Algebraic number theory, serge lang 1

CHAPTER I Algebraic Integers This chapter describes the basic aspects of the ring of algebraic integers in a number field always assumed to be of finite degree over the rational numbers

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60 ARNOLD Mathematical Methods in C1assical Mechanics

61 WHITEHEAD Elements of Homotopy Theory

62 KARGAPOLOV/MERZIJAKOV Fundamentals of the Theory of Groups

63 BOLLABAs Graph Theory

64 EDWARDS Fourier Series Vol I 2nd ed

65 WELLS Differential Analysis on Complex Manifolds 2nd ed

66 W ATERHOUSE Introduction to Affine Group Schemes

67 SERRE Local Fields

68 WEIDMANN Linear Operators in Hilbert Spaces

69 LANG Cyclotomic Fields 11

70 MASSEY Singular Homology Theory

71 FARKAS/KRA Riemann Surfaces

72 STILLWELL C1assical Topology and Combinatorial Group Theory

73 HUNGERFORD Algebra

74 DAVENPORT Multiplicative Number Theory 2nd ed

75 HOCHSCHILD Basic Theory of Aigebraic Groups and Lie Aigebras

76 IITAKA Aigebraic Geometry

77 HEcKE Lectures on the Theory of Aigebraic Numbers

79 WALTERS An Introduction to Ergodic Theory

80 ROBINSON A Course in the Theory of Groups

81 FORSTER Lectures on Riemann Surfaces

82 BOTT/Tu Differential Forms in Aigebraic Topology

83 WASHINGTON Introduction to Cyclotomic Fields

84 IRELAND/RoSEN A C1assicallntroduction to Modern Number Theory

85 EDWARDS Fourier Series: Vol 11 2nd ed

86 VAN LINT Introduction to Coding Theory

87 BROWN Cohomology of Groups

88 PIERCE Associative Aigebras

89 LANG Introduction to Aigebraic and Abelian Functions 2nd ed

91 BEARDON On the Geometry of Discrete Groups

92 DIESTEL Sequences and Series in Banach Spaces

93 DUBROVIN/FoMENKO/NoVIKOV Modern Geometry-Methods and Applications Vol I

94 WARNER Foundations of Differentiable Manifolds and Lie Groups

95 SHIRYAYEV Probability, Statistics, and Random Processes

96 CONWAY A Course in Functional Analysis

97 KOBLITZ Introduction to Elliptic Curves and Modular Forms

98 BRÖCKER/tom DIECK Representations of Compact Lie Groups

99 GROVE/BENSON Finite Reflection Groups 2nd ed

100 BERG/CHRISTENSEN/RESSEL Harmonic Analysis on Semigroups: Theory of positive definite and related functions

101 EDWARDS Galois Theory

102 VARADARAJAN Lie Groups, Lie Aigebras and Their Representations

106 SILVERMAN The Arithmetic of Elliptic Curves

107 OLVER Applications of Lie Groups to Differential Equations

Serge Lang

Theory

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AM.S Classitications: 1065 1250

With 7 Illustrations

Library of Congress Cataloging in Publication Data

Lang, Serge

Aigebraic number theory

(Graduate texts in mathematics; 110)

at Berkeley Berkeley, CA 94720 U.S.A

Originally published in 1970 © by Addison-Wesley Publishing Company, Inc., Reading, Massachusetts

© 1986 by Springer-Verlag New York Inc

Softcover reprint ofthe hardcover 1st edition 1986

All rights reserved No part of this book may be translated or reproduced in any form without written permission from Springer-Verlag 175 Fifth Avenue New Y ork, New York 10010 U.S.A

9 8 7 654 3 2 1

ISBN 978-1-4684-0298-8 ISBN 978-1-4684-0296-4 (eBook)

DOI 10.1007/978-1-4684-0296-4

collec-those of Weber, Hasse, Hecke, and Hilbert's Zahlbericht It seems that

over the years, everything that has been done has proved useful, retically or as examples, for the further development of the theory Old, and seemingly isolated special cases have continuously acquired renewed significance, often after half a century or more

theo-The point of view taken here is principally global, and we deal with local fields only incidentally For a more complete treatment of these,

cf Serre's book Corps Locaux There is much to be said for a direct global

approach to number fields Stylistically, I have intermingled the ideal and idelic approaches without prejudice for either I also include two proofs of the functional equation for the zeta function, to acquaint the reader with different techniques (in some sense equivalent, but in another sense, suggestive of very different moods) Even though areader will prefer some techniques over alternative ones, it is important at least that he should be aware of all the possibilities

New York

June 1970

SERGE LANG

In the course of the Brauer-Siegel theorem, we use the discriminant formula, for which we refer to Artin-Tate where a detailed proof is given At that point, the use of this theorem is highly technical, and is due to the fact that one does not know that the zeros of the zeta function don't occur in a smaU interval to the left of 1 If one knew this, the proof would become only a page long, and the L-series would not be needed at aU We give Siegel's original proof for that in Chapter XIII

conductor-My Algebra gives more than enough background for the present book

In fact, Algebra already contains a good part of the theory of integral extensions, and valuation theory, redone here in Chapters land II Furthermore, Algebra also contains whatever will be needed of group representation theory, used in a couple of isolated instances for applica-tions of the class field theory, or to the Brauer-Siegel theorem

The word ring will always mean commutative ring without zero divisors and with unit element (unless otherwise specified)

If K is a field, then K* denotes its multiplicative group, and K its algebraic closure Occasionally, a bar is also used to denote reduction modulo a prime ideal

We use the 0 and 0 notation If I, gare two functions of a real variable, and g is always ~ 0, we write 1 = O(g) if there exists a constant C > 0 such that I/(x) I ~ Cg(x) for all sufficiently large x We write 1 = o(g) if limz +oo/(x)/g(x) = O We write/"" g if lim", _ ",/(x)/g(x) = 1

vii

Contents

Part One General Basic Theory

CHAPTER I Algebraic In tegers

7 Discrete valuation rings

8 Explicit factorization of a prime

CHAPTER II COInpletions

1 Definitions and completions

2 Polynomials in complete fields

x CONTENTS

CHAPTER IV Cyclotomic Fields

CHAPTER VI The Ideal Function

CHAPTER VII Ideles and Adeles

4 Generalized ideal class groups; relations with idele classes 145

CHAPTER VIII Elementary Properties of the Zeta Function and L-series

CONTENTS

PartTwo Class Field Theory

CHAPTER IX

Norm Index Computations

1 Algebraie preliminaries

2 Exponential and logarithm functions

3 The loeal norm index

4 A theorem on units

5 The global eyelie norm index

6 Applieations

CHAPTER X

The Artin Symbol, Reciproeity Law, and Class Field Theory

1 Formalism of the Artin symbol

2 Existenee of a eonduetor for the Artin symbol

3 Class fields

CHAPTER XI

The Existenee Theorem and Loeal Class Field Theory

1 Reduetion to Kummer extensions

2 Proof of the existenee theorem

3 The eomplete splitting theorem

4 Loeal class field theory and the ramifieation theorem

5 The Hilbert class field and the prineipal ideal theorem

6 Infinite divisibility of the universal norms

CHAPTER XII L-series Again

1 The proper abelian L-series

2 Artin (non-abelian) L-series

3 Indueed eharaeters and L-series eontributions

xii CONTENTS

Part Three Analytic Theory

CHAPTER XIII Functional Equation of the Zeta Function, Hecke's Proof

CHAPTER XIV Functional Equation, Tate's Thesis

CHAPTER XV Density of Primes and Tauberian Theorem

CHAPTER XVI The Brauer-Siegel Theorem

CONTEN'rs

CHAPTER XVII Explicit Formulas

1 Weierstrass factorization of the L-series

2 An estimate for A' / A

3 The basic sum

4 Evaluation of the sum: First part

5 Evaluation of the sum: Second part

PART ONE BASIC THEORY

CHAPTER I

Algebraic Integers

This chapter describes the basic aspects of the ring of algebraic integers

in a number field (always assumed to be of finite degree over the rational numbers Q) This includes the general prime ideal structure

Some proofs are given in a more general context, but only when they could not be made shorter by specializing the hypothesis to the concrete situation we have in mind It is not our intention to write a treatise on commutative algebra

Let A be a ring By a multiplicative subset of A we mean a subset containing 1 and such that, whenever two elements x, y lie in the subset, then so does the product xy We shall also assume throughout that 0 does not lie in the subset

Let K be the quotient field of A, and let S be a multiplicative subset

of A By S-1 A we shall denote the set of quotients xis with x in A and

s in S It is a ring, and A has a canonical inclusion in S-1 A

If M is an A-module contained in some field L (containing K), then S-1M denotes the set of elements vls with v E M and SES Then S-1M

is an S-1 A-module in the obvious way We shall sometimes consider

the case when M is a ring containing Aassubring

Let p be a prime ideal of A (by definition, p ~ A) Then the ment of p in A, denoted by A - p, is a multiplicative subset S = S~ of A,

comple-and we shall denote S-1 A by A~

ring, and m its maximal ideal, then any element x of 0 not lying in m

must be a unit, because otherwise, the principal ideal xo would be tained in a maximal ideal unequal to m Thus m is the set of non-units

con-of o

3

4 ALGEBRAIC INTEGERS [1, §2]

The ring Ap defined above is a loeal ring As ean be verified at onee, its maximal ideal mp eonsists of the quotients xis, with x in lJ and s in A

but not in lJ

We observe that mp n A = lJ The inclusion :J is clear Conversely,

if an element y = xis lies in mp n A with xE lJ and SES, then x = sy E lJ

and s El lJ Hence y E lJ

Let A be a ring and S a multiplicative subset Let a' be an ideal of S-IA Then

a' = S-I(a' n A)

The inclusion :J is clear Conversely, let xE a' Write x = als with

some a E A and SES Then sx E a' n A, whence x E S-I(a' n A)

Under multiplieation by S-I, the multiplicative system of ideals of A

is mapped homomorphieally onto the multiplieative system of ideals of

S-1 A This is another way of stating what we have just proved If a

is an ideal of A and S-la is the unit ideal, then it is elear that anS is not empty, or as we shall also say, a meets S

§2 Integral closure

Let A be a ring and x an element of some field L containing A We

shall say that x is integral over A if either one of the following conditions

is satisfied

INT 1 There exists a finitely genera ted non-zero A-module Me L such

that xM C M

INT 2 The element x satisfies an equation

with coefficients ai E A, and an integer n ~ 1 (Such an equation will be called an integral equation.)

The two eonditions are actually equivalent Indeed, assume INT 2 The module M generated by 1, x, , x n - 1 is mapped into itself by the element x Conversely, assume there exists M = (VI, ••• , v n ) such that

xM C M, and M 7fI= O Then

with coefficients aij in A Transposing XVI, •• , XVn to the right-hand side

is an integral equation for anx over A

Let B be a ring containing A We shall say that B is integral over A

if every element of B is integral over A

Proposition 2 If B is integral over A and finitely generated as an A-algebra, then B is a finitely generated A-module

Proof We may prove this by induction on the number of ring erators, and thus we may assume that B = A[x] for so me element x inte-gral over A But we have al ready seen that our assertion is true in that case

gen-Proposition 3 Let A C B C C be three rings If B is integral over A and C is integral over B, then C is integral over A

Proof Let x E C Then x satisfies an integral equation

with bi E B Let BI = A[bo, ,bn-d Then BI is a finitely generated A-module by Proposition 2, and BI[x] is a finitely generated BI-module, whence a finitely generated A-module Since multiplication by x maps

BI[x] into itself, it follows that x is integral over A

Proposition 4 Let A eBbe ttW rings, and B integral over A Let u

be a homomorphism of B Then u(B) is integral over u(A)

6 ALGEBRAIC INTEGERS [1, §2]

Proof Apply u to an integral equation satisfied by any element x of B

It will be an integral equation for u(x) over u(A)

The above proposition is used frequently when u is an isomorphism and is particularly useful in Galois theory

Proposition 5 Let A be a ring contained in a field L Let B be the set

of elements of L tchich are integral over A Then B is a ring, called the

integral closure of A in L

Proof Let x, y lie in B, and let M, N be two finitely generated

A-modules such that xM C!lf and yN C N Then MN is finitely generated, and is mapped into itself by multiplication with x ± y and xy

Corollary Let A be a ring, K its quotient field, and L a finite separable extension of K Let x be an element of L tchich is integral over A Then the norm and trace of x from L to K are integral over A, and so are the coefficients of the irreducible polynomial satisfied by x over K

Proof For each isomorphism u of Lover K, ux is integral over A

Since the norm is the product of ux over all such u, and the trace is the sum of ux over all such u, it follows that they are integral over A Simi-larly, the coefficients of the irreducible polynomial are obtained from the elementary symmetrie functions of the UX, and are therefore integral over A

A ring A is said to be integrally closed in a field L if every element

of L which is integral over A in fact lies in A It is said to be

integrally closed if it is integrally closed in its quotient field

Proposition 6 Let A be a N oetherian ring, integrally closed Let L be

a finite separable extension of its quotient field K Then the integral closure

of A in L is finitely generated over A

Proof It will suffice to show that the integral closure of A is contained

in a finitely generated A.-module, because A is assumed to be ~ oetherian Let /t'b • ,U'" be a linear basis of Lover K After multiplying each tCi by a suitable element of 1'1, we may assurne without loss of generality that the ICi are integral over 1'1 (Proposition 1) The trace Tl' from L to

K is a K-linear map of L illto K, and is Ilon-degenerate (i.e there exists

an element x E L sueh that Tr(;l;) ~ 0) If a is a non-zero element of L,

thell the function Tr(a.r) on L is an element of the dual spaee of L (as

K-veetor spaee), and induces a homomorphism of L into its dual spaee

Sinee the kernel is trivial, it follows that L is isomorphie to its dual under

the bi linear form

(.l', y) ~ Tr(.ry)

[1, §2] INTEGRAL CLOSURE 7 Let w~, , w~ be the dual basis of Wl, ••• , W n , so that

Tr(w~wj) = Bij

Let c ~ 0 be an element of A such that cw: is integral over A Let z be

in L, integral over A Then zcw~ is integral over A, and so is Tr(czwi)

for each i If we write

with coefficients b i E K, then

Tr(czwD = cb i ,

and cbi E A because A is integrally closed Hence z is contained in

AC-lWl + + Ac-lwn •

Since z was selected arbitrarily in the integral closure of A in L, it follows

that this integral closure is contained in a finitely generated A-module, and our proof is finished

Proposition 7 If A is a unique factorization domain, then A is grally closed

inte-Proof Suppose that there exists a quotient alb with a, b E A which is

integral over A, and a prime element p in A which divides b but not a

We have, for some integer n ~ 1,

whence

Since p divides b, it must divide an, and hence must divide a, contradiction Theorem 1 Let A be a principal ideal ring, and L a finite separable extension of its quotient field, of degree n Let B be the integral closure of

A in L Then B is afree module ofrank n over A

Proof As a module over A, the integral closure is torsion-free, and by

the general theory of principal ideal rings, any torsion-free finitely erated module is in fact a free module It is obvious that the rank is equal to the degree [L: Kl

gen-Theorem 1 is applied to the ring of ordinary integers Z A finite sion of the rational numbers Q is called a number field The integral closure of Z in a number field K is called the ring of algebraic integers of that field, and is denoted by 0K

exten-8 ALGEBRAIC INTEGERS [1, §3]

Proposition 8 Let A be a subring of a ring B, integral over A Let S

be a multiplicative sub set of A Then S-1 B is integral over S-1 A 1f A

is integrally closed, then S-1 A is integrally closed

Proof If x E Band SES, and if M is a finitely generated A-module such that xM C M, then S-IM is a finitely generated S-IA-module

which is mapped into itself by S-I X , so that S-I X is integral over S-1 A

As to the second assertion, let x be integral over S-1 A, with x in the quotient field of A We have an equation

x n + b n - 1 x n - 1 + + bo = 0,

b i E A and Si E S Thus there exists an element SES such that sx is integral over A, hence lies in A This proves that x lies in S-1 A

Corollary 1f B is the integral closure of A in some field extension L

of the quotient field of A, then S-1 B is the integral closure of S-1 A in L

§3 Prime ideals

Let p be a prime ideal of a ring A and let S = A - p If B is a ring

containing A, we denote by Bp the ring S-IB

Let B be a ring containing a ring A Let p be a prime ideal of A and

~ be a prime ideal of B We say that ~ lies above p if ~ n A = P and

we then write ~Ip If that is the case, then the injection

induces an injection of the factor rings

If B is integral over A, then B I~ is integral over Alp (by Proposition 4)

Nakayama's Lemma Let A be a ring, a an ideal contained in all mal ideals of A, and M a finitely generated A-module 1f aM = ~M, then M=O

maxi-[I, §3] PRIME IDEALS 9

Proof Induction on the number of generators of M Say M is erated by Wl, • ,Wm • There exists an expression

gen-with ai E Q Hence

If 1 - al is not a unit in A, then it is contained in a maximal ideal ll

Since al E II by hypothesis, we have a contradiction Hence 1 - al is

a unit, and dividing by it shows that M can be generated by m - 1

ele-ments, thereby concluding the proof

Proposition 9 Let A be a ring, II a prime ideal, and Ba ring containing

A and integral over A Then llB -;tf= B, and there exists a prime ideal '.ß

of elements of B with coefficients in ll,

with ai E II and bi E B Let B o = A[b 1, ••• ,bnl Then llBo = B o and

B o is a finite A-module by Proposition 2 Hence B o = 0, contradiction

To prove our second assertion, we go back to the original notation, and note the following commutative diagram:

(all arrows inclusions)

We have just proved that muBu -;tf= B u Hence muBu is contained in a maximal ideal im of B u, and im n Au therefore contains mu• Since mu is maximal, it follows that

Let '.ß = im n B Then '.ß is a prime ideal of B, and taking intersections

10 ALGEBRAIC INTEGERS [I, §3]

with A going both ways around OUf diagram shows that 9)1 n A = ~,

To prove this, let b E v, b ~ O Then b satisfies an equation

with ai E A, and ao ~ O But ao lies in v n A

Proposition 10 Let A be a subring of B, and assume B integral over A Let '.ß be a prime ideal of B lying over a prime ideal ~ of A Then Iß is maximal if and only if pis maximal

Proof Assume p maximal in A Then Alp is a field We are reduced

to proving that a ring wh ich is integral over a field is a field If k is a field and x is integral over k, then it is standard from elementary field theory that the ring k[x] is itself a field, so x is invertible in the ring Conversely, assume that '.ß is maximal in B Then BI'.ß is a field, which is integral over the ring Alp If Alp is not a field, it has a non-zero maximal ideal

m By Proposition 9, there exists a maximal ideal 9)1 of BIIß lying above

where G denotes the residue class mod p of an element c E A

We contend that there is a natural bijection betu'een the prime ideals '.ß of

B lying above p and the irreducible factoTs P(X) of leX) (having leading

[I, §4] CHINESE REMAINDER THEOREM 11

coefficient 1) This bijection is such that a prime 'l3 of B lying above II responds to P if and only if 'l3 is the kernel of the homomorphism

cor-A[a]-+ Ä[a]

where a is a root of P

To see this, let 'l3 lie above ll Then the eanonieal homomorphism

B -+ B/'l3 sends a on a root of J whieh is eonjugate to a root of some irredueible faetor of f Furthermore two roots of J are eonjugate over Ä

if and only if they are roots of the same irredueible faetor of f Finally, let z be a root of P in some algebraic closure of Ä The map

Remark 1 As usual, the assumption that II is maximal can be weakened

to II prime by localizing

Remark 2 In dealing with extensions of number fields, the assumption

B = A[a] is not always satisfied, but it is true that B~ = A~[a] for all but

a finite number of ll, so that the previous discussion holds almost always locally Cf Proposition 16 of Chapter III, §3

§4 Chinese remainder theorem

Chinese Remainder Theorem Let A be a ring, and al, , an ideals such that ai + aj = A for all i ~ j Given elements Xl, ,Xn E A, there exists X E A such that X == Xi (mod ai) for alt i

Proof If n = 2, we have an expression

for some elements ai E ai, and we let X = X2al + Xla2

For each i we can find elements ai E al and b i E ai such that

ai + b i = 1, i !!i; 2

Then x = XlYl + + XnYn satisfies our requirements

In the same vein as abüve, we observe that if al, , an are ideals of

a ring A such that

Proposition 11 Let A be a ring, integrally closed in its quotient field K

Let L be a finite Galois extension of K with group G Let p be a maximal ideal of A, and let $, D be prime ideals of the integral closure of A in L lying above p Then there exists u E G such that u$ = D

Proof Suppose that $ ~ uD for any u E G There exists an element

xE B such that

x == 0 (mod $)

x == 1 (mod uD), all u E G

(use the Chinese remainder theorem) The norm

NJ«x) = TI UX

uEG

lies in B n K = A (because A is integrally closed), and lies in $ n A = p

[1, §5] GALOIS EXTE~SIO!'i"S 13 But x G!: uO for aH u E G, so that ux G!: 0 for aH u E G This eontradiets the fact that the norm of x lies in p = 0 n A

If one loealizes, one ean eliminate the hypothesis that p is maximal; just assume that p is prime

Corollary Let A be a ring, integrally closed in its quotient jield K Let E be a jinite separable extension 01 K, and B the integral closure 01 A

in E Let p be a maximal ideal 01 A Then there exists only ajinite number

01 prime ideals 01 B lying above p

Prool Let L be the smallest Galois extension of K eontaining E If

(h, O2 are two distinet prime ideals uf B lying above p, and ~b ~2 are two prime ideals of the integral closure of A in L lying above (h and O2

respeetively, then ~1 ~ ~2' This argument reduees our assertion to the ease that E is Galois over K, and it then beeomes an immediate eonse-quenee of the proposition

Let A be integrally closed in its quotient field K, and let B be its integral

closure in a finite Galois extension L, with group G Then uB = B for

every u E G Let p be a maximal ideal of A, and ~ a maximal ideal of B

lying above p We denote by G'I.l the subgroup of G eonsisting of those automorphisms such that O'"~ =~ Then G'I.l operates in a natural way

on the residue class field B/~, and leaves Alp fixed To eaeh 0'" E G'I.l we ean assoeiate an automorphism Ü of B/~ over Alp, and the map given by

induees a homomorphism of GIlJ into the group of automorphisms of B/~

over Alp

The group G'I.l will be ealled the decOInposition group of~ Its fixed field will be denoted by L d , and will be ealled the decomposition field of~ Let B d be the integral closure of A in L d, and let 0 = ~ n B d •

By Proposition 11, we know that ~ is the only prime of B lying above O Let G = UUjG'I.l be a eoset deeomposition of G'I.l in G Then the prime ideals 0'" j~ are preeisely the distinet primes of B lying above p Indeed, for two elements 0'", TE G we have O'"~ = T~ if and only if T-IU~ = ~,

i.e T-1U lies in G'I.l' Thus T, 0'" lie in the same eoset mod G'I.l'

I t is then immediately elear that the deeomposition group of a prime

u~ iSO'"G'I.lO'"-l

Proposition 12 The jield L d is the smallest subfield E 01 L containing

K such that ~ is the only prime 01 B lying above ~ nE (u'hich is prime in

B nE)

14 .\LGEBIL\lC I:\TEGEHS [I, §5]

Proof Let E be us ubove, und let H be the Gulois group of Lover E

Let q = $ nE By Proposition 11, aB primes of B lying above q are conjugate by elements of H Since there is only one prime, namely $,

it means that H leaves $ invariant Hence H C G'1l und E =:J L d • We have already observed thut L d hus the required property

Proposition 13 Notation being as abol'e, lce haue Alp = Bdle (umler ihe canonical injeciion Alp ~ Bd le)

Proof If (J' is an element of G, not in Gtl, then (J''l.~ ,c $ and (J'-1i.j3 ,c $

N~d(y)==.r (mod'l.~)

But the norm lies in 1(, and even in A, since it is a product of elements integral over A This last congruence holds mod e, since both ;(; and the norm lie in B d This is precisely the meaning of the assertion in our proposition

If r is an element of B, "·e shall denote b~' J it,.; image under the morphism B ~ BI'l.~ Then iJ is the automorphism of Bli.j3 satisfying the

homo-relation

iJJ = (J'.r

If feX) is a polynomial with coefficients in B, we dcnote by fCX) its natural

image under the above homomorphism Thus, if

feX) = b"X" -+- -+- b o

[I, §5] GALOIS EXTE",SIOXS 15 then

Proposition 14 Let A be integrally closed in its quotient field K, and

let B be its integral closure in afinite Galois extension L oj K, with group G Let p be a maximal ideal oj A, and $ a maximal ideal oj B lying above p

Then BI$ is a normal extension oj Alp, and the map (J' f -+ iJ induces a homomorphism oj G'i,I onto the Galois group oj BI$ over Alp

Praaf Let 13 = BI$ and A = Alp Any element of 13 can be "Titten

as x for some xE B Let x generate a separable subextension of 13 over A,

and let j be the irreducible polynomial for x over 1( The coefficients of j

lie in A because x is integral over A, and all the roots of j are integral over A

Thus

m j(X) = TI (X - Xi)

i=l splits into linear factors in B Since

There remains to prove that the map (J' f -+ iJ gives a surjective morphism of G'i,I onto the Galois group of 13 over A To do this, we shall give an argument which reduces our problem to the case ,,,hen $ is the only prime ideal of B lying above p Indeed, by Proposition 13, the residue class fields of the ground ring and the ring B d in the decomposition field are the same This means that to prove our surjectivity, we may take L d

homo-as ground field This is the desired reduction, and we can homo-assume K = L d ,

16 ALGEBRAIC !);TEGERS [I, §5]

induced by elements of G operate transitivelyon the roots off Hence they

give us all automorphisms of the residue class field, as was to be shown

Corollary 1 Let A be a ring integrally closed in its quotient field K Let L be afinite Galois extension of K, and B the integral closure of A in L

Let p be a maximal ideal of A Let cp: A ~ Alp be the canonical morphism, and let if;1, if;2 be tlW homomorphisms of B extending cp in a given algebraic closure of Alp Then there exists an automorphism (1 of Lover K such that

homo-if;1 = if;2°(1

Proof The kerneIs of if;1, if;2 are prime ideals of B wh ich are conjugate

by Proposition 11 Hence there exists an element T of the Galois group G such that if;1, if;2 0 T have the same kerne! Without loss of generality,

we may therefore assume that if;1, if;2 have the same kernel \ß Hence there exists an automorphism W of if;1 (B) onto if;2(B) such that wo if;1 = if;2

There exists an element (1 of G'll such that w 0 if;1 = if;1 0(1, by the preceding proposition This proves wh at we wanted

Remark In all the above propositions, we could assume p prime stead of maximal In that case, one has to localize at p to be able to apply our proofs In the application to number fields, this is unnecessary, since every prime is maximal

in-In the above discussions, the kernel of the map

is called the inertia group T'll of \ß It consists of those automorphisms

of GIß which induce the trivial automorphism on the residue class fieId Its fixed field is called the inertia field, and is denoted by V

Corollary 2 Let the assumptions be as in Corollary 1, and assume that

\ß is the only prime of B lying above p Let f(X) be a polynomial in A[X] with leading coefficient 1 Assume that f is irreducible in K[X), and has a root a in B Then the reduced polynomiall is a pmcer of an irreducible polynomial in A[X]

Proof By Corollary 1, we know that any two roots of 1 are conjugate under some isomorphism of 11 over A, and hence that 1 cannot split into relative prime polynomials Therefore, 1 is apower of an irreducible polynomial

Let k be a number field and E a finite extension of degree N A non-zero

prime ideal of the ring of algebraic integers 0k will usually be called a prime

of k We say that such a prime p splits completely in E if there are

[1, §5] GALOIS EXTEXSIOXS 17

exaetly N different primes of E lying above p If Klk is Galois, then p

splits eompletely in K if and only if G'fl = 1 beeause G permutes the primes

'l3lp transitively

When Klk is abelian, then we have the following eharaeterization of the

fixed field of the deeomposition group

Corollary 3 Let Klk be abelian with group G Let p be a prime oi k, let 'l3

be a prime oi K lying above p and let G'fl be its decomposition group Let E

be the fixed field oi G'fl Then E is the maximal subfield oi K containing k in which p splits completely

Proof Let

G = U (J'iG'fl

i=1

be a eoset deeomposition Let q = 'l3 n E Sinee a Galois group permutes

the primes lying above a given prime transitively, we know that 'l3 is the only prime of K lying above q For eaeh i, the prime (J'i'l3 is the only prime lying above (J'iq, and sinee (J'1'l3, ,(J'r'l3 are distinet, it follows that the primes (J'1q, ,(J'rq are distinet Sinee G is abelian, the primes (J'iq are

primes of E, and [E:k] = r, so that p splits eompletely in E Conversely, let F be an intermediate field between k and K in whieh p splits eompletely, and let H be the Galois group of KIF If (J' E G'fl and 'l3 n F = 'l3F, then (J'

leaves ~~F fixed However, the deeomposition group of 'l3F over p must be trivial sinee p splits eompletely in F Henee the restrietion of (J' to F is the identity, and therefore G'fl eH This proves that FeE, and eoncludes the proof of our eorollary

Let k be a number field and let K be a Galois extension with group G

Let p be a prime of Ok and 'l3 a prime of OK lying above p The residue class field odp is finite, and we shall denote the number of its elements by

Np It is apower of the prime number p lying in p By the theory of finite fields, there exists a unique automorphism of oKI'l3 over odp whieh gener-ates the Galois group of the residue class field extension and has the effeet

In terms of eongruenees, we ean write this automorphism {j as

(J'a == aNP (mod 'l3),

By what we have just seen, there exists a eoset (J'T'fl of T'fl in G'fl whieh

induees {j on the residue class field extension Any element of this eoset will be ealled a Frohenius automorphism of 'l3, and will be denoted by

('l3, Klk) If the inertia group T'fl is trivial, then ('l3, Klk) is uniquely

determined as an element of the deeomposition group G'ß

18 ALGEBRAIC INTEGERS [1, §6]

If 0 is another prime lying above P, and " E G is such that ,,'l3 = 0, then the decomposition group of 0 is given by

Go = G~'il = "G'il"-I,

and similarly for the inertia group, and a Frobenius automorphism

This is immediately verified from the definitions Furthermore, if T'il is trivial, we see that ('l3, Kjk) = 1 if and only if P splits completely, mean-ing that G'il = 1

If Kjk is abelian, and if the inertia group T'il is trivial for one of the 'l3lp

(and hence for all 'l3lp), it follows that to each P in k we are able to associate

a uniquely determined element of G, lying in G'il (the same for all 'l3lp),

Let ° be a ring and K its quotient field A fractional ideal of ° in K is

an o-module a contained in K such that there exists an element c ~ 0

in ° for which ca C o If ° is N oetherian, it follows that ca, and hence a,

[I, §6] DEDEKIND RINGS 19 and (h F- a, a2 F- a Since a was maximal with respect to the stated property, we can find products of prime ideals contained in al and a2

Taking the product of these gives a contradiction

(ii) Every maximal ideal p is invertible

Let p-l be the set of elements .1: E J( such that :1'p C o Then p-1 ~ o

We contend that p-1 F- o Let a E p, a F- O Choose r minimal such that there exists a product

PI Pr C (a) C p

Then one of the Pi, say PI, is contained in p, and hence equal to p, since every prime is maximal Furthermore,

P2 Pr rt (a)

and hence there exists an element b E P2 Pr such that b f1 (a) But

bp C (a) and hence ba-1p C 0, so that ba- 1 E p-1 But b f1 ao and hence

ba- 1 f1 0, thereby proving our contention

We obtain pe pp-l Co Since P is maximal, either p = pp-1 or pp-l = o But p-lp = P would mean that p-l leaves a finitely generated o-module invariant, and hence is integral over o This is impossible, since

o is integra11y closed Hence pp-l = o

(iii) Every non-zero ideal is invertible, by a fractional ideal

Suppose this is not true There exists a maximal non-invertible ideal a Wehave just seen that a cannot be a maximal ideal Hence a C p for some maximal ideal p, and a F- p We get

aC ap-l C aa-1 co

Since a is finitely generated, we cannot have ap-l = a (because p-l is not integral over 0) Hence ap-l is larger than a, hence has an inverse, which, multiplied by p, obviously gives an inverse for a, contradiction

(iv) Let a be an ideal F- 0, and c a fractional ideal such that ac = o Then c = a- 1 (the set of elements xE J( such that xa co)

It is clear that ce a- 1• Conversely, if xa C 0, then xac C c and hence

Given two fractional ideals a, b we say that alb if and only if there exists

an ideal c such that ac = b This amounts to saying that a :J b, because

in that case, we take c = a- 1 b

From the definition of a prime ideal, we see that whenever a, bare two ideals and plab then pla or plb (Namely, ab C P implies aC P or bC p.) Given two factorizations

into prime ideals, we conclude that PI divides the product on the right, hence divides some qi, hence is equal to some qi Multiplying by PlI

both sides of the equality, we proceed by induction to prove that T = 8 and that the factors on both sides are equal, up to apermutation

If a is a fractional ideal ~ 0, and c E 0 is such that c ~ 0 and ca C 0,

then (c) = PI Pr and ca = ql q Hence a has the factorization

a = ql q

PI Pr

(writing l/p instead of p-1) If we caneel any prime appearing both in the numerator and denominator, then it is clear that the factorization is unique

A ring satisfying the properties of Theorem 2 is called a Dedekind ring The ring of algebraic integers in a number field K is a Dedekind ring, because it satisfies the three properties stated in Theorem 2 The multi-plicative group of non-zero fraetional ideals of the ring of algebraic integers

we say that it has a pole at p

Let a be a non-zero element of the quotient field of A Then we can

form the fractional ideal (a) = Aa and we apply the above notions of order, zero, and pole to a

[I, §6] DEDEKIND RINGS 21

If a and bare two fractional ideals, then it is clear that a ~ b if and only

if ordp a ~ ordp b for all primes p Thus we have a criterion for an element

a to belong to a fractional ideal a in terms of orders (taking b = (a»)

If ordp a = 0, then we say that a is a unit at p If that is the case, then

a is a unit in the local ring A p•

In what folIows, by a prime ideal, we shall mean a non-zero prime ideal, unless otherwise specified, and we call a non-zero prime ideal simply a

prime

Proposition 15 Let 0 be a Dedekind ring with only a finite number 01 prime ideals Then 0 is a principal ideal ring

Proof Let P1, , Ps be the prime ideals Given any ideal

select an element 'Tri in Pi but not in pl and find an element a of 0 such that

If

(a) = p11 ••• P:'

is a factorization of the ideal generated by a, then one sees immediately that ei = Ti for all i, and hence that a = (a)

Proposition 16 Let A be a Dedekind ring and S a multiplicative subset

01 A Then S-l Ais a Dedekind ring The map

is a homomorphism 01 the group ollractional ideals 01 A onto the group 01 Iractional ideals 01 S-l A, and the kernel consists 01 those Iractional ideals

01 A which meet S

Proof If p meets S, then

because 1 lies in S-l p If a, bare two ideals of A, then

so multiplication by S-l induces a homomorphism of the group of

(fractional) ideals

If S-la = S-lA, then we can write 1 = als for some a E a and sES

Thus a = sand a meets S This proves that the kernel of our morphism is what we said it iso

homo-22 ALGEBRAIC INTEGERS [I, §7]

Our mapping is surjective since we saw in §1 that every ideal of S-l A

is of type S-la for some ideal a of A The same applies of course to tional ideals This proves our proposition

frac-By a principal fractional ideal we shall me an a fractional ideal of type

aA, generated by a single element CI! in the quotient field of A, and a ~ 0 unless otherwise specified

Let A be a Dedekind ring The group of fractional ideals modulo the

group of principal ideals (i.e non-zero principal fractional ideals) is called the ideal class group of A

Proposition 17 Let A be a Dedekind rir.g, and assume that its group

oj ideal classes isfinite Let al, , Il r be representativejractional ideals 01 the ideal classes, and let b be a non-zero element oj A which lies in all the ai Let S be the multiplicative subset oj A generated by the pmcers oj b Then every ideal oj S-lA is principal

Proof All the ideals S-llll' , S-la r mapon the unit ideal in the homomorphism of Proposition 16 Since every ideal of A is equal to some

ai times a principal ideal, our proposition follows from the surjectivity of Proposition 16

If two fractional ideals a, b lie in the same ideal~lass, we write

Il~b and we say that Il, bare linearly equivalent It is clear that every frac-tional ideal is linearly equivalent to an ideal

The assumptions of Proposition 17 will be proved later to be satisfied

by the ring of integers of an algebraic number field

§7 Discrete valuation rings

A discrete valuation ring 0 is a principal ideal ring having a unique (non-zero) prime ideal m It is therefore a local ring If 'Ir is a generator for m, then it must be the only irreducible element of 0, i.e the only prime element (since any prime element generates a prime ideal) up to a unit,

of course Thus the unique factorization in an arbitrary principal ideal ring has a particularly simple form -in this case: Every element a ~ 0 of

o has an expression

with some integer r, and a unit u in o

Every discrete valuation ring is a Dedekind ring, and every Dedekind ring having only one maximal ideal is a discrete valuation ring If A is

a Dedekind ring, and lJ a prime ideal of A, then Ap is a discrete valuation

[1, §7] DISCRETE VALUATION RINGS 23

ring, since it is equal to S-l A (S = complement of II in A) (cf tion 16)

Proposi-Since every ideal of a discrete valuation ring is principal, it must be some power of the maximal ideal

In proving theorems about Dedekind rings, it is frequently useful to localize with respect to one prime ideal, in which case one obtains a dis-crete valuation ring For instance we have the following proposition

Proposition 18 Let A be a Dedekind ring and M, N two modules over A

If II is a prime of A, denote by Sp the multiplieative set A - ll Assume that SplM C SpIN for alt ll Then M C N

Proof Let a E M For each II we can find Xp E N and Sp E Sp such that a = xp/ Sp Let b be the ideal generated by the Sp Then b is the unit ideal, and we can write

1 = LYpsp

with elements YP E A all but a finite number of which are o This yields

a = Lypspa = LYpxp

and shows that a lies in N, as desired

If A is a discrete valuation ring, then in particular, A is a principal ideal ring, and any finitely genera ted torsion-free module M over A is free If its rank is n, and if II is the maximal ideal of A, then M /llM is a free module of rank n

Proposition 19 Let A be a [oeal ring and M a free module of rank n over A Let II be the maximal ideal of A Then M/pM is a veetor spaee of dimension n over A/ll

Proof This is obvious, because if {Xl, , X n } is a basis for M over A, so

M = LAxi (direct sum), then

M/llM ~ L(A/ll)Xi (direct sum), where Xi is the residue class of Xi mod p

Let A be a Dedekind ring, K its quotient field, L a finite separable

extension of K, and B the integral closure of A in L If II is a prime ideal

of A, then llB is an ideal of Band has a factorization

into primes of B It is clear that a prime l.j3 of B occurs in this factorization

if and only if l.j3 lies above p

24 ALGEBRAIC INTEGERS [1, §7]

If S is the complement of p in A, then multiplying the above tion by S gives us the factorization of S-lp in S-lB The primes S-l\.ßi

factoriza-remain distinct

Each ei is called the ramification index of \.ßi over p, and is also written

e(\.ßilp) If we assume that A is a local ring, then p = (11") is principal (Proposition 15) Let Si be the complement of \.ßi in Band let

Denote by I(A) the group of fractional ideals of a Dedekind ring A

Let K, L, B be as above Then we have a natural injection

I(A) ~ leB)

given by a ~ aB We shall define a homomorphism in the other direction

If \.ß lies above p in B, we denote by h or f(\.ß/p) the degree of the residue class field extension BI\.ß over Alp, and call it the residue class degree

We define the norm NJ(\.ß) to be pf~ and extend our map NJ( to the group of fractional ideals by multiplicativity

Proposition 20 Let A be a Dedekind ring, K its quotient field,

K CE C L two finite separable extensions, and A C Be C the sponding tower of integral closures of A in E and L Let p be a prime of

corre-A, q a prime of B lying above p, and \.ß a prime of C lying above q Then

Proof Obvious

e(\.ß/p) = e(\.ß/q)e(q/p) f(\.ß/p) = f(\.ß/q)f(q/p)

From Proposition 20 it is clear that the norm is transitive, i.e if we have a fractional ideal c of C, then

[I, §7] D1SCRETE VALUATION RINGS 25

Proof We ean loealize at p (multiplying A and B by 8;-1), and thus

may assume that A is a diserete valuation ring In that ease, B is a free

module of rank n = [L: K] over A, and BlpB is a veetor spaee of sion n over Alp

dimen-Let pB = ~·11 ••• ~~r be the faetorization of p in B Sinee ~t; :::> pB for eaeh i, we have a well-defined homomorphism

Let rr be a generator of ~ in B (We know from Proposition 15 that ~

is prineipal.) Letj be an integer ~ 1 We ean view ~j/~i+l as an

Alp-veetor spaee, sinee p~j C ~j+l We eonsider the map

B/~ -+ ~j/~i+1

indueed by multiplying an element of B by rrj This map is an

Alp-homomorphism, whieh is clearly injeetive and surjeetive Henee B/~ and

~j 1~i+1 are Alp-isomorphie

The Alp-veetor spaee B/~· has a eomposition series indueed by the inclusions

The dimension of B/~ over Alp is f"ß, by definition From this it follows

that the dimension of B/~· over Alp is e"ßh, thereby proving our tion

proposi-If e"ß = f"ß = 1 for all ~Ip, then one says that p splits completely in L

In that ease, there are exactly [L: K] primes of B lying above p

Corollary 1 Let a be a fractional ideal of A Then

NTc(aB) = Il[L:Kl

Proof Immediate

26 ALGEBRAIC INTEGERS [1, §7]

Corollary 2 Assume that L is Galois over K Then all the e'ß are equal

to the same number e (jor 'I3lp), all the h are equal to the same number f (jor 'I3lp), and if

then

efr = [L:K]

Proof All the '13 lying above p are conjugate to each other, and hence all the ramification indices and residue class degrees are equal The last formula is clear

Corollary 3 Assume again that L is Galois over K with group G, and let '13 be a prime of B Zying above p in A Then

N]('13 B = TI u'13 = ('131 'I3r)e/

uEG

(with e, f, ras in Corollary 2, and the ideal on the Zeft is viewed as embedded

in I(B)) The number ef is the order of the decomposition group of '13, and

e is the order of the inertia group

Proof The group G operates transitivelyon the primes of B lying above

p, and the order of G'ß is the order of the isotropy group Our assertions

are therefore obvious, taking into account Proposition 14 of §5

Proposition 22 Let A be a Dedekind ring, K its quotient fieZd, E a finite separable extension of K, and B the integral closure of A in E Let b be a fractional ideal of B, and assume b is principal, b = (ß), ß#O Then

the norm on the Zeft being the norm of a fractional ideal as defined above, and the norm on the right being the usual norm of elements of E

Proof Let L be the smallest Galois extension of K containing E The

norm from L to E of band of ß simply raises these to the power [L : E]

Since our proposition asserts an equality between fractional ideals, it will suffice to prove it when the extension is Galois over K In that case, it follows at once from Corollary 3 above

P,.oposition 23 Let A be a discrete valuation ring, K its quotient field,

L a finite separable extension of K, and B the integral closure of A in L Assume that there exists only one prime '13 of B lying above the maximal ideal p of A Let ß be an element of B such that its residue class mod '13

generales BI'13 over Alp and TI an element of B which is of order 1 at '13 Then A[ß, rr] = B

[I, §8] EXPLICIT FACTORIZ.\TIOX OF \ PRIME 27

Proof Let C be the ring A[ß, rr] It can be viewed as a submodule of B

over A, and by Kakayama's lemma, applied to the factor module BIC,

it will suffice to prove that

~B+C = B

But pB = l.l3e, and the products ßirrj generate BIl.l3 e over Alp, as in sition 21 Hence every element x E B is such that

Propo-(mod pB)

for some Cij E A This proves our proposition

Finally, we prove one more result, generalizing the arguments of Proposition 21

Proposition 24 Let A be a Declekind ring, and n a non-zero ideal Let

nu = ordp a Then the canonical map

A -+ TI Alp"P

U

induces an isomorphism of Aln onto the product

Proof The map is surjective according to the Chinese remainder

theorem, and it is clear that its kernel is exactly n

Corollary Assume that Alp is finite rar each prime ideal~ Denote by

Na the number of elements in the residue class ring Ala Then

Na = TI (Np)"u

p

\Y e observe tha t the function N can simply be vie,yed as being extended from the prime ideals to all fractional ideals by multiplicativity

§8 Explicit Jactorization oJ a prime

\Y e return to thc discus",ion at the end of §3 and giye more precise information concel'lling the splitting of the prime, due to Dedekind

Proposition 25 Let A be a Dedekinel ring Il'ith quotient fielel K Let E

be ajinite separable extension of K Let B be the integral closure of A in E and aSSllllle that B = A [al for same element a Let f(X) be the irreducible polynomial of a OL'er K Let p be a prime of A Let 7 be the reelucfion of

f mod p, and let

28 ALGEBRAIC INTEGERS [I, §8)

be the factorization of] into powers of irreducible factors over Ä = A/~,

with leading coefficients 1 Then

Proof Let P be an irreducible factor of], let a be a root of P, and let

~ be the prime of B which is the kernel of the map

A[a] - t Ä[a]

It is clear that 'pB + P(a)B is contained in~ Conversely, let g(a) E ~

for some g(X) E A[X] Then g = Ph with so me Ti E Ä[X], and hence

g - Ph, which is a polynomial with coefficients in A, in fact has coefficients

in p This proves the reverse inclusion, and proves the last formula of our proposition

Finally, let e; be the ramification index of ~i, so that

e' e'

'pB = ~ll ••• ~rr, and let di be the residue class degree [B/~i: A/'p] It is clear that di is the degree of Pi Since f(a) = 0, and since

feX) - P1(x)e l ••• Pr(x)e r E 'pA [X],

it folIo ws that

On the other hand, we see that

whence using (*) we find

~11 ·113:r C 'pB + P1(a)e l • •• I13r(a) erB C 'pB = I13r; ·113:~ This proves that ei ;?; e: for all i But we know that

I t follows that ei = e~ for all i, thus proving our theorem

[I, §8] EXPLICIT FACTORIZATIOX OF A PRIME 29

Remark The hypothesis that B = A[a] for some a is not always fied, but if we are interested in the decomposition of a single prime 1>,

satis-then it suffices to look at the localization B~ over A~, and in that case B~

can be generated by a single element except for a finite number of tions See Proposition 16 of Chapter III, §3

excep-Example Let a3 = 2, and let E = Q(a) It can be shown that the ring of algebraic integers OE is precisely Z[a] Let p = 5 Then we have