Introduction to Quadratic Forms (Grundlehren der mathematischen Wissenschaften 117)

O. T. O'Meara Introduction to Quadratic Forms Third Corrected Printing With 10 Figures Springer-Verlag Berlin Heide...

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O. T. O'Meara

Introduction to Quadratic Forms

Third Corrected Printing

With 10 Figures

Springer-Verlag Berlin Heidelberg New York 1973

O. T. O'Meara University of Notre Dame, Department of Mathematics, Notre Dame, ID 46556/USA

C•eschiiftsfiihrende ferausgeber

B. Eckmann Eidgentissische Technische Hochschule Zürich

B. L. van der Waerden Mathematisches Institut der Universitat Ztirich

AMS Subject Classifications (1970) Primary 1002, 10 B 40, 10 C 05, 10 C 20, 10 C 30, 10 E 45, 1202, 12A10, 12A40, 12 A 45, 12A50, 12A90, 13C10, 13F05, 13F 10, 1502, 15A33, 15 A 36, 15A57, 15A63, 15 A 66,20 G 15,20 G 25, 20G 30,20 G 40, 20H 20, 20 H 25, 20 H 30, Secondary 12 A 65, 12 Jxx

ISBN 3-540-02984-2

Springer-Verlag Berlin Heidelberg New York

ISBN 0-387-02984-2

Springer-Verlag New York Heidelberg Berlin

This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under §54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by SpringerVerlag Herlin Heidelberg 1963, I971, 1973. Library of Congress Catalog Card Number 73-10503 Printed in Germany. Offsetprinting and bookbinding: BriihIsche liniversitiitsdruckerei, GieLlen

In Memory of my Parents

Preface The main purpose of this book is to give an account of the fractional and integral classification problem in the theory of quadratic forms over the local and global fields of algebraic number theory. The first book to investigate this subject in this generality and in the modern setting of geometric algebra is the highly original work Quadratische Formen und orthogonale Gruppen (Berlin, 1952) by M. EICHLER. The subject has made rapid strides since the appearance of this work ten years ago and during this time new concepts have been introduced, new techniques have been developed, new theorems have been proved, and new and simpler proofs have been found. There is therefore a need for a systematic account of the theory that incorporates the developments of the last decade. The classification of quadratic forms depends very strongly on the nature of the underlying domain of coefficients. The domains that are really of interest are the domains of number theory: algebraic number fields, algebraic function fields in one variable over finite constant fields, all completions thereof, and rings of integers contained therein. Part One introduces these domains via valuation theory. The number theoretic and function theoretic cases are handled in a unified way using the Product Formula, and the theory is developed up to the Dirichlet Unit Theorem and the finiteness of class number. It is hoped that this will be of service, not only to the reader who is interested in quadratic forms, but also to the reader who wishes to go deeper into algebraic number theory and class field theory. In Part Two there is a discussion of topics from abstract algebra and geometric algebra which will be used later in the arithmetic theory. Part Three treats the theory of quadratic forms over local and global fields. The direct use of local class field theory has been circumvented by introducing the concept of the quadratic defect (which is needed later for the integral theory) right at the start. The quadratic defect gives, in effect, a systematic way of refining certain types of quadratic approximations. However, the global theory of quadratic forms does present a dilemma. Global class field theory is still so inaccessible that it is not possible merely to quote results from the literature. On the other hand a thorough development of global class field theory cannot be included in a book of this size and scope. We have therefore decided to compromise by specializing the methods of global class field theory to the case of quadratic extensions, thereby

VIII

Preface

obtaining all that is needed for the global theory of quadratic forms. Part Four starts with a systematic development of the formal aspects of integral quadratic forms over Dedekind domains. These techniques are then applied, first to solve the local integral classification problem, then to investigate the global integral theory, in particular to establish the relation between the class, the genus, and the spinor genus of a quadratic form. It must be emphasized that only a small part of the theory of quadratic forms is covered in this book. For the sake of simplicity we confine ourselves entirely to quadratic forms and the orthogonal group, and then to a particular part of this theory, namely to the classification problem over arithmetic fields and rings. Thus we do not even touch upon the theory of hermitian forms, reduction theory and the theory of minima, composition theory, analytic theory, etc. For a discussion of these matters the reader is referred to the books and articles listed in the bibliography. O. T. O'MEARA February, 1962.

I wish to acknowledge the help of many friends and mathematicians in the preparation of this book. Special thanks go to my former teacher EMIL ARTIN and to GEORGE WHAPLES for their influence over the years and for urging me to undertake this project; to RONALD JACOBOWITZ, BARTH POLLAK, CARL RIEHM and HAN SAH for countless discussions and for checking the manuscript; and to Professor F. K. SCHMIDT and the Springer-Verlag for their encouragement and cooperation and for publishing this book in the celebrated Yellow Series. I also wish to thank Princeton University, the University of Notre Dame and the Sloan Foundationl for their generous support. O. T. O'MEARA December, 1962.

1 ALFRED P. SLOAN FELLOW,

1960-1963.

Contents Prerequisites and Notation

XI Part One Arithmetic Theory of Fields

Chapter I. Valuated Fields 11. Valuations 12. Archirnedean valuations 13. Non-archimedean valuations 14. Prolongation of a complete valuation to a finite extension 15. Prolongation of any valuation to a finite separable extension 16. Discrete valuations

1 1 14 20 28 30 37

Chapter II. Dedekind Theory of Ideals 21. Dedekind axioms for S 22. Ideal theory 23. Extension fields

41

Chapter III. Fields of Number Theory 31. Rational global fields 32. Local fields 33. Global fields

54

42 44 52

54 59 65

Part Two Abstract Theory of Quadratic Forms

Chapter IV. Quadratic Forms and the Orthogonal Group 41, Forms, matrices and spaces 42. Quadratic spaces 43. Special subgroups of 0(V) Chapter V. The Algebras of Quadratic Forms 51. Tensor products 52. Wedderburn's theorem on central simple algebras 53. Extending the field of scalars 54. The Clifford algebra 55. The spinor norm 56. Special subgroups of 0,, (V) 57. Quaternion algebras 58. The Hasse algebra

82

82 88 100 112 113 118 129 131 137 141 142 149

X

Contents Part Three

Arithmetic Theory of Quadratic Forms over Fields Chapter VI. The Equivalence of Quadratic Forms 61. 62. 63. 64. 65. 66.

Complete archimedean fields Finite fields Local fields Global notation Squares and norms in global fields Quadratic forms over global fields

Chapter VII. Hilbert's Reciprocity Law 71. Proof of the reciprocity law 72, Existence of forms with prescribed local behavior 73, The quadratic reciprocity law

154 154 157 158 172 173 186 190 190 203 205

Part Four

Arithmetic Theory of Quadratic Forms over Rings Chapter VIII. Quadratic Forms over Dedekind Domains 81. Abstract lattices 82. Lattices in quadratic spaces Chapter IX. Integral Theory of Quadratic Forms over Local Fields 91. 92. 93. 94. 95.

Generalities Classification of lattices over non-dyadic fields Classification of lattices over dyadic fields Effective determination of the invariants Special subgroups of 0,2 (V)

Chapter X. Integral Theory of Quadratic Forms over Global Fields 101. 102. 103. 104. 105. 106.

208 208 220 239 239 246 250 279 280 284

Elementary properties of the orthogonal group over arithmetic fields 285 The genus and the spinor genus 297 305 Finiteness of class number 311 The class and the spinor genus in the indefinite case The indecomposable splitting of a definite lattice 321 323 Definite unimodular lattices over the rational integers

Bibliography Index

336 337

Prerequisites and Notation If X and Y are any two sets, then X ( Y will denote strict inclusion, X -- Y will denote the difference set, X -* Y will denote a surjection Y an injection, X Y a bijection, and X of X onto Y, X Y an arbitrary mapping. By "almost all elements of X" we shall mean "all but a finite number of elements of X". N denotes the set of natural numbers, Z the set of rational integers, Q the set of rational numbers, R the set of real numbers, P the set of positive numbers, and C the set of complex numbers. We assume a knowledge of the elementary definitions and facts of general topology, such as the concepts of continuity, compactness, completeness and the product topology. From algebra we assume a knowledge of I) the elements of group theory and also the fundamental theorem of abelian groups, 2) galois theory up to the fundamental theorem and including the description of finite fields, 3) the rudiments of linear algebra, 4) basic definitions about modules. If X is any additive group, in particular if X is either a field or a vector space, then X will denote the set of non-zero elements of X. If H is a subgroup of a group G, then (G: H) is the index of H in G. If EIF is an extension of fields, then [E:F] is the degree of the extension. The characteristic of F will be written z(F). If a is an element of E that is algebraic over F, then irr (x, a, F) is the irreducible monic polynomial in the variable x that is satisfied by a over the field F. If El and E2 are subfields of E, then E1E2 denotes the compositum of E1 and E2 in E. If EIF is finite, then NE/F will denote the norm mapping from E to F; and SE/F will be the trace.

Part One

Arithmetic Theory of Fields Chapter I

Valuated Fields The descriptive language of general topology is known to all mathematicians. The coincept of a valuation allows one to introduce this language into the theory of algebraic numbers in a natural and fruitful way. We therefore propose to study some of the connections between valuation theory, algebraic number theory, and topology. Strictly speaking the topological considerations are just of a conceptual nature and in fact only the most elementary results on metric spaces and topological groups will be used; nevertheless these considerations are essential to the point of view taken throughout this chapter and indeed throughout the entire book.

S 11. Valuations § 11 A. The definitions Let F be a field. A valuation on F is a mapping I 1 of F into the real numbers R which satisfies (V1) 'al > 0

if a + 0, 101 = 0

(V2) 141 = Icel hôI (v3) Ice + 5- lot! ± for all a, 13 in F. A mapping which satisfies (V1), (V2) and (V3')

IŒ+ /31

5- max(I4

will satisfy (Vs) and will therefore be a valuation. Axiom (V s) is called the triangle law, axiom (V3,) is called the strong triangle law. A valuation which satisfies the strong triangle law is called non-archimedean, a valuation which does not satisfy the strong triangle law is called archimedean. Non-archimedean valuations will be used to describe certain properties of divisibility in algebraic number theory. O'Meara. Introduction to quadratic forms

1

2

Part One. Arithmetic Theory of Fields

The mapping a -4- !al is a multiplicative homomorphism of F into the positive real numbers, and so the set of images of F forms a multiplicative subgroup of R. We call the set jF! —

{1J E RI EF}

the value group of F under the given valuation. We have the equations

Il = 1,

= 1'2 1

=

and also

I Icel — 1/1 11 5-

ice —

PI

Every field F has at least one valuation, the trivial valuation obtained by putting Jal I for all a in F. Such a valuation satisfies the strong triangle law and is therefore non-archimedean. A finite field can possess only the trivial valuation since, if we let q stand for the number of elements in F, we have i ceq-11 = = 1 Yc EF. Any subfield F of the complex numbers C can be regarded as a valuated field by restricting the ordinary absolute value from C to F. Conversely, it will follow from the results of § 12 that every field with an archimedean valuation is obtained essentially in this way. A valuated field which contains the rational numbers Q and which induces the ordinary absolute value on Q must be archimedean since 11+ 11-2> I. Now a few words about the topological properties of the valuated field F. First we notice that F can be regarded as a metric topological space in a natural way: define the distance between two points a and 13 of F to be la — fil. If we take this topology on F and the product topology on F x F, then it is easily seen by elementary methods that the mappings (a, #)

cc

fl and (a, fl)

a 13

of F x F into F are continuous. So are the mappings Œ

— cc and

a

cc-11

of F intoF and of .t into F , respectively. These four facts simply mean, in the language of topological groups, that F is a topological field. Hence the mappings (txi, tx2 , • • • CC n ) ± (x2 ± • ' • + Gen and ac2 , • • • , an) -4- 0C1 0C2 • • an of Fx••-xF into F are continuous. Hence a polynomial with coefficients in F determines a continuous function of Fx•••x F into F; and a rational function is continuous at any point of FX•••x F at

3

Chapter I. Valuated Fields

which its denominator is not zero. The inequality I shows that the mapping

— loco' I gJoc — °col

of F into R is continuous. The limit of a sequence and the sum of a series can be defined as it is usually defined in a first course on the calculus. We find that if an and /3„ -->-- fl as n co, then fi , OCn fin —>— oc fi oc-1 if a + 0

± fin --->-- oc 15,741 —>--

Similarly if E a, and E bA converge, then so do

E (a, ± bA) 1

1

aA

1

bA .

The terms of any convergent series must tend to O. The closure 0 of a subfield G of F is again a subfield of F. For we can find an--)-- a and fin > 18 with ay, and fin in G whenever a and /3 are given in 0; then a+ lim (an + E. Hence 0 is closed under addition. Similarly with multiplication and inversion. Hence 0 is a field. Closely related to the concept of a valuation is the concept of an analytic map. An analytic map is an isomorphism ep of the valuated field F onto a valuated field F' such that 197 al = !al holds for all a in F. In other words, an analytic map preserves the valuation as well as the algebraic structure. An analytic map is therefore a topological isomorphism between the topological fields F and F'. Suppose now that F' is just any abstract field, but that F is still the valuated field under discussion. Also suppose that we have an isomorphism g) of F onto F'. We can then define a valuation on F' by putting lfiI = 197'131 for all /3 in F'. When we perform this construction we shall say that 47) has carried the valuation from F to F'. Clearly the valuation just defined makes 99 analytic. We conclude this subparagraph with an important example. Consider the rational numbers CI and a fixed prime number ft. A typical a E Q can be written in the form 1

l'art One. Arithmetic Theory of Fields

4

with

in

and n prime to p. Do this with each a and put ! OEIP = (7)-1)i.

It is easy to show that this defines a non-archimedean valuation on Q. To say that at is small under this valuation means that it is highly divisible by p. (Here is our first glimpse of the connection between valuations and number theory; we shall return to this example in a more general setting in Chapter III.)

§ 11B. Non-archimedean valuations 11:1. A valuation on a field F is non-archimedean if and only if it is bounded on the natural integers of F.

Proof. We recall that the natural integers in an arbitrary field F are the finite sums of the form 1 ± • • • ± 1. We need only do the sufficiency. Thus we are given a fixed positive bound M such that mini M holds for any natural integer in in F. Then la + # in = l(aZ +

13)n i = I te + ( I) cen —' /3 + - • - + 131

5- Ill la 1t1 + MT loeln' A + • • • + 1 1 1 Ifiln 5- /11 {10eln+ loin -1 1A + - • • + li3I"} M (n + 1) {max (lal , II)}".

Hence

M'/"(n + 1 ) 1/n max (lock WI) -

loc + 1:31

If we let n --->-- co we obtain the result.

q. e. d.

This result has two immediate consequences. First, a field of characteristic )5 0 can have no archimedean valuations. Second, a valuated extension field E of F is non-archimedean if and only if F is non-archimedean under the induced valuation. 11:2. Principle of Domination. In a non-archimedean field we have ›

loci+ • - • + cc.1 = 1%1

if la21 < locl i for 2> 1. Proof. It suffices to prove la 4- /31 = loi when oc!> In We have

lal = 1—# + a ± #1 and so jai __ la + ,61. But la + #1 5- max Hence loc + #1 =

loi.

max (1461 , la + i31) ,

(lal 41) = lal q. e. d.


5

00

11:2a. Suppose that f otA is convergent. If 14 < 14 for 2> 1, then 1

Local = 1ccil . 1

11:3. Let EIF be an algebraic extension of fields. Suppose a valuation on E induces the trivial valuation on F. Then the valuation is trivial on E. Proof. For suppose that the given valuation is non-trivial on E. Then we can find a E E with lal > 1. Let us write an+ a1 an -1 ± • • • ± an _i a ± an . 0

with all ai in F. Now all Jail are either 0 or 1 since the valuation is trivial on F. And jet > jail whenever n > i. Hence Joel = lan + a1ces-1 + ' • ' + anl by the Principle of Domination. Hence jell . 0, and this is absurd. I. e. d. We shall see later in Chapter III that the above result does not hold if the extension EIF is transcendental. (

§ 11C. Equivalent valuations Consider two valuations 1 11 and 1 12 on the same field F. We say that I li and I 12 are equivalent valuations if they define the same topology on F. It is clear that equivalence of valuations is an equivalence relation on the set of all valuations on F. 11:4. Let 1 11 and 1 12 be two valuations on the same field F. Then the following assertions are equivalent: (1) The two valuations are equivalent, (2) Iali < 1 .4.r. loc 12‹ 1, (3) There is a positive number e such that kil l ---= ice1 2 for all oc in F.

Proof. (1) = (2). On grounds of symmetry it is enough to consider an a in F with lall 1 if and only if 1a1 2 > 1. Hence lali= 1 if and only if 1212= 1. In particular, if one of the valuations is trivial then so is the other. We may therefore assume that neither valuation is trivial


Take ao in F with 0 < 1001 2 < 1. Then 0 < 10013. < 1 by hypothesis. Hence we have 1 0012 = local 1 where e = log 104012110g WI> O . We claim that 1 0 12 = locl! for all oc in F. For suppose if possible that there is an oc for which 101 2 and 1017 are not equal. Replacing oc by its inverse if necessary allows us to assume that 1 0E12‹ Ioc. Now choose a rational number min with n > 0 such that This gives

1 0e12‹ 1ceolT in = JocniaT1 2 ‹ 1

and

• laniceri > 1

which denies our hypothesis. Hence our supposition about a is false. Hence (3) follows. q. e. d. (3) => (1). This part is clear , 11:4a. Suppose I 6 is non-trivial. Then 6 is equivalent to I 1 2 if lock< 1 => loti2< 1 • Proof. If tali > 1, then 1 012> 1 by taking inverses. It is therefore enough to prove Ia6= 1 = 1012 . 1. Choose i6 E F with 0 < 1/36. < 1. Then

Icenflli< 1= 1 0V12< 1 = lat12 1/312< 1 oo, that 1 012 5 1 . It follows from the last inequality, by letting n Replace oc by oc-1. This gives IaI 2 1. Hence 101 2 = 1. q. e. d. 11:4b. The trivial valuation is equivalent to itself and itself alone. 11:5. Let1 !l and I 12 be two equivalent valuations on a field E and let F be an arbitrary subfield of E. Suppose the two valuations induce the same non-trivial valuation on F. Then 1 and1 12 are equal on E.. Proof. We have a positive number e such that locI!= 101 2 for all oc in E. Choose oco E F with O < 1oc011= icco12< 1 Then Iccol! = Io!. Hence e = 1. q. e. d. •

Consider the valuation I I on our field F and let e be any positive number. We know that I le, if it is a valuation, will be equivalent to 1 I. Of course I IQ need not be a valuation at all; for instance the ordinary absolute value on Q with e> 1 gives 11 ± lle. 2e> 2 = Ille+ Ille. However I Ie is a valuation whenever 0 < e :< 1. To see this we observe that ia 1610 (lad + WV; it therefore suffices to prove that

1/3 1)e .% lode + !Me.


7

But 1 — loci 1-OEFI

±

( loci Lœl-I 1/31 ) e± Ial

) e '

1. So it is true. since 0 < e In the non-archimedean case things are simpler. The strong triangle law must obviously hold for IQ if it holds for I I, even if e is greater than 1. Hence 1 Ie is a valuation if I I is non-archimedean and e> O. It is clear that I IQ is non-archimedean if and only if I is. § 11D. Prime spots

Consider a field F. By a prime spot, or simply a spot, on F we mean a single class of equivalent valuations on F; thus a spot is a certain set of maps of F into R. Consider .a prime spot p on F. Each valuation I I R E' defines the same topology on F by the definition of a prime spot. We call this the p-adic topology on F. If p contains the trivial valuation (in which case it can contain no other) we call p the trivial spot on F. In the same way we can define archimedean and non-archimedean spots. If p is non-trivial it will contain an infinite number of valuations. Two spots on F are equal if and only if their topologies are the same. Suppose a: F F' is an isomorphism of a field F with a spot p onto an abstract field F'. It is easily seen that there is a unique spot q on F' which makes a. topological: the existence of q is obtained by letting a carry some valuation in p over to F', and the uniqueness of of follows from the fact that both a and cr-1 will be topological. In this construction we say that a carries the spot p to F'. The unique spot on F' that makes a topological will be written Pa. To each I I I, E p there corresponds a valuation I I pc, Er such that Apo = la-1 19 1p V 9 EF',

namely the valuation obtained by carrying 1 I I, to F'. 11 : 6. Let F and G be two fields provided with p-adic and q-adic topologies respectively. Let a be a topological isomorphism of F onto G. Then q = pa. And for each I Ep there is a lq E q which makes a analytic. Proof. Clearly q = per by definition of p6. Then I 1,1 is simply the q. e. d. valuation I Ivy defined above, Let 93 be a prime spot on an extension E of F. Each valuation in 93 induces a valuation on F, and all valuations of F that are obtained from 93 in this way are equivalent. Hence 93 determines a unique spot p on F. We say that q3 induces p, or that q3 -divides p, and we write

Tir •


8

Whenever we refer to the spot q3 on F we shall really mean that spot p on F which is divisible by T. Here the 93-adic topology on E induces the p-adic topology on F. We refer to this induced topology as the q3-adic topology on F. Now consider a set of prime spots S on F and another set T on E. We say that T divides S and write T IS if the spot induced on F by each spot (13 in T is in S. It is clear that there is an absolutely largest set of spots T on E which divides a given set S on F; we then say that T fully divides S and we write Ti1S. One often uses the same letter S to denote the set of spots on E which fully divides the given set S on F.

§ 11E. The Weak Approximation Theorem 11:7. Let I IA ( 1 2 n) be a finite number of inequivalent nontrivial valuations on a field F. Then there is an oc E F such that l oc> 1 and n. lock< 1 for 2 5. A Proof. If n = 1 it is simply the fact that 1 11 is non-trivial. Next let n ----- 2. Since I and 1 1 2 are inequivalent we can find b, c in F such that

1b11 ‹ 1 ,

1b1 2

1,

1c11

1

1c1 2 < 1 .

Then oc = clb does the job. We continue by induction to n. First choose b with 11)11 > 1 and IbIl < 1 (2 5_ A 5_ n 1), then c with 1c11 > 1 and 1c1„< 1. If 'bi n < 1 we are through. If lbin = 1, form cbr and observe that for sufficiently large values of r we have Oil > 1 , IcbrIA < 1 (2

2

n) ;

take oc = cbr. Finally consider lb in > 1. Using the fact that 1 + br fbj < 1 we easily see that b" 1 +b'

1cIA {

0

1 if

if A = 1 or 2. n if 2 A n—1.

This time take oc — cbrAi br) with a sufficiently large r. q. e. d. 11:8. Theorem. Let I la ( 1 n) be a finite, number of inequivalent non-trivial valuations on a field F. Consider n field elements ocA ( 1 2 n). Then for each e > 0 there is an oc E F such that k ocl IA ‹ s for 1 2 n. Proof. For each i (1 n) we can find bi EF such that lbi l i > 1 oo we see that and Pi k< 1 when A + 1. If we let r under { 0 under 1

1

+

1

1 la if A

Hence n Cr

=

Œb

If +

-3.—

.

9


I. Then a = cr with a sufficiently large r q. e. d.

under the topology defined by is the oc we require.

§ 11F. Complete valuations and complete spots Consider the distance function cl (oc, /3) = la — 131 associated with the valuation 1 1 on F. We can follow the language of metric topology and introduce the concept of a Cauchy sequence and completeness with respect to cl (oc, 13). Completeness of 1 1 then means, by definition, that every Cauchy sequence converges to a limit in F. 11:9. Example. We have already mentioned that the terms of any convergent series over a valuated field must tend to O. If F is a field with a complete non-archimedean valuation there is the following remarkable converse: every infinite series whose terms tend to 0 is 03

convergent. For if we form the partial sums Al,

.

.

,

A n, . . . of E acA.

we see from the strong triangle law that max aan+1 1,

lA ni — An!

, C0

hence the partial sums form a Cauchy sequence, hence E a2 has a limit

1

in F. Let p be a spot on the field F. We say that F is complete at p, or simply that F is complete, if there is at least one complete valuation in p. Because of the formula = J 12 relating equivalent valuations we see that if F is complete at p, then every valuation in p is complete. By a completion of a field F at one of its spots p we mean a composite object consisting of a field E and a prime spot q3 on E with the following properties: 1. E is complete at q3, 2. F is a subfield of E and q3 p, 3. F is dense in E. We shall often shorten the terminology and just refer to a completion E of a given field F; this will of course mean that we have a certain prime spot p on F in mind and that E is really a composite object consisting of the field E and a prime spot q3 on E. 11:10. Example. A complete field is its own completion. It has no other completion. 11:11. Example. Consider the trivial spot p on F. Here every Cauchy sequence has the form • • •

J.

//V

oc, • • •

OC1

and this converge § to oc. Hence F is complete.

• • •

10


11:12. Example. Let F and G be two fields provided with p-adic and cf-adic topologies respectively. Let a be a topological isomorphism of F onto G. Then F is complete at p if and only if G is complete at cf. 11: 13. Theorem. A field F has a completion at each of its spots. Proof. Consider a spot p on F and a valuation I I E p. It is enough to construct a field E F and to provide it with a complete valuation J I which induces the original valuation on F and is such that F is dense in E. The required spot 93 on E is then the one to which I I belongs. Let d(a, 13) = - PI be the metric associated with the given valuation. We know from topology that the metric space F has a completion, i. e. that there is a metric space E which is complete and contains F as a dense subsetPand such that the metric d on E induces the original metric on F. We have to make this metric space into a valuated field. In order to define addition and multiplication let us consider two typical elements a and 16 of E. Since E is the closure of F we can find oc and b„-›- fi sequences {an} and {bn} of elements of F such that under the metric d. Now these sequences are Cauchy sequences in F; this implies that {an ± bn} and fan bn} are Cauchy sequences too; hence they converge to limits in E. Define a ± fi = lim (an + bn) , oc fi = lim (ab). Take the original 0 and 1 of F as the 0 and 1 of E. One may check that these definitions are independent of the original choice of {an} and {bn}. Clearly the new laws of composition agree with the original ones on F. The field axioms should now be checked for E. For instance, oc + fi= lim (an + bn) =z lirn (bn + an) = fi+ oc proves commutativity of addition; and the limit of the Cauchy sequence {— an} gives the negative of oc. Finally define Ial = d (a, 0) for all oc in E. This gives the original valuation on F. Note that if an -÷- oc then d (an, cc) lani - loi I = Id (an, 0) - d (a, 0)1 so that lad is then lim Ian'. Hence 141 = lim labI = lim Ian ' Ian" = Similarly I a ± /31 loi ± in Hence 1 1 is a valuation on E. And the metric associated with this valuation is d since Foc = lim bnl lim d (an, bn) = d q. e. d. 11:14. Let E be a completion of F and let 97 be a topological isomorphism of F into some complete field G. Then there is a unique prolongation of 92 So a topological isonzorpizism of E into G.

11


Proof. Let 931p and q be the given spots on E, F and G. It follows from Proposition 11:6 that there exist valuations 1 1T E 93 and I iq E which make p analytic on F. We now define Toc for a typical a E E. Approximate to oc by elements of F, gay an---)-- oc with all an E F. Then {an} is a Cauchy sequence, hence {pan} is too. The latter Cauchy sequence has a limit in G; define poc= lim pan . The definition of pa is independent of the choice of the an and it agrees with p on F. If we now consider a typical 13 E E and an approximation bn--->-13 by elements of F., we can easily check that 92 (cc + fl) = POE + 9713 P(Œ13) = (Pa) (97P) , 1970eig= iociv • Hence p is analytic. This proves the existence part of the proposition. Let tp be another prolongation of p. Then p = lim pan = lim tp an = tp

.

n

SO

is unique. q. e. d. 11:15. Let E1 and E2 be completions F at the same spot p. Then there is a unique topological isomorphism of E1 onto E2 which is the identity on F. Proof. This is a special case of the last proposition, obtained by q. e. d. taking p as the identity map on F. The important results of this subparagraph have now been established: a field has a completion at each of its prime spots, and this completion is unique up to a natural topological isomorphism. There are two instances where we can be more specific: First suppose that EIF is an extension with spots 931p, and let E be complete at 93. In this event the closure P of F in E (at the prime spot induced by 93 on P, and hence with the topology induced from E) is a completion of F at p. This is true since a closed subset of a complete metric space is complete. In the second instance consider an extension EIF with spots 931p, but do not necessarily assume that E is complete at T. Suppose there is a subfield El with a spot 931 induced by 93 which is a completion of F at p. Then E1 is absolutely unique (not just up to an isomorphism); indeed El is closed in E since it is complete; but E1 is part of the closure of F in E since F is dense in El ; hence E1 is the closure of F in E. 11:16. Notation. The same letter p will be used for the prime spot of any completion of F at p. We usually use Ft, to denote a completion of F at p. Thus p will also refer to the spot on F. If EIF is an extension with spots 931p we can form a completion Ev with its spot 93. We let F93 denote the closure of F in E. We know that F93 provided with the spot induced by 93 is a completion of F at p; we call it the completion in .E93 of F at p. According to our convention the spot induced by 93 on F93 is 97

12


also written p. We have a natural topological isomorphism FT >—> F. If HIF is a subextension of EIF, and if q3 induces 130 on H, then the closure HT with its spot 93 0 is the completion in ET of H at q30. The T-adic topology on ET induces the T radic topology on 14 since q3p30, hence FT, is the closure of F in HT as well as in ET . Furthermore, the spot p on FT is induced by 930 on HT as well as by q3 on E. Hence FT the completion of F in HT as well as in ET . withspo §11G. Normed spaces over complete fields Let V be a vector space over a valuated field F. A norm on V is a real valued function 11 1 with the following properties: (1) 04 > 0 if x E V, and 11011 := O, (2) ;1ax11. 104;14 Ya EF , x EV, (3) li x Y11 11x11 + 113'11 V x, E V We can introduce a distance function on V by defining 11x — yll to be the distance between two typical points x and y of V; this makes V into a metric space and the various topological concepts are thereby introduced into V. 11: 17. Let V be a finite dimensional vector space over a field with a complete valuation. Then all norms with respect to the given valuation induce the same topology on V. And V is complete under any one of them.

Proof. 1) Let F be the given field, 1 1 the given valuation, and n the dimension of V over F. We consider a typical norm 11 11 on V which we shall call the given norm; the topology associated with this norm will be called the given topology. Fix a base yl, . . , y„ for V. Introduce a new norm 11 11 0 by defining +••

anYnllo

max 1 ccil

for a typical vector in V. We shall refer to this as the constructed norm, and its topology will be called the constructed topology. If n 1 we can easily find a constant K such that 11x11 = Klix11 0 VxEV. The entire result then follows. We proceed by induction to any n> 1. 2) We claim that we can find constants A and B such that A 11x;1 0 5_ B11.4 0 V x EV . For B it is easy: just take B

1iyi1. Now let us find A. Consider the subspace U = Fy2+ • • • ± Fyn. By the inductive hypothesis U is complete under the given norm, hence it is closed in V. Now additive translation in V is clearly continuous, hence U ± y1 is a closed subset of V. There is therefore a neighborhood of 0 which contains no vector

Chapter 1. Valuated Fields

13

whose first coordinate in the base y. . y„ is 1. Hence there is a neighborhood N of O (in the given topology) such that every vector with at least one coordinate equal to 1 falls outside N. In fact we can suppose that N is an open circular neighborhood, of radius A say, in the metric derived from 11 11. Now consider a typical x in V, say = avi d- - • • + cen,yn with ilxiio= A, Then the first coordinate of y = x/Œ1 is 1, hence y is not in N, i. e. Ilyll A loci! = A114 0. So we have our A. hence xj 3) It is clear from step 2) that every neighborhood of one topology contains a neighborhood of the other. So the given topology is the same as the constructed topology on V. 4) Finally the question of completeness. Put W — Fy„ so that V is the direct sum V =WEDU. Consider a Cauchy sequence x„, . . x„,... of vectors under the given norm. Then step 2) says that this is a Cauchy sequence under the constructed norm. Write X

X v = Wv + 14,

,

wvE w ,

st„( u

Then jjw„— wi,11 0 Ilx— xpll 0, so that the w, form a Cauchy sequence of vectors of W under the constructed norm. Similarly with the u,. Hence by the inductive hypothesis, 3 lim wy — w E W , 3 1iinu ,, = u E U. q. e. d. Hence w u. So V is complete. 11:18. Theorem. Let E1F be a finite extension of fields with spots 93 1P. Suppose that F is complete at p. Then (1) E is complete at 93, (2) 93 is the only spot on E which divides p. Proof. Consider two spots 9:31p and 931p on E. Pick j Icp E 93 and IclyE93' in such a way that they both induce the same j 1p E P on F. Regard E as a finite dimensional vector space over F in the natural way; then i v and I Iv become norms with respect to the complete valuation j j p on F. Hence j Iv is complete by the last proposition. And also j and J I v induce the same topology on E, again by the last proposition. q. e. d. So E is complete at 93 and 93.93'. 11:18a. Corollary. Let 92 be an isomorphism of E into a field G with a spot cr. Suppose that 92 is topological on F. Then it is topological on E. Proof. We can assume that 92E G. Let 92-1 carry q to a spot * on E. Then 92: E G is topological under the *-adic topology on E and the q-adic one on G. It is therefore enough to prove that * = 93. Now the restriction 92: F>—). 92F is topological with the *-adic topology on F; it is also topological with the p-adic topology on F by hypothesis. Hence

14


the *-adic and p-adic topologies are equal on F. Hence *1p by § 11D. But 931p. Hence 93 = * by the theorem. q. e. d. 11:19. Let Ell' be a finite extension of fields and let 93 be a spot on E. Then (1) Ev . EFT , (2) [ET : FT] [E:

< co . Proof. Recall the notation: ET is a completion of E at 93, F13 is the completion of F at 93 obtained by taking the closure of F in ET, and EFT is the compositum of E and FT in E. First we prove that [EF T : FT] [E :F EF by the following argument of field theory: take a tower of simple extensions from F to E, translate the tower by thereby obtain a tower from F13 to EF93 in which the successive layers have smaller degrees than before their translation; then [EF T : FT] [E : 1]. We now prove the first part. Since EF931F93 is finite, and since F,43 is complete at 93, we deduce from Theorem 11:18 that EF93 is complete at 93, hence it is closed in E93 under the 93-adic topology. But ET is the closure of E and ES EFT S Ev . Hence EFI3 = ET. q. e. d. The second part is now immediate, ]

12. Archimedean valuations The purpose of § 12 is to show that there is exactly one archimedean spot on Q, namely the one determined by the ordinary absolute value, and that there are essentially two complete archimedean fields, namely the real and complex numbers R and C. We shall use J L for the ordinary absolute value on Q (see § 31 for further discussion of the spot co). 12:1. There is exactly one archimedean spot p on Q. Every valuation I I in p is of the form I I = 1 IL where Joe is the ordinary absolute value Q and 0 < e 1. Proof. 1) Consider any two rational integers m> 1 and n> 1. Suppose we express m to the base n as follows: 0

m = a0 + (ti n -1- • • • + an

with 0 ai < n , ar > 0 , r

Then 'ai l

---

11 < n

Hence Jmj 1. For suppose not. Then we have an n > 1 with II < 1. Consider any rational integer m> 1. Then by step 1) we have

IM1

n (1 + inl + • • • + i ni r+ ")

Ini

I

j is bounded on the natural integers of Q and is therefore nonarchimedean, contrary to hypothesis. Hence we do indeed have Ini >. 1 whenever n> 1. The second conclusion is this: if m> 1 and n> 1 are two rational integers, then i m i < (logni _t_ i)[n ilogrnilogn . Hence

logn '

I =

To see this express m to the base n and use the inequality in step 1). We obtain

1m1

n (r + 1) in[r

since in! 1. But r log m/logn since nr m. The second conclusion then follows on replacing r by logmilogn. 3) Now use the inequality proved in step 2), substitute mk for m, and let k -4- co. This gives

17n1

In1Mgmil°gn •

Take logarithms and then interchange m and n: we get logIm logm

login'

logn •

Hence log inl/logn is a constant, e say, for all rational integers n> 1. Hence Ini= ne for these n. Hence

1x1 =

IxI

VxE

.

But there is at least one rational integer n> 1 such that Ini > 1 since p n. Hence 0 < e 1. is archimedean. For this n we have 1 < In! 4) We have therefore proved that a valuation I in an arbitrary archimedean spot p is equivalent to I . Hence there is just one archiq. e. d. medean spot on Q. 12:2. Lemma. Let F be a field with a complete archimedean valuation

Ico

l. Then 1

1

a EF2 whenever lai < -4- 11•

Proof. Since x(F) must be 0 we can define r — 4 la!/141 < 1. It is enough to prove that the polynomial x2 + 2x — a has a root in F; for if it does, then its discriminant will be a square in F, and its discriminant is actually 4 (1 + a), hence 1 + a E F2.

16


Define a sequence xo, x1,.

EF by means of the formulas

x„,

a X0 =1 , X„ +1 .—

—

2.x,

Then we must have lx„1 -2- 121: for this already holds when 7, = 0 and by an inductive argument we obtain lxv+11=

121 —

xp

21 121 => —

From this follows 141

xvj Hence the Ratio Test says that

r< 1•

E lx,--vi — xyl

is convergent. Hence for each e > 0 there is a yo such that 1xA —

-1 E lxv+i — xr1 < e Ii

tc>. vo. But this means that {x} is a Cauchy sequence, so it must have a limit x 0 in F. But xv+ixy+ 2x„— a = 0 .

Letting y -.0-- co in this equation shows that x EF is a solution to the q. e. d. equation x2 + 2x — a = O. 12:3. Lemma. Let F be a field with a complete archimedean valuation. Then there is a prolongation of the valuation from F to F (1) where i2 = — 1. Proof. We can assume that i is not in F. Let 1 1 denote the given Define valuation on F. Let N denote the norm Nr lorI = iNce1 1/2 Va EF (i) . This is possible since Na EF for all a EF. This new function is a prolongation of the original valuation since l ad !N ap = l oc211/2= j ai cc E F How do we know that the prolongation is a valuation ? Only the triangle law needs checking, and here it is enough to verify that 11 + ce1 1+12I V cc EF(i) . This holds for all or E F. So consider cc E F (i) — F. Let x2 + bx + c = irr (x, cc, F) . Then c = Nor and so 1c1112 = 1a1. If we had 1b1 2 > 4 1cl, the quantity b2 — 4c = b 2 (1 — 4,2c)


17

would be in F2 by Lemma 12:2, and so x2 + bx c would be reducible. Hence 114 2 _< 4 IcI. Now x2 + (b — 2) x + (1 + c — b) = irr (x, 1 + cc, F) .

Hence al 2 . 11 - c

bi

1 ± Ici

Ibl

1 ± Ici + 2 1012 (1 + 1012)2

Hence 11 +

= ( 1 + loc1) 2.

q. e. d.

1 + lcd.

12:4. Theorem. Let F be a complete archimedean field. Then there is a topological isomorphism of F onto either the real or complex numbers.

Proof. 1) Let p be the given complete archimedean spot on F. Consider the complex numbers C at the spot q determined by the ordinary absolute value. We let Q be the prime field of F and R the closure of Q in F under the p-adic topology; thus R is the completion of Q in F. 2) First a reduction of the problem. Suppose we have proved our theorem in the case where 1/1- 1 EF. Let us show how to derive the general case from this. We can consider F. Put E.F (V— 1). Then there is a prime spot 93 on E which divides p by Lemma 12 : 3. And E is complete by Theorem 11:18. But we are assuming that our result holds in this situation. Hence there is a topological isomorphism 92 of E onto C. Now 93R is complete, hence closed in C. But 93R 4- C. Hence TR R. Hence 92F = R. 3) Therefore assume that 1/1--- E F. Put C = R 1) . Let 92 be the natural isomorphism 92: Q Q. This is topological since Q has just one archimedean spot. By Proposition 11:14 there is a prolongation of 92 to a topological isomorphism 93: R R. By field F theory there is a prolongation of 92 to an algebraic isomor- f phism 92: C, and this prolongation must be topological by Corollary 11:18a since R is complete. If F = C we are )? through. We therefore assume that C F and use this to produce a contradiction. Q 4) Fix I I E p. Then by Proposition 11:6 there is a valuation in q which makes 92: C C analytic. We therefore have the following additional information about the field C: the value group ICI is equal to P, and every closed bounded subset of C is compact. Consider a point x E F C, j. e. a point x in F that is not in C. We shall minimize the distance from x to points of C. To do this consider a O'Meara, Introduction to quadratic forms

2

18


closed sphere M which meets C and has center at x. Then the function — x1 of M nC into R is continuous; and M n C is a closed bounded subset of C and is therefore compact; so our function attains its minimum; we therefore have cco EM im C such that

1cco — xl

— x1 V ccEMnC.

So in fact we really have 0 < loco — x1

Vex EC.

— xl

Replace x by oft,— x and then scale by a suitable element of C (recalling that ICI = P) ; in this way we obtain z EF C such that 2 = 1z1 s

— z1 V at EC .

( 1) 5) Consider any z EF — C which satisfies equation (1), and any natural number n. We find 2n

lzni + 1

(1 +

le- 11 k1

tz -

1Z — 11 21' -'

where 1 = and let n

are the n-th roots of unity in C. Divide by 2n-1 oo; this gives lz — 11 S 2, hence lz — 1 = 2. Therefore 2, • • •

CY&

2 = lz — 11

1oe

(z

1)1 V a EC .

So z — 1 EF C has the property of equation (1) whenever z does. Hence so does z — n for any n E N. So for a fixed z and for all n E N we have 2 = Jz — n1 1n1 1z1. Thus 1 1 is bounded on the rational integers in F; this is impossible since the valuation is archimedean. Hence our assumption that F +C is false. So F = C and we are through. q. e. d. Consider a field F and an archimedean _spot p on F. Let Fp be a completion of F at p. We now know that there is a topological isomorphism 99 of Fp onto either R or C. Let 99-1 carry the ordinary absolute value back to F. This gives us a valuation 1 1 E p on F such that 1m1 = m holds for all natural numbers m in F. If 1 1' E p is another valuation with the same property, then 1 1 and 1 1' are equal on the prime field and


19

equivalent on F, hence they are equal on F by Proposition 11:5. So we can make the following definition. 12:5. Definition. Let p be an archimedean spot on a field F. By the ordinary absolute value on F at p we mean that unique valuation I I E p with the property that fini = in holds for all natural numbers m in F. 12 : 6. Let p be an archimedean spot on a field F. Then every valuation in p is of the form I le with 0 < p _‹ 1, where I 1 is the ordinary absolute value on F at p. Proof. Consider typical I l * E p. Then there is a e (0 < e 1) such

that Icci* — tale V cc E

where Q is the prime field of F. Now I 1 4, and I 1e are equivalent on F. q. e. d. Hence I * . I le by Proposition 11:5. 12:7. Remark. Let Ell' be an extension of fields with spots 93 I p which are not necessarily. archimedean nor necessarily complete. However, suppose that p is non-trivial. Then restricting a valuation in g3 to F gives a valuation in p; by Proposition 11:5 this sets up an injection of the valuations in 93 into those in p. In fact this is a bijection: in the nonarchimedean case surjectivity follows easily from the strong triangle law and the formula 1 I * . I le relating equivalent valuations, in the archimedean case it follows from the last proposition. 12:8. Definition. Let p be an archimedean spot on a field F. We call p real or complex according as Fp is isomorphic to R or C. 12:9. Definition. Let p be an archimedean spot on a field F. By the normalized valuation on F (or on Fp) at p we mean the function if p real, 111P= I Icej 2 if p complex, where f is the ordinary absolute value on F (or on Fp) at p. 12: 10. Remark. The normalized valuation is a true valuation at a real spot. It is not a true valuation at a complex spot; there the triangle law must be replaced by or more generally,

la 4- flip 5_

2(j]+ !flip) 2r-1

(Er ICCdp) • 1

Normalized valuations will also be introduced over the local fields of Chapter III. They provide one way of regularizing the behavior of the product formula of Chapter III. 2*

20


12:11. Definition. Let p be a real spot on a field F, and let a be any element of F (or of Fp). We say that a is positive at p if a F. We say that a is negative at p if a E- F. 12: 12. Example. Let p be a real spot on the field F. So there is a topological isomorphism of Fp onto the field of real numbers R. This isomorphism carries the positive elements of Fp onto the positive real numbers, and the negative elements of Fp onto the negative real numbers. The positive elements of Fp are an open subgroup of index 2 in F. The negative elements of Fp are an open subset of Ft,. We have the disjoint union Fr . —FVJOu.n e of Fp into negative elements, zero, and positive elements. 12:13. Example. Let F be a prime field of characteristic O. Then there is exactly one archimedean spot on F. This spot is real. The positive elements of F are the elements of the form min with m and n natural numbers in F. S 13. Non-architnedean valuations § 13A. The residue class field We let F be an arbitrary field, p any non-archimedean prime spot on F. We define o(p) = {cx EFI ki p s 1} u (p) ={ EF I jal p = 1} m (p) = {a E F II,,< 1} ,

where lp denotes a valuation in p; these definitions are clearly independent of the choice of I lp in p. The elements of o(p) are called the integers of F at p, or simply the integers of F when there is no risk of confusion; every rational integer in F is an integer of F at p by the strong triangle law. It is easily verified that o (p) is a subring containing the identity of F, and that F is the quotient field of o (p). We call 0 (p) the ring of integers of F (at p) or the valuation ring of p. If q is some other spot on F, then it follows from Proposition 11:7 that 0 (p) (q) p=q Note that o (p) F if and only if p is the trivial spot on F. Thus o (p) is only exceptionally a field — if and only if p is trivial. The set u (p) is a multiplicative subgroup of o (p) ; it consists precisely of all invertible elements of the ring o (p); accordingly we shall call u (p) the group of units of F at p, or simply the group of units of F when there is no risk of confusion. Now let us comment on m(p). Again we see that (q) ; p q n(p)


21

and m(p) = 0 if and only if p is the trivial spot. It is easily verified using the strong triangle law that m (p) is an ideal in o (p). We call m (p) the maximal ideal of F at p. .This name is justified by the fact that m (p) is the absolutely largest proper ideal in o (p). Why is this true ? Consider an ideal a of o (p) that is not contained in m (p) ; pick a Ea—m (p) S u (p). Then 1 = a a--1 E a. Hence a = o (p) and a is not proper. We are ready to define the residue class field of F at p. Essentially ifis the field o (p)/m (p). But we would like a little more flexibility than the usual definition of the quotient ring o (p)/nt (p) will permit. So we frame our definition as follows: by a residue class field of F at p we mean any composite object (92 , H) consisting of a field H and a ring homomorphism 99 of o (p) onto H which has kernel m (p). The field F will always have at least one residue class field at p, namely the field o(p)fm(p) together with the naturally associated homomorphism of o (p) onto it. Also the residue class field is essentially unique. More precisely: let (99, H) and (99', H') be two residue class fields of F at p; then there is a unique ring isomorphism v of H onto H' such that v o = . (The existence of ip is one of the elementary isomorphism theorems of ring theory.) A residue class field is usually referred to without explicitly mentioning the other half of the composite object, namely the homomorphism 99. The bar symbol is often used for the mapping 99 and when this is done one lets F denote the corresponding field. (Of course F is the image of o (p), not of F, under the bar mapping.) When several spots are under discussion at the same time we can use F (p) for a residue class field of F at p. 13: 1. Example. As an example of the notation let us see what is meant by Fp (p). Here Fp is a completion of F at p; and we have agreed to use the same letter p for the spot on the completion F,; hence Fp (p) means a residue class field of the completion Fp at p. 13 : 2. Example. x (F (p)) x (F) whenever X(F) O. 13:3. Notation. Let a be an additive subgroup of F. For any two elements ce, # E F the congruence oc p mod a takes on its usual meaning, namely cc — 8 E a. This defines an equivalence relation in the usual way. Furthermore, if mod a, and ce' = 13' mod a, oe then mod A a + a' 13 ± 13' mod et and 2, ce for any A E F, where Aa is the additive subgroup /14

={2.xixE a} .

22

Part One: Arithmetic Theory of Fields

For any y E F we shall write cc

/3 mod y

instead of cc =# mod y (p) . § 13B. The fundamental invariants e and f Consider an arbitrary extension of fields EIF provided with nonarchimedean spots 931p. Fix completions Fp and ET , and let F93 be the completion of F at p that is obtained by taking the closure of F in ;3 in the usual way. According to our conventions the spot p can refer to any one of the fields F, Fp or FT ; similarly with 93. Now this presents a problem: does o (p) refer to F or does it refer to F,,? This point should be cleared up right away even though it will not be needed until much later. We therefore make the following convention: o (p) is to be the valuation ring of F at p, while the valuation ring of F at p is to be denoted by the new symbol op. Similarly we introduce new symbols up and mp for the group of units and the maximal ideal of Fp at p. Similarly with E and E at 93. Note that ,,

(93) F = o(p) ç- 0 (93) u (T) n F = u (p) Çu(3) m (93) n F m (p) S_ m(93) . And similar formulas hold between F and Fp. 13:4. In the above situation consider valuations I lp E p, 1 193 E93 and

lpE P. Then

(1) IF 193 Ç 1413, 14= NI, (2) (IE 93 193 : 141 93) = (1E1 93 : IFIq3) (3) (1E1;3 : 1F1;3) = (1E1 93 : 1F1 93). Proof. (1) Clearly every element in the value group 11193 is in the value group 1.E1v. In particular, [Tip Ç IFidp. Conversely, take cc E Fp and write it cc = Jim an with an E F. For large enough n we have lan — alp 1. Since the given valuation is complete and archimedean there is a topological isomorphism of F onto either R or C by Theorem 12:4, hence onto R; so E = F (y - ). But in the proof of Lemma 12:3 we showed that IN'/2 prolonged the valuation from F to E. Hence the archimedean case is established. Therefore assume that the given valuation is non-archimedeanl. Only the strong triangle law really needs proof, and in fact we just have to show that INal 5. 1 IN(1 oc)1 5.. 1. Let (x) = xm + ai xm -1 + • • • + am . irr (x, cc, F) . Now N a is a certain power of Np(/oc, hence

jam ! = INp (co ap al

1. If we now apply the Reducibility Criterion of § 13D to the irreducible polynomial f (x) over F we find that 'ail S 1

for

But f (x — 1) = irr (x, 1 + oc, F), and this polynomial in x has constant term Since the strong triangle law holds in F we have 11 -1- E (± ezA)1;5 1. But N(1 + oc) is a power of the above term. Hence IN (1 ± oc)1 _5 1, as q. e. d. required. 1 Here we use the classical method of deriving the prolongation theorem from Hensel's lemma. It is also possible to obtain this result, in fact a more general result, in an entirely different way. One introduces three new equivalent concepts, general valuations, general valuation rings, and places, then one uses Zorn's lemma to prove a prolongation theorem for places from which one obtains a prolongation theorem for general valuations, and finally one proves that the prolongation of an ordinary valuation is an ordinary valuation and that it satisfies the formula of the theorem. Hensel's lemma can then be derived from the prolongation theorem. For a detailed account of this method see G. WHAPLES, Class field theory, (University of Indiana lectures, 1959).

Pad One. Arithmetic Theory of Fields

30

§ 15. Prolongation of any valuation to a finite separable extension

In this paragraph we consider a finite separable extension EIF of degree n, and a prime spot p on F. The spot in question can be quite arbitrary, but the extension must be separable. For inseparable extensions the results involve more technicalities, and are fortunately unnecessary for our subsequent use. We prove that p is divisible by at least one and at most n spots on E, and we investigate the behavior of the degree and the norm at all the 93 dividing p. § 15A. Construction and notation As we have just said, EIF is a separable extension of degree n and p is a prime spot on F. This is fixed for the entire paragraph. We let T be the set of prime spots on E which divide p. Since EIF is separable there will be a primitive element 6 such that E = F (6); we put 1(x) = irr (x, & F). We fix a completion Fr of F at p; as usual, p will be used to denote the prime spot on F. We define ; to be some fixed splitting field of f(x) over F. Since Fp is complete we know from Theorems 14 : 1 and 11 : 18 that there is exactly one spot 13o on Ep which divides p on F. Since EIF is separable we know from field theory that there are exactly n distinct isomorphisms of E into Ep which are the identity on F. We let Z denote the set of these isomorphisms. For each a Ez the isomorphism a -1 : aE E will carry the spot 13o on aE back to E; in § 11D we agreed to write this spot on E in the form 93r. Clearly 930 'E VaET, = p. In particular T is not empty: since 9313 -1 I pa."' on EIF and every spot on F is divisible by at least one spot on E. For each 13 E T we fix a completion ET of E at 13 ; as usual, we let 13 denote the corresponding spot on E. We take the completion Fq3 of F at p that is contained in E. Then P

9

there is a unique topological isomorphism of Fv onto Fr which is the identity on F. We shall need this map so we agree to denote it by I. Proposition 11:19 informs us that E93 = EFv . F(6).

The map 4 can be used to prove that every 93 E T has the form 3 -1 for some a E l'. Namely, 4: Fq3 )—).1.1, is an isomorphism; so field theory provides a prolongation a: Ev Ep of I. But //3 is topological, hence a is topological by Corollary 11: 18a. Restrict a to E. Then o.-1 : aE >—›- E

Chapter T. Valuated Fields

31

is a topological isomorphism. Hence 93 = 93a0 -1 by Proposition 11:6. This proves the assertion. Incidentally the rule a -->-- 93a0 -1 provides a surjection

. In particular there are at most n spots on E which divide p. Suppose a E L' determines the spot 93 -= 930a 1 . Then a: E crE is a topological isomorphism under the 93-adic topology on E. Hence by Proposition 11: 14 there is a unique prolongation of a to a topological isomorphism of _El:5 into E. We shall denote this prolongation by aq3 . Now a is the identity on F; hence by Proposition 11:14 again we have

a,43 L13 on F93 . 15:1. Example. If (crE) 930 denotes the closure of crE in Ep, then (crE)= cr,43 Eq3 = (Fr) (a E) = Fp (crà) . § 15B. Local degrees and local norms 15:2. 93r= 930r -1 if and only if a (5 and 1- (5 are conjugate over F. Proof. 1) First the necessity. Put 93 = 93a0-1 = 93r. The topological isomorphisms cri3 : Eq3 cri3 Ec43 , TT: ET >—)- TT are both equal to

4 on F,13 , hence TT aV is the identity on F. But TT oV (cià) T(13 (6) —

Hence cr(5 and T6 are conjugate over F. 2) Now the sufficiency. Let us put 93 = 93g-' and

93' = 93.r.

Then

Fp (r (5) . cr,43 .E43 Fp (a (5) , rty E Since cr6 and T6 are conjugate over Fp we can find an abstract isomorphism

17) : cri >-which is the identity on Fp and carries crb to T6. Since Fp is complete, (p will be topological by Corollary 11:18a. Now a typical a C E has the form

= ao + al (5 ± • • ± a„_, (5n -, ai EF, from which it follows that

(cri3 a) -= Xc4y (a)

V aEE.

Hence TV q) 6'13 is a topological isomorphism of E,43 onto Eiy which is the identity on E; but this means that the 93-adic and 93'-adic topologies are equal on E. Hence 93 . 93'. q. e. d. Consider our extension EIF with spots 93 p. By the local degree of the extension at 93 (or at 93 1p) we mean the field degree

n (93 J p) =

F].


32

This quantity clearly depends on 93, but it does not depend on the particular completion E93 that is chosen at the given 91 By the local norm at 93 I p we mean the multiplicative homomorphism

Ni31p : ET that is defined by the equation ip oc

= //3 (NE93/F93 a) V oe E

Similarly we define the local trace at 93 I p to be the additive homomor-

phism STip : ET

Fp

that is defined by the equation

Svip a = 193 (SE/F43 a) V oc E ET . Clearly the local norm and the local trace depend on E93 andFp. The local trace is just mentioned in passing and will not be used in the sequel. 15:3. Theorem. Let EIF be a finite separable extension of degree n, and let p be a spot on F. Then for all oc in E we have the formulas

(1) E n(93 1P) n (2) IT Nvip oc =

NE/p0C

93 iP

(3) E

S

SigirpOC .

VIP

Proof. Define an equivalence relation on Z by saying a if 93`10-1 —Tr Thus a T if and only if a (3 and Tc5 are conjugate over F. Let a denote the coset of a EE under this equivalence relation, and let (a) denote the number of elements in the coset a. Let P be the subset of Z that is obtained by picking exactly one representative from each coset in E. By definition 4*(a) is the number of T E Z for which TC5 and a (5 are conjugate over Fp ; and this number is equal to the number of conjugates of a& over Fp ; hence .

# (a) = [Fp (a 6) : F] = n (Tr I p) . These are the preliminaries to the proof. The three formulas now follow quickly. (1) First the formula for the local degrees:

E ET

n (93 1p) = E n(cP aEP

I p) =crEP E #(0)- n

(2) Next the local norms. Consider a E P, T Ea. Put 93 = 9341 -1 = 93C1. Then as r runs through â we obtain (a) distinct isomorphisms Tv (V


33

of Fp (crb) into Ep which are identity on Fp ; since # (a) = [Fp (a 6) : Fp j we must therefore have all such isomorphisms. So for each a E E, NVJP C;C =

"43(NE43iFT CZ)

= (IT

(NET/FT CX)

Np p (a 6) F p (45 93 CO

= H (x 3 o') (crc43 at) TEFF

TEY

Hence N Tip ot =1-1 T) eX 93 ,P

= IITŒ

crEP(rEô

NEIFOE •

TEE

(3) The result for the local trace follows just as it did for the local

norm. In fact, one need only replace /1 by step (2).

E. and N

by S throughout

q. e. d. 15:4. Let EIF be a finite separable extension and let p be a spot on F. Consider the p-adic topology on F, and a topology on E which is finer than al193-adic topologies for all 931 p. Then NE1F

: E F

is continuous.

Proof. For each cr E Et the mapping cr: E Ep is continuous under the 93r-adic topology on E, hence under the given topology on E. Hence (acc), EE of E into n-space Ep x • • • x Ep is continthe mapping cc uous; but the multiplication map of Ep x • • • x Ep into Ep is continuous since Ep is a topological field; hence

//ace=

N Ell, a

aEE

is continuous, q. e. d. 15:4 a. Let a be an element of E. Given e> 0 there is a 6 > 0 such that INEIFx NEIFtXl p < s holds whenever x E E satisfies lx al v < 6 for all 93 I p. (Here lp and 1 I T are valuations in p and 93 respectively.) Proof. This follows from the proposition by considering the topology on E that is defined by the new distance function d (x, y) = max lx — y1,13

q. e. d.

931P

15:5. Let EIF be a finite extension and let p be a real spot on F. Suppose 0) real spots 93 on E which divide p and at which there are exactly r (r a given a EÈ is negative. Then (— 1) r NE/F is positive at p. O'Meara, Introduction to quadratic forms

3

34


Proof. Let P be the real spots on E which divide p and at which ce is positive, let N be those at which a is negative, and let C be the complex spots dividing p. Then /1/931p ocEN931p E43 S n V93EPvC. If 93 E N, then 93 is real so that ET . F93 , hence

Nup oe E Nv ip (— 43 ) (—F V 93 E N .

Hence NE I F OC = 11 N VII> OC E (- 1)rn . 93IP

q. e. d. 15:6. Example. Take a valuation 1 I p in p and a valuation I IT at each 93 dividing p. Consider oc E E such that loc193.- 1 VTIP • Then it follows easily from the norm formula of Theorem 15:3 that INEIF alp __ 1 and that INEIF 11 p = 1

if and only if f al v = 1 V 931 p.

Example. Recall that /(x) -= irr (x, 6, F). Let

15:7.

/ (x) = A (x) . /2 (x) • • • tr (x) be a factorization of /(x) into irreducible factors over Fp ; all these factors are distinct because of the general assumption of separability. Take exactly one root 6i of fi (x) in Ep for each 1 (1 I r). Let at be the isomorphism of E = F (6) onto F (6i) which is identity on F and carries 6 to bi. Then cri b and aj (3 are conjugate over Fp if and only if I -, j. Moreover a b is conjugate to some cri 6 over Fp for every a E I. Hence 93grl (1 5_ i 5_ 7) are all the distinct prime spots on E which divide the given p. And the local degree at the spot corresponding to ai is equal to

[Fp (à) : F] = deg fi . We have therefore established a correspondence --4. Ii 4-). 6.i 4-). Cri 4-). Toad between the irreducible factors of fix) over Fp and the spots on E dividing a given p, in which the degree of an irreducible factor is equal to the local degree at the corresponding spot.

35


15:8. Example. Suppose F is the field of rational numbers Q and E = Q ) with a E Q. Consider the archimedean spot p on Q. If a > 0 with a Q2 there are two spots 93i p on E, both of local degree 1. If a < O there is just one spot, of local degree 2. /3 \ 2), and again consider Example. Suppose F = Q and E 15:9. the archimedean spot p on Q. Then x3 — 2 has the irreducible decom-

position x3 — 2

—

(x2 + T/2 x + 31/4)

over R = Q. Hence there are two spots on E which divide p, the one of local degree 1, the other of local degree 2. Incidentally the first of these spots must be real and the second complex. § 15 C. The decomposition field

The situation that we are working with throughout § 15 is that of a finite separable extension EIF of degree n. We shall now introduce the concept of a decomposition field under the additional assumption that the extension in question is abelian. (A finite extension is called abelian if it is a galois extension with abelian galois group; a galois extension is one that is normal and separable.) We therefore assume that EIF is abelian and we let 6 6 (EIF) denote its galois group. 'We still have the same fixed prime spot p on F. Let ç'13 be any spot on E which divides p. Then 936 has been defined for all a E 6: it is that unique spot on E which makes the automorphism a of E (under 93) onto topological. We easily see that E (under

v)

93' p for all a 6 and that

V T = 9311 for all a, T E 6 . Consider the completions F(43 Ç E. Then E93 = FE = E-43 (6) is a (

splitting field of / (x) over FT since Eli,- is normal. And Fq3 is a completion of F at p. Hence we can regard Eci3 as the Ep and 93 as the To of our discussion to date. In particular we have the following facts which we shall put to immediate use: if 93 is any spot dividing p, then every spot dividing p is of the form 93- for some cr E 6; and 93a = 93r if and only a-' 6 and -r--'6 are conjugate over F. We define the decomposition group of EIF at p to be the subgroup

3 — {c1 E 6193a= 93} of 6. The decomposition group at p depends only on p, not on 93. For if we start with some other spot on E which divides p, say with the spot 93T where T is in 6, we obtain {a 6 i (TT

)

T=

131 3*


36

and it is easily seen, since G is abelian, that this group is equal to 8. We define the decomposition field of EIF at p to be the fixed field of the group 8. 15:10. Let EIF be an abelian extension with spots 93Ip and 931p, and decomposition field Z at p. Then Z1 n (VIP) (1) n (VP) (2) Z93 = F93 , Z = F93 n E , (3) e (93 I p) = e (93' I p) , f (93 I p) = f (93'1 p) if non-archimedean. Proof. (1) The local degree n (93 1p) is [F93 (6) : F93] and this is the number of conjugates of 6 over FT ; this is clearly equal to the number of a E G for which crà and 6 are conjugate over Fcp. But a is in 8 if and only if cr4 is in 8, hence if and only if a 6 and 6 are conjugate over F93 . Hence the order of 3 is equal to n(93 I p), hence so is [E: Z]. Hence

n (VP) =

"(VIP).

(2) Consider the extension EIZ with its galois group 8. The decomposition group of this at the spot induced by 93 on Z is simply S. Hence by the first part of this proposition, [E' : Z] = [E :Z]. But F93 S_ Zçp has [ET : Fv] = [E : Z]. Hence F93 .--- Z. Since 4. F93 we must have Z F93 n E. But [E:F93 E] [ET : FT ] [E: Z]. Hence F93r E = Z. [ET : (FTr E)93] (3) Now p becomes non-archimedean. Write 93P = 93r with a suitable E G. First let us do the ramification index. Fix J 193E91 Then using the definition of Tr along with Proposition 1 1 :6 we can find I iv ( 93r with the following property: the mapping r of E (under I IT) onto E (under! Iv) is analytic. Then WIT = ItElv= JE çpT and !FIT = IrF193r= !FITT . Hence e (93 j p) = e (93r ip) . Now for the degree of inertia. Clearly T (I) ((43 )) = o (93/ and T (rn ((43 )) = m (Y) • Take a residue class field E (931 and let F (V) be the residue class field of F at p that is thereby obtained by natural restriction. Consider the composite homomorphism o('43) >—). 0 NY) E (931 • This has kernel m (93). Hence the composite homomorphism in conjunction with E (93r) is a residue class field of E at (43; write E ((43 )=E (93r). But o (p) is carried onto F (V) by the composite homomorphism. Hence F (43) F(3'). Hence

((43 1p) = [E

(

(43): F ((43 )]

CE My) F (931] -- f Mr p) •

q. e. d.


37

The preceding discussion shows that the local degree n (931p) of an abelian extension depends only on p and not on 91 (This is not true for general separable extensions; for instance, see Example 15:9.) Accordingly one refers to the common value of the n (931p) as the local degree of the abelian extension EIF at p, and one denotes this common value by fir The same simplification applies to the ramification index and the degree of inertia in the non-archimedean case, and they are denoted by e 4, respectively. 15: 10 a. Suppose dis a factor of tsp. Then there is a field HwithZ CHSE such that the local degree of HIF at p is equal to d. Proof. The galois group G (E931F13) of the galois extension E93/4 is naturally isomorphic (by restriction) to a subgroup of (E1F). Hence of order ttpid. The G (Ev/F93) is abelian ; so it contains a subgroup natural image of in G (EIF) is of order npfd; so the fixed field H of this group has [E:H] = 'tad. Now every element of Z is left fixed by since Z13. F93, hence ZCHS E. But [H: Z] d; [ET : H93] 5_ [E: H] nad , [H 93 : Z93] and we know that [ET : ZI3] = 'sr ; hence [HT : Z93] = d. Thus the field H has all the desired properties. q. e. d. S 16. Discrete valuations

The value group IF of a non-trivial valuation on a field F is clearly infinite. In fact it is either a discrete subset or an everywhere dense subset of the set of positive numbers P; in the first instance it is infinite cyclic while in the second it is not. In order to verify these assertions consider the topological mapping log: P R which sends the multiplicative structure on P to the additive structure on R. The value group 1F1 then becomes the additive subgroup log 1F1 of R. But every nontrivial additive subgroup of R is either a discrete infinite cyclic group, or else an everywhere dense non-cyclic subgroup of R. Hence 1FI has the property stated. We shall call a valuation discrete if it is non-trivial and if its value group is a discrete infinite cyclic subgroup of P • 1 Thus an archimedean valuation cannot be discrete. The formula relating equivalent valuations shows that if a valuation is discrete, then so is every valuation that is equivalent to it. Accordingly we say that a prime spot p on a field F is discrete if it contains at least one discrete valuation. So if p is discrete it is non-trivial and non-archimedean, and every valuation in it is discrete. A discrete valuation does not yield the discrete topology. In fact it is easy to see that Me topology on an arbitrary valuated field is the discrete topology if and only if the valuation is trivial.

38


16: 1. Let EIF be a finite extension of fields with spots 93 p. Then 93 is discrete if and only if p is. Proof. Fix a valuation 1 1 E 93. If 93 is discrete, then p is non-trivial by Proposition 11 :3, and IF' is infinite cyclic being a subgroup of 1E1, hence p is discrete. Conversely, suppose p is discrete. Then 1E16S_ IF' S_ 1E1 where e e ((41p) is the ramification index of the extension. Hence 1E1 must be a discrete subgroup of P. Hence 93 is discrete. q. e. d. Consider a discrete spot p on F. Recall from ring theory that an element in the integral domain o (p) is called a prime element of o (p) if it is a non-unit such that in every factorization = a (3 with a, fl E (p) either a or (3 is a unit: By a prime element of F at p we shall mean a prime element of the integral domain o (p) in the above sense. Suppose n is any element of o (p) ; pick a valuation 1 1 E p; then it is easily seen that z is a prime element of F at p if and only if 1n1 is that element of 1F1 with largest value less than 1; and this is equivalent to saying that In' generates 1F1. In particular, this shows that there is always at least one prime element * of F at p. Now suppose that n actually is a prime element of F at p. Then the following three facts are true: first, n' E F is a prime element at p if and only if al7e is a unit; second, m (p) is a principal ideal, in fact (P) and third, is also a prime element of Fp at p since 1F1 = 11'4 It follows from the description of a prime element n in terms of valuations that every a E F can be expressed in the form CC =

with e a unit at p and y E Z. If z is fixed, then the representation is unique; if the prime element is allowed to vary, then e will vary but y will not. We can therefore define the order of a at p to be ordp oc = y. We formally put ordp 0 cc. The following rules for operating with the order function are evident: < ifltp 4*. ordp a > ordv fi , ordp a (3 = ordp a + ordp fi , ordp a>0«aEm (p) , ordp 0 « a E (p) . 16:2. Let EIF be a finite extension of fields with discrete spots 931p. Then ord93 a =-- e ((43 I P) ord p a for all a E F.

Chapter L Valuated Fields

39

Proof. Fix I E3. Then (14 iFi) = e 1P) (=

Hence if 17,

are prime elements of E,F we must have

Vil e = ini Therefore n

21 He

with A E u(). So en.ordpa =

(8 Aord pcc) pord p o:i

for some s E u (p). This gives the result. q. e. d. Consider o (p). In each coset of o (p) modulo m(p) pick exactly one representative c; always agree to pick c = 0 as the representative of m (p). Call any set C which is so obtained a representative set in o (p) of the residue class field of F at p. Suppose a representative set C has been fixed. For each y in Z pick 7c„ in F with ordp 74= v (for instance the gr„ could be the powers nti of a fixed prime element at p; we choose ny instead of nv in order to provide greater flexibility for applications). The immediate significance of the sets C and {7} is that they give unique power series expansions for the elements of F. 16:3. Let C be a fixed representative set of the residue class field of F at a discrete spot p. Suppose nv EF is chosen at each y E Z with ordp ar,= v. Then every element oc of

fr can be expressed uniquely in the form cc Œ=

ov a,

with c, E C, n = ordp oc, and cn + O. Proof. 1) We can put a = en. for some e E u (p) since ordp oc = n. Choose cn E C with e cn modm (p). Then oc = cn

+ cc' with ordp oc ' > n.

Next apply this procedure to oc' to obtain a", then to oc", and so on. After nt I steps we obtain an expression Œ = c.,,,ff,„± • • • ± c n+ ,„74, 4_,n + ot(m+i) with ordp 0'1+4 > n ± tn. In this way we can define cn + 1 , • • • , Cn + no

• • •

EC•

The partial sums Cnnn+ ' • • + cn-Fin=n+m

clearly converge to Œ. Hence oo

oc

= 11

40


2) In order to prove uniqueness we consider two expressions DO

00

(1, 74

cvny==

with the c, and d, in C. Let i be the first integer for which c == di ; then Jci — dilp 1 since ci and di will fall in different residue classes of o (p) modulo m (p) when they are not equal; hence OD

E (cp— dv) nv by Corollary 11:2a. But

by hypothesis. This gives a contradiction. So we do indeed have ci = clj for all i. q. e. d. 16:4. Theorem. Let EIF be a finite extension of degree n with discrete spots 931p. Suppose further that the spots are complete. Then

e (93 p) f (C43 1p)

n Proof. 1) Write e = e (931p) and / / (931p). Let (-, E) be a residue class field of E at 93, and let (- ,F) be obtained from it by natural restriction. Choose col, (of in o (93) in such a way that ã,. c.7), form a base for E over F. If we consider all elements of the form bl col + • • • ± 'yob. with the b' s in o (p) we fall in each residue class of o (93) modulo m(93) at least once. Hence we can select a representative set C of the residue class field of E at 93 in which every element is of the form

with all bi Eo (p) .

b1 w1 + • • • + b,w1 ,

2) Let 17, a be fixed prime elements of E,F at 93, p respectively. For each xEZ write x=pe+ v with a, y E Z and O y e— 1. Define IT,. nil Hy. Then

ordv /ix = ts ord93

v

ge+v—x.

3) We have already proved in Proposition 13:6 that ef n. It therefore suffices to prove that the cf elements span E over F. Consider a typical a EE. By Proposition 16:3 we can express it in the series expansion 00

a

E C„ /7„ se

Chapter IL Dedekind Theory of Ideals

with ord13 Œ

41

se and the C. in C. If we group the terms we obtain

co CC = E ( Cpell 08+ Cpe+1 17116+1+ • • • + Cp6-1-(0-1) 140+(e-1)) it=

co

co

Cpellpe)+ • • • 4. ( P= 8

Cpe+(e-1) ite+(e-1)) • P=g

It therefore suffices for us to prove that each 00

Er Cps-I-v/48+v—

Cpe-FsarP 11-v P=8

14 = 8

is in the space spanned by the above coA H' over F. But by the choice of the set C we can write oo

oo

E

C14 , + „no=

E E b„c0A v-eg ,i=s A=1

0=8

l

i

ce

E E 1=1

blo 2e)

,

0=8

with all &to in o (p). Hence 00 Cpe+vilite+v

E

F co/ riv •

ii -8

q. e.

d.

Chapter II

Dedekind Theory of Ideals In Chapter I we studied the ring of integers o (p) of a single nonarchimedean spot p. We shall see in § 33J that the set of algebraic integers of a number field F can be expressed in the form o (S) = n 0 (P) pE non-archimedean spots on F.

where S consists of all This exhibits a strong connection between the algebraic integers and the prime spots of a number field, and we shall start to exploit it here. Specifically, we shall use the theory of prime spots to set up an ideal theory in o (S). For the present we can be quite general and we consider an arbitrary field F that is provided with a set of spots satisfying certain axioms. We shall call these axioms the Dedekind axioms for S since they lead to Dedelcind's ideal theory in o (S). The general assumption throughout this chapter is that we have a field F and a non-empty set of spots S on F which satisfies the axioms (DO, (D2) and (D3) given immediately below. We fix I i p E p at each p in the given set S.

42


§ 21. Dedekind axioms for S. We call the non-empty set of spots S on the field F a Dedekind set of spots if it satisfies the following three axioms: (D1)every spot in S is discrete (D2)for each cfc(F we have loch, < 1 for almost all p E S (133) whenever q and q' are distinct spots in S, there corresponds to each e > 0 an a E F such that 11q < ,

lock, < , locj p 5_ I

for all p E S (q vq1). By the ring of integers of F at S we shall mean the subring o (S) = im o(p) pEs

of F. The multiplicative subgroup it (S) = r u(p) pEs

of o (S) consists precisely of the invertible elements of o (S) and so we shall refer to it as the group of units of F at S. At times it is convenient to relax the notation and to write o and u instead of o (S) and u (S). 2 1:1. Example. The set of all non-archimedean spots on the field of rational numbers Q is Dedekind. This will be proved in § 31. The axioms are clearly independent of the choice of t IpE ix By taking inverses we note that (D2) actually implies the stronger assertion of equality: if oc E fr, then ki p = 1 for almost all p E S. If S consists of just a single discrete spot, then (D2) is automatically satisfied while (DO is vacuously true; hence a single discrete spot is always Dedekind. Similarly if S satisfies (D1) and is at the same time a finite set, then it is Dedekind; this follows at once from the Weak Approximation Theorem of § 11E. Approximation is of course the key to the third axiom; we can use this axiom to derive an approximation theorem which, in certain important situations, is stronger than the one given in § 11E; this we now do. 21:2. Strong Approximation Theorem. Let T be a finite subset of the Dedekind set of spots S on the field F. Suppose that ap C F is given, one for each p E T. Then for each e> 0 there is an A E F such that •

flA ocp ir < e VpET 141 VpES—T. Proof. 1) By the Weak Approximation Theorem of § 11E we can

assume that S is an infinite set of spots. We can also assume that itcp lq < 1

Y p ET,

V ciES—T:

Chapter II. Dedekind Theory of Ideals

43

for if necessary we can adjoin to T all those qES— T at which iccpiq> for at least one cep ; and then define ccq . 0 at the new q; the new set T which is obtained by this adjunction is still finite because of the axiom (D2); if we can prove our theorem for the new T we will have it for the original one. Hence we can indeed make the above assumption. In the same way we can assume, by adjoining a single spot to T if necessary, that T consists of at least two spots. 2) Consider a spot p E T that is fixed for the moment. For each E T p we can find an element of o (S) that is arbitrarily close to 1 at p and to 0 at q. Do this for each gET—p and multiply all these answers together. Using the fact that multiplication is continuous in the p-adic topology we can obtain in this way an element of 0 (S) that is arbitrarily close to 1 at p and to 0 at all q E T — p. Let us denote such an element by A. Then obtain an A p for each p E T. A P This element satisfies 3) The rest is easy. Just form E aP A. pET

oc„A,

1 V cf ES — T

PET

by choice of T and the A. And by continuity of addition and multiplication it can be made arbitrarily close to cep simultaneously at all p E T just by making the approximations A p in step 2) sharp enough. q. e. d. 21 : 2 a. Corollary. Let 4 be given in the value group IFlp at each p E S, with almost all 4= 1. Then there is an A E F such that

,, VpET, 1446, 4 VpES. Proof. We can assume that = 1 for all pES— T by enlarging T if necessary. Pick oci,EF for each p E T in such a way that jacp ip = ' A1,,=

Choose A EF with { IA

cerlp < tapip VPET 1 Vp ES— T.

q. e. This A is the required element. 21:3. Let S be a Dedekind set of spots on F. Then o(S)F. And F is the quotient field of 0 (S). Proof. We have o (p) cF for any p E S since all spots in S are nontrivial. Hence 0 (S) (F. Consider a typical a EF which we wish to express as a quotient of elements of 0 (S). Put

T {p E Si lai r > 1} .

We can clearly assume that T is not empty. So T will be a finite set by Axiom (D2). Now we have cc-1 E o (p) for all p E T, hence we can use the

44


Strong Approximation Theorem to find b E o (S) such that Jb—or9— lair is continuous. Consider an e-neighborhood (0 <s < 1) of a typical element a of o(p). Choose A E F such that { IA — alp < 1A1 4 1 VciES—p.

Then A E o (S). Hence every neighborhood of cc meets o (S) and so 1) (p) is its closure in F at p. q. e. d. The second part of the proposition is now clear , 22. Ideal theory

S is a Dedekind set of spots on the field F. We use o and u instead of (S) and U (S). § 22A. Operations with fractional ideals The field F is of course a vector space over itself in the obvious natural way. And it is an 0-module under the induced laws. We define a fractional ideal a of F at S to be a non-zero 0-module a S F which has the following property: there is a non-zero A E o such that Act Ç 0. This property simply asserts that the elements of the 0-module a are to have bounded denominators. We use /(S), or simply I, to denote the set of all fractional ideals of F at S. We note that every finitely generated non-zero 0-module in F is a fractional ideal at S (the converse is also true and it will be established in Corollary 22:5b). In particular, the set ao is a fractional 0-ideal for any a E F. We call any such ideal a principal ideal and we let P (S), or just P, denote the set of principal ideals in I (S). We shall call the fractional ideal a integral if a S 0. Thus the integral ideals are the ideals of o in the usual sense, with the exception of O. We say that a divides b, or that b is a multiple of a, if b Ç a; we denote this fact by writing a l b. We define the greatest common divisor, the least


common multiple, and the product of any two fractional ideals a, follows: g.c.d.: a + = {cc ± /3 I cc E E b}

1.c.m.:

anb,

product:

ab =

45

b as

E cc 131ccE a, fl E b . (finite

•

These new objects are clearly o-modules; and one easily verifies that their elements are of bounded denominator; in order to verify that all are in I we must still show that they are non-zero; for a + b and a b this is clear; for a n b we first choose A, pi in o such that Aa and fib are integral; then a n b 2a rub (Aa) (0) DO . Hence we have proved that

a+b, a nb , ab El. Clearly ma El for any a E 1 . We say that a and b are relatively prime if a + b =0. Product formation provides I with an associative and commutative multiplicative law in which o acts as an identity. We shall see that every a E I has an inverse under this law so that I is in fact a group.

§ 22B. Valuations on I For any a E I and any p E S we define lair

tad P ' = max aEa

Clearly !al p > O. If we take A EF such that Aa 0, then a C A-1 0 and so lalp IA-1p . Hence the maximum in the definition of lalp is attained, so lalp E P. This also shows that lalt, 1 for almost all p E S since it is true for A-1 ; hence lalp = 1 for almost all p ES . Clearly o=

1. And for any a EF we have

One easily proves that

local= 1 0614 •

lablp. Iallb1, la + blp = max (lalr, lbli,) hold for any a, b E I. It is also true that blp= min (! alr, ibir) but this is a little more difficult, so we offer a proof. Clearly la n bip min (lalp, lblp). We must reverse this inequality. Take lai r lblp. Pick a E a and b E b in such a way that

lalp= 145_ lblp.

46


By Corollary 21:2a there is a tt EF such that rip —

Ittl q 5_ min (1,

a b

YqES—p. ,)

Hence /Ab E ay. And tt E o so that also Aub E b. Hence pb

E a n b. Hence

ittbip = ' a l = lalp = min (Ial p, II).

la n bi,,

This proves the assertion. 22: 1. Let S be a Dedekind set of spots on a field F and let a, b be fractional ideals at S. Then (1) a = {ot EF I lair -5- lai r Y p

E S}

(2) a _.ç_ 1) .=. [alp 5. Iblp Yp ES. Proof. 1) The second part follows at once from the first. In order to prove the first we have to show that an a in F which satisfies 'al p .5_ lair for all p E S is in a. This reduces (on replacing a by ac -la) to proving that 1 E a if 1 'al p for all p E S. Now it is clearly enough to prove that 1Eano; and furthermore la n 0Ip = 1. So we have reduced the question to the following: given a C o with 14= 1 for all p E S, prove that 1 E a. This is the form we need to establish the proof. 2) Pick an arbitrary non-zero y in a and fix it. Consider the set T ---- {P E Si !Yip < 1} If T is empty, then y -1 E o and so 1 = y -1 y E a as asserted. Hence assume that T is not empty. So T is a finite subset of S. Consider a p E T, fixed for the moment. Choose oc, E a such that loci4= lal p = 1. Then the Strong Approximation Theorem gives us an element of F which will be written fl, to denote its dependence on the temporarily fixed p, such that

1 ap

—

t flP IP1 — 1v1 P 'lark --- IY1of YgEs — P.

{

Since y E o (S) and lap ip = 1 it is clear that ,84, is in 0(S). Now do this once at each p E T to obtain an ap and a /3p as above. For each q E T we have 1 — E at) /31,1 =1( 1 — ccq Ai) —

pET or And for each q ES — T we have

1

PET

14

E

PE T—q

act,

1 — lylq -

16'1 5. iviq a


47

Hence

flp)

(1— E cep PET

E ° (S)

by definition of o (S). So — z Œf3 E yoCa. pcT

q. e.

But cep E a and 161p E o. Hence I E a .

d.

22 : 2. Let S be a Dedekind set of spots on F. At each p ( S let there be given an ocp (fr with almost all lap i p equal to 1. Then =

EF1

jap i t,

Vp E

is a fractional ideal at S. And

lal p = lap l p VpES. Proof. Clearly a is an o-module. We deduce from Corollary 21:2a that it is not O. Again by Corollary 21:2a there is a A ( F such that

min (1, IceiT'ip) Vp ES. So there is a non-zero A ( o with I

vpcs,

vz(a. This means that Ax o for all x Ca; hence a is a fractional ideal. Clearly 14,5. lar ip holds for all p E S. But Corollary 21:2a gives us an x E a such that jx1 p = locp I p at a fixed p E S. Hence lalp q. e. d. 22:3. Example. The following identities are true for all a, b, c E and all n E N : a(b+ c) = ab+ ac, a(bnc)= abnac (a + 6)n „ an+ ton , (a n 6)", n bn iActplp

an (b + c) = (a n b) + (a n c) , a + (b c) = (a + b) n (a + c) oh = (a + b) (a n b) . Also an— bn for some n

N —> a = b.

All proofs by inspection using the properties of I 1p on I. § 22C. Properties of o and I

The factorization theory of ideals follows quickly from the results of § 22B. For instance / is a group: namely, for any a (I take that b E / for which Ibl p = ja1 17 1 for all p E S ; then la bl p = 1 = jol p for all p ( S and SO ab o. Hence every a / has an inverse and / is a group. The inverse of a will be written a1.

48


22:4. Example. (a + b) a-1 n 1)-1 , (a n b) -1 = a-1 + b-1 . The principal ideals P are a subgroup of I. Hence we can form the quotient group _TIP; this is called the ideal class group of F at S. The class number hF (S) is the order of the ideal class group. Clearly hp(S) . 1 means that every fractional ideal is principal, and this is equivalent to saying that o is a principal ideal domain. The class number is 1 when S is a finite set: given a E I use Corollary 21:2a to pick a EF such that kip= lalp for all p E S then a = ow by Proposition 22:1. It will be shown in § 33H that the class number is finite over algebraic number fields. 22:5. Let S be a Dedekind set of spots on F and let a, b, c be fractional ideals at S. Then there are non-zero field elements a and fl such that c= aa + )6b And in fact 13 can be anything for which fib Ç c. Proof. Take /3 E F such that pbCc (there is always at least one such )6; for instance any 16 E b—lc will do). Let Tibe any finite subset of S with the property that lblp . icip = Ay . 1 V p ES — T. By Corollary 21:2a there is an a EP such that Vp E T VpES— T. liocip 5._ 1 Then Jota )6blp = max (locait, , Ply) JcJ, Vp ES. q. e. d. Hence aa + Pb = c by Proposition 22:1. 22:5a. Every fractional ideal is of the form ao f3c) with cc, fi in F. 22 : 5 b. Every fractional ideal is a finitely generated co-module. 22 : 5c. Suppose b is actually integral. Then there is an integral ideal a' in the ideal class of a such that a' and b are relatively prime. Consider an ideal a with OC aC 0. We call a decomposable if there are integral ideals b and c which are distinct from 0, and hence from a, such that a = bc; if there are no such ideals we call a indecomposable. We recall from elementary algebra that an ideal a of o is called a prime ideal if the residue class ring o/a is an integral domain. We shall prove shortly that the indecomposable ideals, the prime ideals, and the maximal ideals of o are the same (provided of course that we omit the ideals 0 and o). It is convenient to introduce (0); 0 (S) n this is an integral ideal and Corollary 21:2a shows that if lo (S) ntn (Oki= krrir if q P qES— p for any p in S, where 7rp is a prime element of F at p.

49

Chapter H. Dedekind Theory of Ideals

22 : 6. Let a be a fractional ideal with respect to a Dedekind set of spots S on the field F. Suppose Oc a c o. Then the following assertions about a are equivalent: (1) a.onm (p) for some p E S (2) a is a prime ideal in p (3) a is indecomposable (4) a is a maximal ideal in o. Proof. (1) = (2). Consider typical a, 13 in D — a. Then

la 4= 141161p= 1, hence a 19 E o — a, hence via is an integral domain, hence a is a prime ideal. (2) = (3). Suppose we had a factorization a = b c with OCbCo and Oc cc D. Then a c b and a c c. Pick 13 E b — a and y E c — a. Then 13 y E bc= a. And this denies the primeness of a. (3) (4). Suppose we had b El with ac bc D. Then a = b(b - i a) would be a factorization in which b C o and b -1 a co. Thus a would be decomposable. (4) (1). By Proposition 22: 1 there is a p ES such that 'al p < 1. So -

-

!alp inpip= Iv (S) m (p)p with ni, a prime element of F at p. Hence lal q lo (S) nm (p)l q holds for all q E S. Hence a C o (S) n m (p), again by Proposition 22: 1. Hence a = D (S) n m (p) by the hypothesis that a is a maximal ideal in D. q. e. d.

The preceding discussion provides a natural bijection of the set of prime spots S onto the set of proper prime ideals of D (S) that is determined by the rule p -4- o (S) nm (p) . In order to cut down on notation we shall refer to D (S) n m(p) as the prime ideal p of F at S whenever it is convenient to do so, and provided there is no risk of confusing the two p's. When we use this convention we can refer to S as either the given set of prime spots, or the set of prime ideals of D at the given set of prime spots. As an example of the convention let us see what is meant by the equation

if cl = P IPl q if q ES— p. Here np is a prime element of F at the prime spot p; the first p is a prime ideal; all other p's and q's are prime spots. 22: 7. Unique Factorization Theorem. Let S be a Dedekind set of spots f inpip I1

on a field F. Then the fractional ideals under product formation form a free abelian group. The set of prime ideals S is a base for this group. O'Meara, Introduction to quadratic forms

4

SO


Proof. Consider typical a E I. At each p E S choose 'pp E Z in such a way that lai p = 1p1pvP. Then

laig= iffrPlq VqES. Hence a = HpvP by Proposition 22:1. Thus the multiplicative group I is generated by its prime ideals. The uniqueness is also clear from the equation

Hr

vp

1(11 f4

V41

q. e. d. For example consider a,b EI in their prime factorizations

a=

b = fipmr

Then

ab

vr,ftp, VpES.

Also

a + b = Hpmin (vr. PP),

a n b =J7pmax (rr. Pp) .

And a and b are relatively prime if and only if min (vp , eu;,) = 0 for

all p E S. 22:8. Definition. The order ordp a of the fractional ideal a at p is the power of p that appears in the prime factorization

a= 11 pvp Thus ordp a b = ordp a+ ordp b, and ordp ao = ordp a, and ' a lp . 1 7p 1oprap a

22:9. Example. Given fractional ideals a, b, r1, r2 we claim that there are non-zero scalars a, /3 such that aa laa

fib=' 166 r2 =

r2 .

It is enough to prove this with r1 ± r2 = 0. By Corollary 22:5c we can pick a EF such that aa r2 = o, then we can pick fi EF with /3b aa r1 = o. So we clearly must have fib + aa= o. And it follows, for instance by the Unique Factorization Theorem, that a a jgb r2 = o.

§ 22D. Three residue class isomorphisms Consider two fractional ideals a and b with b Ç a. Being ideals they are also additive groups and as such we can form their quotient group

51


a/b. Our purpose in this subparagraph is to demonstrate three additive group isomorphisms. These are F (p) for each p E S. 1. o/p 2. a/b o/a- lb if b a. 3. o/ab (o/a) e (o/b) if a + b = o. F) at p and restrict For the first we consider the residue class field the bar mapping from o (p) to 0. This restriction has ke rn el o(S) (p) =p ; and its image is P since there is a representative of every residue class of F at p in o by Proposition 21:4. One of the elementary isomorphism P. theorems of group theory now provides the desired isomorphism o/p Next consider a/b. Here let bar denote the natural homomorphism of b; this is possible by Proposia onto a/b. Pick a E a such that a = tion 22:5. Define the composite homomorphism X --->-- Ot X

-4--

Ot X

of o into a/b. This composite map is clearly surjective; and its kernel is a-1 b. Hence there is a natural isomorphism o/a-lb ›--)- a/b. In the third instance we let bar denote the natural homomorphism of o onto 0/a b. The restriction of bar to a has kernel a n ab a b. Hence

ofb . )--4- a/ab Similarly 5 ofa. Now = 6. since a and b are relatively prime. Let us show that •:1 n E = O. For any E n E) we have a E a and b E b, such that g = = 5 . Thus x—aEab(a, x bEabCa, b aEa. So b = (b a) + a E a. Hence b Ean b. But a n b (a n b) (a + b) = a b . Hence b E ab. Hence 5 . O. Hence g = O. So n E = O. So )3 = a ED E. —

—

—

Therefore

o/ab

(o/a) e (o/b)

as asserted.

§ 22E. Discrete valuation rings P Suppose S is a Dedekind set consisting of exactly one discrete spot. In this event o (S) = o (p) and we obtain an ideal theory corresponding to the discrete valuation ring o (p). Our conventions now read as follows:

o = 0 (S)

o(p), P =nt

7r0

where a is a prime element for F at p. Every ideal is principal since S is a finite set of spots. But here we can say more. For the Unique Factorization Theorem tells us that every fractional ideal has the form pv for exactly one y E Z. Hence the ideals nro with v in Z are all the distinct fractional ideals of F at p. 4*

52


§ 22F. Definition of a Dedeldnd domain We shall call a subring R of a field F a Dedekind domain' if there is a Dedekind set of spots S on F such that R = o (S). We shall prove in the next proposition that S, if it exists, is unique. We then call S the underlying set of spots of the Dedekind domain R.

22:10. Let S and T be Dedekind sets of spots on the field F. I/ (s) 0 ( T), then S . T. Proof. It is enough to prove that * E S whenever * is a discrete spot on F for which o (S) c o (*) . Let o stand for o (S). Pick 1 1 * in * and define

a

fix E Di

< 11

Then a is clearly an ideal in o since o C o (*). And 0 c a since 1 I * is non-trivial on F and hence on 0. And a Go since 1 is not in a. Hence OcaCo. It is easily seen that a is actually a prime ideal in o. So there is a prime spot p in S such that a is equal to the prime ideal p of F at S. Hence for any a in o we have 'alp < 1 1f and only if 1 cel * < 1. If we can prove that every 13 in F which satisfies 1 1611, < 1 also satisfies 1/31 * < 1 we shall have p = * by Corollary 11:4 a, and we shall be through. So consider such a /3. By Corollary 21:2a there is a y such that Iv -= 1 and IVIcr min ( 1, Ifi'lq) for all q E S p. Here y is in o and ly1 * . 1. Now fi y is also in o with 119 yi p < 1. Hence 1 16 yl * < 1. Hence 1 161* < 1. q. e. d.

§ 23 . Extension fields Consider a finite extension EfF, a set of spots S on F, and a set T on E. Recall from § 11D that T S means that the spot induced on F by each spot in T is in S, and T11 S means that T consists of all spots on E which induce a spot in S on F. Our main purpose here is to show that if S is Dedekind and if T IS, then T is also Dedekind. This will be done under the assumption of separability of EIF since this is how we developed the theory of prolongations in § 15, but as a matter of fact the result is true in general. The reader will naturally ask for the connection between the ideal theory at S and that at T. This is a classical question with classical answers, but unfortunately we cannot go into it here2 23:1. Theorem. Let TiS be sets of spots on the finite separable extension EIF. If S is Dedekind then so is T. Proof. We can assume that T11 S. Fix 1 143 E 93 at each 93 E T, and II E p at each p E S; let this be done in such a way that I t, is the restriction of 1 1,13 whenever 931p. Now every spot in S is discrete, hence .

There are several equivalent definitions of a Dedekind domain. See O. ZARISKI and P. SAMUEL, Commutative algebra (Princeton, 1958) and E. ARTIN, Theory of algebraic numbers (Gottingen lectures, 1956). 2 For a discussion of classical ideal theory see E. HECKE, Vorlesungen über die Theorie der algebraischen Zahlen (New York, 1948). 1

Chapter IL Dedekind Theory of Ideals

53

the same is true in T by Proposition 16: 1. Hence T satisfies Axiom (D1) of § 21. Consider a typical a E E and write am+

am -' + • • • + an,= 0

with all ai E F. Let So be a subset consisting of almost all spots of S such that Ia1 1, 1 V ai, V p E S. . Let Tol I So. So To is a subset of T which consists of almost all spots in T by § 15A. By the Principle of Domination it is impossible to have laIT > 1 at any 93 in To. Hence I aI93 _< 1 for all 93 E To. So Axiom (D2) holds in T. Now we do Axiom (D3). Here we have two distinct spots S31 and 932 of T under consideration, and also a real number e with 0 <e < 1. We ask for an A E E such that

11 — Alvi < , jA193,< and IA IT 1 for all 93 E T. Let p1 and p2 be the spots in S which are divisible by 931 and 932 respectively. First suppose p3. and p2 are distinct. Pick a E o (S) with 1 1 — airt< e, !alp.< e. Then A oc is the element we are looking for. Hence we can assume that !h= p2 = P. say. Define Tp Ç T such that Tpll p. Here Tp is finite by § 15A. By the Weak Approximation Theorem of § 11E we have a B E E such that IBIT .5 1 for all 93 E Tp with 11 BI931 small and IBI93,< 8. Choose oc E (S) in such a way that 11 ocivi is small and. 1.130c193 ,.5 1 for all 93 E T Tr. If all approximations are good enough we will have 11 — ocB1 931 < e by continuity of multiplication. Here A = aB is the element we are looking for. q. e. d. 23:2. Let EIF be a finite separable extension with Dedekind sets of spots TI IS. Then the following assertions for a typical oc E E are equivalent: (1) a Eo (T) (2) irr (x, oc, F) has all coefficients in o (S) (3) oc satisfies a monic polynomial with coefficients in o (S). Proof. (1) = (2). There is no loss of generality in assuming that E = F(a). Let /(x)

(x, a, F) xm al x"1 -1 + • • • ±

We have to prove that

ai Eo (p) V at , VpES. So consider ap ES and perform the following construction of § 15: take the completion Fp of F at p and the splitting field Ep of / (JO over Fp, and let 930 be the unique spot on Ep which divides p on Fr. Let al , . an,

54


be all the roots of 1(x) in E. Let ai be the isomorphism of E into Ep which is the identity on F and which carries cc to oc. Then 930°F1 is a split on E which divides p, hence it is in T, hence a is an integer of E at 9371. Now ai is a topological isomorphism of E (under 93r) into Ep, hence o,. oc is an integer of aiE at 930. So we have proved that al, , 0Cm are all in o (To). But (x) = (x am) • al) Hence all ai are in o (To) n F o(p). This proves the first implication. The implication (2) (3) is obvious. An easy application of the Prinq. e. d. ciple of Domination will show that (3) = (1). Chapter III

Fields of Number Theory The first two chapters have been done in great generality. Before we can move on to the deeper results of number theory we shall have to make additional assumptions about the underlying field F. We shall do this by explicitly stating our fields of interest. They are the field of rational numbers or any field of rational functions in one variable over a finite field of coefficients, all finite extensions of these fields, and all

completions thereof. By restricting ourselves to these fields we obtain two additional properties. Roughly speaking, the first of these properties is one of finiteness of the residue class field and the second is one of dependence among the valuations. These are actually the decisive properties that distinguish the rest of the arithmetic theory from the first two chapters. In fact it is possible to axiomatize these properties 1 and to show that they lead directly to the fields of number theory, but we shall not go into that here. S 31. Rational global fields

By a rational number field Q we shall mean any prime field of characteristic O. Thus the field of rational numbers Q is essentially the only rational number field. By the ring of rational integers. Z of Q we mean the subset 0, +1, +2,... of Q. Similarly the prime numbers of Q are the elements

2, 3, 5, 7, 11, .

.

1 This is done by E. ARTIN and G. WHAPLES, Bull. Am. Math. Soc. (1945), 469-492.

PP.

Chapter III. Fields of Number Theory

55

We call a field Q a rational function field if there is a subfield k con taming a finite number of elements, and an element x of Q which is transcendental over k, such that Q k (x) . Strictly speaking we should call such a field a rational function field in one variable over a finite constant field, but all rational function fields used in this book will satisfy the additional assumptions so we settle with the shorter terminology. By the integers of Q we mean the polynomial ring k [x] contained in Q; by a prime function or prime polynomial we mean a monic irreducible polynomial in k Note that these concepts depend on the choice of x so that we should really refer to them "with respect to x". However k is independent of the choice of x: in fact we shall characterize k as the intersection of all valuations rings in.Q. We call k the constant field of the rational function field Q. A field Q will be called a rational global field if it is either a rational number field or a rational function field. By the ring of integers Z of Q we mean Z in the first instance and k [x] in the second. Similarly a prime is either a prime number or a prime polynomial as the case may be. Each prime fi of a rational global field Q determines a spot called the fi-adic spot at Q. It is defined as follows: take a typical a E Q and write it in the form OC

= pi uiv

with both u and y in Z and with neither divisible by p. Define kip= » where A is some real constant with 0 Na

H

q. e. d.

1

PER l am

Proof. Take the factorization into prime ideals: a=

H pvy pEs

Then

11

(Np)vp.

Na PER

PER

1PIP

§ 33D. Introduction of the Mile group

H1 pEs la1P

q. e. d.

Ji?

D is the set of all non-trivial spots (discrete or archirnedean) on our global field F. Consider the multiplicative group pED

consisting of the direct product of all the multiplicative groups Fp ; a typical element of this big group is defined in terms of its p-coordinates, say = (iP)p ES?

E.frp) ;

and the multiplication in the direct product is, by definition, coordinatewise. An Wee is defined to be an element I of the above direct product which satisfies the following extra condition:

Nip . 1 for almost all. p E Q. The set of all Wits is a subgroup of the direct product called the group of idèles and written J. We let ip denote the p-coordinate of any i E jp at the spot p E D. For any a EP and any i E jp we let a i stand for the id& in jp whose p-coordinate at each p E D is equal to alp. We define the volume of an idèle i to be the positive real number 11 1 11 = H 1i! , pED


69

where I I, denotes the normalized valuation at the spot p; the product used in the definition is essentially a finite one since almost all firjr, are equal to 1. I1 is evident that

hull = 111I ihi And the Product Formula shows that • We say that a subset X of F is bounded by the idèle I E jp if every x E X satisfies

IA)

liplp vp EQ.

As a trivial example we observe that the field element 0 is bounded by every i E J7. We let M(i) stand for the total number of field elements bounded by the idèle i; then M (i) is either a natural number or oo. The mapping X +4 01X

sets up a bijection between the field elements bounded by i and those bounded by at; hence M(i) = M(cci) . § 33E. The "density" M(i)ifill

The purpose of this subparagraph is to find very rough estimates for the "density" of field elements bounded by an idèle. We shall use the Pigeon Holing Principle: if 1 letters are to be placed in h boxes then at or more letters. least one box must receive 33:4. Theorem. Let F be a global field. Then there is a contant D> 0 which depends only on F such that

M (i)

max (1,

ViEJF•

In particular, an idèle can bound no more than a finite number of elements

of F.

Proof. 1) Take a discrete spot q E D and fix it. Let co be the number of archimedean spots in Q; if F is an algebraic function field, this number is 0; if F is an algebraic number field it is finite since F is then a finite extension of its prime field. We claim that D

4°11■1 q

will do the job. So we must consider a typical idèle t E hi and show that it satisfies the inequality of the theorem.


70

2) If we replace i by ai with a E F then neither M(i) nor Pi' is altered. Hence by the Weak Approximation Theorem it is enough to consider an i for which ligi q = 1. Let us do so. If 0 is the only field element bounded by i, then M(i) ---- 1 max (1, Mill) and we are through. So let us assume that there is a set X which is bounded by t and consists of M elements of F with 1 < M < oc; we will be through if we can prove that M Diiiii. The rest of the proof is spent on this inequality. 3) We introduce Dedekind ideal theory at the single spot cf. Put 0 — o (q) and q = m(q). We have X C 0 since X is bounded by the idèle I and liq l q = 1. Form the chain of ideals 0)(I)•-•Dcr)cr+ 1 ) — •,

choose that r

0 for which

(N q)r < M

(Nq)+i.

Now (o: qf) = (N cry by Proposition 33:2. So by the Pigeon Holing Principle, there is at least one coset of o modulo qf which contains two distinct elements a l, a2 E X. Here al — a2 E qr. Hence I al — C(2141 5-

I q 1 cr I = kb —ii j - -. -- M. - - •

But Iiir if p discrete ial — cc2IP 5- 1 4 Nip if p archimedean. Hence by the Product Formula,. 1 = -Mal — agir P

5' 4u) (

H 1iplp)N ctilll

‘P+q

= 4'Nci IliiiiM •

Thus M ..,S. DIN, as we asserted.

q. e. d. 33:5. Theorem. Let F be a global field. Then there is a constant C (0 < C < 1) which depends only on F such that

C ilill <M (i) V i E Jp In particular, an idèle with a big enough volume bounds at least one nonzero field element.

Proof. 1) Since F is a global field it is a finite separable extension of a rational global field Q. Let Z be the ring of integers of Q. Let S denote the set of all finite spots on Q, let co be the infinite spot. Define T C Qp


and T'

71

Pp by

TI IS, T'l I 00 . Then T is a Dedekind set of spots on F by Proposition 31:2 and Theorem 23:1; and T' is a finite set of spots which is either all archimedean or all discrete. Since T is a Dedekind set of spots it introduces Dedekind ideal theory into F. We put 0 = 0 ( T) as usual. Then Z= o(S) Ç 0 ( T) = o. Take a base xi , . . . , xn for F over Q. We can find a non-zero In E Z which is arbitrarily small at any predetermined finite subset of S, and hence at any predetermined finite subset of T; the Product Formula will ensure at the same time that nt is arbitrarily large at oo, and hence at all p E T'. Replace xi,. . xfi by tnxi, . Inx„. Then the new xi, ...,x. still form a base for FIQ; but they also have the following additional properties: Ixi ly 5._ 1, . . 5_ 1 VpET, k11 1,.., IX,n ip 1 VpET'. In particular, X11 . , x. are now in o. We assume that the chosen base

has these additional properties. By the Weak Approximation Theorem we can find a number B E F such that

IBJ for 1 i

n2n+1

n and for all p E T'. Put

C=H

1 . I B41r

This will be our C. Note that IBN, 4> 1 for all p E T'. 2) We have to consider a typical idèle i and show that it satisfies the inequality of the theorem. By the Weak Approximation Theorem there is an a EF which is arbitrarily close to BB/ip at each p E T'; if the approx-

imation is good enough we can achieve

i 1321p< lociplp< 1 134 11, VPEr. in virtue of the continuity of the map

that la ii,tnl y

m. Now pick m E Z such

1 Vp E T.

If we replace by the idèle anti, then neither M(i) nor the volume i is changed. So let us make this replacement. In effect, this allows us to make the following assumption about the idèle under consideration:


72

there is an m E Z such that

liv ir

1 ImB2 11, < Il r< ImB4i p

V p E 7' V p E TI.

3) We must now introduce three new idèles 0, j, t. Once we have done so we will be able to state the idea of the proof and the reason behind the notation. Let us define these idèles by specifying their p-coordinat es : Op =1 VpET, Op = m B j =t

VP E T,

VpET',

jp = mB 2 V p E T' ,

tr . Vp E T, tp = mB 4 Vp ET. The rest of the proof now goes as follows. First get a lower estimate on M(0). Use this in an application of the Pigeon Holing Principle to get a lower estimate for M(j). This gives a lower estimate for M(i) since Orly -5-- lip It, for all p E Q in virtue of the choice of tin step 2). On the other hand f provides an upper estimate on the volume of i since lipi p f tr ip for all p E Q. We therefore obtain a lower estimate on the density. It will be shown that this estimate is greater than C. So much for the idea. Let us now carry out the details. 4) First we want our estimate for M(0). We define X {mi + • - • + m.„ xt, mi E Z with Imiico Suppose we count X. In the number theoretic case we get elements of X if the mi vary through the subset {0, ± 1, ± 2, . . + m} of Z; hence X contains strictly more than elements. In the function theoretic case we do a similar thing, only this time the mi are to be polynomials with degmi degm; here again we find that X contains strictly more than qndeig yit

Iminoo

elements, where q is the number of elements in the constant field of Q. An easy calculation now shows that X is bounded by the idèle 0 (the definition of B is used in this calculation: indeed, B was specifically designed to make this calculation work). Hence we have the desired estimate: M(0) > *1'1 5) Let a be the ideal at T that is defined by Jail,— lip lp for all p Thus a o and by Proposition 33:3

Na= ET

l ir iP •

E T.


73

Now every field element bounded by I) is in o; hence the Pigeon Holing Principle gives at least one coset of o modulo a which contains t field elements Ch2 • • •

that are bounded by with H lipip.

t>

PET

Then C1C1 — °CI) CC 2

21 , • • • , t3tt

°CI

are in a; an easy computation then shows that these t elements are bounded by j; hence

m (i) > Im1:0 H lipip • pET

6) The final calculations. We have

m (t )

114- (j ) > 11(11 >

imB.,,

But Imroo= H *Vial) H 1mIP per PET'

•

Hence

mo) 11111

H

1

PET , 11341P

q. e. d. § 33F. The group of S-units Let S be any non-empty set of spots on the global field F. The set U(S)

{cc

EPI kip= 1 V P E S}

is a subgroup of called the group of S-units; the elements of this group are called the S-units of F. We shall often write u instead of U(S). If S happens to be a Dedekind set of spots, then u (S) coincides with our earlier definition of the group of units of F at S. We shall call u (?) the group of absolute units of F. This group has the following simple description. 33 : 6. The grout of absolute units of a global field is cyclic of finite order. It consists of all the roots of unity in the given field. Proof. Suppose we have shown that u (D) is cyclic of finite order. Then every element of u (D) has to be a root of unity; conversely every root of unity in F must clearly also fall in u (D). Hence the second part

of the proposition follows from the first. It therefore remains for us to prove that u(Q) is a finite cyclic group.


74

Now u (D) is bounded by the idèle (1),? ED . Hence by Theorem 33:4 u (D) is a finite subgroup of P. But every finite subgroup of the multiplicative group of an arbitrary field is cyclic. Hence u (D) is finite cyclic. q. e. d. 33 : 7. Let F be a global field and let K be a positive constant. Then the number of discrete spots p on F such that Np K is finite. Proof. We can assume that K is a natural number. Take K + 1 ccE with aco = O. By the Product distinct elements of F, say ao, Formula there is a finite set T of spots on F such that loci — ; Ir . 1 for 0 I <j K whenever p E DE — T. Let us suppose that T has been chosen large enough to include all archimedean spots on F. Our proposition will be established if we can prove that Np > K for all p E Q,- T. Suppose to the contrary that there is a spot p E Dy, - T such that Np 5 K. Now o(p) has exactly Np cosets modulo m (p) ; and cco, ,OEK fall in o (p); hence there is at least one coset of o (p) modulo m(p) which contains two distinct ces, say oc i and;; this implies that 04— cc5 Em (p). But this is impossible since loci — ; II,. 1. So we do indeed have Np> K for any p E T. q. e. (1. In practice S will be a proper subset of Q consisting of almost all discrete spots in D. (For example S could be the set of all discrete spots

on an algebraic number field.) When we are in this situation we shall let s stand for the number of spots in D — S; so here 1 s < oo. 33 : 8. Let S be a set of discrete spots on the global field F. Suppose that s < oo. Then for eachgED — S S consists of exactly s spots with 2 there is an S-unit e g such that

(S cf) . leglq > 1, isq lp < 1 V p E Proof. 1) We put N q 1 if q is archimedean (N q is already defined if q is discrete). Let C stand for the constant of Theorem 33:5: then any idèle i with C lit!' 1 will have to bound at least one non-zero field element. Let C be chosen so small that there is at least one spot p in S for which Np 5 NqIC. Thus the set W .{p ESINp 5 N qIC}

is non-empty; and this is a finite subset of S by Proposition 33:7. Put X

EF1Tv-czi-5 . 14,5 1 V p ES} .

Note that for each x E X and for each pES—W we have C Np < -j I, leg lp _L

with M and N symmetric matrices over F. This means that V is a

90

Part Two. Abstract Theory of Quadratic Forms

quadratic space with, a base in which V ( 1-1-1O) .

Similarly V J •j with all cei in F means that V has a base x1,.. . , xn in which Q(x1) for 1 n and B (xi, xi) = 0 for 1 i < j n. This prompts us to make the following definition: a base x1 , . . xn for the quadratic space V is said to be an orthogonal base if .

B (xi, xi) = 0 for

-

1S i<j‹ n.

42:1. Every non-zero quadratic space has an orthogonal base. Proof. If 42 (V) = 0 we are through. Otherwise we can pick x E V with 42 (x)+ O. Take a base x, x2, . . . , x„ for V. Then B(x, x2) B(x, x„) X, X 2 X, . . • X n (x) is still a base for V, and the last n — 1 vectors span an (n — 1)-dimensional subspace W for which B (x, W) = O. Now apply induction to W. q. e. d. and W= W, Consider two quadratic spaces V with the subspaces Wi pair-wise orthogonal (do not necessarily assume that the Wit give an orthogonal splitting for W). Suppose that a representation o•: V1 is given for each I. Then we know from linear algebra that there is a unique linear transformation a of V into W which agrees with each ai on V1. In fact the reader may easily verify that a is also a representation a: V W We write this representation in the form -

12(x)

O'

O' j ' 4 ' ar The important case is where V == W, all Vi = W,E, and all a1 are in 0 (V 1) ; in this event ai l • • • I (4E 0 ( 11; if we also take r = r1 I • • • I ; with all ri E 0 (V1) we obtain the following rules: err — ((VI) • • j (arrr) a--1 =aril•-•laT1 det a = deter1 det 0 2 . . . det a,.

where det is the determinant of the linear transformation in question. § 42B. Orthogonal complements Let U be a subspace of the quadratic space V. By the orthogonal complement U* of U in V we mean the subspace U*. {x

E VI B (x, U) = .

Chapter IV. Quadratic Forms and the Orthogonal Group

91

We define the radical of V as the subspace rad V E V 1 B (x, V) = O}. Thus rad V = V*. We shall say that V is a regular quadratic space if rad V= O. 42:2. Let the quadratic space V be a sum of pair-wise orthogonal subspaces, i.e. V = V 1 -1- • • • V, with B (V i,V i) = 0 /or 1 i <j r. Then rad V = rad V1 + • • • + rad V,..

Proof. Take a typical x E rad V and express it as x = Ex A with each xA E V. Then for each i (1 i r) we have B (xj, V = B(1

V1) Ç B (x, V) = O,

so each xi is in Tad Vi, hence x E E rad Vi. Conversely, if we take x= ExA with each xl E rad VA we have B (x, V) S B

171) + - • • + B (x T, V ,.) 0 ,

q. e. d. so x E rad V. 4 2 : 2 a. V is regular if and only if all the V are regular. 42:2b. I/ V is regular, then V = V 1 1- • • 1 V. Proof. It is only necessary to show that the sum is direct. So let us take x1 + • • • + X 0 with xi E V i for 1 i r. Then B (xi, V 1).

0 = B (x1 + • • • + x r, V i)

q. e. d. Hence xi E rad Vi = O. 42:3. The quadratic space V is regular if and only if dV O. Proof. Take an orthogonal base V (Fx1) - • • ± (Fx,i) . Then V is regular if and only if all F x1 are regular. But the line F xi is regular if and only if Q (xi) + O. Hence V is regular if and only if Q (xi) Q (x2) • - • Q (x.)+ O •

q. e. d. This quantity is simply the discriminant d (x11 . . . , x,„). 42:4. Let U be a regular subspace of the quadratic space V. Then U splits V, in fact V U J U*. And if V = U 1W is any other splitting, then W= U*. Proof. Take an orthogonal base for U: U (F xl)

• - • j (F x 2,) .

Since U is regular, all Q(x i) + 0 . A typical z with

Y—

B (z, la)

E V can be written z =y + w

B (z, x„)

Q(z1) x1+ • • • + Q(z) w=z—y.

92


Here y is clearly in U; and an easy computation gives B (w, xi) = 0 for h 1 I p, hence w E U*. This proves that V= U U*. Now Ur U* = rad U = O. So V = U U*. Hence V=U1U*. If V=U±W, then W U. Dut we have just proved that V=U±U*. q. e. d. Hence dim W = dimU*. Hence W = U. Let us show how the given bilinear form B naturally determines an isomorphism 0 of V onto the dual space V', i. e. onto the set of linear functionals on V, when V is regular under B. Fix a vector x in V for a moment. Then B (x, 5') defines a linear functional 92.: V -4- F since B is bilinear. Hence tp„ is in the dual space V'. Now free x. This associates a linear functional 93„ with each x E V, so we have a mapping defined by 0(x) = 9).. The bilinearity of B again shows that 0 is linear. And the regularity of V shows that 0 is injective and hence bijective. Hence we have defined the 0 we were after. We know from linear algebra that every base . , x„ for V has a dual base in V'. If we let 0 carry the dual back to V we obtain a base . . . , y„, for V with the property that B (xi, yi) = 6i i (Kronecker delta). x„ Moreover this base is unique. We shall call it the dual of the base with respect to the bilinear form B. 42:5. Example. Let V be a regular quadratic space, let N be the matrix of V in the base . . x„, and let M be the matrix of V in the dual base y,. . . , yn. We claim that M = N -1. For by § 41B we have M = 'TNT where T = (lb ) is the matrix determine& by yi = f tii xi. A

On the other hand we have E B (xi, h.) tau B (xi, y = A

so that NT is the identity matrix. Hence M ='T. Hence M =MNM.

So M = N -1 as asserted. 4 2 : 6. U is an arbitrary subspace of a regular quadratic space V. Then dim U dimU*. dim V , U** =U Proof. Consider the isomorphism 0 of V onto the dual V'. It is easily seen from the definition of 0 that 0 U* is the annihilator of U in V'. Hence dim U*= dim U* . dim V - dim U . As for the second part, we have U U** by definition of U*, hence q. e. d. U == U** by a dimension argument. .


93

§ 42C. Radical splittings Consider the radical rad V of the quadratic space V and let U be any subspace of V for which V = U rad V. Then clearly V — U rad V and we call this a radical splitting of V. Obviously U is not unique unless V is regular or V = rad V, but we shall see in Proposition 42:8 that it

is always unique up to isometry. The equations rad V = rad U

rad (rad V) = rad U

rad V

imply that rad U = 0, and so U is regular. 42:7. Let a: V —>-- W be a representation of quadratic spaces.

If V is

regular, then a is an isometry.

Proof. Take x in the kernel of a. Then B (x, V) = B (ax, aV) = O .

Hence x Erad V. Hence x = O.

q. e. d. 42:8. Let V = U rad V and V1 = U1 I rad V1 be radical splittings of the quadratic spaces V and V1. Then (1) . V (2) V

Vi if and only

u

U1,

V1 if and only if U U1 with rad V

rad

Proof. (1) Suppose we are given a representation a: U U1. Define a representation T : rad V rad V1 by putting Tx .= 0 for all x in rad V. Then a 1_ r is a representation of V into V1. Conversely let a: V V1 be given. Define the linear transformation U U1 by putting ax

z

with x E U, px E U1,z E radV1, Then Q (x) = Q (ax) =Q x)

and so g) is a representation. (2) Let a: V >—). V1 be the given isometry. Clearly a carries rad V to rad V1 so that rad V rad V1. And we know from the first part that U U1. But every representation of a regular space is an isometry by Proposition 42:7. Hence U >-- U1. Hence U U1 . The converse is clear, q. e. d. One of the main problems in the theory of quadratic forms is the problem of finding invariants that will fully describe the equivalence class of a given form. The interpretation of this in the language of vector spaces is the following: find invariants that will determine whether or not two quadratic spaces over the same field are isometric. This theory has been fully developed over certain fields, notably over the field of real numbers and over the local and global fields of number theory. We shall take this up in Chapter VI. For the present we mention that by Proposition 42:8 it is enough to consider regular quadratic spaces only.

94


For this reason we shall usually assume that all given quadratic spaces are regular. Of course this does not eliminate all mention of spaces that are not regular since, for instance, the subspaces of a regular space are not necessarily regular. § 42 D. Isotropy Let x be a non-zero vector in the quadratic space V: we call x isotropic if Q (x) = 0, we call it anisotropic if Q (x)+ O. Let V bea non-zero quadratic space: we call V isotropic if it contains an isotropic vector, we call it anisotropic if it does not contain an isotropic vector, we call it totally isotropic if each of its non-zero vectors is isotropic. All these definitions will apply to non-zero vectors and non-zero spaces only. We have V totally isotropic if and only if V is non-zero with Q (V) = O. For any quadratic space V we have Q (V)

0 (2). This follows by inspection of the defining matrix of a hyperbolic plane. (2) => (3). Take an isotropic vector x in V and extend it to a base for V. The matrix of V in this base has the form

19,2/ E F.

So d V — —fie. But V is given regular, hence i3 E F . But —fie and — 1 are equal in F/F2. Hence d V = 1.


95

(3) (1). Q (V)4. 0 since V is regular. Take a non-zero field element a in Q (V) and pick x E V with Q (x) = a. Then F x splits V since it is regular, hence V = (F x) J (Fy) for some y E V. Now the information d V = —1 implies that — Q (x) Q (y) is a non-zero square, so we can assume after a suitable scaling of y that Q (y) = — a. Then ..

V=F( x+ 2 1+F( x—Y )

The matrix of V in this base is easily seen to be ri 01 ). Hence V is a hyperbolic plane. q. e. d. 42:10. Every regular isotropic quadratic space is split by a hyperbolic plane. Hence it is universal. Proof. Let x be an isotropic vector in the given space V. Since V is regular there is a y V with B (x, y) + 0. Then U Fx Fy is a regular

binary isotropic space, hence it is a hyperbolic plane. Being regular it must split V. q. e. d. 42:11. Let V be a regular quadratic space, let a be a scalar. Then a

E Q (V) if

and only if < — a> ± Vis isotropic.

Proof. We have to consider W.Fz±V with Q (z) = — Œ. If there is an x in V with Q (x) = a, then Q(z x) = 0 and W is isotropic. Conversely suppose that W is isotropic. If V is isotropic it is universal by the last proposition and we are through. If V is not isotropic we must have a non-zero scalar )3 and a non-zero vector y E V such that Q (fl z y) O. But then —fi2a Q(y) O. So a = Q (y)1 E Q (V). q. e. d. 42:12. Let U be a regular ternary subspace of a regular quaternary space V. Suppose V has discriminant 1. Then V is isotropic if and only if U is isotropic. Proof. We must take an isotropic V and deduce that U is isotropic. Write V = U I with a E F. Then —aEQ(U) by Proposition 42:11. Hence U has a splitting U = P J < — a> with P a plane. So V= Pj. 1_ ,

hence 1 dV = —dP. So dP = —1 and P is a hyperbolic plane by Proposition 42:9. Hence U is isotropic. q. e. d. 42: 13. V is a regular quadratic space, U is a subspace with Q (U) = 0, and x1 . . . , is a base for U. Then there is a subspace H1 ± • • ± H, of V in which each H i is a hyperbolic plane with xi E Hi. Proof. If r . 1 we takeyi E V with B (x1, y1) + 0 and put H1 . F Then H1 is a hyperbolic plane with the desired property_ Let us proceed by induction to any r> 1. Put U,._ 1 . Fx1 + - and Ur = U. ,

96


Then U,._1 C U,., so U: C U* 1 . Pick y E U* 1 — U and put H,.= Fxr + Fy,.. Then B (xi, yr) = 0 for 1 5 I r — 1, hence B (x,., y,.)+ O. Hence H,. is a hyperbolic plane containing x,.. Write V = H,. 1 H. Then H,.S U....1 since x,. E U,?_ 1 and *y,. E hence U,._ 1 S H. Apply the inductive assumption to U,._ 1 regarded as a subspace of H. This gives j_ • • •

with xEH, for 1 desired properties.

.11,*

I

i < r — 1. So Hi I_ • • • d_ H- r -1

H,. has the q. e. d.

§ 42E. Involutions and symmetries The identity l v of the ring 4,(V) is by definition that linear transformation which leaves every vector of V fixed, i. e. 1 v (x) = x for all x in V. And the negative — l v of 1 v reverses every vector of V, i. e. — l v (x) = x for all x E V. Clearly both ± l v are in On (V) since Q (±x)

= Q (x) for all x in V Recall from linear algebra that a linear transformation a is called an involution if a2 = 1 v 42:14. V is a quadratic space and a E On (V). Then a is an involution if and only if there is a splitting V = U 1 W for which a = l u j l w. Proof. If a = — l u j l w, then a2 = 1 u 1_ 1 w = I v and a is an involution. Conversely suppose that a is an involution. Then o 2 = 1(= 1 v). Consider the linear transformations a — 1 and a ± 1 and put U.(a-1)V and W = + 1) V. A typical x E V has the form x — (a— 1) + (a + 1) ÷ EU

W.

So V =U+ W.A typical y in U has the form (a — 1) x, so ay = a2x ax= x ax= —y;

hence ay = —y for all y in U; similarly az = z for all z in W. This shows first of all that U n W = 0, so V = U e W. And secondly it implies that B(U, W) = 0 since for typical y E U and z E W we have B (y , z) = B (oy , a z) = B (— y , z)

O. Hence V= U _LW. And thirdly it says that a is —1 i. e. B (y, z) on U and lw on W, so a = — l u j 1 w. q. e. d. 42:14a. W is the set of vectors left fixed by a, U the set of vectors reversed by a.

The most important involutions on a quadratic space V are the symmetries which we now define. Fix a vector y E V with Q(y)+ O. Define a mapping T.,: V V by the formula t x x

2 B (z, y)

(Y)

y.


97

Then the following facts can be verified directly: T y is linear, it is an involution, it is a representation and hence an element of O, (V), it reverses every vector in the line F y, it leaves every vector in the hyperplane (F y)* fixed. We call rw the symmetry with respect to the vector y or with respect to the line F y; note that Ty = rse if and only if F y and Fy' are the same line. In particular there are exactly as many symmetries as there are anisotropie lines in V. It is easily seen that crz11cr -1 —

rev

whenever a E 0„(V). And every symmetry, being an involution, is its own inverse. 42:15. Example. Let H = Fx Fy be a hyperbolic plane with Q (x) = Q (y) = O. Consider a vector z = x — ay for some oc j= O. Then z is anisotropie, so we can form the symmetry;. An easy computation gives • "G X =

ay , -czy = a - lx .

Now every anisotropic z falls in the line F (x — ay) for some ce E F. Hence every symmetry has the above action on x and y. In particular, every symmetry on a hyperbolic plane intercha.nges the two isotropic lines. Conversely, if a E 0 2 (V) interchanges the two isotropic lines it must be a symmetry: ax = ay with oc E F, hence the equation B(x, y) B(ax, ay) yields ay = cc-lx, hence a = rx....cgy. Finally we note that the action of a a in 02 (V) which is not a symmetry is described by a x ax ,

ay = a-ly ;

hence every such a is of the form a

Tx—yTx— cc v -

§ 42F. Wift's theorem 42:16. Theorem. Let U and W be isometric regular subspaces of quadratic space V. Then U* and W* are isometric.

a

Proof. 1) First suppose that U and W are lines, say U = F x and W F y with Q (x) Q (y) + O. Then Q (x +y) + Q (x y) = 2Q (x) ± 2Q (y) = 4Q (x) ,

hence either Q (x y) or Q (x — y) is not zero. Replacing y • by —y if necessary allows us to assume that Q (x — y) O. We may therefore form the symmetry T„_ • We have 2 B (x, x — y) Q(z O'Meara, Introduction to quadratic forms

y)

y) 7

98


But Q (x — y) = Q (x) Q (y) — 2 B (x, y) = 2 Q (x)

2B (x, y)

= 2B (x, x — y) . W*. In other words U* and W* are Hence r,x= y. Hence isometric when U and W are lines. 2) Now the general case by induction to dim U. Since U and W are given isometric we can take non-trivial splittings U U1 1_ U2 and W W1 1 W2 with Urstz W1 and U2'. " W2. Then the inductive assumption says that U2 I U* is isometric to W2 I W* , hence there is a splitting U2 j U* = X j Y With X W2 and Y W*. But then X W 2 U2, U*. and the inductive assumption again says that Y U*. Hence q. e. d. 42:17. Theorem. Let V and V' be regular isometric quadratic spaces, let U be any subspace of V, and let a be an isometry of U into V'. Then there is a prolongation of cx to an isometry of V onto V'. Proof. Write U = W j rad U and let x11 . , x,. be a base for rad U. By Proposition 42:13 there is a subspace

H = H1 j - • • j H,

of the quadratic space W* in which each Hi is a hyperbolic plane such that xi C Hi. Since H is regular it splits W*, hence there is a subspace S of W* such that V=H±S±W. Put U' — ci U, W' = a W, and x = o.xi for 1 rad U'

< r. So

a (rad U) = Fx1 + • • • + Fx; .

And U'

W' j rad U'.

We can repeat the preceding construction on this arrangement to obtain a splitting V' H S 1_ W' in which H' Hi ± • • - j_ H; '

where the H are hyperbolic planes in which 4, E H. Now there is clearly an isometry of H onto H' which agrees with a on each xi , hence on rad U. Also the given o. carries W to W'. Hence there is a prolongation of cx to an isometry o. of H j W onto H' j W'. An easy application of Theorem 42: 16 now says that S is isometric to S'. Hence there is a prolongation of o• to an isometry of V onto V'. q. e. d.


99

One can use the last two theorems to attach an invariant called the index to a regular quadratic space V. Consider a maximal subspace M of V with the property Q (M) = 0, let M' be another such subspace, and suppose for the sake of argument that dimM < dimilr. Then there is an isomorphism a of M into M', and this isomorphism is in fact an isometry since Q (M) =Q (M') 0, hence there is a prolongation of a to an isometry a of V onto V by Theorem 42:17. Then M M' and so dirnM = dim M'. We But Q (a --1 M1 = O. Hence M = have therefore proved that all maximal subspaces M of V with the property Q(M) = 0 have the same dimension. This dimension is called the index of V and is written ind V. There is another way of looking at the index. If we keep splitting off hyperbolic planes in V we ultimately obtain a splitting V= I_ • • _L H,. Vo with O < 2r < dim V in which each Hi is a hyperbolic plane and Vo is either 0 or anisotropic. By Theorem 42:16 we see first that r does not depend on how the splitting is performed, and then that Vo is unique up to isometry. An easy application of Proposition 42:13 shows that r is actually the index of V. In particular this proves that r ind V satisfies the inequality < 2 ind V < dim V . We call V a hyperbolic space if 0 ax x isotropic in J,

(*)

and we deduce that n 4, n is even, and a is a rotation. It is evident that we cannot have n — I. If we have n 2 we pick an anisotropid x in V; then the condition (*) implies that x and ax are independent with discriminant d(x, ax) equal to 0; this denies the regularity of V. We therefore must have n 3. We shall now deduce from condition (*) that n is even and a is a rotation. We claim that Q (a x x) = 0 holds for all x E V, not just for the anisotropic ones. To see this we consider an isotropic y in V; then there is a hyperbolic plane which contains y and splits V. hence there is a vector z with Q (z)+ 0 and B(y, z) 0. Hence Q(y + E.2)+ 0, hence Q (y + s z) (y + z))

0,

Q (a z z) = 0 ,

by the condition (*), hence Q (ay y) + 2e B (oy — y a z z)

0.

All this holds for any E EF in particular for E = +1. Do it for E ± I, then for E = - 1, then add. We obtain Q (ay — y) = 0 as we asserted. Hence the space W = — 1) V satisfies Q (W) = 0. Now for any x in V and any y E W* we have ,

B(x, ay — y) = B(ax, ay — y) — B(ax — x, ay — y) B(ax, ay — y) = B ((Ix, ay) — B (ax, y) = B(x, y) — B (a x, y) — B (a x x, y)

= 0. I A two or three line proof will show that a is a product of symmetries, even that it is a product of at most 2 n — 1 symmetries. The difficulty is in showing that a is a product of at most n symmetries.


Hence cry — y is in rad V. But V is regular. Hence cry Hence Q (We) = 0 by condition(*). Hence

103

y for ally in W.

W Wik-S W**. W.

Hence W Wt. Hence n is even and dim W = n/2, in other words V is a hyperbolic space and W is a maximal totally isotropic subspace. But a is identity on W = W. Hence a is a rotation by Proposition 43:1. 2) Now we can prove the theorem. The proof is by induction on n. For n = I the result is trivial, so let n > 1. Suppose that there is an anisotropic vector x in V such that ax = x. Then the restriction of a to the hyperplane U orthogonal to F x is an element of 0„_1 (U). By the inductive assumption this restriction is a product of at most n — 1 symmetries taken with respect to lines in U. Now each of these symmetries has a natural prolongation obtained by taking the symmetry on V with respect to the original line in U. The product of the prolongations taken in the original order agrees with a on U, and also on F x where both a and the product are identity, hence they agree on V. Hence a is a product of at most n — 1 symmetries when it leaves an anisotropic vector fixed. Next suppose that there is an anisotropic vector x such that Q (ox x)-+ O. Form the symmetry Tax _ x. Now T_ x o- is easily seen to leave x fixed. And x is anisotropic. Hence To., x cr is a product of at most n — 1 symmetries. Hence a is a product of at most n symmetries. Hence any a which does not satisfy condition (*) of step 1) is a product of at most n symmetries. Consider a a which satisfies (*). Then n is even and a is a rotation. Fix a symmetry T of V. Then Ta is a reflexion, hence it cannot satisfy (4 ) , hence Ta is a product of at most n symmetries, hence a is a product of at most n ± 1 symmetries. Buta cannot be a product of n + 1 symmetries since a is a rotation and n 1 is odd. q. e. d. Hence a is a product of at most n symmetries. If a is any isometry of V onto itself we call the subspace {x E Via x = x} the fixed space of a. For example 0 is the fixed space of —1 and V is the flied space of I v. The fixed space of a symmetry is the hyperplane orthogonal to the line used in defining the symmetry. We must take care not to confuse the fixed space of a with the subspaces left fixed by a; for instance V is always left fixed by a, but it is the fixed space of a only when a is the identity. 43 : 3a. Corollary. If a is a product of r symmetries, then the dimension of its fixed space is at least n — r. Proof. Express a as a product a . . Tr of symmetries and let U1 be the fixed space of ; for 1 j r. Then U1 n • n Ur is contained

104


in the fixed space of a. It is therefore enough to prove that if U 1, . . . , Ur are any r hyperplanes in a vector space V, then

dim (U1 (-1.• • n Ur) n — r. We do this by induction to r. For r = 1 it is the definition of a hyperplane. For r> 1 we have dim ( n • • • n U,) = dim (U1 • - - U r _1) Ur n • • • n Ur -1) + U,)

dim(( (n

r + 1) + (n — 1) — (n)

=n— r.

q. e. d.

43:3b. Corollary. Suppose a can be expressed as a product of n symmetries. Then it can be expressed as a product of n symmetries with the first (or last) symmetry chosen arbitrarily. Proof. Write a as a product of n symmetries, say a = •t1 . . . Tn . Let T be an arbitrary given symmetry. Then by the theorem we can express -cc as a product of at most n symmetries, hence

a=

. . . 74

with r < n I. Here det will be (-1)*. On the other hand det a is (— 1)n by hypothesis. Hence r and n have the same parity. In particular, n. If r < n, put an even number of T'S at the end of the above r expression for a to obtain . a = T-4 This allows us to choose the first symmetry in an arbitrary way. Similarly with the last. q. e. d. § 43 C. Binary and ternary spaces The orthogonal groups of binary and ternary spaces have certain special properties that can be used in the general theory. (Incidentally the theory of the binary orthogonal group is quite different from the theory in higher dimensions.) For instance if V is a regular binary space every reflexion is a symmetry by Theorem 43:3; so the number of rotations, which is always equal to the number of reflexions, is also equal to the number of symmetries and hence to the number of anisotropic lines in V. As another example let us show that (4- (V) is commutative. To see this first consider a typical symmetry T and a typical rotation a. Then ra = is is a reflexion and hence a symmetry. So TTTi T

TI T

a -1 .

Hence for any other rotation e we have apa -l= rri Q Ti T = T ,0 -1 T =

e.

Chapter 1V. Quadratic Forms and the Orthogonal Group

105

So 024 ' is commutative. Later we shall see that O is not commutative when n 3. dim V 3. Let d be the 43:4. V is a regular quadratic space with 1 dimension of the fixed space of an isometry a of V onto V. Then a is a product of n — d, but not of less than n — d, symmetries of V. Proof. It is clear that a is not a product of r < n --- d symmetries for if it were, the fixed space of a would have dimension > n — r>d. We must therefore prove that a has at least one expression as a product 1 the result is clear. of n — cl symmetries. For n Next consider n — 2. If cl — 0, a. is neither the identity nor a symmetry, hence it is a product of two symmetries as required. If d — 1 the fixed space is a line; it cannot be an isotropic line by Proposition 43:2; hence the fixed space is a regular line, hence a is the symmetry with respect to the line orthogonal to this fixed space. If cl = 2, a is the identity. This finishes the case n = 2. Finally n = 3. If cl = 0, a cannot be a product of less than three symmetries. If cl = 1. the fixed space is a line; here it is enough to prove that o. is a rotation; suppose if possible that a is not a rotation; then a is a refiexion and so a is a rotation; hence a keeps a vector fixed, hence a reverses a vector; so a keeps one vector fixed and it reverses another; these two vectors are clearly independent ; hence a2 keeps every vector in a certain plane fixed; but a2 is also a rotation; hence a2 I y by Proposition 43:2; therefore a is both a reflexion and an involution ; by Proposition 42:14 a will either be — 1 v or a symmetry; neither is possible since the fixed space of a is a line; hence a is indeed a rotation, hence a product of two symmetries. If cl --- 2 the fixed space is a plane, hence a regular plane by Proposition 43:2; hence a is a symmetry with respect to the line orthogonal to this plane. The case cl = 3 is trivial , q. e. d. The last proposition shows , that for ternary spaces the fixed space of any rotation other than the identity is a line. This line is called the axis of the rotation. We regard every line as the axis of the identity rotation. The set of all rotations with given axis is a subgroup of 04' (V). lithe axis in question is an anisotropic line L, then there is a natural isomorphism (obtained by restriction) of. the group of rotations having L as axis onto the group a-,4- (L*). 43 : 5. An isotropic line is given in a regular ternary space. Then the multiplicative group of rotations having this line as axis is isomorphic to the additive group of the field of scalars.

Proof. Let Ex be the axis in question. The quadratic space V is split by a hyperbolic plane containing x, hence we can find vectors such that

V — (F x ± Fy) j Fz

106


with Q (x) =-- Q (y) = 0 and B (x, y) = 1. Let U denote the plane U = F x Fz; this plane has radical Fx. In order to establish the isomorphism of F onto the group of rotations with axis F x we consider a typical a in F and we define the linear map by the equations :u

e

X=X,Z

x z

This map is an isometry by Proposition 41:1; hence there is a prolongation of p to an isometry e of V onto V by Witt's theorem. Moreover this prolongation is unique by Proposition 43:2. We have therefore associated a unique isometry of V onto V with the field element cc; we write this isometry as 0 its defining equations are .

;

eccx .7c , Clearly ea+ p= O cc Qp. In particular Elœ . 042 so that e cc is always a rotation; pc, gives an and Fx is the axis of this rotation. Hence the map ac isomorphism of F into the group of rotations with axis Fx. It remains for us to prove that this isomorphism is surjective. Consider a typical a E 0: (V) With axis Fx. Write Or

X=X

ay =

ax+by+cz az =dx+ey+ fz. Then B(ax,ay) =1 so that b = 1. And B(ax,crz)--- 0 so that e = O. But det a =--- 1. Hence f =-- 1. Hence a is a rotation whose action on U is given by ax= x, az—dx-Fz. Hence a and pa agree on U. Hence they are equal by Proposition 43:2. q. e. d.

43:5a. Every rotation on an isotropic axis is the square of a rotation on the same axis. There .is at least one rotation on this axis which is not the identity. § 43D. The commutator subgroup Sin of On We let 1-27„ (V) or simply 14, stand for the commutator subgroup of the orthogonal group On (V) of a regular quadratic space V. Clearly

The groups 0,127, have been. defined for non-zero regular spaces only; any reference to these groups will carry the implicit understanding that the underlying space is non-zero and regular; the same understanding will be made with the group Zn in § 43E.

Chapter 1V. Quadratic Forms and the Orthogonal Group

107

4 3 : 6. Q the square of every element of On . It is generated by the commutators of the form Tx Tv Tx T y = Tx T„Tx--1 T— y 1 where Tx and Ty are symmetries. In particular it is generated by the squares of elements of O;.

Proof. Let G be the subgroup of On that is generated by the commuTax , hence tators T x T y Tx T y . For any a in On we have CrTx a (TxTy Tx Ty)

a

'VG

xTcy yraxTay

hence G is a normal subgroup of O. We may therefore form the quotient group 0,,IG. Let bar denote the natural map of On onto 0/G. Then for typical symmetries Tx and T y we have Tx iy -T-x -Fy ,= 1, hence fx T-y — i,. express them in terms of symSo if we consider typical a, o E O metries by the equations Cr =7 T1 . . . Tm

e = Tm+i ... Tr ,

we have 56

-

• • • fm .7m +- 1 • • • 1:r

fm -H. • • • 'Tr fi • • • Tm =

Hence 07 IG is commutative. Hence Q of G. Hence G = Q. Finally we have 52

= "T-1 • • 7. fi • Ï.,72

G. But G c Q

• • Tm Tm

65 • the definition

1•

Hence a 2 E G , f2n. Thus [4 contains the square of every element of O. But Q generated by the squares (Tr)2 by definition of G — Q. generated by the squares of all elements of Hence Q q. e. d. also the commutator subgroup of when n 43:7. Q Proof. Let Dn+ be the commutator subgroup of Ont at least until we have proved that it is equal to Q. Every commutator of O is a commutator of 0„ so Dn+ Ç Q. We must reverse this inclusion relation. Since generated by commutators of we proved in Proposition 43:6 that Q the form Tx Ty Tx Ty where T x and Ty are symmetries of V, it is enough to prove that every such element Tx T,T x Te, is in Q. We claim that F x Fy is contained in a regular ternary subspace T of V. If Fx ± Fy is regular this fact is clearly true. If F x Fy is not regular its radical is a line Fz, hence Fx Fy = Fx Fz. Then there is a hyperbolic plane H which is orthogonal to Fx and contains Fz. So T F x j. H is the required ternary space. In either case we have our T, hence a splitting V=TW be the symmetries of T For the rest of the proof we let Fx and with respect to x and y, and we let 1 denote the identity of 0,2_3(W).

108


Then To; Tv rx -tv =

C1; 7

11 17

w'r v) -I- 1

= ( — Fx ± 1 ) (— iv I 1 ) ( — Yx ± 1 ) (— fli ± 1 )

Now — rt is a rotation on T since ix is a symmetry, hence (-1;± 1) is a rotation on V. And (—.T. ± 1) -1 . (— Yx I 1). The same with (— fv 1 1). Hence T x T y To rt, is a commutator of elements of 0, hence it is in .Q: . q. e. d. 43:8. Remarks. When n. 1 the group 01 is a group of two elements so its entire structure is trivial. If n -- 2 we know that Ot is commutative so that its commutator subgroup is i; we shall see in Corollary 43:12a that Q2= 1 if and only if V is a hyperbolic plane over a field of three elements; in particular, Proposition 43:7 is only exceptionally true for binary spaces. We know that the commutator subgroup Q2 of 02 is generated by the set of squares of all rotations; but this set is a group since OI is commutative; hence Q2 is the set of squares of all rotations; symbolically, Q2= (0 ) 2. §43E. The center Zn of On The symbol Z ? (V) or simply Zn will denote the center of the orthogonal group of a regular quadratic space. In this subparagraph we assume that a regular n-ary quadratic space V is given as the underlying space and that On, 0:,(2„, 4 refer to the orthogonal group O n (V) of V. 4 3 : 9. Let ci be an isometry of the regular space V onto itself which leaves ,

every line fixed. Then a . ± l v •

Proof. Fix a vector x in V with Q(x)+ O. Then ax = ax by hypothesis, and a2 Q (x) = Q (crx) = Q (x) so that a = ± 1. If y falls in F x we have cry ----- ay. Otherwise x and y will be independent. We have ay — fly for some i8 EF and we have to prove that i8 = cc. Now a (x + y) = y(x+y) for some y E F. Hence ax+ Ay — ax + ay — a(x + y). yx + yy.. Hence a = y = 13. q. e. d. 4 3 : 10. Let V be a regular n-ary isotropic space with n 3 and let Cr be an isometry of Y onto itself which leaves all isotropic lines fixed. Then ci= ± 1 v .

Proof. Since V is isotropic it is split by a hyperbolic plane, say V — H ± W with H the hyperbolic plane. Note that H, indeed that any hyperbolic plane in V is left fixed by a since it is spanned by two isotropic lines each of which is left fixed. Our first claim is that a leaves every line in W fixed. We need only consider the line Fz in W with Q (z) + O. Now H is universal, so it represents — Q (z), hence there is a hyperbolic plane containing the vector z and contained in the space H I Fz; this plane is left fixed by a, hence crz falls in it, hence az falls in H IFz. But


109

crH — H. Hence crz, being orthogonal to aH — H, falls in Fz. Thus our first claim is established. So Proposition 43:9 tells us that a is either l w on W. We can replace the given a by a if necessary, so we can assume that a z = z for all z in W. Let O denote the restriction of a to H. It is enough to prove that d 1H. Now o is in 02 (H); and d does not interchange the two isotropic lines of H, so O is a rotation (see Example 42:15). It therefore suffices to find a single non-zero vector of II which is left fixed by d, since the fixed space of a rotation of H is either 0 or H itself. Pick an anisotropic vector z in W and then an x in H with Q (x) — Q (z). Then x + z is isotropic, so a (x + z) = a (x + z) for some cc in F. Hence —

ax

a z = cr(x+ z)— ax + az = ax+ z.

q. e. d. Then ax+ (oe 1) z = crx E H, hence cc= 1 and crx x. 4 3 : 11. Let V be any regular n-ary quadratic space, other than a hyperbolic plane over a field of three elements, and let a be an isometry of V onto itself which leaves all anisotropic lines fixed. Then a = ± 1 v. Proof. The case n = 1 is trivial so assume n 2. If V is anisotropic we are through by Proposition 43:9. Hence assume thal V is isotropic. First we do the case n = 2. Here V is a hyperbolic plane. If a leaves the two isotropic lines of V fixed, then it leaves all lines fixed and we are through. Suppose if possible that a interchanges the isotropic lines of V. Then a is a symmetry by Example 42: 15, say with respect to the line Fx. Let Fy be the line orthogonal to Fx. Consider any anisotropic vector of the form x + ay with a E F. Then —

ax=—x, ay

for some

y ,(x + ay) ---- y (x + ay)

E F. Hence y x + y ay = cr (x + cey) =

—

x+ ay .

Hence by comparing coefficients a O. In other words Fx and Fy are the only anisotropic lines in V. But there are at least four anisotropie lines in a hyperbolic plane over a field of five or more elements. Hence our assumption that a interchanges the isotropic lines of V is untenable. Hence a = ±1 v. There remains the general case with n 3. Every regular plane in. V is spanned by two anisotropic lines, hence it is left fixed by a. If we can show that every isotropic vector x falls in two distinct hyperbolic planes, then ax will fall in each of the planes, hence in the line Fx; this will imply that every isotropic line, hence every line, is left fixed by a. So we will be through by Proposition 43:9. So let us find two hyperbolic planes containing x. There is at least one hyperbolic plane H = Fx + Fy —

110


containing x since V is regular. Take a non-zero vector z orthogonal to H. Then H' = Fx F (y z) is a hyperbolic plane containing x and q. e. d. distinct from H. So our assertion is established, 43 : 12. 4= {± 1 v } with one exception, when V is a hyperbolic plane over a field of three elements. In the exceptional case Z2 = 02. Proof. Clearly ± l v are in Zn. We must prove the converse, so we consider a typical a in Z. Let Fx be any anisotropic line in V. Then Tx=

ara 1 = Taxi

hence ax is in the line F x. So a leaves all anisotropic lines fixed. Hence a = iv by Proposition 43:11. There remains the exceptional case of a hyperbolic plane V over a field of three elements. Here V contains exactly two distinct anisotropie lines. Hence 02- consists of two distinct symmetries, hence 02 is of order 4. q. e. d. Every %group of order 4 is commutative, 43: 12 a. O is commutative in exactly two exceptional cases, when n = 1 and when V is a hyperbolic plane over a field of three elements. Otherwise it is not commutative.

Proof. The exceptional cases are already known to have commutative groups. If n> I, V has two or more anisotropie lines (for instance the lines of an orthogonal base) and so On has order at least 4. In particular On q. e. d. cannot be {± i v } . 43 : 12 b. 0: is commutative when n is 1 or 2. It is not commutative when n 3. Proof. The commutativity of op is trivial, for 02+ it was proved in 3 we have (4 + I since 0,„ is not commutative, but D. is § 43C. If n the commutator subgroup of 0: , hence 0: is not commutative , q. e. d. 43:13. The centralizer of 12„, in O. is Z„ when n 3. Proof. We must take a typical a in 0„ that commutes with every element of 12,, and we must prove that a is in Z„. First suppose that V is anisotropic. It is enough to prove that every plane is left fixed by a, for then every line is left fixed by a, and so a = lv by Proposition 43:9. So we consider a typical plane U in V and prove aU = U. Let V = U j W be the corresponding splitting. Since U is =isotropic it contains at least three distinct anisotropic lines, hence 0: = 0 (U) has at least three rotations; now a binary space has exactly two involutions that are rotations, namely ± l ; hence there is a rotation e in 0: with p2 + 1. Put j p j. l w. Then W is the fixed space of 62 since 0 is the fixed space of 2 . Hence a W is the fixed space of crj2 cr-1 . But 62 is in D„, hence ap2 a -1 j2 by hypothesis.. So a W is also the fixed space of #2. Hence aW = W. Hence a U U as asserted.


111

We are left with the case of an isotropic space V. By Proposition 43: 10 it is enough to prove that a leaves a typical isotropic line Fx fixed. Now there is a hyperbolic plane, and hence a regular ternary subspace U of V, which contains Fx. Take the splitting V = U ± W. By Corollary 43: 5a there is a rotation in 0: = 0: (U) which has axis F x and whose square is not the identity. Put 6 = e ± 1 w. The fixed space of 62 is Fx J. W since Fx is the fixed space of 02. Hence the fixed space of a02a-1 is F (ax) ± (a W). But #2 is in 14, hence a 62 a ---1 = 62 by hypothesis. So F(ax) ± a W is also the fixed space of 62. Hence

F (ax) ± (a W) = F x ± W. But F (ax) and Fx are the radicals in this equation, hence F (ax) F x. Thus a keeps the typical isotropic line Fx fixed and this is what we asserted. q. e. d. 4 3:13a. Suppose n 3. Then the center of 0: is 0: n Zn. And the

center of .Q„ is 12„ §43F. Irreducibility of Da for n 3 43:14. V is a regular n-ary quadratic space with n 3, and U is a subspace with OCUC V. Then there is a a in fIn such that ca 7+ U. Proof. First suppose that U is a regular subspace of V. Take the splitting V.U_LW and consider the involution — l u I 1 w. There is a a in Q„ which does not commute with this involution since the centralizer of Q 0„ is {± 41. If we had cal = U we could write and 0 E O (W); but then a would commute a =Ti e with with — 1 u-L l w. Hence aU+ U. Now suppose that U is not regular. Replacing U by rad U if necessary allows us to assume that Q (U) = 0. Fix a line F x inside U. If we can move it out of U using an element of 12 we shall be through. To this end we consider a regular ternary subspace T of V which contains x. Then T r U = Fx since Fx is a maximal totally isotropic subspace of T. Take the splitting V=T±W and fix a hyperbolic plane Fx+Fy inside T with Q (y) = 0. By Corollary 43 :5a there is a a in 0: (T) which is the square of a rotation on T and whose fixed space is the line Fy. Put = a I l w. Then 5 is the square of a rotation and hence it is in Q. We claim that ax= Ox is not in U. Now axisin T and T r U = Fx, so it is enough to show that a moves the line Fx. If a did not move the line Fx, then a would leave both the hyperbolic plane Fx+Fy and the vector y fixed, hence it would induce an isometry of this plane onto itself which left y fixed, hence it would be the identity on Fx Fy by Example 42: 15, and this is impossible since Fy is the fixed space of a.

q. e. d.

112


43 : 14 a. .Q.„ n Z.(

„

if n

3.

43 : 15. Let V be any regular n-ary quadratic space other than a hyperbolic plane over a field of three elements, and let cc be a non-zero element of Q (V). Then there is a base V = F x1 -1- «+Fx in which Q (xi) = cc for 1 i n. This holds with cc = 0 if V is isotropic.

Proof. For n = I the result is trivial. If n = 2 we can assume that cc is not zero. Take x in V with Q (x) = cc. There are at least three anisotropic lines in V since V is not a hyperbolic plane over a field of three elements. Then the symmetry T with respect to a line which is neither equal to Fx nor orthogonal to it shifts Fx. So X, TX is a base with the desired property. If n 3 we fix x in V with Q (x) = cc. If cc is 0 we assume that x O. Let U be the subspace of V that is spanned by ax as a runs through Q. Then cr U U for all a in Q. Hence U = V by Proposition 43:14. We can therefore find al , . . . , an in 1'4 such that ai x, ...,crn xi's a base for V. q. e. d. Chapter V

The Algebras of Quadratic Forms Our purpose in this chapter is to introduce three algebras of importance in the theory of quadratic forms, the Clifford algebra, the quaternion algebra, and the Hasse algebra. The Clifford algebra will be developed from first principles and its main use for us will be in the definition of an invariant called the spinor norm. The quaternion algebra and the Hasse algebra play an important role in the arithmetic theory of quadratic forms. The definition of the Hasse algebra depends on some of the structure theory of central simple algebras, in particular it needs Wedderburn's theorem and the theory of similarity of algebras that is normally used in defining the Brauer group. We have therefore included a proof of Wedderburn's theorem and some of its consequences. Also included as a convenience to the reader is a brief discussion of the tensor product of finite dimensional vector spaces'. The general assumption that the characteristic of the underlying field F is not 2 will not be used in the first two paragraphs of this chapter. 1 Our presentation of the tensor product, of the method of extending the field of scalars, and of the Clifford algebra, is strictly for finite dimensional vector spaces over fields. These concepts can also be developed using iuvariant methods for modules over commutative rings. For further information we refer the reader to C. CHEVALLEV, Fundamental concepts of algebra (New York, 195(3).

Chapter V. The Algebras of Quadratic Forms

113

§ 51. Tensor products § 51A. Abstract vector spaces Consider finite dimensional vector spaces T,U,V,W over an arbitrary field F. By a bilinear mapping t of U X V into T we mean a mapping t:Ux which has the following properties: t(u u', y ± e) = t(u, y) t(u' , y) y') t (u' , , t (au, fly) = Œ/9t (u, y) whenever u, 14' E U and y, y'E V and a, # E F. A tensor product of U and V is a composite object (t, T) consisting of a vector space T and a T which satisfies bilinear mapping t: U x V the following universal mapping property: given any bilinear mapping w of U X V into a vector space_ W, there is exactly one F-linear map p /ix V such that po t = w. Sv It is easy to see that tensor products exist, particularly for the finite dimensional vector spaces under discussion here. Fix a base x1,. , xin for U and a base y i, • yn for V, then take an mn-dimensional vector space T over F with a base {zap} where 1 A m, 1 1u n. Define a bilinear mapping t: UxV—)—Tby the equation t

A

ccAxA,

MA1314 2'40

Is

I

A,

Then it is easily seen that the composite object (t, T) satisfies the universal mapping property stated above. Hence tensor products exist. Consider two tensor products (1, T) and (t', T') of U and V. We claim that there is exactly one isomorphism p of T onto T' such that 9) o t The mapping p required in this assertion is already at hand: it has to be the unique linear map io of T into T' which satisfies the equation po t = t', and whose existence is assured by the universal mapping property of tensor products. We just have to prove that p is bijective. To this end consider the linear map q": T' T for which V or t. Then p' 0 q' is a linear map of V T into T such that (p' oq')o t t. By the definition of T as a tensor product there can be just one such mapping p' of T into T; now the identity 12, on T is such a mapping; hence V q= Similarly q' q! 1 T . These two equations imply, by a simple setO'Meara, introduction to quadratic forms

8


114

theoretic argument, that 97 and qf are bijective. Hence is an isomorphism of T onto T'. We have therefore proved our claim. So tensor products are unique. The existence and uniqueness of tensor products allows us to talk of the tensor product of two vector spaces. Instead of the arbitrary symbols (t, T) for the tensor product we use 0 for the bilinear map t and UO V for the space T. The image of (u, y) EUx V under the bilinear map 0 is written u0 v. Thus the tensor product (0, Uo V) is simply a composite object consisting of a bilinear map 0 and a vector space Uo V. The bilinearity of e now reads (u u') (y 4- y') = u® v+ u' ®v u® y' 4- /4'0 y', (au)0 (fiv) And the universal, mapping property says that whenever a bilinear map w:Ux V W is given, there exists a unique linear map (p: U® V W such that p(uo y) = w (u, y) V (u, y) EUx V.

Consider the explicit tensor product (t, T) defined earlier in the construction of the tensor product of U and V. This had dimension mn. And it was spanned by vectors of the form t(u, y). So the uniqueness of the tensor product says that all tensor products have these properties. Hence dim Uo V = dim U • dim V . And Uo V is spanned by the vectors 240 y as u, y run through U, V. So if bases x1 . ,a and y1 ,.. . , yn. are chosen for U and V, the mn vectors x2 elyi, will form a base for U 0V. Hence every vector of Uo V can be put in the form ,

E

Up

Y1, with all up in U.

The reader may easily verify using linear methods that the u p in the above expression are unique. Similarly every vector in UO V has a unique expression in the form with all VA in V. § 51B. Algebras

An algebra' A over a field F is a vector space provided with a ring structure having an identity lA in which scalar and ring multiplication are related by the equations (xy) = (ax) y = x(Œy) 1

Strictly speaking this should be called an associative algebra with identity.

115


for all ac E F and all x, y EA. The algebra is called commutative if it is commutative under its ring structure. It is called a division algebra if it is a skew field under its ring structure. Note the equation (oc 1 A) x = ocx = x(14 ) for all oc in F and all x in A. We make the general assumption that every algebra is finite dimensional over its field of scalars. 51:1. Example. The set of linear transformations L F (V) of a vector space V into itself is both a ring and a vector space over F. In fact it is an algebra over F. (See § 41A.) A mapping cp of the algebra A into an algebra B is called an algebra homomorphism if it is an F-linear ring homomorphism such that T(1 A) = B. Here is a convenient way of checking whether a given mapping is an algebra homomorphism. Let x l , . , xn be a base for the algebra A and let tp be a mapping of A into an algebra B over the same field F. Suppose that cp if T.-linear and that it preserves multiplication on the given base for A, i.. e. that (xixi) = 49 (xi) p (x1) for 1 Then an easy calculation involving linearity gives

cp (E oc,i x,t) • 9) (E Hence cp is a ring homomorphism. So a map cp: A --›-- B which is /7-linear, which preserves multiplication on a given base for A, and which sends

cp(E acA x,1 - E fit, xp)

—

the identity to the identity, will have to be an algebra homomorphism of A into B. The following rule is useful in the explicit construction or definition of an algebra. Let a vector space V with a base x1 ,. . . , xn be given over 2,' n, 1 j n) a vector of V is specified, a field F. For each i, j (1 and this vector is formally denoted as a product x i x i . ( In practice this can be done by specifying n 3 scalars n

4 and then taking x i x; to be the

vector E ccifi xk . We can extend these products by linearity to a law k= i of multiplication on V, i. e. we define multiplication by the formula

if ocit x,IliE fili xis)= f ac,1 Ai (x2 xi, ) . I A, te A 1\P This law is clearly distributive with respect to addition on V. And it satisfies

cc (x y) = (ccx) y x(ccy) for all oc in F and all x, y in V. Suppose further that the law on V satisfies

(x 1 x 1) xk x i (x pxk) 8*

116

Part Two. Abstract Theory of Quadratic Formé

for all relevant i, j, k. A linear argument then shows that multiplication is associative and so V is a ring. If there is also a vector 1 E V such that 1 xi — xi = xi 1 for 1 I n, then 1 is an identity for the ring V and so V is actually an algebra over F. In other words, whenever we define an algebra in this way we have to check two- things: associativity among the defining basis vectors, and the existence of an element which acts as the identity on the defining base. It is clear that the multiplication obtained on V is uniquely determined by the specified values of the xi xJ . Let us extend the concept of a tensor product to algebras. We consider algebras A, B,C over the same field F. We call a mapping

a multiplicative bilinear mapping if it is bilinear and satisfies the equations w(lA , 1B ) =l e , w (x, y) • w (x', y') =w (xx', yy')

for all x, x'E A and all y, y' E B. Now consider a tensor product (t, T) of A and B regarded just as vector spaces over F. Then T is a vector space and t is bilinear. We claim that there is a unique law of multiplication on T .which makes T into an algebra and makes t multiplicative. For consider bases x1,. . x. for A and y . , • • yr, for B. Then the mn vectors t (xi, yp) form a base for T. Define t(xi,yj)- t(x k ,y 1) = t(xi xk , yiy i)

for all relevant i, j, k, 1. Extend this by linearity to a law of multiplication on T. A linear argument shows that t (x ,y) t (x' ,y)

t (x x' , y y')

holds for all x, x' E A and for all y,y' E B. Hence the new multiplication is associative on the above basis vectors for T, and t.(1 A , 1B) acts as the identity on these basis vectors. 'So we have made T into an algebra with identity 1(1 4 , 1B). Furthermore, any law which makes T into an algebra and makes t multiplicative must agree with the law defined above on the basis vectors of T, hence the two laws must be identical. We have therefore proved our claim: there is a unique law of multiplication which makes T into an algebra and makes t multiplicative. We now agree that a tensor product (t, T) of algebras A, B shall always be their tensor product as vector spaces in which T has been made into an algebra with the above property. In the standard notation for tensor products this reads as follows: A 0 B is made into an algebra by a uniquely determined multiplication having the property that (rely) • (x'0y 1 )= xx'syy' for all x, x' E A and all y, y' E B; and 1 4 0 1 B is the identity of A® B.


117

Consider the tensor product A 0 B of the algebras A, B and let w:AxB-4--C

be a multiplicative bilinear map into some third algebra C. We know that there is a unique F-linear map 4p:A0B-).-C

such that T. (x0 y) = w(x,y) V (x,y) EA x B.

In fact 9) must be an algebra homomorphism. For if we take bases xl, . . ., x„, for A and yi, . . . , yn, for B we have 'T ((xi 0 Yi) (xig OY/)) = 92 (xi xk 0 3W

= w (xi xie, Yaz) = 97 (xi ON • tr (xle 0 Yl) , so that 9) preserves multiplication on a base for A 0 B; and furthermore,

3 ) = w( 1 A, 1 B) = lc, so that 9) preserves the identity. Hence 9) is indeed an algebra homomorphism as we asserted. Incidentally this shows that the uniqueness map between two tensor products of A and B is actually an algebra isomorphism. If we have algebra isomorphisms 9): A »-). A' and v: B ›--› B', then there is an algebra isomorphism OA® 1

A® B »-* A' 0 B' .

For w (x, y) . (9)x)0 (tp y) defines a multiplicative bilinear map w: A x B -).- A' 0 B',

hence there is an algebra homomorphism

e : AOB ---)--A'OB' such that e (x0 y ) = (9) x) 0 (Ip y) for all (x, y) E A X B. Now e is clearly surjective, hence it is bijective since A 0 B and A' 0 B' have the same dimension. Let us prove the following algebra isomorphisms: F0A--- A , AoBr_41B0A ,

A 0 (B o C) al (A o B) 0 C .

In the first instance we introduce a multiplicative bilinear map w: F x .A -- -- A by the equation w (a, x) = ax. This gives us an algebra homomorphism q): F0 A - A such that 9) (a® x) = ax for all a EF and all x E A. This map g) is surjective, hence bijective by a comparison of dimensions. Hence Fo A is indeed isomorphic to A. Similarly A 0 B

118


is proved isomorphic to Bo A by using the multiplicative bilinear map w (x, y) = yox of A X B into BOA. The best sway for us to prove the third isomorphism without getting too involved with the general theory of tensor products is to use bases A, and y„ for B, and z1, . . .,z, for C. The mn r X11 .. X vectors of the form x i ® (y,® zk) now form a base for A® (Be C). Make up a multiplication table for this base in terms of the multiplication tables of the three given bases. Now do the same thing with the (xi oyi)ozt in (A0B)0C. It turns out that the two multiplication tables obtained are the same. Hence the F-linear isomorphism determined by xi ® (y5 Zk) (XI: y") Zic, preserves multiplication among these basis elements, hence it is a ring isomorphism, hence an algebra isomorphism of A (Bo C) onto (A0B)0 C as required. Consider algebras A 11 . . A, over F with r 2. We use the symbol Al o • • •

A,.

to denote an algebra that is defined inductively by the equation A1 0 • • • 01)44 7 = (A 1 O • • • cio,A 7_,) ® A 7 . This algebra is uniquely determined up to an algebra isomorphism. It will also be written

o

Ai

•

It follows easily by induction, using the associativity formula for AO(Bo C), that (A1 0 - • • 0.'1 3) o (A 8+1 0 • • • A,.) is algebra isomorphic to Al e • • o A,. Similarly the commutativity formula A ø B Bo A shows that A1 0 • • • 0A,. is independent of the order of the factors. § 52. Wedderburn,'s theorem on central simple algebras § 52A. Central simple algebras A is an algebra over the field F. The mapping cc

ccl A determines

an algebra isomorphism F >--).F1A C A ,

so A naturally contains a subfield FlA isomorphic to the field of scalars F. This subfield is part of the center of A since (cc 1) x = ccx = x (cc I)


119

for all ocE F and allxEA. If F IA is actually the entire center of A we call A a central algebra over F. By a left ideal of A we mean a left ideal of the ring A. The equation ocx = (oc1)x shows that every left ideal of A is actually a subspace of the vector space A. Similarly with right ideals and with two-sided ideals. We call A a simple algebra if it contains no two-sided ideals other that 0 and A itself. For example, every division algebra is simple. A central simple algebra over a field F is an algebra which is central and simple over F. 52: 1. If A and B are central simple algebras over F, then so is A oFB. Proof. 1) First we prove that A B is central, i. e. that a typical element z in the center of AO B is in F (14 0 1 13). Fix a base yi , . • • , yn for B and express z in the form z = f /4),0y2 (1.4,1 E A) .

a Then z(x® 1) = (x® 1) z holds for all x in A, hence (uÂ x) A

yA = E (xua) ®y. A

Now this representation is unique by § 51A, so /4A x = xua for all x in A, hence each isa is in the center of A, so /4A EF 1 A for 1 2 n. Hence z has the form z = 14 0 y for some y E B. Now repeat the argument on B instead of A. This gives y E F 1B. Hence z is in F (1A 0 1B). So A 0 B is central. 2) Here we shall prove that A to B is simple. We must consider a non-zero two-sided ideal a of A o B and prove that it is all of A 0 B. First an observation. Suppose /41 0 y/ + • •

u2,0 y2, (ui E A, NE B)

is an element of a in which ut is non-zero; the ideal generated by u1 in A is all of A since A is simple; hence we can find /41, . u; in A such that /41 = lA and u 0 v1+"'+ i40v 2, E a. Similarly we can arrange to have /41 0v1' + • • • ± ut,® Ea with vl = 1 B when v 1 is non-zero. So much for the observation. Now we can prove that a = AO B. , yn for B, make a choice Of all non-zero z in a and of all bases in which k is minimal in the representation z = %Oh+ • • + ukeYk

(NE A) -

We can assume that we actually have Z=

1 A 0y1 + /42 0h+ • • • + tikeyk

120


by the above observation. If k> 1, then uk F 14 since if it were we could change the base and produce a smaller k. We can therefore take x E A with xuk + uk x since F iA is the center of A. But then (xo 1B) z z (x 1 B)

is a non-zero element of a with a smaller k. This is impossible. Hence k. 1. Therefore z lA oyi. Now go back to the above observation. This gives us a z in a of the form z = 140 113. Hence a = A B. Hence q. e. d. A ei B is simple. 52:2. Let A be any algebra over F and let B,C be two central simple subalgebras of A which contain the identity of A. If B and C commute element-wise, then LI bc with b EB, c EC is a typical element of the subalgebra generated by B and C. And this algebra is isomorphic to BoE C. Proof. Only the second part really needs proof. We can suppose that A is actually the algebra generated by B and C. Define a multi-

plicative bilinear map by the equation w(b, c) = b c. Then by § 51B there will be an algebra homomorphism q): Be C-.--A such that p(bo c) = b c for all b EB, c EC. This map is clearly onto A. And its kernel is 0 since Bo C is simple. Hence (p : BeC >—). A is an algebra isomorphism. q. e. d. § 52B. The algebra RA (a).

This is a preamble to the proof of Wedderburn's theorem. A is an algebra over F, and a, b are left ideals in A. A mapping q): a b is called A-linear if it satisfies the equations (x -1- y) 9) = (x (p)

(y (p), (ax)

a (x

for all x, y E a and all a E A. (All of a sudden the mapping (p appears on the right! This is our only exception to the rule that mappings are always to be written on the left.) We know that a and b can also be regarded as vector spaces over F, and if this is done the equations (0ex) 97 = ((cciA) x)cp = (m 1 (x (p) = cc (x (p) show that every A-linear mapping is F-linear. We let R (a) denote the set of A-linear maps of the left ideal a into itself. Actually we shall regard RA (a) as an algebra over F where the laws are provided in the following way. Given p, tp E R A (a) and a EF we define ço tp, v and cd (p by the formulas x(q) -1- v) = (x(p)

x((p V) = (x94 V x(cep)

oc(xT) ,

(x v)


121

for each x in a. Each of these three mappings is clearly an A-linear mapping of a into a, hence each of them is an element of RA (a). If we

now define zero and identity by the equations (x) 0 = 0, (x) I x then an easy verification shows that we have made RA (a) into an algebra over F. (As a matter of fact RA (a) is a vector subspace of the algebra LF (a) of all F-linear maps of a into itself; it is not a subring of LE (a) since multiplication has been twisted by writing mappings on the right; it is precisely in order to obtain an algebra having this twisted multiplication that the A-linear mappings were written on the right.) Our interest is really in the minimal left ideals of A. As the name implies, a minimal left ideal is a left ideal which properly contains exactly one left ideal, the zero ideal. Minimal left ideals always exist since every left ideal is a subspace of A, and A is finite dimensional. 52:3. If a is a minimal left ideal in an algebra A over F, then RA (a) is a division algebra over F. Proof. All that we have to do is prove that a typical non-zero 92 in RA (a) is invertible. Now (a) g) is a left ideal contained in a and a is minimal, hence (a) g) = a, hence 92 is surjective. And the kernel of g) is also a left ideal contained in a, so it is O. Hence T: a a is bijective. Let tp be the inverse mapping of T. Then tp is A-linear, hence in RA ( a). And 9) = 1 v T. Hence g) is invertible, q. e. d. 52:4. Let a and b be minimal left ideals in a simple algebra A. Then there is an A-linear bijection cp of a onto b. And RA (a) is algebra isomorphic 10 RA (b). Proof. 1) The set of points {aEAlax=0 V xEb}

is a two-sided ideal in A, hence it is 0 since A is simple. So there is a b E b with OC abS b. But a b is a left ideal of A, hence a b = b since b is minimal. Define the surjection 92: a b by the equation (a) T = ab for each a E a. This is clearly A-linear. And its kernel is a left ideal contained in a, hence it is O. So we do indeed have an A-linear bijection 92 of a onto b. 2) Take a typical a E RA (a) and define W (a) = 92-la

where the A-linear map g)-1 : b >--). a denotes the inverse of T. Then 92 maps b into b and it is clearly A-linear, hence it is in RA (b). So we have constructed a mapping W: RA

(a)

->- RA

(b) .

This mapping is clearly bijective. We leave it to the reader to check that it is an algebra isomorphism. q. e. d.


122

§ 52C. The algebra Mn (D) D is a division algebra over the fi eld F. We let M„ (D) denote the set of all n x n matrices with coefficients in D, and we define addition and multiplication of matrices in the usual way. This makes Mn (D) into a ring with identity. For each a E F and each (di j ) E Mn (D) we define the scalar multiple Œ(d,) = (ad j) . This makes M„ (D) into an algebra over F. (To prove finite dimensionality use the finite dimensionality of D over F.) We let ejj denote the matrix with 1D in the (i, j) position and 0 everywhere else. There are n2 of these matrices and they are called the defining matrices of M(D). They satisfy the equations if j — k if j k

0

Also en + • • • + enn is the identity of Mn (D). 52 : 5. Mn (F) is central simple over F. Proof. 1) First we prove it is central. We must consider a typical matrix f alp % in the center and prove that it is diagonal with all 2 ,14

diagonal entries equal. Fix i+ j and take ejj , Then ozo e,„1= z cc,„eip , P

I

hi

and E ccAm e2 e 5 = E (

,

A,

hence E aft,

cxu e2j . A

Comparing coefficients gives alp = 0 whenever y+ j; and also ail = °z ip Hence the given matrix is of the required type. 2) In order to prove simplicity we must show that a non-zero twosided ideal a contains the identity matrix. Take an element E7 oc21 e p in a with aij , say, non-zero. Then

2 , Is

Œzy7. 1 e j (E ccAp eAls)eji,. e„ A, le

is in a for 1 p n. Hence the identity e n ± • • ± enn is in a. 52:6. There is an algebra isomorphism

q. e. d.

M „(F) p M ,.(F) M n ,.(F) .

Proof. Let ei j (1 Mn (F), let fA, / (1

i n, I <j n) be the defining matrices of k Ç r, I 5 1 r) be the defining Matrices of Mr(F).


Each number 2,(1 2

123

2 nr) can be expressed uniquely in the form

(i — 1) r

k

i

(1

n, 1

k

r) .

And each 14(1 1.4 nr) can be expressed uniquely in the form = (j — 1) r /

(1

n, 1

i

r) .

Put eil e fki •

These (nr) 2 vectors form a base for Mn (F) M? (F) since the above defining matrices form bases for Mn (F) and Mr (F) respectively. An easy computation gives {EA e if EAmEre if p+ v0 . But this means that iff(F) M? (F) has a base whose multiplication table is the same as the multiplication table of the defining matrices of M,(F); and these defining matrices form a base for M„(F). Hence Mn (F) Mr (F) is algebra isomorphic to Mnr (F). q. e. d. 52:7. Suppose D is a central division algebra over F. Then there is algebra isomorphism Mn(F) OFD -`="'M(D) .

an

Proof. Let D be the subalgebra of Mn (D) consisting of the diagonal matrices of the form diag (d, . . . , cl) with cl E D. This subalgebra is an algebra isomorphic to D. Let A be the subalgebra of M.„ (D) consisting of all matrices of the form (d") with dij EF 1D. This subalgebra is algebra isomorphic to Mn (F). Now D and A contain the identity of Mn (D), they commute element-wise, and they are both central simple over F. Furthermore, M is the algebra generated by D and A. Apply Proposition 52:2. This says that DO A is algebra isomorphic to Mn (D). Hence D (81 111, (F) is algebra isomorphic to M.„(D). q. e. d. 52:8. Let A be a central simple algebra over the field F. Suppose there is a division algebra D over F such that A is algebra isomorphic to the matrix algebra M,i (D). Then D is central, n is unique, and D is unique up to an algebra isomorphism. Proof. 1) So Mn (D) is central. Take cl in the center of D. Then the matrix c7 diag (cl, . . . , cl) is in the center of M (D), hence it is of the form cl= 1D for some Œ in F , hence cl = a 1D, hence d EF 1D. So D is

central. 2) Put A' = Mn (D). The given isomorphism carries A to A', it carries a typical minimal left ideal a to a minimal left ideal a' of A', and it induces an algebra isomorphism RA (a) RA , (a') .

124


If we can prove that RA , (a') is algebra isomorphic to D for some minimal left ideal a' of A', then it will be isomorphic to D for all minimal left ideals a' of A' by Proposition 52:4, hence RA (a) will be isomorphic to D for every minimal left ideal a of A. This clearly will prove that D is unique up to isomorphism. And the isomorphism

Mn (F) 0D will imply that n is unique. So the whole proof boils down to this: show that RA ,(a') is algebra isomorphic to D for at least one minimal left ideal a' of A' M(D). , d) whenever d is in D. 3) Let d stand for the matrix diag (d, Note that cZe 5 = e15 ti for all i,j. A typical element of M,,(D) now has the form Y ii eij — ei; clii (d, E D) . Let a' be the set of all matrices of the form A en + • ' + t7fl en with di eD j. e. the set of all matrices in which every column but the first is identically zero. Then a' is clearly a left ideal in A' ---- Mn (D). We claim it is minimal. For suppose that A

A en + ' • ' + dflfl1 is an element of a left ideal b contained in a' with di + 0, say. Then d en= d dïl e5i (di e11+ • • • + 1 e„ 1) is in b for any d E D and 1, j n. Hence b = a'. So a' is indeed a minimal left ideal of A'. We prove RA I (a') D for this minimal left ideal a'. Each d CD a' by the equation defines an A'-linear map Tid : a' (X) 9)d= xcl V x E a' . So we have an algebra isomorphism D ›— RA , (a') 9)d. It remains for us to prove this isomorphism is defined by d surjective. So consider a typical v ERA , (a'). Write Then for 1
--)- 1 1I (D) below.) If we fix a base , x„ for V over D, then we can associate an n X n matrix with each a E L (V) in the usual way, i. e. we write

crx,. E xi d il with dij E D, i=1

and we send a to the matrix (do). This gives us an algebra isomorphism. L D (V)

M„(D) .

§ 52E. Wedderburn's theorem 52 : 9. Theorem. Let A be a central simple aigebral over F. Then there is a central simple division algebra D over F such that A is isomorphic to the algebra

M(F)®,D

.

Here n is unique and D is unique up to isomorphism.

Proof. 1) Let a be a minimal left ideal in A and put D = RA (a). Then D is an algebra over F and in fact it is a division algebra by Proposition 52:3. This will be the division algebra of the theorem. Now by the definition of the laws of composition on D —R4 (a) we have (x+y)99.x92-FyT, x(92-Fip)--=xp-Fxv, x(99 ip) = (x 92) el) , (x) 1» = X We have limited our discussion of the Wedderbum theorem to central simple algebras, but essentially the same proof will show that "a simple ring with identity which satisfies the minimum condition on left ideals is isomorphic to a full matrix ring M(D) over a division ring D". For a different proof, and for further references to the literature, we refer the reader to the revised edition of B. L. VAN DER WAERDEN'S Algebra vol. 2 (Berlin, 1959).

126


for all x, y E a and all 9), ip CD. This simply means that a can be regarded as a right vector space over the division algebra D. In order to prove that a is finite dimensional over D we restrict the scalars to F 1D ; it is enough to prove that a is finite dimensional over F 1D; but

ax

Œ (x 1D)

x (cc 1 g),

and a is finite dimensional over F, hence it is over F 1D . So a is indeed finite dimensional over D. We now form the algebra LD (a) of D-linear maps of a into a (see § 52D). 2) Our first task is to find a natural isomorphism of the algebra A into the algebra LD (a). Given a EA let cra : a a be defined by the equation V xCa Œa (X)1 Then cra, is clearly additive; and

= a (x

(ax) = ((Ta x)

,

by definition of D = RA (a); hence c c, is D-linear. So we have a mapping

A -- LD (a) defined by a aa. It is easily verified that this mapping is an algebra homomorphism. Its kernel is a two-sided ideal in A, hence it is 0 since A is simple. So we have found our algebra isomorphism A ›— LD (a) . 3) The difficult step in the proof is showing that the above map is surjective. Before we attempt this we must establish some facts about annihilators in the algebra A. Given a D-subspace U of a we put

U°={aCillau.0 VueU}; this is a left ideal of A. Conversely, given a left ideal b in A we put

b°—.{xEalbx=0 VbEb}; this is a D-subspace of a. We can therefore form the D-subspace U°°. Clearly U Ç Um. Our purpose here in step 3) is to prove that U = U°°. We do this by induction on dimD U. If U = 0 we have U°— A and then O. Now consider a U with OC ETC a, and assume the result for all subspaces of lower dimension. We must prove it for U. Write U = W + zD with W a hyperplane of U and z E U W. Consider y E- U U °°. We must prove that y C U. If y E W we are through, so suppose y is not in W. Then W°y is not 0, it is a left ideal contained in a, hence W°y = a. Similarly W°z a. We define a map a —›— a as follows: take typical / z E a with 1CW° and send it to /y E a. Why is this well-defined ? If 1 z 1 1 z, then (1 — 1') U (1 — 1') (W + zD) O,


127

since l — EW° , hence (l — l') E U°, hence (l — y = 0 since y E U®, hence ly = l'y. So the map Zz —›— ly (1 E W0) is a well-defined map of a into a. It is clearly A-linear. Hence there is a ( R (a) = D such that (h) = /Y V / E W°.

Then l(y — z (p) —0 \H ( W°, hence y — zp E W°°= W, hence yEW+zp= U. We have therefore proved that U00 = U. 4.) Now take a base for the right D-space a:

a = z1 ./3 + - • • + ;D. Consider a typical a E LD(a). We must find an a E A such that o. is equal to the map a a of step 2). Once this is done we will have established the surjectivity of the map A ›—). L» (a). Let Wi bd the hyperplane spanned by all the vectors z1, . • zti except If we had W?zi = 0 we would have W? a = 0 and so Wi = Wr= a, which is absurd. Hence W? zi + 0, hence W? zi = a. Pick ai E W2 such that ai zi = a;; then mi . 0 for j÷ 1. Do this for 1 I n, and put a = a1 + • • • + an. Then aa zi = azi = a i zi = a;.

Hence aa and a agree on the base z1, . . Zn. Hence aa= . And we have established our algebra isomorphism A LD (a) . 5) We therefore have an algebra isomorphism A A Ia (D). Then D is central by Proposition 52:8. It is simple since it is a division algebra.

Hence A M a (F) ®FD

by Proposition 52:7. And the uniqueness follows from the same two propositions. This completes the proof of one of the most remarkable results in modern algebra. q. e. d. § 52F. Similarity of algebras Consider two central simple algebras A and A' over F. By Wedderburn's theorem there are natural numbers r and r', and central simple division algebras D and D' over F, such that AD

illr (F)

and A'

D' 0 Z14 (F) .

128


We also know that D and D' are essentially unique. We say that A is similar to A' and we write A A' if A and A' have isomorphic division algebra components, i. e. if D is algebra isomorphic to D'. It is easily checked that similarity is an equivalence relation. Of course isomorphic algebras are similar. And the concept of similarity coincides with the concept of isomorphism for algebras of the same dimension. Note that A A' if and only if AOM,(F) ad- A' 0111,(F) for some p and q. From this we can deduce the following result: if A i and A.; are central simple algebras over F with A i 214, for 1 I t, then A1 0 • • • 0 A t Af0 - • • 0 A; . We write A 1 and say that A splits if A 1112,(F) for some p 1 ; this is the sanie as saying that the division algebra component of A is isomorphic to F. Obviously, A 1 A0B. B BeA for all central simple B. There is also the cancellation law' A

B A'OB

A A',

but the proof of this must wait until we have made a brief study of the reciprocal of a central simple algebra. The reciprocal of the central simple algebra A is defined in the following way: leave the vector space structure of A unchanged and provide A with the new ring structure determined by the twisted multiplication a* b ba (a, b E . This new ring is again a central simple algebra over F, it is called the reciprocal algebra of A, and it will be written A*. 52:10. A .10 A* splits. Proof. It is enough to prove that A e F A* is algebra isomorphic to L (A) since L F (A) is algebra isomorphic to the matrix algebra M(F) where n dimpA. For each (a, b) E A x A* we define a mapping Ta, b of A into itself by writing 4Ta,b(x)---axb VxEA. The reader who is prepared to talk about the set of similarity classes of central simple algebras over F will see that the tensor product et induces a law of multiplication on this set, and that the resulting object is a group. This group is calléd the Brauer group of F. 1


129

Clearly Ta, b is an F-linear map of A into A, i. e. Ta, b E L F (A). We therefore have a multiplicative bilinear map w : A x A* —>— LF (A) given by w (a, b) Ta, b . Hence there is an algebra homomorphism —

2:A0A* —›—L F (A)

such that 2(uø b) = 97„,b for all (a, b) e A x A*. The kernel of 2 is a two-sided ideal in the central simple algebra A® A*, hence it is O. So 2 is injective. But dim A 0 A* -= n2 = dim LF (A) . So 2 is bijective. Then A: A0A.* >— L(A)

q. e. d.

is the desired algebra isomorphism. Now we can prove the cancellation result A0 B ,,, A 1 0 B

A ,,, A'.

Namely, AOB ,,, A'0134 A0B0B* ,,, A'OBOB* A 0 M „ (I) , A' 0 M a, (F) --,-> A , A'.

§ 53. Extending the field of scalars EIF is an arbitrary extension of fields.

§ 53A. Abstract vector spaces Consider a vector space V over F and a vector space T over E. As usual we assume that V and T are finite dimensional over F and E respectively. If we say that V is contained in T we understand, of course, that V is a subset of T and we also tacitly assume that the laws induced by T on V agree with the given laws on V. We say that an E-space T is an E-ification of the F-space V if V is contained in T, and if every base for V over F is also a base for T over E. It is clear that if T contains V, and if there is at least one base for V that is a base for T, then every base for V is a base for T and so T is an E-ification of V. Every space V over F has an E-ification. The construction is almost trivial. Take an E-space T' which has no points in common with V, and such that dimE T' = dimF V V. Let xi,. . ., ,c,i be a base for T' over E and define the I.-space V' --, F xi + • • • -I- F act,' .

Then T' is an E-ification of V'. Now V and V' are isomorphic I.-spaces. Hence there is an E-space T containing V which is an E-ification of V. O'Meara, Introduction to quadratic forms

9

130


The last step in the above construction is an application of what we shall call the identification process. This is our first direct use of this process and there might be some point in elaborating on it this once. So let us go back to the isomorphic F-spaces V and V', and the E-ification T' of V' with T' disjoint from V. Define a new set T

(T'— V') jV

and prolongate the F-isomorphism a: V >—). V' to a bijection a: by the formula fax if x E V ax = x if x E — V' .

T'

There is a unique E-space structure on T which makes a-1 : T' T an E-isomorphism, namely the one carried from T' to T by a-1. Then T' is an E-ification of a-1 V'. But the F-space structure on V is identical to the F-space structure on (f-1 V' — V since a was chosen as an F-isomorphism. Hence T is an E-ification of V. If T' is any E-space containing V, and if T is an E-ification of V, then it is easy to see that there is exactly one E-linear map of T into T' which is the identity on V. This immediately implies the following uniqueness theorem for E-ifications : if T and T' are E-ifications of V, there is exactly one E-linear isomorphism ep: T T' such that Tx x for all x E V. In view of the existence and uniqueness of E-ifications we use the symbol EV to denote an E-ification of V. We shall often refer to EV as the E-ification of V. Note that dimBE V --- dimr V . And if U is any subspace of V, the subspace of EV that is spanned by U over E is an E-ification of U. If we have V= e • • e

and we take EU SiEV in the above way, then EV= EUi ED - • • e EU,. § 53B. Algebras Now consider an algebra A over F and let the vector space B D A be

an E-ification of the vector space A. By choosing a base for A it becomes clear that there is a unique multiplication on B which agrees with the given multiplication on A and makes B into an algebra over E. Here B will have the same identity as A. The algebra B is then called an E-ification of the algebra A. It is written EA. The E-linear uniqueness isomorphism between two E-ifications is actually an algebra isomorphism.


131

And if C is an algebra over E which contains A and which has the same identity element as A, there is a unique algebra homomorphism of EA into C which is the identity map on A. By taking bases we can readily see that isomorphic algebras have isomorphic E-ifications. And we can obtain algebra isomorphisms

E(A.1 0 1 A2)

(EA 1) OE (EA

,

and hence

A\>—

E(

(E A i ) .

\15ir

§ 53C. Quadratic spaces Now suppose V is a quadratic space over F, and let B and Q be the corresponding bilinear and quadratic forms on V. Let T be an E-ification of V. By considering a base for V we see that there is a unique symmetric bilinear form

B: T x T E which agrees with the given B on V. We then have an associated quadratic form Q on 7'; and T has been made into a quadratic space in a unique way. This quadratic space is called an E-ification of V and is written EV. Clearly V and EV have the same matrix in any given base for V. So the uniqueness map between any two E-ifications is actually an isometry. And isometric spaces have isometric E-ifications. If we have a splitting V = U1 _L • • I_ (1,. and take E-ifications EUi S EV, then EV.EU 1 ±•••±EU,..

We can easily show that radE V — E rad V. As for discriminants, we have d(EV) = CEP) . dV = dV'

§ 54. The Clifford algebra V will denote a regular n-ary quadratic space over the field F, B will be the associated symmetric bilinear form, Q the associated quadratic form. An orthogonal base in which V=Fx1 ± - • • 1 Fx.„ is fixed for V. We say that an algebra A is compatible with the quadratic space V if V is a subspace of A such that

x2 = Q (x) 1 A V x EV , where I A denotes the identity of A. If this holds, then the equation

Q (x ± y) = Q (x) Q (y) + 2 B (x, y) 9*

132


implies that xy yx = 2 B(x, y)1 4, V x, y E V .

In particular xy y x whenever x and y are orthogonal vectors in V. So the given basis vectors satisfy the relations —xi xi if i+ j I. Q (xi) lA if i.j. If a vector x E V has an inverse x-1 E A, then the equations Q (J ) x-1 = x x x-1 = x

show that x must be anisotropic. Conversely every anisotropic vector x E V has .an. inverse in A, namely (Q x) -1 x

Let C be an algebra compatible with V. We call C a Clifford algebra of the quadratic space V if it satisfies the following universal mapping property: given any algebra A compatible with V, there is exactly one A such that Tx = x for all x E V. We algebra homomorphism 92: C shall prove later that V always has a Clifford algebra. In the meantime let us settle the question of uniqueness. 54: 1. Let C and C' be Clifford algebras of the regular quadratic space V. Then there is exactly one algebra isomorphism identity map on V.

92: C C' which is the

Proof. By definition of the Clifford algebra there is exactly one algebra homomorphism 92: C C' which is the identity on V. Similarly we have a 97': C' C. We have to prove that 92 is bijective. Now 99'. 97: C C is an algebra homomorphism which is the identity on V; but there is exactly one such map; hence 92 is the identity map on C; similarly 92 o 92 is the identity map on C' ; a simple set-theoretic argument then shows that 9, must be bijective, q. e. d. 54:2. V is a regular quadratic space with an orthogonal base x 1,. . and A is an algebra compatible with V. Then the subring A' generated by V in A is a subalgebra containing the identity element of A. It is spanned by all products of the form xi'. xt

with

ei = 01.

Proof. A' consists of all finite sums of the form

with x, y,... E V. This is clearly a subspace of A. It also contains (Qx1) -1.4

1A .

Hence A' is a subalgebra containing the identity element of A.


133

Each product in the above sum is a linear combination of products of the given basis vectors, hence a linear combination of products of the form x.. . x:1/4 with mi 0 in virtue of the relation xi xi = — ;xi for i + j, hence a linear combination of elements xei z . 4. with ei = 0,1 in virtue of the relation x = Q (xi) 1 A • q. e. d. 54:2a. dimF A' < 2. 54 :2b. If dim A' = 2", then A' is a Clifford algebra of V. Proof. 1) We have the formula

H (_

4.1 )

. xt)(x'. .

(x

t+di

47 +4)

15j . By Theorem 58:4 this is true if and only if SV (-1, — 1). q. e. d. 58:7. V is a regular quaternary space with discriminant d, and E is the field E = F a). Then V is isotropic if and only if EV is isotropic. Proof. We must consider an isotropic space EV and deduce that V is isotropic. So suppose that V is not isotropic. Then d must be a nonsquare in F and EIF is quadratic. Every element of EV has the form x y with x, y in V; take an isotropic vector of this form. Then Q (x) d Q (y) + 2 V B (x, y) = O. So Q (x) = d Q (y) and B (x, y) = 0 since the extension is quadratic. If Q (y) = 0 we have Q (x) = 0, hence x and y are both 0 since there are no isotropic vectors in V; but then x + Jf y is not isotropic. So in fact Q (x) and Q(y) are non-zero. Let us write Q (y) e, B (x y) O, Q (x) = — de.

Then V -m <e> i < — de> j P .

where P is a plane contained in V. If we use this expression to compute

the discriminant of V we find that P must have discriminant —1. Hence P is a hyperbolic plane. Hence V is isotropic. q. e. d. 5 8 : 7 a. V is isotropic if and only if E(SV) ( -1 E'—1 ) . Proof. Take a regular ternary subspace U of V. By Propositions 42:12 and 58:7 V is isotropic if and only if EU is isotropic. By Proposition 58:6 E U is isotropic if and only if S (E U) ( -1 ' —I . But E (S V) S (EV) S (E U). q. e. d. 58:8. Theorem. Let F be a field with the property that every regular quinary space over it is isotropic. Then two regular quadratic spaces U over F are isometric if and only if

and V

dim U = dim V , dU = dV , SU — SV Proof. We need only do the sufficiency. Let n be the common dimension. For 1 n 3 the result is true over any field by Theorem 58:4. So assume that n 4. Then U ± is isotropic by hypothesis, hence 1 E Q(U) by Proposition 42:11. Similarly 1 E Q (V). So we have splittings

U U' j_ < 1 > ,

V

V' ± < 1> .

But U' and V' have the same invariants. An inductive argument then gives V', Hence U V. q. e. d.

154

Part Three. Arithmetic Theory of Quadratic Forms over Fields

Part Three

Arithmetic Theory of Quadratic Forms over Fields Chapter VI

The Equivalence of Quadratic Forms One of the major accomplishments in the theory of quadratic forms is the classification of the equivalence class of a quadratic form over arithmetic fields. We are ready to present this part of the theory. Roughly speaking it goes as follows: the global solution is completely described by local and archimedean solutions, the local solution involves the dimension, the discriminant, and an invariant called the Hasse symbol, the complex archimedean solution is trivial, and the real archime dean solution is the well-known law of inertia of Sylvester. $ 61. Complete archim.edean fields

Let us consider the theory of quadratic forms over a complete archimedean field F, j. e. over a field which is complete at the archimedean spot p. We know from § 12 that there is a topological isomorphism of F onto either the real field R or the complex field C; in the first instance p is called real, in the second complex. It is best to treat the real and complex cases separately. § 61A.

The real case

If F is a real complete field, then (F:F2) --- 2 and ± 1 are representatives of the cosets of F modulo F2. So a vector x in a regular n-ary quadratic space V over F will satisfy exactly one of the conditions Q (x)

E È2 Q (x) = O, Q (x) E ,

We call V a positive definite quadratic space over F if (x) E .F12 V x E T.7 ;

we call it negative definite if Q (x)

E

.t2

xE

;

we call it definite if it is either positive or negative definite; we call it indefinite if it is not definite. We can refine an arbitrary orthogonal base to a base V = (F xi) ±• - 1_ (Fx,) I (Fyl) j . • • -

j.

(FJr.)

Chapter VI. The Equivalence of Quadratic Forms

in which Q (xi) = 1 for 1 5_ 5...p and Q ( yj) — 1 for 1 j in which — 1> ± • •• I . V I ••• I

155

g, j. e.

Here we have 0 p n and 0 g n. Since a sum of elements of f:2 is again in 11 we can conclude that V is positive definite if and only if p = n; it is negative definite if and only if g = n; it is indefinite if and only if 0 < p < n. Hence V is indefinite if and only if it is isotropic. We have Q (fr) equal to

_fr2

according as V is positive definite, indefinite, or negative definite. Every subspace of a positive definite space is regular and positive definite; similarly with negative definite spaces. The only space which is both positive and negative definite is the trivial space O. Suppose P is a maximal positive definite subspace of V. Then P is regular and P-2_-•1•••1.

Let P' be some other maximal positive definite subspace of V, say with dim P dim P'. Then there is an isometry a of P into P'. By Witt's theorem there is a prolongation of a to an isometry a of V onto V. So a-' P' will be a positive definite space containing P. Hence dim. P -,- dim P'. We have therefore proved: all maximal positive definite subspaces of V have the same dimension. We call this dimension the positive index of V and write it ind+ V. -Similarly define the negative index ind- V. It is Let us return to our orthogonal base x1,...,x„, easily seen that Fx, ± • • • ± Fx„ is a maximal positive definite subspace of V. And Fyi I • • • IFy g is a maximal negative definite subspace of V. Hence ind+ V .= p , ind- V = g . In particular, ind+ V ind- V . dim V . If p

g we have a splitting V=

I • • ± Hg I Vo

with Hi a hyperbolic plane for 1 I g, and Vo positive definite and therefore 0 or anisotropie. Hence in this case the index of V in the sense of § 42F satisfies ind V g. Similarly we have ind V = p when p g. In other words, ind V = min (ind+ V, ind- V) .


156

61:1. Theorem. Let U and V be regular quadratic spaces over a real complete field F. Then U is represented by V if and only if ind+ V, ind- U < ind- V.

ind+ U

For isometry the conditions read ind+ U = ind+ V, ind- U = ind- V. Proof. The proof is almost trivial and the details are omitted.

q. e. d. 61:2. Remark. < — 1> 1 < — 1> does not represent 1. Hence the quaternion algebra (— 1, — 1) is a division algebra by Proposition 57:9. On the other hand an arbitrary quaternion algebra will be isomorphic to (1, — 1) or (— 1, — 1). Hence there are essentially two distinct quaternion algebras over a real complete field F, namely /1, -1, —1 and ( F § 61B. The complex case Here everything is trivial. Since F is topologically isomorphic to C Hence every regular n-ary quadratic space V has a we have È = splitting

fr2

.

V--1•••±. And V is isotropic when n 2. Also

U

V if and only if dim U

dim V,

and

UV

if and only if dim U = dim V.

There is essentially one quaternion algebra, namely (1, F -1 )

§ 61C. Special subgroups of On (V) We conclude this paragraph by giving the structure of the groups

Zn , 0;,/S)„ , 0:1(4, of a regular n-ary quadratic space V over a complete archimedean field F. Recall that we first raised this question over a general field in § 56. In the complex case we can apply the results stated in § 56 and we find that with f2„1--\

4=

10 if n is even iv if n is odd.

Chapter VT. The Equivalence of Quadratic Forms

157

Now suppose-that F is real. Then by Proposition 55 :2 we have

(0:: 0) —

{ 2 if V is indefinite 1 if V is definite.

By Propositions 55 :5 and 55 :6 we hay e

(:);, = Q. Hence by Proposition 55 :7,

zn,

1 v} if dV = 1 with i v otherwise.

it

even

§ 62. Finite fields Next we consider quadratic spaces over finite fields. Let F be a finite field of q elements. Consider the multiplicative homomorphism

99: 14.'

-›-

defined by the equation q.x = x2 . The kernel of go consists of the elements ± 1 since the equation x2 — 1 0 has precisely these roots in the field F, and these two roots are distinct since F does not have characteristic 2. Hence F2 is a group of —2- (q — 1) elements. Hence F/F2 is a group of

2 elements. Thus every element of I: . is either a square or a square times a fixed non-square. 62: 1. A regular n-ary quadratic space over a finite field is universal if n 2. Proof. It is enough to prove this for binary spaces. By scaling the space we can reduce things to the following: prove that a typical binary space V represents at least one non-square. This we now do. Write V <s> _I_ with e, ô E F. If e or 6 is a non-square we are through. Hence we can assume that V-2-f 1 . If 1 is a square in F, then V is a hyperbolic plane and we are through. Hence we can assume that — 1 is a non-square. iF.2 and 1 + .t2 are finite sets containing the same number of elements. These two sets are not equal since 1 is in the first set but not in the second. Hence there is an element a in F such that 1 + oc2 is not in F.2. This element 1 + oc2 cannot be 0, hence it is a non-square in F, and this non-square is clearly represented by V _I_ . q. e. d. 62: la. A regular quadratic space V over a finite field F has a splitting

V .7'f

So there are essentially two regular quadratic spaces of given dimension over F.

158


3. 1 62: lb. V is isotropic n 62:2. Theorem. U and V are regular quadratic spaces over a finite field F. Then U is isometric to V if and only if dim U = dim V ,

dU , dV.

q. e. d. Proof. Apply Corollary 62:1a. 62:3. Remark. I
represents 1 whenever oc, fi EF. Hence every quaternion algebra is isomorphic to (1, —1) by Proposition 57:9. So there is essentially one quaternion algebra over a finite field F, namely

62:4. Remark. The factor groups n

,

of a regular n-ary quadratic space V (n 2) over a finite field are described by the equations (On+ :07,1 ) , 2, On = and + v} if dV = 1 with n even

nZn -

I.iv otherwise.

§ 63. Local fields Now consider a local field F. Let us recall some of the basic definitions and notation of §§ 13, 16 and 32. F is a field with a complete and discrete prime spot p and the residue class fieldF (p) is a finite field of Np elements. We let stand for the ring of integers (p), u for the group of units u (p), p for the maximal ideal m (p), n for a prime element of F at p, ord for the order function ordr , and I I for the normalized valuation We know from § 22E that the fractional ideals of F at p are the powers

pv= nvo

(v EZ) .

Remember that in this part of the book we are making the general assumption that the characteristic of F is not 2. However it is still possible for the residue class field to have characteristic 2. This is what happens for instance in the case of the 2-adic numbers. We shall call F a dyadic local field if its residue class field has characteristic 2; thus for a dyadic field we have z (1 ) = 0 and z(F (p)) 2. We call F non-dyadic if it is not dyadic; here we have (F (p)) >2 and (F) 0 or z (F) (p)) . This was originally proved by DICKSON and generalized by C. CHEVALLEY, Abh. Math. Sm. Hamburg (1935), pp. 73-75, to forms of any degree: every form of degree cl in cl 1- 1 variables over a finite field has Cl non-trivial zero.


159

Note that F is dyadic if and only if

0 < 121 < 1

(or 0 < ord2 < 00) ;

it is non-dyadic if and only if

121 = 1

(or ord2 = 0) .

We saw in § 62 that exactly half the non-zero elements of a finite field of characteristic not 2 are squares; in particular this is true of the residue class field F (p) of a non-dyadic local field. On the other hand if F is a dyadic local field, then its residue class field is a finite field of characteristic 2; since all finite fields are perfect we can conclude that every element of F (p) is a square, i. e. that

(F (p)) 2 = F (p)

if p dyadic.

This has the following important consequence in F: if e, El are units in a dyadic local field, there is a unit 6 such that

s' 862 modn. In particular, in the dyadic case every unit is a square mod p. § 63A. Quadratic defect 63:1. Local Square Theorem. Let a be an integer in the local field F. Then there is an integer 13 such that

1 ± 4n a

(1 + 2n IV. Proof. The polynomial nx 2 + x — oc is reducible by the Reducibility Criterion of Proposition 13:9. Hence we have 13, IT EF such that

(x — 13) (x

j9').

Then the product of the roots is equal to —'oc, hence one of the roots, say j9 must be in e. But 1 ± 11n-2 + 4 .7-1 a) = __(_..,27---1 ,

2

by the quadratic formula. Hence

1 + 47t — (1 + 27c 13) 2 . q. e. d. 63: l a. Corollary. Suppose e, 6 are units in F such that e 6 mod 4.7. Then E E ô u2. 63: 1 b. Corollary. I":-2 is an open subset of F. Consider any element of the local field F. Then 4: has at least one oc in F. We write expression in the form = n 2+ oc with ?I ,

= n2 +

(n,

F)

160


in all possible ways and take the intersection

Then b (e) is either a fractional ideal or O. We call b (e) the quadratic defect of e. Clearly

b • If e is a square in F, then b (e) = O. Using the Local Square Theorem one can easily show that the converse is also true. Hence

E F2 44- b = O. In particular e always has an expression

= 772+ a with ao = b (e) . From this it follows that

b (a2 e) = a2 b (e)

V a, e E F.

We have

b (e) = eo if ordo e is odd. When ordo e is even we can write = n2re with s a unit, and then b (e) 7C2r b (e). So it is enough to study the quadratic defect on the group of units u. For a unit s we can write s 62 + a with 6 Eu and

ao = b(e) S 0. What is the intuitive meaning of the quadratic defect ? We have just seen that having defect 0 is equivalent to being a square. Consider a non-square unit s with defect b (s) = p4 Ç o. Then we can write s = 62 + a with 6 E u, a E pd, while such an expression is impossible with an a in p4 +'. So here the quadratic defect is that ideal pd with the property that 8 is congruent to a square modpd but not modpd +1.

63:2. Let e be a unit in the local field F. I/ F is non-dyadic, then b(s) is 0 or 0; if F is dyadic, then b(s) is one of the ideals 0C40C4p-1 C4p -3 (-••CP3 CP Proof. 1) If F is non-dyadic, then it follows from the Local Square Theorem that b (s) is 0 whenever it is not o, j. e. whenever 1)(0 p. 2) Now consider the case of a dyadic field F . If b (s) C 4o then b(s) ç 4p and so E is a square by the Local Square Theorem, hence b (s) = O. It remains for us to consider an s with b (s) = peg and 4 0 c pd o, and to prove that d must then be odd. Suppose if possible that we have such an s with an even d. Put d 2r. We can write s = 62 + a with 6 E u and ao — pd. Replacing s by 462 gives us a new s of the form

s

1 + sigr2r


161

with si E u, b (s) = Or, and 2o C pr Ç o. By the perfectness of the residue class field there is a unit 61 such that O.= e modn. Then 1 H._ Eiger 1 + *or (1 + th ar)2 modulo

ger+ 1 •

Hence we have an expression 8 = (1 ± (51 7e)2+ al with al E p2r +1 .

This contradicts the fact that b (s) = p2r. Hence d must be odd. q. e. d. 63:3. Let s be a unit in the local field F. Then b(s) = 40 if and only if F is quadratic unramified. Proof. Here it is understood that F (VW) is provided with that unique spot which divides the given spot p on F. By Proposition 32:3, F(re-)

orivF

is also a local field. 1) First let us be given b (s) = 40. This certainly makes the given extension quadratic. Multiplying s by the square of a unit in F allows us to assume that s = 1 mod4, hence that —41 (s 1) E o. Now F can

(fr)

be obtained by adjoining a root a = ,c2 + x

(-1 + fr ) of the polynomial

—4- (1 — E)

to F. But Proposition 32:6 applies to this situation. Hence F (r)/F is unramified.

2) Now suppose F (11i)IF is quadratic unrarnified. We can assume that s is given in the form s = 1 A- oc with b (s) = oco. Since the given extension is quadratic we know that 4o S ao S o. This finishes the proof for the non-dyadic case. Now assume that F is dyadic and prove that oco = 40. Write A = —1+ I/E—. Then A (A -1- 2) = oc. Let 1 1 be the prolongation of the given valuation on F to F (fr). If we had 1A1 > 121 we would have loc1 = 1Al 2 > 141 ,

hence a would have even order in F (VW), hence it would have even order in F since the given extension is unrarnified; so b (s) would be equal to p2r with p2r 4o and this is impossible by Proposition 63:2. Hence 1A1 121. This implies that lai 141• Hence cco Ç 4o and so oco = 40 as required. q. e. d. 63:4. There is a unit s in the local field F with b (s) = 4o. If s' is any other such unit, then s' E s u2. Proof. By Proposition 32:9 there is an unramified quadratic extension EIF. Since x(F)--k 2 we can obtain E by adjoining a square root to F; O'Meara, lotroductigm to quadratic: fonts

11

162


since ELF is unramified this will have to be a square root of an element of even order. Hence we can write E = F (VW) with s a unit in F. Then b (s) = 4o by Proposition 63:3. Now consider e'. Then E' = F (V7) is quadratic unramified over F by Proposition 63:3. By Proposition 32:10 the two fields E and E' are splitting fields of the same polynomial over F, hence yir E F (ri), hence 1117E VF. So E' E 80. q. e. d. 63 : 5. Let E = 1 ± a be a unit in a dyadic field with 141 < ial 121 we would have loci = 1y1 2, contrary to the assumptions. Hence we cannot have b (s) S pd +1. So b (s) is indeed equal to ao. q. e. d. 63:6. Remark. Each of the ideals in Proposition 63:2 will actually appear as the quadratic defect of some unit e. To get defect 0 take E = 1, to get 4o apply Proposition 63:4, to get pd with d odd and 40 c p 4 Ç o take e - -- 1 + ad and apply Proposition 63:5. We conclude this subparagraph with local index computations that will be needed later in the global theory. For any fractional ideal pr with r> 0 the set 1 pr= {1 + curia E 0} is a neighborhood of the identity 1 under the p-adic topology on F. Clearly 1 ± pr is a subgroup of the group of units u and we have 1 pr+ic 1 + prs u

(if r > 0) .

63:7. Lemma. Let 0 be a homomorphism or a commutative group G into some other group, let 0 G be the image of G and Go the kernel of 0. Then for any subgroup H of G we have (G: H) = (0G: 0 H) (Go : H 0) , where the left hand side is finite if and only if the right hand side is.

Proof. By the isomorphism theorems of group theory we obtain (G : H) (G : G H) (G 0 H : H)

(0 G:0(G 0 11))(G 0 :G 0 nH) = (0G: OH) (G o : H0). q. e. d.

163


63:8. F is a kcal field at p, u is the group of units, and 1 + pr is a neighborhood of the identity with r> O. Then (1) (u : 1 -I- pr) = (N p — 1) (N p)r -1 ,

(2) (1 + pr) 2 = 1 + 2pr if prg, 2p . Proof. (1) Consider a residue class field F) of F at p. The restriction of the bar map is a multiplicative homomorphism of u onto the non-zero elements of F with kernel 1 + p. Hence (u: 1 + p) = Np — 1. Now the mapping yo(1 + = a. is easily seen to be a homomorphism of the multiplicative group 1 + pr onto the additive group F with kernel 1 + pr+ 1. Hence (1 + pr: 1 + pr+ 1) = Np. By taking the tower u 21+pDl+p 2 )•--D1+pr

we obtain (u : 1 + pr) (Np 1) (Np)* --- 1 . S 1 + 2pr when pr Ç 2p. We must reverse the (2) Clearly (1 + inclusion. So consider a typical element 1 + 2a rcr of 1 + 2pr with a E 0.

Then

+ 2 OE nr_, (1 + Goe )2 _ oc znar, + aye? (1 + /3 701 fl7eir E -t- p 2 whenever for some /3 E o. it is enough to prove that 1 fl Et' and prS 2 1'. By the Local Square Theorem there is a y E 21, such that (1 -I- y)2 .--- 1 + 9 7r2r. Then 17 + 21 = 121 and so

12 v1= 1Y(7 + 2)1= 113701 5- 127Er+ 1 hence y E pr + 1 Ç. Pr 63: 9. F is a local field at p and u is the group of units. Then

q. e. d.

(P: P2) = 2 (u : u2) = 4 (NWT".

Proof. 1) To prove the first equality apply Lemma 63:7 with G = H and Oa = lal. Then

(P :P2) = (1P1 1P1 2) (u : u2) = 2 (u : u2) . 2) To prove the second equality apply Lemma 63:7 and Proposition 63:8. This time take G = u, H = 1 + pi' for any r > 0 such that prç_ 2p, and 0 a = O. Then (u : 1 + pr) = 2(u 2 : (1 + p1 2) = 2 (u2 : 1 + 2 pr) = 2 (u2 : 1 p r +ord

Hence (u : u2) (u : 1 + pr)

2 (u : 1 + pr -Foed 2) .

Hence (u : u2) -,--- 2 (Np)°rd 2 q. e. d. 11*

164


§ 63B. The Hilbert symbol and the Hasse symbol In this subparagraph F can either be the local field under discussion or any complete archiMedean field. So F is either a local field at p, or p is real and complete, or p is complex and complete. In any one of these situations it is possible to replace the Hasse algebra by a simpler invariant called the Hasse symbol. In the definition of the Hasse symbol the quaternion algebras are replaced by Hilbert symbols which we now define. Given non-zero scalars a, ft in an arithmetic field of the above type the Hilbert symbol t a, 13\

k

P

or simply (a, 16), is defined to be + 1 if ce 4.2 + /3772. 1 has a solution E F; otherwise the symbol is defined to be — 1. 63:10. Example. Put E=F(110 So EIF is of degree 1 or 2. Then a E NE/FE if and only if (

P ) —1 .

Our first results refer to the local case only. 63:11. Let V be a binary quadratic space over a local field F and let the discriminant dV be a prime element of F. Let ZI denote a fixed unit of quadratic defect 4o. If y is any non-zero scalar, then V represents y or y.Z1 but not both. Proof. By scaling V we can assume that y = 1. 1) Our first task is to prove that V represents either 1 or 4. Since cl V is a prime element there is a splitting V <e>1 in which e and 6 are units in F. If F is non-dyadic, e will either be a square or a square times 4 by Propositions 63:2 and 63:4. We may therefore restrict ourselves to the dyadic case. The above unit e can actually be any unit represented by V. In fact let it be a unit of smallest quadratic defect in Q (V). We can assume that this E has the form E 1 ± fi with b (6) = flo c o. Thus

V rz- 1 One of three things can happen. (i) If b (e) = 0 we have 16' = 0 and so 6 = 1 and V represents 1 as desired. (ii) If b (e) = 4o then V represents a unit of quadratic defect 4o and so it represents 4 by Proposition 63:4. (iii) The one remaining possibility is 4o C b (e) C o. Let us prove that this is impossible. If 4o C fi'D C o we write 19 = el nk with el a unit and b pk Here k is odd. By the perfectness of the residue class field we can choose a unit A such that 2 2 6 = — e1 modn. Then V represents 1 ± egr k 2267.Ek 1

mode +1


165

since k is odd. In other words V represents a unit whose quadratic defect is contained in plc +1 • This is contrary to the choice of e. 2) Finally we have to prove that V cannot represent both 1 and Ll. If it did we would have 1 1 for some 8 C u. Hence A 4.2+ 77287/ for some 77 in F. By the Principle of Domination has to be a unit and 77 has to be an integer. Hence there is an element of the form AIP.= 1 + )32 8n (fi. Co) with quadratic defect 4o. This is impossible by Proposition 63:5. q. e. d. 63 : 11 a. Let e be any unit in F. Then —

I.

and

Proof. The first equation is a direct consequence of the proposition. Let us do the second. In the non-dyadic case we can find n in o such

that

e 2 +4 77 2= 1 modn since a binary quadratic form over a finite field of characteristic not 2 is universal by Proposition 62:1. Here has to be a unit since A is a nonsquare, hence we have A E o such that ep(1 A n) ± A lp = 1 Then 1 ± An is a square by the Local Square Theorem. Hence <e> (e, ) — 1. In the dyadic case use the fact < > represents 1, hence that represents 1 modulo 4e together with the perfectness of the residue class field to show that <e> I represents 1 modulo 4p. q. e. d. Then apply the Local Square Theorem. 63:11b. The quaternion algebra (n, A) is a division algebra. All quaternion division algebras over F are isomorphic to it. Proof. The quadratic space I does not represent 1, by the proposition. Hence (n, A) is a division algebra by Proposition 57:9. It is easily seen that every quaternion algebra over F is similar to a tensor product of quaternion algebras of the form (e, 8n) where e and 8 are units. It therefore suffices to prove that this quaternion algebra is isomorphic to (n , Z1 0) where A 0 is one of the elements 1 and A. But

± ce8nA 0 > . Computing Hasse algebras gives (e, 8n) (4 0, A 0 E 6)0 (A 0, n) . <e>


166

But (A 0 , A 0 e 6) 1 by Corollary 63:11a and Proposition 57:9. Hence q. e. d. (e, 6n) (40, a). 63:12. Example. Corollary 63: I I a shows the very simple nature of the Hilbert symbol over non-dyadic fields. For let e, 6 be units in a given ;16 ) — 1 if and — 1 always. And (-non-dyadic local field. Then ( only if 6 is a square. Now consider the local field or real complete field F. There are essentially two quaternion algebras over F. Hence œb Pi ) ", 1

tg) igign

if and only if the number of division algebras appearing in this tensor product is even. But ( /4 ) is a division algebra if and only if (") 1. Hence the given condition is equivalent to —

(ez b p i)

_ 1.

isin So we have 611

k 1Ztign

k F

if and only if (

°Lb

Pil

(

6

i)

iSigm‘

These results are of course trivial when F is a complex complete field. We can carry over to the Hilbert symbol the various formulas of Proposition 57:10. The first two of these formulas are trivial for the Hilbert symbol, the third now reads ice,a13\ . tœ,—P)

and the fourth,

k P/

k

P

fcc , P1(cc,31_(oc,PY) \PMP/ 63:13. Let F be either a local field or a complete archimedean field at

p.

Given any non-square /3 in P there is an a in F such that (a1P) --

1.

Proof. The complex case cannot arise. In the real case take a= — 1. Now the local case. We can suppose that /3 is either a prime element or a unit of the form /3 = 1 ± y with 10) = ye. Let LI 1 mod4 be a.unit of quadratic defect 4e. If /3 is a prime element take a . J. If ye — 4e


take a =

7C. If

167

40C yo Co take a = — p; then

± --A _L ; but

ordp a /3 = ordp a = ordp y

is odd; hence < a> J does not represent 1 by Proposition 63:11. q. e. d. 63:13a. Let E be a quadratic extension of F. Then (F NEIFE) = 2. Proof. Since the characteristic is not 2 there is a non-square /3 EF such that E = F (IT ) . Consider the multiplicative homomorphism ±1) defined by the equation cc,

p

•

This mapping is surjective by the proposition; its kernel consists of those a E.fr for which ( `13 ) — 1, i. e. for which a

E NEIrt. Hence

(F:NEIFE.) — 2 . q. e. d. Now let V be a regular n-ary quadratic space over F (here F is either a local field or a complete archimedean field at p). Take a splitting Vz-_1_ • • •_L and define the Hasse symbol

Sp V =

17 (oc. d 1in

P

where di = ocr .. . ai. This is independent of the orthogonal splitting chosen for V since the same is true of the Hasse algebra. We clearly have Sp U=Sp V , V -21 1 1 E F. Here n m 3. Then the quaternary space L

represents cc, since all regular quaternary spaces over a local field are universal by Remark 63:18. Hence V has a splitting with Pi . ch. It is conjectured that every form of degree d in di+ 1 variables over a local field has a non-trivial zero. See S. LANG, A nn. Math. 55 (1952), pp. 373-390. 1


171

Repeat this m times. We obtain a splitting of the form V

• ± < oc>

1. Let H be the subfield H = F(I/62 , ri0 of F(07). Then H is a global field and el is a non-square in H. By the Global Square Theorem there is an infinite number of spots 93 on H such that ei H.

These spots will induce an infinite number of spots on F. Consider a spot93 on H for which the induced spot p on F is in S. Then 8 n since 1-43. But for 2St:Ss the S-unit ei is a square in H, hence in Hp, hence eiej 4 1-43, hence eiei Fr,. so and eiei are non-square units at the non-dyadic spot p; this implies that ei is a square at p for 2 i s. ,

q. e. d. es be a set of generators of u modu2. Then there is a set of spots p 1, . , p , E S with the following property: each ei is a nonsquare at pi and a square at the remaining pi.

65:18. Let e,. .

Proof. This is an immediate consequence of Proposition 65: 17. q. e. d. 65:18a. Let a EF be a square at all spots in S2 — S and a unit at all p.). Then a is a square in F. spots in S — (pi u • • Proof. Write S' = S (p1v • • p.).. Suppose if possible that there is a non-square O in F which is a square at all pED—S and a unit at all p E S'. Put E = F(14). These suppositions will lead us to the absurd conclusion that /Ft PF IVElpfs ; this conclusion is absurd since (JF : PFNEIFJE) 2 by Proposition 65:14. In virtue of the general assumption that jp= Ppj; it will be enough if we show that J; ç PPNEIFIE. , c, with So consider a typical idèle i in J. Take numbers cl, ci --,- 0 when i is a square at pi, and ci . 1 when i is a non-square at pi. Let E be the S-unit e= ' • 48 -

Then e is a non-square at exactly those p i where i is a non-square. But each pi is non-dyadic. Hence the idèle (e)i is a square at each pi. If we apply Example 65:4 to this idèle and the set of spots S' we find that q. e. d. (e) I is in NElp JE, Hence i E PINE/i4E. 65 : 19. Every element of a global field is a square at an infinite number of spots.

Proof. Let F be the global field, E the element. We can assume that E is a non-square in F. Then e is an S-unit for a sufficiently small set of spots S. Suppose D — S contains at least two spots. Regard e as one of the generators in a set of generators of u mod u2 and apply Proposition 65: 17. q. e. d. 65:20. Let e be an S-unit in F. Then there are distinct spots q and q' in S such that (i) e is a square at q and q', (ii) any S-unit A which is a square at ci is a square at q', and conversely.


185

Proof. We can assume that e is a non-square in F, else we just replace it by one. If s = 1 we let q and q' be any two spots of S at which E is a square. So assume that s > 1. Take generators e, es, E. of u mod u2. By Proposition 65:17 we can pick q ES in such a way that E, Es, • • • Es-1 are squares at q with Es a non-square at q. Pick a second spot q' with the same properties as q. We have to show that any 6 Eu which is a square at q is also a square at cr. Write = esszl ears y with vi = 0,1 and y in u2. Then Es is the only quantity in the above expression which is not a square at q, hence v8 = O. All the remaining quantities are squares at hence 6 is a square at q'. q. e. d. §65D. Norms 65 : 2 1. (h. : .1),NzipjE) = 2. Proof. 1) In virtue of Proposition 65:14 it is enough to prove that ( h: PFIVEtij E) 5_ 2. Take a set of spots S on F which satisfies the general assumptions of this paragraph and is also so small that 0 is a unit at each of its spots. We know from Proposition 65:12 that (JF : Pp . 8' 2) = 28 and from Example 65:4 that PFJP 2 S PFNELFJE. The idea of the proof is now simply stated: put 01 = P, J' 2 and try to find a strictly ascending tower

OICO2C• — ( 08

of subgroups of PFNE/FJE. If this can be achieved we shall have UP PFNEIFJE)

UF: PFJP

(PrNmaJE P.F.8.2)

5_ 26 ÷ (Os 01)

28 ÷ 28 -1 = 2. 2) So we have to find the tower. Since 0 is a non-square in u we have a set of generators 0 , E2, • . • , E,,

of u mode. Select a set of spots pi,

, p 8 as in Proposition 65:18 (with e1 = 0). Then 0 is a square at p2, •p a and so npi = 1 for 2 Sj.:5 s. Hence PjS NRipfs for 2 5_j5.sby Example 65:2. So if we define 01 -= PFJP 2 and 01 = .410j _l for 2 5j5_swe obtain a tower 45 2

••• of subgroups of PFNEtilz . It remains for us to prove that this tower is strictly increasing. If not we would have Pi S 01 _ 1 for somej(2.j5_s). Take an idèle i in 117 which is a prime element at p i. Then i is in hence it is in (Pp.8.2) 11-'t

186


So we have an idèle t which is a square at all pED—S and a unit at all p E S — (p1 j • • u 1) 1 _1), and also a field element a, such that 1= (a) 1. Comparing coordinates shows that a is a square at all p ED—S and a p.). But then a is a square in F by unit at all p E S (pi v.) • Corollary 65: 18a. And this is absurd since the equation i = (a) t also shows that a is a prime element at pi. So we do indeed have q. e. d. as required. 65 : 22. Suppose that 0 is a unit at all spots in S, and that JE . PEA. Then Piln NE/A = NEIP P. Proof. This follows immediately from Propositions 65:13 and 65:21. q. e. d. 65:23. Hasse Norm Theorem. Let E be a quadratic extension of the global field F. An element a of F is a norm in the extension EIF if and only if it is a local norm at all spots on F.

Proof. We need only prove the sufficiency. Consider a non-zero element a of F which is a local norm at all spots on F. Take a set of spots S satisfying the general assumptions of this paragraph and so small that 0 and a are units at all p E S with JE== PEA. Then the idèle (a) is in 11; it is also in NE /E n by Example 65:3; so by Proposition 65:22 (a) E P n NE/F4= NE/F P1 q. e. d. Hence a E NELF U S NE/F E. .

§ 66. Quadratic forms over global fields Now we can describe quadratic forms over global fields". We consider a regular n-ary quadratic space V over a global field F. The corresponding bilinear and quadratic forms will be written B and Q. Completions Fp are taken and fixed at each spot p on F. We use Vp for the Fp-ification Fp V of V (as a vector space or as a quadratic space) and we call Vp a p-ification or localization of V at p. Here p can be archimedean or discrete. We say that V is isotropic at p if its localization Vp is isotropic; similarly we say that U and V are isometric at p if their localizations are isometric. The Hasse symbol Sp Vp will be written Sp V; its value can be computed directly from an orthogonal splitting V 1•••± for V, say by the formula cti)

Sp V= 1 , P 1 < O>. We .can suppose that 0 is a non-square in F. Take localizations L p , Pp inside Vp in the natural way. Since Vp is isotropic we know from Proposition 42: 11 that Pp represents a at each p on F. Hence the equation — Or (4, has a solution at each p on F. This implies that a is a local norm of the extension E = F(1/(7) over F at each spot p on F. Hence by the Hasse Norm Theorem a is a norm in the extension EIF. So we have a solution —

—

—

—

_

—

on 2

E

F)

Hence V is isotropic'. 4) The quaternary case. First suppose the discriminant dV is 1. Take a regular ternary subspace U of V. Then dV p is 1 and Vp is isotropic at all p on F, hence Up is isotropic at all p by Proposition 42:12. Hence U is isotropic by step 3). So V is isotropic and the case dV = 1 is settled. Let us reduce the general quaternary case to the above. So assume V is quaternary with dV +1. Form the quadratic extension E = F(11 aia2a3a4 ) of F and consider the E-ification EV. For each spot93 on E the localization (E V)

Etpx i ± • • • j_ Ev x4

contains FT; xi. ± • • • ± F93 x4 , Thus the binary and ternary cases are essentially interpretations of results of global class field theory. An examination of the proofs of § 66 will show that the rest of the global theory of quadratic forms is derived from these two cases by simple algebraic and arithmetic methods. Needless to say it would be of great interest and importance to have a direct proof of the entire theory. In the classical situation of the field of rational numbers Q the binary and ternary cases can be obtained by elementary methods, but then one runs into difficulty at tt = 4, and this difficulty is usually overcome by using Dirichlet's theorem on primes in an arithmetic progression together with Hilbert's reciprocity law. For other approaches to the theory over Q we refer the reader to C. L. SIEGEL, Am. J. Math. (1941), pp. 658-680, and to J. W. S. CASSELS, Proc. Cam. Phil. Soc. (1959), pp. 267-270. 1

188


and this latter space is easily seen to be isotropic since Vp . Fpx,

- • • 1.

F,, x4

is isotropic at the spot p induced by 93 on F. Hence (E V) is isotropic for all 93 on E. But d (E V) = 1. Hence E V is isotropic. Hence V is isotropic by Proposition 58:7. 5) Higher dimensional case. Here the proof is by induction to n. Assume n 5. Put U = Fx3. 1. Fx2 , W Fx3 ± • • • I. Fxn, so that V --- U1. W. Take the localizations Up •-• Fp x1

Fp x2, Wp = Fp X3 ±

Fp x,„,

inside Vp. Thus Vp = Up l Wp. Put T = {p

E 12/0 I Wp is anisotropic} .

The set T must be finite by Example 63:14. If T is empty we have W isotropic by the inductive hypothesis and we are through. So we consider a finite non-empty set T. There is a pp in Pp at each p in T such that tip

E Q (u p) ,

E Q (T p)

if Up is anisotropic this a consequence of the isotropy of Vp, otherwise it is a consequence of the universality of hyperbolic planes. So we have np E Fp at each p E T such that Q x1 + rip x2) = gad- a2 = . Use the Weak Approximation Theorem to find n in F with close to 4 and 27 close to np at each p E T. Put Q(xi-F n x2) = ,u. By taking good enough approximations we can make is arbitrarily close to ,up, hence we can make I,u /AV— ll p arbitrarily small at all p in T. Since Fi!, is an open subset of Fp we are sure that we can obtain IL

EILpi

V pE T.

Since x1 d- n x2 is in U we have V ^.1- j. W. Then _LW is isotropic at all p in T since then (W p) — 14 E - ,upn

and it is isotropic at all the remaining spots on F by choice of T. Hence q. e. d. j W is isotropic. Hence V is isotropic. ,

66 : 2. Theorem. An n-ary quadratic space over a function field is isotropic whenever n 5.


189

Proof. The quadratic space in question is isotropic at all spots on the function fi eld by Proposition 63:19, hence it is isotropic by Theorem 66: 1. q. e. d. 66:3. Theorem. U and V are regular quadratic spaces over the global field F. Then U is represented by V it and only if Up is represented by Vp for all p. Proof. Suppose V represents an a E F at all spots p. Then < — a >1 V is isotropic at all p, hence it is isotropic by Theorem 66: 1, hence V represents a. Therefore V represents a whenever it does so all spots on F. This proves the theorem in the case where dim U = 1. We proceed by induction on dim U. Take a non-zero a in Q (U). Then a E Q (Up) Q (V p) , hence a is represented by V at all spots, hence it is represented by V. So we have splittings U L 1/ 1 , V = ± V' in which L and K . It follows easily from Witt's theorem and the representation Up-÷- Vp that (4.-+-V; for all p. Hence U'-+--- V' by the induction. Hence U V. q. e. d. 66:4. Hasse-Minkowski Theorem. U and V are regular quadratic spaces over the global field F. Then U is isometric to V if and only if U p is isometric to Vp for all p. Proof. By Theorem 66:3 there exists a representation U V. Since U is regular this representation is an isometry. q. e. d. 66:5. Remark. We have just shown that an arbitrary quadratic space V over a global field F is completely described by its local behavior at the spots of F. Using our earlier descriptions at the discrete and archimedean completions we find the following complete set of invariants for V: (1) the dimension dim V, (2) the discriminant dV, (3) the Hasse symbols Sp V at all discrete p, (4) the positive indices incl.; V at all real p. Here inq- V is used for the positive index of the localization Vp of V at p. Of course the fourth invariant can be omitted over function fields. 66: 6. Example. Let V be expressed in the form V--.'2

such that 4, 4. 1 1 modp„ ,

A,* 1 modp„.

Suppose we had A p 7,- 1 modp, for some ,u v. Then using the binomial theorem with the fact that A, is a power of A p would give us A, 1 modp„ and this is false. Hence A t,* I modp, if This shows first that pp + p,, if p < y, for otherwise we would have A p+1 1 modpp with 1u + 1 5_ y and pp .,- p„. In other words the prime numbers pi, p., . . . constructed above are distinct and therefore infinite in number. Secondly, it shows that A, * 1 modp, , A„ + „ 1 modp, , in other words that a2 * 1 modp, ,

a2v+ 1 1 modp„.

Thus the period of a modp, is a certain power of 2. q. e. d. 71:15. Let p be a discrete spot on F. Then for an infinite number of Prime numbers p there is a p-cyclotomic extension KIP' which is unramified of local degree 2 at p. Proof. By Lemma 71:14 there are infinitely many prime numbers p such that Np has even order modp. We shall show that any p which has this property and is also a unit at p will work. Take L == F (C) with a

198


primitive p-th root of unity. Then LIP is unramified at p since p is a unit there. Let denote the degree of inertia of the extension at p. Then is a root of unity of period prime to Np, hence it is an ((Np)'— 1)°4 root of unity by Proposition 32:8. But the period of is p. Hence p divides ((NW— 1), in other words (Np)' 1 modp Hence is even by choice of p. so LIP is unramified of even local degree at p. Then by Corollary 15:10a there is a field K such that F with KIF unramified of local degree 2 at p. Of course, KIF isp-cyclotomic. So K is the field we are after. q. e. d. 71:16. Let pi and p2 be discrete spots at which the quadratic extension EIF is unramified of local degree 2. Consider inks ii E

Then

41 - NELFJE •

12 E Ij — N ELFIN.

f 2 E P.FNEIFJR

-

Proof. 1) Let C be an algebraic closure of E. By Propositions 71:9 and 71:15 there is a prime number p, which is a unit at pi, and a prcyclotomic extension Iii/F which is unramified of local degree 2 at pi, which is contained in C, and which satisfies [KlE:E] = [K i :F] . In the same way we can find a prime number p, which is a unit at p2, and a p2-cyclotomic extension K 2/F which is unramified of local degree 2 at p2, which is contained in C, and which satisfies [K 2EK I : EK i] = [K 2 : F] . We are actually going to work inside K1 EK2 and we can now forget about C. We have [K 1 EK 2 :F] [E :F] [K 2 :b.] . /16E1(z

/ /6E

Fig. 1

Using this equation in Conjunction with the fact that N the degree of an extension is EK2 not increased by a field translation, we can easily check that the sides with the same ruling in Figure 1 give field extensions of equal degree. The extension K 1EK 2/F is galois since KIT, EIF, K 2IF are all galois. Now the

Chapter VII. Hilbert's Reciprocity Law

199

action of any automorphism of KI EK2/F is completely described by its action on K11F, K 2IF; but each of these extensions is abelian; hence any two automorphisms of K1 EK 2/F commute; hence K IEK 2/F is abelian. So any intermediate extension is galois and abelian. 2) We fix a spot 931 on KiE which divides the given spot pi on F. Now K1v1/F931 and Evi/Fsp, are um-amified extensions of degree 2, and they are subextensions of a common extension (K1 E)sp1fFsp1, hence they are equal by Example 32:11. By Proposition 11:19 we have (Ki E)Ti = (K1 E) Ev1 = Hence K i EfF is unramified of local degree 2 at pi. Let F1 denote the decomposition field of this extension at pi. Then K1 E/F1 is quadratic by Proposition 15:10, and it is unxamified of local degree 2 at 931. We have E 1 ) F931 = F1931 so that E is not contained in F1. Hence KIE =FIE. Similarly Ki E Repeat all this with p2 to obtain a spot 932 on EK 2 and a decomposition field F2 of E K 21F at p2. This parallels the situation at p1 and analogous equations are obtained. In particular EK 2 = EF2 and EK 2 = F2K2. We shall be interested in the compositum F1F2. We have F1 EF2 = KI EF2 = KIEK 2

so that KI EK 2/FIF2 is at most quadratic. Using this fact together with Figure 1 and an easy degree argument, we find that the sides with the same ruling in Figure 2 correspond to extensions of equal degree. In particular, NEKi Kl E K 2IF1F is quadratic. // / 11 3) Since K1EK2/F1F2 is $i // quadratic it follows from the Global Square Theorem and / Proposition 71:7 that there is a discrete spot 93 on KI EK 2 such that KiE K 2IF1Fa is of local degree 2 at 93, such that p1 and p2 are units at 93, and such that K1 E K21F is unramified at 93. The sigFig. 2 nificance of this choice of (43 lies in the fact that F1F2/F1 and F1F2/F2 have odd local degree at 93. (We shall use the same letter 93 for the spot induced by 93 on a subfield of KI EK2.) That this is actually true follows thus: if F1 F2/F1 had even local degree at 93, then we would have Ev Ç (F1F2)13 by Example 32:11, and this would imply that KI EK2 = F1 EF2 had

200


local degree 1 over F1 F2 at 93. This is contrary to the choice of 93. So F1F2/F1 does indeed have odd local degree at 93. Similarly with F1 F2/F2. 4) Take an idèle j1 E al whose 931-component is a prime element of F at pi ; then the 93 1-component of j1 is actually a prime element of F1 at 931 since FIT is unramified at 93/ • Similarly take j2 E I. Similarly j E 1.13-13Fa• Then Npirdr i j is an element of and its 93-coordinate is an odd power of a prime element of F1 at 93 since F1F2/F1 has odd local degree at 91 But K/ E = FI E is a quadratic extension of F1 which is unramffied of local degree 2 at 93, by choice of 93. Hence

a.

E

NFiFzfri j

Nic i E/F,JKiE

by Examples 63:16 and 65:2. Now If1 E = IfiFi is a A-cyclotomic quadratic extension of F1, and p, is a unit at93, and the local degree of the extension at 93 is equal to 2, hence we can apply Proposition 71:12 and we find that PF,Nxix.F,J.K,E

•

We can therefore conclude from the equation n NRiEl-FLJNi.E)

that

1= 2

N717217 PF,N.gi EfF,JKiE • The same sort of reasoning will show that h 4 P.Fi NK i EfFi Ll iE • JKIE) = 2 by Proposition 65:21. Hence

Now ( JF, :

hN.F1P21F2 j

E P.F,NR,EiFi JK,E •

Taking NFi lp gives us (N.Fir./Pj) E PFNE.E/F.TRIE PFNEirJE Now Np1iF j1 is a prime element at pi since PVT' has local degree 1 at pi ; but EIF is unramified of local degree 2 at pi ; hence (NYJF

Nrif.Ph E - NEIFJE by Examples 63:16 and 65:2. But (Pi : r\ NE!? JE) = 2. Hence (NAir

Hence il(NFi

ji) -1 E NEIF JR j)

•

E PENEbpJE

On grounds of symmetry we have

Hence

I (NA PdF i) E PFNE/FJE

E PFNEIFIN •

q. e. d.


201

71:17. Let q be a discrete spot at which the quadratic extension EIF is unramified. Then 111„ .0 PFATEIFJE

if and only if EIF has local degree 1 at q. Proof. We suppose, if possible, that we have

4, s p,,NE,,,TE with

EIF of local degree 2 at q. We shall use this to arrive at a contradiction. Let S denote the set of all discrete spots at which 'Ell' is unramified. So S consists of almost all spots in Q. If p is any spot in S at which EIF has local degree 1, then Pi;,SNRIFJE and so .1;SPFNEIFJE . If EIF has local degree 2 at the spot p in S then we can reach the same conclusion by applying Proposition 71:16 with pi = q and p2 — p. In other words, our assumption leads to the relation 4 Pr NE/FA VP ES• Let us use this to prove something which by Proposition 65:21 must be absurd, namely that JEC PENE/EJE. So consider a typical I in JE. Use the Weak Approximation Theorem to find an a in F such that loci— 1I p is small for all p in Q — S. Since Fl is an open subset of Fp we have a E ipiI2, V pEQ—S. Replacing i by (a) i allows us to assume that i is a square at all p in because of the assumption Q -- S. We can also assume that i is in that I; c przvv ili, for all p in S. But this refined idèle i is a local norm at all p in Q in virtue of Example 63:16; hence it is in NE/EJE by Example 65:2. q. e. d.

4,

§71 D. Conclusion of the proof 7 1: 18. Theorem. Let a and 13 be non-zero elements of an algebraic number field F. Then their Hilbert symbol is 1 for almost all p, and p ED

P

1.

Proof. The first assertion follows from the Product Formula and Example 63:12.

1) We start by proving the following: given any three distinct spots P1 , p 2, p3 at which fl is not a square, there is a y in F such that ( Y1 P ) is —1 at two of these spots and + 1 at all remaining spots on F. To see this we construct the quadratic extension E ---- F 0) of F. Since EIF has local degree 2 at p/ we can pick

ii E 4. — NE/FJE .

202


Similarly pick 12, 13 corresponding to p2, p3. Then one of the idèles 11 12, 12 i3, ta il must be in Pr NELFJE since this group has index 2 in Jp by Proposition 65:21. Let us say

E PFNE/FJE This means that there is a y in F which is a local norm in EIF at all spots except pi and pa where it is not. Hence y, k P

if

—1

=

= Pi, P2

1 . if P

PI , Ps.

2) If the reciprocity law did not hold for a, 13 we would have )

1.

PED

In this event we could use step 1) to find a new a and a single spot q such that if p q k P ) 1 1 if p —q. We wish to make a slight alteration to j9. We take X. —1 when q is a real archimedean spot (q cannot be complex). If q is discrete we let X be an element of F which is a unit with quadratic defect 4oq at q. Then the quadratic space V= 1
± is isotropic at all spots except q where it is not. Clearly V ± is isotropic at q where q is real; by Proposition 63:17 it is also isotropic for a discrete q since the discriminant X is then a non-square at q. Hence V .1 < X> is isotropic by Theorem 66:1. So V represents X. So V < X> _L < — a' X > is isotropic at all spots except q where it is not. Hence I —1 if p = q k P ) 1 1 if p E q. This is impossible for a real q by Proposition 71:13. Hence we may suppose that q is discrete. In this case we consider the quadratic extension E F (ilr) of F. The information about the Hilbert symbol says that a' is a local norm of E IF at all spots except q where it is not. We claim that c P,N„,,,JE. Consider a typical idèle Then i is also a local norm at all spots except q; hence (oc') i is a local norm at all spots on F; so IE

NELFJE

E (d)' NEIFJE Ç

PFNEIFJE


203

and we have established our claim that

c p,NEI,JE On the other hand EIF is unramified of local degree 2 at q since is a unit of quadratic defect 404 at q. This is impossible by Proposition 71:17. q. e. d. 71: 19. Theorem. Let T be a set consisting of an even number of discrete or real spots on an algebraic number field F. Then there are a, p in 1 such that i ct,p\ J-1 if PE T kP) 1 1 if pED— T. -

Proof. fi can be any element of F which is a non-square at all spots in T. Such an element always exists: for instance the Weak Approximation Theorem provides a /3 which is a prime element at the discrete spots in T and a negative number at the real spots in T. Put E = F (11-g) and define a group homomorphism (± 1) by the formula

where i = (ip) pEs2 denotes a typical idèle in Ir. Then NELFJE is in the kernel of by § 65A; and Pip is in the kernel of (7) by the Hilbert Reciprocity Law; hence PFNE/FJE is in the kernel of Ti; now (p is surjective by Proposition 63:13, and (JF:PFNEIFJE). 2 by Proposition 65:21, hence PENE/FIE is precisely the kernel of 9). Take an idèle j E IF which is a local norm at all pED—T and is not a local norm at any p E T. Then 9? (j) = 1 since T contains an even number of elements, hence j E PpNE/Ej.E. But this relation can also be read as follows: there is an a in F which is a local norm at all p in D — T, and at no p in T. This a gives the desired values to the Hilbert symbol. q. e. d. 7 1:19a. Corollary. /3 can be any element of F which is a non-square at all spots in T. § 72. Existence of forms with prescribed local behavior 72:1. Theorem. A regular n-ary space Up is given over each completion F an algebraic number field F. In order that there exist an n-ary space V Up for all p, it is necessary and sufficient that over F such that V (1) there be a do in F with dUp . do for all p, (2) Sp Ur . 1 for almost all p, (3) // Sp Up . 1.

PEI)

204


Proof. 1) Necessity. To obtain the first condition take do = dV. To obtain the second and third conditions consider a splitting Va4±•••1 . cci, aci . r Each of these is equal to 1 for almost all p, hence Sp Up is. Now apply the Hilbert Reciprocity Law. 2) We must prove the sufficiency. If n = 1 we take Y . <do >. So we assume that n 2. Let T be a finite set of spots on F which contains all archimedean spots and also all spots p at which Sp Up = 1. Write

Then Sp Up = Sp V p is a product of Hilbert symbols of the form

j_ • • • 1.
, P' ± .

205

Chapter WI. Hilbert's Reciprocity Law

Then Pp and .13 , are isometric at all archimedean spots by choice of fl; applying Theorem 63: 20 at the discrete spots shows that .131„- P; for all p in D — R, and Pp P; for all p in R. So we have established our claim. 4) Consider Wp at any p in R. This Wp cannot be a hyperbolic plane since Wp and Up are non-isometric spaces with the same discriminant. We are also assuming that dim Wp 2. Hence

—>— (P _L

Vp ER

by Theorem 63:21. At any p in D R we also have such a representation since then P; P , by step 3). Hence there is a representation P'—>— (Pi_ W) by Theorem 66:3. Hence there is a quadratic space V over F with

This V has discriminant do = d W since ci P' = d P. For each p in D — R we have Pp s-:-. .13 , by choice of P, P', hence by Witt's theorem Up Vp ED — R . If p is in R we have Vp Wp sin.ce 134, Pi,; but Up * WI, by definition

of R; and dIgr = dUR = do ; hence by Theorem 63:20 we must have

Sp Vp — SW= SU R ; so Vp L-2_- Up for all p in R. Therefore Vp z'__ Up for all p and the space V has the desired property. q. e. d.

§ 73. The quadratic reciprocity law We conclude this chapter by finding an expression for the Hilbert symbol in terms of the Legendre symbol over the field of rational numbers Q. Recall the definition of the Legendre symbol: if p is an odd prime number, and if a is any rational integer that is not divisible by p, then

the Legendre symbol (f) is defined to be +1

or —1

according as a is or is not congruent to the square of a rational integer modulo p. In other words, (±i-) is 1 if the natural image of

z/p z is a square, otherwise

a in the finite field

(1,-) 6 is — 1. Now for any finite field K of

206


characteristic not 2 we have (K :K2) = 2 by § 62. Hence fa\ Ib\_lab\

UT)

lT)

Note that the Local Square Theorem tells us that

a (—

) --= 1 if and only

if a is a square in the field of p-adic numbers Q. b We shall use ( ) for the Hilbert symbol over Qp . We know from the formulas of § 63B that the Hilbert symbol is completely determined once its values are known, first for all rational integers a and b that are prime to p, and secondly for all rational integers a that are prime to p with b = p. We shall therefore restrict ourselves to these special cases. 73: 1. Let ft be an odd prime number, and let a and b be rational integers prime to p. Then a,b\ a\

‘25 ) —" k15 1 91 . Proof. The prime spot p is non-dyadic since the prime number p is odd. Apply Example 63:12, using the fact that the p-adic unit a is a square in Qp if and only if (7 a, = 1.

q. e. d.

73:2. Let a and b be rational integers prime to 2. Then

ta,b\

k

2)

a-1

(

—

b-1

(a2 -1)

2

1)

(—

Proof. 1) Every odd rational integer is clearly congruent to one of

the numbers 1, 3, 5, 7 modulo 8. Hence by the Local Square Theorem we can assume without loss of generality that a is one of these four numbers. The same with b. Now if u denotes the group of units of the ring of 2-adic integers Z2, then (u:u2) = 4 by Proposition 63:9. But every element of u is a square times 1, 3, 5 or 7 by the Local Square Theorem and the power series expansion of Example 31:5. Hence the numbers 1, 3, 5, 7 fall in the four distinct cosets of u modulo u 2. In particular 5 is a non-square, hence it is a unit of quadratic defect 4 Z2 . So by applying Corollary 63:11a we find that our proposition holds whenever a or b is 5. Of course the proposition holds whenever a or b is 1. We therefore restrict ourselves to a 3 or 7, b = 3 or 7. 2) We have 7 + 2(3)2= 25, hence - -

f7 2 2

k

1 8 -= 1


207

Then (3,2 2 ) ( 5,22 ) 2 ) = ( 7,2

8 •

Hence the second formula of the proposition holds for all a. 3) It is easily seen that (33\_ 2 )2= ( 3,7 ) _ (7,7 2 ) .

We will be through if we can prove that these three quantities are — 1. Now by Proposition 63 : 13 there is a 2-adic number c such that (--f 7 ) =— 1 2 since 7 is a non-square in Q2. But (1 ) = 1 by step 2), hence we can assume that c is a 2-adic unit, hence that c is 1, 3, 5 or 7. But c cannot be 1 or 5, hence c is 3 or 7. In either event we have our result. q. e. d. We cannot resist giving a proof of the famous Quadratic Reciprocity Law. This is obtained instantly from the Hilbert Reciprocity Law and the above formulas. Here then is the Quadratic Reciprocity Law with its first and second supplements. 73:3. Theorem. Let p and q be distinct odd prime numbers. Then p-1

P . ( I,q ) = (— 1) (-0 p-1 ---i-

-1

(T) — (

1)

q--1

2

2

f

/2

'

Proof. By the Hilbert Reciprocity Law we have

iii (12m _ 1 where 1 runs through all prime spots including 00. But ( Poe"' ) — 1 since p and q are positive real numbers. And if 1 is any prime number distinct from p, q, 2 we have — 1 by Proposition 73:1. Hence ( 111 / q

)

(P q\ p ) ( P yq )— — (—2 P q) Apply Propositions 73:1 and 73:2. This proves the reciprocity law. The first supplement is obtained in the same way from the equation ,

H ( ---1, ' P ) _

I

,

i

the second from the equation

H i2,p1_ 1. 1k 11

q. e. d.

208

Part Four. Arithmetic Theory of Quadratic Forms over Rings

Part Four

Arithmetic Theory of Quadratic Forms over Rings Chapter VIII

Quadratic Forms over

Dedekind Domains

The rest of the book is devoted to a study of the equivalence of quadratic forms over the integers of local and global fields. Our first purpose in the present chapter is to state the nature of this problem in modern terminology and in the general setting of an arbitrary Dedekind domain. Our second purpose is to develop some technique in this general situation. The more interesting results must wait until we specialize to the fields of number theory. . We carry over the notation of Chapter II. F is a field, 0 = 0 (S) is a Dedekind domain defined by a Dedekind set of spots S on F, I = I(S) is the resulting group of fractional ideals, u — u (S) is the group of units of F at S. So here y is the ring of integers of our theory. As in § 22C we shall allow the same letter p to stand for a spot in S and also for the prime ideal of o that is determined by this spot. V will denote an n-dimensional vector space over F. In the second half of the chapter we will make V into a quadratic space by providing it with a symmetric bilinear form B: V x V -->- F. The general assumption that the characteristic of the underlying field F is not 2 will not be used in the first paragraph of this chapter.

§ 81. Abstract lattices § 81A. Definition of a lattice Consider a subset M of V which is an o-module under the laws induced by the vector space structure of V over F. We define

FM = lax! a EF, x EMI.. Since M is an y-module, and since F is the quotient field of o, we have

FM --, {ce-liel cc Eo,

oz+0,

x EM} .

From this it follows that FM is a subspace of V, in fact the subspace of V spanned by M. Given a C F and a E l we put

ocM = {cxxlx EM} ,

aM=afixiflEa,xEMI VIII

Chapter

VIII. Quadratic Forms over Dedekind Domains

209

These are again e-modules and the following laws are easily seen to hold:

a(M n N)=-- (OEM) n (OEN) (ao) M = OEM, (aa) M — (a + b) M = aM a (M

(a M)

(ab) M — a (6M)

bM ,

N) = QM aN

F(M N)— FM -PFN. We call the above o-module M a lattice in V (with respect to o, or with respect to the defining set of spots S) if there is a base x1 , . . x n for V such that M o x, • - • I- o x, ; we say that M is a lattice on V if, in addition to the above property, we have FM V. In particular, 0x 1 + • • • + ox a lattice on V. The single point 0 will always be regarded as a lattice. 8 1 : 1. Let L be a lattice on the vector space V over F. Then the o-module M in V is a lattice in V if and only if there is a non-zero a in o such that OEM L.

Proof. 1) First suppose that M is a lattice in V. So there is a base xn for V such that xi +

M

• • +

Oxn.

Since L is on V we can find n independent elements y 1 ,.. . , yn € L. Write

a1y j= 1

(ai F) •

These a" generate a certain fractional ideal, hence there is a non-zero a in such that aa15 (o for all 1,j. Hence

ocx5 (oy1 +»+ øyÇL , hence OEM C L. 2) Now the converse. We have a non-zero a in such that OEM c L. Since L is a lattice there is a base z1,...,z„ for V such that L Ç o z1 + • • •+ o Zn . Then

m

cc—iL

Hence M is a lattice in V.

o ( z.1 )

•••+

z:) . (-q. e.

d.

8 1 : I a. Let U be a subspace of V with Mc UC V. Then M is a lattice in V if and only i/it is a lattice in U. O'Meara, introduction to quadratic' forms 14

210


Proof. Take a base z11 . .

x, for U and extend it to a base x1, .. . , for V. Put L'= x 1 + • • + ox, and L = oxi + • • • + ox.. If M is a lattice in U, then OEM Ç L'C L for some non-zero a in o, and so M is a lattice in V. If M is a lattice in V, then OEM CL for some non-zero a in o, hence OEM CLr\U=L', and so M is a lattice in U. q. e. d. It follows immediately from the definition that every submodule of a lattice is a lattice. In particular L n K is a lattice whenever L and K are lattices in V. And Proposition 81:1 shows that aL, aL, L K are lattices for any a E F, a E I. Clearly o x is a lattice for any x in V, and ax is also a lattice in V. Hence a1 z1 -1- - a, z,. is a lattice for any ai E I, zi EV. In particular, every finitely generated o-module in V is a lattice.

§ 81B. Bases Consider the lattice L in V. For any non-zero vector x in FL we define the coefficient of x in L to be the set a.= {a E F lax E L} . This is clearly an o-module in F, and it follows from the fact that L spans FL that it is not zero. Now

axx =LnFx, hence a. x is a lattice in Fx, hence a (a.x) Z ox for some non-zero a in o. Therefore aa.0 o, so that a. is actually a fractional ideal in F. Note that It is clear that

ccacco= (to aa; D

O

VocEiX

EL .•

We say that x is a maximal vector of .L if a= o. So x is a maximal vector of L if and only if L n Fx = ox

Every line in FL contains a maximal vector of L when the class number of F at S is equal to 1, i. e. when every fractional ideal is principal. For consider the line Fy in FL. Put ay . ao with a E F, then put x ---- ay; we have a.= o, hence x is a maximal vector of L that falls in the line Fy. 81:2. Given a lattice L on V, a hyperplane U in V, and a vector x o in V — U. Then among all vectors in x o U there is at least one whose coefficient with respect to L is absolutely largest. Let this coefficient be a. Then for any vector x o uo (uo E U) with coefficient a we have L = a (x0 /40) (L n U) .

Proof. 1) We claim that the set a . {a EFlax0EL

Chapter VIII. Quadratic Forms over Dedekind Domains

211

is a fractional ideal in F. It is clearly a non-zero o-module in F. And bY Proposition 81:1 there is a non-zero )3 in o such that /3L oxo + U. Hence (pa) xo _Ç /3L + U oxo + U. So fia cZ o. Hence a is indeed a fractionalideal as claimed. Now the coefficient of any vector in x0 + U is contained in a by definition of a. Hence the first part of the proposition will be proved if we can find a vector u in U such that a (x0 + u) Ç L. Since aa-'= o we can find an expression alfli+ • • • + GerPr= 1 Now each ai provides an expression mi x° = li + ui

E crl)

(ociE ftiE

by definition of a. Then x0=E

But ili a Ç o for 1

Pili+ E Pi ui•

r. Hence a (x0— E

piu)c L.

So we have found u E U such that a (x0 + u) Ç L and the first part of the proposition is proved. 2) We are given that a is the coefficient of x0 + uo, hence

a (x0 + 140) + (L r U) Ç L. . We must reverse this inclusion relation. So consider a typical vector in L which has the form ex (x0 + u) with a EF, uE U. Then axo EL + U and so a E a by definition of a. Hence ex (u

uo) = cc (xo + u) — ex (xo + uo)

Therefore

•

a (x0 + u) = oc (x0 + /40) +

EL•

(u — u0) E a (x0 + uo) + (L

81:3. Theorem. L is a lattice on the vector sfiace V and xl, a base for V. Then there is a base . . , y , with

y; EF xi + • • • + F

q. e. d. , x, is

(1 j n) ,

and there are fractional ideals a11 . . . , a., such that

L= + • ' + anYn • Proof. Let U be the hyperplane U = F xi + • • Proposition 81:2 L = (L U) + ayn

F x._1. Then by

14*

212


for some fractional ideal a n and some y„ E V — U. Proceed by induction

on n — dim V. 4. e. d. 81:4. Example. Let ,x„ be a base for V and letL = %xi + .. . + a„xn with ai E I. Then the coefficient with respect to L of any vector of the form CC1X1+ • • • + rX i

is equal to

(Cti

E F)

(a1 czT1) n • • n (a? ce71) In particular, the coefficient of x i is equal to ai. 81:5. Let L be a lattice on the vector space V. Then there is a tractional ideal a and a base z1, . . zn tor V such that L = azi + 02.2 + • • • + oz,,. Proof. Let us write L = aih+ • • • + any,, in the manner of Theorem 81:3. If n 1 we have L = alyi and we are through. The case of a

general n 3 follows by successive applications of the case n = 2. So let us assume that n = 2. By Proposition 22:5 we can find ch, tx2 E P such that + oc2 ail= o. Put x = z1 y 1 + oc 2y2. Then the coefficient of x in L is equal to

((Ilan (a2cV) = ( ohari. ce2a n-i. = by Examples 22:4 and 81:4. Hence L = ox + by by Theorem 81:3. q. e. d. 81:6. Example. An 0-module in V is a lattice if and only if it is

finitely generated. We say that the base z1, , z„ for V is adapted to the lattice L if there are fractional ideals al, .. . , a„ such that L = arzi d- • • + an zn .

Theorem 81:3 asserts that, there is a base for FL that is adapted to L, where L is any lattice in V. Consider a lattice L in V. It follows immediately from the fact that F is the quotient field of o that a set of vectors in L is independent over o if and only if it is independent over F. Hence a set of vectors of L is maximal independent over o if and only if it is a base for FL. In particular, any two such sets must contain the same number of elements. This number is called the rank of L and is written rank L. Thus rankL dimFL A set of vectors is called a base for L if it is a base in the sense of o-modules, I. e. if it is independent and spans L over o. So xl , . , x i. is a


213

base for L if and only if it is a base for FL with L ------ 0x 1 + - • • +oxr .

A lattice which has a base is called free. Any two bases of a free lattice L contain the same number of elements; this number is called the dimension of L and is written dim L; we have dimL = dimFL = rankL .

Every lattice L is almost free in the sense that it can be expressed in the form L = axi + ox2 -F • • • + ox7 with a a fractional ideal and xl, . . . , x, a base for FL. Clearly every lattice is free when the class number is 1, i. e. when every fractional ideal is principal. § 81C. Change of base Consider two lattices L and K on the same vector space V and let xl , . • . , xn and Yi'. . . , yin be bases in which L = ai x, ± • • • + anxn (ai E I) K= boi ± • • • -{- bnyn

Let

yi = E aii xi ,

(bf E I) .

xi = E biiyi

be the equations relating these two bases. So (ail) is the inverse of the matrix (k J). 81:7. K c L if and only if a ii bi Ç ai for all i,j. Proof. We have K S L if and only if bly, S L, i. e_ if and only if bi (aii xi + • • • + aii xi + • • •) S %xi + • • • + ai xi + • - • n. This is true if and only if ail bl S ai for 1 S i ‹ ti, 1<j‹ n. q. e. d. 81:8. Suppose K c L. Then K. L if and only if

for 1 ‹ j

a1 . • • an bi • • . b. • det (ail) . Proof. First consider K = L. Then by Proposition 81:7 we have c ai for all i., j, hence —

det (ail) = E ± atx . . . an .

E E (al bc71) • • • (an bT01) =

(al - - • an) Oh • • • bn) -1 •

Hence (b1 • . . b n ) det (ail)

(ai. • - an)

214


Similarly (a1 . . an) det (k J) g (t.

.

Now det (a15) is the inverse of det (1)15). Hence the result follows. Now the converse. Since K L we have ai ,E ai br for all relevant 1,1. The cofactor A i; of a15 is equal to A ii — E

in which the first index avoids i and the second j. Hence A ij a i

(a1 .

an)

. . bn) -1 = 0 • det (ai i)

Therefore b a

A il —

det(a ii)

ab -

• 2

This is true for all i, j. Hence L S K by Proposition 81:7. Hence L

K.

q. e. d.

Recall that the elements of 0 are the integers of our theory. Accordingly we say that an n X n matrix (a15) with entries in F is integral (with respect to o, or with respect to the defining set of spots S) if each of its entries is in 0. We shall call (a15) unimodular if it is integral with det (a15) a unit of F at S. By looking at cofactors one sees that the inverse of a unimodular matrix is integral, and hence unimodular. The defining equation of the inverse of a matrix shows that if an integral matrix has an inverse, and if this inverse is integral, then both the matrix and its inverse are unimodular. In other words, an integral matrix is unimodular if and only if it is invertible with an integral inverse. , x„ and consider 81:9. Suppose L is a free lattice with base vectors Yi' . . . , y ,, determined by yi = E aijx i (aiJEF) Then these vectors form a base for L if and only if the matrix (a15) is uni-

modular.

Proof. This is an easy application of Proposition 81:8. q. e. d. . 8 1:10. Example. Let v,. be vectors in V, let s be a unit in 0, and let oc2 , . . a,. be elements of 0. Then Ovi + 0v2 + • • • +

02. + 0v 2 + • - • 0v,.

where 2-11 = svl -F ot2 v2 + • • • + § 81D. Invariant factors 81:11. Theorem'. Given lattices L and K on the non-zero vector space V. 1 This theorem can be used to derive structure theorems for finitely generated modules over the Dedekind domain O. These structure theorems reduce to the Fundamental Theorem of Abelian Groups when 0 is the ring of rational integers Z.


215

Then there is a base x1,. . xn for V in which L = al xi + • • • + an xn K= ai r,xi + • • • + afi r„xn where ai and r i are fractional ideals with

r2 D • • • D tn

•

The ri determined in this way are unique. Proof. 1) We can suppose that K S L (if necessary we can replace K by ŒK where cc is a suitable non-zero element of o). For any x in V we let a. denote the coefficient of x in L and 6. the coefficient of x in K. Then put rx = b./a.. Since we are taking If S L we have b. S a2, and so

rx Ç 0 . 2) We can therefore take a yEÙ for which ry is maximal (though we are not yet sure it will be absolutely largest). By Theorem 81:3 there is a hyperplane U such that L = a„v + (L U) . We claim that b„+„ S b„ for any u E U. If not, then by Proposition 81:2 we can find auEU for which b i,+„ by. But ay+ .(v u) g L = a„,v + (L r U) , so that ay+ . S ay. Hence by+. b. r„ D tv+u — av+u and this contradicts the choice of v. So we do indeed have by+ .c b„ for all u in U. Now apply Proposition 81:2 again. We obtain K b„v + (K r\ U) .

3) An inductive argument gives us expressions L a„v + (a w w + • • + a„z) K a„r„v + (a w r w w + • • • + a z rz z)

with rw • • • D rx. We shall therefore be through with the first part of the theorem if we can prove that r„ r iy. By Example 22:9 we can pick cc, /3 E fr in such a way that + #(07,Ir.7,1) o I cca7, 1 + flawl = r.„ + rw . Put x ccv + 13w. By Examples 22:4 and 81:4 we obtain ox= (ovcc i) r (atufl -i) = (acql + '30;9' = (re + rw) -1 . And similarly b. = o. Hence rx r„ + r„). So ry„ S r„ by choice of v, and this is what we required.

216


4) Now the uniqueness. Let us call ri the i-th invariant factor of K

in L (a formal definition will be made once the theorem is proved). Our purpose is to prove that the invariant factors are indeed invariant, i.e. that they are independent of the base used in defining them. We make a start by remarking that the product of the invariant factors is invariant: the reader may easily verify this by using Proposition 81:8. , rn in the given base and let Consider the invariant factors r1, ,r„' be the invariant factors with respect to some other base. Suppose if possible that the second base gives rise to a different set of invariant factors. Take the first i (1 i n) for which ri r:. We can suppose that we actually have ri + r Dr;. We put =K

(rL) .

A i 1 the 2-th Consider the invariant factors of J in L. For 1 invariant factor is rl in either base; for i A n it is r2 + r in the first base and el; in the second. But this means that the product of all invariant factors in the first base is strictly larger than the product in the second base. We have already remarked that these products must be equal. So we have a contradiction. Hence r2 = ri for 1 A n. q. e. d. 81:12. Definition. The invariants r1 D • • • D rn of the last theorem are called the invariant factors of K in L. And ri is called the i-th invariant factor of K in L for 1 i n). Suppose we have lattices K and L on V with K S L. In this event the invariant factors of K in L are all integral ideals. Referring to Theorem 81:11 and § 22D we find —

(L:K)

11

(a i :airi) =

.11

Igign

(o:ri) .

In the important situations (e.g. local fields and global fields) the indices (o :ri) are all finite; hence (L:K) — L if and only if M —›— N (over t) if-and only if M N (over t) . (2) . K L

q. e. J. Proof. The proof parallels the proof of Proposition 41:2. , xn of the Consider the discriminant dB (x1, . . xn) of a base xi., x„' for L the equation free lattice L on V. If we take another base xi, 1 T/VT shows us.that N' dB (x;, . . x) = 62 dB (x 1, . xn) , xn) in for some unit e in o. Hence the canonical image of dB (x 1, 0 u ( 1 /u2) is independent of the base chosen for L, it is called the discriminant of L, and it is written dBL or dL . When L consists of the single point 0 we take dL — 1. We shall often write dL = a with cc in F; this will really mean that dL is the canonical image of cc in 0 u (F/112). It is equivalent to saying that L has a base L = °x1 + - - • ± ox, in which .

.

§ 82C. The class of a lattice

Consider a regular non-zero quadratic space V over F, and lattices K, L,... on V. We say that K and L are in the same class if K aL for some a E 0(V) . This is clearly an equivalence relation on the set of all lattices on V, and we accordingly obtain a partition of this set into equivalence classes. We use cisL to denote the class of L. The fundamental question of § 82A can now be regarded as a question of characterizing the class clsL. We define the proper class cisf to be the set of all lattices K on V such that K ciL

for some a E 0+ (V) .

The proper classes also put a partition on the set of all lattices on V, and this partition is finer than the partition into classes. In fact it is easily verified, using the fact that 0+(V) has index 2 in 0(V), that each class contains either one or two proper classes. Of course the class and


223

the proper class depend on several factors, such as the underlying set of spots S, the supporting vector space V, and the underlying bilinear form B. We define the group of units of L to be the subgroup 0 (L) = {a

0 (V) a L =

of 0(V). We put 0±(L) --= 0(L) r 0+(V)

.

The determinant map

det: 0(L)

(± 1) has kernel 0+ (L) , hence 0+ (L) is a normal subgroup of 0(L) with

1

(0 (L) : 0±(L))

2.

We shall see later that it is possible for this index to be either 1 or 2. It is 2 when there is at least one reflexion on V which is a unit of L; otherwise it is 1. We define the set 0- (L) 0 (L) n 0(V)

.

Then 0(L) = 0±(L) u 0- (L) , 0+(L) r 0- (L) 0.

82:2. Example. a 0 (L) a-1 = 0 (a L) for any a in 0(V). 82:3. Example. clsL = cls+L if and only if (0(L): 0+ (L)) , 2. 82:4. Example. clsL = cls4 L if dim V is odd, since —1 v is in 0(L) but not in 0+(L). 82:5. Example. Suppose L is free and let x1, with L = oxi + • • • + oxn

.

.

xn, be a base for V

Let M denote the matrix of V in this base. According to § 43A there is a group isomorphism of 0 (V) onto the group of automorphs of M, obtained by sending an isometry a onto its matrix T in the base x1,. This isomorphism carries rotations to proper automorphs. What does it do to 0 (L) ? It carries the units of L to the integral automorphs, Le. to the automorphs with integral entries. And to 0- '- (L)? These elements are carried to the proper integral automorphs of M. Thus (0 (L) : (L)) = 2 if and only if M has an integral automorph of determinant —1. So clsL = cls+L if and only if there is an improper integral automorph of M. 82:6. Example. Let K and L be free lattices on the same quadratic space V with matrices M and N respectively. Then clsK = cisL if and only if clsM = clsN. § 82D. Orthogonal splittings Consider the quadratic space V provided with its symmetric bilinear form B and its associated quadratic form Q. Let L be any lattice in V.

224


We say that L is a direct sum of sublattices L1 , .. L,. if it is their direct sum as o-modules, i.e. if every element x E L can be expressed in one and only one way in the form x=

+••-

(xi E L) .

x,.

We write L=LI ED••••L,.

for the direct sum. We know, for example, that L = L1 ED

L2

if and only if

L.4-1-L 2 with L1 nL2 =0. Suppose L is the direct sum L =L1 e • • • e L,. with = 0 for

B(Li,

1

11/1 0 1— O/sT) . • • 1. 1. Clearly B (x, y)

E E ai aj B (xi, z1) , ‘i

for any x, y in L, hence

E ai ai B (xi, z).

5 L ._Ç

i,

Conversely, for any oc iE di and any pi E al we have oci fli B(zi, xi) — B(aizi ,

thzi) E sL ,

hence ai aj B(zi, zi) S sL , hence f d i al B(zi, xi) = sL .

i,i 3) Finally the norm equation. For a typical vector

x = oh; + - - - ± ar zr in L we have

(aci

E ai)

Q (x) = f 04 Q ( z1) + 2 f oc; B (z 1 , 2 .1) ,

hence tIL S

E alQ (xi) + 2sL . t

On the other hand we have 2sL S nL, and by step 1) aN (zi) . n (aizi) nL , hence nL — E aN(zi) + 2sL . i

q. e. d. 82:8a. For any fractional ideal a we have s (a L) . a 2 (5 L) and n(aL) ----- a2(nL).


229

82:9. Let K be a regular non-zero lattice in the quadratic space V. Then there is a lattice L on V which is split by K and has the same scale

and norm as K. Proof. Since K is regular we have a splitting V = (FK)1 U. Take any lattice J on U. Now sK and nK are non-zero, 'fence they are fractional ideals, hence there is a non-zero a in o such that s (c J) = oJ) Ç sK ,

and n (ccj) = a2 (nj) SnK

Then L = K I (aJ) is a lattice on V with the desired properties. q. e. d. If L is not 0 there is a base xi, x,. for FL such that L a1 x1 + • • • + arx,.

with the ai in I. We define the volume to be L = a? . . . 4. • d

, x,.) .

This quantity is 0 when L is not regular, it is a fractional ideal in F when L is regular. Is it independent of the choice of base for FL? Consider another adaptation L = b 1 y1 + • • • + by

with yi

E aii xi .

Then by §41B we have d•, y,.) = (det a15) 2 • d (x1, . . „

,

hence by Proposition 81:8 we obtain 1)? . . .

• d(y 1 , • • • ,

= a? . . . 4. • d

, x,.) .

So bi- is indeed well-defined. If L is the lattice 0 we define 1.11, = 0. We note that »L = • »K when L has a splitting L =J I K. 82:10. Let L be a non-zero lattice in the quadratic space V. Then

uL C (en? where r denotes the rank of L. Proof. Take an adaptation L a ixi + • • • ar x,.

in the usual way. Put gii = B(xi, xj).Soa i aj gij Ç sL by Proposition 82:8.

230


By definition of the determinant we obtain »L= ... 4.) • E

g„)

(al aAŒ) . . . (ar aco g,..)

ç_ (51,) ? . q. e. d. 82:11. Let K and L be non-zero lattices on the regular quadratic space V with K L. Let a be the product of the invariant factors of K in L. Then a is an integral ideal and a2 (»L) , aL

Proof. Let is a base

K

rn be the

invariant factors of K in L. Then there xn for V and there are fractional ideals ., an suchta

I

L . alx,_ + • - • + anxn K= mi x,. ± • • • + an rn x,z .

Here we have all ri ç o since K C L, hence a C 0. If we compute volumes with respect to the above base we find UK . a2 (bL). Finally, we have q. e. d. a S ri for 1 < i < n, hence aL S K. 82:11a. »KC»/. if K C L.

Example. Consider the lattice L on the regular quadratic space V and let a be an element of 0(V) such that aL C L. Then a, being an isometry, will preserve volume. Hence aL . L, i.e. a is actually in 82:12.

0(L).

dual of a lattice Consider a lattice L in the quadratic space V. Suppose that L is regular. We define the dual of L to be § 82F. The

L4t . {x EFL [ B (x, L) Ço}.

If L is the trivial lattice 0, then 1.4* . O. Suppose L is not O. Then we have a base xl , . . ., x,. for FL and fractional ideals al., .. ., at such that L= al x„. + • • • + af xr .

We claim that L* =aj y1 +•••+ ç l y

where y„ . . y? is the dual base of 0 if i == j, and B (ai xi, B (ai xi, a-ph)

x,. on FL.

Now

C o otherwise. Hence

I-4f • On the other hand, if we take a typical vector z in L 4* we must have ar l Yi + • • • + (171 Yr

Thai

B

13iYi+ • • • + 13,.Yr

ai xi) = B (z, a i xi) c B (z, L) s; o,


231

and so /3, E aTl. Hence we have established our Claim. An incidentally this also shows that L 4 is a lattice. As immediate consequences we have FL 4 = FL,

L4 # L ,

and

(a L)* = a- L# for any fractional ideal a. In virtue of the fact that

d(xl, .

xr) d (yl, . .

yr)

1

(see Example 42:5) we must have

(DL) -1 .

DE*

If L has a splitting L J I K, then L4 = j4 K4 . Finally, it is easily seen that if L, J,K are all on the same space FK, then

L# K 4

L DK and (J

K ) -It =

n K4 .

§ 82G. Modular lattices Consider a lattice L of rank Y in the quadratic space V. Suppose that the scale of L is the fractional ideal a. Then we know from Proposition 82:10 that 5L — a and DL Ç ar. Suppose that L actually satisfies 6L,= a

and UL =a'.

In this event we call L a-modular, or just modular. For any oc in I we call L oc-modular if it is a-modular with a oco. We call L unimodular if it is a-modular with a = o. It is clear from Proposition 82:10 that L is a-modular if 5L a and 1)L = ar An a-modular lattice is regular, non-zero, of scale a, and of volume DL = If L is a-modular, then ocL is oc2 a-modular for any oc in F, and bL is b2 a-modular for any b in I. If we take a non-trivial splitting L J I K, then it is easily seen that L is a-modular if and only if both J and K are. The lattice ox with x an anisotropic vector of V is Q(x)-modular. 82: 13. The free lattice L in the quadratic space V has the matrix M. Then L is a unimodular lattice if and only if M is a unimodular matrix. Proof. Take a base x1 ,. . x,. for L in which L has the matrix M. So L = oxi + • • - +ox,. and DL = (det M)o. If L is unimodular, then all B(x I , x3) are in o since 5L — o, hence M is an integral matrix. Further-

232


more (detM)o .----- o and so detM is a unit. Hence M is unimodular. Now let M be unimodular. Then M is integral and so sL = f B(x i,xi)o S o. ci q. e. d. And detM is a unit so that 13./. = o. Hence L is unimodular. 82: 14. L is a non-zero regular lattice in a quadratic space V. Then L is a-modular if and only if al.* = L. Proof. First suppose that aL4* , L. Then B(L, L) = B(L, aL 4t) C a ,

and so sL C a. Furthermore b.L = » (a./,) = a2 ? (»L4) _, a2r 03 4-1 , hence u.L = ar. Therefore L is a-modular. Now assume that L is a-modular. Then B(L, a--1 L) C o so that a-1 /- C 1.4 t. On the other hand we have I) (a-1 L) -, a-2 r (» L) = (a)-1 = »L*. q. e. d. Hence a-'/, — L 4t by Corollary 82:11 a , 82:14a. Suppose L is a-modular. Then L = {x EFL f B(x, L) S a} . Proof. Since L is a-modular we have B(L, L) S sL = a and so L S {x EFL 1B (x, L) _Ç a} . Conversely consider an x in FL with B(x, L) .0 a. Then B (x, 1-#) = B (x, a --3-1.) C o , hence x E L# 4t = L. 82:14b. L is unimodular if and only if .1.4t = L.

q. e. d.

82 : 15. L is a lattice in a quadratic space V and I is an a-modular sublattice of L. Then J splits L if and only if B(J, L) C a. Proof. If j splits L we have L— JIK and then

B (j, L) = B (J, J) _Ç a.

Conversely suppose j satisfies the condition B(J, L) C a. Since J is modular it is regular, hence by Proposition 42:4 we have a splitting FL = (F J) _i_ U. We claim that L = J _I_ (L r\ U) . It is enough to show that a typical x in L is also in J + (L n U). Write x=y+z (y EFL z E U) Then B(y, J) = B(x, J) _ç_ B (L, J) s; a .


233

But y is in F J. Hence y is in / by Corollary 82:14a. So z is in L and we are through. q. e. d. 82:15a. If J. is an a-modular sublattice of L with sL a, then j splits L. 82:16. Let L be an a-modular lattice and let x be an isotropic vector in L. Then there is a binary lattice J which splits L and contains x. Proof. By Theorem 81:3 there is a base for FL which is adapted to L and includes x among its members, say L=bx+••••

Then by § 82F we have 1.4* = b-ly + • • •

where y is a vector in FL with B(x, y) = 1. Now a/At = L since L is a-mQdular. Hence = bx + b-iay

is a sublattice of L which contains x. But is easily computed and found to be a2. And 15/ S sL S a. Hence J is a-modular and therefore splits L. q. e. d. 82: 17. Suppose is a principal ideal domain. Then a non-zero lattice L in a quadratic space V is a-modular if and only if B(x, L) — a for every maximal vector x in L. Proof. First suppose L is a-modular. Then by Theorem 81:3 we can find a base x, . . . for L in which

L = ox + • • • .

So L 4t oy + • • •

where y is a vector with B (x, y) = 1. Now aL4* = L since L is a-modular, hence a B(x, L) B(x, ay) 2 a , so B(x, L) = a. Conversely suppose B(x, L) = a holds for every maximal vector in L. Every vector y in L falls in ox where x is a maximal vector of L in the line Fy, hence L is regular and s L -= a. So B (a-1 L, L) S 0. This proves that a-1 /, S LA and hence that L S aL4*. On the other hand B(aL 4t, L)S a. If y is a vector of FL—L, then y = ocx with x maximal in L and a not in o, hence B (y , L) = B(x, L) = oca

and this is not contained in a; so no vector y of FL — L can fall in a/ * . Hence aLA .0 L. So aL 4t = L and L is a-modular. q. e. d.

234


§ 82H. Maximal lattices

Let L be a non-zero lattice on the regular quadratic space V, and let a be a fractional ideal. We say that L is a-maximal on V if nL S a and if for every lattice K on V which contains L we have nKÇ a => K--=--L.

We say that L is maximal on V if it is a-maximal for some a. If L is an a-maximal lattice on V, then it is easily seen that ccL is ea-maximal for j_ K is a every cc in F , and bL is 62a-maximal for every b in /; if L non-trivial splitting of the a-maximal lattice L, then J. and K are a-maximal on F/ and F K respectively. 82 : 18. Let L be a lattice on a non-zero regular quadratic space V and suppose nL C a where a is a fractional ideal. Then there is an a-maximal lattice K on V with K D L. Proof. Ler r denote the dimension of V. First observe that for any L on V with nL C. a we have 1 )r L (sL)f The Unique Factorization Theorem shows that the number of fractional

ideals between bL and (a)i. is some finite number v. If L is not a-maximal 21 we have a lattice Li on V with L (LI and nL1 a. If LI is a-maximal we are through. Otherwise repeat the preceding step. Continue in this way to obtain a chain LCLI C• • •CLt

with each nLi c a. This gives rise to the chain of fractional ideals )r

So t v. Thus the process must terminate before 3, steps. In other words we obtain an a-maximal lattice Lt before y steps. This L t is our K.

4. e. d. 82: 18a. Every non-zero regular quadratic space contains an a-maximal lattice for every fractional ideal a. 82: 19. Let L be a lattice on the r-dimensional regular quadratic space V

and let a be a tractional ideal such that nL S a. Then the ideal 2r 03 ÷ (a)*

is integral. If this ideal has no integral square factors, then L is a-maximal. 1 . Proof. The ideal 2' (»L) Mr is integral since (D L) C (8 L) ( -1 Suppose this ideal has no integral square factors. If L is not a maximal we can find a lattice K on V with L (K and nK c a. Then there is a proper integral ideal b with DL = b2 (DK) by Proposition 82:11. And -


235

2f (bK)/ar is integral since nK C a. This implies that 2? Oa L)/ar= 62 2? (» K)/a'

has an integral square factor, and this is contrary to hypothesis. q. e. d. 82:20. Let L be an a-maximal lattice and let x be an isotropic vector in L. Then there is a binary lattice J . which splits L and contains x. Proof. Let V denote the regular quadratic space on which L is a-maximal. Let b denote the coefficient of x in L. We have 1 1 sL ç— --2 (nL) S -2a. Hence B (2 a-1 bx , L) S B(2a -1 L, L) So , hence (2 a-lb) x C 1,4, so 2a-1 6 S c where c denotes the coefficient of x in 1,4. We claim that 2a-1 6 = c. Suppose not. Then 2a-1 6 C c, so that 1 a cx + L is a lattice in V which properly contains L. Now an easy computation gives 1 Q ( 2 a cx ± L) S a . it So — 2 OCX + L is a lattice which properly contains L and has its norm contained in a. This denies the maxirnality of L. So we do indeed have 2a-1 b = c. By Theorem 81:3 we have a base for V which includes the vector x and such that L4 = 2a-1 bx +... . Then by § 82F there is a vector y with B (x, y) = 1 such that L . 1, 4*4*. -21 a 6-13, + • • - -

So we have a sublattice 1 ab-1 y I —bx-F-T

i a-modular. of L with s j S sL S -I1a and » J. = -4-1a2 ; therefore J. is T q. e. d. By Proposition 82:15 j will split L. 82:21. Let L be a lattice on the hyperbolic plane V. Then L is 2 amaximal if and only if L is a-modular with nL S 2a. Proof. Take x, y in V with Q (x) = Q(y) = 0 and B (x, y) . 1. First suppose that L is 2a-maximal on V. By Theorem 81:3 we can write L. bx+ c (ccx + y) for some a in F and some 6, c in I. Now b c S sL S a by Proposition 82:8. If we had I) c C a, then ac--1 x + c(ccx + y)

236


would be a lattice of norm 2a which strictly contained L and this would deny the maximality of L. Hence b c = a and L bx ab -1 (ccx y) .

So 6L Ç a and »L = 02. Hence L is a-modular. And nL S 2a since L is 2 a-maximal. Conversely, suppose that L is a-modular with nL S 2a. Then 22 (» L) -4 (20) 2 =0, hence L is 2a maximal by Proposition 82:19. q. e. d. 82:21a. Suppose either of the equivalent conditions of the proposition -

-

is satisfied. Let F x and F y be the isotropic lines of V. Then the base x, y for V is adapted to L. Proof. We can assume that B (x, y) . 1. In the proof of the proposition

we obtained Now by Proposition 82:8 a2 b-2 (2a) ç n.L Ç 2a, hence a b-la b, hence L bx q. e. d. 82:22. Example. Suppose every fractional ideal is principal, and let a be an element of P. Consider a lattice L on the hyperbolic plane V. Then L is 2 ao-maximal if and only if (0 ci L

ce 0 Hence in this situation any two a maximal lattices on V are isometric. 82:23. Let V be an isotropic regular quadratic space and let K and L be maximal lattices on V. Then there is a splitting V = U I W in which U is a hyperbolic plane and -

L

n U)

(L n W) ,

K = (K n U)

(K n W) .

Proof. Let L be a-maximal, let K be b-maximal. For each non-zero vector x in V we let a. denote the coefficient of x with respect to L and we let b. denote the coefficient of x with respect to K. We put r=

, rx = bxfax .

Now alf S L for some non-zero a in F, hence r.S a-lo; hence we can pick an isotropic vector x in V for which r. is maximal (among all the isotropic vectors of V). By Proposition 82:20 and Corollary 82:21a there is an isotropic vector y such that B (x, y) 1 and L

(axx + av y) 1 • • •

Since ax (ivy is a-maximal its scale ax at, must be equal to -I a by Proposition 82:21. Now bx x

by is contained in K, hence bby s-7, 2 b.


237

Therefore rx ry S t.

Using Proposition 82:20 and Corollary 82:21a again, we find an isotropic vector z such that K= (be z + b ey) ± • • • . This time we obtain I), bv =

hence rz ry D r. Therefore

and a,a,

r xr y Ç r Ç rz ry . But r , was chosen maximal. Hence rx rv = r and so

bb y = (a.ay) (try) = Therefore b xx Hence

by has scale

1

1

a bia = b.

and volume — 4 b2 ; so it is -l b-modular.

K (b z x ± be y) ± • • • L = (a x x + (iv y) 1 • • • q. e. d. Then U=Fx+Fy gives the desired splitting of V. § 821. The lattice La Consider a lattice L in the quadratic space V, and a fractional ideal a in I. Suppose that L is regular. We define La as the sublattice

1

La= Ix E LJB(x, L) Ç a} of L. For the trivial lattice we have La= 0, so let us assume that L is not 0. We immediately have B (La, L) S a , sLa S a , and La= L 1. By suitably scaling V we can assume that sL = 0. Consider a typical ci in O (L) . This a must be expressed as a product of symmetries in 0 (L). Fix y in L with Q (y) equal to some unit e. Then (31 - r + + crY) = 48 and this is a unit since the field is non-dyadic. Hence either Q(y — cry) or Q (y cry) is a unit. In the first instance the criterion of § 91B shows that the symmetry r , , is a unit of L, and we have Tv–avY =

crY •

In the second instance ri, and ry +„ are elements of 0(L), and we have r11+citf

TO/

cry.

So in either case there is an element p which is a product of one or two symmetries in 0(L) such that py = ci y. By Corollary 82:15a we have a splitting L = oy I K. Then p-la induces an isometry on FK; this isometry is a unit of K and hence by the inductive hypothesis it is a J. Starting the induction at n = 2 will show that at most 2n —2 symmetries are needed when n 2.

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

249

product of at most 2n — 3 symmetries in 0 (K). Then g-lcr = (1 ± xi) (1 I r2) . .

Hence a=

e ( 1 I Ti) ( 1 ±x2 ) • •

is a product of at most 2n — 1 symmetries in 0(L). q. e. d. 92 : 5. Let L be a modular lattice on a quadratic space over a non-dyadic local field with dimL

2. Then 0(0+ (L)) = uF.2 . Proof. By suitably scaling the space we can assume without loss of generality that L is unimodular. Take a typical symmetry in 0 (L). This symmetry has the form Ty with y a maximal anisotropic vector in L. Since r is a unit of L we have B (y, L) Ç Q (y) 0 by the criterion of 0 since L is unimodular. Hence Q (y) E u. So § 91B. But B (y , L) O (r) Ç 14'2. Therefore (0± (L))

We then obtain equality here by applying Corollary 92: lb to find a symmetry -t- in 0 (L) with 0 (T) = EF2 for any e in u. q. e. d. 92:6. Example. Let L be a lattice on the regular quadratic space V over the non-dyadic local field F. Consider a maximal anisotropic vector y of L. The criterion of § 91B says that the symmetry is a unit of L if and only if B (oy, L) Ç Q (y) o. By Proposition 82:15 we know that this condition is equivalent to saying that oy splits L since v y is Q (y)modular. Hence for any maximal anisotropie vector y in L we have Tv

E

(L)

oy splits L.

92:7. Example. Suppose V is a regular quadratic space over a nondyadic local field with invariants dV = 1 and S V = 1. We claim that there is a lattice L on V with 0 (0 (L)) =

By Theorem 63:20 we have V=-,- ± • ± .

Hence there is a lattice L on V with Lc j.

...

< 212 (n-1)>,

By Theorem 92:2 we know that if oy splits L, then Q (y) ( 21 U2 for some i (0 i n 1). Hence by Example 92:6 all symmetries in O (L) have spinor norm .P 2. Hence by Theorem 92:4 all elements of 0(L) have spinor norm F 2.


250

92:8. Example. Suppose V satisfies the conditions of Example 92:7, and suppose further that dim V is even. Let e be any unit in o. Then there is a lattice L on V with 0(0- (L))

e 1 "2

.

92:9. Example. Let L be a lattice on a regular n-ary quadratic space

over a non-dyadic local field, with 1

8L (O

n(n-1)

and bLpr

We claim that 0(0.1- (L)) u .

This follows immediately if we can prove that the number t in the Jordan splitting L= I • • • _L L t

is less than n: for then we will have dim L i 2 for at least one 1(1 i t) and we can apply Proposition 92:5 to this L i . Suppose if possible that = n. Then b Li

= sL i S

,

hence „ -2 n(n-1) —

and this is contrary to hypothesis.

§ 93. Classification of lattices over dyadic fields Throughout this paragraph F is a dyadic local field. Thus F has characteristic 0 2 and the residue class field of F at p is a finite field of characteristic 2. So the residue class field is perfect, and the congruence e'

ei52 modn

has a solution i5 for any given units e and e'. For any given a, 13 in 1 we shall write ocfi

if 413 is an element of u2. This defines an equivalence relation. And a 13 if and only if affl is a unit with quadratic defect b (4/3) = 0. For any fractional ideal a we write =.-.P 13 mod a if a/fl is a unit and cc = 13e2 moda for some unit e. This also defines an equivalence relation. And a fi moda if and only if al 13 is a unit with 1 See C. H. SA11, Am. J. Math. (1960), pp. 812-830, for the integral theory over local fields of characteristic 2.

Chapter 1X. Integr -.-1 Theory of Quadratic Forms over Local Fields

251

b (a/fl) Ç a/ 13. In virtue of the perfectness of the residue .class field we have cc 13 mod a p when ai i3 E u. The letter A will denote a fixed unit of quadratic defect 4o. It is assumed that 4 has the form A —1 + 4o for some fixed unit e in F. 1 mod40. Of course A V will be a regular n-ary quadratic space over F, L will be a non-zero regular lattice in V. As usual 2(6L) Ç nL sL But we now have 2(5L) Cs/. , and it is this strict inclusion that makes the dyadic theory of quadratic forms distinctly different from the non-dyadic theory of the last paragraph. 93:1. Notation. We let o 2 denote the set of squares of elements of 0. The symbol 0 2 already has a meaning in the sense of ideal theory, namely oi2 is the product of the fractional ideal o with itself. However this product is equal to o so there is never any need to use the symbol 0 2 in this sense. For us then 0 2 will be the set 02

cx2

0}

93:2. Notation. Given a non-zero scalar a and a fractional ideal a we shall write a mod a dL a mod a. This is the same if dL = 13 for some /3 E F which satisfies 13 in which as saying that L has a base x1 , . .

dB

a mod a .

. . , xn)

Given two lattices L and K we write

dLidK

a mod a

if there are non-zero scalars 13, y such that dL = 13, dK = y, and , #/y a mod a. This is equivalent to saying that there are bases and y•• ym for L and K such that

dB (xj., . . . , x n)IdB (yi, . .

ym )

a mod a .

§ 93A. The norm group gL and the weight tvii It is easily seen that the set Q (L) + 2(6 L) is an additive subgroup ofF. We shall call this subgroup the norm group of L and we shall write it

L Q (L) + 2 (6 L) The norm group gL is a finer object than the norm ideal nL which it generates. We have 2(6L) qL ( nL ,

252


and nL = ao holds for any element a of gL with largest value. Given two regular non-zero lattices K and L in V, then K L gK gL , and g(K L) = gK + gL . The sets Q(L) - and gL don't have to be equal. For instance if L = o x, then Q (L) = 0 2 Q (x) contains no fractional ideals and so it cannot be all of gL. On the other hand, if L is any lattice in V with L o and if L 0 contains a sublattice H of the form H ( 1 0) then we do indeed have gL = Q (L) . In order to prove this we must show that a typical element Q (x) 1 2 (x EL, a ( 0) of g L is also in Q (L). Now we have a splitting L = H j . K since H is unimodular and sL = 0, hence we can write x h k with liEH,kE K. Since Q (H) =--- 20 we can find h' E H with Q (h') = Q (h) + 2a. Then - -

Q (151 + k) Q (h') + Q (k) = Q (h) + 2 oc Q (k) = Q (x) + 2oe . Hence Q (x) + 2 cc is in Q (L) as required. So Q (L) and gL are sometimes equal. As a matter of fact the result that we have just proved will enable us to arrange gL = Q(L) whenever we please. 93 : 3. Let L be a lattice of scale 0 on a hyperbolic space V over a dyadic local field. Then there is a unimodular lattice K on V with LSKSV such that gK gL. Proof. As we ascend a tower of lattices on V we obtain an ascending tower of volumes in F. Hence we can find a lattice K on V with LSK, sK=o, gK=gL,

and such that there is no lattice K' on V which has these properties and strictly contains K. This K will be the lattice required by the proposition. We assert that every isotropic vector x in V which satisfies B (x, K) So is actually in K. For consider K' ox + K. Then sK' = 0 and Q (K') Q (K) + 20 = gK, so that

L S K' , 8K 1 = o , gK' = gL Hence K' = K by choice of K. Therefore x is in K as asserted. Hence every maximal isotropic vector x of K satisfies the equation B (x, K) = o. Pick y E K with B (x, y) = 1. Then ox + oy is a unimodular sublattice of K, so we have a splitting K =- x oy) j.. J. But F J is isotropic since F K is hyperbolic. Hence J has a splitting J = (o x' + o y')1_ J' with x' isotropic and ox' oy' unimodular. Repeat. Ultimately we


253

obtain a splitting of K into binary unimodular lattices. So K is uniq. e. d. modular. 'We let mL denote the largest fractional ideal contained in the norm group g L. Thus 2(sL) SmL S gL SnL.

Let us show that is even. Suppose not. Pick a E gL with al) = nL and write mL = a Or + with an r O. We claim that ap2r S gL. We have to show that any element of the form a s 7L2r with e E u is in g L. By the perfectness of the residue class field we can solve the congruence ordp nL ordp mL

a 8 7c 2r

a 62 7g2r mod

a p2r +1

for some unit 6. But a p2r +1 Ç gL by definition of m L. And a (a a ) 2 is in g L. Hence ac 7c2r is in g L. So ordp nL ordp mL is even. We define the weight toL by the equation ti, L = p (m L) ± 2 (s L) . So mL depends only on gL, while troL depends on L. We have mL ,

p(mL)

2(sL) cwLnL

.

Also ordp nL ordp tvL

is odd

tr•L = p (mL)

and ordp nL ordp toL

is even = toL = mL = 2 (sL) .

Hence ito L = n L =

2 (s L)

YLL

It is easily seen that KL=

tr•IC SID/.

and that

gK =gL with sK=sL => toK=toL. We call the scalar a a norm generator of L if it is a scalar of largest value in gL. Thus a is a norm generator of L if and only if aEgLSao, j. e. if and only if a E gL with ao = nL We call the scalar b a weight generator of L if it is a scalar of largest value in to L. Thus b is a weight generator of L if and only if bo = wL . If a and b are norm and weight generators of L, then 1b1 < 154 , 1b1 lai , 12(L)1 ordp a + ordp b is even =1b1 = 12(641 lbl = lai la' .12(5L)I .

254


Let us prove that every element of gL whose order has opposite parity to the order of nL is in to L. In other words, if a is a norm generator of L, and if fi E gL with ordp a + ordp 46 odd, then /3 E ito.L. It is enough to prove that fib S g L, for then flo S m L by definition of mL, hence S p (mL) S toL. Consider a typical non-zero y in -46o. We have to show that y E g L. Now by Proposition 63: 11 we have E F such that + /3 772 mod4 y. By the Principle of Domination both and 12 are in 0, since y E /3 o S ao. But gL stands multiplication by elements of 02, by definition of g L. And 4 y oLS 4 (nL) S 2 (s L). Hence y E g L, as asserted. 93:4. Let L be a non-zero regular lattice over a dyadic local field with norm and weight generators a and b. Then

gL a02 bo Proof. The set gL stands multiplication by elements of o2, hence ao2 ç g L, hence ao2 roL is contained in g L. Conversely consider a typical element oc of g L. We wish to express a as an element of a 0 2 + to L. By Proposition 63: ii we have scalars and 21 such that a + a n 272 mod4 . By the Principle of Domination we see that E 0, since a E ao. Then oc and aV are in g L, and 4 a o S 2 (sL) S g L, hence an n 2 in in gL, hence it is in to L since its order has opposite parity to the order of nL. q. e. d. 93:5. Example. Let us give a general method for finding norm and weight generators for L (computable methods will be given later in § 94). To find a norm generator simply take any element a of largest value in Q (L). To find a weight generator first take 1,0 E Q (L) of largest value

such that

ordi, a + ordr,b0 is odd.

If boo D 2 (s 4, then to L = boo and bo is a weight generator of L. If boo ç 2 (s L), or if bo does not exist, then to L = 2(s L) and we can take any b for which bo = 2 (s L). 93:6. Example. Let a be a norm generator of L, and let a' be some other scalar. Then a' is a norm generator of L if and only if a a' mod toL. 93:7. Example. Consider a non-zero scalar oc. Then g (OEL) -= (x 2 (g L). in (cc L) cc 2 (m L) , and to (cc L) = a2 01,4 If a is a norm generator of L, then a2 a is a norm generator of (xL. 93:8. Example. What happens when we scale the quadratic space by a non-zero scalar oc? We obtain g (D‘) = oc( L), in (D') = cc (m L) , and to (Di) = oc (to L) . liais a norm generator of L, then ct a is a norm generator of LI. If oc = 1 we obtain gL-i = gL , = tilL , tvL - 1 ttIFL , —

and a is a norm generator for L --1 as well as for L.


255

§ 93B. The matrix A (x, /3) We shall use the symbol A (cc, 13) to denote the 2 x 2 matrix A (a, fl) =

1,3) 1

whenever a, fi are scalars which satisfy the conditions

ac,flEo,

—I -1-afiEu.

These conditions simply guarantee that the matrix A (of, 13) is unimodular. Whenever the symbol A (a, fi) appears it will be understood that a, fi satisfy the above conditions, even if this is not explicitly stated at the time. We use A (a, fl) to denote ordinary multiplication of the matrix A (oc, 13) by the scalar Thus

93:9. Example. If L A (a, 0), then L A (a+2 f3, 0) for any )3 in o. 'For if we take a base L ox oy in which L has the matrix A (oc, 0), then L = o (x + fly) 4- o y also gives a base for L and the matrix of L in this base is A (a+2fi , 0). 93:10. Example. Let L be a binary unimodular lattice and let a be a norm generator of L that is also in Q (L) . We claim that (i) L A (a, fi) for some /3 E tvL, (ii) if n, 2o then the /3 in the above matrix for L is a weight generator of L. To prove this we pick any x c L with Q (x) — a. Then x is a maximal vector in L, so there is a vector y in L with L= ox io y . Now a is a norm generator of L, hence by Proposition 93:4 there is a in o and an a/ in tr•L with Q (y) Then L ex (y ± x) and ?I .

—

71 ) + (va) n B (y , x) E tvL Q (y + x) — (a But mi.( sL o, hence B (x , y ± x) is a unit since L is unimodular. Put z = (y x)1B (x, y + • x). Then L has a matrix of the desired type in the base L =ox + oz. This proves the first part of our contention. Now the second part. Here we are given tvL D 2e. We recall from Example 93:5 that there is a number b E Q (L) such that be = tvi, with ordp a ± ordi, b odd. Then ibl, and 12f < ibl, and

Ifli

b = a42 -I- n

+ #272

with E o. If we had fil < 1bl we would have Ibi = laVi by the Principle of Domination, and this is absurd since ordr a ordp b is odd. Hence lb!, so Iv", fie and f3 is a weight generator of L. 93:11. Example. Let L be a binary unimodular lattice with nL Ç 2o. We say that L A (0, 0) or L A (2, 2o) .

im

We know that L is 2o-maximal by Proposition 82: 19. So if FL is isotropic we will have L A (0, 0) by Example 82:22. Now suppose that

256


A (2a, 2 /3) by Example 93:10, and both FL is not isotropic. Then L cc and /3 will have to be units since otherwise we would have d (FL) =—1

by the Local Square Theorem. Construct a lattice K on a quadratic space F K with K A (2, 2e). Then a direct calculation of Hasse symbols shows that Sp (FL) = S (FK), hence FL and F K are isometric by Theorem 63:20, hence L and K are isometric by Theorem 91:2, hence L

A (2, 2e). 9 3 : 12. Let L be a lattice on a regular quadratic space V over a dyadic local field. Suppose that L has a splitting L = J I K and that J has a base J = x oy in which J A (a, 0) with a E 0. Put

r

+ +0 y

with

zE

Then there is a splitting L = J' K'. And K' is isometric to K. Proof. J' is unimodular and B (J', L) c o since z c K°. Hence we have a splitting L = I K' by Proposition 82:15. For any u EFK we define

Tu=u

B (u, z) y Then Q (9)u) = Q (u) since Q (y) = 0 and B (u, y) 0, hence 97: F K -4– V is a representation; but FK is regular; hence 99: FK>--V is an —

isometry by Proposition 42:7. Now B (9) u , x z) = B(u — B (u, z) y, x z) = B (u, z) — B (u, z) B (y, x) = 0. And similarly B (gm, y) = 0. Hence B(9) (F K), F J') = 0 and we have FK'. an isometry ep: FK Now B(u,z) E o whenever u E K since z E K°. Hence 9).K S L. Hence S F K' r L = K'. But 99 preserves volumes since it is an isometry.. q. e. d. Hence 99K K'. So we have found an isometry of K onto K'. 93:13. Example. Suppose the lattice L on the regular quadratic space V has a splitting L A (o 0)) K

a

with

E F and

cc E 0. Let /3 be any element of gi.K4 °. Then L<M(a+ 6-1 13,0)> _LK.

This is obtained from Proposition 93:12 by scaling. In particular, if L has a splitting L

K

with oc E o and sK S o, then for any fl in Q (K). § 93C. Two cancellation laws 93:14. Theorem. A lattice L on a regular quadratic space over a dyadic local field has splittings L= JIK and L =J' J. K' with J isometric to J' and J A (0, 0). Then K is isometric to K'.'


257

Proof. 1) Take bases J == ox + y and J' =ox' + Dy' in which A (0, 0) and j' A (0, 0). Note that j and j' are contained in L°. Hence j + j' L° and B( j, j') o. 2) First we do the special case where x= x'. We can express y' in the form y' = ocx + y + z (oc E 0, z since y' is in L =J j K and B (x, y') = 1. Then z = y' — ocx y is in (J + J') K Ç L° r K, hence z is in K°. We can therefore apply Proposition 93:12 to the sublattices =ox±o(y -Fax) J' =ox + o (y + cxx + z) of L. This gives us an isometry of K' onto K. 3) Next we do the case where B (J, J') = o. We can suppose that B (x, y'), say, is a unit. Making a slight change to x', y' allows us to assume that B(x, y') = 1. Put J". ox + Dy'. Then ra.-- A (0, 0) is a unimodular sublattice of J J' S L*, hence there is a splitting L = J" ±K". But we can apply the special result of step 2) to J and J", and also to J" and J'. Hence K K" and K" K'. Hence K K'. 4) Finally we consider B (J, J') S p. Here we put j"= ox +0 (y+y'). Then J" is a unimodular sublattice of L° with n J" S 2o. Hence J" A (0, 0) by Example 93:11. And we have a splitting L J" j K". But here B (J, J") o and B (J", J') = 0. Hence K K" and K" K' by step 3). So K K'. q. e. d. By a hyperbolic adjunction to a lattice L on a quadratic space we mean the adjunction of a lattice J of the form

T

al A (0, > 1 • • • _L j. K1 .

q. e. d. K1 by Theorem 93:14. The general cancellation law for lattices on quadratic spaces over non-dyadic fields (Theorem 92:3) does not extend to the dyadic case. For instance there is an isometry (A (0,0)) ± (1 ) (A (1,0)) ±

Then K

by Example 93:13; but (A (0,0)) and (A (1,0)> are not isometric since their norms are not equal.

§ 93D. Unimodular lattices 93:15. A unimodular lattice L in a quadratic space over a dyadic local field has an orthogonal base if and only if nL = sL. If nL c sL, then L is an orthogonal sum of binary sublattices. Proof. If L has an orthogonal base it contains a 1-dimensional unimodular lattice; any such lattice has norm o; hence nL= o = sL. Conversely let us suppose that nL =sL=o. Then there is an xinL with Q (x) in u. The lattice ox is a unimodular sublattice of L and therefore splits L. Hence we have a splitting 1_ • • • 1

<Er> ±

(oh, 181)> .L • ' • 1 (A (oct,

13t) >

in which r 1 and t 0, by § 9IC and Example 93:10. Consider the 3-dimensional sublattice K (8) I (A (a, (I)). If neither or nor 18 is a unit, then A (a ± e, fl) is unimodular and we obtain a new splitting K <e'> j (A (a ± e, 13)) in which a + e is a unit.We may therefore assume that cc, say, is a unit.Then C

(8)

C•r5 >

By successively applying the 3-dimensional case to L we ultimately obtain (el ) as required.

• • • _L

<en>

To prove the last part we take a splitting of L into 1- and 2-dimensional components. If a 1-dimensional component appeared in this splitting it would be unimodular and hence would have norm 0. This is impossible since nLc sL = o. Hence L splits into binary components.

q. e. d.


259

9 3:16. Theorem. Let L and K be unimodular lattices on the same quadratic space over a dyadic local field. Then clsL = clsK if and only if g L g K. 1 Proof. By making the same unimodular hyperbolic adjunction to L and K we can assume, in virtue of Theorem 93:14 and § 93A, that Q (L) = gL = gK = Q (K). Now adjoin the lattice L ± L -4 to each of L and K, and let the resulting lattices be denoted L' and K' respectively. So =gL=gK. Hence by Corollary 93: 14a it will be enough Now g (L 1 to prove that L' is isometric to K'. But the component L --1.1 K of K' is a unimodular lattice on a hyperbolic space, hence K' has a splitting K' = L

(A (al, 0) )1 • • • j (A (cc„., 0))

with all ai in gL = Q (L). Then L L (A (0, 0)) ± • • • j. (A (0, 0)) by Example 93:13. Similarly we find K'

L'

L j.. (A (0, 0)) 1 • • • 1 CA (0, 0)) • q. e. d. Hence K' L'. Hence K L. In other words, clsK = cls L. 9 3:17. Example. Let L be a binary unimodular lattice on a quadratic space over a dyadic local field, let a be a norm generator of L that is in Q (L), let b be a weight generator of L, let the discriminant dL be written in the form dL = — (1 ± oc) with b (1 + oc) = oco. We claim that aa--1 (bo

and

L

A (a, —cca-1).

By Example 93:10 we have L A (a, b2) with 2 E 0, hence —(1 — ab 2) = —62 (1 ± oc) with e E u, hence

txo = b (1 + a) = b (1 — ab S ab So oc a-1 E bo. We must prove that L K where K is a lattice with K A (a, —oca-1). Now , (—a(1 FL (a) and a is a norm generator of both L and K, hence by Theorem 93:16 it suffices to prove that wL = wK. If toL = 2o, then oca-1 E 2o S roK and so tr•K = 2o by Example 93:10. If w L )20, then A is a unit by Example 93:10. We have b(1 — ab 2) = ab Ao since here ordp a + ordp b is odd with 'al p > iblp > 121 r, and so oca-10 = bo. Hence wK aa --1 0 20. Hence toK = cca-lo by Example 93:10. So wK = la- as required. Hence clsL clsK if and only if Q (L) = Q (K), i. e. if and only if L and K represent the same numbers. 1

17*

260


93:18. Example. (i) Let us describe the unimodular lattices of dimension 3. We consider a unimodular lattice L on a quadratic

space V over a dyadic local field. Let a be any norm generator, let b be any weight generator, let d be the discriminant of L. Regard d as an element of u. Note that bo = 2o when ordp a ordp b is even, by § 93A. (ii) First we dispose of all cases with dim L 3 and ordp a ± ordp b even. We claim that (A (0-, 0)) ± ••• .

To see this we take a splitting L—J±K in which / is binary and nK = nL. Let ai be an element of Q (K) such that aio = nK. Then al a norm generator of L and so gL = a1o2 ± 2o. Hence by Example is 93:10 we have a splitting L (a1 61 271, 2 ) )1 K (A (27?', 2C)) 1. K' with 77, c, n P in o. But K' represents an element a2 such that a2o = nL, hence it represents an element of the form 28 with e a unit, hence by the perfectness of the residue class field we can write L

with n in o. Hence by Example 93:11, "

L

I • • • . (vi) As a special example let us consider the case dim L 4 with d L = 1. Then FL is isotropic if ordp a + ordp b is even, and we actually have

L

(A

(a, 0)>I

262


by (ii) and Example 93:9. If ordp a + ordp b is odd, then L (A (a, O)) J ± ( A (0, 0)) K gK = g K and, by Theorem 93:14, (A (a, ,8)) L (A (a + 2, 13)) K

q. e. d. 93 : 2 O. Example. Let L be a modular lattice on the quadratic space V over the dyadic local field F, and suppose that dim V 3. We claim that 0 (01 -

2 u.

By scaling V we can assume that L is unimodular. Let b denote any weight generator of L. By examining the different cases in Example 93:18 we see that there is always a maximal vector y in L with Q(y) = b. This vector satisfies 2 B (y, L)

(Y)

b

°

hence the symmetry T., is a unit of L by § 91B. So there is always a symmetry in 0 (L) with spinor norm bF2. But eb is also a weight generator of L, for any e in u. Hence e E 0 (0+ (L)). Hence 0(0+ (L)) D u. § 93E. The fundamental invariants Consider the non-zero regular lattice L in the quadratic space V over a dyadic local field. Let L have the Jordan splitting L=

• • • ± Lt


263

Put s i = sL i

for 1

i t. Thus s L s. And L i is s i-modular with 1

Note that L 6i

) 2) • • • D st.

s i by § 91C. We define g 1 =gL, w i wPi

Thus g 1 -1)- g 2

•'•

gt

1:D2 ----) • • • :2 rvt

since L 5i D L 6-I when si fix it. Thus we have

D

si. We take a norm generator ai for Pi and we

ao

• • • _D

at ])

and also = a10 2 +

Other relations among ai, si, wi can be deduced from § 93A. The invariants t, dim L i , s i, wi, a i t) will be called the fundamental invariants of the lattice L. The number t is, of course, the number of modular components of any Jordan splitting of L. We shall call the dim L i the fundamental dimensions, the s i the fundamental scales, etc., of the lattice L. The norm group of L is equal to the first fundamental norm group gi of L. All the fundamental invariants other than the ai are unique for a given L. By Example a; will be fundamental norm generators for L if 93:6, scalars a;, and only if ai mod tv i for 1 i t

(1

i

.

Now consider another lattice L' in a quadratic space V' over the same field. Let L' = L I • • •'1_. L, denote a Jordan splitting of L', and let t', dimg, s;, rv'i , a;

be a set of fundamental invariants of L'. Let g i' be the fundamental norm groups of L'. We say that L and L' are of the same fundamental type if t = t', dimL i = dim4 6 i = gi

264


t. This is equivalent to

for 1 _‹ I

t = t', dimL i =dimL,s i =

wi

,

modw

ai

for 1 I z t. It is clear that isometric lattices are of the same fundamental type and that an isometry preserves the fundamental invariants. Suppose L has the same fundamental type as L'. Then the fundamental norm generators satisfy a1 .-1-2 a modwl and so {a1,. . ., a} can be regarded as a set of fundamental norm generators of L'. When this is done we say that we have chosen the same set of fundamental invariants for L and L', or simply that L and L' have the same set of fundamental invariants. The lattice L i is the first component in some Jordan splitting of L 6i. Hence nLi = s i if and only if ni.61= s i, i. e. if and only if a i o = 6i . So lattices of the same fundamental type are also of the same Jordan type. Let us introduce some additional notation. We put si = ordp s i, u i = ordp ai for 1 j t. These quantities clearly depend just on the fundamental type. We define fractional ideals fi for 1 1 and proceed with the induction. By making suitable hyperbolic adjunctions we can assume that i t, in virtue of Theorem 93:14. By adjoining dimLi 3 for 1 the lattice = -LT1. we see that we can assume that each space FL, is hyperbolic, in virtue of Corollary 93:14a. By making suitable hyperbolic cancellations we i t. By scaling V we can can assume that dimLi dimKi = 4 for 1 assume that L1 and K1 are unimod-ular. It is enough to find a new Jordan splitting K .1q If3 I •• I K t L1. For then we can assume that K1 = L1 by Witt's in which K theorem. So the lattices L2 ± L 3 ± • • j_ L, and Iq K3 ' • • I K t are on the same quadratic space, and they clearly satisfy the conditions of the theorem. The inductive hypothesis then asserts that these lattices are isometric. Hence L K. First consider the case û2 = 2o. Then Li . But

± (A (2, 2 i3)> with a, 9 E 0, hence by Proposition 93:19 K1 ± K2 (.11 (2u1, j_ • • • T(cc4 > Q4

in such a way that each of the symmetries appearing in this equation is with respect to a line in U. But 0(U) Q3 (U). Hence a.(24 = Q4 . SO a E .f24, as required. 3) From now on we can assume that F is non-dyadic. Let A denote a fixed non-square unit in F. Take a typical a in 04 and express it in the form r r r r(ccj> Q4 — Q4 -

Otherwise we can write Cr 124

TO)

r Q4 •

282


Hence 0 ri consists of at most two cosets of Q4, namely Q4 and To> T(A> T(,> Tot A>Q4.

Hence (0'4 : Q4) = 2. 4) Finally we must prove that (0,1: Q4) = 2 for the non-dyadic case now under discussion. In order to do this it will be enough to produce an element of 0 41 which is not in Q4. Since V is anisotropic there is exactly one o-maximal lattice on V by Theorem 91: 1. Let L denote this lattice. By Proposition 63:17 we can take a lattice L1 J L 2 on V with

I (--/1>

and

L2

(n> 1 - V by means of the equation ax V x EL;

r;

it is easily seen that 5 is an isometry of V onto and that 19-1cr2 =51 52 for all cri , a2 in 04. Hence we have a natural homomorphism

04 (V)

02 (F) .

By Corollary 92: lb we can find vectors x, y (x) -- 1 , 42 (y) = A ,

(14) = T

L1 and u, V E. L 2 with (v)


283

Define = Tx Ty Tu Tv E

.

We assert that a is not in Dd. Once we have proved this we shall be through. Suppose if possible that a E Dd. Then o E 122 (7) 1%(7'). On the other hand it is easily seen from the defining equation of a symmetry that O = Ty . Hence (9 (a) = Q Q(37) = J. But 71 is not a square in P by the Local Square Theorem. So O is not in 49 (F). This is absurd. Hence (0 D4) =-- 2 as required. q. e. d. 95: la. In the exceptional case TO> r 'V W "r 124 and DA are the two distinct cosets of al modulo Dd. Here Li can be any non-square unit of F.

95:2. Let V be a regular n-ary quadratic space over a local field with n 3, and let a be an element of 1.2n of the form a = "Coo . . . Too with r < n and all ai in F. Then there is a regular ternary subspace of V which represents

Proof. If V is isotropic we take any regular ternary subspace which contains an isotropic vector of V. We may therefore assume that V is a quaternary anisotropic space and that r = 4. In the dyadic case we can use the argument used in step 2) of the proof of Proposition 95:1 to find a regular ternary subspace of V which represents x1,. cc4. So let us suppose that F is non-dyadic. If xi , . 24 fell in distinct cosets of F modulo F2 we could not have ci in 124, by Corollary 95: la. Hence we can assume that 2ia2 E P2 and o 324 E F2, say. There is clearly a ternary (in fact a binary) subspace of V that represents al and as. This space represents al, as. q. e. d. If V is any regular quadratic space over a local field other than an anisotropic quaternary space over a non-dyadic local field, then the condition for 1-4 n Zn to be {1 l v} is the same as the condition for —1 v to be in 49;„ namely dV = 1 with n even. Let us settle the exceptional case. We claim that in the exceptional case we will have Dd n Zd = ± 1 v} if and only if —1 is a non-square in F. For suppose that —1 is a non-square in F. Then —4 E ff'2. Hence

V ±

± ,

therefore , iv =. TO> TO> .r<x> 2. E and so 124 n Z4 = { ± l v}. Conversely suppose that —1 v E124. If-1 were —

284


a square we would have l'o.) t(4)

T(s) T(vrA) 94 T(1) 1..(—A) T r) T(---svA) 94 — (

94 = 94

and this would contradict Corollary 95: la. So —1 is a non-square. Finally we recall from Proposition 91:6 that the group the form

has

for n 3. The same applies if n = 2 with V isotropic. If n = 2 with V anisotropic, then one can use Example 63:15 to show that 0/0,,,' is isomorphic to a subgroup of index 2 in

We have therefore fully described the groups rA Z, 0114, 0110;

over local fields. Chapter X

Integral Theory of Quadratic Forms over Global Fields We conclude this book by introducing the genus and the spinor genus of a lattice on a quadratic space over a global field, and by studying the relation between these two new objects and the class. We shall use these relations to obtain sufficient conditions under which two lattices axe in the same class. We continue with the notation of Chapter VIII, except that the field F is now a global field and S is a Dedekind set of spots which consists of almost all spots on F. We let o be the ring of integers o (S), u the group of units u (S). As usual we let p stand either for a prime spot in S or for the prime ideal which it determines in o. (There will be one exception to this notation: in § 101A we shall let F denote an arbitrary valuated field.) Ar or Q will stand for the set of all non-trivial spots on F, I 1 p will be the normalized valuation on Fp at a spot p in Q. If p is discrete we let op, up, mp stand for the ring of integers, the group of units, and the maximal ideal of Fp at p. ap will be the localization at p of the fractional ideal a of F at S. V will be a regular non-zero n-ary quadratic space over F with symmetric bilinear form B and associated quadratic form Q. We shall consider lattices L, K, ... on V, always with respect to the underlying set of spots S. We let Vp denote a fixed localization of the quadratic space V at a spot p in Q. The lattice L p will be the localization of L in Vp at any spot p. in S. The notation 0 (V), 0+ (V),. .. for the subgroups of the orthogonal group will be carried over from Chapters IV and V.

Chapter X. Integral Theory of Quadratic Forms over Global Fields

285

§ 101. Elementary properties of the orthogonal group over

arithmetic fields § 101A. The orthogonal group over valuated fields In this subparagraph F denotes an arbitrary valuated field, not necessarily the global field F under discussion in this chapter. Let 1 I or I lp be the given valuation on F, and let p be the spot which it determines. V is an n-dimensional vector space over F. A base x1,. . . , x„ is taken and fixed for V. The norms II II which we are about to define are with respect to the same base xj., . . . , x,,, unless otherwise stated. Recall our earlier notation: 1,7 (V) denotes the algebra of linear transformations of V into itself, and /12(F) denotes the algebra of n x n matrices over F. Practically no proofs will be given here. All assertions can be verified either by inspection or by simple direct calculation. First we define the norm on V. Given any vector x in V, express it in the form x — oci xi + • • • + cc„x,,, (cci E F)

and then define the norm of x by the equation

11xIl p = max lai lp . i Use II II instead of 11 Il y whenever convenient. So II 11 is a real-valued function with the following properties: (1) 11x11 > 0 if x E 1,., and 11 0 11 = 0

V a EF, x E V (2) Ilaxil = 1 0:1 11x11 (3) 11x ± Yll .- 11x11 ± 11Y11 V x, y E V. In other words 1 II is a norm in the sense of § 11 G. And we can make V into a metric (topological) space by defining the distance between the vectors x and y to be IIx — yll. As usual,

1 1 14 - 11311 .__ 1 x In the case of a non-archimedean field we have

11x ± Yll 5 max (11x11 , bill) V x, y E V with

lix ± yll = max (Ix°, 11Y11) if 11x11 * 11Y11 • In particular, in the non-archimedean case there is a neighborhood of any given point x0 + 0 throughout which

11x1 1 - 1 x011 .


286

Each of the mappings (x, y) ->- x ± y of V x V into V, x of V x into V, (a, x) ax of F x V into V, is continuous. This means, to use the language of topological groups, that V is a topological vector space over the topological field F. The map (Y1 , • • • , Yr) -›- Yi + ' • ' + of Vx•••x V into V is continuous. So is the map x ->- 11x11 of V into R. Now do the same thing with LE (V). Consider a typical ci in LE (V), write axi = E oci; xi (mu EF)

for 1 j

n and define the norm of a by the equation

lloll

= max

1 = max Ilax5 11 p .

lip whenever convenient. Then 11 11 makes L(V) into a normed vector space, i. e. we have (1) IJciJ > 0 if crEL E (V) with ci== 0, and 110II = 0 V aEF, ci E (V) (2) 11'2 4 = 1 01 11011 Use II II instead of

(3) 1 0' + Tli

1 011 + liT11

V a, 'r E LE (V).

And L (V) is provided with a metric topology in which the distance between a and r is defined to be Ila — T11 . As usual, I 11 0II — II TII I -5 2.11 • In the case of a non-archimedean field we have TII

with

max (11 4 114)

V a, T

E LE (V)

1 0' + Til = max (11011 , linii)

if 1 011 I 114 In particular, in the non-archimedean case there is a neighborhood of any given co == 0 throughout which liall coil • We again have continuity of addition, of taking negatives, and of scalar multiplication, so that LE (V) is also a topological vector space over the given topological field. All this parallels the discussion for V. But we also have multiplicative laws to consider. We find that

lic xil

n 11 011 14 1 01 114

in general if non-archimedean,


287

for all a in Lp (V) and all x in V. Similarly for a, r in L p (V) we have Til 5- I.

n 110 lirO

IiolI uni

in general if non-archimedean.

Hence the mapping an of Lp (V) X L p (V) (a, n)

into Lp (V)

is continuous. This makes LF (V) into a topological ring as well as a topological vector space (as the name suggests, a topological ring is a ring with a topology in which addition, the taking of negatives, and multiplication, are all continuous). The mappings (al,

• • • C rr) ->-

± " • • ±

and

(a1 , • • ar) ->- al • • • ar are continuous. So are the mappings (a, x) -›- ax of LAV) x V into

det a of L p (V)

V,

into F.

The continuity of the determinant map shows that the general linear group G L (V) is an open subset of L F (V) . If we restrict ourselves to GL (V) we find that the mapping

a

a-1 of GL (V)

into GL (V)

is continuous. on the algebra of n x n matrices We can introduce a norm if 111-11 (F) by defining II (aii)11 p = max lad y for a typical matrix (a id) over F. We shall write 11 a.0 1 or liciA instead of II (aii)iip. Note that the norm hail of a linear transformation a is equal to the norm of its matrix in the base x 1 , . . xn. 10 1 : 1. Example. What happens to the norm under a change of base ? Take a new base xi, . . .4 for V with

E ail xi and X5 = f bo xi Let

denote the norm with respect to the base xi, . .

general we have

11z11 nIlb5I1 11x11

and

nhlai

jhl

5 -

II< < n2 au Il °bo11

11all

Then in

288


for any x in V and any Cl in LF (V). In the non-archimedean case we have

ilzil

11a„11

III

Ilxii

and

loi'

1Ia1I Ii

11a11

To conclude we suppose that V has been made into a regular quadratic space by providing it with the symmetric bilinear form B and the quadratic form Q. Then there is a positive constant 2 such that

Allx11 113111 V x, y E

1B(x,

V,

so

(x)i

214 2 VxE V.

The mappings (x, y)

B (x, y) of

x —>— Q (x)

V X V into F,

of V

into F ,

are continuous. 1 0 1 :2. Example. The continuity of the map x —>— Q (x) shows that the set of anisotropic vectors of V is an open subset of V. Let u denote an anisotropic vector of V. Then the mapping u

of the set of anisotropic vectors of V into On (V) is continuous (here t„ denotes the symmetry of V with respect to u). To prove this one considers the defining equation rts x = x

2 B (u,

(u)

of a symmetry. First one shows, using the continuity of the maps u 2 B (u, x), u Q (u), etc., that the mapping u 7.x is continuous for each fixed x in V. One then deduces the continuity of u -L.'. at u0 from the equation — tu0 x511 !iris — risoil = max Hence u -4— T. is continuous. 10 1:3. Example. The continuity of the determinant map tells us that 0+ (V) and OE (V) are closed subsets of 0(V). Hence 0+ (V) is an open and closed subgroup of 0(V). 101:4. Example. (i) Suppose that the field F under discussion is actually a local field. Then for any ci in 0(V) we have deter = ± 1, hence det a is a unit, hence 114711 >_ 1. Now let M be the lattice M ox1 + • • • +1) xn , where x11 .. , xn is the base used in defining 11 11. Then pall = 1 if and only if Ili 1, this is equivalent to 47/1/ S M, and hence to aM = M. So the elements of 0 (M) are precisely the isometries of V with 'loll -= 1. In particular the set of isometries with 11cV = 1 is a group.


289

(ii) Consider a second lattice L on V. We claim that a L L holds for all a in 0(V) which are sufficiently close to l v. Take a base 4, for L, and let 1 -11' denote norms with respect to this new base. Then by Example 101:1 we see that all a which are sufficiently close to i v satisfy 11o.— 1 v 11' < 1. Each such a satisfies 'loll' = 1, hence a L = L. (iii) Consider a third lattice K on V, and suppose that K = AL for some A in 0(V). We claim that a L = K holds for all a in 0(V) which are sufficiently close to A. By choosing a sufficiently close to A we can make 11 2-l a vii i ii 2-1 Ail arbitrarily small. But all A-' ci which are sufficiently close to i, make A-iaL= L. Hence all a which are sufficiently close to A make aL= AL—K. § 101B. The orthogonal group over global fields We return to the situation described at the beginning of the chapter: quadratic forms over global fields. F is again a global field and V is a quadratic space over F. Since each Vp is a vector space over the valuated field Fp we can introduce norms 11 11 p on V and LFp(Vp) with respect to any given base of Vp, in particular with respect to any given base for V over F. We shall always assume that all norms under discussion are with respect to a common base for V. We let xl , , xn denote the vectors of this base. If a new base xi, .. x t, is taken for V, and if we consider the corresponding norms 4, at each p on F, then it follows from Example 101:1 and the Product Formula of § 33B that we have for almost all p.

11 11p = III]

Consider a typical linear transformation a in LE (V). By considering the effect of a on a base for V we see that there is a unique linear transformation op on Vp which induces a on V. We call ap the p-ification or localization at p of a. It is easily verified, again by considering a base for V, that we have the rules (o.

.r) r

ap

(Occf)p = xa,,

)

(a

p

ap Tp

detap = deter

for all a,1** in LF(V) and all a in F. In particular, the mapping a gives us an injective ring homomorphism LF(V)>-- LFp (Vp) . If a is an isometry, then so is 0.p. If a is a rotation, then so is 0.p. If tu is the synunetry of V with respect to the anisotropic line F u, then a geometric argument shows that (r.)4,, is the symmetry of Vp with respect to the line F4,14; we express this symbolically by the equation (Tu)p = Tu • O'Meara, Introduction to quadratic forms

19

290


In keeping with functional notation we let A 9 denote the image of a subset A of Lii, (V) in LFp (Vp) under localization at p. We have Ç On (V) p S On (Vp) , 0„+ (V) p S On+ (Vp) , 1.07,(V) p g_ 1-4(Vp) , 0, (V) _ç_. 0 (V,,). 101:5. Conventions. In some situations things become clearer if the notation is relaxed and a is used for the localization ap of a, in other situations the strict notation ap is preferable. We shall use both. We shall also use 99p (or ci,,) to denote a typical element of LFp (Vp); of course this does not necessarily mean that 999 is the localization of a linear transformation 99 of V. 101:6. Example. Let S be a Dedekind set of spots on F, let L be a lattice in the vector space V over F, and let a be an element of LF (V). We claim that ap L p = (a L) 0 Vp ES. To prove this we express L in the form L = ai yi + - • • + a r yr where the ai are fractional ideals and the yi are elements of V. Then ap L p = ap (a/p yi ± • • • + arpyr) —alp (aYi) + • • • + an, (TYr) — (ai (aYi) + ' • • + ar (aYr))p

= (aL) p .

This proves our claim. As an immediate consequence we have S 0„(4) and 0(L) p for all p in S. 101:7. Weak Approximation Theorem for Rotations. Let V be a regular quadratic space over a global field F and let T be a finite set of spots on F. Suppose çop is given in 0+ (V,,) at each p in T. Then for each e> 0 there is a a in 0+ (V) such that iic — 99pilp< 8 VP ETProof. Express each (Pp as a product of symmetries, 99p = -cti,;;

-

-

where the ur are anisotropic vectors in Vp. Here the number r is even, and we can suppose that the same r applies for all p by adjoining squares of symmetries wherever necessary. Using the Weak Approximation Theorem of § 11E on the coordinates of the vectors /41 in the underlying base x1 ,. . ., x„ we can obtain a vector u, in V such that ilui — Will y is arbitrarily small at all p in T. If the approximation is good enough we


obtain an anisotropic vector u1 such that

Tui

is arbitrarily close to -c•

291 p

at

each p in T, in virtue of the continuity of the map u -4— ru. Now do all this for i 1, 2, . , r and in this way obtain anisotropie vectors U1 , . . , it of V with Tui ruP p

arbitrarily small for all p in 7' and for 1 i r. If good enough approximations are taken all around we can arrange to have

ru, • • VuTuP

.. • r

•

er P
T(i) to> r E 124 (Vp) • And Vp is isotropic at each of the remaining spots p on F, hence .Q4 (Vp) q. e. d. = 01 (Vp), hence ap E ‘24 (V,,) . 101:10. Example. V is a regular n-ary space over the global field F and S is a Dedekind set consisting of almost all spots on F. Consider lattices L,K,... on V with respect to S. We always have cls+L cIsL when n is odd (see Example 82:4). This is also true for even n over local fields (see Corollary 91:4a). It is not true in general for global fields. In fact we claim that there is a lattice K on V with cls+K C clsK whenever n is even. Let us construct such a K. We start with an arbitrary lattice L on V. Write dV s with s in F. We know from Example 66:6 that the Hasse symbol S p V is 1 at almost all p on F. Let W denote a set of


293

non-dyadic spots on F which consists of almost all spots in S, such that s is a W-unit and Sp V = 1 at all p in W, such that JE := PEJY, and finally such that L p is unimodular at all p in W. Pick spots q and q' in W such that s is a square at q and q', and such that a W-unit is a square at q if and only if it is a square at cr. The existence of such a pair of spots was established in Proposition 65:20. By Proposition 81:14 and Example p E S— (q 'u q') and 92:8 there is a lattice K on V with Kr = L

0 (0- (IC q))

ti q

,

0 (0- (K q,))

where 4q is a non-square unit in Fq. We say that this K has the desired properties, L e. cls÷K C clsK, i. e. 0+(K) a= 0(K). For suppose not. Then there is a reflexion a of V such that a K = K. Write 0 (a) = oc with oc in P. If p E W (q q'), then a E 61(0 (4)) = uF. Similarly, a Ezi gn and cc E F. so ac is a non-square at q, it is a square at q', and ordp a is even at all p in W. Since IF = PE JI-v, there is a fl in F with 2 ordp/3 = ordp a at all p in W. Then oc//32 is a W-unit which is a square at q' but not at q. This is impossible by choice of q and cf. So 0+ (K) = 0(K). Hence cls+K ( clsK.

§ 101C. Special subgroups of On (V) Once again we return to the questions raised in § 56. This time we must describe the groups 12„ n Z,, 0:,11-2„ , 010.;, over the global field F which is now under discussion'. Of course if F is a function field and n 5, then V is isotropic and all is already known. However, some extra effort will be needed before we obtain the general case. 101:11. Lemma. Let X be a finite subset of fr and suppose that at each spot p on F there is a regular ternary subspace of V I, which represents all of X. Then there is a regular ternary subspace of V which represents all of X. Proof. 1) For n = 3 the result is an immediate consequence of Theorem 66:3. Suppose n = 4. Let T be a finite set of spots on F which contains all archimedean and dyadic spots and is such that, at each spot p outside T, the discriminant dV p is a unit, the Hasse symbol S, V,is 1, and every element of X is a unit. By hypothesis there is a scalar yi, E Fp and a regular ternary space Up at each p in T such that

Vp I U,, , X ç Q (Up) . I

For the structure of .(2.112n Z„ (n 5) over an algebraic number field see M. KNESER, Crelle's J., 196 (1956), pp. 213-220.

294


By the Weak Approximation Theorem and fact that F is open in Fr there is a y in fr such that y E y r F for all p in T. Then Ur VpET. Now V p represents y at each discrete spots p on F.by Remark 63:18, hence V represents y by Theorem 66:3, hence there is a regular ternary subspace W of V and a splitting V I W. At each p in T we have Up by Witt's theorem, and so X S Q (W 1,) ; at each of the remaining p on F we have X Q (Wp) by Example 63:24; hence X Ç Q (W). This proves the case n 4. 2) Now the case n 5. Here let T be any finite set of spots on F which contains all archimedean spots and is such that every element of X is a unit at all spots outside T. Using the Weak Approximation Theorem one can easily construct a regular binary space U over F which is isotropic at each p in T at which VI, is isotropic, which is positive definite at each real p at which V positive definite, and which is negative definite at each real p at which V p is negative definite. Then there is a representation Up —>-- V p at each p on F by Theorem 63:21, hence there is a representation U -3-- V by Theorem 66:3. In effect this allows us to assume that U S V. Let W be any regular quaternary subspace of V which contains U. Then at each p in T we have X S Q (U p) ; and at each remaining spot p on F there will be a regular binary subspace of Wr which represents all of ur (Example 63:15 and Proposition 63:17) and hence all of X. Hence there is a regular ternary (in fact binary) subspace of WI, at each p on F which represents all of X. Hence by step 1) there is a regular ternary subspace of W (and hence of V) which represents all X. q. e. d. 101:12. Let V be a regular n-ary quadratic space, over the global field F and let a be an element of 01(V). Then a is in,f2„ (V) i/ and only if a p is in f2„(V p ) at each spot p on F. Proof. Only the sufficiency really needs proof. So let us assume that E û (Vp) at each p on F and let us deduce from this that a E (2,, (V). Express a as a product of symmetries with respect to vectors yi, . . , of V: (r n) . Then a E 07,1 (Vp) at each p, hence Q Global Square Theorem we must have Q(311) • • • Q(Yr) So a

- • Q (y ,-)

E F, hence by the

E

E On (V). In particular this proves the proposition when 1 n _‹ 3.

Let us now assume that n >, 4. Since a p E (V r) we can conclude from Proposition 95:2 at the discrete spots and from the archimedean theory of quadratic forms at the archimedean spots that there is a regular


295

ternary subspace of Vp at each p on F which represents all the scalars Q(311), . . Q(y,.). By Lemma 101: 11 there is a regular ternary subspace W of V which represents all these scalars. Pick xi E W with Q (zi) = Q (y1) for 1 i r. By Proposition 55:3 it is enough to prove that T„ . . • Tzr is in Q„ (V). And this follows easily from the fact that rz, " • Tz, induces an element of Q 3 (W) = 03 (W) on W. q. e. d. Now we can describe the groups (2„ n Zn, etc. We have Q7, n Zt, = {±1 } if —lvp is in Q. (Vp) at each p on F, otherwise 12„ n = 1 v. The group 0,112„ is isomorphic to the group

HO"„ (Vp)1(2„ (Vp) (this follows easily from Propositions 101: 9 and 101:12). In particular Q„ when n + 4, while if n = 4 the group 0 41 /Q4 is a direct product of a finite number of groups of order 2. Finally the group OI/0;, can be destribed by Proposition 101:8 (for n 3): it is the group of elements which are positive at all real spots p at which V p is anisotropic, modulo the group F 2. § 101D. The group of split rotations jv We have to work with idèle groups again, particularly with the groups T p

j F3 11 1

TS pS E,

TS, 2

j

defined in §§ 33D, 331 and 65A. Here we are assuming that S is a Dedekind set consisting of almost all spots on the global field F. The idèle concept can be extended to the orthogonal group in the following way. Start with a regular quadratic space V over the global field F. Take a base x11 . . x„ for V and let all the norms IIIlp on Vp and LFp (Vp) be with respect to this fixed underlying base. Consider the multiplicative group H O+ (V) p ED

consisting of the direct product of all the groups 0- F (V,). A typical element of this group is defined by its coordinates, say

(Ep)pED (Ep (°+ TO) , and multiplication in the direct product is, by definition, coordinatewise. If we are just told that E is a typical element of the direct product, then Ep will denote its p-coordinate. For two such elements E, A we have (E A) p EpA p (E-1) p = EF,1 , for all p in Q. We shall call E a split rotation of the quadratic space V if E is an element of the above direct product with. the property llEPlp = 1 for almost all p.

296


This definition is independent of the underlying base since any two systems of norms agree at almost all p by Example 101:1. The set of all split rotations is a subgroup of the above direct product by Example 101:4, it is called the group of split rotations, and it will be written / v. 1 The set of all split rotations Z with the property Er 0' (Vp) YpED is clearly a subgroup of hi; we shall denote this subgroup with the letter Jr. It is evident that liv contains the commutator subgroup of jv ; in particular, .rv is a normal subgroup of Jv and the quotient group JviTv is abelian_ Consider a typical element a of 0+ (V). Then a has a localization up p in D. And Ijap il p = 1 for almost all p, by the Product Formula. atech Hence a determines a split rotation (a) = (ap)p ED • The rule a —>-- (a) therefore provides a natural multiplicative isomorphism of 0+ (V) into Iv. We shall call the split rotation Z principal if there is a rotation a such that Z = (a). The principal split rotations form a subgroup Pv of /v . And we have 0+ (V) >---> P. We shall let D stand for the subgroup 0(04- (V)) of F, and we let PD be the group of principal idèles of the form (a) p ED with a in D. In other Pp . words, P D is the image of D under the natural isomorphism F 101:13. Example. Suppose n 3. Proposition 101:8 tells us that D is then the set of all elements of È which are positive at those real spots p at which Vp is anisotropic. The Weak Approximation Theorem of § 11E shows that (F D) is finite. Hence (Pp : PD) < co. We know from Corollary 33:14a that there is a Dedekind set of spots So which consists of almost all spots on F and is such that IF= PFI: whenever S S So. By considering a finite set of representatives of pp, mod PD we see that there is a set So of the above type such that = PD J;,,g whenever S Ç So . Now consider lattices L, K, . . . on V with respect to the set of spots S under discussion. We define the subgroup h of jv by the equation IL = {E E Tv Ep E 0+ (LO V p E S) . If a is an element of 0+(L), then ap E O nf (L p) holds for all p in S by Example 101:6, hence (a) is an element of Pv n h. On the other hand, if (a) denotes a typical element of Pv n IL , then (a L) a p L p = L p holds for all p in S, hence aL — L by § 81E, hence a E 04- (L). Hence the natural isomorphism of 04- (V) onto Pr carries 04 (L) onto P v n IL . So For a discussion and application of the, topology on J v see M. KNESER, Math. Z. (1961), pp. 188-194.


297

we have the diagram

0+ (V) 1

0+ (L)

Pv

Pv r\ h.

We define the subgroup n of IF by the equation JP, = {i E

IF I ir E 0(0+ (Lp)) V p E .5} .

Take a typical split rotation E and a typical lattice L on V. We know that EL p is a lattice on V r at each p in S, we claim that there is exactly one lattice K on V with Kr ELp for all p in S, and we then define EL to be this lattice K. In order to prove the existence of K it is enough, in view of Proposition 81:14, to show that ELr = Lp for ox,. Then the condition ilEplip— 1 almost all p in S. Put M = o ± • • is equivalent to Er Mi, = M. hence EMr = Mr for almost all p in S. But Mp = Lr for almost all p in S by § 81E. Hence Ev il, = L I, for almost all p in S. Hence K exists. It is unique by § 81E. So the lattice EL is defined; its defining equation is (EL) =ELr Vp ES. Incidentally, note that EL p = Lp for almost all p E S. We have (E A) L = E(AL) V E, A E jv

If a is a rotation of V, then a L = (a) L . The group h can be described as

= {E E Jv I

L = L} .

102. The genus and the spinor genus § 102A. Definition of gen L and spn L We define the genus genL of the lattice L on V to be the set of all lattices K on V with the following property: for each p in S there exists an isometry Ep E O (Vp) such that Kr = Ep L p. The set of all lattices on V is thereby partitioned into genera. We immediately have genK = genL •:* clsKp = clsLr Vp ES. The proper genus can be defined in the same way: we say that K is in the same proper genus as L if for each p in S there is a rotation 4, c O+(V) such that Kr = ErL r. This leads to a partition of the set of all lattices on V into proper genera. The proper genus of L will be written gen+L. We immediately have gen-FK = gen+L cls+Kr = cls+Lr Vp ES.

298


But we already know that the class and the proper class coincide over local fields. Hence we always have genL = gen-FL . The genus can be described in terms of split rotations:

K E genL

K = EL

for some

in I v

almost all p in S). (to prove this use the fact that K = L We say that the lattice K on V is in the same spinor genus as L if there is an isometry a in 0 (V) and a rotation E in 0' (V p) at each p in S such that = Vp(S. This condition can be expressed in the language of split rotations: there is a a in 0(V) and a X in j'v such that K = L' L. We shall use spnL to denote the set of lattices in the same spinor genus as L. It can be verified without difficulty that the set of all lattices on V is partitioned into spinor genera. It is an immediate consequence of the definitions that lattices in the same class are in the same spinor genus, and lattices in the same spinor genus are in the same genus. So the partition into classes is finer than the partition into spinor genera, and the partition into spinor genera is finer than the partition into genera. We have clsL spnL çT genL . We let h(L) be the number of classes in genL, and g(L) the number of spinor genera in genL. We shall see later that h(L) and g(L) are always finite. We say that K is in the same proper spinor genus as L if there is a 0' (V p) at each p in S such that rotation a in 0+ (V) and a.rotation X K1,

Lp

VpES.

This condition can be expressed in the language of split rotations by saying that there is a a in 0+ (V) and a E in Ty such that K = ci X L, or

gen L spn L

cis L gen#L spi) c/8"L

equivalently by saying that there is a A in Pv and a X in Tv such that K A EL. We shall use spn+ L to denote the set of lattices in the same proper spinor genus as L. It is easily seen that the set of all lattices on V


299

is partitioned into proper spinor genera. Again we find that the partition into proper classes, is finer than the partition into proper spinor genera, and the partition into proper spinor genera is finer than the partition into genera. We have cls+L C spn -FL S gen+L .

We let h+ (L) be the number of proper classes in gen L, and g +(L) the number of proper spinor genera in genL. We shall see that h+ (L) and g+ (L) are always finite. All lattices in the same genus have the same volume. For consider K E genL. Then Kr -.".# L t, for all p in S, hence (t3K) 1, = »Kr = 13 L i, = (bL) 1, Vp ES, hence D K = b L. We define the volume of a genus to be the common volume of all lattices in the genus. In the same way we can define the volume of a proper class, of a class, of a proper spinor genus, or of a spinor genus, since each of these sets is contained in a single genus. Similarly we can define the scale and norm of a genus, proper class, etc. If a genus contains an a-maximal (resp. a-modular) lattice, then every lattice in that genus is a-maximal (resp. a-modular). 102:1. Example. If E is any element of jv, then E spnL = spnEL , 2' spn-FL = spn+EL . 102:2. Example. Each class contains either one or two proper classes, hence h (L) h+ (L) _. 2h (L) . It is easily seen that each spinor genus contains either one or two proper spinor genera, so g (L) ._. g+ (L) 2g (L) . But we can actually say more, namely that g+ (L) is either g (L) or 2g(L). For suppose that spnL contains two proper spinor genera. Then spn+L C spnL, hence by Example 102:1 we have spn+EL C spn EL for every E in Tv, hence every spinor genus in genL contains two proper spinor genera, hence g-F (L) = 2g (L). In the same way we find g+ (L)= g(L) when spn+L = spn L. 102:3 Example. Consider the genus genL of an a-maximal lattice L on V. We have already mentioned that every lattice in genL is also a-maximal. Now consider any a-maximal lattice K on V. Then Kr and are ap-maximal at all p in S, hence Kr -2! LI, by Theorem 91:2, hence K ( genL. So the genus of an a-maximal lattice on V consists precisely of all a-maximal lattices on V. In particular, all a-maximal lattices on the same quadratic space have the same scale, norm and volume. 102:4. Example. Suppose K ( genL. Consider a finite subset T of the underlying set of spots S. We claim that there is a lattice K' in

300


cls+K such that IC; = L all p in T. By definition of the genus there is a rotation ipt, E O+(V) at each p in T such that (pr Kt,= L. By the Weak Approximation Theorem for Rotations there is a rotation a in 0+ (V) such that lic — Ta p is arbitrarily small at each p in T. If the approximations are good enough we will have, in virtue of Examples 101:4 and 101:6, (47K) 1, ap K y 991,Kt, = L t,

for all p in T. Then aK is in cls+K, hence it is the desired lattice K'. 1 0 2:5. Example. Let L be a lattice on the quadratic space V, let K be a regular lattice in V. Suppose there is a representation K r -4— Li, at each p in S. We claim that there is a representation K L' of K into a lattice L' in genL. In fact we shall find a lattice L' in gen L such that K L'. To do this we take a finite subset T of S such that Kt, S L all p in S T. Since there is a representation Kr L t, at each p in T, there is an isometry (p p E O (Vr) such that T rL p D Kr. Define L' to be that lattice on V for which VpEST — L' — tinp Ly VpET. —

Then 4D Kt, for all p in S, hence L' D K. Clearly L' E genL. Hence we have proved our assertion. We have the following special case: suppose that the scalar a EF is represented by V and also by L each p in S; then a is represented by some lattice L' in the genus of L. 1 0 2 :6. Example. It is possible for a scalar in Q (V) to be represented by L all p in S without its being represented by L. For instance, consider the set S of all discrete spots on the field of rational numbers Q. Let L be a lattice with respect to S on a quadratic space V over Q with L ± . Then the equation 3= (8/5) 2 + 11(1/5)2 shows that V represents 3, and also that L, represents 3 for all p + 5; but 1.5 also represents 3 by Corollary 92: lb. There is clearly no rational integral solution to the equation E2 + 11772 3. Hence V represents 3, L, represents 3 at all spots p, yet L does not represent 3.

§ 102B. Counting the spinor genera in a genus 1 02: 7.

Proof. 1) First a remark about abstract groups. Let G be any group and let H be any subgroup of G which contains the commutator subgroup of G. If x, y are typical elements of G, then the normality of H in G implies that xH Hx, and the fact that H contains the commutator subgroup of G implies that xyH= yxH, hence the set Hxy is independent of the order of H, x, y. From this it follows for any subgroups X, Y


301

of G that the set HXY is independent of the order of H, X, Y, and that this set is actually the group generated by H, X, Y. In particular, this applies to the subgroups Pv, h of J. So the group generated by these three groups is equal to Ti7Pv.11, this is a normal subgroup of j v, we can form the quotient j virv Pv h, and we can write down the index (Iv: ji,P v j b). Our next claim is that this index is equal to g+ 2) Consider two typical proper spinor genera in genL. They can be written spn+El L and spri.-1-E2 L with E, E E Iv since EL runs through gen L as E runs through J. We then have spn+Ei L = spn+E2 L if and only if L E spn+EVE2 L, and this is equivalent to saying that L = AT Er' Ez L for some A E Pv , T E Tv. Hence spni- El L = spn+E2 L

E2 E EirvPv.h -

Therefore, if we let E run through a complete set of representatives of distinct cosets of jv modulo ji,P v IL we obtain each proper spinor genus spn+EL in genL exactly once. So we have the formula

g+ (L) -- (Iv: ii7Pvh) for the number of proper spinor genera in a genus. 3) We are going to construct a group homomorphism of Iv into IF/PD jt, in a certain natural way. Take a typical element E in J. Then EL p = L almost all p, hence 0(4) S up F21, for almost all p by Proposition 92:5. We can therefore choose an idèle I in JE with ip

E (4) YPED•

If j is any other idèle that is associated with E in this way, then j E in by definition of the spinor norm. But .7 .1,c n c IDA . Hence the natural images of i and j in J.FIPD J.-k, are equal. We therefore have a welldefined map

0 : Tv obtained by sending E to the natural image of I in jp/PA. It is immediately verified that dis is a homomorphism. 4) Let us show that 0 is surjective. We must consider a typical idéle and we must find E E Iv such that it, E 0(4) for all p in D. By definition of we can assume that it, = 1 for all p in D — S; we define 4, to be the identity map on Vp for all p in D— S. What about the remaining p? Since n 3 we have 0(0+ (V)) = F all p in S by Proposition 91:6 and 0(0+ (4)) urFti for almost all p in S by Proposition 92:5; we can therefore choose E E 0+(V1,) at each p in S, with almost all E in Q 4- (L), such that 11, E 0(Er) for all p in S. Then IiEJI = 1 for almost all p in S. Hence E = (E0 1, 02 is a split rotation. And 0E is the natural image of t in jp/P D.3. So 0 is a surjective homomorphism. .3

302


5) It is clear that Pv, Pv, j.L are all part of the kernel of ao, hence Pv Pv jj; is. We leave it to the reader to verify that Tv Pv IL is the entire kernel. Hence

g+(L) =

rvPv.h) = Up: PD.* •

q. e. d.

102:7a. g+ (L) = (Jv : ji,Pv h), and g+ (L) divides (JF : 1. for any n Proof. The assumption n 3 in Proposition 102:7 is used only in q. e. d. showing that the map in the proof is surjective. 102 : 8. Theorem. V is a regular quadratic space over a global field with dim V 3, and L is a lattice on V. Then the number of proper spinor genera in genL is of the form 2r with r 0. Any value of r can be obtained by taking a suitable L on V. Proof. 1) Write the discriminant dV in the form dV e with e in P. We know from Example 66:6 that the Hasse symbol Sp V is 1 at almost all spots p on F. We fix a non-dyadic set of spots W on F which consists of almost all spots in S, such that e is a W-unit and Sp V = 1 at all p in W, such that hp = Ppir, and finally such that L I, is unimodular at all p in W. Proposition 92:5 tells us that. Jr 2) Let us prove two formulas that will be needed in the course of the proof. Consider a lattice K on V with respect to S, and let T be any set of spots on l; with T S and hp = Pill:. Then

g+(K) =

Pilf) =. (114: Pilf.) PD.* = (Png (.11: g n PD.* •

This is our first formula. Now let M be some other lattice on V with respect to S, and suppose that J, Then Then g+ (K ) =

CIF:PDTP

= (IF: PD.11)(PDPI P.Dip = g+ (m) (PD1f, : PD.T1P • Hence

g+ (K) = g+ (M) (Jr : j.y r if

PD.*

)1 c

J. This is our second formula. 3) The proof that g+ (L) is a power of 2 is at hand. We have g+ (L) = (Jr :

PA)


303

by the first formula in step 2). But rir. 2 S it,. Hence

Tr) '2 n g+ (L) (17 .17 12) This is a power of 2 by Proposition 65:7. 4) Next we construct a lattice K on V with g+ (K) = 1. Start with the given lattice L and Use Propositions 81:14 and 91:7 to obtain a lattice K on V with K p = Li, for all p in W and 0 (0+ (KO) =

Then

VpES—W.

frir jf, since

0(0+ (K))=u ,, 1 VI) EW by Proposition 92:5. Hence g n PDJ =Jf. Hence g+ (K) = 1 by the first formula of step 2). 5) We must digress for a moment in order to prove the following: let M and K be lattices on V, let 44 be a non-dyadic spot in S at which

0(0+ (M4)) = u4P, (0+ (Kg)) =.g& , and suppose that Mp = K p for all p in S — cr, then g+ (K) is either equal to g+ (M: or to 2g+ (M). By definition of M and K we have su c nt, hence by the second formula of step 2) it is enough to prove that 2, p n ni hence that uy, : 2. But this last inequality follows easily from the fact that (u4 : = 2. Hence our contention is proved. 6) Now we can complete the proof. It is enough to show how to obtain a lattice K from L with g+ (K) = 2g+ (L). For L is arbitrary, so we can start with an L with g+ (L) = 1, then obtain a new L with g+(L) then an L with g+ (L) = 4, and so on. So consider the given lattice L. Pick q and cr in W in such a way that e is a square at q and g', and such that a W-unit is a square at q if and only if it is a square at g'. The existence of such a pair of spots was established in Proposition 65:20. By Proposition 81:14 and Example 92:7 there is a lattice K on V with Kp = Li, for all p in S (q q') and with

Pal.)

:

0(0+ (K4)) = ft, 0 (0+ (K c()) =.

Clearly 1.15 c J. We claim that TL

nPD.!: C Tfi •

Suppose not. Then .3 S Pal,. Take i E n with ip = 1 for all p ED— q and 14 a non-square unit in Fq. Then i E Pill, hence there is an a in D and a j in If with i = (a) j. So we have a field element a which is a square at cr, a non-square at q, and of even order at all p in W. Now h = 1371y, hence there is an element /3 in F with 2 ordp fi = ordp a for all p in W.

304

Part Four. Arithmetic Theory

of Quadratic Forms over Rings

Then och92 is a W-unit, it is a square at cr, it is not a square at q. This contradicts the choice of q and q'. Hence we do indeed have nrA PD J-1,c c J. So by the second formula in step 2), g+ (K) 7->, 2g+ (L). all p in S — cr, and Let K' be the lattice on V with K'p = L K cr . Then by step 5), g (K) =

(K 1)

Hence either g (K')

or

2g±(K') ;

g+ (K1) = gl (L)

or

2 g+ (L) , or else g+ (K) = 2 g+ (L).

2g+ (L) . q. e. d.

1 0 2:8a. Corollary. The number of proper spinor genera in genL 1. is a power of 2 in general, i. e. when dim V Proof. The case n 3 is covered by the theorem. The case n = 1 is trivial since then genL = L. There remains n = 2. By Corollary 102:7a it is enough to show that (JE : D is a power of 2, and this follows as in steps 1)--3) of the proof of this theorem, q. e. d. 102:9. Theorem. L is a lattice on the regular quadratic space V over the global field F. Suppose that the underlying set of spots S satisfies the I equation J1 PD 0(0+ (Lp)) 2 up Vp ES, then spn 1 L = genL.

Proof. We must show that g+ (L) = 1. We use the fact that g+ (L) divides (IF: Pil). Then J Ç J since 0 (0+ (4)) D u p for all p in S. Hence IF , PD J_Z Ç PD /t. Hence g+ (L) = 1. q. e. d. 1 0 2:1 0. Example. The last theorem can be used to give sufficient conditions under which spn+L genL. Consider the lattice L on the regular quadratic space V over the global field F with dim V 3. (i) First let us examine the condition IF = ppjp:. This is satisfied in the function theoretic case whenever the class number liF (S) is 1 since then IF = PFIlis, with F.D, hence IF = PD .a. It is also satisfied when F is the field of rational numbers for then IF = ./4/4 with F ( ± 1) D, hence I) F _L 1) PD, but (-1) E ji„ hence JE = PDJ. (ii) The local theory has provided sufficient criteria for testing the condition 0(0 -1- (4)) D U p . For instance this condition is satisfied at all p in S whenever L is either modular or maximal on V (see Propositions 91:8 and 92:5, and Example 93:20). Again, it is satisfied at p if there is a 2 Jordan splitting for L p with a modular component of dimension when p is non-dyadic and of dimension 3 when p is dyadic. Example 92:9 shows how to derive from this a simpler, but weaker, criterion involving volumes (the formula given there is for the non-dyadic case only; but a similar, though not identical, result can be obtained at the dyadic spots).


305

§ 103. Finiteness of class number This paragraph is devoted to a single purpose: the proof that the number of proper classes of integral scale with given rank and volume is essentially finite'. This will imply that the number of proper classes in a (L) of § 102A genus is finite. Hence all the numbers h (L), h+ (L), g(L), are finite. We shall need the counting number Na introduced for fractional ideals in § 33C. For any non-zero scalar a we define N a = N (ao). Thus N (a fl) = (Na) (N )3). By Proposition 33:3 and the Product Formula we have

H,

Noe

pED—S

PCS

104.

103: 1. Let L be a lattice on the abstract vector space V over the global field F, and let 99 be a non-singular linear transformation of V into V such that 99L L. Then (L: 99 L) = N (det 99) .

Proof. 971, is a lattice on V since 99 is non-singular. By applying the Invariant Factor Theorem to L and 921, we can find a base x1 . . for V and fractional ideals a 1 ,. . an and 1)1. , bn, such that L

-1- • • • + bn ien

99L = al x/ ± • • • + a„,cn .

Now write Yi

99X J =

aiixi

E F) .

Then + • • • + bnYn • 97 /By comparing the above expressions for 99L we obtain, from Proposition 81:8,

. . . an = b1 . . . bn det (a ii) . Then by Proposition 33:2,

(L: 6, L) =

al) • • (b.: an)

— (Nal . . Nan)/(Nbl . Nb 7 )

= N(det 92) . q. e. d. 103:2. There is a positive constant y which depends only on F and S, and which has the following property: given any n x n matrix (aii) with 1 Using "reduction theory" it is possible to give a short and elementary proof of the finiteness of class number in the classical situation where the ring of integers in question is the ring of rational integers Z. See G. L. WATSON, Integral quadratic forms (Cambridge, 1960). O'Meara, Introduction to quadratic forms 20

306


entries in F and det(a ii) Eu, there are elements $1, . . ., them 0, such that faiii + • - - ± ain flip ..- 7

en in o, not all of

i n and for all p in D — S. Proof. 1) Fix a spot go in D — S. By Theorem 33:5 there is a constant C (0 < C < 1) such that the density M (i)1 II ill of the set of field elements bounded by any idèle i satisfies for I

M(i)lliiii > C •

We can assume, by taking a smaller C if necessary, that C is in the value group IFool go . Put y = 4C-1 . This will be the constant of the proposition. 2) Consider the given matrix (a11). Take a non-zero scalar A in o such that all the elements bii = Aa ii are in o. We have to find elements E, in o, not all of them 0, such that .

.

.

,

ibiii. ± ' • ' ± binnip for 1SiSn and for all p in D— S. Form the cartesian n-space V=Fx•-• x F over F and let L be the lattice L . {(a1, .. . , an)lai E o

for

1s 1

n}

on V. Define a linear transformation 9, E L.F. (V) by the equation

where

97 (a1, - • • , an) — (111'. — , i3n)

fi i= f bu m,

Then

(1 .. i

n) .

i

det 92 --- det (b i 1) — An det (a11)

and 92L S L, hence by Proposition 103:1 (L: 9 2 L) . N (det9,) . (N A)n .

3) Construct an idèle i with if p E S Nip= { 1C-1 lAlq. if P = cio if p E D — (S Li q 0) . Thus 11,4 5 C-1 IA fp for all p in D — S since 0 < C < 1. The volume of i is given by ilill = C-1 H Ai, = C-1 (NA) . IAlp

Then by Theorem 33:5 the idèle I bounds strictly more that NA field elements. Hence there are strictly more than (N A)n vectors (oci, ... , oc,) in L which satisfy kilt>

C-1 !Alp


307

for 1 i S n and for all p in D — S. But (L: 924 . (N A)". Hence at least two of these vectors, say (oh, . . . , an) and (oci, . . . , oc), are congruent modulo 92L. Put ni = ai — cci for 1 5= 1 n. Then (271, • • • , ?In)

E (FL ,

hence we can find El, . . .,

in o such that ni . Li bij el (1 .. i

n) .

i

On the other hand, — fai — ccilp

2 (icxiip ± iceiip) .. 4C -1 !Al p

for! i n and for all p in D— S. In other words, jba + • • • + bi n &lp ,.

71A1p •

as required. q. e. d. 103:3. Lemma. Let V be a regular quadratic space over the global field F, let c be a given fractional ideal. Then there is a finite subset 0 of P such that Q (L) n (1 + 0 for every lattice L on V which satisfies sL C o, DL D c. Proof. 1) We have c S DL S o since sL So. Now the number of fractional ideals bet*een c and o is finite. It therefore suffices to prove the lemma for all lattices L on V which satisfy the condition DL = c (instead of the condition OL D c). 2) First suppose that V is isotropic. All o-maximal lattices on V have

the same volume, let it be b. Consider any lattice L on V of the type under discussion in this lemma. Then L is contained in an 0-maximal lattice M since nL S sL S o. By Proposition 82:11 there is an integral ideal a such that c = a2 b and a M ç L. The ideal a obtained in this way will be the same for all lattices L under discussion since a2 = cb-1. Take a non-zero scalar a in a. Then M S a-1 /, S a-lL . It therefore suffices to prove the following: there is a non-zero field element which is represented by all 0-maximal lattices on V. What is this field element to be? Take a complete set of representatives a1 ,. . ., ak of the group of fractional ideals modulo the subgroup of principal ideals, i. e. of the ideal class group of F at S. Let fl be a non-zero scalar which belongs to all the ideals a1 ,. . . , ak and or', . . . , ail. We claim that every o-maximal lattice M on V represents 132. By Proposition 82:20 there is a splitting M =--- K ± - - • in whichFK is a hyperbolic plane. By Proposition 82:21 and its corollary there is an ideal ai for some i (1 i k), and there is a base x, y for F K with 12 (x) = Q (y) = O, B (x, y) = 1, 20*

308


such that K a i x •y (Er y /32 Then fix -F. •y y is in K and hence in M. But Q x + y) So M represents )62 as asserted. 3) Now let V be anisotropic. By Proposition 81:5 every lattice L on V can be written in the form o x2 + • • • + ox,, L= x„ is a base for V and ai is one of a finite number of where fractional ideals al, ak. We may therefore restrict ourselves to the following situation: given a fractional ideal a prove that there is a finite subset 0 of F such that Q(L) n 0 + 0 holds for every lattice L on V which satisfies L o, » L = c and which has the form L ax, + o x 2 + • • • + oxn in some base , xn for V. We can assume that a D o. Take a lattice K of the above type, write it in the form K= azi + o z2 + • • • + ozn , then fix it and fix the base z1 .. z„ for the rest of the proof. We have (ID°, sifSo, K=c. Let t be an idèle with iv = 1 for all p in S and 2ti' ft 2 y2 max I B (zi, liv I r ,

for all p in D— S, where y denotes the constant of Proposition 103:2. The idèle i bounds just a finite set of field elements by Theorem 33:4. It therefore suffices to show that every lattice L of the type under consideration represents at least one non-zero field element that is bounded by t. So consider the lattice L expressed in the form L ax l -F o x2 + • - - + oxn

with a o , sL co , oL= c Let 97: V V be the linear transformation defined by the equations 92z5 = ; for 1 S j ft. Thus 99K = L. Put 921 = f aii zi

E F) .

Now bK.c.b L, hence det(B(zi, zi)) is a unit times det (B (x1, xj)), iiuncc det (aii) is a unit. By Proposition 103:2 we can find elements in o, not all of them 0, such that E1 .. . 1 4111 ei

'•'

ain enip

y


for 1

i

309

n and for all p in Q — S. Put

z = $1 zi. + • • • + $„z„ . Then Q (n) + 0 since g) is non-singular and V is anisotropic. And z is in K, hence fin is in L, hence Q(g)z) 5L Ç o, hence IQ(cpz)l p

1 V p ES .

Now we also have

pz = ni zi where Here we have Ini l t, i y for 1 calculation then gives

1•2( 9)2.)1,,

i

n, =

n and for all p in Q — S. A direct

2 ' n y2 max IB (zi,

for all p in D — S. Therefore Q ((pz) is a non-zero scalar which is bounded by I and represented by L. q. e. d. 103 : 4. Theorem. Let V be a regular quadratic space over a global field. Then the number of proper classes of lattices on V with integral scale and given volume is finite. In particular, the number of proper classes in a genus is finite. Proof. 1) Let c be an integral ideal. We shall actually prove the following: the number of classes of lattices on V with integral scale and with volume containing c is finite. This of course gives the theorem since each class consists of either one or two proper classes. The proof is by induction on n dim V. For n — 1 the result is trivial. Assume it for n — 1 and deduce it for the given n-ary space V. In virtue of Lemma 103:3 it is enough to prove that the lattices L on V which represent a fixed non-zero scalar a and which satisfy sL Co, »L c fall into a finite number of classes. 2) Fix a vector y in V with Q (y) — a and take the splitting V = Fy ± U. By the inductive hypothesis we can find lattices K1, . . K r on U such that every lattice K on U with 5K c 0 1) K 11 2n c

is isometric to one of them. Define lattices L1 ,. , L. on V by the equations oy K i (1 r) . We claim that for each of the lattices L under consideration there is a lattice L i (1 5 i y) and a a in 0 (V) such that crL D L i . Once this has been demonstrated we shall be through for the following reason: we will have B (crL , L i) ci B (c , L) B (L, L) o, hence L Ç aL

; but the number of lattices between L i and L@ is

310


finite; hence L will be isometric to one of a finite number of lattices; hence the lattices L under consideration will fall into a finite number of classes. 3) So we must find Li and a. By Witt's Theorem we can assume that y E L. Define the sublattice K' = {cix B(x, y) ylx L) of L. Clearly K' is a lattice on U. Put L' == a„y

IC'

where ay is the coefficient of y in L. Thus ay D e since y E L. For each x in L we have ocx B (x, y) y

(ocx B (x, y)

E Li ,

hence ceL L' cL.

Therefore Cit2n C = a:CC (bK 1). But a:a C 6L S o. So bK 1 D 12211 C. Now K' has integral scale since L does. There is therefore an isometry r). Hence there of U onto U which carries K' to Ki for some i (1 j is a a. in 0 (V) such that a L' = avy Iff D Li Then ci L D

D L.

We have therefore found the desired L i and a.

q. e. d. 103:5. Remark. Suppose the global field F and the underlying set of spots S are kept fixed. Let c be a given integral ideal, let n be a given natural number. Then the number of quadratic spaces V of dimension n which can support a lattice L with integral scale and with volume c is finite (at least up to isometry). For let us take a set of non-dyadic spots T which consists of almost all spots in S, such that cp = op for all p in T, and such that hp = PE TT. Consider an n-ary quadratic space V over F and suppose there is a lattice L on V with integral scale and with = c. Put dV = a with cc in F. Then Lp is a unimodulax lattice on Vp at each p in T. Hence the Hasse symbol Sp V is 1 and the order ordp cc is even at all p in T. The information JEzz-- Pill; gives us a fl in F with 2 ordp fi = ordp a at all p in T. Hence we can write dV = s for some s in u ( T), L e. for some T-unit s. But u (T) modulo u (T) 2 is finite by Proposition 65: 6.Therefore there are just a finite number of possibilities for the discriminant dV of a quadratic space V with the given properties. Consider those quadratic spaces V which have the given properties and have fixed discriminant dV = s with s in u (T). Then Sp Vp = 1 at each p in T, hence Vp is unique up to isometry at each p in T by Theorem 63:20. Now the number of quadratic spaces of given dimension and given discriminant over a local field or over a complete archimedean field is clearly finite


311

(up to isometry). In particular this is true over the fields Fr at each p in D — T. Hence by the Hasse-Minkowski Theorem there are only a finite number of possibilities for V. § 104. The class and the spinor genus in the indefinite case 1 0 4 : 1. Lemma. L is a lattice on the quadratic space V under discussion. Suppose dimV 3. Let T be a finite subset of the underlying set of spots S. Then there is a scalar ts in o which is a unit at every spot in T and has the following property: every element of to n Q (V) which is represented by L 9 at each p in S is represented by L. Proof. By enlarging T if necessary we can suppose that S — T contains only non-dyadic spots and that Lp is unimodular at each p in S— T. Hence Q (L p) . op at each p in S— T by Corollary 92: lb. Take lattices LI, . . . , LA on V, one from each of the classes contained in genL, and let these lattices be chosen in such a way that L ip . Lp for all p in T and for 1 S i 5._ h (this is possible by Example 102:4). Using Corollary 21:2a we can find a A in o which is a unit at each spot in T and such that AL i C L for I i h. Put its -,-, A2. This will be our i.z. To prove this we consider an element a of tto which is represented by V and also by Lp at each p in S, and we must prove that a is represented by L. Now at each p in S — T we have

mitt E p .0 op — Q (Lp) And at each p in T we have 422 EQ (L p) since a is represented by Lp and A.2 is a unit at p. So cep, is represented by V and also by Lp at each p in S. Hence 422 is represented by some lattice in genL by Example 102:5, hence a/.1,2 EQ (Li) for some i (1 i 5_, h). Then a EQ(AL i) .s Q (L) . q. e. d. 1 0 4:2. Definition. Let S be a Dedekind set consisting of almost all spots on the global field F. Let V be a regular quadratic space over F. We say that S is indefinite for V if there is at least one spot p (archimedean or discrete) in D — S at which Vp is isotropic. If Vp is anisotropic at each p in D — S, then we say that S is a definite set of spots for V. 1 0 4 :3. Theorem. V is a regular quadratic space over the global field F with dimV 4, S is an indefinite set of spots for V, and T is a finite subset of S. Let a be a non-zero element of Q (V) and suppose that at each p in S there is a xi, in Vp with Q (zp) -, a such that

iizpiip _.1. YpES—T.

312


Then for each s > 0 there is a z in V with Q (z) = a such that

z

I

Vp ES—T

and

liz

zpiip<E VPE T.

Proof. 1) By scaling we can assume that a = I. We may take 0 < € 1. By step lb) we can find a in (4 (V) such that IkiJi = at all p in S (T1 u T), with a arbitrarily close to e „ at each p in T1 , and arbitrarily close to e p 99p at each p in T. Then at each p in S — (T1 u T) ,,

,,


317

we have 11'o 5- 11 Q-111p 114 p 1) hence 11 e1 ; at each p in T1 we have e-1 1 Vp11 p Ile' (0. — lip I e'ff p i — elf p and this is arbitrarily small, hence li e-1 o p 1 ; similarly we can obtain Ile-1 — iPp p < e at each p in T. Hence a is the required element. 2 a) Now the case n = 3. By scaling V we can assume that the discriminant dV is 1. All norms are determined by the base xi., x2, x3 for V. Let 4 4, x; be an orthogonal base for V, and let 11 11 pi denote norms with respect to this new base. Take a finite set of spots T' with T T'ES such that 1 lip II lip' and Q (xl) E up at each p in S —T' for 1 5_ I 3. Define rpp = 1 -pp for all p in T' —T. If we can prove the theorem for the new set T' and the new system of norms, then we can find a a in 0' (V) with 114 piioIi1 pf for all p in S— T' and 994; arbitrarily small for all p in r n particular we can make — 1v pil p < 1 for all p in T' —T and lla (pa p < e for all p in T; this means that = 1 for all p in T'— T, and hence for all p in S — T. In other words, it is enough to prove the theorem for the enlarged set of spots T' and the new system of norms. In effect this allows us to make the following assumption: the base xi., x2, x3 used in determining the norms 11 11 p is an orthogonal base for V in which V

--> which induces the identity map on V. This isometry determines an algebra isomorphism LFp (Vp)›.-0- LFp (Vg) in a natural way. The algebra isomorphism so obtained preserves isometries, rotations, symmetries, spinor norms, norms 1 li p, and localizations of elements of L7 (V). It sends 0 ( Vp) 0(n) and 0' (Vp) to 0' (V). From this it follows that the theorem holds for the given localizations Vp (p ED) if and only if it holds for some other system of localizations v,o, (p E D). Hence we can assume that each Vp is taken in the localization Cp of a quaternary quadratic space C which is defined in the following way: fix a 4-dimensional F-space C containing V, fix a vector 1a in C — V, and make C into a quadratic space over F in such a way that C

Fla V with QM. 1 .

The localization Cp of C is taken and fixed at each p in D. The space Cp is regarded as a quadratic space over F. and Vp is the subspace of Cp spanned by V. A norm 11 ll p is put on Cp with respect to the base 1a, x1, x2, x3. This induces the given norm on vp.

318


Recall that in the theory of quaternion algebras (§ 57) we started with a 4-dimensional vector space and a base, we fixed them, we put a multiplication on the vector space by means of a multiplication table, and we called the resulting object a quaternion algebra. The initial choice of vector space and base was quite arbitrary. Now do all this starting with C = Fla ± ± F x2 F and make C into the quaternion algebra ( —alF1—a2 ) in the defming base l a, x1, x2, 4Similarly regard Cr as the quaternion algebra ( —31Fp ' —a2 ) in the defining base la, xi, x2, h. Clearly the quaternion algebra Cr is the Fr-ification of the quaternion algebra C. Let bar stand for conjugation in C and in each Cr. The space C can be regarded as a quadratic space in two different ways. First we have the quadratic structure used in defining C, namely C = F1a 1 V with Q (l c) = 1. Secondly, we have the quadratic structure associated with the quaternion algebra C in the manner of § 57B. It is easily seen that these two quadratic structures are the same, namely B (x, y) lc = -y (xj", + yi)

V x, y E C

The same holds for each Cr. 2c) So much for the logical niceties. We see from the multiplication table used in defining Cr that iixYlip 5- iixiip Nip holds for any x, y in Cp at any p in S — T. And there is a positive constant r such that frYlip 5 r IlYlip for any x, y in Cr at any p in T. By Proposition 57:13 there is a vector zy in Cr at each p in T with Q (zr) = l and Tr x = x.:17,1 Apply Theorem 104:3 to the quadratic space C. We obtain a vector z in C with Q (z) = 1 such that lizii r 5_ 1 at all p in S — T and Ilz—zrli r < s' at all p in T, where e' is a positive number with < ilZplip ., el < 81112 for all p in T. By Proposition 57:13 there is an element a in 0' (V) with crx=zxz -1 VxEV.

Then for any x in V and any p in S — T we have

(1 0'4 p = Ilz x

iilp

Piip 5-


319

hence Ilaxill p 5_ 1 for 1 i 3, hence kril l, 5_ 1, hence Ilall p = 1. A similar argument (essentially an argument of continuity of multiplication) gives fc — < e at all p in T. q. e. d. 1 0 4:5. Theorem. V is a regular quadratic space over the global field F with dimV 3, S is an indefinite set of spots for V, and L is a lattice on V with respect to S. Then

= spn-FL and clsL = spnL

Proof. We shall prove clsL = spnL. The equation cls+L = spn+L is done in the same way. So we consider a lattice K in spnL and we must prove that K E cIsL. By the definition of the spinor genus we have a in 0(V) and Er in 0' (V r) at each p in S such that K=iEL Vp ES.

Then (0.-1 K) t, = L'p L I, for all p in S. But 0.-1 K E cisK. Hence we may assume that a = l y. So we now have Kt, = Ep L p Vp ES. Fix a base x„ for V, let norms lip be defined with respect to this base on VI, at each p in S, and let M be the lattice M =0 x1+ . . . Take a finite subset T of S such that K r = L p = Mr V p ES —T .

Then by the Strong Approximation Theorem for Rotations there is a rotation e in 0 (V) such that !la p = 1 for all p in S — T, with Ile — Evil p arbitrarily small for all p in T. The first condition informs us that (eL)=eM=K

p

VP EST.

The second condition informs us, in virtue of Example 101:4, that we can obtain er Li, = K t, at all p in T by taking good enough approximations in the selection of e Then L) r . K r for all p in S. Hence e L = K. Hence K E clsL. q. e. d. 1 0 4:6. Example. The purpose of this example is to show that the condition dim V 3 in Theorem 104:5 is essential. Let F be the field of rational numbers, let S be the set of all discrete spots on F, let V be a hyperbolic plane. So S is certainly indefinite for V. And o is the ring of rational integers. Take a lattice L = ox oy on V and suppose that (e

.

L

Put K

o (4 x)

0 (4 y).

10 9\ inx,y.

So

K10 k9

9\

m 321

1

— 4

x' 4y.

320


We claim that but K E spn+L . It is easily seen that L does and that K does not represent 2, hence K cls+L. Let us show that K E spn+L. In fact we shall find rpr E 0' (Vr) at each p in S sucli that K r = rppi.r. If p is non-dyadic this is trivial since then K r — LI, by definition of K and L. So consider the 2-adic spot p. Then Kr and Lr are unimodular with nKr = nLp = 2or, hence by Corollary 82:21a there is a vector w in Vr with Q (w) = 0 and K

cls+L

L r = o r x or w ,

K r = or x)

or (4w) .

By Example 55:1 we have a rotation Tr of V (prx=

x and

itself such that

Irpr w, = 4w , •

and the spinor norm of this rotation is equal to T N. Hence Tr E0' (V r). Moreover epr L r = K. Hence K E spn+L. 1 0 4 :7. Remark. The reader will be able to use Theorem 106:13 to show that the assumption of indefiniteness in Theorem 104 :5is essential. 1 0 4 :8. Remark. Theorem 104:5 tells us that we have cls+L = spn+L in the indefinite case in 3 or more variables. On the other hand, Theorem 102:9 and Example 102:10 give sufficient conditions for spn+L and genL to be equal. Hence we have sufficient conditions for determining when cls+L and genL are equal. Now the genus is completely described by the local theory. Hence we have certain sufficient conditions that can be used to describe the proper class (in general these conditions are not necessary). 104:9. Theorem. V is a regular quadratic space over the global field 3, and S is an indefinite set of spots for V which satisfies F with dim V the condition jp .1)D 11, where D = 0(04-(V)). Suppose that L is either a modular or a maximal lattice on V with respect to S. Then

cls+L = genL

Proof. We have cls+L spn+L by Theorem 104 : 5. But spn+L = genL by Theorem 102:9 and Example 102:10. Hence cls+L = genL. q. e. d. 104:10. Theorem. V is a regular quadratic space over the field of rational numbers Q with dim V 3, and S is an indefinite set of spots for V. L and K are lattices of the same norm and scale on V with respect to S. Suppose that L and K are either both modular or both maximal. Then

cls+L = ds+K

Proof. In the modular case we have genL = genK by Theorems 92:2 and 93:29. In the maximal case we note that both L and K must be

321


a-maximal where a is the common norm a = nL = nK, hence genL = genK by Theorem 91:2. So in either case we have genL = genK. And jr, =PDJ by Example 102: 10. hence by Theorem 104:9,

cls+L genL = genK = cls+K

q. e. d. 1 0 4: 11. Example. Let L be a unimodular lattice on a binary space V over the field of rational numbers, with respect to the set of all discrete spots S on Q. Suppose S is indefinite for V. (Thus L is a Z-lattice and V cr, is isotropic.) Prove from first principles that

L

I or L

1\ 0) '

and hence cls+L gen L. 104: 12 Example. Let L be a unimodular lattice with respect to Z on a quadratic space V over the field of rational numbers Q. Suppose that L has norm Z and that V co is isotropic. Use the Hilbert Reciprocity Law, Theorem 63:20, and the Hasse-Minkowski Theorem to show that

V a'

< 1>

•

.

Deduce that

L

± • ±

• •

(- 1

>•

§ 105. The indecomposable splitting of a definite lattice Consider a lattice L in a quadratic space V. We say that L is decomposable if there exist non-zero lattices K1 and K2 contained in L such that

L = KIL

K2

.

If L is not decomposable we call it indecomposable. It is clear that every lattice L is the orthogonal sum of at most n indecomposable components, where n is the rank of L. A splitting of this sort is called an indecomposable splitting of L. I 0 5 : 1. Theorem. L is a lattice on the regular quadratic space V over an algebraic number field F. Suppose that the underlying sets of spots S is definite for V and also that it contains all dyadic spots on F. Then the components L 1 , . . L. of an indecomposable splitting L = L 1 _1_ • • • I L. are unique (but for their order). Proof. 1) By scaling V we can assume that s L Ç o. We shall again need the counting number Na of § 33C. As in § 103 we put N a = N (ao) for any a in F. We have

Na= 11-T , = H PCS

O'Meara,

Introduction to quadratic forms

I OE!i,

pED—S

21

322


The assumption that sL Co implies that N (Q x) is a natural number for all non-zero x in L. . 2) We shall call a vector x in L reducible if there are non-zero vectors y and z in L with B (y , z)= 0 such that x = y + z. We call x irreducible if it is not reducible. Our purpose here is to show that every vector in L is a sum of irreducible vectors of L. First consider the sum y + z of non-zero vectors y and z of V with B (y, z) = O. Then Q (y z) = Q (y) + Q (x). Here dim V 2, so definiteness implies that the spots in D — S are either real or discrete. If p is any real spot on F, then VI, is anisotropic and so Q (y) and Q (x) are either both positive or both negative at p, hence IQ (y + z)l p > IQ ()lp. Now consider a discrete spot p in D — S. We say that [Q (y + z)l p IQ (y)p. Suppose not. Then IQ (y + z)I p < IQ (y)k, and so 0 mod Q (y) m p .

Q (y) + Q (z) Q (y

Hence by the Local Square Theorem Q (y) is a square times —Q (z). This is of course absurd since Vp is not isotropic. We have therefore proved that IQ (y + z)l p (y)l p holds at all p in D — S with strict inequality at least once. Hence N (Q (Y + 4) > N(Q(Y))

Similarly N (Q (y x)) > N (Q (x)). The proof that every vector x in L is a sum of irreducible vectors of L is now done by induction on the natural number in = N (Qx): if x is reducible write x = y + z with y and z in L and 1 5 N (Q y) in — 1, 1 N (Q m — 1.

3) We put an equivalence relation on the set of non-zero irreducible vectors of L as follows: write x y if there is a chain of irreducible vectors x

z2,

=y

(q

1)

in which B zi+i) =1= 0 for 1 i E q— 1. Let C1, C2, . . . denote the equivalence classes associated with this equivalence relation. Let K1, K2, . . . denote the sublattice of L that is generated by the vectors in 0 since B (C 1, = 0 for I j. Hence the C1, C2, . .. Then B(K i, number of equivalence classes is finite, say C1 , , C. And the sum of the lattices K1 , . . , K t is actually an orthogonal sum: IC1 J - • • ± K . Now we proved in step 2) that every vector in L is a sum of irreducible vectors of L. Hence L=

j_ • • j_ K

4) Consider x in Then x is in L , e. x is in Li j_ • • j_ L,.. But x is an irreducible element of L. Hence x falls in exactly one of the above components of L, say x E LI. It follows from the definition of that G1 c 4. Hence K1 .0 L1 . Hence each K i is contained in some Li. Since L


323

is also equal to Ki j. • • ± K i we therefore see that each L i is the orthogonal sum of all the Ki contained in it. But L i is indecomposable. q. e. d. Hence each L i is a Kj . 105:2. Example. Let V be a quadratic space over the field of rational numbers Q and suppose that V has a base x1,. . . , x4 in which V (1> ± (1> ± (1> ± (1> . Consider an underlying set of spots S consisting of all non-dyadic spots on Q and let L be the lattice L = ox1 j. • • • ± øx4 . It follows easily from the local theory that S is a definite set of spots for V. And the four vectors 1 1 -2- (x ± x2 + x3 ± x4) , -2- (x1 -± x2 — x3 + x4)

form a base for L in which L (1> ± (1> 1 (1> 1 (1> . So the assumption made in Theorem 105:1 that S contain all dyadic spots cannot be relaxed. The reader may easily verify that L also has a splitting in which L ± (2> j_ ± (1> . It is also easily verified that the assumption of definiteness in Theorem 105: 1 is essential.

§ 106. Definite unimodular lattices over the rational integers We conclude with some very special results on the class of a unimodular lattice of small dimension over the ring of rational integers Z. If the -underlying set of spots is indefinite for the quadratic space in question, then the class is equal to the genus and all is known. This is no longer true in the definite case (for instance we shall see that there is a unimodular lattice of dimension 9 whose genus contains two distinct classes). We shall confine ourselves to the definite case. The situation then is this. F is now the field of rational numbers Q, S is the set of all discrete spots on Q, and Z is the ring of integers o (S) of F at S. Lattice theory is with respect to S. As usual we use the same letter p for the prime number p and the prime spot 5 which it determines. V is a regular n-ary quadratic space over Q and it is assumed that S is a definite set of spots for V. This is equivalent to saying that the localization Voc, is either positive definite or negative definite since S consists of all discrete spots on Q. By scaling we can assume that Vœ is actually positive definite. We shall assume that there is at least one O'Meara, Introduction to quadratic forms

21*

324


unimodular lattice on V. Now the discriminant of any unimodular lattice over the ring Z is either + 1 or —1, so in the situation under discussion it has to be +1. In particular, dV = 1. Furthermore there is a unimodular lattice on the localization V„ at each discrete spot p, hence Sp V = 1 for p 3, 5, 7, . . . ; but S,,,, V = 1 since Vc„ is positive definite; hence S2 V = 1 by the Hilbert Reciprocity Law. Hence by Theorem 63:20 and the Hasse-Minkowski Theorem we have

V

± • • • ± .

We therefore assume throughout this paragraph that V has the above form. The symbol 4 will denote the n x n identity matrix. Thus we have V ,.==- In . We call a lattice D on V completely decomposable if it splits into an orthogonal sum of lattices of dimension 1. Thus in the situation under discussion the unimodular lattice D on V is completely decomposable if and only if it has the matrix 4. § 106A. Even and odd lattices

Consider a unimodular lattice L with respect to Z on the given quadratic space V over Q. Then 6L = Z and 2 Z Ç nL S Z so that nL is either Z or 2Z. We call the unimodular lattice L odd if nL Z, we call it even if nL = 2Z. Thus L is even if and only if Q(L) c 2Z. An analogous argument leading to an analogous definition can be employed for unimodular lattices over Z2 (but there is no distinction between odd and even over Z„ when 5> 2). It is easily seen that the unimodular lattice L over Z is even if and only if the localization L2 over Z2 is even. 106:1. V is a regular quadratic space with matrix In over Q. Then there is an even unimodular lattice with respect to Z on V if and only if 0 mod8. Proof. 1) In the course of the proof it will be found necessary to use the 2-adic evaluations of the Hilbert symbol, also the fact that 1, 3, 5, 7 are representatives of the four square classes of 2-adic units, and finally the fact that 5 is a 2-adic unit of quadratic defect 4 Z2. All these things were established in the statement and proof of Proposition 73:2. As in n

§ 93B, we let A (a, /3) stand for the 2-adic matrix (II 131) . 2) First suppose there is an even unimodular lattice on V. Then there is an even unimodular lattice on the localization V2, hence by the local theory (Examples 93:11 and 93:18) we must have either V2

(A (0, 0)) j - • • ± (24 (0, 0)>

V2

(A (0, 0)) ± • • • j_ (A (2, 2)) .

or


325

But d V2 = + 1 and each of the numbers —1, —3, +3 is a non-square in Q2, hence we must actually have V2 2_-'

(A (0, 0)) ± • • - j..

Introduction to Quadratic Forms (Grundlehren der mathematischen Wissenschaften 117)

Introduction to Quadratic Forms (Grundlehren der mathematischen Wissenschaften)

Introduction to Quadratic Forms (Grundlehren der mathematischen Wissenschaften)

Introduction to Modular Forms (Grundlehren der mathematischen Wissenschaften)

Introduction to Modular Forms (Grundlehren der mathematischen Wissenschaften 222)

Introduction to Modular Forms (Grundlehren der mathematischen Wissenschaften 222)

Introduction to Analytic Number Theory. (Grundlehren der mathematischen Wissenschaften 148)

Group Theory I (Grundlehren der mathematischen Wissenschaften)

Constructive Approximation (Grundlehren der mathematischen Wissenschaften)

Constructive Analysis (Grundlehren Der Mathematischen Wissenschaften)

Primzahlverteilung (Grundlehren der Mathematischen Wissenschaften 91)

Univalent Functions (Grundlehren der mathematischen Wissenschaften 259)

Stochastic Analysis (Grundlehren der mathematischen Wissenschaften)

Primzahlverteilung (Grundlehren der Mathematischen Wissenschaften 91)

Categories and Sheaves (Grundlehren der mathematischen Wissenschaften)

Differentiable Manifolds (Grundlehren Der Mathematischen Wissenschaften)

Hyperbolic Complex Spaces (Grundlehren der mathematischen Wissenschaften)

Riemann-Roch Algebra (Grundlehren der mathematischen Wissenschaften)