!J o --+!!J is clearly characterized by the commutativity of the diagram 0*

--+

.0*

! !!J 0

! --+

!!J

that is, by the fact that the two composite homomorphisms 0* --+ .0* --+ !!J and 0* --+!!J o --+!!J coincide (the vertical homomorphisms denote the homomorphisms of the multiplicative semigroups of the rings onto the semigroups of principal divisors). Let p be any prime divisor of the ring 0, vI' the corresponding valuation of the field k, and V~l' "" v~m all extensions of vI' to the field K (\.l31' ... , \.l3m are the corresponding prime divisors of the ring D). Let el ... , em denote the respective ramification indices of the valuations V~l' ." , v~m with respect to vI" Since v~,(a) = e;vp(a) for all a E 0*, then the factor p'p(u) of the principal divisor (a)k E!!J o will become (\.l31 e, ... \.l3m em yp(U) in the principal divisor (a)K E!!J. This means that the isomorphism from !!J o to!!J defined by the mapping (5.1) (for all p) satisfies the requirements of Theorem 3. It is easily seen that the isomorphism !!J o --+!!J, satisfying the requirements of Theorem 3, is unique (Problem 5).

) , ,

I \'

~I

I! :

"'

196

THEORY OF DIVISIBILITY

[Chap. 3

By means of the isomorphism !!J o --+!!J we can identify the semigroup !!J o with its image in!!J. But prime divisors in!!J o will not, in general, remain prime in !!J. For by (5.1) each prime p has the decomposition (5.2) in the semigroup !!J. Using the embedding !!J o --+!!J we may speak of the divisibility of divisors of 0 by divisors of .0. From (5.2) we see that a prime divisor p of the ring 0 is divisible by a prime divisor 'l.' of the ring .0 if and only if the valuation v~ is an extension of the valuation vp • It is further clear that relatively prime divisors in !!J o remain relatively prime in !!J. Definition. Let 'l.'lp. The ramification index e = e~ of the valuation v~ with respect to the valuation vp is also called the ramification index of the prime 'l.' relative to p (or relative to k).

The ramification index is thus the largest natural number e such that 'l.'el p. If ct. is any element of .0, its norm N(ct.) = NK/k(ct.) lies in o. The mapping ct. --+ N(ct.), ct. E .0*, is a homomorphism from the multiplicative semigroup .0* to the semigroup o. Since the norm of any unit of the ring .0* is a unit in 0, this homomorphism induces a homomorphism (ct.)K --+ (N(ct.))k of the semigroup of principal divisors of the ring .0 to the semigroup of principal divisors of the ring o. We shall show that this homomorphism can be extended to a homomorphism of the entire semigroup !!J to !!J o . Theorem 4. There is a homomorphism from the semigroup of divisors !!J --+ !!J 0 , such that

to !!J 0 , N:!!J

(5.3) for any ct. E .0*. We can express (5.3) by saying that the diagram

is commutative. For a fixed prime divisor p E!!J o we denote the ring of the valuation vp by op and its integral closure in the field K by .op . By Theorem 7 of Section 4 the prime divisors 'l.'l' ... , 'l.'m of the ring .0 which divide p correspond uniquely to a complete set of pairwise-nonassociate prime elements 7t 1 , ••• , 7t m of the

Sec. 5]

THEORIES OF DIVISORS FOR FINITE EXTENSIONS

ring .01" The correspondence element ct. E K, if

197

'l.'i +-+7ti has the property that for any nonzero (5.4)

where e is a unit of the ring .01" then ki

=

v~,(ct.).

Let 'l.' be one of the prime divisors 'l.'i which divides 1', and let corresponding prime element of the ring .01' . Set

(5.5) 7t

be the (5.6)

It is clear that d~ does not depend on the choice of 7t. Taking the norm in (4) and comparing with (5) and (6) we obtain the relation vp(N K/k(ct.))

=

Ld~v~(ct.)

(5.7)

~Ip

('l.' runs through all prime divisors of the ring .0 which divide 1'). We can now construct the desired homomorphism N: ~ ~ ~o . A divisor m: = 'l.'t At •• , 'l.'r A r of the semigroup ~ can be conveniently written as an infinite product

m: = TI 'l.'A(~) ~

over all prime divisors 'l.' of~, in which, however, only a finite number of the exponents A(~) are nonzero. [A('l.') equals Ai' if ~ = 'l.'i' and equals zero if the divisor 'l.' is not one of 'l.'t, .,. , ~r'] We can write divisors of the ring 0 in an analogous fashion. If (ct.)K is the principal divisor corresponding to a nonzero element ct. of .0, then we have the representation (5.8) From (5.7) we see that if (N(ct.))k

= TI

pe(p),

(5.9)

I'

then e(p) must satisfy e(p) = Ld~v~(ct.).

(5.10)

~Ip

This suggests the following definition. Definition. Let ~r = TI~'l.'A(~) be a divisor of the ring D. For any prime divisor l' of the ring 0, set a(p)

=

Ld~A('l.').

~Ip

198

THEORY OF DIVISIBILITY

[Chap. 3

The divisor TIp pQ(p) of the ring 0 is called the norm ofthe divisor'll with respect to the extension Klk and is denoted by NK/k('ll), or simply by N('ll). Since A('l.') equals zero for almost all 'l.' (that is, all but a finite number of a(p) also equals zero for almost all p, and hence the expression TIp pQ(p) actually is a divisor of the ring o. From the definition it is clear that

'l.'), then

N('llIB) = N('ll)N(IB)

for any two divisors'll and IB of D. The mapping'll --+ N(~l) is thus a homomorphism of the semigroup .0 to the semigroup ~o. In the case of a prime divisor'll = ~ we clearly have (5.11)

!

j'

i

In view of (5.10), the norm of the divisor (5.8) equals the divisor (5.9) and hence we have proved the existence of a homomorphism N: ~ --+ ~o, which satisfies the requirements of Theorem 3. As in the case of the isomorphism ~o --+~, it can be shown (Problem 4) that the homomorphism N: ~ --+ ~o is uniquely determined by condition (5.3). One of the central problems of the theory of divisors is to determine a rule for the decomposition of the prime divisor p of the ring 0 into prime divisors of the integral closureD of 0 in some finite extension field. In the general case this problem is still unsolved (however, see the end of Section 8.2). Each decomposition (5.2) is characterized by the number m of prime divisors and by the various ramification indices e i = e~;. The natural numbers ei' however, cannot be taken arbitrarily (for a given extension Klk). For they are related to the numbers d~ [see (5.6)] by the formula

L d~e~ = n = (K:k),

(5.12)

~Ip

for the proof of which it suffices to apply formula (5.7) to the case of a prime element p of the ring op [recall that v~,(p) = e;]. 5.3. The Degree of Inertia

The definition of the homomorphism N: ~ --+ ~o depended on the numbers d"" which were defined in a rather formal manner in (5.6). We now clarify the deep arithmetical significance of these numbers. Let 'l3lp. Let op and .o~ denote the rings of the valuations vp and v~, and p and 7t the prime elements in these rings. Since for elements a and b of op the congruences a == b (mod p) in the ring op) and a == b (mod 7t) in the ring .0", are equivalent, each residue class in op modulo p is contained in a single residue class modulo 7t in .o~. This determines an isomorphic embedding of

Sec. 5]

THEORIES OF DIVISORS FOR FINITE EXTENSIONS

199

the residue class field ~p = op/(p) of the valuation vI' into the residue class field ~~ = D~/(7t) of the valuation v~. Using this isomorphism we assume that ~p c ~~. For any ~ E D~ we denote the residue class modulo 7t which contains ~ by ~. The subfield ~p of the field ~'ll then consists of those residue classes of the form a, where a E 01' . Let the residue classes WI' ... , wm of ~~ (Wi E D~) be linearly independent over the field ~p' We now show that then the representatives WI' ... , W m of these classes are linearly independent over the field k. Assume that this is not so, that is, that for some coefficients a i E k, not all zero, we have

Multiplying this relation by a suitable power of p, we may assume that all ai lie in the ring 01' and that at least one of them is not divisible by p. Passing now to the residue class field ~~, we obtain

aiw i + ... + amwm= 0, in which not all coefficients ai E ~p are zero. This contradiction proves our assertion. From the linear independence of WI' ... , W n over the field k it follows that m ~ n = (K: k). Thus the residue class field ~~ is a finite extension of the field ~p for which

Definition. Let the prime divisor I.l3 of the ring D divide the prime divisor p of the ring o. The degree f = f~ = (~~: ~p) of the residue class field of the valuation v~ over the residue class field of the valuation vI' is called the degree of inertia of the prime divisor I.l3 relative to p (or relative to k).

As in Section 5.2 we denote by D p the integral closure of the ring 01' in the field K. In analogy with the definition of a fundamental basis of an algebraic number field, we make the following definition. Definition. A basis WI' ... , W n of the extension K/k is called a fundamental basis for the ring D p relative to 01' if all its elements lie in D p and every element IX E D p is represented by a linear combination

(5.13)

II

I I,

\

with coefficients

0i

in

01' .

We shall see below that in the case of a separable extension K/k, a fundamental basis for the ring D p (for any p) always exists. On the other hand, by Problems 11 and 12, for nonseparable extensions K/k it may occur that the ring D p has no fundamental basis relative to 01' .

I

r

/

200

[Chap. 3

THEORY OF DIVISIBILITY

The value of the concept of a fundamental basis is indicated in the following theorem. Theorem 5. Let 'l.' be a prime divisor of the ring D which divides p, and let n be a prime element of the ring D p which corresponds to it. If the ring D p has a fundamental basis relative to 01" then f~

=

d~

= vp(N K/k(n).

Proof. The prime element n

E D p is clearly also a prime element of the ring We shall show that each residue class ~ of the ring D~ modulo n contains a representative in D p ; that is, for any ~ E D~ there is an element IX E D p such that ~ == IX (mod n).

D~.

Let 'l.' = 'l.'1' 'l.'z, ... , 'l.'m be all prime divisors of the ring D which divide p. By Theorem 6 of Section 4, Y E D p if and only if v~,(Y) ~ 0 for all i ='1, ... , m. Hence the element IX must satisfy V~(~ V~.cIX) ~

-

0

IX)

~

1,

(i = 2, ... , m),

and the proof of its existence is given by Theorem 4 of Section 4. Now let W 1 , ... , W n be a fundamental basis of the ring D p relative to 01" By the above, every element of ~~ can be represented in the form a1w1 + ... + anwn, where ai E 01" and hence ai E ~p. This means that the residue classes W1' ... , wn generate ~~ as a vector space over ~p • Iff = (~~: ~p) = f~ , then we can choose from among themf elements which are linearly independent over ~p' Let these be W1' ... , wf . It is clear that the congruence a1w1

+ ... + afwf == 0 (mod

n),

with ai EOI" holds in the ring D p if and only if ai == 0 (mod p), p being a prime element of the ring 01' . Since each of the residue classes Wj E ~~ forj = f + 1, ... , n can be expressed in terms of W1' ... , wf' then f

w j ==

for some b js from

01"

2: bjsws(mod n) s=j

(j

= f + 1, ... , n)

Set for i = 1, ...

ej =

f

-

2: bjsws + Wj

s= 1

for j = f

,f,

+ 1, ... , n.

Sec. 5]

THEORIES OF DIVISORS FOR FINITE EXTENSIONS

201

It is clear that el , ••• , en also form a fundamental basis of .0" relative to 0" (since all W s can be expressed in terms of es with coefficients in 0,,). Each element ef + l , ... , en is divisible in the ring .0" by n, and therefore the congruence

holds if and only if al == ... == af

.=

0 (mod p).

Consider the set 9R of all elements of the ring .0" which are divisible by n. By what was just proved, the set 9R consists of all linear combinations of the elements (5.14) with coefficients in 0". On the other hand, it is clear that 9R also coincides with the set of all linear combinations of the elements (5.15) with coefficients in 0". Let C denote the transition matrix from the basis (5.14) to the basis (5.15). Since every element ne j can be expressed in terms of the basis (5.14) with coefficients from 0", then det C is an element of 0". By symmetry this also holds for det C- l . Hence det C is a unit in the ring 0,,; that is, v" (det C) = o. If we multiply the first f columns of the matrix C by p, then we clearly obtain a matrix A = (a i ) , for which

ne

n

i

=

L aije

j •

j=l

Therefore, N K/k(n)

= det A = pf det C,

so that

and Theorem 5 is proved. Theorem 6. If the extension Kjk is separable, then.o" always has a fundamental basis relative to 0" .

Before starting the proof of this theorem we note that it is analogous to the proof of Theorem 6 of Section 2, Chapter 2. Since every element of K, after multiplication by a suitable power of a prime element of the ring 0", becomes integral with respect to 0", the extension Kjk

I

~

: !

!

THEORY OF DIVISIBILITY

202

[Chap. 3

has a basis, Ct l , .,. , Ct n , all elements of which lie in .0" . Consider the dual basis *, ... , Ctn * (see the Supplement, Section 2.3; here we have already assumed that Klk is separable). If Ct E .0" and

Ct l

,j,

~I I II i

(5.16)

!' '

:!

Ii

I

,I

!

Ii

where Ci E k, then Ci = Sp (CtCtJ, and this means that C i EO" (since CtCt i ED,,). For each s = 1, ... , n we consider in the ring .0" those elements which when expressed in terms of the basis Ct l *, ... , Ctn * have the form

1

,i' I' 'I

I

I I

csCt.*

+ ... + CnCt n *

(5.17)

and we choose among these an element Ws

= CssCt.* + ... + CsnCt n*

such that v,,(cs ) ~ vicss) for all coefficients Cs of elements of the form (5.17) of .0". It is clear that c"~ #- 0 for all s, so that the elements WI' ... , W n of .0" are linearly independent over k. Let Ct be any element of .0". If we represent it in the form (5.16), then CI = clla l , where al EO", by choice of WI' For the difference Ct - a l WI we have the expansion

and here C2 ' = c 22 a 2 , where a 2 E 0" by choice of W2 • Continuing this process n times we finally arrive at the expansion (5.13), in which all coefficients a i belong to 0". The basis W j is hence a fundamental basis relative to 0", and Theorem 6 is proved. From Theorems 5 and 6 and formula (5.12) we easily obtain the following assertion. Theorem 7. If the extension Klk is separable and p is a fixed prime divisor of the ring 0, then the ramification indices e~ and degrees of inertiaf~ of the prime divisors 'l.' of the ring .0 which divide p are connected by the relation

L e~f~ = n = (K: k).

~I"

Hence for separable extensions Klk formula (5.7) can be written in the form (5.18) Remark. For nonseparable extensions, Theorem 7 is no longer necessarily valid. However, the inequality L~I" e~f~ ~ n always holds (see Problem 13). It can further be shown that, in general, f~ ~ d~ .

Sec. 5]

THEORIES OF DIVISORS FOR FINITE EXTENSIONS

203

5.4. Finiteness of the Number of Ramified Prime Divisors Definition. The prime divisor p of the ring 0 is called ramified in the ring .0 if it is divisible by the square of some prime divisor of the ring .0, and is called unramified otherwise.

Hence p is unramified if and only if all ei in (5.2) are equal to 1. Under the assumption that the extension K/k is separable, we shall obtain an important condition for p to be unramified. Assume that the ring.o p contains some primitive element efor the extension K/k, such that the discriminant D(f) of its minimum polynomial f(t) is a unit in 01" We now show that in this case the powers 1, e, ... , en-I, where n = (K: k) form a fundamental basis for the ring .01' over 01" Let w 1 , ... , W n be any fundamantal basis for .01" and let C be the matrix of transition from the basis Wi to the basis ej • Then D(f) = D(1, e, ... , en- 1) = (det C)2 D(W1' ... , wn) [see the Supplement, formula (2.12)]. Since D(f) is a unit in 01" and both terms on the right belong to 01" then det C is a unit in 01" and hence 1, e, ... , en - 1 is also a fundamental basis. Let p be a prime element of the ring 01' and ~p the residue class field of the valuation vI" For any polynomial g(t) with coefficients in 01' we denote by g(t) the polynomial obtained by replacing all coefficients of g(t) by their residue classes modulo p. Since the discriminant D(J) E ~p of the polynomial J(t) E ~p[t] is equal to the residue class modulo p of the discriminant D(f) E 01" then by our assumption the discriminant D(f) is nonzero. Hence, all factors in the decomposition (5.19) into irreducible factors in the ring ~p[t] are distinct (here ij5i is some polynomial of op[t]). If we denote the degree of ij5i by d;, then we clearly have d1

+ ... + dm = n = (K:k).

(5.20)

Theorem 8. If the discriminant of the minimum polynomial f(t) of a primitive element e E .01' is a unit in 01' , then the prime divisor p is unramified in .0 and the prime divisors 'l.'i of the decomposition p

= 'l.'1 .. , 'l.'m

can be put in one-to-one correspondence with the irreducible polynomials ij5; E ~p[t] of the decomposition (5.19) in such a way that the degree of inertia fi of the prime divisor 'l.'i coincides with the degree d i of the corresponding polynomial ij5;(t). i 1,

204

THEORY OF DIVISIBILITY

[Chap. 3

Proof. Let g(t) be any polynomial of op[t]. We shall show that if the polynomials 9 and qJi are relatively prime in the ring ~p[t], then the elements g(O) and CPi(O) are relatively prime in the ring D p • For then there exist polynomials u(t), v(t), and /(t) in the ring op[t] such that g(t)u(t)

+ CPi(t)V(t) = I + p/(t).

If g(O) and cp;(8) were divisible in the ring D p by some prime element n, then since nip (Theorem 7 of Section 4), from the preceding equation (for t = 0), it would follow that nil. This contradiction proves our assertion. Since the irreducible polynomials qJi are distinct, then the elements CP1 (0), ... , CPm(O) are pairwise relatively prime. Assume that cp;(O) is a unit in D p , that is, that cp;(O)~ = I, ~ E D p • Since 1,0, ... , on-1 form a fundamental basis for D p over op, then ~ = h(O), where h(t) E op[t]. The equation cp;(O)h(O) = I implies that cpi(t)h(t) = I + l(t)q(t), where q(t) E op[t] [since the leading coefficient ofl(t) equals I]. Passing to the residue class field ~p we obtain qJih = I + qJ1 ... qJmq, and we again have a contradiction. Hence none of the elements CP1(O), ... , CPm(O) are units in D p • For each i choose in D p a prime element n;lcp;(O). Since we have proved that the cp;(8) are pairwise relatively prime, the prime elements nl' ... , n m are pairwise-nonassociate. Let 'l.'1'"'' 'l.'m denote the corresponding prime divisors of the ring .0, and 11' ... ,1m denote the degrees of inertia of these divisors. In the residue class field ~~i of the valuation V~i' the residue classes I, {}, ... , {}d,-l are linearly independent over ~p (d i is the degree of qJ J For if there is a polynomial g(t) E op[t] of degree < d i for which g({}) = 0, then the element g(O) is divisible by ni in the ring D p and hence g(O) and CPi(O) are not relatively prime. But we saw at the beginning of the proof that then g(t) must be divisible by qJ;(t), and this can happen only if all coefficients of g(t) are zero. We have shown that (i

= I, ... , m).

Comparing this inequality with (5.20) and considering Theorem 7, we see that 'l.'1' ... , 'l.'m are the only prime divisors which divide p, that their ramification indices ei all equal I, and that d i =/;. This proves Theorem 8. Finally, we note that since cp;(e) is divisible by n i but not divisible by any other prime element n j ' then n i can be determined as the greatest common divisor in the ring D p of the elements cp;(O) and p. Corollary. If the extension Kjk is separable, then there are only finitely many prime divisors p of the ring 0 which are ramified in D. Let 0 be any primitive element of the extension Kjk which is contained in D. The discriminant D = D(l, 8, ... , en -1) is an element of 0*. If p,r D,

I'IIi.

Sec. 5]

THEORIES OF DIVISORS FOR FINITE EXTENSIONS

205

then by the theorem p is not ramified in D. Thus only those prime divisors of the ring 0 which divide D can be ramified in D.

PROBLEMS

1. Let 0 be a ring with a theory of divisors. k its quotient field. and k eKe K' a tower of finite extensions. Let C and C' denote the integral closure of the ring 0 in the fields K and K'. If 1.l3' is any prime divisor of the ring C'. denote the prime divisor of the ring C which is divisible by 1.l3' by W. and the prime divisor of the ring 0 whiCh is divisible by 1.l3' by \.1. Show that the degree of inertia of 1.l3' relative to k equals the product of the degree of inertia of 1.l3' relative to K and the degree of inertia of I.l3 relative to k. Formulate and prove an analogous assertion for the index of ramification. 2. Let the ring 0 with quotient field k have a theory of divisors with only a finite number of prime divisors, and let the prime divisor \.1 correspond to the prime element p of the ring o. Show that the residue class ring o/(p) is isomorphic to the residue class field ~" of the valuation "". 3. Let "" be a valuation of the field k. 0" its ring. K/k a finite separable extension.C" the integral closure of the ring 0 in the field K. and WI • .•.• W n a basis for the field Kover k. all elements of which lie in the ring D". Show that if the discriminant D(wl> .... w n ) is a unit in the ring 0". then W, • ... , W n is a fundamental basis for the ring C" over 0". 4. Show that the homomorphism N : §J ->- §Jo. satisfying the conditions of Theorem 4. is unique. 5. Show that the isomorphic embedding §J 0 -+ §J. satisfying the conditions ofTheorem 3, is unique. 6. Let a be a divisor of the ring o. Considering it as a divisor of the ring C (using the embedding §J ->- §Jo) show that (n = (K:k».

7. Let K/ k be a separable extension of degree n. Show that if the divisor a of the ring 0 becomes a principal divisor of the ring C. then an is a principal divisor of o. 8. Let K/k be separable. Show that the norm NK/.(a) of a divisor a of the ring C is the greatest common divisor of the principal divisors (NK/.(ex» •• where ex runs through all elements of C divisible by a. 9. The polynomial f(t) = t n + alt n -I + ... + an with coefficients in the ring 0 is called an Eisenstein polynomial relative to the prime divisor \.1. if al> .... an are all divisible by i • ' and an. while divisible by \.1. is not divisible by \!'. If the ring C contains a primitive element 8 for the extension K/k of degree n, with the minimum polynomial of 8 an Eisenstein polynomial relative to \.1. show that \! is divisible by only one prime divisor 'll of the ring C and that (the degree of inertia of I.l3 relative to \.1 hence equals 1). 10. Under the same hypotheses show that the basis 1.8..... 8n -I is a fundamental basis of the ring £" relative to 0". 11. Let k o be any field of characteristic p and k = ko(x. y) a field of rational functions in x and y over the field k o . Consider the valuation "0 of k, which was defined in Problem 9

I

I

THEORY OF DIVISIBILITY

206

[Chap. 3

of Section 4, where for the series g(l) E k o {I} [transcendental over ko(t)] we take a series of the form

By Problem 8 of Section 4 there is a unique extension of the valuation I' to the purely inseparable extension K = k({/y) of degree p over k. Show that the ramification index of I' relative to 1'0 equals I, and that the residue class field of the valuation I' coincides with the residue class field of the valuation 1'0 (under the inclusion isomorphism). It now follows from Theorem 5 and (5.12) that the ring C of the valuation 1', which is the integral closure in K of the ring of the valuation 1'0, does not have a fundamental basis relative to o. 12. Under the same assumptions as in the preceding problem, give a direct proof (without involving Theorem 5) that C has no fundamental basis over o. 13. Let be a ring with a theory of divisors, k its quotient field, K/ k a finite extension of degree n, C the integral closure of in K, \1 a prime divisor of the ring 0, '13" ... , '13m the prime divisors of the ring C which divide \1, e" ... ,em their ramification indices, and flo ... ,f.. their degrees of inertia relative to \1. For any s = I, ... ,m we denote by a'll' the residue class in the field L'll' which contains ct E C'll,' Choose elements W" E Cp (I ~ i ~ f,) so that the classes 0; that is, p == 0 (mod 1'). If the prime number q is different from p, then vp(q) = 0, and therefore vp(q) = 0; that is, q ;ji 0 (mod 1'). The residue class ring .011', being finite and without divisors of zero, is a finite field (Supplement, Section 3). Since for any a E .0 we have pa == 0 (mod 1'), then the characteristic of this field is p. Hence we have Theorem 2. Any prime divisor l' of an algebraic number field divides one and only one rational prime p. The residue class ring .011' is a finite field of characteristic p.

A theory of divisors for an algebraic number field hence has the property that the residue class ring modulo a prime divisor is a field. In general, this is not the case. For example, in the ring of polynomials k[x, y] in two variables over a field k the residue class ring of the prime divisor (x) is isomorphic to the ring of polynomials k[y] and hence is not a field. The residue class ring .011' is a field if and only if the congruence a~ == 1 (mod 1') is always solvable when a ¥= 0 (mod 1'). Hence only under this assumption can we expect to construct a completely adequate theory of congruences in the ring D.

Sec. 6]

DEDEKIND RINGS

209

6.2. Congruences in Dedekind Rings Definition. A ring .0 is called a Dedekind ring if it has a theory of divisors .0* --+ ~ and for every prime divisor l' E ~ the residue class ring .0/1' is a field. Examples of Dedekind rings, other than the maximal orders of algebraic number fields, can be obtained by taking the integral closure of the polynomial ring k[x] in a single variable in a finite extension of the field of rational functionsf(x) (Problems I and 2). The valuation ring D. of any valuation v is also a Dedekind ring (see Section 4.1), as is any ring which has a theory of divisors with only a finite number of prime divisors (Problem 3). Lemma 1 . If .0 is a Dedekind ring and Ct E .0 is not divisible by the prime divisor 1', then the congruence Ct~ == 1 (mod pm) is solvable in .0 for any natural number m. Proof. For m = 1 the congruence is solvable by the definition of a Dedekind ring. The lemma will be proved by induction on m. Suppose that for some ~o ED we have Ct~o == 1 (mod pm). Choose an element w in the ring .0 for which vp(w) = m. The principal divisor (w) has the form (w) = pm a, where a is not divisible by p. Choose an element Y E .0 for which vp(Y) = 0 and y == 0 (mod a). The product y(Ct~o - 1) will be divisible by pm a = (w), and hence y(Ct~o - 1) = Wll with Il E D. We now try to solve the congruence Ct~ == 1 (mod pm+l), taking as ~ an element in the form ~ = ~o + WA, where A is to be chosen suitably in D. Since y(Ct~ - 1)

=

y(Ct~o - 1)

+ YCtwA = w(1l + YIXA)

and W == 0 (mod pm), then we shall achieve our goal if A satisfies the congruence AIXY == - Il (mod 1'). But since Cty is not divisible by 1', this congruence is solvable. Hence there is an element ~ ED for which y(Ct~ -1) == 0 (mod pm + 1) and since vp(Y) = 0, dividing by y, we obtain Ct~ - 1 == 0 (mod pm+l). Lemma 1 is proved. l

Theorem 3. If PI' '" , Pm are distinct prime divisors of the Dedekind ring .0, and PI' ... ,Pm are any elements of .0, then there is an element ~ inD which satisfies

II I.

! ~ == (k 1 ,

... ,

Pm (mod Pm

I

km )

k m are any natural numbers).

Proof. For each divisor

(i

= 1, ... , m)

I

Ii

I Ii , ,

II

210

THEORY OF DIVISIBILITY

[Chap. 3

we can find an element Ct. i E .0 which is divisible by 0 1 but not divisible by PI' Lemma 1 guarantees that we can solve the congruence Ct.i~1 == f3i (mod p/ i ) in ~ i E D. It is easily seen that the element

satisfies the requirements of the theorem. Theorem 4. If Ct. # 0 and 13 are elements of a Dedekind ring .0, then the congruence Ct.~

== 13 (mod 0)

(6.1)

is solvable if and only if 13 is divisible by the greatest common divisor of the divisors (Ct.) and o. Proof. We first assume that the divisors (Ct.) and

0 are relatively prime, and will show that in this case the congruence (6.1) is solvable for any 13. Let I km o = p/ '" Pm = P/'Oi' where PI' '" ,Pm are distinct prime divisors. By Lemma 1 for each i = 1, ... , m in the ring .0 there is an element ~;' such that Ct.~;' == 13 (mod p/ ' ). By Theorem 3 we can find for each i an element ~I for which ~i == ~/ (mod p/') and ~i == 0 (mod oJ It is now clear that the sum ~I + ... + ~m = ~ will satisfy the congruences Cf.~ == 13 (mod 1'/') for i = 1, ... , m, and hence will also satisfy (6.1). We now prove the theorem in general. Let b = Pill ... Pm 1m be the greatest common divisor of the divisors (Ct.) and o. If (6.1) holds modulo 0, then it also holds modulo b, and since Ct. == 0 (mod b), then we must also have 13 = 0 (mod b). This proves the necessity of this condition. Assume now that 13 is divisible by b. By Theorem 3 of Section 4, there is an element J1. E K for which

(i=l, ... ,m).

(6.2)

We shall show that we can choose J1. so that also (6.3)

!

I

~I II

I:

I, I

Ii

for all prime divisors q, distinct from Pb ... , Pm' Suppose that J1. does not satisfy condition (6.3), and let ql' ... , qs be the prime divisors, different from 1'1> ... , Pm' for which Vqj(J1.) = -r j < O. Choose in .0 an element Y such that Vqj(Y) = r j (l ~j ~ s) and vP1(Y) = 0 (l ~ i ~ m). It is clear that the element J1.' = J1.Y satisfies both conditions (6.2) and (6.3), and our assertion is proved. Let the divisor b be determined by 0 = db. If J1. satisfies conditions (6.2) and (6.3), then the element Cf.J1. belongs to .0 and is relatively prime to b. Since we have assumed 13 divisible by d, then f3J1. also belongs to D. We have already

Sec. 6]

DEDEKIND RINGS

211

proved that then there exists an element ~ in the ring D such that CtI1~ == {311 (mod b). For i = 1, ... , m we have vp,(Ct~ - {3)

and this means that

~

=

Vpi(CtI1~

-

{311)

+ Ii

~ k i - Ii

+ Ii =

k i,

satisfies (6.1).

6.3. Divisors and Ideals In this section we show that there is a one-to-one correspondence between divisors and nonzero ideals in a Dedekind ring. For each divisor 0 denote by a the set of all elements of the ring D which are divisible by o. It is clear that a is a nonzero ideal of the ring D. Theorem 5. In a Dedekind ring D the mapping 0 -> ii (0 E .@) is an isomorphism of the semigroup of divisors .@ onto the semigroup of all nonzero ideals of the ring D.

We first verify the following lemma. Lemma 2. If Ct l , ••• , ct. s are any nonzero elements of the Dedekind ring D and b is the greatest common divisor of the principal divisors (ct. l ), '" , (ct. s ), then any element Ct E D which is divisible by b can be written in the form

The proof )fthe lemma is by induction on s. For s = 1 the lemma is obvious. Let s ~ L. Let bl denote the greatest common divisor of the divisors (Ct l ), ... , (Cts - l )· Then b is the greatest common divisor of the divisors b l and (ct..). Let Ct be divisible by b. By Theorem 4 the congruence Cts~ == Ct (mod b ) has a l solution ~ E D. By the induction hypothesis there are elements ~1> ••• , ~s-l in the ring D such that Ct - ~Cts = ~ICtI + '" + ~s-ICts-I' Lemma 2 is proved. Proof of Theorem 5. By condition (3) of the definition of a theory of

divisors the mapping 0 ...... a takes distinct divisors to distinct ideals. Let A be any nonzero ideal of the ring D. For each prime divisor l' set

a(1') = min vp(Ct). aEA

It is clear that a(1') will be nonzero for only a finite number of prime divisors 1'. Hence the prodJlct 0 = 1'0(1'), in which l' runs through all prime divisors for which a(1') # 0, is a divisor. We shall show that 0 = A. Let ct. be any element of o. It is clear that we can find a finite set of elements ct. l , ... , Ct s in A such that a(1') = min (vp(ct. l ), ... , vp(Cts)) for all 1'. This means that the divisor 0 is the greate:,' common divisor of the principal divisors (Ctd, ... , (ct. s)' By Lemma 2

TIp

I !

II

I 1 i

I

:

I

I

THEORY OF DIVISIBILITY

212

[Chap. 3

the element rx can be represented in the form rx = ~1 rxl + ... + ~srxs with coefficients ~i ED. It follows that rx E A, and hence that G c A. Since it is clear that A C G, we obtain A = G. We have thus proved that a -+ a is a one-to-one mapping of the set of all divisors of the ring .0 onto the set of all nonzero ideals of D. We shall now show that this mapping is an isomorphism, that is, that for any two divisors a and b we have

ab = abo

(6.4)

Denote the product ab by C. Since C is a nonzero ideal of .0, there is a divisor c such that c = C. We must prove that c = abo Let the prime divisor I' enter into the divisors a and b with exponents a and b. Then min vvCy) = )lEe

min vp(rx{3) = min vp(rx) rIEo,peb

aEa

+ min vp({3)

=

a

+ b.

pel,

Since this is true for all prime divisors 1', then c = ab and (6.4) is proved. From the fact that the mapping a -+ a is an isomorphism, it follows, in particular, that the set of all nonzero ideals of the Dedekind ring .0 form a semigroup with unique factorization under the operation of multiplication. To construct a theory of divisors in Dedekind rings (in particular, in the maximal order of an algebraic number field), we could take the semigroup of nonzero ideals for the semigroup ~. The image of the element rx under the homomorphism .0* -+ ~ would then be the principal ideal (rx) generated by this element. This construction of a theory of divisors is due to Dedekind. I

.~

6.4. Fractional Divisors If we construct a theory of divisors .0* -+ ~ for the ring .0, then we obtain some information on the structure of the semigroup .0*. It is natural to try an analogous procedure with the multiplicative group K* of the quotient field K. To do this we need to extend the concept of a divisor. Following an established tradition, we shall reserve the term" divisor" for this broader concept, and will call divisors in the earlier sense "integral divisors. " Definition. Let .0 be a ring with a theory of divisors, with quotient field K, and let 1'1' ... , Pm be a finite system of prime divisors. An expression

(6.5)

with integer exponents k 1 , ••• , k m (not necessarily positive) is called a divisor of the field K. If all the exponents k i are nonnegative, then the divisor is called integral (or a divisor of the ring D). Otherwise it is called fractional.

Sec. 6]

DEDEKIND RINGS

213

It is sometimes convenient to write a divisor (6.5) as a formal infinite product (6.6)

over all prime divisors p, in which almost all exponents a(p) are zero. Multiplication of divisors is determined by the formula (Q pQ(P»)(Q pb(P»)

=Q

pQ(p)+b(p).

For integral divisors this definition coincides with the definition of multiplication in the semigroup :?2. It is easily seen that under this operation the set of all divisors of the field K is an Abelian group, which we shall denote by fh. The unit element of this group is the divisor e for which all exponents a(p) in (6.6) are zero. Since every nonzero element ~ E K is the quotient of two elements of .0, it follows from condition (I) of Theorem 4 of Section 3 that for all but a finite number of the valuations vp ' which correspond to the prime divisors p, we have vi~) = O. We denote this finite set by vP1 ' ... , vp~' The divisor

is called the principal divisor corresponding to the element ~ E K*, and is denoted by (~). When applied to elements of the ring .0, the new concept of a principal divisor coincides with the previous one (Section 3.4). By condition (2) of Theorem 4 of Section 3 the principal divisor (~) will be integral if and only if ~ belongs to D. From the definition of a valuation (Section 3.4) it easily follows that the mapping ~ --+ (~), ~ E K*, is a homomorphism K* --+ fh of the multiplicative group of the field K to the group of divisors ;g. By Theorem 2 of Section 3 this homomorphism maps onto the entire group jJ) (that is, is an epimorphism) if and only if .0 has unique factorization. The kernel of this map clearly is the group of units of the ring .0, and this means that for elements ~, 11 of K* we have (~) = (11) if and only if ~ = 118, where 8 is a unit of the ring D. We now define a concept of divisibility for arbitrary divisors. Let 0 = nppQ(p) and b = n P pb(p) be two divisors (not necessarily integral). We say that 0 is divisible by b if there is an integral divisor e such that 0 = be. In other words, o is divisible by b if and only if a(p) ~ h(p) for all p. For any 0 and b set d(p) = min (a(p), b(p)). Since the rational integer d(p) is equal to zero for almost all p, then the product b = np pd(p) is a divisor. The divisor b is called the greatest common dh'isor of the divisors 0 and b (0 and b are both divisible by band b is divisible by every common divisor of

, I

I

I

214

THEORY OF DIVISIBILITY

[Chap. 3

o and b). The least common multiple of the divisors 0 and b is defined analogously. The element CI. E K is cal1ed divisible by the divisor 0 = TIp 1'a(p), if CI. = 0 or the principal divisor (CI.) is divisible by o. In terms of valuations this is characterized by vp(a) ~ o(p) for al1 p. The correspondence of the preceding section between integral divisors and ideals of a Dedekind ring can be extended to fractional divisors, providing the proper generalization of the concept of ideal is used. As in Section 6.3, we denote the set of al1 elements of the field K which are divisible by the divisor 0 by a (these elements may now be nonintegral). From condition (3) in the definition of a valuation (Section 3.4) it fol1ows that if CI. and P are divisible by 0, then CI. ± P are also divisible by o. This means that the set a is a group under addition. Further, if CI. E a and ~ E .0, then the product ~CI. also belongs to o. We now verify the fol1owing formula: (y)o = yo

For the element

(6.7)

(YEK*, oED).

~

is divisible by (y)o if and only if any of the fol1owing hold: for all 1'; vp(~/Y) ;?; a(p) for all 1'; ~/y E 0; ~ E yo [here a(1') denotes the power to which p appears in the divisor oj. It is clear that for any divisor we can find an element y E .0* such that the divisor (y)o is integral. FOP'1Ula (6.7) shows that for such a y we will have yo c D.

vpW ;?; vp(y) + a(1')

I I

I 1

!

I I r

: 1:: !I I

~I

I I

I

iI I

II

il

Definition. Let D be a Dedekind ring with quotient field K. A subset A c K, containing at least one nonzero element, is called an ideal of the field K (relative 1.0 D), if it satisfies: (l) A is a group under the operation of addition. (2) For any CI. E A and any ~ E .0, the product ~CI. lies in A. (3) There is a nonzero element y of the field K such that yA s D. The ideal A is called integral if it is contained in .0 and otherwise is called fractional. An integral ide'al in K is clearly just a nonzero ideal in D. If A and B are two ideals of the field K, then by their product AB we mean the set of al1 elements y E K which can be represented in the form y

=

Cl.1Pl

+ ... + Cl.mPm

(m;?; 1)

Cl. i E

A,

Pi E

B

(l ~ i ~ m).

It is clear that the product of two ideals of a field K is again an ideal of the field K. (When the ideals are integral, the definition of product coincides with the usual notion of the product of two ideals in a commutative ring.) We have already verified that for any divisor 0 of K, the set a is an ideal in K. Assume that for two divisors 0 and b we have a = b. Choose a nonzero element y so that the divisors (y)o and (y)b are both integral. From formula (6.7) we

Sec. 5]

DEDEKIND RINGS

215

have (y)a = (y)b, so that (y)a = (y)b and hence a = b. Hence the mapping a ..... a is a monomorphism. Now let A be any ideal of the field K. If the element l' i' 0 is chosen so that yA c .0, then yA will be a nonzero ideal of the ring .0, and hence by Theorem 5 there exists an integral divisor c such that c = yA. Set a = C(y)-I. Then yA = (y)a = ya, so that A = a. Thus each ideal of the field K is the image of some divisor under the mapping a ..... a. If a and bare two divisors, then, taking elements l' i' 0 and 1" i' 0 so that (y)a and (y')b are integral divisors, we have [from Theorem 5 and formula (6.7)]

dab = (y)a'(y')b = (y)a'(y')b = ya'y'b = yy'ab, so that ab = abo The mapping a ..... a hence is an isomorphism. It follows that the set of all ideals of the field K is a group under multiplication. The unit element in this group is the ring .0 = e. The inverse of the ideal a will be the ideal a-I. We formulate this generalization of Theorem 5. Theorem 6. Let .0 be a Dedekind ring with quotient field K. For every divisor a, denote by a the set of all elements of K which are divisible by a. The mapping a ..... ais an isomorphism of the group of all divisors of the field K onto the group of all ideals of the field K. This mapping takes integral divisors to integral ideals and conversely.

I,

r J

I PROBLEMS

1J

i

1

1. Show that the ring k [x 1of polynomials in one variable over a field k is Dedekind. 2. Let 0 be a Dedekind ring with quotient field k. Show that the integral closure C of the ring 0 in any finite extension of the field k is also Dedekind. 3. Show that any ring which has a theory of divisors with a finite number of prime divisors is Dedekind. 4. Show that a system of congruences

I

I Ii I ~

=

in a Dedekind ring is solvable if and only if IXI IXj (mod 01), i #- j, where hI) is the greatest common divisor of the divisors 01 and OJ. 5. Let C be a Dedekind ring and 0 a divisor of C. Show that the set of those residue classes in Cj 0 which consist of elements relatively prime to 0 is a group under the operation of multiplication. 6. Let f(x) be a polynomial of degree m with coefficients in the Dedekind ring D, with not all coefficients divisible by a prime divisor~. Show that the congruencef(x) == 0 (mod ~) has at most m solutions (noncongruent modulo ~), in C.

l' I

[Chap. 3

THEORY OF DIVISIBILITY

216

7. Let (; be a Dedekind ring, ~ a prime divisor of D, and f(x) a polynomial with coefficients in D. If for some element et E D we have feet)

show that for every

m;;;'

II

.1 !

I I

I'

f'(et)~

2 there exists an element f(O

I! .

= 0 (mod ~), == 0 (mod

~m),

g in

0 (mod

~),

the ring 0 such that

g:="" et (mod

~).

8. Show that in a Dedekind ring every ideal is either principal or is generated by two elements. 9. Let (; be a Dedekind ring with quotient field K. Show that under the isomorphism a ->- a of the group of divisors of the field K onto the group of ideals of the field K, the greatest common divisor of divisors corresponds to the sum of the corresponding ideals, and the least common multiple of divisors corresponds to the intersection of the corresponding ideals. (By the sum A + B of the ideals A and B we mean the set of all sums IX + {J, where IX E A and (J E B.) 10. The ring D = k[x, y I of polynomials in two variables over the field k has unique factorization and hence has a theory of divisors. Show that the ideal A = (x, y) of the ring l: which is generated by the elements x and y does not correspond to any divisor. 11. Show that if 0 is a ring with a theory of divisors D* ->- ~ in which every nonzero ideal of D is of the form a (where a E ~), then D is Dedekind. 12. Let D be a ring in which the nonzero ideals form a semigroup with unique factorization under multiplication. Show that 0 is Dedekind. 13. Let £: be a Dedekind ring with quotient field K. If A and B are ideals of the field K (relative to C), we say that A is divisible by B if there is an integral ideal C such that A = Be. Show that A is divisible by B if and only if A c B. 14. Let 0 be any ring with a theory of divisors and let ~ be a prime divisor. Show that the set ~ of all elements IX E l: which are divisible by \J is a minimal prime ideal of the ring O. (An ideal P in a ring 0 is called prime if the quotient ring OJ P has no divisors of zero, that is, if the product of any two elements of D, which do not lie in P, does not lie in P. The prime ideal P is called minimal if it does not contain any other prime ideal except the zero ideal.) 15. If C is a ring with a theory of divisors, show that any nonzero prime ideal P of D contains a prime ideal of the form j), where ~ is some prime divisor of the ring D.

l

7. Divisors in Algebraic Number Fields

'j, :

7.1. The Absolute Norm of a Divisor

1I

I

By Theorem 2 of Section 5 the maximal order D of any algebraic number field K is a ring with a theory of divisors. Further, we saw in Section 6.1 that the residue class ring Dip modulo a prime divisor p is a finite field, and hence that the ring D is Dedekind. Consider the algebraic number field K as an extension of the field of rational numbers R Coffinite degree). Since the divisors of the ring Z are in one-to-one correspondence with the natural numbers, we can assume that the group of

Sec. 7]

DIVISORS IN ALGEBRAIC NUMBER FIELDS

217

all divisors (integral and fractional) of the field R coincides with the multiplicative group of positive rational numbers. In Section 5.2 we defined the concept of the norm of a divisor of the ring D relative to a given extension Kfk. If a is a divisor of the order D of the algebraic number field K, then we call the norm N(a) = N K/R(a) the absolute norm of a. We extend the concept of absolute norm to fractional divisors by setting

N(~)

=

lJ

N(m) ,

N(lJ)

where m and lJ are integral divisors. The mapping a ..... N(a) will then be a homomorphism from the group of all divisors of the field K to the multiplicative group of positive rational numbers. The absolute norm of a principal divisor (~), ~ E K*, equals the absolute value of the norm of the number ~: N«~))

=

IN(~)I.

Indeed, if ~ is integral this is just (5.3). If ~ = rxf Pwith integral rx and

N(rx)

N(e)

(7.1)

P, then

IN(rx)1

= N(P» = IN(P)I = IN(~)I.

The degree of inertia f of a prime divisor l' of the field K relative to R is called the absolute degree of inertia of l' (or simply the degree of 1'). The ramification index e of the divisor l' relative to R is called the absolute ramification index of p. If l' divides the rational prime l' and if l' has degree f, then by (5.11), N(p) = p'.

(7.2)

Let PI' ... , Pm be all prime divisors of the field K which divide p, and let el' .,. , em be their ramification indices. Then in the field K we have the decomposition By Theorem 7 of Section 5 the ramification indices ei and degrees fi of the divisors Pi are connected by the relation (7.3) Theorem 1. The absolute norm of an integral divisor a of the algebraic number field K is equal to the number of residue classes in the maximal order D modulo a. Proof. We first prove the theorem for a prime divisor p. Let p be the rational prime which is divisible by p. The degree of inertiaf of the divisor l' (by the

I I Ii ! I

I;

I.

[Chap. 3

THEORY OF DIVISIBILITY

218

definition of Section 5.3) equals the degree of the residue class field ~1' of the valuation v 1' over the residue class field ~1' of the valuation v 1" Since L1' clearly consists of p elements, ~1' is a finite field with pI elements. Hence it suffices to show that the fields ~1' and DIp are isomorphic, that is, that the inclusion isomorphism DIp --+ ~1' maps the field DIp onto the entire field L1' • To do this it suffices to show that for any ~ E K for which v1'(O ~ 0, there exists an element IX E D such that v1'(~ - IX) ~ 1. We denote by ql' qs all those prime divisors of the field K for which Vqi(~) = -k i < O. By Theorem 3 of Section 6 there is an element "I in the order D such that 00"

"I == 1 (mod p), "I == 0 (mod q/i),

I;

O=l,oo.,s).

It is clear that IX = y~ ED and V1'(IX - ~) ~ 1. Hence Theorem 1 is proved in the case of prime divisors. To prove the theorem in general it suffices to prove that if it holds for integral divisors a and b, then it also holds for their product abo By condition (3) of Theorem 4 of Section 3 there is an element "I #- 0 in the maximal order D such that alY and the divisor ("I) a -I is relatively prime to b. Let lXI' ... , IX, [r = Na)] be a complete set of residue-class representatives of D modulo the divisor a, and PI' ... , Ps [s = N(b)] be a complete set for b. We shall show that then the rs numbers (7.4) form a complete system of residue-class representatives modulo abo Let any number of D. For some i (1 ~ i ~ r) IX

==

IXi

(mod a).

IXi

(mod ab).

IX

be

Consider the congruence y~

==

IX -

(7.5)

Since by choice of "I the greatest common divisor of the divisors ("I) and ab equals a, and IX - IXi is divisible by a, then by Theorem 4 of Section 6 this congruence has a solution ~ E D. If ~ == Pi (mod b) for some j (1 ~j ~ s), then y~ == yP i (mod ab). Along with (7.5) this shows that IX

==

IXi

+ yP i

(mod ab).

We have proved that every residue class modulo ab has a representative of the form (7.4). We now must show that the numbers (7.4) are pairwise-noncongruent modulo abo Let IXi

+ yP j == IXk + "1131

(mod ab).

I Sec. 7]

DIVISORS IN ALGEBRAIC NUMBER FIELDS

II

219

Since this congruence also holds modulo a, andy == 0 (mod a), we obtain == Ct. k (mod a), and this means that i = k, and we obtain

Ct.!

y(f3j - 13,)

== 0 (mod ab).

(7.6)

Let the prime divisor p occur in the divisors a and b with exponents a and b > O. Since vl'(y) = a, it follows from (7.6) that vl'(f3 j - 131) ?: b. Since this is true for all prime divisors p which occur in b with positive exponent, then f3 j == 131 (mod b), so thatj = I. Hence the numbers (7.4) for a complete set of residue representatives modulo abo The number of residue classes of the ring .0 modulo ab hence equals rs = N(a)N(b) = N(ab). Theorem I is proved. If a is any divisor of the field K (integral or fractional) as in Section 6.3, we denote by a the ideal of the field K, consisting of all Ct. E K which are divisible by a. Let the number "I be chosen so that ya c D. By the corollary of Theorem 2 of Section 2, Chapter 2, the set ya is a module of the field K (a submodule of the ring D). But then the ideal a also is a module of the field K. If Ct. E n, Ct. #- 0, and W l , ... , W n is a basis for the ring .0, then all the products Ct.W l , ... , Ct.W n lie in a, and hence a contains n = (K: R) linearly independent (over R) numbers of the field K. Hence, for any divisor a, the ideal a is a full module of the field K. Its coefficient ring will clearly be the maximal order D. Conversely, if A is a full module of the field K, whose coefficient ring is the maximal order .0, then A fulfills all the conditions of being an ideal (Section 6.4). Thus the set of all ideals a coincides with the set of all full modules of the field K which belong to the maximal order D. In Section 6.1 of Chapter 2 we introduced the concept of the norm of a full module of an algebraic number field. We can therefore speak of the norm of the ideal a. We shall show that the norm of any divisor coincides with the norm of its ideal: (7.7) N(a) = N(a).

i

I.

1

I I J

For integral divisors this follows from Theorem I of this section and Theorem I of Section 6, Chapter 2. If the divisor a is fractional, then we can find a "I E K* such that the divisor ("I -l)a = b is integral. Then by Theorem 2 of Section 6, Chapter 2, we have N(a) = N(b)IN(y)1 = N(b) IN(y) I = N(yb) = N(yb) = N(a),

and (7.7) is proved for all a. As a simple application of the concept of norm we give a more precise estimate for the number w(a) of nonassociate numbers in the maximal order

I I

1!

1

i

220

THEORY OF DIVISIBILITY

[Chap. 3

I

whose norm has absolute value equal to a (in the proof of Theorem 5 of Section 2, Chapter 2, we showed that w(a) ~ a"). Let ljJ(a) denote the number of integral divisors with norm a. Since the numbers a and p are associate if and only if the principal divisors (a) and (p) are equal, then from (7.1) we have w(a)

~

ljJ(a).

We will find an estimate for (a). Let a

=

p/t ... p/.,

where the Pi are distinct primes. If N(a) = a, then a = at ... as' where a i consists of those prime divisors V which divide Pi' By formula (7.2) and the multiplicativity of the norm, we have N(a i ) = p/', and this means that ljJ(a) = ljJ(p/ t ) ... (V/ s ). It therefore suffices to obtain an estimate for ljJ(pk). Let Vt, ... , Vm be the distinct prime divisors which divide p, and letft, .. · ,fm be their degrees. Since N(VtXt ... VmXm)

, I·

I'

the problem reduces to the determination of all solutions of the equation ftXt

°

I'

i II :1

'II,

I, ,f I'

= pJtXt+ '" +JmXm

+ ... + fmxm

= k

in nonnegative Xi' Since ~ Xi ~ k, then the number of solutions cannot exceed (k + 1)m. But m ~ n = (K: R), and thus ljJ(a) ~ «k t

+ 1) '"

(k s

+ 1))".

The expression in parentheses on the right equals, as is well known, the number r(a) of all divisors of a. We have hence obtained the estimate w(a)

~

ljJ(a)

~

(r(a))".

(7.8)

To compare our estimate (7.8) with the previous estimate w(a) ~ a", we note that for any e > 0, the quantity r(a)/a' converges to zero as a--+ 00. 7.2. Divisor Classes

f, .

I'

1

I

I

,I

I: il

Definition. Two divisors a and b of the algebraic number field K are called equivalent, and we write a - b, if they differ by a factor which is a principal divisor: a = b(a), CI. E K*. The set of all divisors of K which are equivalent to a given divisor a, is called a divisor class and denoted by [a].

In the terminology of group theory, the equivalence a - b denotes that the divisors a and b belong to the same coset of the subgroup of all principal divisors, and the divisor class [a] can be defined as the coset of the subgroup

Sec. 7]

DIVISORS IN ALGEBRAIC NUMBER FIELDS

221

of all principal divisors which contains the divisor a. We clearly have [a] = [b] if and only if a ~ b . For any two divisor classes [a] and [b] set [a]'[b]

= [ab].

It is easily verified that this definition is independent of the choice of a and b in these divisor classes, and that under this operation the set of all divisor classes becomes a group, the divisor class group of the field K. The unit element is the class [e], consisting of all principal divisors. The inverse of the class [a] is the class [a-I]. In the terminology of group theory the divisor-class group is the factor group of the group of all divisors by the subgroup of principal divisors. The divisor-class group, and particularly its order, is an important arithmetic invariant of the algebraic number field K. If the number of divisor classes equals I, then this means that every divisor is principal, which is equivalent to the maximal order of the field K having unique factorization (Theorem 2 of Section 3). The question of whether the algebraic integers of the number field K have unique factorization is hence a part of the problem of determining the number of divisor classes of this field. We shall now show that this number is always finite. Theorem 2. The divisor class group of an algebraic number field is a

finite group. Proof From the definition of equivalence of divisors it easily follows that

the divisors a and b are equivalent if and only if the corresponding ideals a and Ii are similar (in the sense of similarity of modules; Section 1.3 of Chapter 2). The partitioning of divisors into classes of equivalent divisors hence corresponds to the partitioning of ideals of the field K (that is, of full modules whose coefficient ring is the maximal order of the field K) into classes of similar ideals. By Theorem 3 of Section 6, Chapter 2, the number of classes of similar modules with given coefficient ring is finite. Thus the number of classes of similar ideals, and the number of classes of equivalent divisors, are also finite. Remark 1. Theorem 2 was obtained as a simple corollary of Theorem 3 of Section 6, Chapter 2. The proof of the latter theorem was based on geometric considerations, in particular, on Minkowski's lemma on convex bodies. Hence, the proof of Theorem 2 is also based on Minkowski's lemma. Remark 2. From the proof of Theorem 3 of Section 6, Chapter 2, we can deduce the following strengthening of Theorem 2. In each divisor class

222

[Chap. 3

THEORY OF DIVISIBILITY

of an algebraic number field K of degree n = s + 2t, ther.e exists an integral divisor with norm (2jrrY where D is the discriminant of the field K (that is, the discriminant of the ring of integers of the field K). Let [b] be any divisor class. Then there is an ideal A = ab -1, similar to the ideal r;-=t, for which A ::::J .0 and (A : D) ~ (2jrrY (see the proof of Theorem 3 of Section 6, Chapter 2). Since the ideal A contains .0, its inverse will be integral: A = a=t with integral a. From a -1 = aP it follows that a(a) = b, that is, that the integral divisor a is contained in the class [b], and here (Problem 2)

v'IDI,

v'1DT

N(a)

N(e)

-

_

= N(a_1)=(a-1:e)=(A:.o)~

(2)1 !InI ; ylDj.

Theorem 3. If the divisor-class group of the field K has order h, then the hth power of any divisor is principal. Proof. The assertion of the theorem is a simple corollary of elementary

group theory. The order of every element of a finite group divides the order of the group. Let a be any divisor. Since [a]h is the unit element of the divisorclass group, then [a h ] = [e], and this means that the divisor a h is ·principal.

1

I

I' I

I

I

,I,

I:

I: f !

'I' ,

j

Corollary. If the number h of divisor classes of the field K is not divisible by the prime number I, and if the divisor a' is principal, then a is also principal.

Since 1 and h are relatively prime, we can find rational integers u and v such that lu + hv = I. Since the divisors a' and a h are principal (the first by assumption, and the second by Theorem 3), it follows that a1u and ahv are also principal. But then so is the product a1u + hv = a. By Problem 20, every algebraic number field K can be embedded in a larger algebraic number field K, so that every divisor of the field K will be principal in K. We cannot, however, assert that every divisor of the field K is principal. Moreover, it has recently been shown (by Golod and Shafarevich) that there exist algebraic number fields, for example, K= R(v'-3·5·7·IC·-i3·17·19), which are not contained in any extension field with h = I. The following question is still open: Are there infinitely many algebraic number fields with h = I? Examination of tables show that such fields occur rather frequently (see the tables of h for real quadratic fields and totally real cubic fields). For certain classes of fields (for example, for quadratic and cyclotomic fields, see Chapter 5) formulas for the number of divisor classes have been found, but in the general case little is known about h and the divisor-class group. Among the few general theorems about the number h is the theorem of Siegel and Brauer, which asserts that for all fields with a fixed degree n,

Sec. 7]

DIViSORS IN ALGEBRAIC NUMBER FIELDS

223

the number h of divisor classes, the regulator R, and the discriminant Dare related by the following asymptotic formula:

In (hR) InJID/

for IDI .....

--.,...,= ..... 1

OCJ

(*)

[R. Brauer, On the zeta functions of algebraic number fields, Am. J. Math, 69, No.2, 243-250 (1947)]. Since for imaginary quadratic fields the regulator of the field is equal to 1, then it follows from (*) that as IDI --+ 00, so does h --+ 00. In particular, we may deduce that there are only a finite number of imaginary quadratic fields with h = 1. In tables we see nine imaginary quadratic fields with h = 1 (their discriminants are - 3, - 4, - 7, - 8, -11, -19, - 43, - 67, -163). It is known that there is at most one more imaginary quadratic field with h = 1. It is not known whether or not it exists. In the general case we can say almost nothing from (*) about the behavior of the number h, since we know very little about the value of the regulator R.

7.3. Applications to Fermat's Theorem The results of the preceding section allow us to prove the validity of Theorem 1 of Section 1 for a much wider class of exponents I. Theorem 4. Let 1 be an odd prime and let ( be a primitive Ith root of 1. If the number of divisor classes of the field R(O is not divisible by I, then the first case of Fermat's theorem holds for the exponent I. Proof. Assume that, contrary to the theorem, there exist rational integers

x, Y, and z, not divisible by I, and satisfying the equation Xl

We may further assume that ring of integers of the field

X,

+ yl = Zl.

y, and z are pairwise relatively prime. In the

Reo our equation can be written in the form 1-[

n (x +

(k y )

=

Zl.

kzO

Since X + Y == Xl + yl = Zl == z (mod I) and z is not divisible by I, then x + y is also not divisible by I. Then, as we proved in Lemma 5 of Section 1, for m t= n (mod I) there are numbers ';0 and rJo in the ring Z[(] such that

+ ("y)';o + (x + (mY)rJo = 1. divisors (x + (ky ) (k = 0, 1, ... ,I (x

Hence the principal 1) are pairwise relatively prime. Since their product is an Ith power [of the divisor (z)], then each of them separately must be an lth power. In particular, (x

+ (y)

= ai,

224

THEORY OF DIVISIBILITY

[Chap. 3

where a is an integral divisor of the field R(O. Since we have assumed that the number of divisor classes of the field R(O is not divisible by I, it follows from the corollary to Theorem 3 that the divisor a is principal; that is, a = (a), where a belongs to the maximal order .0 = Z[(] of the field R(O. From the equality

it now follows that

x where

G

+ (y = ea l ,

is a unit of the ring D. Analogously we obtain

x - (z = (a 1 ED,

~

•

;

. j I

i

I'

I·

. I

I

I

I.

r i

I

Glal1

Gl a unit in D). We have reached equations which, as was shown in Section 1.3, lead to a contradiction (in that part of the proof of Theorem 1 of Section 1 unique factorization was not used). Hence Theorem 4 is proved. Those odd primes I for which the number of divisor classes of the field R(O, (I = I, is not divisible by I, are called regular primes, and all others are caned irregular. By very beautiful number-theoretic and analytic arguments Kummer obtained a fairly simple criterion (which we shan present in Section 6.4 of Chapter 5), which allows one to check easily whether a given prime I is regular or not. Using this method it can be verified that among the prime numbers < 100 only three, 37, 59, and 67, are irregular, and all the rest are regular. To show how much broader the class ofexponents for which Theorem 4 holds is than the class for which Theorem 1 of Section 1 holds, we note that among the prime numbers < 100, only for the first seven, 3, 5, 7, 11, 13, 17, 19, do we have unique factorization in the ring .0 := Z[(] where (I = 1. In his first paper Kummer stated the hypothesis that the number of irregular primes was finite. He later retracted this and hypothesized that regular primes occur twice as frequently as irregular ones. With the aid of electronic computers, it has been shown that of the 550 odd primes ~ 4001, there are 334 regular ones and 216 irregular ones. A table of an irregular primes ~ 4001 is given at the end of the book. Jensen (see Section 7.2 of Chapter 5) showed that the number of irregular primes is infinite. It is not known whether there are infinitely many regular primes; there are no indications that there are only finitely many. The first case of Fermat's theorem for the exponent I is also connected with the number ho of divisor classes of the field R«( + C 1) = R[2 cos (2njl)]. It is easily seen that R«( + C 1 ) consists of all real numbers of the field R(O. Vandiver showed that if the number ho of divisor classes of the field R«( + (-1) is not divisible by I, then the first case of Fermat's theorem is valid for the exponent I [H. S. Vandiver, Fermat's last theorem and the second factor in

Sec. 7]

DIVISORS IN ALGEBRAIC NUMBER FIELDS

225

the cyclotomic class number, Bull. Am. Math. Soc. 40, No.2, 118-126 (1934)]. It is not known if there exist prime numbers / for which the number h o is divisible by /. It has only been verified that there are none among numbers ~ 4001. We note here some other results which bear on the first case of Fermat's theorem. Wieferich showed that the first case of Fermat's theorem is valid for all primes / such that 21- 1 1= 1 (mod /2) [A. Wieferich, Zum letzen Fermatschen Theorem, J. Math. 136,293-302 (1909)]. To indicate the strength of this result we note that only two prime numbers / ~ 200, 183, namely, 1093 and 3511, satisfy the condition 2 1- 1 == 1 (mod /2) [Erna H. Pearson, Math. Camp. 17, No. 82, 194-195 (1963)]. However, it is not known whether there are infinitely many such /. Several other authors have shown that the first case of Fermat's theorem holds for all/ such that ql-1 1= 1 (mod /2) for some prime number q ~ 43 (D. Mirimanoff, H. S. Vandiver, G. Frobenius, F. Pollaczek, T. Morishima, J. B. Rosser). This has made it possible to verify the first case of Fermat's theorem for all prime numbers 1. S follows:

Hinl: In the

e

25. Let be polynomial 9l(1 that, modulo p

where 9l" ... , ' Show that the I

:Chap.3

luding e

Use integral

t D.

I

-7,

Sec. 7]

DIVISORS IN ALGEBRAIC NUMBER FIELDS

19. Let a be a divisor of the algebraic number field K, such that am = (a) is a principal divisor. Show that the divisor a becomes principal in the field K(m~Fi). 20. Show that for any algebraic number field K there is a finite extension K/ K such that every divisor a of the field K becomes principal in the field K. 21. Let Kbe a cubic field andp a prime which factors as a product of three distinct prime divisors in K: p = \)\)'\)". If IX is an integer in K with Sp(IX) = 0 and \)\)'IIX, show that \)HI IX and hence pllX. 22. Show that the field R(8), 8 3 = 6, has only one divisor class. [By Problem 24 of Section 2, Chapter 2, the numbers 1,8,8 2 form a fundamental basis for the field R(B).] 23. Let K be the cubic field K = R(B), 8 3 = 6. Show that there is no number IX #- 0 of K of the form IX = X + y8, x and y relatively prime rational integers, for which N(IX) = 10z 3 (z a rational integer). Deduce that the equation x 3 + 6y 3 = IOz 3 (and hence also the equation 3x 3 + 4y 3 + 5z 3 = 0) has no nontrivial solution in rational integers. Hint: Assume that the number IX exists and show that it must have the form IX = IXog3, where g is an integer of the field K and IXo is one of the following six numbers: AfL, AfLE, AfLE 2, AV, AVE, AVE 2.

-s,

Dot of 1,

Here A=2-8 [N(A) = 2]; fL=8-1 [N(fL) = 5]; v=(8 2 +8+1)2=13+S8+38 2 [N(v) = 5'5 3 ]; and E = 1 - 68 + 38 2 is a fundamental unit of the field K (Problem 4 of Section 5, Chapter 2). For the proof use Problem 21, applied to the number IX8, Problems 17 and 22, and also prime factorizations in the field K of the numbers 2, 3, and 5. Further, setting g = u + v8 + w8 2 , write

:ction 2, K 2 , and

= \)\)'\)H.

visors in K" K 2 ,

lative to l1e index

IX = IXoe =

if Hint: In the case d

.lgebraic >t divide ==IX s of the

e

+ 'Y8 +

n8 2 ,

where , 'Y, and n are integral cubic forms in the variables u, v, and w. Show that for any of the six values of IXo the equation n(u, v, w) = 0 has only the trivial solution in rational (and in 3-adic) numbers.] 24. Let a and b be natural numbers which are square-free and relatively prime, and let d = ab 2 > I. Show that in the field R(f/d) the number 3 factors into prime divisors as follows: if do;t ±I (mod 9),

field K. w thatp

ramified r) if and

233

d== ±I (mod 9).

== ± 1 (mod 9) consider the norms w

1

=

"3 (I

+a

~

~ ab 2

+T

N(w - 1), N(w), N(w

+ I),where

~ a 2 b),

a=±l, T=±l, aa==Tb==1(mod3).

25. Let 8 be an integral primitive element of the algebraic number field K, with minimum polynomial ep(t), and let p be a rational prime which does not divide the index of 8. Suppose that, modulo p, we have the factorization

where ep" ... , epm are distinct irreducible polynomials modulo p with degrees Show that the decomposition of the number p in the field K takes the form

I" ... .fm.

234

THEORY OF DIVISIBILITY

[Chap. 3

where the distinct prime divisors \)" ... , \)m have degrees II> ... ,1m and where lptCO) "'" 0 (mod \)1) for i = I, ... , m. Hint: Use the fact that every integer of K is congruent modulo \)1 to a linear combination (with rational integer coefficients) of the powers 0' (s ;;;. 0). 26. Let 0 be an integral primitive element of the field K, and p a prime number which does not divide the index of O. For any rational integer x, show that the number 0 + x is not divisible by any prime divisor which divides p and is of degree greater than I. Further, show that 0 + x is not divisible by the product of any two distinct prime divisors which divide p. 27. Under the same assumptions, let \)" ... , \), be distinct prime divisors which divide p, with degrees I" ... ,Is, and let r < I, + .., +Is. If xo, ... ,xr - , are any rational integers, show that the number or + xr_IOr -' + ... + Xo is not divisible by the product \)1 ... \),.

8. Quadratic Fields

In this section we consider the somewhat simpler theory of divisors for the case of quadratic fields. We start with a description of the prime divisors. , I

I'

8.1. Prime Divisors

Ii I;

Since every prime divisor divides one and only one rational prime, to give a complete description of the set of all prime divisors it suffices to show how each rational prime p factors as a product of prime divisors. For quadratic fields it follows from (7.3) that there are only three possibilities for the numbers m,li' ei: (1) m = 2, it =12 = 1, et = e 2 = 1; (2) m = 1, 1= 2, e = 1; (3) m = 1, 1= 1, e = 2. Corresponding to these we have the following types of decomposition: (1) p = 1'1", N(p) = N(p') = p, l' #- 1"; (2) p = 1', N(p) =p2; (3) p = lJ2, N(p) =p. Our problem is to discover what determines the type of factorization for any prime p. The answer will be easily derived from Theorem 8 of Section 5. In Section 7.1 of Chapter 2 we showed that every quadratic field has a unique representation in the form R( .jd), where d is a square-free rational integer. First, let p be an odd prime. If p does not divide d, then it also does not divide the discriminant of the polynomial x 2 - d, a root of which generates the field. We then deduce from Theorem 8 of Section 5 that p has a decomposition of either the first or the second type, depending on whether the polynomial x 2 - d is reducible modulo p or not. This in turn depends on whether d is a quadratic residue modulo p or not.

Ii

I

j

I

Sec.B]

If pld, then d It follows from

QUADRATIC FIELDS

235

= pd1 , where d 1 is not divisible by p, since d is square-free.

that all prime divisors which divide p occur with an even exponent in its factorization, which is possible only for the third type of decomposition. Thus for odd p we have the first, second, or third type of decomposition in the following three cases, respectively: (1) p,r d, (dip) = 1; (2) p,r d, (dip) = - 1; (3) pld. Note that since the discriminant D of the field R( .Jd) is either d or 4d (Theorem 1 of Section 7, Chapter 2), then in each of these conditions we could replace d by D. The case p = 2 remains. Assume first that 2,r D. By Theorem 1 of Section 7, Chapter 2, this means that D = d == 1 (mod 4). It is clear that R( .Jd) = R(co), where co = (- 1 + .J D)/2. The minimum polynomial of co is

1-D x 2 +x+ _ _ 4

(8.1)

Since the discriminant of the basis 1, co is odd, we obtain from Theorem 8 of Section 5 that the number 2 will have either the first or the second type of decomposition, depending on whether the polynomial (8.1) is reducible or not. But the polynomial x 2 + x + a is reducible modulo 2 if and only if 21a. Thus for 2,r D we obtain the first and the second types of decomposi tion in the respective cases D == 1 (mod 8) and D == 5 (mod 8). We now show that if 21 D, then 2 always has the third type of decomposition. If 21d, then d = 2d', 2,r d', and from

just as in the case of odd p, we see that 2 has the third type of decomposition. If 2,r d, then d == 3 (mod 4) (Theorem 1 of Section 7 of Chapter 2) and we have (1 with the integer et = (1

+ d)/2 +

+ .JJ)2 = 2et

.JJ relatively prime to 2, since its norm

is not divisible by 2. Thus we again obtain the third type of decomposition for 2. We have obtained the following theorem.

THEORY OF DIVISIBILITY

236

[Chap. 3

Theorem 1. In a quadratic field with discriminant D the prime number p has the decomposition

p = p2,

N(p) = p,

if and only if p divides D. If p is odd and does not divide D, then p

= pp', p #- p',

N(p)

=

p=p,

N(p')

=

P

N(p)=p2

for (;) = 1; for

(~)

= - 1.

If 2 does not divide D [and hence D == I (mod 4)], then p #- p',

2 = pp',

N(p) = N(p') = 2 2

= p,

N(p)

=4

for D = I (mod 8); for D = 5 (mod 8).

;I t

'Ii

I'

8.2. Rules of Decomposition

Theorem 1teIls us that the type of decomposition of the odd prime p is determined by the residue of D (or d) modulo p, and even by the Legendre symbol (Dip) = (dip) as a function of p. It is natural to ask if the theorem can be reformulated so that the decomposition depends on the residue of p with respect to some modulus (the modulus depending only on the field). To achieve such a formulation, we use the reciprocity law for the Jacobi symbol. It is weIl known that the Jacobi symbol (clb) is defined for odd c and positive odd b, relatively prime to c. The reciprocity law for this symbol states that

(~)

= ( - I)[(b - 1)/2]' [(c -

1)/2]C:I)

(the proof for c < 0 is easily reduced to the positive case). Let p be any odd prime. If d = D == I (mod 4), then

(~) = (~)

= (-

1)[(p - 1)/2]' [Cd -

1)/2](1~1)

=

C~I)'

(8.2)

since (d - 1)/2 is even. If d == 3 (mod 4), then

(f) = (~) = ( - I)[(P-l)/21'[(d-l)/21C~I) = ( -

I)(P-l)/2C~I)'

(8.3)

since (d - 1)/2 is odd. FinaIly, for d = 2d', 2,td', we have

(f)

=

(~)

=

e)(~)

= (_1)[(P -1)/Sl+[(P-l)/2 1[(d'-I)/2 1C:'I)' 2

(8.4)

Sec.B]

QUADRATIC FIELDS

237

The value of the Jacobi symbol (plldl) [or (p/ld'l)] depends only on the residue of p modulo jdl (or Id'l). If d == I (mod 4), so that d = D, then (Dip) depends only on the residue of p modulo Idl = IDI. If d == 3 (mod 4), so that D = 4d, then (Dip) depends not only on the residue of p modulo Idl, but also on the number (_1)(P-l)/2, that is, on the residue of p modulo 4; hence (Dip) depends onthe residue ofp modulo 41dl = IDI.Finally,if d = 2d',D = 4d=8d', then (Dip) depends on the residue of p modulo Id'l, (_1)(P-l)/2 depends on the residue ofp modulo 4, and ( - I )(p2 - 1 )/8 depends on the residue ofp modulo 8. Hence in this case the value of (Dip) depends on the residue of p modulo 81d'l = IDI· Thus in all cases the type of decomposition of the prime number p depends only on its residue modulo IDI, so that all prime numbers having the same residue have the same decomposition. This conclusion, which is completely nonobvious a priori, is the most important property of the decomposition rules for prime numbers in quadratic fields. To make this new form of the decomposition rule more clear, we introduce a new function. For all x relatively prime to the discriminant D, we set

X(x)

=

( _

(1:1)

for d == 1 (mod 4),

1)(X-l)/2(~)

for d == 3 (mod 4),

Idl

( -

2

1)[(x -1)/8]+[(X-l)/2]' [(d'-l) I2J

C;,,)

(8.5)

for d = 2d'

[in case d == 2,3 (mod 4) the expressions (_1)(x-l)/2 and (_1)(x 2 -1)/8 make sense since the discriminant D = 4d is even and so x is odd]. In the arguments above which showed that for odd p the value of (Dip) depends only on the residue of p modulo IDI, we never used the fact that p was prime. Hence by the same arguments it follows that X(x) depends only on the residue of x modulo 1DI. Further, it is easily verified that if (x, D) = I and (x', D) = I, then X(xx') = X(x)X(x'). This means that the function X can be considered as a homomorphism of the multiplicative group of residue classes modulo 1DI, relatively prime to IDI, to the group of order 2 consisting of + I and - I. If we extend such a function by giving it to the value 0 on all numbers not relatively prime to D, it is called a numerical character. Definition. The numerical character X with modulus IDI, where the value of X(x) for x relatively prime to D is given by (8.5), is called the character of the field R( .Jd).

Returning to (8.2), (8.3) and (8.4), we see that the decomposition of an odd prime p which does not divide D will be of the first type if X(p) = + 1 and of

238

IiiI ~

Ii

H

II. II

II

I'

[Chap. 3

the second type if X(p) = -1. This result remains true for p = 2. For if 2,r D, then D == 1 (mod 4) and this means that X(2) = (2f1D!), which equals + 1 for D == 1 (mod 8) and - I for D == 5 (mod 8). Hence we have the fol1owing new formulation of the rule for decomposition in quadratic fields.

I

~f ' ,

Ii

THEORY OF DIVISIBILITY

Theorem 2. If X is the character of the quadratic field R( .jd), then the decomposition of a prime pin R( /d) is given by the conditions: p

= pp', p #' p', p=p,

P

= p2,

N(p)

= N(p') = P

N(p)=p2, N(p)

= p,

if X(p) = 1; ifX(p)

= -1;

if X(p) = O.

AI1 rational integers are partitioned into three sets, depending on the value of X. Each of these sets consists of certain residue classes modulo IDI. By Theorem 2 the type of decomposition of p depends only on which of the three sets contains p. A law of decomposition like that in quadratic fields, where -the type of decomposition depends only on the residue of the prime p with respect to a certain fixed modulus, also occurs for certain other fields. This is the case, for example, for cyclotomic fields (see Section 2.2. of Chapter 5). But it is far from being the case in general. Since the knowledge of such a law of decomposition al10ws us to solve many number-theoretic problems (see, for example, the fol1owing section and Section 2 of Chapter 5), it would be interesting to know for precisely which fields we have such a simple law of decomposition. The answer to this question leads into class field theory. It can be shown that any such field is a normal extension of the field of rational numbers, the Galois group of which is Abelian. Among such fields lie, of course, quadratic fields, which have a cyclic group of order 2 as Galois group. The simplest examples of non-Abelian fields are cubic fields whose discriminant is not a perfect square. An example is the field R(O), where (J3 - 0 - 1 = O. Hence for this field there does not exist any integer M such that the type of decomposition into prime divisors of the prime number p depends only on the residue of p modulo M. Class field theory solves much more general problems than those we have mentioned. It al10ws one to describe the law of decomposition for prime divisors of an arbitrary algebraic number field k in some extension K/k, provided that the Galois group of this extension is Abelian (we spoke above of the special case when k = R). Class field theory has many number-theoretic applications. It al10ws us to carryover the theorem on quadratic forms with rational coefficients, proved in Chapter 1, to the case of quadratic forms with

i

240

THEORY OF DIVISIBILITY

[Chap. 3

of strict divisor classes is also finite, and is related to the number h of divisor classes in the usual sense by ford jo, then

Sec. 1 ]

FIELDS COMPLETE WITH RESPECT TO A VALUATION

261

Hence v(y) ~ min v(aijw j 1t j ) > jo. j*jo

This contradiction shows the linear independence of the elements (1.10) over the field k. Now let r:J. be any element of .0. By the corollary of Theorem 1 we have r:J.

==

~o

+ ~11t + .. , + ~e_l1te-1

(mod

e 1t ),

where ~ j are elements of some fixed system S of residues, which we may take to consist of numbers of the form (1.9). Since 1to and 1t e are associate in .0 [see (1.8)], then congruences in .0 modulo 1to and modulo 1t e are equivalent. Hence we have f.-I 0 )W 1t j ( mo d 1to, ) r:J. - " L, " L,(aij i i= 1 j=o

=

and this means that

Analogously,

Continuing this process indefinitely, we obtain a sequence of equations fe-I _"

r:J... -

"(n)

L, L, a ij W i 1t i= 1 j=O

j

+ 1t Or:J. n + b

(a..n+lE.o,

For fixed i andj we have an infinite sequence {aiJ}. Consider the series 00

"L,

a!n)~ n IJ "0 .

n=O

Since the aWare integers, this series converges and its sum element of k; that is, aij E o. We shall show that

aij

is an integral

(1.11 ) By the construction of the elements

r:J. 1 , r:J. 2 , •••

from which it follows that the difference

we have

262

LOCAL METHODS

[Chap. 4

is divisible by non (in the ring .0). Since this holds for all n, this difference must be zero and (1.11) is valid. If f3 is any element of K, then for some m, the element f3nom will be integral. Representing it in the form (1.11), we see that f3 is a linear combination of the elements (1.10) with coefficients in k. Hence the system (1.10) is a basis for K over k, and since for integral a. E K all coefficients aij in (1.11) lie in 0, this basis is fundamental. Theorem 4 is proved. Since there arefe elements in the basis (1.10), we also have the following result. Theorem 5. The product of the ramification index and the degree of inertia is equal to the degree n = (K : k); that is,

fe= n.

Set NK/k(n) = nomu, where u is a unit of the ring o. Taking the norm of (l.8), we obtain

!

I

II

where v is also a unit in o. It follows that n = me (and v = 1), and hence m = f Hence the degree of inertiaf of the extension Klk could also be defined by (1.12) where n is a prime element of the ring of integral elements of K. From this it easily follows that for any a. of the field K we have (1.13)

I I

'I

if I

I I

!

I

Note that Theorem 5 and (1.12) are both immediate corollaries of Theorem 5 and (5.12) of Section 5, Chapter 3. Definition. If e = 1, the extension Klk is called unramified. If e Klk is called totally ramified.

= n, then

It follows from Theorem 5 that for unramified extensions the degree of inertia coincides with the degree of the extension. For totally ramified extensions the residue class fields ~ and ~o coincide; that is, every integral element of K is congruent modulo n with an integral element of k. It can be shown (Problem 12) that if the residue class field ~ of the field K is separable over the residue class field ~o of the field k, then there is a uniquely determined intermediate field T, such that the extension Tlk is unramified and KITis totally ramified. The field Tis called the inertiafield of the extension

Klk.

Sec. 1 ]

FIELDS COMPLETE WITH RESPECT TO A VALUATION

263

1.5. Fields of Formal Power Series

Fields of formal power series are fields, complete with respect JO a valuation. These fields are constructed in the following manner. Let k o be any field. The set 0 of all formal power series of the type (1.14) becomes a commutative ring with unit when the operations of addition and multiplication are defined as is usual for power series. This ring has no divisors ofzero, and the units are precisely those series (1.14) in which ao -::j. O. The quotient field of 0 is called the field offormal power series in t over the field k o . It is denoted by ko{t}. Just as for the field of p-adic numbers (Section 3.3 of Chapter 1), every nonzero element ~ of the field ko{t} has a unique representation in the form

where m is some integer (positive, negative, or zero). Setting v(~) = m for ~ -::j. 0 and v(O) = 00, we obtain a valuation v, and the field ko{t} is easily shown to be complete under v. The ring of the valuation v coincides with the ring 0 of series of the type (1.14). As prime element in 0 we may take t. Since two series of the form (1.14) are congruent modulo t if and only if their initial terms coincide, then each residue class of modulo t contains a unique element of k o . Hence the residue class field lo of the field k o{ t} is canonically isomorphic to the field k o . It is easily seen that the field of formal power series ko{t} is a completion of the field of rational functions ko(t), under the valuation corresponding to the irreducible polynomial t of the ring ko[t] (see Problem 7 of Section 4 of Chapter 1). Since k o c k o{t} and k o ~ lo, the characteristic of a field of formal power series coincides with the characteristic of its residue class field. It can be shown that this property characterizes formal power series fields among all fields which are complete under a valuation. Namely, if k is complete under a valuation and its characteristic equals the characteristic of its residue class field, then k contains a subfield k o , where the elements of k o form a complete system of residues modulo the prime element 71:. But for such a system of residues, the operations with series (1.3) reduce to the usual operations with power series, and hence k will be a field of formal power series in n with coefficients from k o . The proof of the existence of the subfield k o in the general case is rather difficult, and we shall not give it. (Two special cases, for which the proof is relatively easy, are indicated in Problems 7 and 11.) If k o' is an extension of k o , then ko'{t} is an extension of ko{t}, and if k o' /k o is finite, then k o'{t} /k o{t} is also finite and has the same degree. Another

LOCAL METHODS

264

[Chap. 4

method for constructing finite extensions of the field ko{t} is to map it isomorphically into the field ko{u}, with t --+ un (n a natural number). If we identify k o{ t} with its image under this mapping, that is, if we set t = un, then ko{u} will be a finite extension of ko{t} of degree n. It is clear that ko{u} is obtained from ko{t} by adjoining an nth root of t. For fields of characteristic zero, all finite extensions can be reduced to these two types. Theorem 6. Let k o be a field of characteristic zero. If K is a finite extension of the field k = ko{t}, with ramification index e, then K is a subfield of an extension of the form ko'{u}, where k o' is a finite extension of k o and ue = t. ~o and ~ denote the residue class fields of k and K, let f denote the degree of inertia of Kfk, let n be a prime element of K, and for any ~ E K let ~ be its residue class in~. The elements of the field k o , as we have seen, form a natural system of representatives for the residue classes of ~o. We first show that there is a subfield S of K which contains k o and is a complete system of representatives for the residue classes of~. Since any finite extension of a field of characteristic zero is simple, then ~ = ~o(~), where ~ is some residue class of ~. Let F be the minimum polynomial of ~ over ~o. If we replace all coefficients of F (which are residue classes of ~o) by the corresponding elements of k o , we obtain an irreducible polynomial F over k o of degree f, for which

Proof. Let

F(~)

== 0 (mod n)

and

F'(~)

:f. 0 (mod n).

By the second remark at the end of Section 1.2, there is an element () in the field K with () = ~ and F«(}) = O. Consider the subfield S = k o«(}) of the field K. Since () is a root of an irreducible polynomial over k o of degreef, then (S : k o) =f, and every element of S has a unique representation in the form ao+a 1() +"'+G f - 1(} f- 1

(

GiE k) '0'

the corresponding residue classes modulo n are given by ii o + ~ = ~o(~) and (~: ~o) =f, these linear combinations run without repetition through all residue classes of~. This shows that the elements of the subfield S (which is a finite extension of the field k o) form a complete system of representatives for the residue classes of~. By Theorem 1 the field K is the field of formal power series in n with coefficients in S; that is, K = S{n}. Theorem 6 would be proved (in a stronger form) if we could show that n can be chosen so that it is an eth root of t. However, it is not always possible to choose n in this way, and therefore we must pass to some finite extension k o' of the field of coefficients S. By (1.8) we have Since () = iil~

~,

+ ... + iif_l~f-l. Since

I

=

nee,

(1.15)

Sec. 1 ]

FIELDS COMPLETE WITH RESPECT TO A VALUATION

265

where E is a unit in the ring of integral elements of K. Let IX denote that element of S for which IX == E (mod n) and denote by k o' the field S(~a) (if IX = y' for some YES, then k o' = S). The field of formal power series K' = ko'{n} clearly contains K as a subfield and is a finite extension of k. We show that it can be represented in the form ko'{u}, where u' = t. Consider the polynomial G(X) = xe - E. Since in K' we have G(y) == 0 (mod n)

and

G'(y)

"I- 0 (mod n),

where by y we denote the root ~a., then in K' there exists a unit 1'/ with 1'/ == y (mod n) and 1'/' = E (here we again apply the remark at the end of Section 2.2). Now replace the prime element n of the field K' by the element u = n1'/. Then K' can also be considered as the field of formal power series in U over the field k o', that is, K' = ko'{u}, where u' = t by (1.15). The proof of Theorem 6 is complete. Remark. Theorem 6 is no longer valid for arbitrary finite extensions of formal power series fields k = ko{t} of characteristic p "# O. However, it is easily seen that it remains true in the case of extensions Kfk, for which the residue class field ~ is separable over ~o and the ramification index e is not divisible by p.

PROBLEMS 1. A nontrivial metric rp of a field k is called discrete if the only limit point for the set of values rp(x), x E k, is zero. Show that any discrete metric is induced by some valuation v of k by (1.1). 2. Let k be complete under a valuation and let K/k be a finite extension with a fundamental basis 8" ... ,8•. Show that the elements

also form a fundamental basis for Kover k if and only if all a'J are integral and the determinant det(a'J) is a unit in k. /

3. Using the notations of Theorem 4, let oc

=

e-l

L L a'Jw,TTJ (al) E k) 1=, J=O

be any element

of K. Set m = min vo(a'J)' Show that if jo is the smallest value of the indexj for which there exists i = i o with vo(a'oJo) = m, then v(oc) = jo + em, where v is the valuation of the field K. 4. Show that every element of the field of formal power series k o {t}, not lying in k o • is transcendental over k o .

266

LOCAL METHODS

[Chap. 4

5. Under the assumptions of Theorem 6, show that the subfield S c K, which contains k o and is a complete set of representatives for the residue class field of K, is uniquely determined. 6. Let k be algebraically closed and of characteristic zero, and let k = k o {t}. Show that k has one and only one extension of degree n, for all natural numbers n, namely, k(':jt) (that is, show that any two extensions of degree n are isomorphic under an isomorphism which is the identity on k). 7. Let K be complete under a valuation with residue class field 1: of characteristic zero. Show that K contains a subfield S which is a full system of representatives for the residue classes of 1:, and hence that K = S {1T}, where 1T is a prime element of K. (Use the fact that any field can be obtained from the prime field by a pure transcendental extension followed by an algebraic extension.) 8. Under the assumptions of Problem 7, assume also that the> residue class field 1: is algebraic over the prime field. Show that the subfield S is then unique. 9. Let K be complete under a valuation with residue class field 1:. If 1: is a perfect field of characteristic p (which means that every element is a pth power, and the map which takes every element to its pth power is an automorphism), show that there is a unique" multiplicatively closed" system S of representatives of the residue classes of 1:, so that if IX E Sand f3 E S, then IXf3 E S. (Dete!:.mine IX E S, representing the class (E 1:, by IX = lim IX. P·, where IX. belongs to the class

e-..)

10. Under the same notations, assume that 1: is a finite field with pf elements. Show that in the field K the polynomial t Pf - t factors into linear factors and that the set of its roots is a multiplicatively closed system S of representatives of the residue classes of 1:. 11. Assume that the field K of Problem 9 also has characteristic p, and that its residue class field 1: is perfect. Show that then the multiplicatively closed system S will also be additively closed, so that it will be a subfield of K, and K = S {1T}, where 1T is a prime element of K. 12. Let k be complete under a valuation and let K be a finite extension. Assume that the residue class field 1: of K is separable over the residue class field 1: 0 of k. Show that among the intermediate fields L, k c L c K, which are unramified over k, there is a largest such field T (which contains all other intermediate fields which are unramified over k). Verify that the residue class field of T coincides with 1:, and that the degree (T: k) equals

(1: : 1: 0 ). 13. Let f(X) = x m + a,xm -1 + ... + a m be an irreducible polynomial over a field complete with respect to a valuation. If the constant term am is integral, show that all other coefficients a" ... , am -1 are also integral. 14. Let ~ be a primitive root of degree p' of 1 (s ~ 1). Show that the degree of the field over the field R p of p-adic numbers is (p - I)p' -1, and that the extension Rp(~)/R p is totally ramified. Rp(~)

p-l

15. Let ~ be a primitive root of degree p of 1. Show that Rp(~) = RpCv' - p). 16. Let k be complete under a valuation, K/k a finite extension, and 1: and 1: 0 the residue class fields of K and k. Show that if 1:/1: 0 is separable, then K/k has a fundamental basis consisting of powers of a single element (that is, D = ()[8], 8 E D, where D and 0 are the rings of integral elements of K and k). Hint: Show that if 1: = 1: 0 (8), then a representative 8 E D can be chosen so that f(8) is a prime element in D. Here f(t) E o(t) is chosen so that i(t) E 1: 0 (t) is the minimum polynomial of the element 8 E 1:.

Sec. 2]

FINITE EXTENSIONS OF FIELDS WITH VALUATIONS

267 <X>

17. In a field complete under a valuation, show that the infinite product II (I

+ an),

11=1

an # -1. converges if and only if an -+ 0 as n -+ 00.

2. Finite Extensions of Fields with Valuations

Let k be a field with a valuation vp and let Kjk be a finite extension. The ring u = up can be considered as a ring in which we have a theory of divisors with a unique prime divisor p. By Theorem I of Section 5, Chapter 3, if.o is the integral closure of the ring u in the field K, then .0 has a theory of divisors with a finite number of prime divisors ~1' .... ~m (all of which divide p). Let ~ be a prime divisor of the ring .0 and let K~ be the completion of the field K with respect to the valuation v~. Those elements of K~ which are limits of elements of k form a subfield, which is topologically isomorphic to the completion k p of the field k with respect to the valuation vp ' Using the embedding k p --+ K~ we may assume that k p is a subfield of the field K~. Let K = k(r1.1' ... , r1. r ). The elements r1.i E K also belong to K~, and since they are algebraic over k, they are also algebraic over k p • Hence the extension k p(r1. 1, ... , r1. r )jk p is finite (with degree not exceeding the degree of Kjk), and then by Theorem 2 of Section I the field k p(r1. 1• ... , r1. r ) is complete. Every element of K~ is the limit of a sequence of elements of K, and since K c k p(r1. b ... , r1. r ) and k p(r1. b ... , r1. r ) is complete, then K~ c k p(r1.b ... , r1. r ). Since the reverse inclusion also holds, we have K~ = k p(r1. 1, ... , r1. r ). We have proved that the extension K~jkp is finite and that (K~

: kp)

~

(K: k).

Since the residue class fields of the valuations vp and v~ coincide with the residue class fields of the completions k p and K~ (see Section 1.1), then the degree of inertiaf~ of the divisor ~ relative to p coincides with the degree of inertia of the extension K~(kp . It is also clear that the ramification index e~ of the divisor ~ relative to p coincides with the ramification index of K~(kp . By Theorem 5 of Section I the numbers f~ and e~ are related to the degree n~ = (K~ : k p ) by f~e~

=

n~.

For the rest of this section we shall assume that the extension K(k is separable, and we will study the relations between the various completions K~ , ... , K~n of the field K under all extensions of the valuation vp ' Let OJl, ... , OJ n be a basis for the extension K(k. If for some r1. E K, all the coefficients aj in the representation (2.1)

[Chap. 4

LOCAL METHODS

268

are small relative to p (that is, small relative to the valuation vp), then the element a. will clearly be small relative to each of the prime divisors ~s' The following converse also holds.

lemma 1. For any integer N there exist an M such that whenever vIll.(a.) ~

M for all s

= 1, ... ,m, then all coefficients a j in (2.1) satisfy

vp(a j ) ~ N.

Proof. Let

Wi *, .,. , W n * be the dual basis of the basis Wi> ... , W n (here we are already using separability; see Section 2.3 of the Supplement). Then aj

= SPK/k(a.W/) = Sp a.w/.

Let es be the ramification index of ~s relative to p and p be a prime element in the ring up of the valuation vp ' so that es = vlll. (p). Set

M

= max (esN

- vlll,(w/)).

s.j

If vlll,(a.)

~

M for all s, then for fixed j we have

vlll.(a.w/) ~ esN = Vllls(pN), and this means that a.w/ = pN y, where vIlls(Y) ~ 0 (1 ~ s ~ m). By Theorem 6 of Section 4 of Chapter 3, the element y belongs to the integral closure of the ring up in the field K, and therefore Sp y E up , that is, vp(Sp y) ~ 0, so that vp(a j )

= vp(Sp(a.w/)) =

Vp(pN Sp y) ~ N,

and Lemma 1 is proved. Corollary. If a sequence {a.d of elements from the field K is a Cauchy sequence relative to each of the prime divisors ~s (s = 1, ... , m), then all the sequences {all}:;' l ' defined by

a.k = a /k l wl + .. ,+

an (k l wn

(a /k) E k),

are Cauchy sequences relative to p. Now consider the completions Kill,' ... , K Illm of the field K with respect to each of the prime divisors ~l' ... , ~m and form the direct sum Kill, EEl .,. EEl K Illm , which we denote by K p • The elements of this direct sum are sequences e = (el' ... , em), where el E Kill' ... , em E K Illm • We define addition and multiplication in K p component-wise, so that K p becomes a ring. For any y E k p , set y(el' ... , em) = (yel, ... , yem)' The ring K p then becomes a vector space over the field k p • If we denote the degree of Kills over k p by ns , then the dimension of K p over k p is given by (2.2)

Sec. 2]

FINITE EXTENSIONS OF FIELDS WITH VALUATIONS

269

There is a natural way of defining convergence in the ring K p • We say that the sequence {(e 1(kl, ... , em (k»))~ 1> e/ k) E K~s' converges to the element (e 1, ... , em) if for each s the sequence g/k)} converges to es relative to convergence in the field K~s' It is easily seen that, under this definition of convergence, the operation of multiplication of elements of K p by elements of k p is continuous. In other words, if "I = limk-+ 00 y(k), y(k) E k p , and = lim k-+ 00 ~):(k) ,~):(k) E K p' then lim y(k)e(k) = "Ie. (2.3~

e

k-+

00

We now define an imbedding K .... K p , setting

a=

(ex, ... , ex)

E

Kp

,

(ex

E

K).

Since K c K~. for all s, the sequence (ex, ... , ex) is an element of K p • It is clear that the mapping ex .... a defines an isomorphism of the field K into the ring K p • We denote the image of K under this isomorphism by K. To avoid confusion we note that the components of the product

ya = ("lex,

... , "lex)

which in this representation look identical, actually may be different, since the product "lex depends on the field K~. in which it is taken, even when "lex E k p • Theorem 1. If WI' ... , W n is a basis for the separable extension K/k, then is a basis for the ring K p as a vector space over k p •

(01' ... , (On

Proof. We first show that K is everywhere dense in K p ' that is, that every element of K p is the limit of a sequence of elements from K. Let = (el' ... , em) be any element of Kp , es E K~s (s = 1, .. , , m). Since K is dense in K~s ' then for any natural number k there exists an element (ex?» E K for which v~.(es exs(k» ~ k. By Theorem 4 of Section 4, Chapter 3, there is an element ex(k) in the field K for which V~.<ex(k) - ex/ k» ~ k for s = 1, ... ,m. The element ex(k) satisfies v~.(es - ex(k» ~ k (s = 1, ... , m),

e

and this means that the sequence {a(k)};,= 1 of elements of K p converges in the ring K p to the element Represent each element ex(k) in the form

e.

ex(k)

= a 1(k)W 1 + ... + an (k)w n ,

(a /k)

E

k).

Since the sequence {ex(k)} is a Cauchy sequence relative to each prime divisor ~s, then by the corollary of Lemma 1 the sequences {ap)} are all Cauchy sequences relative to p, and hence they converge in k p • Set "I j = lim k-+ oo a/ k ) (j = 1, ... ,n). Since for any a Eke k p and any E K p ,

e

ae = de,

(2.4)

[Chap. 4

LOCAL METHODS

270

then II

II

12k = L il/k)Wj = L ap)w j . j=1 j=1 Passing to the limit as k

--+ 00

and considering (2.3), we obtain II

~

= lim 12k = LyjWj . j=1

k-oo

This proves that the elements wj generate the vector space K p • We still need to show that they are linearly independent over k p • Let

· k'IS d ense In . k p' t hen Yj = l'Imk=oo SInce a:(k)

aj (k)

= a l (k)W I + .,. + all(k)w

Then lim 12(k) = lim" L, a J(k)W.J

, ;

:I II

ii'

;! . J.. ,

k .... oo

h ,were II

E

aj (k) E

K.

=" ,,·W. = 0

k .... oo j

L, IJ j

k . Set

J

•

This means that the sequence {a:(k)} is a null sequence in K relative to all the prime divisors ~. (s = 1, ' .. ,m). Then by the corollary of Lemma 1 all the sequences {all} in k are null sequences relative to p, and this means that 'II = 0, .. , , 'III = O. The proof of Theorem 1 is complete. Remark. In terms of the tensor products of algebras, Theorem 1 shows that the algebra K p over the field k p is isomorphic to the tensor product K flh k p , that is, that it can be obtained from K (regarded as an algebra over k), by extending the ground field from k to k p •

We have shown that ~he dimension of the vector space K p over k p equals n = (K: k). On the other hand, this dimension is given by the sum (2.2). Since n. = nllls = elll.JIlls' we arrive at

L elllflll = n ~

(~ running through all prime divisors of the ring -0). Hence we have obtained another proof of Theorem 7 of Section 5, Chapter 3.

Theorem 2. Let q>(X) denote the characteristic polynomial of the element a: E K, relative to the separable extension Klk, and let q>Ill(X) denote the characteristic polynomial of a: relative to KIlllk p • Then q>(X)

= TI q>Ill(X), III

Sec. 2]

FINITE EXTENSIONS OF FIELDS WITH VALUATIONS

271

Proof. Consider the linear mapping ~ --+ a~ of the vector space Kp to itself.

If a.W k =

I?= 1 ak,w

ak1

"

E

k, then by (2.4), rXWk =

L ak,w, . I

This means that the characteristic polynomial of our transformation coincides with the characteristic polynomial of the matrix (a k1 ), that is, coincides with q>(X). We now take another basis for Kp over k p • Let PS j (j = 1, ... ,ns) be any basis for the extension K'(j.lk p (s = 1, '" ,m). If we denote by PSj that element of K p which is zero in all components except the sth one, and whose sth component equals PS j ' then the set of all elements Psj

(s=I, ... ,m;j=l, ... ,n,)

(2.5)

is a basis for the ring K p over k p • Let n.

(S)P sl' a. Psj - " L. Yjf 1= 1

so that q>dX) is the characteristic polynomial of the matrix (yj,), It is now easily seen that the matrix of the linear mapping ~ --+ a~ with the basis (2.5) will be a block-diagonal matrix with the blocks (yj,) on the main diagonal. Theorem 2 follows immediately. For elements a. E K we introduce the concepts of local norm N'(j(a.) and local trace SP'(j(a.): N'(j(a.) = N K'(lIkp(a.), From Theorem 2 we deduce that N K/k(a.)

= TI '(lip

N'(j(a.),

SPK/k(a.) =

I

SP'(j(a.)·

(2.6)

'(lip

From the first of these formulas and (1.13) we deduce vp(NK/k(a.» = If'(jv'(j(a.),

(2.7)

'l1IP

which we proved by other methods in Section 5 of Chapter 3.

e,

Theorem 3. Let K(k be a separable extension with primitive element so that K = k(e), and let q>(X) be the minimum polynomial of e over k. Let p be a prime divisor of k, and let

q>(X) = q>1(X)", q>m(X)

by the factorization into irreducible polynomials in kp[X]. Let \.01' ... , \.Om be the prime divisors of the field K which divide p. There is a one-to-one correspondence between the divisors \.Os and the factors q>.(X), such that the polynomial q>.(X) which corresponds to \.Os coincides with the minimum polynomial of the element e E K'(j, over the field k p .

272

Proof. Since q>(X) is the characteristic polynomial of () relative to K(k, by Theorem 2 we have the factorization q>l(X)", q>m(X), where q>j(X) is the characteristic polynomial for () relative to K~)kp' Hence the factors q>.(X) correspond to the prime divisors ~s' But we saw in the preceding section that K~. = k p «(}), () EKe K~., and therefore each of the polynomials q>.(X) is irreducible over k p ' and Theorem 3 is proved.

Remark. Let 0 be any ring with a theory of divisors (and quotient field k) and let V be a prime divisor in o. If K(k is a finite separable extension, then Theorem 3 gives a description of all prime divisors of the integral closure .0 of 0 in K which divide V (at least it gives their number m and the products e~f~)·

3. Factorization of Polynomials in a Field Complete with Respect to a Valuation

In view of Theorem 3 of Section 2 it is important to have a method for factoring polynomials into irreducible factors in a field complete under a valuation. In this section we shall show that in such fields the decomposition of a polynomial with integral coefficients is completely determined by its decomposition modulo some power of the prime element. lemma. Let

be a subring of the field k, and let g(X) and h(X) be polynomials of degree m and n with coefficients in o. If the resultant p = R (f, g) of the polynomials f and 9 is nonzero, then for any polynomial/(X) E o[X] of degree ~m + n - 1 there exist polynomials q>(X) and IjJ(X) in o[X] of degrees ~ n - 1 and ~ m - 1 such that 0

L aru s + r+s=i L brvs = pc;

(i

= 0, 1, ... , m + n

- 1).

(3.1)

r+s=i

Proof. Set

'I'

n

m

g(X) =

L a;X m- i,

L b;X n-;,

h(X) =

i=O

m+n-l

l(X) =

;=0

q>(X)

=

"-1

L UiX n- 1 - i,

L

Cixm+n-l-i,

i=O

IjJ(X)

;=0

=

m-l

L ViX m- 1 - i.

i=O

To determine the m + n unknowns U o , ... , Un-I; V o , ... , V m - b we equate the coefficients of like powers of X in (3.1), obtaining a system of m + n equations

L

r+s=i

1

1

I 1,: I,! ! :1

I

[Chap. 4

LOCAL METHODS

.• ,,'

Ii

aru.

+

L brvs =

r+s=i

pc;

(i

= 0, 1, ... , m + n - 1).

Sec. 3]

FACTORIZATION OF POLYNOMIALS

273

The determinant of this system equals

(3.2)

.....

n

m

(with zeros everywhere else); that is, it equals the resultant p = R(g, h). Since p "# 0, this system has a unique solution, and since all the constant terms PC i are divisible by p, then the values of U i and Vi will belong to the ring u. The lemma is proved. Now let k be complete under the valuation v, with u the ring of integral elements of k and n a prime element of u. Two polynomials f(X) and h (X) of u[XJ are called congruent modulo n\ and we write f(X) == h (X) (mod nk ), if this congruence holds for the coefficient of each power of X. Theorem 1. Let f(X) E u[XJ have degree m + n. Suppose that there exist polynomials go(X) and ho(X) in u[XJ of degrees m and n such that (1) f and goho have the same leading coefficients; (2) the resultant R(go , h o) is nonzero; and (3) if r = v(R(go, ho)), then

(3.3) Then there are polynomials g(X) and h(X) in u[XJ of degrees m and n, for which f(X) = g(X)h(X), g(X) == go(X),

h(X) == ho(X) (mod nr+l)

and the leading coefficients of g(X) and h(X) coincide with those of go(X) and ho(X), respectively. Proof. For each k ~ I we construct by induction polynomials

({>k E

u[XJ

of degree ~ m - 1 and t/Jk

i

I

, I

•I I i

[Chap. 4

LOCAL METHODS

274

o[X] of degree ~ n - 1 so that the polynomials gk =go + nr+1({J1 + + nr+k({Jk' r 1 hk = ho + n + t/Jl + + nr+kt/Jk

will satisfy

E

f== gkhk (mod n 2r +H 1).

I

(3.4)

Assume that we have constructed the polynomials ({Jl' ... , ({Jk-l and t/Jl' ... , t/Jk-l with the required properties, so that f = gk-lhk-l + n 2r +kl, (3.5) where leX) E o[X]. The polynomials go and gk-l' as well as h o and hk- b have the same leading coefficient, so leX) has degree ~m + m + n - l. Further, gk-l == go, hk- 1 == ho(mod nr+1) and therefore R(gk-b hk- 1) == R(go, ho) (mod nr+1), which means that V(R(gk-l' hk- 1)) = r. By the lemma there exist polynomials ({Jk and t/Jk in o[X] with degrees ~m - 1 and ~n - 1, for which nrl=gk_It/Jk+hk_l({Jk'

(3.6)

We shall show that (fJk and t/Jk satisfy our requirements. Since gk = gk-I

+ nr+k({Jk' hk- 1 + nr+kt/Jk'

then by (3.5) and (3.6) f - gkhk = n 2r +kl_ n r+k(gk_lt/Jk

+ hk- 1(fJk) ~ k + 1).

2r 2k n 2r +2k q,kt/Jk = - n + (fJkt/Jk'

so (3.4) also holds (since 2k Now consider the polynomials 00

g(X) = go

'I'

+

00

L nr+k({Jk'

heX) = ho +

k=1

L nr+kt/Jk>

k=1

whose coefficients (except the leading ones) are defined as the sums of convergent power series. Since 9 == gk and h == hk (mod n Hr + 1), then gh == gkhk (mod n r+H I), so that by (3.4)

f== gh (mod n r + H1 ).

Since this holds for all k, thenf = gh, and Theorem 1 is proved. Remark. From the proof of Theorem 1 it easily follows that if go and ho satisfy the condition f== goh o (mod n S ~ 2r + 1, instead of (3.3), then 9 and h can be chosen so that S

),

Sec. 3]

FACTORIZATION OF POLYNOMIALS

275

We consider an important special case of Theorem 1. The polynomial f(X) E u[X] will be called primitive if at least one of its coefficients is a unit in u. Let 1: be the residue class field of the ring u modulo the prime element 71:. If we replace the coefficients off by the corresponding residue classes in 1:, we obtain a polynomial j with coefficients in the field 1:. Assume that in the ring 1:[X] the polynomial jhas a decomposition (3.7) in which the factors iJo and no are relatively prime. We may choose polynomials 90 and ho in the ring u[X] so that the degree of 90 coincides with the degree of iJo and the degree and leading coefficient of the polynomials f and 90 h o coincide. Consider the resultant R(90, h o) of the polynomials 90 and h o , that is, the determinant of (3.2). If we replace each entry in this determinant by the corresponding residue class modulo n, we obtain the resultant ·R(iJo, no) of the polynomials iJo and no (here the leading coefficient of no may be zero). The resultant R(iJo, no) is nonzero, since by choice of 90 the leading coefficient of iJo is nonzero, and the polynomials iJo and no are assumed relatively prime. (Recall that the resultant of two polynomials is zero if and only if the two polynomials have a common divisor, or both of them have leading coefficient zero.) Hence R(90, h o) ¥= 0 (mod n)p that is, v(R(90, ho)) = r = O. Equation (3.7) means that f == 90ho (mod n). Thus we see that 90 and h o fulfill all the conditions of Theorem 1 (with r = 0), and we have the following result. Theorem 2 (Hensel's lemma). Let f(X) be a primitive polynomial with coefficients in the ring u of integral elements of a field complete under a valuation. If in the residue class field 1: the polynomialjE 1:[X] has a factorization (90' h o E u[X])

with iJo and no relatively prime, then there exist polynomials 90 and u[X] such that

ho in

f(X) = 9(X)h(X),

with iJ = iJo' n = no, and the degree of 9 equal to the degree of iJo . With the aid of Theorem 1 we can solve the problem of the factorization of polynomials over a field complete under a valuation. We need consider only polynomials with integral coefficients and leading coefficient 1 (if the leading coefficient of a polynomial of degree n in u[X] is a, then we can multiply the polynomial by (/'-1 and take aX as a new variable). Since Gauss's lemma on the factorization of polynomials with integer coefficients also

1,1 '

II

I i

,I

I

,I

, I

I

" I

.' t

,

.

II ,

276

LOCAL METHODS

[Chap. 4

holds for the ring o[X], then every irreducible factor of f(X) with leading coefficient 1 will lie in 0 [X]. If the polynomialf(X) does not have multiple roots (in any finite extension of the field k), then its discriminant D(f) = ± R(f, 1') is nonzero. Let d = v(D(f)), and consider any factorization f== s, p = 7T e e - I , then f3 c= (l + 7Tkye)P (mod 7T e +k+I). Then apply Problem 17 of Section 1.) 3. Under the conditions of Problem 2 assume that Cl is congruent to 1 modulo 7Tm but is not a pth power in k. Show that k(r;/rx)/k is an unramified extension of degree p. [Find the characteristic polynomial f(t) of the element y = 7T -saj~ - 1) and verify that !'(y) is a unit; now apply the last assertion of Problem 1.) 4. Retaining the conditions of Problem 2, assume that Cl E k is integral and satisfies h the conditions: Cl == 1 (mod 7T ), Cl c= 1 (mod 7T h 1, (h, p) = 1, h < m = ep/(p - 1). Show that Cl is not a pth power in k and that the extension k(r;l~)/k is totally ramified. (Consider the exponent with which the prime element of the field k(J;j

f(x i

+ cn) = fc(x t ,

... , X n

for all points of the region vex;)

~ N

... ,

x n ),

[fc also converges for vex;) ~ N].

5.2. Exponential and Logarithmic Functions In this section we assume that k is a finite extension of the p-adic field R p • We denote by the valuation of k by v, the ramification index over R p bye, and a prime element of the ring of integral elements of k by n. Consider in k the power series exp x = I log(l

X

X

Z

xn

+ -l! + -2! + ... + -n! + ... '

+ x) = x

X

Z

__

2

(5.6)

n

x + ... + (_I)n - t _ + ...

(5.7)

n

We shall find the region of convergence of the series (5.6). Since the prime number p occurs in n! with exponent [nip] + [nlpz] + "', then yen!) = e([!2J P

+

[nzJ P

+ ... )
n(v(x) _ _ e_). n! P- 1

(5.8)

1

I I

286

LOCAL METHODS

[Chap. 4

If vex) > ej(p - 1), then v(x"jn!) -+ 00, and the series (5.6) converges. On the other hand, if vex) ::;; ej(p - 1), we have for n = pS,

v(:~) = nv(x) -

e(ps-1

= nv(x) _

+ ... + P + 1)

e n - 1 = n (V(X) _ _ e_) p-l p-l

+ _e_ p-l

::;; _e_ , p-l

and hence for such x the general·term in (5.6) does not converge to zero. This shows that the series (5.6) converges precisely for those x for which vex) ~ K, where K=

[_e1+ p-l

1.

Formal multiplication of the power series exp x and exp y is easily seen to give the series exp(x + y), and hence for vex) ~ K and v(y) ~ K we have the formula exp (x

+ y) = exp x·exp y.

(5.9)

We now turn to the series (5.7). If vex) ::;; 0, then v(x"jn) does not converge to infinity as n -+ 00, and hence for such x the series (5.7) does not converge. Now let vex) ~ 1. If n = pan1 , (n 1, p) = 1, then pa::;; nand yen) = ea ~ e(ln njln p), so that v(X") n

=

nv(x) _ yen)

~ nv(x) _

e In n , ~p

and this means that v(x"jn) -+ 00 as n -+ 00. Hence the series (5.7) converges if and only if vex) ~ 1. If vex) ~ 1, then the element 8 = 1 + x is a unit in the ring 0 of integral elements of the field k, with 8 == 1 (mod n). Conversely, if a unit 8 satisfies this congruence, then it has the form 8 = 1 + x, where vex) ~ 1. We can such a unit of the ring 0 a principal unit of the field k. The series (5.7) hence defines a function log 8 on the multiplicative group of an principal units of the field k. We shan show that for any two principal units 81 and 82 , we have the formula (5.10)

Let 81 = 1 + x, and

82

= 1 + y, and let

v(y) ~ (x), so that y

=

tx with t integral

+ x)(1 + y) = 1 + (t + l)x + tx 2 • We shan consider the expression (t + l)x + tx 2 as a power series in x, for which an terms lie in the region of convergence of the series log(l + z). Since the (l

Sec. 5]

ANALTYIC FUNCTIONS IN COMPLETE FIELDS

formal substitution of this expression in the series 10g(1 + 10g(1 + tx), then by Theorem 2 10g(1

287

+ z) gives log(l + x)

+ (t + I)x + tx 2) = 10g(1 + x) + log(l + tx),

which proves (5.10). The formal substitution of the series (5.7) in (5.6) and of the series exp(x - I) in (5.7) give us the following formal identities: exp log (I log

+ x) = I + x; exp x = x.

(5.11) (5.12)

Since these are formal identities, to verify them we can assume that x is a complex variable and use the theorem on substitution of series in series for complex power series (see, for example, K. Knopp, "Elements of The Theory of Functions," Sections 41 and 45, Dover (New York, 1952). To see under what conditions the formal identities (5.11) and (5.12) can be considered as equations in k, we turn to Theorem 2. By this theorem (5.11) will hold provided the terms of the series log(l +x) satisfy v(x"(n) ~ K. For n = I this gives us v(x) ~ K. But if v(x) ~ K, then v(x"(n) ~ nK ~ K for I :s;; n :s;; p - I and v(X") _ K

n

~ (n

_ I)K _ V(II)

> (n _ I) _e_ _ e In n p-I

~p

= e(n

- 1)

In p

(~

_

p - I

~) ~ 0

n - I

for n ~ p ~ 2 [here we are using the fact that the function In t(t - I) for t ~ 2 is monotone decreasing]. Hence (5.11) is valid under the condition v(x) ~ K. Further, under this condition, v(log(1 + x)) ~ K. We turn to formula (5.12). It follows from (5.8) that if v(x) ~ K, then every term of the series exp(x - I) is contained in the region of convergence of the series 10g(1 + x), and this means that (5.12) holds for all x for which exp x is defined. We denote by A the additive group of all x E k for which v(x) ~ K, and by M the multiplicative group of units of the form 8 = I + x, X E A. We have shown that the mapping 8 -+ log 8 (8 E M) is a homomorphism from the group M to the group A. We now show that the mapping x -+ exp x is a homomorphism from A to M. In view of (5.9) we need only show that v(x"(n!) ~ n for all x E A and all n ~ I. Let p" :s;; n < p" + 1. Then

vG~) - ~ (n - I)K - e([~] + [;2] K

~

(11 - I)e en p" - I p- I p" p -·1

~O,

+ ... +

[~])

[Chap. 4

LOCAL METHODS

288

which is what is required. Formulas (5.11) and (5.12) now show that the mappings log: M -+ A and exp: A -+ M are both one-to-one and are inverses of each other. Hence we have the following result. Theorem 3. The mapping x -+ expxis an isomorphism of the additive group of all numbers of the field k which are divisible by nrc (K = [e(p - 1)] + 1) onto the multiplicative group of all principal units e which are congruent to 1 modulo nrc. The inverse isomorphism is given by e -+ log e [for e == 1 (mod nrc)].

The mapping e -+ log e is, in general, not an isomorphism on the whole group of principal units (Problem 5). Also, the value of log e is not necessarily integral. In real analysis one also considers the exponential function aX = eX In a. Its analog in the field k is the function r(

= exp(x log 1"/),

(5.13)

where 1"/ is a principal unit of the field k. This function is defined provided that K - v(log 1"/). Therefore if 1"/ == 1 (mod nrc), then I"/x will make sense for all integral x of k, and the value of I"/x will satisfy I"/x == 1 (mod nrc). If 1"/ == 1 (mod nrc) and x and yare any integral elements, then we have the formulas

v(x) ~

PROBLEMS 1. Let f(x) be an analytic function in the region v(x);;o}-L (in a complete field with valuation v). If f has an infinite number of zeros in the region v(x);;o }-L, show that f is identically zero. 2. Let k be a field of characteristic zero, complete under a non-Archimedean metric r:p (Problem 4 of Section 4, Chapter I). Assume that the metric r:p satisfies r:p(p) < 1 for some rational prime p. Show that the region of convergence of the series logO + x) is the set of all x for which r:p(x) < I, and the region of convergence of the series exp x is given by r:p(x)


3, the numerator of 1T must be divisible either by an odd prime p, or by 2 2 (in the latter case set p = 2). From this it follows that the series sin x and cos x converge in the p-adic field R p 1T

Sec. 5] for

ANALYTIC FUNCTIONS IN COMPLETE FIELDS

x = "TT.

289

But in view of the formula sin (x + y)

it follows from sin

"TT =

=

sin x cos y + cos x sin y

0 that sin

n"TT=

0

for all natural n. The function sin x thus has an infinite number of zeros in its region of convergence. But then, by Problem I, it would be identically zero, which is a contradiction. 5. Let k be a finite extension of the p-adic field R., and let e be a principal unit of k. Show that log e = 0 if and only if e is a root of degree p' (s;;. 0) of I. 6. Keep all notations of Section 5.2. The principal units e, for which e """ I (mod TT'), form a multiplicative group, which we call M •. The integers of k which are divisible by TT' form an additive group A•. Show that for k;;. K the mapping e -+ log e is an isomorphism of the group M. onto the group A. (the inverse mapping being x -+ exp x, x EA.). 7. In any complete field with valuation, show that the region of convergence of a power series f(x) = a.x· is contained in the region of convergence of its derivative f'(x) = na.x· Give an example in whieh f and f' have different regions of convergence (with the field k of characteristic zero). 8. Show that in the ring of 2-integral rational numbers the sum

2::= 0

2::=0 -I.

2· 22 23 2+2"+3+ ... +-;;can be made divisible by any given power of 2, by taking n large enough. 9. Show that all coefficients a. of the series 2

X'

E.(x)=exp ( x+ -

p

) + x· 2 -+ ...

p

are p-integral rational numbers. (Hint: Show that the number

equals the number of elements in the symmetric group of nth degree which have order a power of p. Use the theorem which says that if d divides the order of the finite group G, then the number of elements U E G which satisfy the equation ud = I is divisible by d.) 10. Show that E.(x)

=

TI

(I - x'") -p(.. l/r.

(m,p)=!

(m runs through all natural numbers relatively prime to p; f-L(m) is the Mobius function). 11. Let TJ be a principal unit in a finite extension field of the field R., and let x be a p-adic integer. Choose a sequence of natural numbers {a.} which converges to x. Show that lim._ oo TJa. exists and that it is independent of the choice of {a.}. Further, show that the function TJa = lim TJa. '_00 of the p-adic integer x coincides with the function (5.13).

II' ~ 'I :1 "I

,I ,I ,

LOCAL METHODS

290

[Chap. 4

6. Skolem's Method

In this section we study the application of Skolem's method to equations of the form

I

(6.1 )

·1

:1

where F is an irreducible, decomposable, nonfull form (Section 1.3 of Chapter 2), and c is a rational number. This method is based on some simple properties of local analytic manifolds over p-adic fields, which will be proved in the next section. An example which illustrates the idea of Skolem's method was given at the beginning of this chapter.

I

6.1. Representation of Numbers by Nonfull Decomposable Forms

i i ,i

,I

In Section 1.3 of Chapter 2 we saw that (6.1) can be written in the form

,>; .:\

(6.2)

'1"

or ,"·

I' 1

if l'

N(rx)

= a,

(rx EM),

(6.3)

where Ill' ... ,11m are numbers of some algebraic number field k, and M = {Ill' ... ,11m} is the module generated by these numbers (a is a rational number). Replacing, if necessary, the form F by a form integrally equivalent to it, we may assume that the numbers Ill' ... ,11m of the module M are linearly independent over the field R of rational numbers. Since Mis nonfull, m < n = (k: R). In Chapter 2 we saw how to find all solutions to (6.3) when M is a full module of k. It is thus natural to embed M in a full module M, and to use the methods of Chapter 2 to find all solutions of the equation N(rx) = a, rx E M, and then to pick out those solutions which lie in M. It is clear that any module of k can be embedded in a full module. To do this it suffices to extend the linearly independent set Ill' ... ,11m to a basis Ill' ... ,Iln of the field k and to set M = {Ill' ... ,Iln}· If all rx E M for which N(rx) = a have already been found, then we shall obtain all solutions of (6.3) if we can isolate those solutions for which in the representation

rx = the coefficients X m+l

X m + 1, ... 'X n

Xllll

+ ... + xnll n

are equal to zero. To express the conditions

= ." = X n = 0 directly in terms of rx, it is convenient to use the dual basis

III *, ." , Iln * (see Section 2.3 of the Supplement). Since the trace Sp Iljlli * is 0 for i t= j and 1 for i = j, then Xi = Sp rxlli* (1 ~ i ~ n). It follows that the

Sec. 6]

SKOLEM'S METHOD

291

numbers IX E M which lie in the submodule M are characterized by the conditions Sp IXlli* = 0

(i = m

+ I, ... , n).

(6.4)

By Theorem 1 of Section 5, Chapter 2, all solutions of the equation N(IX) = a, IX E M, can be written in the form (1

~j ~

k),

(6.5)

where "11' ... ,Yk is a finite set of numbers of the module M with norm a; is a system of independent units of the field k; and U 1 , •.. , Ur are arbitrary rational integers. From (6.4) we see that every solution to (6.3) corresponds to a solution in one of the k systems of equations

Cl' ... , Cr

(i = m

+ 1,

... , n)

(6.6)

in rational integers Ul> '" , Ur (here "I is one of they j)' Let K be an algebraic number field which contains all fields conjugate with k, and let 0"1' ... , O"n be all isomorphisms of k into K. Since Sp ¢ = 0"1(0 + ... + O"n(O for any ¢ E k, then the system (6.6) can be written n

L O"/yp/)O"iCt)"' ... O"icr)"' = 0

j=1

(i

=

m

+ 1,

... ,n).

(6.7)

It is clear that to prove the finiteness of the number of solutions to the equa-

tion (6.3) it suffices to show that each system of the form (6.7) has only a finite number of solutions in rational integers U 1 , '" , u.' Remark. The set of all numbers of the field k of the form cl"' ... Cr "', where u1 , ... , U r run through all rational integers, is a multiplicative subgroup of k which we shall denote by U. The solutions of (6.3) then coincide with numbers of the intersections

M n yjU

(j

= I, ... ,k).

(6.8)

Instead of the set (6.8) we may also consider the similar set "I j -1 M n U. Then the problem of finding all solutions to (6.1) is reduced to the problem of finding the intersection of a module and a multiplicative subgroup of the field k. We note also that we may replace the module M by the vector space L (over R) which is spanned by Ill> ... , Ilm' For yjU c M and L n M = M, so L n "I j U = M n Yj U. 6.2. The Relation to Local Analytic Manifolds

The idea of Skolem's method is that in some cases we can prove the finiteness of the number of solutions of (6.1) by proving that the system (6.7)

I

. I··

292

,1

[Chap. 4

LOCAL METHODS

has only a finite number of solutions even when the variables U 1, ... , U r take on ~-adic integral values (that is, they take integral values in the completion K",), where ~ is any prime divisor of the field K. After such an extension of the domain of possible values for the variables, we may consider the set of all solutions of the system (6.7) as a local analytic manifold in r-dimensional space, and then apply properties of such manifolds. When we allow that variables U 1, ... ,Ur in (6.7) to take ~-adic values, we encounter the obstacle that the exponential function SU = exp(u log s) is only defined for all integral ~-adic u when s == 1 (mod ~") (I( is an integer which depends only on the field K",; see the end of Section 5). We avoid this difficulty in the following manner. By Problem 6 of Section 7, Chapter 3, there exists a natural number q such that any integer rx E K which is not divisible by ~ satisfies

rx q == 1 (mod Each exponent

(6.9)

in (6.5) can be written

Ui

,: ; I

~").

Ui

= Pi

and hence the unit s =

+ qv

SI u,

...

j ,

(0::;; Pi

< q,

Vi

EZ),

s/r has the representation (l

= 1, ... , qr),

where fJ I is one of the q' numbers (0::;; Pi < q).

Hence we obtain a new representation for numbers rx of the form (6.5), in which Si is replaced by Si\ and the finite set of numbers "I j by the set of numbers "IjfJ I • Since the Si are units, then the congruence (6.9) holds for all of the numbers a j(sJ, and hence the functions a /Siqt are defined for all ~-adic integers U E K",. We have proved the following result. Lemma 1. After making new choices, if necessary, for the numbers "Ij and Si in (6.5), the functions aj(sJu are defined for all integers of the field

K",. In the future we shall assume that this condition holds without special mention. We turn to the system (6.7). In view of (5.9) and (5.13), we can put this system in the form n

LA j;!

, Ii

ij

exp L/u!, ... , u.)

=0

(i

=

rn

+ 1,

... , n),

(6.10)

Sec. 6]

SKOLEM'S METHOD

293

where r

Lj(u l ,

... ,

2: Uk log aiek),

ur ) =

k=1

A ij

=

aj(yp/).

Since the left side of (6.10) consists of power series which converge for all integral U 1 , ... , Un and hence represent analytic functions, then the set of all solutions to (6.10) can be interpreted as a local analytic manifold (in a neighborhood of any solution) in the sense of the definition of Section 7. The system (6.10) consists of n - m equations in r variables. It is natural to expect that the manifold defined by this system will consist of only a finite number of points, provided that n - m ~ r. Recall that the number r comes from the Dirichlet theorem on units, and r = s + 1 - 1, where s is the number of real, and 1the number of pairs, of complex embeddings of the field k in the field of complex numbers. Since n = s + 21, then n - m ~ r if and only if 1 ~ m - 1. In the simplest interesting case m = 2, and the condition reduces to 1 ~ 1. This means that there should be at least one pair of complex embeddings of k. This case leads to Thue's theorem, and will be considered in the next sections. Assume that the system (6.10) has an infinite number of solutions (ulS' ... , urs ), S = 1,2, .... Since the ring of ~-adic integers is compact (see Theorem 6 of Section 3, Chapter 1, and the second remark at the end of Section 1.2 of this chapter), we can choose from this sequence a convergent subsequence, the limit of which we denote by (u 1 *, ... , U r *). It is clear that the point (u 1 *, ... , ur*) is also a solution of (6.10), that is, that it lies on the manifold determined by this system, and that in any neighborhood of this point there are an infinite number of points of the manifold. We now change variables by the formula ~-adic

(1::;;;::;;r).

The system (7.10) then becomes

2:" A~ exp L/v

l , ... ,

vr )

=0

(i

=

m

+ 1, ... , n),

(6.11)

j= I

where A i/

= Aij expLj(ul*'

... , ur *).

The constant terms of the series on the left of(6.11) are all zero. We denote by V the local analytic manifold (see Section 7) determined by (6.11) [in the neighborhood of the point (0, ... , 0)]. Since this manifold does not consist of a single point (any neighborhood of the origin contains an infinite number of points of the manifold), then by Theorem 2 of Section 7 the manifold V

294

LOCAL METHODS

[Chap. 4

contains an analytic curve; that is, there is a system of formal power series

(not identically zero and without constant terms) with coefficients from a finite extension of K<JJ' such that the series (6.12) identically satisfy the relations n

L1A;j exp Pit) =

0

(i

j;

=

m

+ 1, ... , n),

Hence we have the following result. Theorem 1. If the equation (6.1) has an infinite number of solutions, then at least one local analytic manifold of the type (6.11) [for some I' = I'j and some point (u 1 *, '" , u:)] contains an analytic curve.

This theorem is the heart of Skolem's method. It reduces the question of the finiteness of the nnmber of solutions to (6.1) to the proof that the systems (6.11) do not have solutions in formal power series, that is, that the corresponding local analytic manifolds do not contain analytic curves. Note that there are n - r linear relations on the n series Pit) defined by (6.12): n

LB j;l

1j

Pit) =

o.

(1 ::;; i ::;; n - r),

since they are linear combinations of the r series Wk(l). Thus the existence of an analytic curve on V implies the solvability [in power series P;(t) without constant term] of the system n

L A;j exp Pit) = 0 j;l

(m

+ 1 ::;; i

::;; II),

(6.13)

n

L BijP/t) = 0 j;l

(1 ::;; i ::;; n - r = t

+ 1),

in which both groups of equations are linearly independent. [The linear independence of the equations of the first group follows from the fact that the determinant det a/I'Il;*), whose square is the discriminant of the basis I'lli*' is nonzero, and hence the rank of the matrix (Ai) (m + 1 ::;; i ::;; n, 1 ::;;j::;; n), and, consequently, of the matrix (Aij*)' isn - m.] If we assume that n - m ~ r, then the total number of equations in (6.13) will be ~n.

295

SKOLEM'S METHOD

Sec. 6]

6.3. Thue's Theorem

Thue's theorem states that if the formf(x, y) = aox· + a 1 x"-l y + + a.y· in two variables with rational integral coefficients is irreducible and has degree n ~ 3, then the equation (6.14) f(x,y) = c 00.

has only a finite number of solutions in integers. Since forms in two variables are always decomposable, and when n > 2 they are nonfull, then the equation (6.14) is a special case of (6.1). Here m = 2, so to apply Skolem's method we must have t ~ I; that is, the equation f(x, I) = 0 must have at least one complex root. In such a case we shall say that the formf(x, y) has complex roots. We shall prove Thue's theorem by Skolem's method under this assumption. In other words, we shall prove the following assertion. Theorem 2. If the formf(x, y) has integral coefficients, is irreducible, has degree ~ 3, and has at least one complex root, then the equation f(x,y)

=c

has only a finite number of solutions in integers. Proof. We assume that the coefficient a o of x· equals I [otherwise we can multiply (6.14) by ao·-] and replace aox by x]. Setk = R(8), K = R(8], ... ,8.),

where 8 = 8J, 82 ,

00'

,

8. are determined by the decomposition f(x, I)

= (x + ( 1 ) " , (x + 8.).

For eachj = I, n we denote by aj the isomorphism of k to K which takes 8 to 8j ' Since f(x, y) = N(x + y8) (N denoting the norm of kj R), then (6.14) can be written in the form (6.3), where by M we mean the module {I, 8}. Hence in this case /1] = I, /12 = 8 (m = 2). Assume that the equation (6.3) has an infinite number of solutions IX = x + y8 in the module M = {l, 8}. Then for some I' = I'j E k, an infinite number of these solutions will be of the form (6.5), where the independent units e1' ... , e, of k satisfy Lemma I. The exponents u 1, ... , u, in (6.5), corresponding to each solution IX, will satisfy the system (6.10). We choose among the so that the corresponding points solutions IX a sequence lXI' IX2 , 00.

,

00.

(s=I,2,oo.)

(6.15)

converge to some point (u] *, ... , u:). In Section 6.2 we saw that the local analytic manifold V, defined by (6.11), contains an analytic curve (01(1), (O,(t), and for any such curve on V the series (6.12) satisfy some system of the form (6.13). The rest of the proof of Theorem 2 is based on the following important auxiliary result. 00'

,

296

[Chap. 4

LOCAL METHODS

Lemma 2. Let there be given a system of equations

L• a

ij

exp P j = 0

j= 1

(6.16)

•

L bijP

j

= 0

j=l

in which each group of equations is linearly independent. If n1 = n - 2, n2 ~ 2 and if the system has a solution in formal power series P1 (t), ... ,p.(t) without constant term, then Pk(t) = Pit) for at least two distinct indices k and j. [The coefficients aij and b ij , as well as the coefficients of the series Pit), lie in some fixed field of characteristic zero.] We give the proof of this lemma below, but now we show how the lemma implies Theorem 2. By Lemma 2, for any curve Wl (t), ... , wr(t) on V, Pk(t) = Pit) for at least two distinct indices j and k; that is, (6.17)

I

In r-dimensional space consider the points (Vl' '" ,vr ) of the manifold W which are determined by

1

:.~; 1

I

J

i:l '.'1·: :1 t

It follows from (6.17) that any curve which lies on the local analytic manifold V also lies on W. But then by Theorem 3 of Section 7, V c W; that is, all points of the manifold V, contained is some sufficiently small neighborhood of the origin, also belong to W. But on the other hand, we shall now show that among the points (Vis' ... , vrs ) E V, which are obtained from the points (6.15) by U is = u i * + Vis and which converge to the origin, only a finite number lie on the manifold W. This contradiction will prove Theorem 2. Let a = x + yO and a' = x' + y'O be two points of the sequence {as}, for which the corresponding points of V lie in the manifold determined by L k = L j • If a = yet"! ... er"' and Ui = u i * + Vi' then Ui

=

Ui*

+ Vi'

u/a) = u/y)u/e 1)"!· .. · U/er)"'·u/et)"! ... U/e r)"'

=

Cj

exp L/v 1 ,

... , Vr )

(with cj independent of a) and analogously uk(a) = Ck exp L k (v 1 ,"',

Vr),

SKOLEM'S METHOD

Sec. 5]

297

so that oia) uk(a) --=--

In precisely the same fashion we find that uia') = uk(a') cj Ck

From the last two equations we obtain x+Y(}j X+Y(}k = , x' + y'(}j x' + y'(}k

so that (xy' - X'Y)«(}k - ()) = 0,

and since ()j ¥-

(}k,

then xY' - x'y = 0.

This means that x + Y() = d(x' + y'(}) for some rational d. Taking norms and using the fact that N(a) = N(a'), we obtain d n = I, so that d = ± I, and a' = ±a. Thus each of the n(n - 1)/2 manifolds given by L k = L j , whose union is W, contains not more than two points of V which correspond to numbers of the sequence {a.}. Then W contains at most n(n - 1) such points. Thus any neighborhood of the origin contains points of V not lying on W, so V (as a local analytic manifold) is not contained in W, which contradicts our earlier conclusion that V c W. As we have noted, this contradiction proves Theorem 2. Proof of Lemma 2. Since the first group of equations linearly independent (and n1 = n - 2), we can, after changing the numbering if necessary, express exp Pi (i = I, ... ,n - 2) in terms ofexp Pn - 1 and exp Pn :

exp Pi = a i exp P n -

1

+ hi exp Pn •

(6.18)

If a i = 0, then from the absence of constant terms and the equation exp Pi = hi exp Pn , we deduce that hi = I and Pi = Pn • Hence we may assume that all a i are nonzero. Set

(i=I, ... ,n-l) and assume that all Qi are nonzero. By (6.18) we have exp Qi

= a i exp

Qn-l

+ hi'

(6.19)

298

LOCAL METHODS

[Chap. 4

so that by differentiation with respect to t (Problem 10) we obtain Q;' exp Qi = aiQ~-l exp Qn-1'

(6.20)

From (6.19) and (6.20) we deduce

Q;'=Q~-lexpQn-1

il !

,

!

I

ci

(i = 1, ... , n - 2),

1

+ exp Qn-1

(6.21)

where Ci = biai-1. We now use the second group of equations of (6.16). By assumption there are at least two linearly independent equations in this group. Hence we can find a nontrivial relation among the Qi : n-1 I diQi = 0. i= 1 Differentiating this identity and replacing Q;' by (6.21), we obtain

,

Qn-1 exp Qn-1 and since Q~-l ¥-

(n-2 I

di i= lei + exp Qn-1

+

dn - 1 ) = 0, exp Qn-1

°and exp Qn-1 ¥- 0, then n-1

d.

I

I

i=l Ci + exp Qn-1

=0

(6.22)

(here we take Cn- 1 = 0). We claim that (6.22) can hold only if the rational function n-1

d.

I-' i= 1 C + Z

(6.23)

i

is identically zero. Assume that it is not, that is, assume that (6.23) equals rp(z)fljJ(z), where rp(z) ¥- 0. Then since rp(exp Qn-1) = 0, the nonconstant formal power series exp Qn -1 is the root of a polynomial, which contradicts the assertion of Problem 4 of Section 1. It is clear that the function (6.4) can vanish identically only when Ck = cj for at least two distinct indices j and k. Since C = ba, we then find from (6.19) that

from which it easily follows that Pk = Pj • Lemma 2 is proved. Remark. Skolem's method allows us to prove that (6.14) has only a finite number of integral solutions. But it does not give an algorithm for finding

Sec. 6]

SKOLEM'S METHOD

299

these solutions. The reason is as follows. After proving that the system (6.7) has only a finite number of 'P-adic solutions, it is easy to find an algorithm for the computation of the coefficients in the expansions of these solutions in power series of the prime element. However, there is no algorithm which allows us to judge from a finite number of coefficients whether or not we are dealing with a rational solution. This defect is shared by all known proofs of Thue's theorem. Even when the equation (6.14) is of third degree, there is no known algorithm for finding all integral solutions, or even for determining if there exist any solutions. 6.4. Remarks on Forms in More Variables The following question now arises: Under what conditions does an equation of the type (6.1) with a nonfull decomposable form have only a finite number of solutions in integers? Such equations sometimes have an infinite number of solutions. As an example consider the equation x4

+ 4 y4 + 9z 4

-

4X 2 y2

-

6X 2 Z 2

-

12 y2 z2

= N(x + yJ"2 + zJ3) = 1

[the norm is taken in the extension R( .)2, .)3)/R]. This equation has two infinite sets of solutions, given by the formulas

x

+ y.)"2 =

±(l

+ .)2)"

(z = 0),

x

+ z.)3 =

±(2

+ .)3)"

(y

= 0).

The reason for this is that, setting z = 0 or y = 0, we obtain from our form the square of a full form: (x 2 - 2y 2? or (x 2 - 3Z 2 )2. This occurs because the module {l, .)2, .)3}, which corresponds to our form, contains full modules of smaller fields, namely, {l, Ii} c R( .)2) and {l, .)3} c R( .)3). We describe a general class of forms with analogous properties. We write (6.1) in the form (6.3) and consider the vector space L (over R) which is generated by the numbers of the module M. The module M is called degenerate if the corresponding space L contains a subspace L' which is similar to some subfield k' c k, where k' is neither the field of rational numbers nor an imaginary quadratic field. We show that for a degenerate module the equation (6.3) has an infinite number of solutions (at least for some a). For if L' = yk' (y E k) and M' = L' (\ M, then y- 1 M' is a full module of the field k'. By the assumptions on the field k' the number of fundamental units in any order is nonzero, and therefore the equation (6.24) has an infinite number of solutions (provided it has at least one solution). Set

300 al

LOCAL METHODS

= Nk/R(y)d, where r = (k : k').

[Chap. 4

Since

Nk/R(~Y) = (Nk'/RWYNk/R(y) = a

and ~y E M' c M [for any ~ which satisfies (6.24)], then the equation Nk/R(f/) = al' f/ E M, has an infinite number of solutions. The basic conjecture on equations of the form (6.1) is that any such equation has only a finite number of integral solutions, provided that the associated module is not degenerate. Apparently the only known approach to this hypothesis lies in Skolem's method (application of which, as we have seen, requires the additional restriction that t ~ m - I). The basic stage at which the condition m = 2 was used in the proof of Theorem 2 was Lemma 2. The generalization of Lemma 2 to the case nl + n z ~ n (instead of nl = n - 2 and n z ~ 2) is apparently the principal obstacle to a proof of the above hypothesis (in the case t ~ m - 1). Skolem proved this generalization in the case n = 5, nl = 2, n z = 3 and thus deduced that the number of solutions to (6.1) is finite when n = 5, m = 3, t = 2 [T. Skolem, Einige Satze uber p-adische Potenzreihen mit Anwendung auf gewisse exponentielle Gleichungen, Math. Ann. 111, No.3, 399/424 (1935)]. This indicates the validity of our hypothesis in the case n = 5 (under the condition t ~ m - I; the nondegeneracy of the module does not arise here since the field k has prime degree and hence has no subfields). The validity of the hypothesis was proved for m = 3 (and hence under the restriction that t ~ 2) by Chabauty [C. Chabauty, Sur les equations diophantiennes liees aux unites d'un corps de nombres algebriques fini, Ann. Mat. Pura Appl. 17, 127/168 (1938)]. His method, however, avoids consideration of the system (6.16) by introducing other more refined techniques. Hence the generalization of Lemma 2 remains unproved even in the case n l = n - 3 (except for the case n = 5 considered by Skolem).

PROBLEMS

1. Let the series f(t) = ao + alt + a 2t 2 + for all p-adic integral values of t. If

...

withp-adic integral coefficients converge

(k = 2, 3, ...),

show that the equationf(t) = 0 has precisely one solution in p-adic integers if I'p(ao) ;;. I'p(a l ), and has no solution in p-adic integers if I'p(ao) < I'p(a l ). 2. Let d> 1 be a square-free natural number, and let (a, b) and (a, b) be two nontrivial (distinct from (1,0» integral solutions of the equation

x3

+ dy 3 = 1.

Sec.6] Set

13 =

SKOLEM'S METHOD

a + b-tld and

13, =

a,

+ b, -tid in

. .

the cubic field K

301 =

R(-tld). Show that then

13 =13,

for rational integers U and v, at least one of which is not divisible by 3. 3. Keeping the notations of the preceding problem, assume that d", ± 1 (mod 9). Then in K we have the decomposition 3 = \)' (Problem 24 of Section 7, Chapter 3), and hence the completion Kp of the field K is of degree 3 over the field R, of 3-adic numbers. Assuming that v", 0 (mod 3), set t = u/v. Show that the number t, considered as a 3-adic integer, is a root of the equation

2:'" ant n =

(*)

0,

11=2

where an = (1 /n!) Sp«log 1])n), 1] = 13'. (Here Sp denotes the trace for the extension K p / R, .) Show that the series in (*) converges when t is any 3-adic integer. (Hint: Show that Sp(log 1]) = 0 and Sp 1]' = 3, 1], = 13 ,'.) 4. Show that the coefficients an in (*) satisfy v,(a2) = v,(a,) = flo + 3, v,(an) > flo + 3 where flo = 1I,(a'b'd) (II, being the 3-adic valuation). [Hint: Use the fact that if 1] = 1 + 3x, x = ab-tld 13, then

log 1]

== 3x -

for n> 3,

9 - x 2 + 9 x' (mod 34 +U),

2

and also the fact that the trace of any element of the ring 0,[ -tI d] is divisible by 3 (0, is the ring of 3-adic integers).] 5. Using Problems 1 to 4, show that the equation x' + dy' = 1, with d '" ± 1 (mod 9). has at most one nontrivial solution in rational integers. 6. Prove the assertion of the preceding problem in the case d == ± I (mod 9). [Hint: Recall that the number 3 factors in K = R(-tId) in the form 3 = \)2 q (Problem 24 of Section 7, Chapter 3), and carryover the considerations of Problems 3 and 4 to the direct sum K, = Kp lOB K q (see Section 2). The logarithmic function on K, is defined just as on a field; the series converges for all g = (a, f3) E K" where a and f3 are principal units in Kp and K q • The trace Sp(g) is defined as the trace of the matrix of the linear mapping f gt (f E K,), and therefore for any elements of i (Section 2) it coincides with the trace of the corresponding number of K.] 7. Let the seriesf(t) = ao + a,t + a2t2 -!- ... have p-adic integralcoefficients,andconverge for all p-adic integral values of t. If an is a p-adic unit and a, == 0 (mod p) for all s> n, show that the equation f(t) = 0 has at most n solutions in p-adic integers. 8. Let the sequence of integers -)0

Uo, Ul, ... , U II

,

•••

(**)

satisfy the recurrence relation Un = a 1 Un _, + '" + amUn_ m (am # 0) with rational integer coefficients a" ... ,am' Assume that the polynomial g;(x) = x m - a,x·' - I - ... - a., has no multiple roots. Show that there exists a natural number M with the following property: For each residue class modulo M, either all Un (with n in that class) are equal, or no number occurs infinitely often among the u. (that is, show that the sequence (**) is either periodic or else assumes any given value only a finite number of times). [Hint: Use the formula u. = A,a,· + ... + Ama mn [a, are the roots of (g;(x)] and the fact that for any prime p and natural number M the function aiM. = exp (x log aiM) will be an analytic function for all p-adic integral values of x.]

Sec. 7]

LOCAL ANALYTIC MANIFO LDS

303

a function on the points of V lying in some e-neighborhood of the origin (the neighborhood depending on the function). Hence we call the factor ring D the ring of analytic functio ns on V. Defini tion. The local manifold V is called irreducible if the ring of functions DjW. v on V has no divisors of zero. Otherwise it is called reducib le. The investigation of local analytic manifolds is based on three simple facts, one from algebra and two dealing with the properties of power series. We state them without proof, giving references.

Lemma 1. Let gl(t), ... ,gm(t) be polynomials of k[t] with leading coefficient 1. There is a system hI' ... , hr of polynomials in several variabl es, one for each coefficient of the gj' and with integer coefficients, such that if the coefficients of gl (t), ... , gm(t) are substituted for the corresponding variable s in hI' .. , , h" then hI = ... = hr = 0 if and only if gt> ... ,gm have a commo n root in some extension of k. If m = 2, then r = I and hI is the resultant of the polynomials gl and g2 . The general case is easily reduced to this case. The proof is given in " Modern Algebr a" by B. L. van der Waerden, Vol. II, Section 77, Ungur, New York, 1950. Lemma 2. Suppose that the power series f(x , .. , ,x ) is such that all l n terms of degree ... , CPk are power series in Xl' ... 'X - with zero constan t n l term. This corollary to the Weierstrass prepara tion theorem is proved by O. Zariski and P. Samuel, in "Comm utative Algebra," Vol. II, p. 145, Princet on University Press, Princeton, N.J., 1960. Note that the condition that the coefficient of x/ be nonzero always can be obtained after a linear change of variables. Further , it is easily checked that if .we have a finite setfl' ... ,fm of power series, then a linear change of variables can be found so that they all satisfy this condition simultaneously. Lemma 3. Any ideal W. of the ring D has a finite set of generators; that is there exist power series hI' ... , hs in W. such that any hEW. can be represe nted in the form

for some gt> .. , ,gs of D.

i !

304

LOCAL METHODS

[Chap. 4

The proof of this lemma can also be found in "Commutative Algebra," Vol. II, p. 148. We need Lemma 3 for the proof of the following theon:lil. Theorem 1. Every local manifold is a finite union of irreducible manifolds. Proof. Let the manifold V be determined by (7.1). If V is reducible, then

there exist power seriesfand 9 in .0, which do not vanish on the points of V in any neighborhood of the origin, but such thatfg is identically zero on V in some neighborhood of the origin. Let VI and VI' be the manifolds obtained by adjoining to the system (7.1) the equations f(X) = and g( X) = 0, respectively. It is clear that VI and VI' are submanifolds of V and that

l

t

°

If the manifolds VI and VI' are irreducible, then the theorem is proved. If one of them is reducible, then we can, in the same way, represent it as the union of two proper submanifolds. Continuing this process, we either represent V as a finite union of irreducible submanifolds, or we obtain an infinite sequence of manifolds V

= Vo

# VI # V # ....

(7.2)

~Vi

be the ideal of the variety

2

We show that the second case is impossible. Let Vi' From (7.2) it follows that ~v

# ~lvl # ~V2 # ....

(7.3)

Denote by ~ the union of the ideals ~Vi' By Lemma 3 the ideal ~ is generated by a finite system of series hI' ... , h s • Since each series of ~{ is contained in some ideal ~Vi' then there is an integer k such that all the series hi> .. , , hs are contained in ~Vk' But then ~ c ~Vk and hence ~Vk = ~{Vk+ 1 = "', which contradicts (7.3). Theorem I is proved. We now describe a general method for studying local manifolds, based on a reduction to manifolds in spaces of lowest possible dimension. Let the manifold V in the space f(n be defined by the equations (7.1). Assume that V is different from kn, so that the series fI' .. , ,fm (m ~ 1) are not identically zero. Also assume that we have made a linear change of variables so that the polynomialsfi all satisfy the conditions of Lemma 2. Then, by this lemma, we can find power series eI(X), ... , em(X) in .0 with nonzero constant term, such that (7.4)

where qJij = qJij(XI' .•. ,xn - I ) are power series in n - 1 variables with zero constant term. Since ei(X) #- in some e-neighborhood of the origin, then the

°

J

Sec. 7]

LOCAL ANALYTIC MANIFOLDS

305

local manifold V is also given by the system of equations 91(X) = 0, ... ,9m(J{) = 0,

(7.5)

where each 9j is a polynomial in X n with leading coefficient 1. We now apply Lemma I to these polynomials. The corresponding polynomials hI' '" , h, in the coefficients of the polynomials 91' ... ,9m will be power series in Xl' ... , Xn- l without constant term, and since all the qJij converge in some e-neighborhood of the origin, then the series hI' ... ,h, will converge in the same neighborhood. Consider now the local manifold W in the space [(n-l defined by the equations hl(Xl , ... ,xn- l )

= 0, ... ,h,(xl , ... ,xn- l ) = 0.

It is clear that a point (al> , an-I) E [(n-l belongs to W if and only if there , an-I' an) E V. Thus W is a projection of the exists an an such that (at. manifold V into the hyperplane X n = 0. Here each point (aI' ... , an-I) E W is the projection of a finite set of points (aI' ... ,an-I' an) E V, since an is a common root of the polynomials 9i(a l , ... ,an-l,xn). The passage from the manifold V to its projection W gives us a method for investigating local manifolds. ,

Definition. By a curve in the space [(n we mean a system of n integral formal power series wl(t), .. , , wn(t) which have zero constant term and have coefficients in k or in some finite extension of k (and are not all identically zero).

For our purposes it is not necessary to assume that the series Wi(t) = + a i ,t 2 + ... converge, and it is simpler not to do so. Thus a curve does not consist of a set of points, but only of a collection of series Wi(t).

a i ,!

Definition. We shall say that the curve wl(t), ... , wn(t) lies on the manifold V if for any series I(x l , ... , x n) of the ideal ~v the power series I(wl (t), ... , wn(t)) is identically zero.

Our basic result on local analytic manifolds is the following. Theorem 2. Any local manifold either coincides with the origin or contains some curve. The proof will be given by induction on the dimension n. By Lemma 3 the ideal ~v has a finite set of generators. Hence we may assume that the system (7.1), which defines the variety V, consists of a set of generators for the ideal ~v . For n = I the variety V consists only of the origin if at least one of the series Ii is not identically zero, and coincides with P

If

306

LOCAL METHODS

[Chap. 4

if allfi are identically zero. In the second case any series w(t) satisfies the system (7.1). Now let n > 1. The assertion of the theorem is clear if allfi are identically zero (or if m = 0). Therefore we assume that none of the series fl' '" ,fm (m > 0) are equal to zero. Further, assume that these series are in the form given by Lemma 2, so that instead of the equations (7.1) we have V given by (7.5), where 9i is determined by (7.4). Let W be the projection of V in k·- I . By induction, we assume the theorem valid for W. If W coincides with the origin, then the local manifold V will be defined by the equations 9;(0, .. , ,0, x.) =

°

(1

~

~

i ~ m),

~

i

p

F

(7.6)

whose coefficients lie in the field k l {t} of formal power series in t over k l . Further, these polynomials have a common root X. = ~ in some finite extension n of the field kl{t}. By Theorem 6 of Section 1 the field n is contained in the field of formal power series k'{u}, where u e = t for some natural number e, and k' is a finite extension of k l • Hence the element ~ can be represented as a power series ~ = w(u) with coefficients in k'. Since ~ is a root of the polynomials (7.6), which have leading coefficient 1 and integral coefficients in the field k l {t}, then the series w(u) is an integral element of the field k' {u}, that is, it does not contain any term with negative exponent. In the representation (7.4) all the series qJij have zero constant term. Substituting the series wl(u e ), •.• , w._I(u e ) for XI' ... , X._I in (7.4), and substituting w(u) for X., we see that the series w(u) has zero constant term and that (1

c(

o

w

i ~ m),

and it will also coincide with the origin. If W is different from the origin, then W contains a curve wl(t), ... , w.-I(t). Let k l denote a finite extension of the field k which contains all coefficients of the power series WI' ... , w._ I . From the definition of W it follows that if we substitute the series WI (t), ... , W. -I (t) in the series 91' ... ,9m for XI' ... , X._ I then we obtain m polynomials in X., (1

S

~

"

T L

cc

al

si s~

h T1 ,( ,(

Sl m).

Since the series WI' .. , , W. -I are not all zero, then the set of series WI (u e ), .•. , w._I(u e), w(u) is a curve in k·. By assumption the seriesfl' ... ,1m' and thus also the series 91' ... ,9m generate the ideal ~[y. Hence for any seriesf(xl , •.• , x.) of my we have

a

f

a and this means that the curve wl(u V. Theorem 2 is proved.

e

), '"

,

w._I(u e ), w(u) lies on the manifold

Sec. 7]

LOCAL ANALYTIC MANIFOLDS

307

Theorem 3. If V and V' are two local manifolds in j(n, where V is not contained in V', then there is a curve in i(n which lies on V and does not lie on V'. Proof. We can assume that the manifold V is irreducible, since otherwise we may replace V by one of its irreducible components. Let the manifold V' be defined by the equations

where F j is a series of the ring.o. Since V¢ V', then at least one of the series F j does not vanish on the points of V (in any neighborhood of the origin). We denote this series by F(X) and will show that there is a curve wl(t), '" , wn(t) on V for which F(w l (t), ... , wn(t)) ¥- 0.

The proof will proceed by induction on n. We can clearly assume that the series F(X) satisfies the conditions of Lemma 2, so that there exists a series e(X) = e(xl , ... , x n) ED with nOnzero constant term so that e(X)F(X)

= G(Xl>"" x n) = X/ + !/JIX/- I + ... + !/Jk'

(7.7)

where !/JI, ... , !/Jk are series in XI' •. , ,xn - I • In the case V = j(n (in particular, if n = 1) Theorem 3 is proved, for example, by taking wl(t) = '" = wn-I(t) = 0, wn(t) = t. If V¥- j(n, then we consider the projection We j(n-I of the manifold V (here we assume that the series II' ... ,fm, as well as F( X), satisfy the conditions of Lemma 2; as we have seen, this can be achieved by a linear change of variables). The manifold W is also irreducible, since the ring of functions on it, that is, the factor ring .on-I/'lI w = .on-I, is a subring of the ring of functions .o/'lI v =.6 on V (as .on -I c .0 and 'lI w c 'ltv). For each series lED we denote by J the corresponding function of .6. It follows from (7.4) that

x:' + iPiIX/,-1 -* ... + iPik, = 0, and this means that the function xn of the ring .on -I' It follows that the functions G

.6 is integral over the subring

.7: - k - I .7: = X-nk + '!'IX + ... + '!'k n

also are integral over .6n-I' We take an equation

G + LI G S

I

S -

+ '" + L s =

°

(7.8)

LOCAL METHODS

308

[Chap. 4

with smallest possible s. It is clear that Is ¥- 0, since then we could obtain an equation for G with smaller s. Hence the series L s E D n - 1 does not vanish on the points of W (in any neighborhood of the origin). By the induction hypothesis there exists a curve w 1(t), ... ,Wn -1(t) in the space f(n-1 which lies on W and is such that L s(w 1(t), ... , Wn -1(t)) ¥- 0. In the proof of Theorem 2 we saw that there exist curves of the form W1(U'), ... , w n - 1(u'), w(u) which lie on the manifold V. We shall show that for such a curve (w(u) ¥- 0)

and hence that this curve does not lie on the manifold V'. For if the series on the left were identically zero, then by (7.8) we would have L s(Wl(U'), ... , w n - 1(u') = 0,

and after replacing u· by t, Llwl(t), ... , W n -l(t) = 0,

which is impossible by choice of the curve Wl(t), ... ,Wn -l(t). Theorem 3 is proved.

>.4

an on

)0-

W aw he

CHAPTER

5

Analytic Methods

Dn

3

In Chapter 3 we saw how important a role the number h of divisor classes of an algebraic number field played in the arithmetic of the field. Thus one would like to have an explicit formula for the number h, in terms of simpler values which depend on the field K. Although this has not been accomplished for arbitrary algebraic number fields, for certain fields of great interest (such as quadratic fields and cyclotomic fields) such formulas have been found. The number of divisor classes is a characteristic of the set of all divisors of the field K. Since all divisors are products of prime divisors and the number of prime divisors is infinite, then to compute the number h in a finite number of steps we must use some infinite processes. This is why, in the determination of h, we shall have to consider infinite products, series, and other analytic concepts. The apparatus of mathematical analysis can be applied to solve many problems of the theory of numbers. In this chapter we give an example of the application of this apparatus by using it to compute the number of divisor classes. 1. Analytic Formulas for the Number of Divisor Classes

1.1. The Dedekind Zeta Function The determination of the number h of divisor classes of the algebraic number field K is based on consideration of the Dedekind zeta function (K(S), defined by (1.1) 309

[Chap. 5

S

where a runs through all integral divisors of the field K, and N(a) denotes the norm of the divisor a. We shall show that the series on the left side of (1.1) converges for I < s < 00, and is a continuous function of the real variable s on this interval. Further, we shall obtain the formula

e:

ANALYTIC METHODS

310

= hK,

lim (s - IKK(s)

(1.2)

5-1+0

where K is a constant which depends on the field K in a simple manner, and which will be computed in the course of the proof. Formula (1.2) becomes valuable because the function (K(S) also has a representation as an infinite product (K(S)

=

I) 1 -

[ljN(p)'J ,

(1.3)

where a runs through all integral divisors of the given divisor class C, and the exterior summation is taken over all h classes C. To prove that the series (1.1) converges, it suffices to show that each of the series I

=

VI

VI

a

I

carried out over all prime divisors p of the field K, this representation being called Euler's identity. If for the field Kwe have a good knowledge of the prime divisors (that is, if we know how rational primes factor into prime divisors in K), then we can obtain an explicit expression for h from formulas (1.2) and (1.3). By this route we shall obtain formulas for h in later sections when K is a quadratic or a cyclotomic field. We break the series (1.2) into the sum of h series

fcCs)

d

L N( a )5

(1.4)

aEC

o s a r

~

t s

c s t

r ~

t t

E

converges for s > I. Further, if we show that for each class C the limit lim (s - l)fc(S) 5-1 +0

exists and has the same value K for each divisor class C, then we will have obtained formula (1.2). We now transform the series (1.4) into a series over certain integers of the field K. In the inverse divisor class C- I we choose an integral divisor a'. Then for any a E C the product aa' will be a principal divisor: aa'

= (c -2 Ix,1 e .

316

ANALYTIC METHODS

[Chap. 5

The fundamental domain X hence consists of the points indicated in Figure 7 (the boundary rays which lie closest to the positive x-axis are not included in X).

FIG. 7

, i FIG. 8

Now let K be an imaginary quadratic field. Since here s = 0, t = 1, then r = s + t - 1 = O. Hence the fundamental domain X consists of all points X = Y + iz for which N(x)

.= y2 + Z2 =1=

[see Figure 8 for the case K = R(

0, (0 ~ arg

.J- 3), with m =

x

< 2;)

6].

1.3. Computation of the Volume We now turn to the computation of the n-dimensional volume of the set T, which consists of all points of the fundamental domain X for which IN(x) I ~ 1.

~p.5

e7 ed

Sec.1]

FORMU LAS FOR NUMBE R OF DIVISOR CLASSES

317

It will be shown in the course of the compu tations that the volume exists

and is nonzero. (For quadra tic fields, the set Tis indicated in Figures 7 and 8.) We first show that the set T is bounde d. In every ray which is contain ed in the cone X, there is one and only one point x for which IN(x) I = I. Denote the set of all such points by S. It is clear that T consists of all points ~x (0 < ~ ~ I), where x runs through all points of S. Consid er the formula (1.7) for any point x E 9l" with nonzer o norm. We compu te the sum of the compo nents of this vector. By (3.15) of Chapte r 2 the sum on the left equals InIN(x)l. By (3.18) of Chapte r 2 the sum on the right is ~(s + 21) = n~. This means that ~ = (Ijn) InIN(x)l, and (1.7) can be written in the form (1.13)

Now if XES, then InIN(x)1 = 0, and hence the point lex) = (II (x), .. , , Is+rCx)) E 9l,+t is represented in the form lex) = ~Il(el) + ... + ~/(er), where ~ ~i < I. It follows that there is a constan t p such that lix) < p, and then IXkl < eP for I ~ k ~ sand IXs+jl < eP/ 2 for I ~j ~ 1 for all XES [see (3.13)

o

and (3.12) of Chapte r 2]. This shows that the set S, and hence also the set T, is bounde d. We shall replace the set T by anothe r set which is easily obtaine d from T, and which has the advanta ge that it is defined by a simpler set of conditi ons. We first note the following almost obviou s lemma. Lemm a 2. If e is a unit of the field K, then the linear transfo rmation of the space 9l" given by x -+ xx(e) is volume-preserving. Under any nonsing ular linear transfo rmation the volume of a set is multiplied by the absolut e value of the determ inant of the matrix of the transfo rmation [see (4.2) of Chapte r 2]. We showed in Section 3.1 of Chapte r 2 that the determ inant of the transfo rmation x -+ xx(e) equals N(x(e)), that is, equals N(e) = ±1. As before, let ( denote that mth root of I for which 0"1(0 = cos(2nj m) + i sin(2njm). Consid er the sets T k (k = 0, I, ... , m - I), obtaine d from Tby the linear transfo rmation x -+ XX((k) (To = T). By Lemma 2 we have v(T k) = veT) (provided the volume of one of the sets exists). Since IN(xx((k))1 l(xx((k) )

= IN(x)N((k)1 = IN(x)l,

= lex) + l((k) = lex),

arg (XX((k))1 = arg x I

2n k, m

+-

then (by the definition of the fundam ental domain X) the set T consist s of all k

ANALYTIC METHODS

318

points x

E

[Chap. 5

91" for which:

(1) 0 < IN(x) I ~ 1. (2) The coefficients in (1.13) satisfy 0 ~ ~ i < 1. (3) 2nkjm ~ argx 1 < (2njk)(k + 1). Thus To, T 1 , '" ,Tm - 1 are pairwise-nonintersecting and their UnIon Uk'';o' T k is defined by conditions (1) and (2) [without condition (3)]. Let f denote the set of all points x E Uk'=-Ol T k , for which Xl > 0, ... , X s > 0 [see (3.2) of Chapter 2]. We fix a set of 5 signs 15 1 , •• , , bs (15; = ± 1). Ifwe multiply all points in 91" by the point (15 1, ... , bs ; 1, ... , 1) E £s.t =91", we obtain a volume-preserving linear transformation of 91". If we apply a1l2 s such linear transformations to f, we obtain 2S pairwise-nonintersecting sets whose union coincides with Uk'':-Ol T k • If we can show that f has nonzero volume ii, then it will follow that T has a well-defined volume, which is given by v(T)

"

'.

I

2s

=-

m

(1.14)

ii.

(For real quadratic fields f is that part of T which is contained in the first quadrant, and for imaginary quadratic fields f coincides with the unit disc minus the origin, see Figures 7 and 8.) The vector equation (1.13) yields the following system:

(j

'

where ej = 1 if 1 by the formulas

~j ~ 5,

and ej = 2 if 5 (k

Yj

= 1, ... , s + t),

+ 1 ~j ~ 5 + t.

We change variables

= 1, ... , s),

= Ps+j C.OS C{Jj

}

(j

= 1, ... , t).

Zj=ps+jsmC{Jj (Here the real numbers Y j and Z j are given by X s + j = Y j + iz j ' 1 ~ j ~ t, see Section 3.1 of Chapter 2.) The Jacobian of this transformation is easily computed to be Ps+1·"Ps+t. Since l/x) = Inp/ i and N(x) = flj~t1 p/i (we assume that Xl> 0, ... , X s > 0), then in terms of the variables P1' ... , Ps +" C{J" '" , C{J" the set f is given by the conditions: (1) P1 > 0, ... ,Ps+t > 0, flj~\ p/ i ~ 1.

(2) In the equations

11

e. (s+t ) In p/ i = ~ In p/'

(j

= 1, .. , ,x + t)

the coefficients

~k

r

+ k~l ~klj(£k)

satisfy 0

~ ~k

< 1 (k = 1, ... ,r).

Sec.1]

FORMULAS FOR NUMBER OF DIVISOR CLASSES

319

Since these conditions do not impose any restrictions on the variables then they independently take on all values in [0,2n). We now replace PI' ... ,Ps+ t by the new variables~, ~l' ... ,.~, by the formulas qJI' .. , , qJt,

(j

= 1, '" , s + t).

(1.15)

Adding these equations and noting that s+t

s+t

Le

j

L liCk) = 0,

= n,

j~1

j~l

(1.16)

we obtain s+t

~

= [lp/J.

(1.17)

j~l

The set Tis determined by the conditions (k

It is now clear that the volume

v = veT) exists.

= 1, ... , r).

Since

the Jacobian of the transformation (1.17) equals

J= Ps+ t

n~

Ps+tl () Ps+tl () s+t Cl ... s+t C, es + t es + t

In the last determinant we add all rows to the first row. Considering (1.16) and (1.17) and recalling the definition of the regulator R of the field K (see Section 4.4 of Chapter 2), we obtain R

IJI=-t---2Ps+I"'Ps+t

'!

320

[Chap. 5

ANALYTIC METHODS

v:

It is now easy to compute the volume

v = r·-j dX 1 ... dxsdy1dz l '" dy,dz, (T)

= J ... J Ps+ 1 ... Ps+' dpi ... dps+, d({J1 ... d({J, (T)

= J:"d({J1 ... J:"d({J, roo J Ps+ 1

oo'

Ps+' dpi ... dps+,

= 2'n' r·-jIJIPs+I'" Ps+t d~

d~I'" d~r

_ 'Rfl d~Jld~I'" -n 0 0

n'R.

fld~r = 0

Substituting this value of v in (1.14) we finally obtain,

2s n'R

v(T)=- . m

1

::1.

,

104. Dirichlet's Principle II ;

We first consider the function (K(S) when K is the rational field R. Since integral divisors in R can be identified with natural numbers and N(n) = n, then (1.18) Hence the Dedekind (-function for R coincides with the Riemann (-function (s). We shall show that the series (1.18) converges for s > 1. Since the function IIx' is decreasing for X> 0, then

J

"+I

,

dx 1 .. < ---;< x n

J'

,-I

dx --;, x

where the inequality on the left holds for n ~ 1, and that on the right for n ~ 2. Hence for any natural number N > I we have

N+ 1 dx I1

"

i !

N1

- 0, the equation F(~x) = ~nF(x) holds. (2) The set T, which consists of all points of X for which F(x) ~ 1, is bounded and has nonzero n-dimensional volume v = v(T). The points of the cone at which F(x) = 1 form a surface which intersects each ray of the cone in precisely one point, and which separates from the rest of the cone a bounded subset with nonzero volume. It is clear that the giving of such a surface in X is equivalent to the definition of such a function F(x). Let Wl be an n-dimensional lattice in illn with the volume of a fundamental parallelepiped denoted by ~. Consider the series

_

((S)

=

,,1 L. xE!lJlnX

(1.20)

(s> 1),

-

F(X)5

taken over all points x of the lattice Wl which are contained in the cone X. This series depends on the cone X, the function F, and the lattice Wl. Theorem 3. Under the assumptions given above, the series (1.20) converges for all s > 1 and

lim (s - 1)C(s) 5-1 +0

= .!:.. .

(1.21)

~

Proof. For any real r > 0 we denote by Wl r the lattice which is obtained by contracting Wl by a factor of r. The volume of a fundamental parallelepiped of Wl is then given by ~/rn. If N(r) is the number of points of the lattice Wl r which are contained in the set T, then by the definition of volume we have . ~ . N(r) v = v(T) = hmN(r)--;; = ~ hm - n r-oo

r

r-oo

r

.

(1.22)

Consider the set rT, obtained by expanding T by a factor of r. It is clear that

,

I

322

[Chap. 5

ANALYTIC METHODS

N(r) also equals the number of points of the lattice Wl contained in rT, and that this is equal to the number of points x E Wl (\ X, for which F(x) ~ r". The points of Wl (\ X can be arranged in a sequence {xd so that 0< F(x l ) ~ F(x 2 )

~ ••• ~

F(x k) ~ ., ..

Set z!i(Xk) = rk ·

The points Xl' ... , Xk belong to the set rkT, so N(r k) ~ k. But for any e> 0, the point Xk does not belong to the set (r k - e)T, so N(r k - e) < k. Thus N(r k - e) < k

~

N(r k).

so that N(rk - e)(r k - e)" k - - < ---;; e) (rkrk rk

--:---"----~"

Taking the limit as k

-+ 00,

that is, as rk -+

~

N(r k) rk

--" .

00, and

considering (1.22), we obtain

k v lim--=-. k~oo F(Xk) L\

(1.23)

We compare the series (s) = Lk=l IjF(xk)S with the series (1.18). Since limk~oo [kSjF(xSl = (vjL\)' =I 0, then along with the series (1.18) the series (1.20) also converges (if, of course, s> 1). Let e be any positive number. By (1.23) we have _1 < (~ + e)~ (~L\ _e)~k < F(Xk) L\ k

for all k greater than some sufficiently large k o . Hence

V )S 00 1 00 1 (V )S 00 1 L -< LS L - - < -+e (--e L\ k=ko k k=ko F(xkY L\ k=ko k S for all s > 1. We multiply this inequality by s - 1 and let s ten l to 1 from the right. Since lims~l(s - I)Li,,=~/(IjkS) = 0, then by (1.19), kns~l+O(S - 1) D>')=ko (Ijk S) = 1. Since also Ijms~l(s - l)LZo=-ll[ljF(xk),l = 0, we obtain the inequality

~

. 1

l:

L\

- e

~ lim (s - 1)~(s) ~ lim (s - l)~(s) ~ ~ + e. S~o

s~l+O

L\

and since e was arbitrary, this proves Theorem 3.

,,;,I' ,

II Iii

'Ii

Remark. There is a certain similarity between (1.21) and (1.22). To make this similarity more precise, we assume that the volume L\ of a fundamental parallelepiped of the lattice Wl is equal to 1, and we write them in the form

Sec. 1 ]

FORMULAS FOR NUMBER OF DIVISOR CLASSES

lim (s - 1)~(s)

= V,

323

(1.21 ')

s-! +0

lim r-oo

, I

, I

~N(r) = v.

(1.22')

r

Both limits have the same value, the volume of the set T. The volume is determined in (1.22') in the following way. The lattice Wl is shrunk by a factor of r, and the number N(r) of points of Wl r which are contained in T is determined. Then the number N(r) is multiplied by the volume Ijr n of a fundamental parallelepiped of the lattice Wl n , and finally we pass to the limit as r -. 00. The same idea is involved in (1.21 '). Here the sum (s) plays the role of the number N(r), and the factor (s - 1) corresponds to the factor Ijr n • We take the limit as s -. 1 from the right instead of as r -. 00. Turning to the fundamental domain X of an algebraic number field K, we see that the function F(x) = IN(x) I satisfies conditions (1) and (2). Hence we may apply Theorem 3 to the series (1.8), and this means that it converges for s > 1 and that relation (1.9) holds. We have now proved all assertions which were used in the first paragraph, and hence have completed the proof of Theorem 2.

1.5. Euler's Identity To use the formula (1.2) to compute the number h, we must have some other method for determining the limit lims~l(s - lKK(s). In some cases this can be done by using the representation of (K(S) as an infinite product, known as Euler's identity. Theorem 4. For s > 1, the function (K(S) can be represented as a convergent infinite product 1 (K(S) = 1 - [ljN(p)'J ,

IJ

,

where P runs through all prime divisors of the field K. Proof. For every prime divisor P we have

1 1 1 - [ljN(p)'J = 1 + N(p)'

I'

1 + N(p)2s

+ ....

(1.24)

Let N be any natural number, and Pb '" , Pr all prime divisors with norms not exceeding N. Multiplying the absolutely convergent series (1.24) for P = PI' ... , Pr' we obtain

)-1 =

1 1 - --s ( N(p)';;N N(p)

fl

I kr~O N(PI 00

k,

k, .....

1

1 s = I'--s , ... Pr) Q N(a) kr

I !

ANALYTIC METHODS

324

[Chap. 5

I',

where in the sum a runs through all integral divisors of the field K which are not divisible by a prime divisor with norm exceeding N. Comparing the . series with the series CK(S) = IljN(a)', we obtain

I'

1-)-1 - CK(S) I < I 1 n (1 - -N(a)S N(p}' I since the series I' contains all terms corresponding to integral divisors with N(a»N

N(p)';;N

norm

~ N.

Since for

S

> I the series (1.1) converges, then 1

I --+0 N(a»N N(a)' as N -+ 00, and this proves the theorem. Theorem 4 will be valuable because, along with Theorem 2, it establishes a connection between the number h and prime divisors of the field K. As we remarked in Section 1.1, if we know all prime divisors of the field K, then using Theorem 4, the left side of (1.2) can be computed, and this allows us to obtain a formula for h. On the other hand, since Kh =F 0, Theorem 4 gives an important property of prime divisors in a field K. For example, we shall use it in Section 3 to obtain the celebrated theorem of Dirichlet on the distribution of rational prime numbers in arithmetic progressions.

PROBLEMS

1. Use the convergence of the series series

~:=l (l/n')

(s> 1) to show that when s> 1, the

where II runs through all prime divisors of the field K, also converges. 2. Use Problem 1 to prove the convergence of the product

I] 1 -

1 [1/N(Il)']

(s> 1).

Deduce that the series

converges. 3. Let at and bt (k;;. 1) be positive real numbers with lim t _., bt/at = c. Show that if the series ~:': 1 at' converges for s > 1 and lim'_l + 0 (s - 1):= 1 at' = A, then the series ~:=l b t ' also converges for s> 1 and

.,

lim (s - 1) &-+1+0

Lb.' = cA.

k=l

5

h

Sec. 2]

DIVISOR CLASSES OF CYCLOTOMIC FIELDS

325

4. Let C be any divisor class of the algebraic number field K. Denote by Z(t, C) the number of integral divisors a of the class C for which N(a) ~ Show that

t.

.

1~~

Z(t, C)

-t- =X=

2'+'1T'R

mvlDj'

5. Let .p(a) denote the number of integral divisors of the algebraic number field K with norm a. Show that

~K(S) = ~(s)

i

~

"= 1

n' '

where

[/L(a)

is the Mobius function].

2. The Number of Divisor Classes of Cyclotomic Fields

Let m be a natural number, and let ( be a primitive mth root of 1. Since the mth roots of 1 in the complex plane divide the unit circle into m equal parts, the field R«) is called the mth cyclotomic (circle-dividing) field. In this section we shall use Theorems 2 and 4 of Section 1 to find a formula for h, the number of divisor classes in a cyclotomic field. To do this we must determine the factorization of rational primes into prime divisors in such fields. We first determine the degree of the field R«).

2.1. Irreducibility of the Cyclotomic Polynomial The degree of the field R(O equals the degree of the minimum polynomial of the number (over the rational field R. In this section we shall show that the minimum polynomial of ( is the polynomial m = m(t) =

TI (t _ (k) (k, m)= 1

(the product is taken over the indicated residue classes modulo m), which has as roots all primitive mth roots of 1. Since the degree ofm equals the value of the Euler function cp(m), it will follow that (R(O : R) = cp(m). The polynomial m(t) is called the mth cyclotomic polynomial. We first show that the coefficients of m(t) are rational integers. For m = 1, this is clear (1 = t - 1). We proceed by induction on m. Since every mth root of 1 is a primitive root of some degree dim, then t

m

-

1 = TI d , d

where d runs through all divisors of the number m. By the induction assumption the polynomial F = TIumd has rational integral coefficients and its

I·

326

ANALYTIC METHODS

[Chap.5

leading coefficient is I. Hence the coefficients of ", = (t'" - l)jF are also rational integers. As usual, Z denotes the ring of rational integers, Zp the field of residue classes modulo the prime number p, and for a E Z we denote the corresponding residue class in Zp by ii. If /(t) is a polynomial with rational integer coefficients, we denote by J(t) the polynomial obtained from / by replacing all coefficients by their residue classes modulo p. It is clear that the mapping / -+ Jis a homomorphism of the ring Z[t] onto the ring ZAt]. Since (J + gY = jP + iY, and aP = ii, then in the ring Zp[t] we have the formula

(J(t»P = J(tP).

(2.1)

Set h = t'" - I. If p does not divide m, then the polynomial Ii of ZAt] is relatively prime to its derivative and hence has distinct roots. Noting that ill", divides Ii, we have the following assertion.

I.

Lemma 1. If the prime number p does not divide m, then the polynomial

ill E Zp[t] has no multiple roots. If/(t) is the minimum polynomial of (, then ", = /G, where G and/both belong to the ring Z[t]. If p is any prime not dividing m, then (P is a primitive mth root of I; that is, ",«(P) = O. We shall show that (P is a root off Otherwise we would have G«(P) = O. Then consider the polynomial H(t) = G(tP). Since H(O = G«p) = 0, then H is divisible by f, that is, H = /Q, where Q E Z[t]. Passing to Zp we obtain H = JQ. But by (2.1), H(t) = G(t P ) = (G(t»p, so that

GP =JQ. Let l/i be any irreducible factor of J (in the ring ZAt]). It follows from the last equation that G is divisible by l/i. But then it follows from ill", = JG that ill", is divisible by l/i2, contradicting Lemma I. Thus (P cannot be a root of G(t), and hence is a root of/(t). If (' is any root of "" then (' = (k, where k is relatively prime to m. Let k = PtP2 '" Ps' We have just shown that (PI is a root of /(t). Analogously, replacing ( by (PI, we find that (PIP2 is a root of /(t). Continuing this process, we find that (k is a root of /(t). We have shown that any root of ", is also a root off, and hence ", = f We formulate this result in the following theorem. Theorem 1. For any natural number m, the cyclotomic polynomial ", is irreducible over the field of rational numbers. Corollary. The degree of the mth cyclotomic field is qJ(m) [where qJ(m) is Euler's function] .

1

.1

:1. .1 i 1'1,\ I

Sec. 2]

DIVISOR CLASSES OF CYCLOTOMIC FIELDS

327 I , I

2.2. Decomposition of Primes in Cyclotomic Fields

Since the mth cyclotomic field R(O has degree qJ(m), then the numbers 1, (, ... , ('I'(m)-l

(2.2)

form a basis for R(O over R. Lemma 2. If the prime number p does not divide m, then it does not divide the discriminant D = D(I, (, .. , , ('I'(m) -1) of the basis (2.2). Proof. The discriminant D equals the discriminant D(m) of the cyclotomic

polynomial m' The residue class D(m) E Zp of the number D(m) clearly coincides with the discriminant D( 0 and represents a continuous function on the interval (0, (0). For if X i' Xo, then LX(k) = 0, where k runs through a complete set of residues modulo m. Representing the natural number n in the form n = mq + r, o ~ r < m, we find An = 1 x(k) = 1 x(k), SO that IAnl ~ r < m. Turning to the function CK(S), we multiply (2.12) by s - 1 and take the limit as s -+ 1 from above. By (1.19) we find that

Iz=

lim (s -

I;; =

lK K(s) = F(1)

,-1+0

where L(l, X)

=

f

TI L(I, X),

x(n)

n= 1

(2.14)

x*xo

n

(2.15)

Note that since the series (2.15) does not converge absolutely, we must keep in mind that its terms appear in order of increasing n. Comparing (2.14) with Theorem 2 of Section 1, we obtain the following formula:

wJfDJ

h = 2'+' tR F (1)

n

TI

1.*Xo

L(1, X)

(2.16)

332

ANALYTIC METHODS

[Chap. 5

(here w denotes the number of roots of 1 contained in K). The expression (2.16) for the number of divisor classes of a cyclotomic field is not definitive, since it still contains the infinite series £(l, X). The summation of these series will be carried out in the next section. 2.4. Summation of the Series £(1, X).

Assuming that X is a nonunit character modulo m, we turn to the series (2.13). Omitting those summands which are zero and noting that if n1 == n 2 (mod m), then x(n 1 ) = x(n 2 ), we arrive at the following form (valid for s> 1): £(s, X) = L: X(x) L: ~1 (x,m)=l

nEx(modm)n

The inner series can be written in the form

~~s '

L,

n=l

n

where for n == x (mod m),

= {I0 n

C

for n =1= x (mod m).

To find a convenient way of writing the coefficients Cn , we consider the following formula: for r == 0 (mod m), for r

=1=

0 (mod m),

where 2n

C= cos - + i m

. 2n SIn -

m

is a primitive mth root of 1. We stress that when considering the algebraic properties of a cyclotomic field, it does not matter which primitive mth root of 1 is denoted by " but for analytic computations we must fix a definite complex root. Hence we have 1 m-1 cn = - L: c(x-n)k.

I I

Ii

I

m k=O

Thus £(s, X)

L:

=

00

X(x)

(x.m)=l

1 m-1(

= -

m

L:

k=O

1 m-1

L: - L:

1

c(x-n)k

n

n=lmk=O

L:

(x,m)=l

X(x)CX

k)

s

L: 00

n=l

C- nk -s .

n

!

Sec. 2]

DIVISOR CLASSES OF CYCLOTOMIC FIELDS

333

We have already encountered the expression in parentheses in the case m = p in Section 2 of Chapter 1, where it was called a Gaussian sum. We now define Gaussian sums for arbitrary m. Definition. Let ( be a fixed primitive mth root of 1, and let Xbe a numerical character modulo m. The expression 'u(X)

=

L

X(x)(UX,

xmodm

where x runs through a full (or a reduced) system of residues modulo m, is called the Gaussian sum corresponding to the character Xand the rational integer a. The Gaussian sum 'u(X) depends not only on Xand the residue of a modulo m, but also on the choice of the root (. In'the future we shall always assume that ( is cos(2nlm) + i sin(2nlm). The Gaussian sum with this choice of ( is called normed. The sum 'l(X) will also be denoted by ,(X). If X is not the unit character, then 'o(X)

=

L

(x,m)=l

x(x) = O.

Hence our expression for L(s, X) can be written in the form 1

L(s, X)

=-

m-l

00

L 'k(X) L

mk=l

0=1

(-nk -S •

n

We can apply Lemma 4 to the series L:'=l(CO k ln S )(C k ;61 for k ;6 0, so L~~l cnk = 0). By this lemma our series converges for 0 < s < 00 and represents a continuous function of s. Hence we may set s = 1 in this last equation and obtain

To find the sum of the inner series, we turn to the power series L:'=l(zOln). It is well known that it converges for Izi < 1 and represents there that branch of the function -In(l - z), the imaginary part of which (that is, the coefficient of i) is contained in the interval (- n12, nI2). Since this series also converges at the point z = C k (on the unit circle), then by Abel's theorem 00 (-ok - = -In(l-C k ), 0=1 n and hence

L

(2.17)

i

'!

i

ANALYTIC METHODS

334

[Chap. 5

Hence we have obtained a finite expression for the series L(l, X). Substituting it in (2.16), we obtain a formula for the number of divisor classes of a cyclotomic field which does not contain any infinite series. The formula (2.17) can be further investigated and considerably simplified. In the next section we shall do this, but only for the case when X is a primitive character. In Section 5 we shall apply these results to the further study of the formula for h in the case of the lth cyclotomic field, with I a prime. In this case the formula has particularly important applications. 2.5. The Series L(l, X) for Primitive Characters

We shall show that if X is a primitive character modulo m and (a, m) = r> I, then

Set m = rd. It is clear that Ca is a primitive dth root of I, and therefore Ca = Ca , provided that z == I (mod d). We take as z a number for which (z, m) = I, z == I (mod d) and X(z) i' I (the existence of such a d is guaranteed by Theorem 4 of Section 5 of the Supplement). As x runs through a complete system of residues modulo m so does zx, so that %

Ta(X)

=

L:

X(zxW:

x

= X(z)

x mod m

L:

X(xW

X

= X(z)ra(X)·

x mod m

Since X(z) i' I, it follows that Ta(X) = O. Further, if (a, m) = I, then

=

Ta(X)

x(a)-IT(x)·

Indeed, as x runs through a complete system of residues modulo m, so does ax and thus x(a)Ta(X)

=

L

x mod

X

x(axW t= TI(X) m

= T(X)·

\

Hence if X is a primitive character we can write (2.17) in the form

'I

i

L(l, X) = - T(X) L: i(k)ln(l - C k ). m (k,m)= I

(2.18)

We turn to the study of the sum Sl.

=

L:

i(k)ln(l - C k )

(2.19)

(k,m)= I

(k running through a reduced system of residues modulo m). The study of the

sum Sl. leads to two essentially different types of behavior. To distinguish these cases we need the following definition.

l.5

109 10-

:d. ve he lis

Sec. 2]

DIVISOR CLASSES OF CYCLOT OMIC FIELDS

335

Defini tion. The numerical charact er X is called even if X( -1) = 1 [and hence X( -x) = X(x) for all integers x], and is called odd if X( -1) = -1 [and xC-x) = -X(x)].

Since (x( - 1))2 = X(( - 1)2) = x(I) = 1,

then X( -1) = ± 1, and thus every charact er X is either even or odd. The numbe r 1 - C k (for 0 < k < m) can be represented as 1- C where

nk) = 2 sin nk( ~ cos (n "2 - ~

k

-n12 < nl2 -nklm < n12; k

In(l - C Further , since 1 -

C k and 1 -

(k

. (n"2 - -;; nk)) ,

Sin

therefore

= Inl1

)

+ i

~).

- Ckl + inG -

are conjugate, then

(1 k)

In(l - ( k ) = Inl1 - ( k I - .m 2" - m . (Note that the last two formula s are valid only when k lies between 0 and m.) Now assume that the charact er X (and hence also i) is even. Interch anging k and - k in (2.19), we obtain Sl.

=

L

X(k)ln(I - (k),

(k,m)= 1

and with (2.19) this yields

L

2Sl.=

(k,m)= 1

iCk)[ln(I -Ck)+ ln(I _(k)]

L

=2

(k,m)= 1

iCk)lnl1 - (kl = 2

L

X(k)ln 2 sin

(k,m)= 1 o 1. Note that the values of In £(s, Xo) will be real for the unit character Xo' We estimate the function R(s, X): «)

IR(s,

1

1

x)1 < L:pn=Zp L: -;; < L:pPP(

1)

v f

1

«)

v c s

< n=1 L: nn ( + 1) = 1.

t

s

Thus IR(s, X)! < 1 for all s > 1. Along with the numerical character X we consider the corresponding character of the group Gm (which consists of all residue classes modulo m which consist of numbers relatively prime to m), which we also denote by X. Let C run through all classes of the group Gm • Since X(p) = X(C) for p E C, then

[recall that X(p) = 0 if p divides m]. Setting f(s, C) =

L: peC

1

~

P

,

we can put (3.7) in the form In £(s, X) =

L: x(C)f(s, C) + R(s, X).

(3.8)

C

Since there are I + O. 3. The rational integer a is called an nth power residue modulo the prime p, if the congruence x· ;= a (mod p) is solvable. For any a and any n, show that there are infinitely many p such that a is an nth power residue. 4. Let the integers aI, .. , , a. be such that a 1 Xl ... a. X. is a square if and only if all X, are even. For any choice of e" ... ,e. (e, = ± I), show that there exist infinitely many primes p (not dividing a" ... ,a.) for which

Hint: Consider the sum

4. The Number of Divisor Classes of Quadratic Fields

4.1. A Formulafor the Number of Dipisor Classes Let K = R( Jd) be a quadratic field (d a square-free rational integer). By Theorem 2 of Section 8, Chapter 3, a rational prime p has the following factorization into prime divisors in K: (1) P = PI", I'

:' I

(2) P = 1', (3) p

= 1'2,

Pi' p',

N(p)

=

N(p')

=

N(p)

=

p2,

if X(p) = - 1;

N(p)

=

p,

if X(p) = 0;

p,

if X(p) = 1;

where X is the character of the quadratic field K (see the definition of Section 8.2 of Chapter 3). Hence in the product

.1

5

19

1-

Y

Sec. 4]

DIVISOR CLASSES OF QUADRATIC FIELDS

343

the factor corresponding to p will be one of the following: (1)

(2) (3)

( pS1)-I( pI)-I. S' 1)-1_( - 1--pS1)-I(1+-pS1)-1 (1-p2s 1--

1--

1

1--.. p

In all three cases this can be written

(1- ?f (1- X~)fl l

e

p

Since [lp(l - 1jp S)-1 representation

= (s) v

~K

()

S

(Theorem 4 of Section 1), then (K(S) has the

= ~V() s [l ( 1 -

x(p)) -S

P

P

I

.

(4.1)

The infinite product on the right is the L-series L(s, X) for the character X(with modulus ID I, where D is the discriminant of the field K), and since this character is not the unit character, then L(s, X) is a continuous function on the interval < s < 00 (corollary of Lemma 4 of Section 2). Multiplying (4.1) by s - 1 and taking the limit as s -+ 1 + 0, we obtain [by (1.19)]

°

lim (s - l)(K(s)

= L(1, X).

s-I +0

Now we use Theorem 2 of Section 1. For real quadratic fields s = 2, t = 0, m = 2, and R = In e (e> 1 a fundamental unit of the field); for imaginary quadratic fields s = 0, t = 1, and R = 1, and hence

JI5

for d > 0,

h = [2lne L(l, X)

rnJIDI 2rr

L(l

,x)

for d < 0.

[By Section 7.3 of Chapter 2 the number m of roots of 1 contained in K equals 4 for K = R( equals 6 for K = R( -3), and equals 2 for all other imaginary quadratic fields.] In the next section we shall show that the character of a quadratic field with discriminant D is a primitive character modulo ID I (see the definition of Section 5.3 of the Supplement), and also that it is even for real fields and odd for imaginary fields. Therefore we can use formulas (2.20) and (2.21) to find

J=1),

J

I'

344

ANALYTIC METHODS

[Chap. 5

L(l, X). To find a closed formula for h we must still find the value of the normed

Gaussian sum ,(X) = '1(X), In Section 4.3 we shall see that the sum ,(X) equals for real fields and equals i JI DI for imaginary fields. Noting also that for real fields X(D - x) = X(x), we can formulate the following theorem [to simplify the formulas we eliminate the fields R( J-=T) and R( J - 3), which have discriminants - 4 and - 3, and for which m equals 4 and 6; for these fields h = I].

JJ5

i

r.

Theorem 1. The number of divisor classes of a discriminant D is given by 1 h= - L: X(x) In sin In e (x.D)= 1 O<x I is a fundamental unit of the field; for an imaginary quadratic field with discriminant D < - 4 we have the formula h

=- -

1

IDI

L:

X(x)x.

(4.3)

(x,D)=1

O<x I (since e > I). Hence we have the following theorem. Theorem 2. Let K be a real quadratic field with discriminant D and character X. The number" given by (4.4) is a unit in K, and is related to the fundamental unit e > 1 by h

e = ", where h is the number of divisor classes of K.

In spite of its simple formulation, there has never been an elementary proof of Theorem 2. Further, it has not been proved by purely arithmetic methods even that" > I. From the inequality" > 1 we can deduce some consequences on the distribution of quadratic residues modulo a prime p == I (mod 4).

5

Sec.4]

:i

The quadratic field R( has discriminant p and its character X(x) coincides with the Legendre symbol (xlp). Therefore we have the inequality

)

DIVISOR CLASSES OF QUADRATIC FIELDS

345

Ji>

)

. 7T.b TI' 7T.Q TI SIO-p> aSIO-, p

1

b

where Q and b run through the quadratic residues and nonresidues, respectively, in the interval (0, pI2). Since the function sin x is monotone on the interval (0, 7T./2), it follows from this inequality that the values of the 7T.blp are "on the average" greater than the values 7T.Qlp, that is, that the quadratic residues modulo p "cluster" at the beginning of the interval (0, pI2), and the nonresidues at the end [when p == I (mod 4) precisely half of the numbers in the interval (0, p12) are quadratic residues]. For prime numbers p == 3 (mod 4) we can obtain information on the distribution of residues and nonresidues by considering formula (4.3) for the field R(J -p). First, we put formula (4.3) in simpler form in the general case. We denote IDI bym. First, assume that m is even. It is easily verified (Problem 9) that in this case X(x +mI2) = -X(x) and formula (4.3) gives us

hm

= -

= -

L

X(x)x -

L

X(x)x

L

X(x),

o <x<mj2

O<x<mj2

m

= -

+

+~)(x +~) 2 2

L

x(x

L

X(X)(X

L

X(x).

0 <x<mj2

O<x<mj2

+ ~) 2

20<x<mj2

so that 1

h=-

2 0 <x<mj2

Note that since m is even, X(2) = 0. Now let m be odd. Since the character X of an imaginary quadratic field is odd, that is, X( -1) = -1 (we have already noted that this will be proved in the following section in Theorem 6), then it follows from (4.3) that hm

= -

L

o <x<mj2

= - 2

X(x)x -

L

x(m - x)(m - x)

0 <x<mj2

L X(x)x + m0 <x<mj2 L X(x). 0< x<mj2

(4.5)

ANALYTIC METHODS

346

[Chap. 5

On the other hand,

hrn = -

L

X(x)x -

O<x<m x even

=

-4

L

x(rn - x)(rn - x)

O<x<m x even

L X(2x)x O<x<m/Z

+ rn

L X(2x), O<x<m/Z

so that

hrnx(2) = - 4

L X(x)x o <x<m/Z

+ rn 0

L X(x). <x<m/Z

(4.6)

Equating the sums LX(x)x in (4.5) and (4.6), we obtain the equation

h(2 - X(2») =

L X(x). 0< x< m/Z

Since this equation also holds for even m [when 21 m, then X(2) = 0], we have the following theorem. Theorem 3. For an imaginary quadratic field with discriminant D < - 4 and character X we have the following formula:

h=

1

L

2 - X(2) 0 <x< IDI/Z (x,D)= 1

X(x).

(4.7)

We now apply Theorem 3 to the case of the field R( J -p), where p is a prime of the form 4n + 3. Since -p == 1 (mod 4), in this case D = -p and the value of the character X(x) coincides with that of the Legendre symbol (x/p). The number of summands in Lo<x N. Thus if p is a prime of the form 4n + 3, then the quadratic residues outnumber the nonresidues on the interval (0, p/2) [by a number divisible by 3 if p == 3 (mod 8) and p i' 3]. This assertion, despite its simplicity, lies among some very deep results of number theory. It was obtained by us as a simple corollary of the fact that the number h, and hence the expression on the right in (4.7) is positive. However,

DIVISOR CLASSES OF QUADRATIC FIELDS

347

1.5

Sec. 4]

6)

the sign of this expression depends on knowledge of the value of the Gaussian sum 'l(X), and we shall see in Section 4.3 that the determination of the sign 'l(X) is a very difficult problem. If D == 1 (mod 8), the formula for the number h for an imaginary quadratic field can be proved by purely arithmetic methods. This was done by B. A. Venkov. His proof was based on the theory oLthe representation of binary forms by sums of squares of linear forms and on some delicate properties of continued fractions [B. A. Venkov, On the number of classes of binary quadratic forms with negative determinant. I and II, Izv. Akad. Nauk SSSR Ser. VII, No. 4-5, 375-392 (1928); No. 6-7, 455-480 (1928)]. In the case D == 1 (mod 8), as in the case of real quadratic fields, a purely arithmetic derivation of the formula for h has never been obtained. Also there is no known elementary proof of the fact that for a prime p of the form 8n + 7 the interval (0, p12) contains more quadratic residues than nonresidues.

re

4

a :i

II

Remark. Letp be a prime of the form 8n + 7. It can be shown by elementary means (Problem 7) that the interval (0, p12) contains just as many odd quadratic residues as odd nonresidues. Hence the number h for the field R(J - p), p == 7 (mod 8), is also given by

h= V* - N*,

where V* and N* denote the number of quadratic residues and nonresidues among the even integers in the interval (0, pI2).

I,

I

4.2. The Character of a Quadratic Field

We shall prove those assertions about the character of a quadratic field which were used in Section 4.1. Theorem 5. The character X of a quadratic field with discriminant D is primitive (with modulus ID I). Proof. By Theorem 4 of Section 5 of the Supplement it suffices to show that for any prime number p which divides D there is an x such that (x, D) = 1, x == 1 (mod ID lip) and X(x) = -1. First, consider the case p i' 2. Choose any quadratic nonresidue s modulo p and pick x as a solution to the system of congruences

x == s (mod p),

348

ANALYTIC METHODS

[Chap. 5

Using formula (8.5) of Chapter 3 it is easily checked that X(x) = (xlp) = = -1. Now let p = 2. If d == 3 (mod 4), D = 4d, then, solving the congruences

(sip)

x == 3 (mod 4), x == I (mod 2Id!),

we shall also have X(x) = (_I)(;c-l)/2 = -1. If d = 2d', D = 4d = 8d', then for the number x, given by x == 5 (mod 8), x == 1 (mod 4Id'!),

we shall have X(x) = = -1. We have shown that Xis primitive. (_1)(;(2- 1 )/8

Theorem 6. The characters of real quadratic fields are even and the characters of imaginary quadratic fields are odd. Proof. Let X be the character of the quadratic field R( .Jd). We compute X( -I), using (8.5) of Chapter 3. If d == 1 (mod 4), then

X( -

I)

= (~/) = (_1)(ld l -l)/2 = (-

1)l(d-l)/2 J +[(ld l -llf2].

If d == 3 (mod 4), then X( - I)

= _ (~II) = - (-

l)s~O

= TI 1/r+s TI r>s

=

r>s

(2i sin _(r_---.-:s)_n) P

'p(p-l)/22 P (p-l)/2

I

s)n TI S . ( I'i-n , p

r>s

since

J:,

p-l

p-I r - I

(I' + s)

= r~1 s~o (I'

1/-(r-s»)

r>s

1'(1'

P-I(

+ s)

= r~1

1'2

+

-1)) = 2p (P-2_

1)2

2

is divisible by 2p. We compare this expression for det A with (4.17). Since siner - s)rrlp > 0 for 0 ~ s < I' ~p - I, we must have the plus sign in (4.17). Thus det A = i P(P-I)/2 p P/2. On the other hand, we have det A =

P

TI Ak = (- I)bn -

i)d p P/2 = i 2b + C - dpp/2.

k=1

This yields the congruence

2b

p-I

+ C - d == P-2- (mod 4),

from which, in view of (4.14) and (4.16), we deduce a - b

= P + 1 - 2b 2

p+1

p-I

== - - - - - = 1 (mod 4) 2 2 C -

I

: I

I: '

for p == 1 (mod 4),

d == _ p - 1 + 2b 2

p-l

p+l

2

2

= - - - + - - = 1 (mod 4)

for p == 3 (mod 4).

i

Sec. 4]

DIVISOR CLASSES OF QUADRATIC FIELDS

353

These congruences show that the differences a - band c - d in (4.16) both equal + 1, and by (4.10) this finally yields

Jp iJP

rex) = {

for P

=1 (mod 4),

for P = 3 (mod 4).

This completes the proof of Theorem 7 for the case of a prime modulus m=p. To prove the theorem in the general case, use the result of Problem 4 of Section 2. If X is a primitive quadratic character modulo m, this problem shows that the normed Gaussian sum rex) can easily be expressed in terms of the normed Gaussian sums for the nonunit modulo 4, the two primitive characters modulo 8, and quadratic characters of odd primes p. Since we know all these Gaussian sums (see Problems 10 and 11 for the moduli 4 and 8), then the formula of Problem 4 of Section 2 allows us to give an explicit expression for rex). Suppose, for example, that we have the character X(x)

= (_1)[(X2-1J/8]+[(X-1J/2](~),

(x, 2r)

=1

modulo m = 8r, where r is an odd square-free natural number. If r = PI ... Ps' then X has the representation x(x)

= (_1)[(X'-lJ/8

+[(X-lJf2](;J ... (;J

Let (X be the number of primes among PI' ... ,Ps which are congruent to 3 modulo 4. Then

rex) = 2iJ"i iaJ;( _1)[(r2-1J/8]+[(r-1J/2] (~) TI (Pk) r

k*i Pi

= ja+ 1~( _1)[(r-1J/2]+C~2 = J;'ja+ 1 +2a+a(a-1J

if X( -1) = ( _ly-1 = 1, if X( - 1) = ( -

1)a + 1 = -

1.

The Gaussian sums for the other primitive quadratic characters are treated analogously. This proof of Theorem 7 (for prime moduli) is due to Shur. Another proof, found by Kronecker, is given in Problems 13 to 16. PROBLEMS

1. Knowing 1ha1 (I + V5)/2 = 2 cos 7T/5 is a fundamental unit for the field R(V5), use formula (4.2) to compute the number h for this field.

II ,

354

ANALYTIC METHODS

[Chap. 5

2. Compute the number h for the fields R( V - 5) and R( V- 23). 3. Show that a quadratic field with discriminant D is a subfield of the mth cyclotomic field, where m = IDI. 4. Let p be an odd prime, and let' be a primitive pth root of 1. Show that the cyclotomic field R(') contains one and only one quadratic subfield. This subfield is R(Vp) if p == 1 (mod 4), and R(V -p) if p == 3 (mod 4). (To solve this and succeeding problems, use the fundamental theorem of Galois theory.) 5. Let p == 1 (mod 4) be a prime, and define the number

. 7Tbj TIsm-, . 7Ta TIsmp

b

P

•

where a and b run through the quadratic residues and nonresidues modulo p in the interval (0, p/2). Without using Theorem 2 show that this number is a unit of the quadratic field R( Vp), and that its norm is -1. 6. Using ,the second assertion of Problem 5, show that the number of divisor classes in the field R( Vp), p a prime, p == 1 (mod 4), is odd and that the norm of a fundamental unit of this field is - 1. 7. Let p be a prime number of the form 8n + 7. Show that precisely half of the even numbers in the interval (0,p/2) are quadratic residues modulo p. 8. If there is a primitive quadratic character with modulus m, show that m is of the form r, 4r, or 8r, where r is an odd square-free natural number. Further, show that every primitive quadratic character is of the form ! ,

X(x) =

X(x)

U}

for the modulus r,

(x, r) = I

= (-l)(~ -il / 2(:;), (x, 2r) =

1

X(x) = (_l)(.>-ll/S(n,

for the modulus 4r,

} (x, 2r) = 1

X(x) =

(_li(~>-')/84[('-1l/2l(~),

for the modulus 8r.

9. If X is a primitive quadratic character with even modulus m (m show that X( x

+

i)= -

(_l)(~2-11/8

and

XH(x) =

4r or 8r with odd r),

X(x).

10. Let X be the character modulo 4 given by X(x) the normed Gaussian sum 'T,(X) equals 2i. 11. Consider the primitive characters x'(x) =

=

= (-l)(~ -Il / 2, (x,

(_l)[(~2-11/8l+[('-')/2J

Verify that the normed Gaussian sums are 'T,(X') = 2v2, and 'T'(X 12. Give the proof of Theorem 7 for arbitrary moduli.

2)

=

1. Show that

(21' x) modulo 8.

H )

=

2iV2.

Sec. 5]

DIVISOR CLASSES OF PRIME CYCLOTOMIC FIELDS

355

13. Let p be an odd prime and set' = cos 27r/p + isin 27r/p. Let

(k,l-l)=1

where k runs through a reduced system of residues modulo l - I.

364

ANALYTIC METHODS

[Chap. 5

We shall now show that B is divisible by l'" -1. In the ring of integers of the field R(O), we have the congruence 1-2

F(ok)(l - gOk)

= L (gOk)'(l -

gOk)

.=0

= l_(gOky-l = l_gl-1 =O(modl), so FW)(l - gOk) is divisible by I. From the above results we see that F(Ok) isdivisibleby/if(k,I...,-I» l,andby/qk-1if(k,I-I)= I.If(k,I-I» 1, let qk denote the unit divisor. Then F(Ok) is divisible by Iqk -1 for all k. Hence the product B = F(0)F(03) ... F(01-2) is divisible by 1m

fl

qk- 1 = 1m

k=1.3 •...•1-2

fl

(k.l-l)=1

q; 1 = 1m -1 ,

which completes the proof that h* is an integer. 5.4. The Relative Primality of h* and I In Section 7.3 of Chapter 3 we saw how important it is to have criteria for determining whether h and I are relatively prime, that is, for determining whether the prime I is regular. Since h = hoh m , then I will be a regular prime if and only if neither of the factors h o , h* is divisible by I. In this section we shall find a condition which is necessary and sufficient for I not to divide h*. In the next section we shall show that if I does not divide h*, then it also does not divide h o , so our condition will turn out to be a criterion for the regularity of I. Preserving the notations of Section 5.3, consider the expression B

Im -

1

fl

k=I.3 .....1-2

(5.16)

[here we identify the principal divisor «(X) with the number (Xl In view of (5.6) the number h* is divisible by I if and only if the rational integer (5.16) is divisible by all prime divisors q.. with (s, 1- 1) = 1. In particular, the number (5.16) will be divisible by ql _ 2 = q-1' so that at least one of the integral divisors FW)qkl-1 (k = 1,3, ... ,1- 2) is divisible by q-l' For this to happen it is necessary and sufficient that the divisor F«(Jk)qk be divisible by q:l' We shall show that this cannot happen for k = 1- 2 = - 1 (mod 1- 1). By (5.15), 0-l g = 1 (mod q-l), so that 1-2

F(O-I) = L(0-l g )'=1-1= -1 (mod q-l), r=O

that is, F(O-I) is not divisible by q-l' and this means that F(O-I)q_l is not

p.5

Sec. 5]

the

divisible by q:l' Thus for h* to be divisible by I it is necessary and sufficient that FW) be divisible by q:l for some k = 1,3, ... ,I -4. Up to this time we have not imposed any restrictions on the choice of the primitive root g modulo l. Now we assume that g satisfies the congruence

365

DIVISOR CLASSES OF PRIME CYCLOTOMIC FIELDS

i-I == 1 (mod 12) (if g does not satisfy this congruence, replace g by g + xl with suitable x). Since the congruence (5.14) will now hold modulo fl, then 1 - (ig will be divisible by q/ for any k relatively prime to 1- 1. In particular,

() == g (mod

q:l)'

With this choice of g the condition for q:l to divide F«()k) can easily be found. Indeed, since 1-2

FW)

1- 2

= L gs()'k == L gsgsk (mod

q:

1), s=o s=o then the number F«()k) is divisible by q:l if and only if 1-2

L gsgSk == 0 (mod 12 ). s=o

'or tlg

In order to put (5.17) in more convenient form, consider the congruence.

ar

gs == gS

[n

.I

*,

:-

[)

+ las (mod 12 )

(0 ~ s ~ 1- 2),

(5.18)

where as is an integer. If we raise both sides of (5.18) to the power k (k = 1, 3, '" , 1- 4), then we find that

+ (k + I)gsk[a s == gs(k+l) + (k + l)gs\gs _

+1

g/+1 == gs(k+l)

that is, g/+1 == (k

i)

i) s e

(5.17)

+ l)gsgSk -

gS) (mod 12 );

kg s(k+l) (mod [2).

Summing (5.19) for s = 0, 1, ... , 1- 2 and noting that if + 1 k + 1 ~ 1- 3 and gl-1 == 1 (mod 12 ), we obtain 1- 2 g (1- 1 )(k + 1) I L gS(k+ 1) = - == 0 (mod [2) s=o l+I_1 and hence 1-2

y

1-2

L g/+ 1 == (k s=o

But k

+ 1 '" 0

+ 1) L

gsgsk (mod [2).

s=o

(mod I), and therefore (5.17) is equivalent to 1-2

1-1

Sk+l= Lgsk+1 = Ln k+1 ==0(mod[2). s=o n= 1

Hence we have proved the following theorem.

(5.19) '"

1 (mod I) for

ANALYTIC METHODS

366

[Chap. 5

Theorem 3. In order that the number h* not be divisible by I, it is necessary and sufficient that none of the numbers

(k = 2, 4, ... , 1- 3) be divisible by

(5.20)

F.

Note that each of the numbers Sk [k '" 0 (mod 1- 1)] is divisible by I [see (8.10)]. We reformulate Theorem 3' in terms of Bernoulli numbers (Bernoulli numbers will be defined and studied in Section 8). Since the numbers 2, 4, .. , , I - 3 are not divisible by I - 1, then by the theorem of von Staudt (Theorem 4 of Section 8) the Bernoulli numbers B 2 , B 4 , '" , Bt - 3 are I-integers (I does not appear in their denominators). Further, we have the congruence

(k = 2, 4, .. , , I - 3)

(5.21)

[in the ring of I-integral numbers; see (8.11)]. Hence the following theorem is valid. Theorem 4. In order that h* not be divisible by I, it is necessary and sufficient that the numerators of the Bernoullt numbers B 2 , B 4 , ••• , B t - 3 not be divisible by I.

For example, since the numerators of the numbers B 2 , B 4 , B 6 , B g , B lO

,

B12 , B14 are not divisible by 17, then 1= 17 is regular. Remark. To determine whether h* and I are relatively prime, it is not necessary to find the precise value of the Bernoulli numbers. It suffices to consider the recurrence relation (8.2) as a congruence modulo I and to use these congruences to compute the sequence B 2 , B 4 , ... , B t - 3' The number h* will be relatively prime to I if and only if none of these numbers is divisible by I.

PROBLEMS

1. Let K o be the subfieId of the Ith cyclotomic field R(~) which consists of all real numbers in R('). Show that K o = R(' + '-1) and is of degree (1- 1)/2. Further, show that the field K o has discriminant 1(1-3)/2, and that its regulator R o is related to the regulator R of the field R(') by R = 2(1-3)/2 R o • 2. Let p be a prime different from I, and let I be the smallest natural number for which p' == 1 (mod I). Show that the number p factors in the field Ko as the product of (1- 1)/21 prime divisors of degree I when I is odd, and as the product of (1- 1)/1 prime divisors of degree 1/2 when I is even.

Sec. 6]

A CRITERION FOR REGULARITY

367

3. Show that the '-function '''o(s) of the field K o satisfies lim (s - 1)'"o(s) =

n

L(l, X),

X*Xo

s-+1+0

X(-l)= 1

where X runs through all even numerical characters modulo /, except the unit character Xo . 4. Show that the real subfield R(' + ,-I) of the Ith cyclotomic field has h o divisor classes, where ho is the factor of the number of divisor classes of the field R('). 5. Show that h* is given by h*

=

(2/;m_Ildet(gm+I+) - gl+)O", I, Hm-,l,

where g, is the smallest positive residue of the number g' modulo / = 2m + I (g is a primitive root modulo I). 6. Compute the factor h* for / = 7. 7. Show that the prime number 37 is irregular.

6. A Criterion for Regularity

Our goal in this section is to show that when the factor h* of the number of divisor classes of the Ith cyclotomic field is not divisible by I, then the factor ho is also not divisible by I, and hence the prime I is regular. En route we shall also show that when I is regular, every unit of the field K = R(O which is congruent modulo I to a rational integer is an lth power. On this assertion, known as Kummer's lemma, is based the proof of the second case of Fermat's theorem for regular primes. Both the regularity criterion and Kummer's lemma will be found as simple corollaries of the following result. If I ,r h* and K. is the I-adic completion of the field K = R(O where 1 = (1 - 0, then the numbers log 0k l - 1 [k = 2, 3, ... , (1- 1)/2] form a basis for the set of all .. real" l-adic integers with zero trace [the units Ok are defined by (5.10)]. 6.1. The Field of I-adie Numbers

We know that the cyclotomic field K = R«), (= cos 2n/1 + i sin 2n/l, 3 a prime, has degree 1- 1 and that the number I has the factorization 1= 11- 1 , where 1 = (1 - 0 is a prime divisor of first degree. We consider the I-adic completion K[ of the field K. The elements of this completion are called l-adic numbers. The complete field K 1 contains a subfield which is canonically isomorphic to the field R, of I-adic numbers (this subfield coincides with the completion of the field R in K.). Using this canonical isomorphism we shall assume that R 1 c K[ . Since 1 is the only prime divisor which divides I, then by Theorem 1 of Section 2, Chapter 4, the degree of the extension Kd R 1 equals I - 1 = (K: R). Hence [see (2.6) of Chapter 4] for any r:x E K we have I

~

NK/R(r:x)

= NK[/RI(r:x)·

(6.1)

368

ANALYTIC METHODS

[Chap. 5

Lemma 1. There is an element A in the ring of r-adic integers such that:

+ 1 = 0, A == ( - 1 (mod

(1) A'-1

(2)

A2 ).

The element A is uniquely determined by (1) and (2). In view of (1.5) of Chapter 3 we have (1 _

1

0'

1

, 2 = (1 + ()(l + ( + ( 2 ) .,. (l + ( + ... + ().

We now consider congruences modulo the prime element 1 - ( of the field K 1 [recall that vI (1 - 0 = 1]. Since (== 1 (mod 1 - 0 and (1- I)! + 1 == o (mod I) (Wilson's theorem), then (l _

1

0

1

'

== 2·3 ... (1 -

1)

== - 1 (mod 1 - O.

We shall show that the r-adie unit

-1 a = .,..-,(1------:::-0"' I

i'

, I

'

which is congruent to 1 modulo 1 - (, can be represented in the form a = y'-t. Consider the polynomial F(X) = a. Since F(l) == 0 (mod 1 - 0 and P(1) ;;j:. 0 (mod 1 - 0, then there is a unit y in K for which F(y) = 0 (see 1)y, Section 1.2 of Chapter 4). Hence a = y'-t, as was claimed. Setting A = we obtain a prime element A with the desired properties. Any other number At, satisfying the first condition of the lemma, has the form A.8, where () is a (1- 1)th root of 1. From A.8 == A(mod A2) it follows that () == 1 (mod A). If the root () were different from 1, then 1- 1 would be divisible by A, which is impossible. Hence () = 1 and At = A. Lemma 1 is proved. From now on A will denote that prime element of the field K which is uniquely determined by the conditions of Lemma 1. For each k which is relatively prime to 1 the correspondence ( ~ (k determines an automorphism Uk of the extension K/R. If u is any of these automorphisms, then the function v'(a) = vI(u(a)), a E K, is a valuation of the field K, and is an extension of the I-adic valuation V, of the field R. But there is only one extension of V, to the field K, namely VI • Hence v' = VI' and this means that vI(u(a)) = vI(a) for all a E K. It follows from this that the automorphism u takes any Cauchy sequence of elements of K (relative to the metric which corresponds to the prime divisor I) to another Cauchy sequence in K. This allows us to extend the automorphism u = Uk to the field K I • Namely, if = limn .... ooa n (an E K), then we can set

x,-t -

«( -

e

uW = lim u(an)

1

Sec. 6]

A CRITERION FOR REGULARITY

369

[it is easily verified that u(e) does not depend on the choice of the sequence {an}, and also that the mapping -+ u(e) is an automorphism of the extension KdRd· Since the extension KdR, has degree of inertia 1 and ramification index l- 1, then by Theorem 4 of Section 1, Chapter 4, all I-adic integers can be uniquely represented in the form

e

ao + alA

+ ... + a l _ 2 A' - 2

(6.2)

where the a j are l-adic integers. The subfield of real numbers of the field K consists of all a E K which are left fixed by the automorphism u -1 : , -+ ,-1. We shall determine which I-adic numbers are invariant under the automorphism u -1 . Since A' - 1 = -l, then also .(U_1(A))'-1 = -l, and this means that U_1(A) = AO, where 0 is an (l- l)th root of 1. By Problem 4 of Section 3, Chapter 1, the root 0 is contained in R so that "

U_1 2(A) = U-1(U-1(A)) = U-1(OA) = OU_1(A) = 02 A,

and since also u _1 2 (A) = A, then 0 = ± 1. If 0 = 1, then any I-adic number which can be represented in the form (6.2) with l-adic coefficients ai, would be left fixed by the automorphism u -1' and this is not the case. Hence 0 = -1, and u -1 (A) = - A. Hence when the automorphism u -1 acts on the field K[, the I-adic numbers which are left fixed are those of the form m-I

I: biA

2i

i=O

( biE R "

m

1) .

1= -2-

(6.3)

The set of all such numbers is a subfield of K( of degree m = (/ - 1)/2 over R, . It will be convenient to call them" real" I-adic numbers. We compute the trace of the I-adic number (6.2) (relative to the extension KdR,). For any i = 1, ... , l - 2, the matrix of the linear transformation 2 -+ Aie E K() with respect to the basis 1, A, ... ,A - will have zeros on the ' main diagonal (since l'-l = -l), and therefore SPKIIR,(A i ) = 0 (for i = 1, ... , l - 2). It follows that the trace of the number (6.2) equals a o(/ - 1). The I-adic numbers with trace (over R,) equal to zero are thus characterized by having the coefficient a o equal to zero in (6.2). We shall be interested in the set 9Jl of all "real" I-adic integers with zero trace. It is clear from the above remarks that 9Jl coincides with the set of all linear combinations of the form

e

(e

m-1

I: b A

2i

i

(6.4)

i= 1

where the b i are l-adic integers. We consider the functions log e and exp a over the field K(, which are

370

ANALYTIC METHODS

[Chap. 5

defined by power series (see Section 5.2 of Chapter 4). Since the ramification index e of the extension KdR, equals 1 - 1, then the number [el(l- 1)] + 1 equals 2, and this means that the series exp IX converges for all integers IX E K[, divisible by ).2. As we know, the function log e is defined for all principal units of the field K I • If e is a principal unit of the field K, that is, e == 1 (mod ).), then for any automorphism Uk we again have uk(e) == 1 (mod ).), and this means that log uk(e) is defined. But then (Corollary 1 of Theorem 11, Section 2 of the Supplement), I-I SPKI/R, log e = Uk(lOg e) = log (uk(e)

L

L

k= 1

k

= log(I} Uk(e)) = log(N KI/R,e). Now assume that e is a unit of the field K. It is clear that e is also a unit in the field K, but log e is not necessarily defined, since e will not, in general, be a principal unit. But for some rational integer 0 which is not divisible by 1 we shall have e == 0 (mod ).). From 0 1- 1 == 1 (mod ).) it follows that el - 1 == 1 (mod ).); that is, el - 1 is a principal unit in K[. The logarithm log el - 1 is thus defined, and by formula (6.1) SPK(/R,(logel -

1

)

= log(NK[/R,e ' - 1) =

log(N K/Rel - 1)

= 0;

that is, the l-adic integer log /:;'-1 has zero trace. If e is a real unit of the field K, then it is clear that log el - 1 will be "real." Hence for any real unit e of the field K the l-adic number log el - 1 belongs to the set 9Jl; that is, it can be represented in the form (6.4). In particular, this holds for the units. Ok (k = 2, 3, ... , m = (1- 1)/2) defined by (5.10). Thus we have log 0k l -

1

=

m-l

Lb

;=1

ki

).2i

(2

~

k

~

m)

(6.5)

where the coefficients bki are I-adic integers. Our problem is now to show that when 1 does not divide h* (the factor of the number of divisor classes of the field K), then the l-adic numbers log 0k l - 1 form a basis for 9Jl over the ring of I-adic integers in the sense that any ~ E 9Jl has a unique representation as a linear combination of the log 0k'-1 with I-adic integer coefficients. To do this it clearly suffices to show that det(bkJ is an I-adic unit, that is, that det(bkJ 1= 0 (mod I).

6.2. Some Congruences The series for exp x in the field K( converges only for those integers x

Sec. 6]

A CRITERION FOR REGULARITY

371

which are divisible by A2 • We also consider the polynomial

obtained from the series for exp x by deleting all terms with degrees ~ I. Since the coefficients l/k! for k ~ 1- 1 are I-adic integers, then E(x) will be a principal unit of the field K, for all integral x == 0 (mod A). We know that the formal product of the series exp x and exp y equals the series exp(x + y). It follows that E(x)E(y)

= E(x + y) + F(x, y),

(6.6)

where F(x, y) is a polynomial with I-adic integral coefficients in which all terms have degree ~ I. Lemma 2. The congruence

E(A)' == 1 (mod A21 -

1

)

holds in the ring of I-adic integers. Set E(x)

where g(x) = 1 + x/2! + integer coefficients. Then E(x)'

= 1 + xg(x),

... + x ' - 2 /(/_

= 1 + Ct' xg(x) = 1 + lh(x)

1)! is a polynomial with I-adic

+ ... + C,'-I(xg(x»'-1 + xlg(X)'

+ xlg(X)I,

where hex) is also polynomial with I-adic integer coefficients. On the other hand, by (6.6) we see that E(X)I

= E(lx) + x'M(x),

and this means that lh(x)

Ix

=-

I!

(Ix) 2

+-

2!

(lX)'-1

1

+ ... + - - + x H(x) (I-1)!

'

(6.7) i I

where H(x) = M(x) - g(X)I. Looking at the coefficients for the various powers of x in this equation, we see that all coefficients of H(x) are I-adic integers divisible by I. Dividing (6.7) by I, we arrive at hex)

=

IX 2

x

+ 2! + ... +

11- 2 X 1-' (1- I)!

1

+ x G(x),

I;

372

ANALYTIC METHODS

[Chap. 5

where G(x) has l-adic integer coefficients. If we set x = A, we obtain the equation h(A) == A(mod AI), and this means that Lh(A) == LA (mod A21 - 1). (6.8) Further, since g(A) == 1 (mod A), then g(A)1 == 1 (mod AI) so that AIg(A)1 == AI(mod A21 ).

(6.9)

From (6.8) and (6.9) we obtain

(since lA

E(A)' = 1 + Lh(A)

+ Alg(A)' == 1 + LA + AI =

+ AI = 0), which

proves the lemma.

1 (mod A2I - 1)

, I

Lemma 3. The following congruence holds for any natural number k: E(kA) == Ck (mod AI).

It follows from formula (5.6) that E(kA) == E(A)k (mod AI), so it suffices to prove the lemma for the case k = 1. By the definition of the prime element A we have C== 1 + A (mod A2 ). On the other hand, E(A) == 1 + A(mod A2), and therefore

C 1 E(A) == 1 (mod A2 ). Set where y is an I-adic integer. Raising this equation to the Ith power and using Lemma 2, we obtain the congruence

Y(IA2 + L(l

~ 1) YA4 + ... + yl-1 A21) == o(mod A21-1).

The expression in parentheses is divisible by AI+ 1 (and by no higher power of A), so y == 0 (mod AI- 2), and

C 1E(A) == 1 (mod AI), which proves the lemma. We also consider the polynomial x2

L(l

+ x) = x -"2 + .,. + (-

obtained from the series log(l

+ x)

XI- 1

1)'-2

L_

(6.9')

1'

by deleting terms of degree

~

1.

Sec. 6]

A CRITERION FOR REGULARITY

373

Lemma 4. If the I-adic integer rx is divisible by A2 , then

L(1

Indeed, for n

~

+ rx) == log(1 + rx) (mod A').

1 we have

~

I

+ (n

- l)

l)n ( In I In n ) -- - In I 1- 1 n - 1

+ (l -

~

I

(see Section 5.2 of Chapter 4). Lemma 5. Let e1 and e2 be principal I-adic units. Then L(e 1 e2 )

Since the series log(l and log(l + y), then L(I

== L(e l ) + L(e 2 ) (mod A').

+ x + Y + xy) equals the sum of the series log(l + x)

+ x + Y + xy) = L(I + x) + L(I + y) + G(x, y),

where the polynomial G(x, y) contains only terms of degree ~ 1 and has I-adic integer coefficients. The assertion of Lemma 5 follows from the fact that G(x, y) == 0 (mod A') if x and yare divisible by A. Lemma 6. The congruence

L(O == A(mod A.' ) holds in the ring of I-adic integers. To prove this we use the formal equality log exp x easily that L(E(x))

=

x

= x. From this it follows

+ H(x),

where H(x) is a polynomial in which all terms have degree ~ I, and which has I-adic integer coefficients. Setting x = A and using Lemma 3 for k = 1 we obtain the desired congruence. Remark. Let U be the multiplicative group of cosets modulo A' in the group of all principal I-adic units, and let X be the additive group of cosets modulo A' in the group of all I-adic integers which are divisible by A. It is now easily seen that the mapping e -+ L(e) induces an isomorphism of the group U onto the group X. The inverse isomorphism X -+ U is induced by the mapping rx -+ E(rx) [rx == 0 (mod A)].

374

ANALYTIC METHODS

[Chap. 5

S.

6.3. A Basisfor the Real I-Adic Integers in the Case (h*, I) = 1

al

We return to the question which was raised at the end of Section 6.1. To determine whether the determinant det(bkJ is divisible by I, it is only necessary to consider the coefficients bki modulo I. It is clear that two I-adic integers of the form (6.2) are congruent modulo I if and only if their corresponding coefficients in the expansion (6.2) are congruent modulo I (in the ring of I-adic integers). Hence to find the bki modulo Iwe may replace the numbers log 0k l - I by any I-adic integers congruent to them modulo I (that is, modulo

w

AI-I).

We use the notations of Section 5.2. The principal unit 0/- 1 is real, hence congruent to 1 modulo A2 , so that, by Lemma 4, log 0k l We now compute L(Ok ' -

I

).

I

=L(Ok

l

-

l

)

(mod AI).

B

(6.10) Si

Since

Zl

1 o -~1'1 I-k , vk

k-v ~

-

then

A

0k l = (1 + (+ ... + (k-I)I( _1)I-k.

But (

=1 (mod A), so that

=k (mod A), (1 + (+ ... + (k-I)I =k (mod AI). ' 1+

and hence Since k l

( + ... + ,k-I

=k (mod AI-I), then also

(1 + (+ ... + (k-I)/

(t

Ie tl

=k (mod A'-I).

Thus

al

or

T

By Lemma 5 we have

L(Ok

1- I

- 1) ) =L -A-1) - L ('k ~ + (k (' -

1

B

I + L(~)(mod A1- I ). 1) -2-

But by Lemma 3, (k - 1 = E(kA) - 1 ( d "-I) kA kA mo A. ,

y

5

Sec. 6]

375

A CRITERION FOR REGULARITY

and, therefore, using Lemma 6, we obtain

I.

y

c

LUh'-I) ==

1) _~ _L(E(k~~ - 1) + k2.1. (mod ..1.'-1).

L(E(..1.~ -

We now show that

1) _: = mf X +X' k=I(2k)!2k 2k

x

s

)

B 2k

L(E(X) -

g

1R(x)

(6.11)

'

2

where the polynomial R(x) has l-adic integer coefficients and B 2k are the Bernoulli numbers (see Section 8). We use the identity

I

_x_= B nxn. ex-I n=on! Since B 1 = -1, and all remaining Bernoulli numbers with odd index equal zero, then our identity can be written in the form

~_!_!= 2

eX - l

x

I

B 2k x 2k k=I(2k)!

I•

After integrating we obtain eX - 1 x ;. B 2k 2k I n - - - - = L.. x x 2 k=1 (2k)!2k

(6.12)

(the constant term of the series equals zero, since for x = 0 the function on the left vanishes). The formula (6.11) now follows from (6.12). If we substitute the value k..1. for x in (6.11), we find that L

(

E(k,1,) k..1.

1)

k..1.

m-I

-"

-"2 = if-I

B .k 2i . 1. 2i

(d..1.' - 1) mo ,

21

(2i)!2i

and hence L(e~-I) ==

m-I

L

B (1 2i

k 2i )..1. 2i

~,.

(21).21

i= 1

(mod . 1.' -

1

(6.12')

).

This shows that the coefficients b kl in (6.5) satisfy bki ==

B2i (l - k (2i)! 2i

2i

)

(

(mod I)

2~k ~m

I- 1

= -2-' 1 ~ i ~

But then det(b ki ) is congruent modulo I to the determinant 2 2 - 1 2 4 -1 m-I

i[1

(-l)m- I B 21

32

-

1

34

-

1

2' - 3 _1 3'

3 -

1

(2i)! 2i

m 2 -1

m 4 -1

m' - 3

-

1

)

m - 1 .

376

ANALYTIC METHODS

[Chap. 5

We easily evaluate this determinant by reducing it to the Vandermonde determinant. It equals the product

Il

(r 2

S2)

-

l~s~,~m

= Il (r + s)(r -

s),

s 0; that is, e E E. From the congruence: e' - 1 == a ' - 1 == 1 (mod I) it follows that log e' - 1 == 0 (mod I), and therefore by Theorem 1, m

log e' - 1 = LIck log (h'-1,

(6.16)

k=2

with I-adic integral Ck • On the other hand, since the subgroup Eo is of finite index in E, then e E Eo for some natural number a, and hence Q

(6.17) with rational integers dk • We can assume that the set of numbers a, d2 , ••• , dm has greatest common divisor 1 (since the group E has no elements of finite order). Raising (6.17) to the power 1- 1 and taking the logarithm (in the field K 1), we obtain

=

a log e'- 1

m

Ld

k

log (h'- 1 •

k=2

Comparing with (6.16), we arrive at

dk = lack

(k

= 2, ... , m).

Since the numbers aCk are I-adic integers, then it follows that all dk are divisible by I, and this means that e is an lth power: e =e 1', where e1 E Eo. Since (a, d2 , .,. , dm ) = 1, then a is relatively prime to I, and by picking rational integers u and v such that 1 = au + lv, we find Q

Q

e = (e

Q )"(

eV )'

= (e 1"e v )',

which proves the theorem. PROBLEMS

1. Let p be a prime number of the form 4n + 1, ~ m = (p - 1)/2. Set

,= n

p-I

t= 1

Ol -CliP),

=

cos 27T/p

+ i sin 27T/p, .\ =

~ -

1,

378 where O.

ANALYTIC METHODS =

[Chap. 5

sin (kTr/p) [sin (7T/p»)-t, 1 .,; k.,; p - 1. Show that the congruence

holds in the pth cyclotomic field. Here L denotes the function defined by (6.9') and B m the Bernoulli number. [Use (6.12') and Problem 14, Section 4.) 2. Let E = T + uviJ> 1 be a fundamental unit in the quadratic field R( VP), where 1 (mod 4), and let h be the number of divisor classes of this field. Using the preceding problem and Theorem 2 of Section 4, show that

pc-=;

, ! !

hU oo=TBm (mod p)

!

(

1)

_ p m --2

(in the ring of p-integral rational numbers).

7. The Second Case of Fermat's Theorem for Regular Exponents

7.1. Fermat's Theorem Theorem 1. If the prime number I X'

~

3 is regular, then the equation

+ y' =

(7.1)

Zl

has no solution in nonzero rational numbers

X,

y, z.

Proof. Assume that

!

(I

I

j

I't '

I ll"

"

( ;:

~ I, I

I, j

j I

,I , I! 't !

1

X, y, z are relatively prime (nonzero) integers which satisfy (7.1). Since the first case of Fermat's theorem has already been treated in Section 7.3 of Chapter 3, we may assume that one (and only one) of these numbers is divisible by I. We shall let I divide z [if, for example, y is divisible by I, then we can write (7.1) in the form x' + (-Z)' = (- y)']. Let z = Ikz o , where (zo , I) = I, k ~ I. In the Ith cyclotomic field R(O, the number I has the factorization I = (I - 0 ' - 1 e, where e is a unit in R(O (Lemma I of Section I, Chapter 3). Hence we can put (7.1) in the form

x' ,

+ y' = e(l

- o'mzo',

(7.2)

where m = k(/ - 1) > O. To prove the theorem it suffices to show that an equation of the form (7.2) is impossible. We shall actually show somewhat more. Not only will we show that an equation of the form (7.2) is impossible in rational integers x, y, and Zo relatively prime to I, but even that it is impossible in integers of the field R(O which are relatively prime to I - C. Assuming the converse, we take that solution of (7.2) in which the exponent m ~ I is smallest. To avoid introducing new notation, we shall assume this solution to be given by (7.2). Hence x, y, and Zo denote integers of R(O which are relatively prime to I - C, and e is some unit of the field R(O.

1.5

Sec. 7]

379

FERMAT'S THEOREM FOR REGULAR EXPONENTS

Rm.

8m =re

ng

As in Section 6, I denotes the prime divisor (1 - 0 of the field We factor the left side of (7.2) into linear terms and then pass to the corresponding equation in divisors. We obtain 1-1 (x + y ) = lima', (7.3) k=a

n

,k

where the divisor a = (za) is relatively prime to I. Since 1m ~ I> 0, it follows from (7.3) that at least one of the terms on the left is divisible by I. But since

x

+ ,iy = X + 'k y

then all the numbers x

+ ,k y

are divisible by I. If for some x

+

°

°

~

'k y ==

_ (k(1 _ ,i-k)y,

(0 ~ k ~ 1 - 1)

(7.4)

k < i ~ 1- 1,

+

X

,iy (mod 1

2

),

2

then also 'ky(l - ,i-k) == (mod 1 ), and this is impossible, since ,ky is relatively prime to I, and 1 - ,i-k is associate with 1 - ,. Hence the numbers (7.4) are pairwise-noncongruent modulo t2, so the expressions x

1 -,

l)

h d

:e

e

"

e

)

+ ,k y

(k

= 0, I, ... ,1 - 1)

are pairwise-noncongruent modulo I. Since N(I) = I, these expressions form a complete set of residues modulo I, and hence one of them is divisible by I. It follows that one (and only one) of the numbers (7.4) is divisible by 12 • Since we may replace y by any of the numbers y in (7.2), we may assume that x + y is divisible by 12 , and that all the other numbers x + y are divisible by I but not divisible by 12 • Then the left side of (7.3) is divisible at least by 1'-11 2 = 1'+1, so that m > 1. Now let m denote the greatest common divisor of the divisors (x) and (y). Since x and yare not divisible by I, then m is not divisible by I. Then it is clear that (x + y ) is divisible by 1m, and (x + y) is divisible by II(m-1)+l m . We set

'k

'k

,k

(x

+ y) =

II(m-1)+l mCa , (k = 1, ." , 1- 1),

11

and we shall show that the divisors Ca, C1' '" ,CI-1 are pairwise relatively prime. Indeed, if Ci and Ck (0 ~ i < k ~ 1- 1) had common divisor p, then x + ,iy and x + ,ky would be divisible by Imp, so that ,iy(l - ,k-i) and x(1 - ,k-i) would also be divisible by Imp. But this would imply that x and y were divisible by mp, contradicting the choice of m. Writing (7.3) in the form m'I'mcaC1 ...

C'-l

= Ilmal,

,I ~

II I,

I r

I,

380

ANALYTIC METHODS

we deduce (since the

Ck

[Chap. 5

are pairwise relatively prime), that (0:::;; k :::;; 1- 1),

and this means that (7.5) (1 :::;; k :::;; 1 - 1).

(7.6)

Solving (7.5) for m and substituting in (7.6), we obtain (x

+ ,ky )I ' (m-1) =

(x

+ Y)(OkOo -1 y,

(7.7)

from which it follows that the divisors [(OkOO -1)/ are principal (since I = (1 - 0]. Now we use the regularity of I. Since the number of classes of divisors of the field R(O is not divisible by I, then by the corollary to Theorem 3 of Section 7, Chapter 3, the divisors 0kOO -1 are also principal; that is, °kOO -1

=

G:)

(1 :::;; k :::;; I-I),

(7.8)

where (Xk and Ih are integers of the field R(O. The divisors Ok (1 :::;; k :::;; 1- 1) and 00 are relatively prime to I, so we may assume that (Xk and 13k are not divisible by I. Principal divisors are equal if and only if the corresponding numbers differ only by a unit factor. Therefore by (7.7) and (7.8) we have (x

+ 'k

y)(1-

ol(m-1)

= (x +

y)(;:)'ek

(1 :::;; k :::;; 1- 1),

(7.9)

where ek is a unit of the field R(O. Now consider the following obvious equation: (x

+

Ifwe multiply it by (1 (x

+

'y)(1 + 0 - (x + '2 y)= '(x + y). o'(m-l)

and use (7.9) with k = 1 and k = 2, we obtain

y)(;:)'e (1 + 0 - (x + y)(;:)'e 1

2

= (x +

yK(1 -

o'(m-1),

so that

Hence we have shown that (7.10)

where (x, 13, and yare integers of R(O, not divisible by I, and eo and e' are units of the field RC'). We shall transform (7.10) to the form (7.2).

,I:

Sec. 7]

FERMAT'S THEOREM FOR REGULAR EXPONENTS

We have seen that m> 1, so that m - 1 > 0 and I(m - 1) means that a l + eo{31 == 0 (mod 11).

381 ~

I, and this

Since {3 is relatively prime to 1, there is a number {3' such that {3{3' == 1 (mod 11). Multiplying the last congruence by {3", we obtain eo == WI (mod II),

where w = -a{3' is an integer of the field R(O. Since N(I) = I, then any integer of R(O is congruent modulo I to a rational integer. If w == a (mod I), then l Wi == a (mod II), and this means that the unit eo is congruent modulo II to a rational integer. By Kummer's lemma (Theorem 3 of Section 6; here we again use the fact that the prime I is regular) the unit eo is an lth power in R(O, that is, eo = rJ l , where rJ is another unit of the field R(O. The equation (7.10) then takes the form l a + (rJ{3)' = e'(1 - Ol(m-l)yl. We have obtained an equation of the same type as (7.2), but the exponent m has here been replaced by m - 1. But this is impossible, since we chose m to be as smal1 as possible. This contradiction shows that the equation (7.1) has no solution in nonzero rational integers x, y, and z, one of which is divisible by I, that is, that the second case of Fermat's theorem holds for regular primes. Theorem 1 is proved.

7.2. The Infinitude of the Number of Irregular Primes In all existing tables the irregular primes are outnumbered by the regular ones. However, it is not known whether this is true for all intervals (1, N). Further, the infinitude of the set of regular primes is stil1 an open question. Hence the following theorem is of considerable interest. Theorem 2. There are infinitely many irregular prime numbers.

The proof of Theorem 2 is based on some properties of Bernoul1i numbers. These properties are formulated and proved in the following section. Let Pi> '" ,Ps be any finite set of irregular prime numbers. Theorem 2 will be proved if we can find an irregular prime number P, different frompl' ... ,Ps' Set n = r(Pl - 1) .. · (Ps - 1). Since the Bernoul1i numbers B lk have the property B 2k -

I2k l

-+00

as k -+

00

"

382

ANALYTIC METHODS

[Chap. 5

(see the end of Section 8), then we can choose r large enough so that Bnln has absolute value greater than 1. Let p be any prime number which divides the numerator of this number (in lowest terms). If (p - 1) I n, then by von Staudt's theorem (Theorem 4 of Section 8), p would divide the denominator of Bn , and this is not the case by choice of p. Hence (p - 1) I n, and p is different from Pl, ... , Ps (and different from 2). Let n = m + a(p - 1), with 2:::;; m :::;; p - 3 (note that m is even). We now use Kummer's congruence (Theorem 5 of Section 8), and obtain the congruence Bm

Bn

m

n

- == - (mod p)

in the ring of p-integral rational numbers. But Bnln == 0 (mod p), so Bmlm == '" ,p - 3, it follows from the corollary of Theorem 2 of Section 6 that the number p is irregular. Theorem 2 is proved.

o(mod p) and Bm == 0 (mod p). Since m is one of the numbers 2, 4,

PROBLEMS

1. Show that the equation x 3 + y3 = 5z 3 has no solution in rational integers with z#O. 2. Show that there are infinitely many irregular primes of the form 4n + 3 (use p- -l,lems 9 and 10 of Section 8).

8. Bernoulli Numbers

In this section we shall prove those properties of Bernoulli numbers which have been used in the preceding sections. All the power series to be considered converge in some neighborhood of the origin, but their radii of convergence could easily be computed. But we shall not worry about questions of convergence, since for our purposes it suffices to consider all series formally (except in the proof of Theorem 6). Definition. The rational numbers B m (m t

--= 1 +

e' - l

co

L

~

1), defined by

B

~tm

m=lm!'

(8.1)

are called Bernoulli numbers. We use the following notations. Iff(x) = a + ax + ... + ax is a polynomial, then by f(B) we mean the number a o + alB l + ... + anBn . Analogously, if (x, t) is a power series of the form L~=ofn(x)tn, where fn(x) is a polynomial,

3S3

5

Sec.S]

7

then by f(H, t) we mean the series r..';';ofiH)t". Using this notation, the expansion (8.1), which defines the Bernoulli numbers, can be written in the form

S

BERNOUlli NUMBERS

1

t

r

-t - = e

e

Bt

1

-

.

It is easily seen that for any number a

(to prove this it suffices to multiply the series on the left). Theorem 1. The Bernoulli numbers satisfy the recurrence relation for m ~ 2, (I + B)m - B m = 0

(8.2)

which in expanded form becomes 1+

m-I

r.. CmkBk = 0

(m ~ 2)

k;1

(the Cm k are the binomial coefficients). To prove this theorem we write (8.1) in the form t

=

e( 1+ B)t _

eBt.

Comparing the coefficients of the terms t m1m! (m ~ 2), we obtain the relation (8.2). For m = 2 formula (8.2) gives us I + 2HI = 0, and this means that

HI

= -1·

Theorem 2. All Bernoulli numbers with odd index, except HI' equal zero: H2m + 1

=0

for

m

~ 1.

(8.3)

The equality (8.3) is clearly equivalent to the fact that the function t

t

-t--

e-l

+- = 1+ 2

B m ~t

r.. 00

m;2m!

is even, and this is easily verified. We give the values of the first 12 Bernoulli numbers with even index: I' B 6' 4

B =2

1 30'

= --

691 B I2 = - 2730'

6

6'

---no'

B 22 =

1 30'

B = --

7

B I4 =

174611

B 20 = -

1 42'

B =-

B I6 = -

854513 138 '

8

3617 510' B 24 = -

B

10

5 66'

=-

43867 B I8 =

798'

236364091 2730

384

ANALYTIC METHODS

[Chap. 5

Bernoulli numbers are connected with the sums of series of natural numbers. Set

Theorem 3. The sums Sk(n) satisfy the formula (m

- B m+1,

+ l)Sm(n) = (n + B)m+l

m ~ 1,

(8.4)

or in expanded form m

(m

+ l)Sm(n) = L C~+ IBkn m+1-\

m ~ 1

k~O

(B o = 1).

(8.5)

In fact, the expression on the right in (8.4) equals the coefficient of tm + 1/ + I)! in the series e("+B)t - e Bt • On the other hand,

(m

= nt +

("-IL r L m~1 r~1 00

m

m 1 )t + - , = nt m.

+ L 00

m~1

(m

+ l)Sm(n)tm+1

,

(m+1).

which proves the formula (8.4). Note that for n = 1 the formula (8.4) coincides with (8.2). Theorem 4 (von Staudt's Theorem). Let p be a prime and m and even integer. If (p - l),r m, then B m is p-integral (that is, p does not appear in the denominator of B m ). If (p - 1) I m, then pBm is p-integral, and (modp).

We prove Theorem 4 by induction on m, using the relation (m

+ I)Sm(P) =

m-I

(m

+ 1)BmP + L

C~+IBkpm+I-\

k~O

which is obtained from (8.5) by substituting p for n. We write this in the form (8.6) and we shall show that all terms under the summation sign are p-integers which are divisible by p (in the ring of p-integral numbers). The term pBk for k < m is a p-integer by the induction assumption. We consider the terms 1 k m-k - - 1 Cm + 1 p .

m+

(8.7)

385

BERNOULLI NUMBERS

Sec. 8]

If p = 2, then since m + I is odd, this number is a 2-integer and is divisible by 2 (since k < m). If p =1= 2, we write (8.7) in the form _1_ C m+ i - k m-k = m(m - 1) .. · (k + 1) m-k m+l m+i P (m-k+l)! P . The number p occurs in (m - k

[~] +

+ I)! = r!

[;2] + ... < ~ + ;2 + ... =

with exponent

P

~1~~~r -

1 = m - k,

and hence [I/(m - k + I)!]pm-k is a p-integer and is divisible by p. Hence we have shown that pBm is p-integral and that (8.8)

pBm == Sm(P) (mod p) in the ring of p-integral numbers. On the other hand, we have the congruences Sm(P) == -l(mod p)

if (p - 1) I m,

(8.9)

Sm(p)==O(modp)

if(p-l),rm.

(8.10)

Indeed, if(p - I) I m, then X" == I (mod p), for I ~ x ~ p - I, and hence p- 1 p-i Sm(P) = L X m == L 1 = P - 1 == -1 (mod p). x;i

x;i

If (p - I),r m, then, taking 9 to be a primitive root modulo p, we shall have p-i p-2 g(p-i)m_l Sm(P) = L x m == L gmr = m == 0 (mod p), x; 1 r;O g - 1 since gr i == I (mod p) and gm ¢. I (mod p). Comparing (8.8) and (8.10), we see that if (p - I),r m, then pBm == 0 (mod 0), and this means that B m is p-integral. The second assertion of Theorem 4 follows from (8.8) and (8.9). In the case m ~ P - I the number p - I does not divide any number k < m, so that all Bk for k < mare p-integral. Hence every term on the right in (8.6) is divisible by p2, and we have the following assertion. Corollary. If p

=1=

2 and m ~ p - I (m even), then

pBm == Sm(P) (mod p2).

(8.11)

Theorem 5 (Kummer's Congruence). If p is prime and (p - I),r m (m positive and even), then the number Bm/m is a p-integer, and Bm + p -

i

m+p-l

_B- ( mod) = p. m

m

(8.12)

386

ANALYTIC METHODS

[Chap. 5

In other words, the expression Bm/m (for (p - l),r m) has period p - I modulo p. Proof. Consider the function

(8.13) where 9 is a primitive root modulo p, I < 9 < p. We set e t F(t) =

-

I

= u. Then

gt t - - = tG(u) (1+u)9-1 u '

where

g G(u) - (-1-+-u-)-9---1

1 u

g gu

+ '" + u 9

U

It is clear that the numbers Ck are p-integral. We shall show that in the expansion of the function G(u) in powers of t: (8.14) all the coefficients Am are p-integral, and that they have period p - I modulo p (for m > 0). It is clear that if this latter property holds for some collection of series, then it also holds for any linear combination of these series with p-integral coefficients. Hence it suffices to verify it for the functions (e t - l)k. But these functions are in turn linear combinations of the functions err with r ~ 0, and

so by the small Fermat theorem r"+ p-l

==

r" (mod p)

(n > 0).

Hence the functions ert have the desired property, and our assertion about the coefficients Am is proved. Comparing the coefficients in (8.13) and (8.14), we see that

Bm(gm - 1) m!

Am - 1 (m-l)!'

so that

Since gm - 1 1= 0 (mod p) (because (p - l),r m), then the sequence of numbers gm - 1 also has period p - 1 modulo p, by the small Fermat theorem.

Sec.S]

3S7

BERNOULLI NUMBERS

It now follows from what we have proved about the numbers Am that the numbers Bm(m, when (p - 1) %m, are p-integral and have period p - 1 modulo p. Theorem 5 is proved. Theorem 6. The Bernoulli number B 2m is given by the formula B 2m = ( -1)

m_1 2(2m)! (2n)2m (2m),

(8.15)

where (2m) is the value of the Riemann (-function (s) for s = 2m. To prove this we use the expansion of the function l(e t - 1) into partial fractions 1

1

1

+00

-t = - - + L e -1 2 n=-oot-2nin 1

1

2t

00

= - 2" + t + n~l t 2 + (2nn)2 .

(8.16)

This expansion can be derived from the familiar expansion for the cotangent 1 cot z = - + Z

00

L1 Z 2 -

n=

2z

(nn

)2 '

by using the fact that

eiz + e- iz 2i cotz=i. e'z _ e-'z. =i+ e2''z - 1 • It follows from (8.16) that t t 00 t2 -t - - 1 - - + 2 " 2 e - 12 n~l t + (2nn)2 ' and since

L 00

m=l

(_l)m-l ( - t ) 2m , 2nn

then

Comparing this equation with (8.1) and equating the coefficients of the various powers of t, we obtain (8.15).

388

ANALYTIC METHODS

[Chap. 5

From the formula (8.15) we obtain an estimate for the growth of the numbers IE2m l with increasing m. Since '(2m) > 1 and (2m)! > (2m/e)2m (by Stirling's formula), then m)2m IB 2m l > 2 ( ne . In particular, we find that

I~~I-+oo

-+

as m

00.

PROBLEMS

1. Show that (x

+ B)m =

m ~ 1.

(x - 1 - B)m,

2. Show that

(~2 +B)m

=

(_1 -l)Bm. 2m -

1

3. Let p be an odd prime number. Show that

x.f

(p-1)/2 x(p -1)/2

l

== 2

((2)) P2

B(P+1)/2

(mod p).

4. Let p > 3 be a prime number of the form 4k + 3. If h denotes the number of divisor classes of the imaginary quadratic field R(V -p), show that h satisfies the congruence

== -2B(p+1)/2 (modp).

h

5. If p

> 3 is a prime number, show that 1 1 1 +- +- + 2 3

... + p-1 -1- == 0 (mod p 2 ).

6. Prove the formula (kx + B)m = k m-1

t-1( L

'=0

x

+ -S + B)m k

(k and m are natural numbers).

7. The function tan x has the expansion X2n-1

<Xl

tanx=

LT (2n-I)! n

n=l

where

To

n

=

2 2n (2 2n _ 1) IB2n l 2n .

Show that all the coefficients Tn are natural numbers.

,

Sec. 8]

BERNOULLI NUMBERS

389

8. If m > 1, show that 2B 2m

"'"

1 (mod 4).

9. Let q be a prime number such that 2q + 1 is composite [for instance, q "'" 1 (mod 3)]. Show that the numerator of the Bernoulli number B 2 • is divisible by a prime of the form 4n+ 3. 10. Let p" ... , P, be prime numbers greater than 3, and let q be a natural number such that q =< 1 (mod M), where M = (PI - 1) ... (P, - 1). Show that none of the prime numbers p" ... ,P, divide the numerator of the fraction B 2 ./"U[.

Algebraic Supplement

1. Quadratic Forms over Arbitrary Fields of Characteristic #- 2 In this section we describe some of the general properties of quadratic forms over arbitrary fields. We shall state some well-known results without proof. Throughout, K will denote an arbitrary field whose characteristic is not 2. For any matrix A, we shall denote the transpose by A'.

1.1. Equivalence of Quadratic Forms

By a quadratic form over the field K we mean a homogeneous polynomial of degree 2 with coefficients in K. Any quadratic form f can be written n

f=

L

aijXiX j ,

i,j= 1

where aij = a ji • The symmetric matrix A = (aij)

is called the matrix of the quadratic form! If the matrix is given, the quadratic form is completely determined (except for the names of the variables). The determinant d = detA is called the determinant of the quadratic form! If d = 0, the form f is called singular, and otherwise it is called nonsingular. If we let X denote the column vector of the variables XI' X2 , •. , , X n , then the quadratic form can be written f= X'AX. 390

Sec.1]

QUADRATIC FORMS OVER ARBITRARY FIELDS

Suppose we replace the variables according to the formula

XI' .•. , X n by

391

the new variables Yl' ... ,Yn

n

Xi

=

L CijYj

(1 ,;;; i ,;;; n,

Cij E

K).

j=1

In matrix form this linear substitution becomes

X=CY, where Y is the column vector of the variables YI' ... , Yn' and C is the matrix XI' .. , , X n infby the corresponding expressions in YI' ... ,Yn' then (after carrying out the indicated operations) we shall obtain a quadratic form 9 (also over the field K) in the variables YI' ... , Yn' The matrix Al of the quadratic form 9 equals

(cij)' Ifwe replace the variables

(1.1) Two quadratic forms f and 9 are called equivalent, and we write f '" g, if there is a nonsingular change of variables which takes one form to the other. From formula (1.1) we obtain Theorem 1. If two quadratic forms are equivalent, then their determinants differ by a nonzero factor which is a square in K.

Let 'I be an element of K. If there exist elements f(ct lo

.. , ,

ct n )

ct l , ... , ctn

in K for which

= 'I,

then we say that the form f represents y. In other words, a number is represented by a quadratic form if it is the value of the form for some values of the variables. It is easily seen that equivalent quadratic forms represent the same elements of the field K. We shall further say that the formf represents zero in the field K if there exist values ct; E K, not all zero, such that f(ct 1 , ... , ct n ) = 0. The property of representing zero is clearly preserved if we pass to an equivalent form.

ct

Theorem 2. If a quadratic form f in n variables represents an element '# 0, then it is equivalent to a form of the type ctXI

2

+ g(X2'

... ,

x n ),

where 9 is a quadratic form in n - I variables. Regarding the proof of this theorem we note only the following. If f(ct 1 , .. , , ct n) = ct, then not all ct; are equal to zero, so we can find a nonsingular matrix C, whose first row is ct l , .,. , ct n • If we apply to f the linear substitution whose matrix is C, we obtain a form in which the coefficient of the square of the first variable is ct. The rest of the proof is carried out as usual.

392

ALGEBRAIC SUPPLEMENT

If the matrix of a quadratic form is diagonal (that is, if the coefficient of every product of distinct variables equals zero), then we say that the form is diagonal. Theorem 2 now implies Theorem 3. Any quadratic form over K can be put in diagonal form by some nonsingular linear substitution. In other words, every form is equivalent to a diagonal form. In terms of matrices, Theorem 3 shows that for any symmetric matrix A there exists a nonsingular matrix C such that the matrix C'A C is diagonal. 1.2. The Direct Sum of Quadratic Forms Since the names of the variables are not of any significance, we can assume that two given quadratic forms f and 9 have different variables. In this case the form f + 9 is called the direct sum of f and g, and is denoted by f 9 (this must not be confused with the usual addition of quadratic forms when the forms have the same variables). It is clear that if 9 ~ h, thenf 9 ~ f h. We shall now show that the following converse holds.

+

+ +

Theorem 4 (Witt's Theorem). Letf, g, and h be nonsingular quadratic forms over the field K. If the forms f 9 and f h are equivalent, then the forms 9 and h are also equivalent.

+

+

Proof. Let fo be a diagonal form equivalent to f Then, as noted above, f 9 ~ fo 9 and f h ~ fo h, so that fo 9 ~ fo h. Hence we may assume thatfis a diagonal form. It is now easily seen that to prove the theorem it suffices to consider the case f = ax 2 , a -# 0. Let A and B denote the matrices of 9 and h. Since the forms ax 2 9 and ax 2 h are equivalent, there exists a matrix

+

+

+

+

+

+

+

+

such that

y (S'

T')

Q'

(aO A 0) (yT QS) = (a° B' 0)

(Here S is a row matrix and T is a column matrix.) From this equation we obtain (1.2) yaS S'aS

+ T'AQ = 0,

(1.3)

+

(1.4)

Q'AQ = B.

Sec.1]

f

QUADRATIC FORMS OVER ARBITRARY FIELDS

393

We must show that there exists a nonsingular matrix Co such that Co' A Co = B. The matrix Co will be found in the form

CO =Q where the element

~

+ ~TS,

must be suitably chosen. By (I.2) and (1.3) we have

+ ~S'T')A(Q + ~TS) Q'AQ + ~S'T'AQ + ~Q'ATS + eS'T'ATS Q'AQ + a[(1 - y2)e - 2y~]S'S.

Co'ACo = (Q' = =

e e 'I 11

In view of (I.4) the last expression will equal the matrix B, provided that (I - y2)e - 2y~ = I. This equation, which can also be written in the form (y~ + 1)2 = 0, always has a solution ~o E K for any y E K (recall that the characteristic of K is not 2). Hence we have found a matrix Co = Q +~oTS, for which Co' A Co = B. Since the matrix B is nonsingular, then Co is also nonsingular. Theorem 4 is proved.

e-

!.

1.3. Representation of Field Elements e

Theorem 5.. If a nonsingular quadratic form represents zero in the field K, then it also represents all elements of K.

..,

Proof. Since equivalent forms represent the same field elements, it suffices

y

n

:s

:s

Ie

2)

3)

4)

to prove the theorem for a diagonal form f = al Xl 2 + ... + anx/. Let 2 2 al et 1 + .. , + anet n = be a representation of zero, and let y be any element of K. We can assume that etl ;6 0. We express the variables Xl> ... , X n in terms of a new variable t:

°

= et l (1 + t), X k = etk(l - t) Substituting in the form f we obtain f* = f*(t) = 2a l et 1 2 t - 2a 2 et/t - ...

(k

Xl

If we now set t = y/4a l et 1

2

,

= 2,

- 2a net n2 t

... , n).

= 4a l et/t.

we obtain/* = y.

Theorem 6. A nonsingular quadratic formf represents the element y ;6 in K if and only if the form - YXo 2 f represents zero.

+

°

Proof. The necessity of the condition is clear. Assume that

-yet o2

+ f(et l , ... , et n) =

0,

where not all eti equal zero. If et o ;6 0, then y = f(etl/eto,"" etn/et o). If et o = 0, then the form f represents zero, and hence by Theorem 5 it represents alI elements of the field K.

394

ALGEBRAIC SUPPLEMENT

Remark. From the proof of Theorem 6 it is clear that if we determine all representations of zero by the form -yx/ ! (only those in which Xo #are relevant), then we have also determined all representations of y by the form! Hence the question of the representability of an element of the field K by a nonsingular form can be reduced to the question of the representability of zero by a nonsingular form in one more variable.

°

+

Theorem 7. If a nonsingular form! represents zero, then it is equivalent to a form of the following type:

YlY2

+ g(Y3'

, Yn)'

, ct n such thatf(ct l , ... , ct n) = 1. By Theorem 2 we can now put f in the form x 1 2 + fl (X 2, ... ,xn). Since the form x/ fl represents zero, we can find P2' ... , Pn such thatfl(p2' ... Pn) = -1. Again applyin§ Theorem 2, we can put!l in the form -x/ + g(Y3' ... , Yn). Setting Xl - X2 = Yl' and Xl + X2 = Y2' we obtain the desired result.

Proof. Using Theorem 5, we first find ct l ,

+

Remark. If we know some representation of zero by the formj, then all the operations described in the proof of Theorem 7 can be carried out explicitly, and the form g(Y3"'" Yn) can be determined. Now assume that for any quadratic form which represents zero over the field K, an actual representation of zero can be found. Then any nonsingular form can be transformed to a form of the type

YlY2

+ '" + Y2s-lY2s + h(Y2s+1'

... , Yn),

(1.5)

where the form h does not represent zero. In any representation of zero by the form (1.5), at least one of the variables Yl' Y2' ... , Y2s+l' Y2s must be nonzero. To determine all representations of zero in which, say, Yl = ct l #- 0, we note that we can give Y3' .. , ,Yn arbitrary values ct3 , ... , ctn and then determine Y2 by the condition ct l Y2

+ ct3ct4 + ... + g(ct2s+l'

... , ctn) = 0.

This gives us an effective method for finding all representations of zero by a nonsingular quadratic form over the field K, provided that we have a method for determining whether or not a given form represents zero, and, in case it does, an algorithm for finding some specific representation of zero. Theorem 8. Let the field K contain more than five elements. If the diagonal form

represents zero in the field K, then there is a representation of zero in which all the variables take nonzero values.

Sec. 1 ]

395

QUADRATIC FORMS OVER ARBITRARY FIELDS

Proof We first show that if

°

ae

= A #- 0, then for any b #there exist nonzero elements rx and p such that arx 2 + bp2 = A. To prove this fact we consider the identity (t - 1)2 4t ----=+ = 1. (t + 1)2 (t + 1)2

Multiplying this identity by a(

ae = A, we obtain

t 1) 2 c-+at (-2~- )2 =A. t+1 t+1

(1.6)

Choose a nonzero y in K so that the value of t = to = by2/a is not ± 1. This can be done because each of the equations bx 2 - a = and bx 2 + a = has at most two solutions for x in K, and the field K has more than five elements. Setting t = to in (1.6), we obtain

°

°

to - 1)2 (2~y)2 +b - - =..1., to + 1 to + 1

a ( ~--

and our assertion is proved. We can now easily complete the proof of the theorem. If the representation a 1 ~12 + .,. + a.~. 2 = is such that ~1 #- 0, ... , ~r #- 0, ~r+1 = ... = ~. = 0, where r~ 2, then we have shown that we can find rx #- and p#-O such that ar~/ = arrx 2 + ar+ 1P2, and this yields a representation of zero in which the number of nonzero variables is increased by one. Repeating this process, we arrive at a representation in which all the variables have nonzero value.

°

°

1.4. Binary Quadratic Forms

A quadratic form in two variables is called a binary quadratic form. Theorem 9. All nonsingular binary quadratic forms which represent zero in K are equivalent.

Indeed, by Theorem 7, any such form is equivalent to the form Y1Y2 . Theorem 10. In order that the binary quadratic form f with determinant d#-O represents zero in K, it is necessary and sufficient that the element -d be a square in K (that is, -d = rx 2 , rx E K). Proof. The necessity of the condition follows from Theorems 1 and 7. Con-

versely, if f

=

ax 2 + by 2 and -d = -ab

=

rx 2 , thenf(rx, a)

=

arx 2 + ba 2 = 0.

Theorem 11. Letf and 9 be two nonsingular binary quadratic forms over the field K. In order that f and 9 be equivalent, it is necessary and sufficient

396

ALGEBRAIC SUPPLEMENT

that their determinants differ by a factor which is a square in K, and that there exist some nonzero element of K which is represented by both f and g. Proof. Both conditions are clearly necessary. To prove sufficiency, let rx #- 0 be an element of K which is represented by both f and g. By Theorem 2 fand 9 are equivalent to the formsf1 = rxx 2 + fly 2 and gl = rxx 2 + fl'y2. Since rxfl and rxfl' differ by a square factor, then fl' = fJ'/, y E K, and this means that

f1-g1 andf-g·

PROBLEMS

1. Show that a singular quadratic form always represents zero. 2. Show that Theorem 5 does not hold for singular quadratic forms. 3. If the binary form x 2 - cxy2 represents the elements y, and y2 of K, show that it also represents their product. 4. Show that Theorem 8 is not valid for fields with not more than five elements. 5. We shall decompose the set of all nonsingular quadratic forms over Kin n = 0, 1,2, ... variables into the so-called Witt classes. (We treat the zero form as a nonsingular form on the empty set of variables, and consider it to represent zero.) Two forms I, and 12 belong to the same Witt class [Id = [/2], if the corresponding forms h in (1.5) have the same number of variables and are equivalent. We add Witt classes by the formula [Id + [/2] = [I, +/2]' Show that these definitions make sense and that under 1his operation the set of Witt classes becomes a group. 6. Determine the group of Witt classes for the real and complex fields. 7. Show that a quadratic form over a finite field (characteristic oF 2) in three or more variables represents zero.

2. Algebraic Extensions

Many theorems of this section are given without proof. For the proofs the reader may consult, for example, "Modern Algebra," by B. L. van der Waerden, Vol. 1, Chap. 5, Ungar, New York, 1950. 2.1. Finite Extensions

If the field n contains the field k as a subfield, then we say that n is an extension of the field k. To denote that n is being considered as an extension field of k, we write njk. If K is a subfield of n which contains k, that is, k eKe n, then K is called an intermediate field for the extension njk. For any extension njk, we may consider n as a vector space over k. Definition. The extension Kjk is called finite if K, considered as a vector space over k, is finite-dimensional. This dimension is called the degree of the

Sec. 2]

ALGEBRAIC EXTENSIONS

397

extension and is denoted by (K: k). Any basis for K as a vector space over k is called a basis for the extension Klk. If the extension Klk is finite, then for any intermediate field K o , the extensions Kolk and KIKo are clearly both finite. The following converse also holds. Theorem 1. Let K o be an intermediate field for the extension Klk. If the extensions KIKo and Ko/k are finite, then Klk is also finite, and its degree equals the products of the degrees of the extensions KIKo and Ko/k:

(K: k) = (K: Ko)(Ko : k). Proof. Let 01 , ... Om, be a basis for KIKo , and Wi' ... , W n be a basis for Kolk. Then every element of K can be represented as a linear combination (over k) of the products wiO j , so the extension Klk is finite. Further, it is easily checked that these products are linearly independent over k, so (K:k) = mn. For any field k we denote the ring of polynomials in the variable t with coefficients in k by k[t]. Let nlk be an extension of the field k. An element rx En is called algebraic over k; it is the root of some nonzero polynomialf(t) of k[t]. Among all such polynomials we take that polynomial cp(t) "# 0 which is of lowest degree and has leading coefficient 1. Since all polynomialsf(t) which have rx as a root are divisible by cp(t) (otherwise the remainder after division offby cp would be a polynomial of lower degree with rx as a root), then the polynomial cp(t) is uniquely determined. It is called the minimum polynomial of the algebraic element rx over the field k. The minimum polynomial cp E k[t] is always irreducible, since from cp = gh it follows that rx is either a root of g(t) or of h(t). Any element a E k is algebraic over k, and its minimum polynomial is t - a. An element ~ E n which is not algebraic over k is called transcendental over k. The extension nlk is called algebraic if every rx E n is algebraic over k.

:, j :' :1

Theorem 2. Any finite extension is algebraic. i

Theorem 3. Let the element rx of the extension n/k be algebraic over k, and let its minimum polynomial cp(t) E k[t] have degree m. Then the elements 1, rx, ... , rx m - 1 are linearly independent over k and the set of all linear combinations (2.1) with coefficients ai in k is an intermediate field, denoted by k(rx). The extension k(rx)lk is finite of degree m. To add two elements of the field k(rx), written in the form (2.1), we simply add the corresponding coefficients. To put the product of the elements

398

ALGEBRAIC SUPPLEMENT

~ = g(rx) and '1 = h(rx) (g and h are polynomials in k[t] of degree the form (2.1), we must divide gh by cp with remainder:

g(t)h(t) = cp(t)q(t)

~

m - 1) in

+ r(t),

where the degree of r(t) does not exceed m - 1; since cp(rx) = 0, then ~'1 = r(rx). Hence the operation of multiplication in the field k(rx) is determined by the minimum polynomial cp(t) of the element rx. Let rx1' .,. , rxs be a finite set of elements of n, which are algebraic over the field k, and let m 1 , ••• , m s be the degrees of their minimum polynomials over k. The set of all linear combinations of the elements

with coefficients in k is an intermediate field. It is denoted by k(rx1' ... ,rxs ) and is called the field generated by the elements rx1, .,. , rx s ' Its degree over k does not exceed the product m1 ... ms • Any finite extension Kjk, contained in n, can be represented in the form k( rxl> ... , rx.) for some rx1' ... , rx s ' Definition. A finite extension Kjk is called simple if there is an element () such that K = k«()). Any element () E K, for which K = k«()), is called a primitive element of the extension Kjk.

The primitive elements of Kover k are those elements whose minimum polynomial has degree equal to the degree of Kjk. Theorem 4. Let njk and n'jk be two extensions of the field k, and let nand ()' E n' be algebraic elements over k with the same minimum polynomial cp(t). Then there is a unique isomorphism of the field k«()) onto the field k«()') for which () -+ ()' and a -+ a for all a E k. () E

Let m be the degree of the polynomial cp(l). The isomorphism k«()) of Theorem 4 coincides with the mapping ao + a1 ()

+ ... + am + 1 ()m - 1 -+ ao + a1 ()' + ... + am + 1 ()'m - 1

-+

k«()') (2.2)

(a1' ... , am -1 are arbitrary elements of the field k). So far we have considered finite extensions Kjk which are contained in a given extension njk. We now turn to the question of the construction of finite extensions over a fixed field k. Theorem 5. Let k be a field, and let cp(t) be an irreducible polynomial of k[t] of degree n. Then there exists a finite extension Kjk of degree n in which the polynomial cp has a root. The extension Kjk is unique (up to an isomorphism which is the identity map on k). If cp«()) = 0, () E K, then K = k«()).

Sec. 2]

ALGEBRAIC EXTENSIONS

399

The field K (in the case n > 1) is constructed in the following manner. We choose some new object e and consider the set K of all formal linear combinations (2.3) with coefficients in k. If we denote the polynomial ao + alt + ... + an_lt n- 1 by get), then the expression (2.3) can be written gee). Let ~ = gee) and '1 = h(e) be two linear combinations of the type (2.3) (g and h are polynomials in k[t] of degree ~ n - 1). Let s(t) denote the sum g(t) + h(t) and let r(t) denote the remainder after division of the product g(t)h(t) by (X~ K) is a linear transformation of K (considered as a vector space over k). The characteristic polynomial j~(t) of this transformation is also called the characteristic polynomial of the element (X E K, relative to the extension Kjk. If W l , .. , , W n is a basis for the extension Kjk and (~E

n

(XW j

=

L aijw

j

(aj j

E

k),

(2.4)

j=l

then faCt)

= det(tE -

(aij»'

where E is the n by n identity matrix. Theorem 6. The characteristic polynomial j~(t) of an element (X E K relative to the extension Kjk is a power of its minimum polynomial ... , O'n' If 11. is any element of K, then the characteristic polynomial fa(t) has the factorization

in the field

n.

The elements 0'1(11.), ... , O'n(l1.) (which lie in the field n) are called the conjugates of the element 11. E K. The images O'l(K), ... , O'n(K) of the field K under the isomorphisms O'i are called the conjugate fields of the field K. If () is a primitive element of the field Kover k, then it is clear that O'i(K) = k(O'i«())' Corollary 1. Using the above notations we have

N K/k(l1.) = 0'1(11.)0'2(11.)'" O'n(I1.), SPK/k(l1.) = 0'1(11.)

+ 0'2(11.) + ... + O'n(I1.)· ~

Corollary 2. There are precisely n isomorphisms of any finite extension of the rational numbers of degree n into the field of complex numbers.

(

s

Sec. 3]

FINITE FIELDS

405

Let WI' '" , w n be a basis for K/k. Since Sp(wiw) = L~= 1 Us(Wi)Us(W), then the matrix (Sp(wiw)) is the product of the matrices (u;(w))' and (u;(w)). Hence we have the following formula for the discriminant of the basis Wi: (2.15)

PROBLEMS

1. Let n = k(x) be the field of rational functions in x with coefficients from k. Show that any element of n which does not lie in k is transcendental over k. 2. Let k eKe L be a tower of finite extensions. If 8 is any element of L, prove the formula N K/1 (NL/ K(8» = NL/K(G).

[First assume that L for K/k.l

=

K(8), and consider the basis w/8J for L/k, where w/ is a basis

3. Find a primitive element for the extension R(V2, V3) of the field R of rational numbers, and express it in terms of and 4. Show that a finite extension K/k is simple if and only if there are only a finite number of intermediate fields. 5. Let k be any field of characteristic p 'f O. Show that the polynomial 1(/) = IP - 1 - a (a E k) is either irreducible or factors completely into linear factors in k[/]. Further, in the former case show that the extension k(8)/k, where 1(8) = 0, is separable. 6. Let k o be a field of characteristic p 'f 0, and let k = ko(x) be the field of rational functions in x with coefficients from k o . Show that the polynomial 1(/) = IP - x is irreducible in k[/]. Further, show that the extension k(8)/k, where f(8) = 0, is inseparable. 7. Let K/k be a finite extension of degree n. If there exists some extension n/k for which there are n isomorphisms of K imo n which leave every element of k fixed, show that the extension K/k is separable. 8. Let k be a field of characteristic 'f p which contains a primitive pth root of I (that is, an element f: with f:P = I and f:> 'f I for 0 < k < p). If the element Cl: E k is not the pth power of some element of k, show that (k({/~) : k) = p. 9. Let K/k be a finite separable extension and let g; be a linear mapping from K (considered as a vector space over k) to k. Show that there is a unique element Cl: in the field K such that

vi vi

(~E K).

3. Finite Fields A field 1: is called finite if it has only a finite number of elements. The field Zp of residue classes in th~ ring Z of integers modulo a prime number p is an example of a finite field. Every finite field is of finite characteristic and, if the characteristic of the finite field 1: equals p, then this field contains a prime subfield (a subfield not containing any proper subfie]d) which is isomorphic

406

ALGEBRAIC SUPPLEMENT

to Zp- Hence we may assume that Zp C 1:. The extension 1:/Zp is clearly finite. Ifit is of degree m and if WI' ... , W m is a basis for 1:/Zp,then every element ~ E 1: has a unique representation in the form ~ = CIW I + ... + CmW m , where Ci E Zp. Since there are pm such linear combinations, we have proved that the number of elements of any finite field is a power of its characteristic. The multiplicative group 1:* of a finite field 1: is a finite Abelian group. We consider its structure. Lemma 1. A finite subgroup G of the multiplicative group K* of any field K is al'Nays cyclic. Proof. We first show that if an Abelian group G contains elements of

orders m and n, then it contains an element whose order equals the least common multiple r of m and n. Let the elements x and y of G have orders m and n, respectively. If(m, n) = I, then it is easily seen that the product xy has order r = mn. In general, by considering the prir,le factorizations of the numbers m and n, we can find factorizations so that (m o , no) = I and r = mono. The elements x m , and y"' have orders m o and no, and their product xm'y"1 has order r = mono. Now let G be a finite subgroup of order 9 of the multiplicative group of the field K. If m is the maximum of the orders of the elements of G, then clearly m ~ g. On the other hand, it follows from what we have just shown that the order of every element divides m; that is, every element of the group G is a root of the polynomial t m - I. Since a polynomial of degree m can have at most m roots, 9 ~ m. Hence 9 = m, and this means that G is cyclic. Applying this lemma to the case of finite fields, we obtain the following fact. Theorem 1. The multiplicative group of a finite field which has pm elements is a cyclic group of order pm - I.

e t s tl a n

a;

oj

t .

fie fo dl

Corollary. Any finite extension of a finite field is simple.

Indeed, if 8 is a generating element of the group 1:*, then it is clear that Zi8) = 1:. Hence for any intermediate field 1: 0 we have 1: 0 (8) = 1:. It also follows from Theorem I that all elements of 1: are roots of the polynomialJP~ - t, and since the degree of this polynomial equals the number of elements in 1:, then in the ring 1:[t] we have the factorization t P'"

-

/

= Il (I ~eI:

(~

runs through all elements of the field 1:).

- ~)

fin

be ax ele an~

fiel

Sec. 3]

rly :nt

~re

FINITE FIELDS

407

Theorem 2. For any prime number p and any natural number m there exists one and only one (up to isomorphism) finite field with pm elements.

:he

Proof. By the corollary to Theorem 5 of Section 2 there is an extension Q/Z in which the polynomial t pm - t factors into linear factors. Let 1: denote the :et of all roots of this polynomial (in Q). Since in any field of characteristic

LIp.

p the formula

K

of st m as he

10

)f Iy

Ie

a t

holds, then the sum and difference of any two elements of 1: also belongs to L It is clear that the set 1: is closed under the operations of multiplication and division (except division by zero). Hence 1: is a subfield of the field Q. The pm polynomialt pm - thasnomultipleroots(sinceitsderivativepm t - 1 - I = -I never vanishes) and hence 1: consists of pm elements. The existence of a finite field with pm elements is proved. Let 1: and 1:' be two extensions of Zp of degree m. Choose a primitive element 0 in 1: (corollary of Theorem I) and denote its minimum polynomial by cp(t). Since cp(t) divides the polynomial t pm - t, and the latter polynomial splits into linear factors over L', then ({J(t) has a root (J' E L'. The degree of the extension Zp(()')/Zp equals the degree of the polynomial cp(t), that is, m, and therefore Zp(()') = 1:'. The existence of an isomorphism of 1: onto 1:' now follows from Theorem 4 of Section 2. The finite field with pm elements is often denoted by GF(pm) (and finite fields are often called Galois fields).

Corollary. For every natural number n there is an irreducible polynomial of degree n over the finite field Lo = GF(p'). Indeed, p' - I divides pn, - I, so that the set of all roots of the polynomial t - t in the field 1: = GF(p,n) forms a subfield which is isomorphic to the field 1: 0 , Hence we can assume that 1: 0 c 1:. If () E 1: is a primitive element for the extension 1:/1: 0 , then the minimum polynomial of () will be an irre-

ducible polynomial in 1: o[t] of degree n, since . ~ ) _ (1: : Zp) _ rn _ ( 1: . ~o - - - n. (1: 0 : Zp) r

f

In conclusion we note that in order to show that a given finite commutative ring is a field, it suffices to show that it has no divisors of zero. Indeed, let 0 be a finite ring without zero divisors, and let a be a nonzero element of O. If aXl = aX2' then a(x 1 - X2) = 0, and Xl = X2' Thus as X runs through all elements of the (finite) ring 0, ax also takes on every value in O. Then for any b in 0 the equation ax = b is solvable in 0, and this means that 0 is a field.

408

ALGEBRAIC SUPPLEMENT

PROBLEMS

1. Let r(m) denote the number of distinct irreducible polynomials of degree m with leading coefficient 1 in the ring Zp[t]. Show that r(m) =

.!. L JJ.(~)pd

m

dl'"

d

JJ. denotes the Mobius function). 2. Find all irreducible polynomials of degree 2 over the field Z, = GF(5). 3. Show that the field GF(p'") is contained in the field GF(P") (up to isomorphism) if and only if min. (d runs through all divisors of m, and

4. What is the degree over Zp of the splitting field of the polynomial t" - 1? 5. Let ~ = GF(p'"). Show that each mapping a, : g--'" g-', g E ~ (i = 0, I, ... , m - I) is an automorphism of the field ~ and show that every automorphism of ~ coincides with one and only one of the a,. 6. Let ~o = GF(p') and let ~ be a finite extension of ~o of degree n. Show that each mapping g --",gP" (i = 0, I, ... , n - I) is an automorphism of the field ~ which leaves every element of ~o fixed. Further, show that these n automorphisms are distinct and that every automorphism of ~ which is the identity on ~o coincides with one of them. Let f(t) be the characteristic polynomial of the element g E ~ relative to the extension ~/~o. Show that we have the factorization h(t) = (t - g)(t - g.) .. ; (t - en-I),

in the field ~, where q

=

p' (use Theorem 8 of Section 2). Deduce that

7. Show that a finite extension of a finite field is always separable. 8. Using the above notations, show that every element of the field ~o is the norm of Sl'me element of ~. 9. Let ~ = GF(q'), where p'" = q, and let IX E~. Show that the equation is solvable in ~ if and only if IX + IX· + ... + IX.' - I = O.

e- g

=

IX

'0. Let e be a primitive pth root of 1 (over the field of rational numbers). Let ~o = GF(p) and ~ = GF(p'"). Since the elements of the field ~o can be considered as residue classes of integers modulo the prime p, then the expression eSP!Y) makes sense for any y E ~ (the trace is taken relative to the extension ~/~o). Show that for

IX

'f 0,

for

IX

=

O.

11. Let X be a character of the multiplicative group of the field ~ = GF(p'"), and set p'" = q (for the definition of a character, see Section 5). Extend X to the Whole field ~ by setting X(O) = O. The expression (IX E

~),

Sec. 3]

FINITE FIELDS

409

which is a complex number, is called the Gaussian sum of the finite field 1:. Assuming that the character X is not the unit character, prove the formulas 0:=1-0,

!T'<X)!

=yi

2 T,<X) =

0:=1-0,

O.

• ;