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 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.

Chapter I. Valuated Fields

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 and to prove that 1a1 2 < 1. Let N= {x EFI1x12 ‹ 1} .

This set N is a neighborhood of 0 under the topology induced by either valuation. Now jai l = 1a17 can be made arbitrarily small by choosing n large enough, in particular we can choose an n such that an E N. But then lall . < 1. And so 1/12‹ 1 . (2) 4 (3). By taking inverses we deduce from (2) that Jail > 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

Part One. Arithmetic Theory of Fields

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.

Chapter I. Valuated Fields

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 •

Part One. Arithmetic Theory of Fields

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

Chapter I. Valuated Fields

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

Part One. Arithmetic Theory of Fields

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

Chapter I. Valuated Fields

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

Part One. Arithmetic Theory of Fields

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

Part One. Arithmetic Theory of Fields

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
0.

15

Chapter I. Valuated Fields

2) We can draw two conclusions from the above inequality. The first is that I ni 1 holds for any rational integer n> 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

Part One. Arithmetic Theory of Fields

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)

Chapter I. Valuated Fields

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

Part One. Arithmetic Theory of Fields

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

Chapter I. Valuated Fields

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

Part One. Arithmetic Theory of Fields

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)

Chapter I. Valuated Fields

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
23

Chapter I. Valuated Fields

We go through a somewhat similar procedure to get the degree of inertia f p). Consider the composite object (-, E) comprising the residue class field of E at 91 Here E is the image of o (I3) under the bar mapping. Let F be the image of o (p) = o ((43) r F in R. The kernel of the bar mapping on o (p) ism (p). Hence (-, F) is a residue class field of F at p. We shall call the residue class field obtained in this way the residue class field of F at p that is obtained by natural restriction from the residue class field of E at 93. We introduce the following convention along the lines of the Ev/E43 convention for completions: if E (13) denotes a residue class field of E at 93, then F (93) will denote the residue class field of F at p that is obtained from E ((43) by natural restriction. We know that we have a natural ring isomorphism F (p) F (13) . In particular, a residue class field of F at p can be naturally imbedded in any residue class field of E at 93 when 93 lp; in other words there is a E (93) . natural isomorphism F (p)

13:5. (1) F(p) = Fp (p) , (2) [4(93) : F v(93)] = [E(3) : F (3)]. Proof. (1) Here it is understood that F (p) is obtained from F(p) by natural restriction. Use bar for the natural homomorphism of op onto Fp (p). Then (P) = F(P), op= Fr (P) We have to prove that o p Ç o (p). To this end consider typical a Then a lim a,, with a„ E F So there is an a EF with la — or < 1. Then a — a E mp. So a E o (p), and Ft = a . Hence ;3; Ç o (p). (2) It is understood that F93 ( (43) is obtained from E43 33) and that F (V is obtained from E (93) by natural restriction. Now it is easily seen that E (93) can be replaced by any other residue class field of E at 93 without changing the degree [E 33) : F Mil, hence we can assume that E ( (43) is obtained from E93 33) by natural restriction. So all residue class fields under discussion are now obtained from 4 ( 3) by natural restricET(q3)

E(43) F

F(43)

tion. The result follows from the above diagrams since, by the first part of this proposition, we have F (13) =Rp(43) and E ( 3) E (3). q. e. d.

24

Part One. Arithmetic Theory of Fields

We define the degree of inertia of the extension EIF at 931p to be the field degree

R93 1P) = this is either a natural number or co. The last proposition shows that it has the same value for the extension Ec43/F9? at 13 I p. 13:6. Let EIF be a finite extension of degree n with non-archimedean spots 93 p. Then e(93 1P)/(93 1P) n Proof. Fix 1 1 ET. Let (-, E) be a residue class field of E at q3 and let (-, F) be obtained from it by natural restriction. Put e= (1E1: 1F1) , f= [E:F] .

n. We must prove ef c7),,. of E which are independent over F. Consider r elements Choose representatives co l, ., co,. of these elements in o (93). Then all these representatives must be in u (13). Consider a non-zero element A = E aA coa with the ez2 in F and suppose that lad = max 14- Then IAI =

lad

• C01+ --i eza CO2+

• • • ± —a tz .LWrI•

Now we cannot have col

a 032 + + _l

corl <

li

for this would lead to a dependence relation among the (Ws over Hence 1A1 --= 1a11. In other words, we have lai coi + • • • + a r co,.1 = max 1c/A 1

F.

E IF!

Now consider representatives .. E iEi taken from s different cosets of 1E1 modulo 1F1. Here the ni are chosen in E. We claim that the rs elements {owei } are independent over F. Suppose we have a relation

E f a2o co2 ni,. O,

a20 EF F.

0 with Ao = E ctAo coi. We just saw that 141 E 1F1 for any such A p+ O. Therefore, since the Igrol are in different cosets of 1E1 modulo IF1, no two lAigro l can be equal unless they are O. Applying the Principle of Domination to the equation EA,= 0 Write this in the form E

then shows that all A i, must be O. Hence

A

a,to coA . 0 forall p.

But max iczo i = E a4c0/1 • Hence all alp = O. Thus the {coA no} are independent, as asserted.

Chapter I. Valuated Fields

25

This proves that both e and f must be finite. Hence we can take s = e and r = f. This proves the proposition. q. e. d.

§ 13C. Transcendental prolongation of a valuation There are two points worth remembering when constructing a valuation on, an abstract field. First, if a function ! ! satisfies the first two valuation axioms (V1) and (V2) of § 11 A, and if Il+fX15- 1, then must satisfy (Vs') and it is therefore a non-archimedean valuation. Secondly, consider a field F which is the quotient field of one of its R be given and suppose subrings Z. Let a real valued function ! 1: Z that it satisfies the valuation axioms (V1), (V2), (V31) of § 11A for elements of Z. Then there is a unique prolongation of this function to a non-archimedean valuation on F. This prolongation is defined as follows: take a typical cc E F, write it oc= alb with a, b E Z, and define loci = lal/Ibi. We leave it to the reader to verify that the new function is well-defined, that it is a prolongation of the original function, and that it satisfies the valuation axioms (V1 ), (V2), (V r) It is always possible to prolongate a valuation from a non-archimedean field F to the field of rational functions F (x), i. e. to the simple transcendental extension F (x) of F. Here is one important way of doing it. For a typical polynomial f (x) = aoxn + a1xn-1 + ' • + an E F [x] define If (x)! = If! = max !eh! .

IocI 5 1

We start by verifying the valuation axioms for the ring F [x]. To do this we must consider some other typical polynomial g(x) = b O x'n b i x'in -1 + • • • + b„,EF [x] . Axioms (V1) and (V3 ) are immediate. So is the inequality If .g1 ifI.Igi. In order to establish equality here we consider the last coefficient ai for which if! !ad, and the last bi for which 1g! = !M. It follows from the Principle of Domination that the coefficient of 0 +5 in /(x) g (x) has value Hence if • gi = VI • igi as asserted. Hence the valuation 1 6051 iti axioms hold in F [x]. But F (x) is the quotient field of F [x]; so there is a unique prolongation of ! ! from F [x] to F (x) given by !fig! =Ifl/Igl. This is the valuation we are after. We shall call a polynomial with coefficients in the non-archimedean field F an integral polynomial if all its coefficients are integers in F (under the given valuation). Thus Ifl 5 1 f integral where I I is the preceding prolongation of the valuation on F to F (x).

26

Part One. Arithmetic Theory of Fields

13 : 7. The monic irreducible factors of a manic polynomial with integral coefficients in a non-architnedean field also have integral coefficients. Proof. Let JI be a valuation determining the given non-archimedean prime spot on F, and prolongate it to the valuation I I on F (x). Consider the integral polynomial / = f (x) EF [x] and let = PiP2 • • • Pr, pi EF [x] ,

be its factorization into irreducible monic factors. Then all i. And III 1 1P1 1 1 P21 • • IPrl Hence all IN = 1. Hence all p, are integral,

IN

1 for

q. e. d.

§13D. Complete refinement of an approximation.

In certain problems it is important to have systematic methods for refining approximate equations to precise equality. The first time one encounters this is usually with Newton's method in an elementary calculus course. In fact we have already used this procedure to show that 1 + a is a square for sufficiently small a in § 12. Newton's method for roots of polynomials can be carried over to complete non-archimedean fields, but we shall not go into that here. Instead we give an analogous procedure that concerns itself with the factors, not the roots, of polynomials. Refinements of approximations will turn up again in Chapter VI in connection with the arithmetic theory of quadratic forms. 13:8. Hensel's lemma. Let F be a complete non-archimedeatz field and consider the polynomial f (x) with integral coefficients. Suppose that 1(x), when read in the residue class field, factors into two relatively prime polynomials one of which has degree i> O. Then f (x) has a factor of degree i in F [x]. Proof. 1) Let p be the underlying prime spot. Fix a valuation I I F) denote the residue class field of F at p. For any integralLet polynomial g(x) let g(x), or simply g, denote the polynomial obtained by reading g (x) in F, i. e. by barring all coefficients of g (x).

The given factors of j(x) can be obtained from integral polynomials over F; in fact we can write f(x) = 9-51(x) I/71(x) where gh (x) and th (x) are integral polynomials over F with deg 99i = deg (751 . 1, deg vi= deg i, with the leading term of ch a unit, and with deg wi + deg vi .5 deg f . We then bave f= (71 lyi = 991 /pi and, since rpi and /T), are relatively prime, there are integral polynomials A, B such that A q5i -F B1251 . 1. If we

27

Chapter I. Valuated Fields

now read this last sentence back in F we get If —

<1 , 1A 991 -1- B

11 < 1 ,

where the valuation in question is the valuation on F(x) that was introduced in § 11C. We therefore have a fi eld element a such that

If - 921 v1I 5IA 921 +

< 1

(1

vh— 11 .5 In1 < 1.

)

2) So far we have just interpreted the given conditions as an approximation 9)1 tp to f. The idea of the proof is to keep finding polynomials 99 and ip which give better and better approximations to f, and then ta obtain a precise factorization by taking the limits of the approximating polynomials. The completeness of the field ensures that the limits in question do indeed exist. With this in mind let us consider integral polynomials 9) and v such that W VI 5- IV (n E N) (2) — SP11 Ini , IV — Inl deg 99 = , deg 9) deg ip deg / We show how to sharpen 9) and p, i. e. we find integral polynomials w' and vi such that If — Inln+1 (3) IV — 9'11 5- 12zI IV' — Pi deg 9? -= 1, deg gl + deg V.S. deg f. And the sharpening will be done in such a way that 6491 -5- Inln, Iv/ VI 5- lain

(4)

By equations (1) and (2) we can find integral polynomials C and D with Afp-f-Bv=1-FaC, f=991p-FarlD.

Multiplying the first of these by anD and substituting in the second gives

v + n"DA

anDB v — zn-F1DC

(5) By equation (2) the leading coefficient of 9) must be a unit. Therefore if we divide w into any integral polynomial we obtain an integral quotient and remainder. Hence we can find integral polynomials )3, a, y, T such that DB-1392+a, DC— yrp+T , where deg a < deg go and deg T < deg g). Then a computation gives f=g9v±genEcp+znav—alt -ElT an-laE)}

(6)

Part One. Arithmetic Theory of Fields

28

where E = DA + fi v ny. If we now put w'= (p nna and tp' =v nnE we have integral polynomials V, v' which satisfy equation (4) and the first part of equation (3). The second part of equation (3) is immediate. Now deg V= deg(?) since dega < deg T. By inspecting every term in the sum in equation (6) except nn.E g) we see that this exceptional term must itself have degree .5 degf. Hence the last part of equation (3) is proved. 3) In step 2) we have given the refinement procedure to be used at each stage of the approximation. We must now apply it all the way. Starting with qh., th we successively improve our estimates and we obtain two sequences of integral polynomials (9)„} and {TN} by the preceding process. These polynomials will satisfy the following inequalities: If — rn Vni 199n — (pu i ini deg 92,,-= 1, deg 94,-F deg v„ 5 degf,, kon+i — 927&15 k", Hence the two sequences are Cauchy sequences of polynomials; and their degrees are bounded; from which it readily follows, in virtue of the completeness of F, that both sequences converge to polynomials in F [x]. We define

0 = lim çon , We have deg D i since deg p,,= i for all n. On the other hand I 99i — 92„1 < 1 and, for large n, I 92„ — 01 < 1. Hence 19h— 01 < 1. But the leading coefficient of Ti is a unit. So we get deg 0 deg 99/ . i. Finally = ilim(f — rn Vn)1 = lim it Pn Vni = 0 • I/ q. e. d. So f = 0 P. We have the desired factorization. 13 : 9. Reducibility Criterion. The polynomial f (x) = ao + a1 xn -1 -1- • - - -1- a„ over a field with a complete non-archimedean valuation is reducible if 0 < ia o l < [ai l holds for some i < n. Proof. By scaling f we can assume that VI = 1. We choose the smallest i for which fai l = = 1. So 0 < I < n. Put 99(x) = ax" '+ • • • + an and let v (x) be the polynomial identically 1. In the residue class field we have f = fp with deg 475 = n — i. But F/5 and IT are relatively prime. Hence f has a factor of degree n i. So f is reducible. q. e. d.

§ 14. Prolongation of a complete valuation to a finite extension 14 : 1. Theorem. Let E a complete valuation

1.

be a finite extension of degree n of a field F with Then there is a unique"' prolongation' of the valua-

Chapter I. Valuated Fields

29

ation from F to E and it is defined by the formula

= W imp cellin for all oc E E. The prolongation is complete.

Proof. Let us use N for the norm NE/F. We know from Theorem 11:18 that the prolongation, if it exists, is unique and complete. Furthermore the function bd = War", if it is a valuation at all, will be a prolongation of the original valuation since N oc == cc" for all oc E F. It remains for us to prove that this new function is indeed a valuation. First the archimedean case. We can assume that [E:Fj> 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].

Part One. Arithmetic Theory of Fields

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

Chapter I. Valuated Fields

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

Part One. Arithmetic Theory of Fields

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

Chapter I. Valuated Fields

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*

Part One. Arithmetic Theory of Fields

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.

Chapter I. Valuated Fields

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

Part One. Arithmetic Theory of Fields

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

Part One. Arithmetic Theory of Fields

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

Part One. Arithmetic Theory of Fields

§ 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

Part One. Arithmetic Theory of Fields

Strong Approximation Theorem to find b E o (S) such that Jb—or9
E T.

But then Iblp = alp for all p E T. Hence ab E o (S). Then a = (ab)lb expresses oc as a quotient of two elements in 0 (S), so o (S) has F as a q. e. d. quotient field. 21:4. Let S be a Dedekind set of spots on F. Then (p) is the closure of o(S) in F at each p E S. In particular o(S) contains a representative set of the residue class field at p.

Proof. o(p) is closed since the mapping oc -->— 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

Chapter II. Dedekind Theory of Ideals

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

Part One. Arithmetic Theory of Fields

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

Chapter II. Dedekind Theory of Ideals

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

Part One. Arithmetic Theory of Fields

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

Part One. Arithmetic Theory of Fields

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

Chapter II. Dedekind Theory of Ideals

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

Part One. Arithmetic Theory of Fields

§ 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

Part One. Arithmetic Theory of Fields

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
I al = A

—

(deg

I

—

deg g)

where A is a real constant with 0 < A < 1 and deg stands for the degree of a polynomial. This definition of the spot 00 is again independent of the choice of 2. Note that 00 is archimedean in the rational number case, and non-archimcdean in the rational function case.

56

Part One. Arithmetic Theory of Fields

We shall prove in a moment that every non-trivial spot on a rational global field is either a p-adic spot for some prime p, or else the spot oc; the first type is called a finite spot, the second infinite. But note that in the function field case a spot may be regarded as infinite under one choice of x, but finite under another; for instance the infinite spot under x is a finite spot under y . 1/x. Whenever we refer to a spot p without further qualification it can be either finite or infinite. 31:1. The finite p-adic spots and the spot on are all the non-trivial spots on the rational global field Q. All these spots are different. Proof. 1) If p and q are distinct primes in Q then p Ern (p) and p E u (q) so that the spots p and q are distinct. Can the spot p be equal to the spot oc ? This is clearly impossible in the number theoretic case since oc is archimedean and p is not. It is also impossible in the function theoretic case since x f o (00) and x Eo (p). This proves the second part of the

proposition. 2) Now let a non-trivial spot p be given on Q. We want to prove that p is one of the spots under consideration. If p is archimedean we are in the number theoretic case and so p .---- oo by Example 12:13. Hence assume that p is non-archimedean. Pick a valuation I 1 in p. If Q is a rational function field with x ( o (p), then for any 1(x) in k [x] we have Ifix)1 ---- ixi degf

by the Principle of Domination since the valuation is trivial on the finite field k, hence f(z)

I

, g (x) I —

I1

deg g— deg

/

17

with 111x1 < 1, so p= 00. We are therefore left with the following possibility: p is non-archimedean and Q is either a rational number field or a rational function 1. In either event we have Z S o (p). This, together field in which Ix! with the fact that p is non-trivial, implies that 0 Cm (p) n ZC Z. Hence m (p) n Z =

pz

for some p E Z which is neither 0 nor a unit of the ring Z. We claim that p can be chosen a prime: for if we have a factorization p = rs with r, s E Z then irsI <1 implies that In <1 1 say; hence r Em(p) n Z . pZ; hence there is an r1 E Z such that 1 = rl s; in other words s is a unit of the ring Z, so p is a prime times a unit. We can therefore assume that p is a prime. A typical a E Q can be written a ---- piulv with u, y in Z — pZ. Then lui = Iv! = 1, hence locI .-- ipli with !pi < 1 , hence 1 1 is in the spot p, hence p . p.

q. e. d.

Chapter III. Fields of Number Theory

57

31:2. The set of all finite spots Son a rational global field Q is Dedekind.

And Z = o (S). Proof. The Dedekind axioms (D1) and (D2) of § 21 are immediate. Let us verify -(D3). Consider the spots determined by two distinct primes p and q of Z. For each r E N write 1 . pr with , 77 EZ , and put cer=

Then lad,

or , i — ocr= ?W.

1 for all primes 1 since ocr E Z. And

iccrip 5- 1P1; , 1 1 — Gerle

lqlrq -

If r is large, flir; and iql'are small. Hence (1)3) holds. Clearly Z S o (S). Suppose a EQ—Z. Then there is a prime p such that oc = p-iu/y with i E N and both u and y prime to p. But then q. e. d. 104= I piii › 1. So a f o (S). Hence Z . o (S). 31:3. Example. Let Q be a rational function field. Then the only polynomials that are integral at 00 are constants. Hence k — n o (p) , p

the intersection being taken over all spots on Q. Thus the field of constants k is independent of x, as we asserted earlier. Incidentally this shows that the set of all spots on Q, even though it satisfies (D1) and (D2), is not Dedekind.

31:4. Remark. Consider the finite spot p determined by the prime p of the rational global field Q. A representative set of the residue class field at fi can be found in Z by Proposition 21:4; this representative set can then be refined to the following: 1. the elements 0, 1, . . . , p— 1 of Q when Q is a rational number field, 2. the set of all polynomials over k (include 0) of degree < degp when Q is a rational function field. The number of elements in the residue class field of Q at p is therefore equal to 1.p in the first case, 2. gcleg 25 in the second, where q is the order of k. In the function theoretic case one easily shows that Q has k as a representative set of the residue class fi eld at 00. 31:5. Example. Consider the field of rational numbers Q and a prime number p in Q. The prime number p determines the fi-adic spot p on Q. Take a completion Q, of Q at p and fix it. As usual let p stand for the prime spot associated with Q,. The field Q, provided with its spot p is called the field of p-adic numbers. The ring of integers of Q, is called the ring of p-adic integers and is written Z,. The spot p on Q or Q, is

58

Part One. Arithmetic Theory of Fields

discrete with prime element p. And 0, 1, . , p 1 is a representative set of the residue class field of Q or Q, at p. Hence by Proposition 16:3 every p-adic number has the form

cypv (0

c„.< p, c,„ + 0)

where the sum is taken in the p-adic topology. 31:6. Definition. An algebraic number field, or just a number field, is a finite extension of a rational number field. 31:7. Definition. An algebraic function field, or just a function field, is a finite extension of a rational function field. 31:8. Definition. A global field is a finite extension of a rational global field. In other words, the global fields are the algebraic number fields and the algebraic function fields. 31:9. Every global field is a finite separable extension of a rational global field. Proof. 1) Let F be the given global field. By definition F contains a rational global field Q such that FIQ is finite. In the number theoretic case all extensions are separable. Hence we can assume that Q is a rational function field Q k(x). We shall show that for a suitable choice of x the extension FIQ is separable'. The method of finding the new x will be the following: keep taking p-th roots of x in F until x no longer has a p-th root in F. Here p denotes the characteristic of the field F. 2) The next step is in fact a lemma: prove that every function field F satisfies [F :Fvv] = pv for any natural number y, where FPv is the field consisting of the p'-th powers of the elements of F. First consider the rational function field Q k (x). Then Qv is the rational function field QP =-- k (xv) since kv= k. And clearly [Q:1221 .‹ p. Now any relation of the form

to(xl + xh(xl + • • • + = with all f. (xv) E k [xv] would make x algebraic over k, hence xv -- ' are independent over QP, hence [Q :Qv] = p. From this we can deduce that [F:Fv] p: let w be the isomorphism of F onto Fv that is defined by we = 4.9; then 92(Q) = Qv; hence [ F : Q] = [Fv:Q 11; but [F:Fv] [FP:QP] [F:(21 [Q :Qv] ; hence

[F:Fv] = [Q:Q 21 = p as asserted. F is then said to be separably generated over k. See O. ZARISKI and P. SAMUEL, Commutative algebra (Princeton, 1958), p. 105, for the corresponding result in an algebraic function field in several variables over a perfect ground field.

Chapter III. Fields of Number Theory

59

Finally we observe that FP is a finite extension of the rational function field QP, hence it is an algebraic function field, hence [FP:FP'] =fi; etc.; hence [17 :F9] pv.. This proves the lemma. 3) x is clearly not a p- th power in k(x). If x is in FP, replace x by its p-th root, thereby obtaining a larger Q. Repeat. Ultimately we can assume that x F. We claim that (x) is separable for this choice of x. Suppose not : let Fo be the separable part of the extension FIQ; thus QCF0 cF; now LF:Fo] = fie for some e 1; but we know from galois theory that the fie-th power of every element of F is in Fo , i. c. FPe c F0. hence FPe = Fo by step 2); hence x FPe Ç FP, and this is a contradiction. q. e. d. So FIQ is indeed separable.

§ 32. Local fields § 32A. The normalized valuation Let p be any non-archimedean spot on a field F. We define Np to be the number of elements in the residue class field of F at p. Suppose that p is discrete and Np < 00 . Under these assumptions it is possible to canonically select a valuation at the spot p. (The fact that such 'a canonical choice is possible is an important part of the development of algebraic number theory; in particular it makes possible the Product Formula of § 33B.) How is the canonical choice to be made ? Each valuation in p is completely determined by the generator of its value group ; we choose that valuation I h, in p under which Np becomes a generator of the value group IFI p ; iii other words, I Ip is defined by the equation Pad (

2-Np

lordp cc

Voc(F.

The valuation defined in this way is called the normalized valuation at p. From now on, whenever the spot p in question is discrete with finite residue class field, we let I stand for the normalized valuation at p. If p is a spot having the above properties, then so is the spot p on the completion Fp . And Np has the same value on the completion as on F. In particular, the normalized valuation on Fr induces the normalized valuation on F at p. We let! I t, stand for both. 32:1. Definition. A local field is a composite object consisting of a spot p on a field F such that (I) p is complete and discrete, (2) the residue class field at p is finite. 32:2. Example. The field of p-adic numbers Q is a local field.

60

Part One. Arithmetic Theory of Fields

32:3. Let EIF be a finite extension of fields with spots 931p. Suppose F is a local field at p. Then (1) E is a local field at 93 (2) N93 (Np)/ (931P) Va EE (both valuations normalized). (3) 112193 --- INE/Y. Proof. E (93)IF (93) is a finite extension of degree f (931p), and F ( (43) contains Np elements, hence E (93) contains N93 . (Np)/ (9311') elements. In particular E ((43) is finite. But E is complete by Theorem 11: 18, and it is discrete by Proposition 16:1, hence it is a local field at 93. We are left with the third part of the proposition. Before proving this let us first consider a typical oc in F; then the equation e ((431p) f ((431p) n = [E:F] of Theorem 16:4 gives us I )ord93 a j' _1 tt (TIP) e ("i31p) ordp a 2717) Nti:5 1°4 1°4 •

y

Then by Theorem 14:1 we have for any oc E E 1 04— 1 NEIF I

ii'TEIF

P

q. e. d. 32 : 4. The completion of a global field at any one of its non-trivial nonarchimedean spots is a local field. Proof. Let F be the field, p the spot. Take a rational global field Q c F over which F is finite. Form the completions Fp and Qp c Fp . Then Q is discrete with finite residue class field at the spot induced by p by Remark 31:4, hence Qp is a local field. But F1,/Q1, is finite since FIQ is. Hence Fp is a local field by Proposition 32:3. q. e. d. 32:5. Example. Let F bé a local field with maximal ideal p. Then by § 22E every integral o-ideal is of the form pv for some y E N. Use the results of § 22D to show that the group index (o: p') is (N p)'. § 32B. Ramified and unramified extensions of local fields' We consider a finite extension EIF of local fields at the spots 931p. We shorten the notation and put

e = e((43 1p) ,

f((431p) ,

n = [E:fi] .

We know from Theorem 16:4 that

et = n. We shall say that the extension EIF (of local fields at (431p) is unramified if e=1,f=n. If EIF is not unramified we call it ramified. We call EIF fully ramified if e=n,=1. For a general theory of ramification see, for instance, O. F. G. SCHILLING, The theory of valuations (New York, 1950)..

Chapter HI. Fields of Number Theory

61

Thus a fully ramified extension is ramified when n> 1. Let us make the following observations: given finite extensions F SE SI/ of local fields, then HIF is unramified •=. HIE and E1F are„ HIF is fully ramified 4v:. HIE and EIF are;

these facts follow easily from the multiplicative behavior of e, f, n in a tower of fields. 32:6. Let EIF be a finite extension of local fields. Suppose that E = F (oc) for some oc which satisfies a manic integral polynomial 1(x) over F with 1' (oc) a unit. Then cc is integral. And EIF is unramified. And the residue class

E is equal to F (). Proof. 1) The Principle of Domination says that oc must be an integer of the local field E. Let g(x) = irr (x, oc, F). Write / (x) = g(x) h (x) with g (x) and h (x) monic ; then g (x) and h (x) have integral coefficients in F by Proposition 13:7; and the formula for differentiation gives f' (a) g' (cc) h (oc) ; so g' (oc) is a unit. We have therefore shown that the given polynomial 1(x) can be taken irreducible over F. 2) Next we claim that f(x) is irreducible over F. Suppose not. Let ço (x) = irr (x, Since J(&) = 0 there is a polynomial y (x) over F such that J(x) = (x) y (x) and field

r).

1< deg q) < deg/

deg/ .

Now f' ( a) + 0 since f (oc) is a unit of E; hence El is not a multiple root of f(x); hence cp (x) and y (x) are relatively prime. Then Hensel's lemma implies that 1(x) has a factor of degree equal to deg T.. But this is impossible since /(x) is irreducible. Hence J(z) is indeed irreducible over F. 3) We have ire E E since oc is an integer of E. But jr(&) ---- O. Hence

[E: But [E:F] nle

[F() :F] = deg/. deg/ = n. n. Hence [E:F]

n,

e

1.

So E1F is unramified. And also by comparing degrees in the above q. e. d. inequalities we obtain E F(). 32:6a. Suppose m is a unit in F. Then the local field F() obtained by adjoining an m-th root of unity to F is unramified over F.

32:7. Lemma. Let C be a cyclic subgroup of order m of F with m relatively prime to Np. Then distinct elements of C tall in distinct cosets of (p) modulo m (p).

Part One. Arithmetic Theory of Fields

62

Proof. Let C generate C. Then C consists precisely of all m-th roots of unity 1, C, .. . , Cm -4 in F. Hence m-1 xm — 1 = H (x — Ci) . o

Computing derivatives at x ffici(n,-1)

gives =

H (ci _ 41)

But the left hand side is a unit since m is prime to Np, j. e. since m is not 0 in the residue class field; and every term on the right is an integer; hence every term Ci— 5 is a unit. Thus V and V fall in different cosets q. e. d. whenever i +j. 32 : 8. Let F be a local field at p. Then the group C of (Np V& roots of unity in F is cyclic of order Np —1. So F contains all (Np — 1)th roots of unity. It contains no other roots of unity of period prime to Np. Proof. Consider any extension F (C)IF where C is a root of the polynomial xNP -4 — 1. By Proposition 32:6 we obtain [F (C):F] = [F But Np = ; and being a finite field of Np elements, consists of all solutions of x'= z; hence EP; hence [F() :F] = 1 and C is in F. So F contains all roots of the polynomial XNP -1 - 1, and these roots are Np — 1 in number since the polynomial in question has distinct roots over F. Hence C has order Np — 1. Since C is a finite subgroup of F it is cyclic'. The second past of the proposition is now obvious. Let us do the third. Consider a root of unity n EF of period m where m is prime to Np. Then Lemma 32:7 tells us that the elements 1,•, nm --1 of the cyclic group generated by n fall in distinct cosets of o(p) modulo m (p). Hence fi has period m in the residue class field of F at p. But there are Np — 1 elements in the multiplicative group of F. So m divides Np — 1. q. e. d. Hence n is an (Np — 1)t 1 root of unity, 32:8a. The (Np — 1)" roots of unity plus 0 form a representative set in F of the residue class field F (p). Proof. The elements of the set 0 j C fall in different cosets of o (p) modulo m(p) by Lemma 32:7. But the number of elements is equal to the number of cosets. q. e. d. 32:9. Let EIF be a finite extension of local fields. Suppose that E is a splitting field of

r,

r

over F. Then EIF is unrarnified of degree n. It is a known result of galois theory, easily deduced from the Fundamental Theorem of Abelian Groups and the fact that xn--= 1 has at most n solutions in a field, that every finite subgroup of the multiplicative group of an arbitrary field is cyclic. 1

Chapter III. Fields of Number Theory

63

Proof. E contains all ((Np)n— 1)8 h roots of unity and these are (N p)n — 1 in number since the polynomial x(NP)n— x has distinct roots over F. All these roots have period prime to N p, and hence to N93. But by Proposition 32:8 there are N93 — 1 elements in E which have — 1 N93 — 1. Hence (N p)" period prime to N93. Hence (N N93.

Now every finite subgroup of F is cyclic. Hence E F (C) with 1,-(N On. C. Proposition 32:6 tells us that EIF is unramified with residue class field E = F (C). Here C N PYZ C, hence every a in E satisfies the equation a(N Pr= a. So N93 (Np)n. Hence (Np)n= N93. This gives f (93 I p) = [E :F1 and proves that EIF is unramified of degree n.

q. e. d. 32: 10. Let E1F be a finite extension of local fields. Suppose the extension is unramified of degree n. Then E is a splitting field over F of the polynomial

x(NPr — x .

Proof. N93 . (Np)n. So E contains all ((Np)"— i)th roots of unity. Hence it contains a splitting field H of the given polynomial. But [H:F] =n by Proposition 32:9. So H = E. q. e. d. 32: 11. Example. Consider a finite extension Ell' of local fields. If Km is a subfield of E such that Km1F is unramified of degree tn, then Km is the splitting field in E of x over F. Hence Km is unique. Consider another such field K,. and suppose r divides m. Then every root of x(NP)' — x = 0

is also a root of AN Pr = x(N . . .

x.

Hence K,.0 Km whenever r divides m. 32: 12. Any unramified extension EIF of local fields at 93 J p dis galois with cyclic galois group G (E1F). There is a unique a E G (EIF) such that ax= xNP mod m (T) V x E o (93) -

This o. generates G (E1F). Proof. EIF is galois since E is a splitting field of the separable polynomial xN 13 -1 - 1 over F. Let E.) be a residue class field of E at 93. Consider a typical e E G (E1F). Then e is topological by Corollary 11: 18a, hence e sends o() onto o(3) and m() onto m(93), hence it induces an automorphism on E that satisfies v x E 0 (93) •

Part One. Arithmetic Theory of Fields

64

Clearly

g is in G (E/F). Hence the rule e

provides a mapping

G (EIF) -4- 6 (E/F) . This mapping is clearly a group homomorphism. If = -1, then

e x x Em (93) vx

Eo (93) ,

hence e c C for all (N93 - 1)th roots of unity C in E by Corollary 32:8a, hence e 1. In other words, the above mapping is injective. But G (EIF) and G (E/F) have the same number of elements since the given extension is unramified. Hence the above mapping is bijective. Now the galois group of an extension of finite fields is cyclic. Hence (EIF) is cyclic. In fact G (R/F) is generated by an automorphism a o such that 0.0x0=

So if ci ls chosen in G (EIF) with a ax

x0EB ao, we have

xNP mod m (93) Y x E o (T) •

Hence a is the generator of G (E1F) with the required property. The uniqueness of a follows from the injectivity of the map e q. e. d. 32:12a. crC = CNp for every (N93 - V' root of unity in E. Proof. Apply Corollary 32:8a. q. e. d. 32:13. Definition. The automorphism a of Proposition 32:12 is called the Frobenius automorphism. 32:14. Let DIF be a finite extension of local fields with subextensions EIF and H1F. Then EIF unransified = EHIH unramified.

Proof. Write E F(C) with C satisfying the equation x(NP)"- x= 0; this is possible by Proposition 32:10. But then EH = H (c) ; and H (C)IH q. e. d. is unramified by Corollary 32:6a. 32 : 15. Let EIF be an extension of local fields with spots 931p. Suppose E = F(a) with cc a root of the so-called Eisenstein polynomial xr+ a1 xr -1 + - • - + ar which has all eh Em (p) and a, a prime element at p. Then EIF is fully ramified of degree r, and ac is a prime element at 93.

Proof. Pick a valuation J I E 91 Then a satisfies ar+ a' -1 + • • • + ar = 0

with all Ictif lad < 1. Hence loci < 1 by the Principle of Domination. But then Icti oer - il < lad for i 1, 2, . . . , r - 1. Hence loci = lad, again by the Principle of Domination. So loci =1/14 But a, is a prime element

Chapter III. Fields of Number Theory

65

at p. Hence e r. However n r since a satisfies an equation of degree r over F. Hence n r e n. Hence r e = n and the given extension is fully ramified of degree r with a a prime element at 93. q. e. d. § 33. Global fields § 33A. The normalized valuations

Consider a global field F. Let Q =QF stand for the set of all nontrivial spots on F. If p is a non-archimedean spot in D, then p is discrete, and Ft, is a local field, and a normalized valuation I i t, has been defined on F and Ft, (see § 32A). What about the remaining spots in Q? If F is an algebraic function field there are no remaining spots. If F is an algebraic number field there will be a finite number of archimedean spots in Q, some of them real and some of them complex; for each of these a normalized valuation J i t, has been defined in § 12. We make the rule for this entire paragraph on Global Fields, that It, will always denote the normalized valuation (discrete, real, or complex) on the global field F (or on the completion Fp ) at the spot p E D. It is easily seen that a topological isomorphism of one local field onto another is analytic with respect to the normalized valuations. The same applies to complete archimedean fields. In particular, the topological isomorphism between two completions of a global field at the same spot is analytic with respect to the normalized valuations on these completions. Suppose we have to consider a finite separable extension E1F of global fields and let 93 J p be two spots on E1F. Then j l w denotes the normalized valuation on Ei3 at 93, and J i p the normalized valuation on Fp at p. We also let i p denote the normalized valuation on F94-3 at p. We claim that ice193 .-. 'Nut, alp Vcx E Eçp where Np p : Fp is the norm function of § 15B. To prove this in the discrete case we apply Proposition 32:3 and obtain 1 04193= INET/F93 (x1P= 'NAP alP . If p and 93 are both complex or both real the result is trivial; if p is real and 93 complex we have lock3= 1 041 2 = 1NE93/FV

= INVPŒIP

where J J denotes the ordinary absolute value on O'Meara, Introduction to quadratic forms

ET

and F. 5

Part One. Arithmetic Theory of Fields

66

§ 33B. Product formula 33:1. Theorem. Let a be a non-zero element of the global field F. Then lafr . 1

for almost all p ED ,

and

p cp

Proof. 1) First suppose that F is a rational number field Q. Write oe

where p,, . . . , p,. are distinct prime numbers in Q. The non-archimedean spots are the fi-adic spots for all prime numbers fi; now N fi p by Remark 31:4; and ordin a = yi ; hence Pi

1 Nvi k-) pi •

, pr . The Similarly 1(4= 1 for all prime numbers other than p1, only remaining spot is the one determined by the ordinary absolute value which in this case is the normalized valuation' l op at oo. This gives Finally, loclœ= Pli• • • 1 vi PH EQ i ccir =. kij

v

r pi.) (Pyl ..P1=

• • (--\ r

This completes the proof of the first case. 2) A similar proof works for a rational function field Q. Write Q = k (x) by choosing some suitable transcendental x over the constant field k. Write where e Ek and the pi are irreducible monic polynomials over k. The finite spots are determined by the prime polynomials p; now here N fi = qdegP where q is the number of elements of k, again by Remark 31:4; hence /1 ‘rviegp, Similarly al,. 1 for all other prime polynomials. At the infinite spot we have N oo q, and so \ —(dega) k100=

where

q)

deg a = v . degpi + • • • + y . degpr . ,

Hence

H 1 04= 1 pED

67

Chapter III. Fields of Number Theory

3) Finally the general case of any global field F. By Proposition 31:9 there is a rational global field Q over which F is finite and separable. Let S be the set of all finite spots on Q, oo the single infinite spot on Q. Let T, T' be sets of spots on F with TIIS , T'lloo

Then S is Dedekind by Proposition 31:2; so T is Dedekind by Theorem 23:1 since FIQ is finite and separable; hence lai r = 1 for almost all p in T by the definition of a Dedekind set of spots. But T' is a finite set of spots since FIQ is finite and separable; and Qp= TiT'. Hence 14 == 1 for almost all p E Q. The product formula has already been established for Q; we shall use this fact to establish it for F. Consider a typical a E P. Each p E IJp induces a spot p E QQ ; then by § 33A and Theorem 15:3 we obtain

( 11 1Np li,a1 2,) = 17 (1.1 vivoz12,) =. 1

11 ( .17 1041= P

I

\PIP

1)

\PIP

P

q. e. d.

§ 33C. Comment on ideal theory Suppose S is a Dedekind set of spots on the global field F. Let us follow the convention of § 22 and write o for o (S) and p for the prime ideal p = o (S) r m(p) determined by the spot p. Every ideal a El (S) can

be expressed uniquely in the form a= 11 prP. pCS

We define Na =

11"

(Np)vp

ES

Note that N (a b) = (Na) (Nb), and Na is a natural number if a c o. The significance of Na is given by the following proposition. 33:2. Let S be a Dedekind set of spots on the global field F, and let a Ç b be fractional ideals at S. Then Na/Nb. Proof. The various isomorphisms used in the proof are obtained from § 22D. We can assume that b = o since there is an additive isomorphism

b/ct ›—› ofb -1 a

and since N (b - la) = Na/Nb; we now have to prove that (o : a) = Na. . . . pry, into distinct prime factors; then Take the factorization a = 0/a

(0/1);') e • • • e (010); 5*

68

Part One. Arithmetic Theory of Fields

so this reduces things to a power of a single prime ideal; accordingly let a = pv with y E N. Form the chain 0)p)•• .Dpv.

Then 0/P

+1

Hence Hence (o

(N p)" .

33:3.

F (p)

(Pi :Pi+1) = NI> 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

Chapter III. Fields of Number Theory

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.

Part One. Arithmetic Theory of Fields

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

Chapter III. Fields of Number Theory

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:

Part One. Arithmetic Theory of Fields

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.

Chapter III. Fields of Number Theory

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.

Part One. Arithmetic Theory of Fields

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
Chapter III. Fields of Number Theory

75

On the other hand, at each p E W the function Pcip takes on just a finite number of values as x runs through X. Hence the set of S-tuples of the form (Ix1p);, Es with x EX is finite. Hence we can find xl, . . ., xr E X such that

X=

U Xi U (S) . 15.it.

2) The second step of the proof consists in defining an idèle I in jp with certain properties. The p-coordinate iv of I can be any element of Pp

which satisfies

1 if p ES liP lp { < lxilp if li_r,pED—(Sug). ----

This still leaves the g-coordinate unspecified; here 1,1 can be any element of Pc( which makes the volume satisfy the inequality

1

Ng .

C Hill

Why is there such an iq ? If q is archimedean it is because the value group of Fq is all of P; if g is discrete one takes any iq EFq with I iq i q = (Ng)--m where nt is a number in Z for which

(Ng)14 5. C

H lip i, 5.

(N g)m+1 .

Thus our definition is in order and we have our Wile I. 3) Since C Hill idèle I. Then

1 there is a non-zero field element x bounded by the

H ixl, 5 H PED-S

lip !, = 114 5. N qIC .

PED-S

Hence for every p' E S we have

1=

H

ix', 5 lxip, N qIC ,

PED

hence

CIN q 5_ ki p,

1,

in other words x E X. So by step 1) we can find an xi and an S-unit E such that x --- ext. We claim that E is the S-unit .required by the proposition. For P < liPiP <1 lei ' — lxI kit' = ixiiP holds for all p E D — (S ■..) g); and so Jelq > 1 by the Product Formula.

q. e. d. 33:9. Let S be the set of spots given in the last proposition, and let T be a subset of s — 1 elements of D — S. An S-unit E q which satisfies leglq> I, leg lp<1

V p E S2 - (S-v (I)

76

Part One. Arithmetic Theory of Fields

is given for each q E T. Then these S-uitits are multiplicatively independent and they generate a subgroup of finite index' in u (S). Proof. Let q' be that spot in D S which is not in T.

1) If we had a dependence relation among the

Bq

we could write it

H rrIcr e = 711" eqvg 44 E 2 —P gEP with vq 0 for all q in T and vq > 0 for all q in P, where P is some nonempty subset of T. Put ri= H e q= Hep. riEP

gET—P

Then loch,. 1 for all p ES. And for each p E P we have

H 671 1 cr(T—P p And similarly 'al p I for all p E T — P. But ictip=

ialcr = grip el e < 1 This contradicts the Product Formula for a. Hence there can be no dependence among the eq . This proves the first part. 2) Define an idèle i in jp by the equations

ip = 1 Vp ES

eq

Let ai,. . ay be all the S-units that are bounded by the idèle i; we have r 1 since, for instance, 1 is bounded by i; and we have r < co by Theorem 33:4. Let H denote the subgroup of u that is generated by the given eq (q E n. It is enough to prove that uS -

igigr

cci H ,

for then (u:H) r 1 for each q ET; for if we have •a q E T with Ink 1, then eq n is an element of H which is bounded by i and has and this denies the minimal property in the choice of n . 18q 7/14' < Hence we have exhibited an n E H which is bounded byt and satisfies iniq'‹ 1 , Inlq >1 Vq-ET. Consider a typical a E u. Then there is an n fE H such that VqE For this reason the construction of the sq is sometimes called the construction of "enough" units.

Chapter III. Fields of Number Theory

77

(for instance, a sufficiently high power of T7 -1). Choose an n of the above sort for which ]ccri'l q, is smallest (why is this possible?). We then repeat an argument already used in this proof to show that iceni cr < 1, locnik> 1 V q E T . But this means that an' is bounded by the id& 1. Hence an' = cc i for some i. So a E cei H. Hence u ai H. q. e. d. 1ZiZr 33: 10. The Dirichlet Unit Theorem. Let S be a set of discrete spots on a global field F, and suppose that the number of spots in Q S is equal to s with 1 s 1. We can then apply the two preceding propositions. They give us a free commutative group H of rank s — 1 which has finite index in u. Hence u is a finitely generated abelian group. Since H contains s — '1 independent elements we must have rank u s — 1. Now ui H where i is the index (u :H) <00, and furthermore any s elements of H are dependent, hence rank u= s-1. So u is a direct product of its elements of finite order with s 1 infinite cyclic groups. But the elements of finite order are the roots of unity, and these are the absolute units by Proposition 33:6. q. e. d. i

—

§ 33G. The very strong approximation theorem 33:11. Theorem. Let T be any finite set of spots on the global field F and let q E D — T. Suppose that a p EF is given, one for each p E T. Then for each E> 0 there is an A EF such that {

IA cep fp <e Vp ET VpE.Q—(Tvq).

Proof. 1) We can take it that 0 < e < 1. Suppose we enlarge T to some finite set of spots which still does not contain q; and suppose we define al, to be 0 at each of the new spots introduced; if we can prove the theorem for the new set of spots, then we shall have it for the original T. Accordingly we can assume that the given set T has the following additional properties: T v q contains all archimedean spots, and S = (T q) is Dedekind. 2) Now apply the Weak Approximation Theorem. This gives an element cc EF which is arbitrarily close to cep at each p E T; this is still not the required A since in general there will be a finite set of spots W S at which fah, > 1. We shall however be through if we can prove that, given any finite subset W of S, there is a /3 EF which is arbitrarily small at all p E W, which is arbitrarily close to 1 at all p E T, and which is integral at all p E S. For if this can be done, and if we take good enough

Part One. Arithmetic Theory of Fields

78

approximations all round, then a fi will be close to ap at all p E T, and it will have la fik, 1 at all p E S, so A = cc fl will be the element we are after. We therefore have to prove that such a /3 does in fact exist. 3) Let (3 be the degree of approximation needed for fi. Since S is a Dedekind set of spots on F we can introduce Dedekind ideal theory. Put o = o (S) and let a be an integral ideal at S for which

lai<

Then

YPEW.

r (o : a)

Na
Pick a representative set x,. . a,. of o modulo a. Now Proposition 33:8 provides us with an S-unit E which is arbitrarily small at all p E T, say IE °cl i p < 6 (1 r, E T) . Then E cep . . . , E cc,, still fall in distinct cosets of o modulo a since E is an S-unit; hence these elements are also a representative set of o modulo a. Let E al, say, fall in the same coset as —1. Put p = Eoci d- 1. Then fi E o so that /3 is integral at all p E S; and /3 = E al + 1 is in a since E is in the coset of — 1, so and

Ply -5, 'alp< Ifi_ 1Jp

VPEW;

JE0tjJp <Ô

Hence fi has the required properties. 33 : 11 a. Corollary. The A of the

YpET. q. e. d.

theorem can be made to satisfy the

additional inequality IA 1 4 > 11e.

Proof. Fix a cr E S.)

14). Choose A E F in such a way that

(T

IA 1P 5-

11

71,-

VisET

1

YpES2— (Tuquql

IA IQ' < -1Ti

P

where n is a natural number. If n is big enough, the Product Formula will force AI4 > 1/e. q. e. d. § 33H. Finiteness of class numberl 33:12. Theorem. Let S bea set of discrete spots on the global field F. Then S is Dedekind if and only if SC D. Our approach to the theory of S-units and the finiteness of class number is based on the Pigeon Holing Principle and the Product Formula. This is the modern elementary approach of E. ARTIN and G. WHAPLRS, Bull. Am. Math_ Soc. (1945), pp. 469-492. For the very elegant topological approach based on fundamental properties of locally compact groups see K. IWASAWA, Ann. Math. 57 (1953), pp. 331-356.

Chapter H. Fields of Number Theory

79

Proof. If we are given S c D, then S is Dedekind by the Very Strong Approximation Theorem. Conversely, let a Dedekind set S be given. Suppose if possible that S = D. Then o (S) must be u (S) v 0 by the Product Formula. So o (S) is a finite set. But this is impossible since F is the quotient field of o (S) by Proposition 21:3. Hence we cannot have S= Q. SoSCD. q. e. d. 33:13. Theorem. Let S be a Dedekind set of spots on the global field F. Then the ideal class number hp (S) is finite. Proof. 1) Fix g E Q — S. Put N q = 1 if g is archimedean (N g is already defined if g is discrete). Consider the constant Ng/C where C is the constant of Theorem 33:5. We make the following claim: given any fractional ideal a at S, there is a non-zero a E a such that

N (cco)

(N gIC) Na. To see this, consider an idèle i in hr whose coordinates satisfy lipip= IaI and whose volume satisfies

Vp ES

1 5. C11111

Ng. (Such an i exists by an argument used in step 2) of Proposition 33:8.) Since gill 1 the idèle j bounds at least one non-zero field element a by Theorem 33:5. By Proposition 33:3 and the Product Formula, N(Œo) — 11 !al p ED—B

H Ik

pED—s

= 11 111 (plis 14,1ip ) = lhhi (N a) (N qIC) Na. lip ip talp for all p ES . Hence cc E a. So we have proved our

But 14 claim. 2) We immediately deduce from the result of step 1) that we have a constant K (namely K = N crIC) which depends only on F and S and has the following property: given any fractional ideal a at S, there is an integral ideal a' in the same ideal 'class as a such that 1 Na' K. But 1 The only thing that matters here is that K is constant. But the number K is obviously of significance when calculating class numbers of specific fields since it allows you to restrict yourself to integral ideals a which satisfy 1 S Na K. From this point of view the smaller the K the better. For class number computations in number fields and for Minkowski's value of K we refer the reader to E. ARTIN, Theory of algebraic numbers (Gtittingen lectures, 1956).

80

Part One. Arithmetic Theory of Fields

it follows from Proposition 33:7 that the number of integral ideals b with Nb K is finite. Hence the total number of possible a' is finite. In other words, the ideal class number hp (S) is finite, q. e. d. 33:13a. Corollary. There is a Dedekind set So consisting of almost all spots on F suck that hp (S) = 1 whenever S Ç S o.

Proof. Start with a Dedekind set T consisting of almost all spots on F. Let al, . . a,. be a representative set for the ideal class group at T. Take a subset So consisting of almost all spots in T such that lai l y = 1 for 1S r and for all p E So. We say that hp (S) = 1 for any S Ç So. To see this consider a typical fractional ideal b at S. Take a E I (T) with Ialp = Ibip for all p E S. Then a --- ccaj for some ai and some a E P. Hence Ibip= 'alp= Iccaiip= kip VP ES.

Hence b

ao. So hF (S)

q. e. d.

1.

§ 331. More about Mies

Let us return to a discussion of the id& group f7 of a global field F. If a is any non-zero field element we know from the Product Formula that 14, = 1 for almost all p E D. Hence we can associate an idèle (Œ)p ED

with every such a. An idèle which can be expressed in this form is called a principal Mae; the set of all principal idèles is a subgroup of jp, which will be written PIT. The rule cc —>— (a) p ED provides a natural isomorphism F

Pp.çJip.

Now fix a set of spots S Ç Q (in practice S will be a Dedekind set consisting of almost all spots in 0). We call the idèle i an S-idèle if 11;4= 1 Y p ES .

The set of S-idèles is a subgroup J of the full Mae group Jp. We put Pi;

J n 137-

Thus restriction of the natural map u (S)

)--o. Pp produces Pj

33:14. Theorem. Let S be a Dedekind set of spots on the global field F. Then

UP:

= hP (.9) <00

.

Chapter III. Fields of Number Theory

81

Proof. We define a mapping cp: JF ->- I (S)

as follows: given an id& i let w I be that fractional ideal a at S for which 1a1p — lipip at all p E S. Thus ITilp= Nip

VP ES

is the defining equation of 9). This mapping is a homomorphism since IT(ti)I= iipipip= Irilp Rri) (0)I holds for all p in S. It is clearly surjective with kernel J. And the inverse image of the group of principal ideals P (S) is PEA. So the kernel of the compoqite map

I (S)

I (S)I P (S)

is PFJj. Hence JEIPFB)----> I (S)IP(S) . q. e. d. 33:14a. Corollary. There is a Dedekind set So consisting of almost all spots on F such that (JE : PEA = 1 whenever S Ç S. § 33J. Note on classical ideal theory

In the classical approach to algebraic number theory one works with a subfield F of the complex numbers C which is finite over the rational numbers Q. The algebraic integers are those complex numbers a which satisfy a polynomial equation am+ al am-1 + • • • ± am = 0 (all ai E Z) . And the ring of integers R of the algebraic number field F consists of all those algebraic integers which fall in F. How is this related to the modern approach which we have given here ? Of course we expect it to be a special case, and in fact it is. For instance, consider the set of all discrete spots S on Q and let TI I S on F. Then Z = o (S) by Proposition 31:2. Hence R o (T) by Proposition 23:2. So the ring of integers in the classical sense is that ring of integers in our sense which corresponds to the set of all discrete spots on F. In particular, the Dirichlet Unit Theorem gives the structure of the group of units of R. And the ideal class group of the fractional ideal theory determined by the ring R is finite.

82

Part Two. Abstract Theory of Quadratic Forms

Part Two

Abstract Theory of Quadratic Forms Chapter IV Quadratic Forms and the Orthogonal Group We leave the arithmetic theory of fields in order to develop a different subject, the abstract theory of quadratic forms. In the latter half of the book we shall combine these two subjects into the arithmetic theory of quadratic forms. Our immediate purpose is to introduce a quadratic form and an orthogonal geometry on an arbitrary finite dimensional vector space and to study certain groups of linear transformations that leave the quadratic form invariant. We must make the assumptionl from now on that the field of scalars F does not have characteristic 2. As we indicated, our vector spaces are assumed to be finite dimensional. § 41. Forms, matrices and spaces § 41A. Quadratic spaces Let V be an n-dimensional vector space over an abstract field F of characteristic not 2. Consider a symmetric bilinear form2 B on V, i. e. a mapping B: V x V —>-- F with the following properties: B(x, y + z) = B (x, y) + B(x, z) B(ax, y) = aB(x, y) , H(x, y) = B(y, x)

for all x, y, z E V and all a EF. If we put Q (x) = B(x, x) we obtain a mapping 1 There will be two exceptions to this assumption about the characteristic: the Wedderburn theory of §§ 51 and 52 and the abstract theory of lattices of § 81 can be over fields of any characteristic. 2 We restrict ourselves entirely to symmetric bilinear forms over fields of

characteristic not 2, and to the so-called orthogonal geometry determined by such forms. If the bilinear form is alternating instead of symmetric the geometry is called symplectic, if the bilinear form is hermitian the geometry is caned unitary. There is also a theory of quadratic forms and a corresponding orthogonal geometry over fields of characteristic 2. We refer the reader to J. DIEUDONNA, La géométrie des groupes classiques (Berlin, 1955) for information about these geometries.

Chapter IV. Quadratic Forms and the Orthogonal Group

83

A quadratic map on the vector space V is any mapping Q: V F which can be obtained from a symmetric bilinear form B by the substitution x =-y in the above way. (This is not the definition in characteristic 2.) The following identities hold: Q (a x) = oc2 Q (x) Q (x ± y) = Q (x) Q (y) + 2 B (x, y) Q (E oci xi) = 2; Q (xi) ± 2

E (xi .; B (xi, xi)
The second identity shows that the symmetric bilinear form B associated with the quadratic map Q is unique. A quadratic space is a composite object consisting of a (finite dimensional) vector space V, a quadratic map Q, and the symmetric bilinear form B associated with Q. We say that a quadratic space is binary, ternary, quaternary, quinary, . . . , n-ary, according as its dimension is 2, 3, 4, 5, . . . , n. An n-ary space is usually assumed to be non-zero. The quadratic space V is said to represent a field element a if there is a vector x in V such that Q (x)-- a, in other words if a C Q (V). We say that V is universal if Q (V) = F. Two subsets X and Y of V are called orthogonal if B (X, Y) = O. For any fixed scalar a we define BΠ(x, y) = aB(x, y)

V x, y E V

.

This provides V with a new symmetric bilinear form Ba: V X V -÷ F ,

and hence puts a new quadratic structure on V. We let (Pc denote the quadratic map associated with B. When this is done we say that we have scaled by a and we let VI denote the vector space V provided with its new quadratic structure. We should not confuse scaling of the quadratic space V with scaling of a vector x E V; by the latter we mean that the vector x is replaced by ax. Thus (x) =

Q (x) ,

while Q (a x) = a2 Q (x) .

Note that Qœ (V) = 002 (V)

Consider two finite dimensional vector spaces V and W over F. We know from linear algebra that the set of all linear transformations of V into W is a finite dimensional vector space over F in a natural way. This set will be written LF(V, W)

or simply L (V, W). When V = W the composition of transformations provides L (V, V) with a multiplicative law which makes it into a ring 6*

84

Part Two. Abstract Theory of Quadratic Forms

with identity; we then use L ip (V)

or 4, (V) or just L (V) instead of LF(V , V). Scalar and ring multiplication in L (V) are related by the equations (97 V) = (Œ99) = P(or V) • for all E F and all Top E (V). (See § 51S for the significance of these equations.) The invertible linear transformations in LF (V) form a group

called the general linear group and written GL F (V) , G Li; (V) , or G L (V) . Suppose now that V and W. are quadratic spaces and let Q and B denote the quadratic map and bilinear form on each of them. A linear transformation a E L (V, W) is called a representation of V into W with respect to the quadratic maps Q on V and W if Q (x) V X E V . If a is a representation, the identity 12 (a x)

B (x, y) = T {Q (x + Y) - Q (x) - Q (y)}

implies that B (a x, y) = B (x, y)

Y x, y

EV

.

An injective representation is called an isometry of V into W. And V and W are said to be isometric if there exists an isometry a of V onto W. We let V -).-W denote a representation, V W an isometry of V into W, and V >-). W or VW an isometry of V onto W. The set of all isometries of V into W will be written 0 ( V, W) .

If V W we write On (V)

or 0(V)

instead of 0 (V, V). In this situation every isometry is invertible, so 0,, (V) S G L 7,(V) .

Moreover, the composition of two isometrics is an isometry, so is the inverse of an isometry, hence the set O, (V) is actually a subgroup of GL 7„(V). This group is called the orthogonal group of V with respect to the quadratic map Q. If a is a non-zero scalar we have On, (V) = On (Vag) . 41:1. Let a be a linear transformation of the quadratic space V into the quadratic space W, and let xl, .. . , x„ be a base for V. Suppose

B(axi,axi)=-. B(xj , xi)

Chapter IV. Quadratic Forms and the Orthogonal Group

85

/or 1 S

n and 1 j n. Then a is a representation. Proof. Express a typical x E V in the given base: x = E ai xi. Then Q (ax) =Q if oc i a;)

= =E

Q (a xi) + 2 E ai ai B (a xi, ax5) <1 4 Q (xi) + 2 E ai B (xi, x5) '

<j

Q (x) .

q. e. d. §41B. Symmetric matrices Let M be an in x in symmetric matrix and N an n X n symmetric matrix over F. We write N (over F) M

and say that M is represented by N if there is an n x in matrix T with coefficients in F such that M = tT N T , where tT stands for the transpose of T. If this can be done with an invertible T over F we say that M is equivalent to N and write MN (over .F) .

Equivalence of matrices is clearly an equivalence relation on the set of all n X n symmetric matrices over F. Now consider an n-dimensional quadratic space V with bilinear form B and quadratic map Q. We can associate a symmetric matrix N with each base x„ for V by taking the n X n matrix whose i,j entry is B (xi, x5). We call N the matrix of the quadratic space V in the base . . x„ and write V N in xi,

, x,„ .

If there is at least one base x1,. . . , x„ for which this holds, then we say that V has the matrix N and we write V

N

What happens to N under a change of base ? Suppose that V N' in the base xi,. . 4, and let T = (45) be the matrix that carries the first base to the second, i. e. let X; =

tAiX da. A

Then B

B (E

toiXo

E 12 JX,a) A

EtB le, A

AX X ) tAi

86

Part Two. Abstract Theory of Quadratic Forms

Hence N'= ' TNT . In other words, N' and N are equivalent over F. And it is now clear that if N" is equivalent to N, then there is a base , x', V in which V x„, and if N is a given If V is just a vector space with base xj., . n x n symmetric matrix, is there a bilinear form B or V with respect to which V has the matrix N in x1 , . . xn ? Yes. Just define

B (facx

,

f

=ocA131 axi,, P

for typical vectors Ev xl and f 16p xi, of V, where ail, is the A, pg entry of N.

Given a symmetric n X n matrix N we let < N > stand for an n-dimensional quadratic space which has the matrix N. For example, if a is a given field element then < a > denotes a line which contains a non-zero vector x for which Q(x). a. We shall find it convenient to use < M > for different quadratic spaces having the same matrix M. 41:2. Let U and V be quadratic spaces with matrices M and N respectively. Then N (1) U -4-- V if and only if M (2) U V if and only if M N. Proof. (1) Put M = (a 15) and N (b 15). Take a base x1,.. . , X for U in which B(x1, x5) = ai 5 and a base . . , yn for V in which B (yi,y5) b i 5. First suppose that we have a representation a: U -›- V. Then B x5) = aij , where xi = axe . Write xi = f toyi, and let T be the

n x in matrix T = (to). Then f tuy9

a15 = B(z11 x5) B •P

A

tp PI

Hence M == 'T NT as asserted. Conversely, suppose that we are given an n X in matrix T such that M = 'T N T. Define zi = E tpap and use the preceding calculation to observe that B x5) can now be proved equal to a •5 . Construct the xi for 1 i tn. Then linear transformation a E L(U, V) given by

ai5 . B(xi, x.).

Hence a is a representation by Proposition 41:1. So U V. (2) The second part of the proposition is done in the same way. q. e. d.

Chapter IV. Quadratic Forms and the Orthogonal Group

87

Suppose we take m vectors 21, . . . , z. in the quadratic space V. The determinant det

zi))

of the in x in matrix (B (z, xi)) is called the discriminant of the vectors 21, . . . ,2m and is written dB (Zip

• • •

2.)

or simply d (h, In particular, if if V N in this base, then

xfi is a base for V and

d (x1, . . . , xn) = detN N.

If we take another base x,

.

.

the equation N' = 4 T N T shows that ) = a2 d(x1, . . , xfi)

for some a E F. Hence the canonical image of d (x11 . . xn) in 0 u (FI1 2) is independent of the base, it is called the discriminant of the quadratic space V, and it is written dB V or dV . If V consists just of the single point 0, the discriminant is defined to be d V = 1.

The set 0 (F/F2) used in the definition of the discriminant is formed in the obvious way: take the quotient groupadjoin 0 to it, and define 0 times anything to be O. We shall often write d V = a with a in F ; this will really mean that (1V is the canonical image of a in 0 \..) (F/F2). It is equivalent to saying that V has a base V =Fx1 + • • • -I- Fx. in which dB 41:3. Example. If

. xfi) = a .

V ne N and a EF, then Vcg

aN. Hence

dVm= afldV. 41:4. Any in vectors , x„, in the quadratic space V with d (x11 . . . , x,n) + 0 are independent. Proof. Adependence E oct x O yieldsf Œ 1 B (xi, xi) = 0 for I j m. This is a dependence among the rows of the matrix (13 (xi, xi)); such a q. e. d. dependence is impossible since d (x11 . . x.,)+ O. ,

§ 41C. Quadratic forms Let us briefly indicate the connection between quadratic spaces and the general definition of a quadratic form. In general a linear, quadratic, cubic, . . . , d-ic form over an arbitrary field is a homogeneous polynomial of degree d in a polynomial ring in sufficiently many variables, say in F PC1, . , X„,

.] .

88

Part Two. Abstract Theory of Quadratic Forms

For instance, 32(1 + 5X2 + 22(3

is a ternary linear form. And /(X1, X2) = X? -I- 32(1 2(2 + 724 is a binary quadratic form. Now if the characteristic is not 2, every n-ary quadratic form can be expressed with symmetric cross terms, for instance here we have

f (Xl, X2) = 2q+ X 1X 2 + --X 2 X1 + 7X. In this way we can associate a symmetric matrix, here it would be 3

72 ) (3

T

with a given quadratic form. Hence we can associate a quadratic space with a given quadratic form. The quadratic space V associated with / (X1, X2) would have a base x1, x2 in which V had the above matrix, and (oel + 0:2x2) =

+ 3041a2+ c4;

in other words the quadratic map Q is obtained from f by substituting ce2 for X1, X2. This brings out the connection between quadratic forms and quadratic maps and it also shows why we must not expect the definitions of § 41A to work when the characteristic is equal to 2. One could define the representation and equivalence of quadratic forms using polynomials. These definitions, when correctly interpreted, correspond to the ones given here in terms of matrices and spaces. From now on the term quadratic form will be used for the quadratic map Q on a quadratic space V.

§ 42. Quadratic spaces § 42A. Orthogonal splittings Consider the quadratic space V with its symmetric bilinear form B and associated quadratic form Q. We say that V has the orthogonal splitting V= V1 1_ • • - _L V,. into subspaces V1, . .

V,. if V is the direct sum V= V1 . • » ED V,.

with the V,/ pair-wise orthogonal, i. e. with B (Vi,V j) = 0 for

1

<j r

Chapter IV. Quadratic Forms and the Orthogonal Group

89

We call the V1 the components of the splitting. We also use the notation and 9 Vi

Vi

for orthogonal splitting and direct sum respectively. We formally define Vi — 0 and 9 Vi = 0 , 0

where 0 denotes the empty set. We say that a subspace U splits V, or that it is a component of V, if there exists a subspace W such that V= Uj W. By taking a base for each component we see that the discriminant satisfies d(Vi j_ • - • j_ V r)

=

d

dV„,

• .

the multiplication being performed of course in 0 i (F/F2). If X1, . . X, are quadratic spaces which are not necessarily subspaces of V we write X, to signify that V has an orthogonal splitting V= Vj • • • IV,. in which each subspace Vi is isometric to Xi for 1 î r. Suppose that V is just an abstract vector space, but that it is also a direct sum V = V1 9 • • • 9 V,. , V,. Then there is a unique symmetric of quadratic subspaces VI, bilinear form on V which induces the given bilinear forms on the V1 and under which V = V _L • • • _L V,..

For if B1, ..

B, are the respective given bilinear forms, define B (Lixa,

ZB A (xA, ya)

for typical vectors Exit, Zya of 9 VA; it is easily seen that B has the A

required properties. Incidentally, this shows that if we are given any quadratic spaces V1,. . V„ not necessarily contained in a given space, there exists a quadratic space V such that V

V1 ± • • • _L V,..

As an example of the notation consider the equation V <M> _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 <Œ1 >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

Part Two. Abstract Theory of Quadratic Forms

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. .

Chapter IV. Quadratic Forms and the Orthogonal Group

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.

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Part Two. Abstract Theory of Quadratic Forms

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 <4. B (V , V) — 0 ,

IQ (x y) Q (x) Q (y)}. Do isobecause of the identity B (x, y) tropic spaces exist ? Yes. For instance, any space having a matrix with a .zero on the diagonal is isotropic. The simplest and most important example of a regular isotropic space is the hyperbolic plane: a space V is called a hyperbolic plane if it has the matrix V Ci 01 ) in one of its bases. Thus all hyperbolic planes are isometric, regular, and with discriminant — 1. Suppose the hyperbolic plane V is written V=Fx+Fy with Q (x) Q (y) = O. Then the lines F x and Fy are isotropic. And the equation Q (ax ± Ay) = 2 oc fi B (x, y) shows that these are the only isotropic lines in V. This equation also shows that Q (V) = F, in other words every-hyperbolic plane is universal. 42:9. The following assertions are equivalent for a binary quadratic space V: (1) V is a hyperbolic plane, (2) V is isotropic and regular,

(3) d V = — 1. Proof. (1) => (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.

Chapter IV. Quadratic Forms and the Orthogonal Group

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. <—a>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

Part Two. Abstract Theory of Quadratic Forms

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.

Chapter IV. Quadratic Forms and the Orthogonal Group

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

Part Two. Abstract Theory of Quadratic Forms

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.

Chapter IV. Quadratic Forms and the Orthogonal Group

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 <2r =. dim V. Thus V is hyperbolic if and only if it has a splitting V = H1 • • _L in which r 1 with all the Hi hyperbolic planes. § 42G. Effective orthogonalization A good practical way of specifying a quadratic space is by giving a symmetric matrix. Suppose a space V is determined by a given symmetric matrix M. The fact that V always has an orthogonal base is the same as saying that there is a diagonal matrix equivalent to M. Is there an effective procedure for finding this diagonal matrix ? Such a procedure is implicit in the proof of the existence of an orthogonal base (Proposition 42:1) and we formalize it here. Start with a symmetric matrix M and consider the following transformations on M: (a) multiply column I by a non-zero scalar a, then multiply the resulting row ï by the same a, (b) interchange columns I and j, then interchange the resulting rows i and j, (c) add a times column i to column j, then a times the resulting row to row ] (a any scalar, 1+ j). 7*

100

Part Two. Abstract Theory of Quadratic Forms

Call each of these three transformations an elementary transformation on the matrix M. An elementary transformation can always be interpreted in terms of a change of base on the space V. To see this let us consider a base x1, . . . , xn for V in which the associated matrix is M. Then the matrix obtained by the first elementary transformation is actually the matrix of V in the base obtained by replacing xi by axi and leaving all other xi fixed. Similarly the second elementary transformation corresponds to interchanging xi and xi. And the third corresponds to replacing xi by xi + ocxi. Hence the matrix obtained by an elementary transformation, or by any series of elementary transformations, is equivalent to M. We leave it to the reader to show that a series of elementary transformations can always be explicitly found which will transform M into a matrix diag

..

0, . . , 0)

with all ai non-zero. Incidentally the number r is the dimension of the radical of V. 43. Special subgroups of O(V) The given quadratic space V is non-zero and regular throughout this paragraph. We show that the symmetries of V generate O, (V), and we make a preliminary study of certain subgroups On+ , D. and zn of On (V). § 43A. Rotations and reflexions Consider the orthogonal group On (V) of the regular non-zero quadratic space V. Fix a base xn for V and let M denote the matrix of V in this base. Consider an isometry a of V and let T = () be the matrix of a in the above base, i. e. axj = E ti ,xi .

Then axi, . . ax, is also a base for V and the matrix of V in this base is M since a is an isorhetry. Hence by § 41B we have M = IT M T. Now the determinant of the linear transformation a is equal to det T, hence 1 if a E 6412 ( det a = since V is regular. We call a a rotation if det a = ± 1, and we let 0: (V) or 04- (V) or 0: denote the set of rotations of V. We call a a reflexion if det - 1 and we let 0; (V), etc. stand for the set of reflexions of V. It is clear that 0: (V) is a subgroup of O,, (V), and that Ow (V) is not. We have On = 0: On- , 0: n On- = 0 . 0: is not empty since, for instance, the square of any isometry is a rotation. 0; is not empty since, for instance, every symmetry is a reflexion (see § 42E). The map det : -› ± 1

Chapter 1V. Quadratic Forms

and the Orthogonal Group

101

is a surjective group homomorphism with kernel Ot, hence Ot is a normal subgroup of 0„ with (0„ : Ott) 2. The preceding discussion shows that the base , x„ determines a group isomorphism of 0„(V) onto the group of matrices {T} with the property that M.iTMT

The matrices T with this property are called the automorphs of M. Since detM 0, every automorph has determinant +1. The automorphs of determinant +1 are called the proper automorphs of M; they correspond to the rotations of V under the above groUp isomorphism. Since every symmetry is a reflexion, and since there are as many symmetries as there are anisotropic lines, the number of reflexions is at least equal to the number of anisotropic lines. Now there are just as many rotations as there are reflexions. Hence the number of rotations is at least equal to the number of anisotropic lines. 43: L V is a hyperbolic space and a is an isometry of V onto V which is identity on a maximal totally isotropic subspace of V. Then a is a rotation. Proof. Let M be the given maximal totally isotropic subspace of V, let r be its dimension, so 2r is the dimension of V. By Proposition 42: 13 there is another maximal totally isotropic subspace N of V such that V= MEDN. For any x M, y E N we have ax x and B (x, ay — y) = B (x, ay) B (x, y) = B (x, ay) — B (x, ay) = O, so ay — y E M*. But M S M*, so by comparing dimensions M = M*. Hence ay — y E M for all y E N. Take a base x1,. . X,. for M and a base yl, , yr for N. Computing det a in the resulting base for V gives q. e. d. • deta = 1. Hence a is a rotation. 43:2. Let U be a hyperplane of the regular quadratic space V and let a be an isometry of U into V. If U is regular there are exactly two prolongations of a to On and the one is a symmetry times the other. If U is not regular there is exactly one Prolongation of a to O. Proof. (1) First do the case of a regular hyperplane U. By Witt 's theorem there is a prolongation ai of a to On . Let 0 2 be any element of On . Then 02 is a prolongation of a if and only if ç' a1 is identity on U. This means that al and alt.,/ are the only prolongations of a, where Fy is the line orthogonal to the hyperplane U. (2) Now let us suppose that U is not regular. We know that there is at least one prolongation of a to On . If there were another we would have a e in 0„ which is identity on U but not on V. Let us prove that this is impossible. U* is a line since U is a hyperplane, hence rad U = U n U* is a line F x say; a radical splitting of U therefore leads to a splitting (F x Fy) Q (x) = Q (y) = 0 , W U V=

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Part Two. Abstract Theory of Quadratic Forms

Now e leaves W fixed. Hence it leaves Fx +Fy fixed. But ex = x. Hence ey ay for some scalar a. But B (x, y) = B (ex, ey ) = B (x, ay) = ccB (x, y) .

Hence a . 1. Hence e is identity on V. Hence every e which is identity on Û is identity on V. So a has just one prolongation, q. e. d. §43B. Generation of On by symmetries 43:3. Theorem. Every isometry a of a regular n-ary quadratic space onto itself is a product of at most n symmetries.

Proof. 1 1) The first step is in fact a lemma. We suppose that the isometry a on the space V satisfies the condition x anisotropic in V => 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.

Chapter IV. Quadratic Forms and the Orthogonal Group

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

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Part Two. Abstract Theory of Quadratic Forms

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

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Part Two. Abstract Theory of Quadratic Forms

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

Part Two. Abstract Theory of Quadratic Forms

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

Chapter IV. Quadratic Forms and the Orthogonal Group

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

Part Two. Abstract Theory of Quadratic Forms

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.

Chapter IV. Quadratic Forms and the Orthogonal Group

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

Part Two. Abstract Theory of Quadratic Forms

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

Part Two. Abstract Theory of Quadratic Forms

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

Chapter V. The Algebras of Quadratic Forms

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.

Chapter V. The Algebras of Quadratic Forms

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

Part Two. Abstract Theory of Quadratic Forms

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)

Chapter V. The Algebras of Quadratic Forms

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 (u 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

Part Two. Abstract Theory of Quadratic Forms

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)

Chapter V. The Algebras of Quadratic Forms

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.

Part Two. Abstract Theory of Quadratic Forms

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).

Chapter V. The Algebras of Quadratic Forms

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

Part Two. Abstract Theory of Quadratic Forms

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 <

(en) V = A en + • + dn e 1 •

dien. = (Aeu + • • + d 7 e) = (elien) = O. Hence (e11) (li en. And (e11) V = (e11 en) V = e • l (en. V) =

Hence (x) v = xcil for all x

a'. Hence v

q. e. d.

Chapter V. The Algebras of Quadratic Forms

125

§ 52D. The algebra LD (V)

As a final preparation for the proof of Wedderburn's theorem let us recall some facts about vector spaces over skew fields. Let D be a division algebra over F, and let V be an n-dimensional right vector space over D. We use L (V) to denote the set of all D-linear maps of V into V. We return to the practice of writing mappings on the left. Then L (V) is a ring in a natural way, multiplication being the composition of maps. It is also a vector space over F under the following scalar multiplication: (oc97) x = (Tx) (oc 1»)

for all cc EF, E L D (V) x EV. And cc(92 V) = (x92) V = (cc V) for all a EF and epop E L » (V). So L (V) is actually an algebra over F.

(Finite dimensionality can be proved in several ways; for instance it follows from the isomorphism L (V) >--)- 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

Part Two. Abstract Theory of Quadratic Forms

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

Π(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,

Chapter V. The Algebras of Quadratic Forms

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

Part Two. Abstract Theory of Quadratic Forms

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

Chapter V. The Algebras of Quadratic Forms

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

Part Two. Abstract Theory of Quadratic Forms

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.

Chapter V. The Algebras of Quadratic Forms

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

Part Two. Abstract Theory of Quadratic Forms

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.

Chapter V. The Algebras of Quadratic Forms

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
whenever di = 0,1 and ei = 0,1 for 1 i n. The proof of the formula follows from the relation xi xi = —xixi for i+ j in this way: move xlz step by step to 4z and count the number of changes of sign, then move x€2. to .4., and so on, then count the total number of changes of sign. 2) We now prove that A' is a Clifford algebra of V under the assumption that dimF A'. 2. This assumption guarantees that the products .

(ei = 0,1)

form a base for A'. Consider a typical algebra A* compatible with V and let * denote its law of multiplication. We can define an F-linear map 9): A' -4— A* such that

x1.

1•

• • Xe:

Xe1 * • ' ' * 41.

This map preserves the identity; and it is easily seen to preserve multiplication on the above base for A' because of the formula of step 1) ; hence it is an algebra homomorphism; clearly it is the identityinap on V. Furthermore, any F-linear map with all these properties must be 9a itself. Hence A' is indeed a Clifford algebra of V. q. e. d. 54:3. Every regular quadratic space has a Clifford algebra.

x„ a fixed orthogonal Proof. 1) V is the space in question, base for V. We must introduce some notation. (The usefulness of this new notation is confined to the actual construction of the Clifford algebra and it should be forgotten once the construction has been completed and the proposition has been proved.) K2= {0,1} is a fixed finite field of two elements and (5, E, are typical elements in K2. For each ri EF and each s E K2 we define ce=

{ 1 if s --- 0 a if e-=l.

Part Two. Abstract Theory of Quadratic Forms

134

We then have the following rules: le. 1 , (oc 'V= a' fle, 008+4 —cce ce if a = ± 1 , ces.+:6 a,c. We let K be the cartesian product

K = K2 x • • • x

K2

(n times)

made into an additive group by component-wise addition: thus for typical = (61, , , E = (sp , en ) in K we have d + E = (61 + el, . . . , 6n + e,). 2) Take a vector space C' of dimension 2" over F which is disjoint from V. Fix a base for C' and label it with elements of K. So X4 denotes a typical vector in this base. Define

Et

.2c4 xE .

H (— 1)41 ( qfiei X4+ E in vgi
on these basis vectors; clearly the identity relation

X4 X 0

X4= X 0X4

is true. Hence (§ 51B) C' is an algebra over F. And by definition of C', dim Ci

2".

3) Put 11.

X (1 1 Of • • • 0) • • •

X(0, • • • 00 1)

then let V' be the n-dimensional subspace

V 1 = Fxi + • • • ± F 4, of C'. It follows by inspection from the definition of multiplication that if j+ i 0 ic.'1 I; + z; ei = {2qi l c, if j = i . Hence xy + yx E F 1c, for all x, y E V'. So for each x, y E V' we can define B (x, y) such that

B (x, y) 10.=

(xy y x) .

135

Chapter V. The Algebras of Quadratic Forms

Then B: V' x V' F is a symmetric bilinear form on V'. So V' is naturally a quadratic space with quadratic form Q, say. We have

Q(x) l=x2 V x CV' ; and C' is clearly the subring generated by V'; and dim C' = 2, hence C' is a Clifford algebra of V'. Now

V'

(F

I (Fx) x)

with Q = qi = Q (x i) for 1 i < n. Hence V is isometric to V'. The identification process (see § 53A for a detailed application of the identification process) then gives us a Clifford algebra C containing V. q. e. d. 54:3a. The Clifford algebra C of a regular n-ary quadratic space V is generated by V. And dim C 2. Proof. This is true for at least one Clifford algebra of V, namely the one just constructed in the last proof. Hence by Proposition 54: 1 it is true for C. q. e. d. Since by Proposition 54:2 the 2n vectors xei t . . . x

(e i , 0,1)

of the Clifford algebra C span C, and since we have just proved that dim C = 2, it follows that these 24 vectors form a base for C. We shall call this base the base derived from xl , , x„, or simply the derived base of C. Any element of the Clifford algebra C of V which can be expressed in the form

2: (x y • • .)

finite

with each term (xy . . .) a product of an even number of elements y,... of V, is called an even element of C. The set C+ of all even elements of C is a subalgebra containing the identity of C. Clearly C+ is spanned by all even products of the basis elements x i , . . . , xn of V, hence Q (x i) l o it follows from the relations x i x j — — xi xi (for i4 j) and that C+ is spanned by the derived basis vectors of the form x e, ,v e„ " n with e i — 0,1 and el. + • • • e11 0 mod2. Now there are 2 vectors of this sort. Hence dim C+ = 2n —1 . '

54:4. a is an even element of the Clifford algebra C of the regular quadratic space V. If a commutes with every element of V, then a is in F l c. Proof. Express a as a linear combination

a -

ccAX2,

Part Two. Abstract Theory of Quadratic Forms

136

where the Xa are distinct even vectors of the base for C that is derived from the given orthogonal base x1,. . . , xn for V. For any such ICA we have xi X2 - ,-- -1- XA xi for 1 i n. Furthermore, if we consider a particular X; in which 2Cia -=-- xe,..

. . . xee . . . xe:

with ei = 1, say, then XAe xi . — xiXAs. Now axf = xi a. So (ccAe xii.+ E cca x-,) xi .-- xi (cras X,19 -1— E oca Xa) a+ A. a+a• = (-- Ma eX,16-1- E ±ez2 29 ; . 2+4

q. e. d. Hence al.= O. Hence a EF lc. 54:5. C is a Clifford algebra of a regular quadratic space V. Then there is exactly one algebra anti-isomorphism C >7.4- C which is the identity map on V.

Proof. Let C* be the vector space C provided with the new twisted

multiplication a * b = ba V a, b

EC.

Then C* is again an algebra over F. It is easily seen to be a Clifford algebra of V. We therefore have exactly one algebra isomorphism a >—). a of C onto C* which is the identity on V. But to say that the map a >—). a of C onto C* is an isomorphism is the same as saying that it is an anti-isomorphism of C onto C in virtue of the equations

q. e. d. 54:6. V is a regular quadratic space and T, . . . , Tur are symmetries such that rn, . . . vn,-- 1 v • Then Q(u1) . . . Q (u r) Eil. Proof. Take a Clifford algebra C of V. Then for any anisotropic u ( V and for all x E V we have 2 B (z, u) Q (u) U = X — (Q u) -1 (X U + U X) U =X—X—UXU -1 =—UX

u-4 .

Now r is even since I v is a rotation, hence the given equation rul ...Tur = I v implies that (ui . . . u,.) x (ill . . . u,.) -1 . x VxEV. So (141 ... itr) commutes with x for all x E V, hence u1 . . . ur = a1 a for some a EF by Proposition 54:4. Apply the anti-isomorphism bar

Chapter V. The Algebras of Quadratic Forms

137

of Proposition 54:5: Q (141) • • • Q (14,-)

la = (u1

• Ur) (Ur • •

= (U1 • • • Ur) (ar • • •

U1)

ail

= (u1 • - • ur) (141 • • - ur)

= (a l e) (a l e) = ex2 1 0

q. e. d.

§ 55. The spinor norm We consider a non-zero regular n-ary quadratic space V with its corresponding symmetric bilinear form B and the associated quadratic form Q. Consider any a E On (V). By Theorem 43:3 we can express a as a product of symmetries, say Suppose this is done in some other way, say O. = Tv, • • . Tv, •

Then acr4 = I v so that • Tuv, Tv,

. . Tv), = 1 yr

hence by Proposition 54:6 Q (ut) E Q

. .. Q (v We can therefore attach a well-defined invariantl to any a E On, namely the canonical image of Q (u) Q(ur) in F1F2. This invariant is called the spinor norm of a and is written Q ( h)

••

0(a)

E

Clearly 0 (a T) = 0(a) 0 (T) , so we have a group homomorphism •

•

0: 0„ FIP.

Our interest in the spinor norm of a reflexion is just a passing one and we shall usually confine ourselves to the restriction 0: 0 4,-,

The kernel of this restriction is written 0;,, (V). Thus an' (V) = {o. 0: (V) I 0 (a) = i}. Professor ZASSENHAUS has shown me some of his unpublished results on the spinor norm in which well-definedness is proved without making use of the Clifford 1

algebra. As a by-product he obtains the explicit formula. .0 (a) = det the spinor norm of a rotation a which satisfies det (1, a)4: O.

1v0 + - 2 )

for

138

Part Two. Abstract Theory of Quadratic Forms

A commutator clearly has spinor norm 1, so

as c 0:z c 0;1-, c On . In fact each of these groups is a normal subgroup of O. Suppose that the quadratic space V is scaled by a non-zero oc in F. Let us consider the isometries of Vœ. We have already mentioned the obvious fact that O,, (V) = On (Vcc). The anisotropic lines of V and Pc are the same, and the symmetries r„ are the same linear transformations whether u is regarded as an element of V or as an element of Va since X

2 E (x , u)

U—X—

2 Ell (x, u)

QŒ(u) U. 42 (u) However the spinor norm of T„ is changed by the factor a on passing from V to Pc. On the other hand a rotation, being a product of an even number of symmetries, will have its spinor norm changed by an even power of a and hence not at all. In particular, 0,', (V) = The other subgroups Zn, 14, On+ of 0„ are clearly unchanged on scaling V by a. We are not going to be too rigid in our use of the 0 notation, although it will always be clear from the context just what we have in mind. The equation Oa = oc with cc Ei• will often appear; this really means that Oa is the canonical image of cc in FfF.2. Occasionally we regard Oa as the full coset ak2 taken as a subset of È. More generally, consider any subset X of 0„; strictly speaking the symbol OX is the image of X in F/F2 under 0; but we shall also regard it as the full set of images 0 X -- u 0 (a) 12 aEX

of X in P; if X is a subgroup of 0„, then OX (in the relaxed notation) is a subgroup of P. 55:1. Example. Let us go back to the hyperbolic plane H =Fx + Fy of Example 42:15. In fact suppose -e- y(30) in the base x, y. H--— If -c is a symmetry it must interchange the two isotropic lines, in fact it must have the form TX = czy , ry . or -lx .

What is the spinor norm of the symmetry described by these equations? We have seen that T is actually the symmetry with respect to the line X — oty, i. e. T = T z — ccy . Hence 0 (r) = Q (x — ay)--- — 2 a y .

Chapter V. The Algebras of Quadratic Forms

139

From this we can derive the spinor norm of the rotation described by the equations 17 X = OC X , ay = orl y For a has the form a = Tz _ yr,, hence 0(a) = (-2 y) (-2 oc 7), hence .

55:2. V is a regular n-ary quadratic space with n>. 2. Then 0(0,4 ) is the subgroup of F consisting of all non-zero scalars of the form aci

•

Œ2r

with 2r < n and oil, . . ce2 , in Q (V). Proof. The proof is immediate. q. e. d. 55:2a. If V is isotropic, then 0(0:) = P. 55:3. Let u„, u, and v1, , v, be anisotropic vectors in a regular quadratic space V, and suppose that {Q (u,), . Q (at,)} is a permutation of {Q (v„), . . .,Q 04». Then

DI% Proof. Let w1, . v.', be a rearrangement of y1,. y, in which Q (w1) = Q (u1) for 1 r. Let bar be the natural homomorphism of On onto 0nfl2n . Then Onig, is commutative, so Tu, • • • Ts,.

Hence

E Tv„ •

••

•

frr • • • Yvi Yui • • • t:14 = (-7WiL

Tyr • • • Tv, TUI • • • rtg, E (Ty),

;AI)

' ( Ywr itir) •

• • • (rWr TUt) pn •

It therefore suffices to prove that E 12„ whenever Q (w) = Q (u) By Theorem 42:17 there is a a (On such that crw = u. Now a T to 0-1 = "ra ="--hence TW

=

a TTV1

a-1 E £l

O.

.

q. e. d. 55:4. V = U _LW is a splitting of the regular quadratic space V, and a is an element of 0(V) with the splitting a=r Ie where t E 0(U) and e E 0(W). Then 0(a) = 0(r) e (0 . Proof. Express T as a product of symmetries with E 0 (U). Then 1 w= (xi ± 1 w) • • - (41 1 w) , and so 0(r = H (ri) 0(T). = 110(r

Similarly 0(1u 1 & = e (e). Then 0(a) = 0(r 1 1 w) 0( 1 u

0 (T) 0 (e) •

q. e. d.

Part Two. Abstract Theory of Quadrati Forms

140

55 : 5. V is a regular n-ary space with 1 n

3. Then

(Ott ) 2 = = . Proof. Here (0:) 2 stands for the set of squares of rotations of V. We saw in Proposition 43:6 that 11„ is generated by (Or) 2 , in particular First we prove that 12„= 0,,' . Let us consider a typical a E On' . Since a is a rotation and n 3, we can write a as a product of two symmetries, a = xcer„ say. Then 0(a) — 1, so Q (x) Q (y) EF2, so Q (x) E Q (y)F2 , so we can assume after suitably scaling the vector y that Q (x) =Q (y). But then T y E -cœl2„ by Proposition 55:3. Hence a . t Ty E Now we show that (0,1) 2 =14. If n = 1 this is obvious. If n = 2 then 0: is commutative, so (0:) 2 is a group, so this group must be 12n (see Remark 43:8). Consider the case n =3. Take a typical a E O. If a is a rotation with an isotropic axis, then a is the square of a rotation by Corollary 43:5a, j. e. a E (01) 2. Otherwise a will be a rotation with an anisotropic line L as axis. We can then write

e

a = 1L 1

where e is a rotation on the plane L* orthogonal to L. But 0 (g) = 0 (a) = 1 by Proposition 55:4 since a is in O. Hence e E (0-2F (L*)) 2. Hence a E ( 19-3 ) 2 . So in general 0; (0-3') 2. q. e. d. 55:6. U is a regular subspace of a regular quadratic space V. If Q (U) = Q (V) and 12 (U) = 0' (U), then 12 (V) = 0' (V). Proof. We have to consider a typical a E 0' (V) and prove that-it is in 12(V). Express a as a product of symmetries: a=

Take u11 .

Tv,

Tv?

.

u,.E U with 02 ( 14i) = Q (7)1) for 1

E

I

r, and put

(To

9 Tu, • • - Tur Then a E e (2(V) by Proposition 55:3, so it is enough to prove that E D(V). If we take the splitting V Uj W, then e l w with E O (U) since each ui is in U. But 0(j) = O() = 1, so j E 0' (U), hence j ED (U), hence e E Q (V). q. e. d. 55:6a. If V is isotropic, then D (V) = 0' (V). And we have group isomorphisms

0+10'. 0+1D

klfra

Proof. Take a hyperbolic plane 11 Ç V. Then Q (H) F =Q (V), and S2 (H) = 0' (H) by Proposition 55:5. Hence 12 (V) — 0' (V).

Chapter V. The Algebras of Quadratic Forms

141

We have O (0,t) F by Corollary 55:2a. Hence the spinor norm 0:0;1; F/1 2 is surjective. Its kernel is On by definition of On' . Hence q. e. d.

55:7. Let V be a regular n-ary quadratic space. Then 0 (-1v) is equal to the discriminant dV. And —1 y is in 0,', if and only if n is even with dV = 1 Proof. 1) Take an orthogonal base . . xn for V. Then rxi Txn 1 v, hence O (— iv) Q (xi) • • Q (xn) F:2 — dV .

.

.

— —

2) If — iv is in 0;, it is a rotation with with dV 1. Similarly with the converse ,

(

—

1 y ) = 1, hence n is even q. e. d.

§ .56. Special subgroups of On(V) The subgroups 19;,' ,O'n , Q, Zn of the orthogonal group On of a regular n-ary quadratic space V give rise to a normal series

< 1 < .(27, n Zn < Q < O'„ < and it is natural to ask for a description of the factors of this series, say in terms of F and V. A study of the factor n would take us too far afield so here we must content ourselves with a statement of one of the known basic results: the group .S2„1.S2, n Zn is simple when V is isotropic and n 5 • 1 However we can conveniently study the remaining factors of the series. First the obvious remark that On/On+ is a cyclic group of order 2 allows us to concentrate on the factors pn n Zn 0;11•Qn W 1- 10 ;1 • We can use the results of §§ 43E and 55 to obtain the following n, 3, or when V is isotropic descriptions when V is arbitrary with 1 with n 2: 1, otherwise it is i, 1. .52„ n4 is {±1 17} if n is even with dV

2. 0J.(2 1, 3. 0,1/011 is isomorphic to the group {a 43 E FI oc, 13 EQ (V)} modulo F 2 when 1 n 3, 3'. 0:/0:, is isomorphic to F/F . 2 in the isotropic case. We are therefore left with the factors Q Zn, If4, 0,1-10,', for an anisotropic V with n 4. The structure of these groups depends strongly on the nature of F and V. We shall discuss this part of the theory over certain fields of interest in subsequent chapters of the book.

I see

E. ART1N, Geo.melric algebra (New York, 1957) for a proof of this result n 4. and for results on the structure of O,( V) when 1

142

Part Two. Abstract Theory of Quadratic Forms

§ 57. Quaternion algebras . § 57A. The quaternion algebra Let a field F and two scalars oc, f3 in F be given. Take a 4-dimensional space V and a base 1, x1 x2, x3 for V. Thus (

4',

,

V = Fl F xl + F xa + F x3 .

Define a multiplication on these basis elements by the multiplication table I

X1

X2

X3

1

1

Xi

X2

X3

X1

Xi

cc 1

X3

CC X2

Xs

X2

—z3

P1

- Pi

Xs

x3

-x 3'c s e Pzi

—cefil

and extend this by linearity to a multiplication on V by the method of § 51B. By inspecting the table we find that the basis vectors satisfy the associativity relations, and the element 1 satisfies the identity relation, hence V is an algebra. The vector space V and the base 1, x1, x2, x3 used in the above construction can, of course, be quite arbitrary. In order to be more specific we assume that, for each pair oc, 13 in F, a space V and a base 1, x1, x2, x3 are taken in some way and then fixed for the rest of the discussion. The algebra obtained by the preceding construction is then called the quaternion algebra «,, p

k

F '

and the base 1, x1, x2, x3 is called the defining base of the algebra. The elements of a quaternion algebra are called quaternions, the elements of Fi are called scalar quaternions, the elements of Fx2 + F x2 + F x3 are called pure quaternions, the set of pure quaternions is written cc,

fir

k F — Fxi + F x2 + F x3 . If E is any extension field of F, then the E-ification E

p F ) has a base

cc, 13 with the same multiplication table as i, E , hence there is an algebra isomorphism E

P

When there is no danger of confusion we write (a, 16) instead of ( (ci'f ). It is convenient to remember that xi xj = — E F Xk , 4 E P 1 for any permutation i., j, k of the digits 1, 2, 3.

Chapter V. The Algebras of Quadratic Forms

143

Consider an arbitrary quaternion x = E0 ! ± E X J ± eg X2 ± E3 X3 We define the conjugate of x to be the quaternion eo 1 — 4x2— E3 X3 • A direct computation gives. .

nx—ng , x+y=g+9, 757-9g,

for all x, y E c4F4 ) and all 71 E F. We define the norm N and trace T of x by the equations Nx=xg,

Tx= x+g.

An easy computation gives

($20 -n._

+ accf3)1, Tx = 2 41 . In particular this shows that both N x and Tx are scalar quaternions satisfying Nx=

Ng=Nx, Tg ----. Tx .

For any quaternions x and y we have N(xy)=Nx•Ny, , T(x+ y). Tx+ Ty.

We see that x is pure if and only if g= — x, and x is scalar if and only if g= x. We see that —N x ff x is pure x2 N x if x is scalar. Hence x2 is a scalar quaternion whenever x is either a scalar quaternion or a pure one. From this it follows that x2 is not a scalar quaternion if x is neither scalar nor pure. Hence x2 E Fl if and only if x is either scalar or pure. 57:1. x is any element in a quaternion algebra. Then x is invertible if and only if Nx EF 1. If this condition is satisfied, then x-4 = (N x) -1 R. Proof. Suppose x-1 exists. Then x-1 (Nx) =

xg = g+ O,

hence Nx+ 0, hence NxEF 1. Conversely, suppose Nx E F 1. Then ((Nx) -1 g) x = (N

g = (N x) -1 N x = 1,

and similarly x ((Nx) -1 g) = 1. q. e. d. 57:2. ( Ff ) is central simple. Proof. 1) First let us prove that the given algebra is central. We must consider a typical x = E0 1 ± e, x, + E2 x, + E3 x3

Part Two. Abstract Theory of Quadratic Forms

144

in the center and prove that it is scalar. Let i, j, k be any permutation of the digits 1, 2, 3. Then

0 = xx k — x k x = 2 (e

4:i x; = O. Hence

Multiplying by xrl we get Similarly

ei xi) x k . 0 and = O.

eic 7-= O. Hence x is scalar. So (--œ-/-) is central.

2) We have to show that a non-zero two-sided ideal a is the entire quaternion algebra. It is enough to find an element of F 1 in a. Take a non-zero quaternion x eo l 1 x1 d3 x3 in a. If 4:1 , e2, e3 are all 0, then x El; 1 and we are through. Hence we may assume that ei + 0 for some j with 1 Ç j 3. Let i,j,k be a permutation of the digits 1, 2, 3. Then

xx k — x k x = 2

j .xj) xk

is in a, hence

= eixi SO

yx ; dis in a. But this element is in F 1. Hence (

) is simple.

q. e. d.

57 : 3. y and z are two elements of an arbitrary algebraA over F such that

y 2 . a1A , z 2 . fl 1 A , yz — zy

with a, fl in F. Then the subspace F 1A + Fy +17 z+Fyz . is an algebra isomorphic to ( a Proof. Let C denote the giv en subspace. A linear argument using the given relations shows that C is actually a subring, hence a subalgebra of A. Take the F-linear surjection a, p : F defined by

(pl = 'A'

y , (px 2 . z , (px 3 — yz

A simple argument involving the associativity of multiplication shows that 99 preserves products. Hence q) is an algebra homomorphism. Its kernel is a two-sided ideal in (2c2,-) , hence it is 0, hence 99 is a bijection, hence an algebra isomorphism.

q. e. d.

145

Chapter V. The Algebras of Quadratic Forms

57:4. Example. Let us prove that (

_1)

is algebra isomorphic to the

matrix algebra M2 (F). Consider the defining matrices en, en, e , e22 of M2 (F). Then en + e22 is the identity matrix 1 of M2 (F). Put y = + Z = en — e12, so that y2 = 1 , z2 = — 1 , yz = — zy . Apply Proposition 57:3. Then

M2 (F) = Fl +

+ Fz + Fyz (

1, ---1 .

57:5. Example. The Clifford algebra of the quadratic space V ± is isomorphic to the quaternion algebra (a, /3). 57 : 6. An algebra isomorphism cp of one quaternion algebra onto another sends the pure quaternions to the pure quaternio.ns and satisfies 97X ipg N(fpX) = Cp(NX) , T(fpX) ==. cp(Tx) for all x. Proof. 1) It is enough to characterize pure quaternions in a form that is invariant under isomorphism. But we have such a characterization on hand: a non-zero x is pure if and only if x2 is in the center F1 and x is not. 2) Write x = e0 i + y with 4:0 E F and y pure. Then cpy is pure. Hence

And so

Tx = c■ 1 +

1 — 99Y = T(eo 1 — Y) = 992 -

=

N(x) = (99 x) rg.).X) = (9) x) (g) =

q. e. d.

Similarly T(Tx) =

§ 57B. The quadratic form of a quaternion algebra Consider the quaternion algebra (a, (3) over F. Then -2- (x5;

y ) = -2- T (x9) EF 1.

For each pair of quaternions x, y we let B(x, y) denote that uniquely determined element of F for which B (x, y) 1 = -2- T (x5;) . This provides (a, 13) with a symmetric bilinear form B so that (a, /3) can be regarded as a quadratic space in a natural way. The associated quadratic form satisfies Q (x) 1 = N x for all x

If x is a scalar quaternion and y is pure, then x51 is pure so B (x, y) 1 = O'Meara,

1111 rocluctiori

to

•ie forl iLs

T(xji) — O. 10

Part Two. Abstract Theory of Quadratic Forms

146

Hence

1 ) 1 (oc,

(cc, /3) =

P )° .

If x and y are both pure, then

B (x, y) 1 —

(x9 + y ) = —

B(x, y) = 0

xy— yx

(xy yx) ,

hence (if

x, y pure) .

If x is scalar then Q (x) 1 = x2 ; if x is pure then Q (x) 1 , x2. In particular,

(a,13):.-,' <1>1 <—a> 1 <-13> 1 . Hence the discriminant of (a, (3) is equal to

d(a, fl)

1.

57:7. Example. Let C be a quaternion algebra and let V be a binary subspace of the quadratic space Co of pure quaternions such that V <— a> 1 <- 13> with a, /3 in F. So there are x, y in C° such that

x2 . al , 372,- 13 I ,

xy

yx

Hence by Proposition 57:3 there is an algebra isomorphism

C

(a, fi).

57:8. Let C and D be quaternion algebras over the field F. Then the following assertions are equivalent: (1) C is algebra isomorphic to D (2) C is isometric to D (3) Co is isometric to Do. Proof. Let C be the quaternion algebra (a, 13), let D be the quaternion algebra (y, 6). If C and D are isomorphic as algebras they will be isometric as quadratic spaces since an algebra isomorphism preserves norms by Proposition 57:6. If C is isometric to D, then C° is isometric to Do by Witt's theorem (§ 42F) .If C° is isometric to D°, then D° contains a binary subspace of the form <— a> 1 < —fl>, hence D is algebra isomorphic to C by Example 57:7. q. e. d. 57 : 9. Let a and i6 be non-zero elements of a field F. Then the following assertions are equivalent: (1) (a, fi) is algebra isomorphic to (1, — I) (2) (a, 13) is not a division algebra (3) (a, 13) is isotropic (4) (a, IV is isotropic (5) I <16> represents 1

(6) a E NF,IFE where E = F (0).

Chapter V. The Algebras of Quadratic Forma

147

Proof. (1) implies (2). Let 1, x1, x2, x3 be the defining base of the quaternion algebra (1, — 1). Then (x1 + x2 = 0. So (1, — 1) is not a ) 2

division algebra. Hence (a, fl) is not a division algebra. (2) implies (3). If (a, /3) is not a division algebra, then there is at least one element x+ 0 in (a, fi) which does not have an inverse. By Proposition 57: 1, N x must be 0; hence Q (x) = 0; hence (a, /3) is isotropic. (3) implies (4). Apply Proposition 42: 12. (4) implies (5). We are given that <—a> <-13> _L is isotropic, hence < a sp > < 132 > < cc2 132 > is isotropic, i. e. 1 1 < — 1> is isotropic. So < a> 1 < 16> represents 1 by Proposition 42: 11. (5) implies By Proposition 42: 11 the given information implies that 1 <13> 1<-1> is isotropic, hence <—a> J <-13> 1 <1> is isotropic, hence <1> 1 < —fi> represents a by Proposition 42:11. So there exist EF such that a — Then in E =F (II fi) we have (E) . 1117) = '1 VP) = NEIF (6) implies (1). We have E F such that

e, (e +

— the. (e + n

(e —

NEIER

+

fi') •

n 2 /3, hence <—a> 1 <-13> 1 <1> is isotropic, hence <— fi> 1<—a> 1

So a =

is isotropic. So there is an x in (a, fir with x + (30 and Q (x) 0; this x has N x = 0 and is therefore not invertible. So (a, fi) cannot be a division algebra. Hence by Wedderburn's theorem (a, i3) is algebra isomorphic to M2 (F), hence it is algebra isomorphic to (1, — 1) by Example 57:4. q. e. d. 57 : 10. Given a, #, y, A, z EP. Then we have the following algebra isomorphisms:

(1) (1, a)

(1, —1)

(a, — a)

(a, 1 — a),

(0e, fl)

(0c 22, 13 to), (3) (cc, « /3) (cc, — 13) (4) (cc, fi) OP (cc, y) (cc, /3 y) 0 ( 1 ; —1 ). (2) (fi, oc)

10*

148

Part Two. Abstract Theory of Quadratic Forms

Proof. The first three parts follow immediately from Propositions 57:8 and 57:9. The fourth part is a little harder. Let 1, x1 , x2, x3 be the defining base of (a, fl), and let I, y2 , y3 be the defining base of (a, y). Consider the resulting base of sixteen elements of the form xi oy i (with x o = 1, yo = 1) for A = (a, p) ø (a, y). Define X = F (10 1) + 1; (x1 01) + F(x2 0 y2) + F (x3 0 y2)

Y — F (101) + F(10y 2)

F (xi y3) +

(— y (xi Oyi)) .

An easy application of Proposition 57:3 gives X -212 (cc, /3 y) ,

Y

( y, — a2 y)

( 1, — 1 ) .

The basis elements used in defining X commute with those used in defining Y, hence X commutes element-wise with Y. Also X and Y contain the identity 101 of A. And X and Y are central simple since they are quaternion algebras. By Proposition 52:2 the algebra generated by X and Y is isomorphic to Xø Y. Comparing dimensions gives Af...f Xe Y as desired, q. e. d. 57:11. Remark. For trivial reasons we can replace the isomorphism symbol by the similarity symbol throughout the statement of the last proposition. We also have (a, /3) 0 (a, 7i) —

/3 y)

since (1, —1) splits by Example 57:4. If we take

p

y we obtain

(a, /3) 0 (a , /3) — 1 so that (a, p) is isomorphic to its own reciprocal. § 57C. The rotations of (,/J)0 57:12. Let u be an anisotropic pure quaternion in the quaternion algebra (a, )3) over F. Then u is invertible and Tu x =

u x u -1 V x

E (cc,

,8)°.

Proof, u is invertible by Proposition 57: I. Then for all x we have 2B (x, u) (u) = x (xa

X X

E (cc,

IV

It

(Nu) - iu

x (x u u x) (N u) --1 = — uxu -1

57:13. Let y be an invertible quaternion in defined by p)0 au x yxy - i V x c

fi).

(1. e. d. Then the map Cf

Chapter V. The Algebras of Quadratic Forms

149

is a rotation of (a, fir . Every rotation of (a, IV has this form. And 0( a,,) = Q(Y). 1 Proof. 1) Clearly ay is an F-linear injection, hence bijection, of (a, fir

onto itself. It is an isometry since N (ct x) = N (y x y -1) = (N y) (N x) (N y) -1 — N x for all x in (cc, i3)° • 2) Next we prove that cr„ is a rotation. Suppose not. Then a, is a product of three symmetries of (a, fi)°, hence by Proposition 57 :12 there is an invertible quaternion y such that yxy - i. — vxv -1 V x E (a, . Hence there is an invertible quaternion w such that wx = xw V x E (a, 13)°. Write w wo with e0 E F and wo pure. Then taking x = wo we get ww o = — w o w. But wwo= (4 1 + wo) wo— wow • Hence wwo = O. But w is invertible. Hence wo . O. Hence w = 41. Hence (41) x = — x (41) for all x E (a, fi)°. Hence eo. OE Hence w = O. And this is absurd since w is invertible. 3) A typical rotation a of (a, fl)° is a product of two symmetries cr = t„,tn, with u1, u2 pure. Then by Proposition 57:12 we have ox (u1 u2) x (u1 u2) -1 for all x E (a, fi) 0. Hence a ----- (ruin.. So every rotation of (oc, fl)° has the desired form. 4) Finally we must compute the spinor norm 0 (cry). Write my = ;girth ul , u2 pure. Then with x (142u2) -1 V x E Yx3n1 = fir Hence (ui u2) -ly is in the center Fl. Hence y a ul u2. So Q (y) 1 = Ny= oc 2Nu1 Nu2 = a2 Q (ui) Q (u 2) 1.

But 0 (ay) =Q (u 1) Q (u2). Hence 0 (ay) = Q (y) .

q. e. d.

§ 58. The Hasse algebra Our purpose in this paragraph is to study an invariant called the Hasse algebra of a regular quadratic space. So let us consider a regular n-ary space V over the field F. Take an orthogonal base for V and fix it for the moment. Suppose that Va2 1— "I This result is the basis of the 3-dimensional part of the proof of the Strong Approximation Theorem for Rotations (§ 104).

150

Part Two. Abstract Theory of Quadratic Forms

in this base. We define the Hasse algebra SFIT =

(cti,

where di = x1 . . . ai. When there is just one field under discussion we

write S V instead of SF V. The Hasse algebra is central simple since it is a tensor product of quaternion algebras. We can therefore apply the theory of similarity of algebras to it. But before we proceed further we must be sure that S V is uniquely determined, at least up to an algebra isomorphism. 58: 1. Lemma. Let X0 and X* be two orthogonal bases for V. Then there is a chain of orthogonal bases

Xo • • 5 -,'-. Xi-2. ->--- 9Ci ' ' -0-- X* in which X i is obtained by altering at most two adjacent basis vectors of X i_ l. Proof. We can assume that n> 3. Put X,i,= bli, • • • 15inj•

1) First we prove the following: there is a chain X0 --)-- • • • of the required type in which yl. is the first basis vector of X. Of all bases % which can be obtained by such a chain we choose an 9C in which :y1 has most coordinates O. In fact we can take X = {x1,. . ., xn} with yi = %xi + • • • ± an xn where oci + 0 for 1 I fi. If fi -=, 1 we are through. We will derive a contradiction for _I)> 1. If p 3 we cannot have { Q (ceixi) + Q (cc2x2) = O

Q (cc2x2) ± Q (oc3 x3) . 0 Q (0e3x3) + Q (cc-ixi) — O,

for then we would have Q (a 3x3) — 0 after a suitable elimination. We can therefore assume that Q (cci xi) ± Q (a2 x2) + 0 when p 3. For p ---. 2 this is automatically satisfied since Q (yi) + O. Defifie g1 =

Gelx1+ oe2X2 ,

22= X2

B pl. , zsi) , Q (k) xi. •

This gives a new base in a chain of the required type with a smaller p.

This is a contradiction. Hence p = 1. 2) The lemma can now be proved by induction to n. Run a chain of the required type from 2C0 to a base X= {hi z2, . . ., zn}. Then (Fz2) ± - • • ± (Fzn) = (Fy2) I • • • ± . (Fyn) ,

so we can run a chain of the required type from {zo, .. . , .z„} to fyo, ... , y,j, hence from 9C to 94, hence from 9C0 to SV* . q. e. d.

Chapter V. The Algebras of Quadratic Forms

151

58:2. S V is well-defined, at least up to an algebra isomorphism. Proof. We shall use the rules established in Proposition 57: 10 throughout the proof. If n . 1 the result is immediate. So assume n> 1. The

lemma shows that it is enough to compare S V in two orthogonal bases x, and x;, . . . , x. in which x,1 -=.- xl for A+ i, i + 1 where i is some fixed integer with 1S iS n— 1. So here we will have aa = a'A for A+i,i+ 1 where al . Q (x,1) and al . Q (xi). We also have Fxi IFxj+1 = Fxi IF 4+1 so that cil = cil for 1 ..‹. A < i and A E dil i'2 for i + 1 ..‹. À • to show that 0 (ai, di) -.':- 0 (al, d3 . 122.

n. We have

15.irt

This reduces to proving .(ai, di) 0 (ai+i , di+i) -Al (o4, 43 0 (a4 1 , d 1). A simple calculation using the rules shows that the left hand side is similar to (ccicci+1 , — di-1) 0 (ccii , oci+i) Similarly with the right hand side. So we are reduced to proving (cti , cci+1) '' (4 04+i) -

But the Clifford algebra of F xi I F xj+1 . F xii, ± F 4+1 is isomorphic to both (ai, ai+i) and (04, a'i+i ) by Example 57:5. Hence (ai,aii_ i) -.-• (oc, 4+1). Hence S V is well-defined up to an isomorphism. q. e. d. 58:3. Remark. Now that the Hasse algebra is well-defined we should give some rules for operating with itl. So consider our regular n-ary space V over F. If V has a splitting V -_- < al > 1 • • • 1_ < an > (cci E F) , then 0 ((Xi, 0 ) . 1.gin

For a non-trivial splitting V-- UI W we obtain S V , SU 0 (du, dW) 0 SW;

here dU denotes the discriminant of U or, strictly speaking, a representative of the discriminant in F, similarly with dW. If we scale V by a 1 See E. WITT's paper on quadratic forms in Crelle' s J. 176 (1937), pp. 31-44, for the connection between the liasse algebra and the Clifford algebra.

152

Part Two. Abstract Theory of Quadratic Forms

non-zero a in F we find S

(a, (— 1)71 (+ i)/2 (d V) + 1)

SV .

The gehavior under E-ification is described by S(EV)-- ES(V). If V is ternary we have (— 1, — 1) SV (—

cc, as) .

If V is quaternary we have (— 1, — 1) S V ( — a1 oc2 , — ma) e (a4, dV) . 58:4. Theorem. V and W are regular n-ary quadratic spaces with 1 n 3. Then V is isometric to W if and only if dim V = dim W , dV = dW , SV SW . Proof. We need only prove the sufficiency. For n = 1 it is trivial, in fact the discriminant is enough. Next let us do the ternary case. Let dV = a with ccEP. Then S Vac— S WŒ by Remark 58: 3, and dVcc dWac = 1. In effect this allows us to assume that dV = dW = 1. Hence we have splittings <—cc>1 <—fi> I W <—y> I <-6> I <1 6> for suitable a, fl, y, 6 EF. . Hence V (a, fir , and

(y, 6)°. On the other hand, if we apply Remark 58:3 to the above splitting for V we obtain (— 1, — 1)e S V ( — a 13, fl) (a, 13) W

and similarly

(— 1, — 1)

SW

(y, (3 ) . Hence (a, /3) (y, 6). Hence (ac, IV and (y, (3 )° are isometric by Proposition 57:8. So V is isometric to W. The binary case follows from the ternary case. If V and W have the same invariants, then so do VI< 1> and W_i_< 1>. Hence VI <1> and WI< 1> are isometric by the ternary case just proved. Hence V W by Witt's theorem. q. e. d. 58:5. Example. Over certain fields the invariants dim V, dV , SV are enough to characterize V under isometry even when n 4. However this is not true in general. For instance the real quadratic spaces V

4 j_ <1>

4

and W ± <-1>

have the same invariants but are not isometric.

Chapter V. The Algebras of Quadratic Forms

153

5 8 : 6. A regular ternary space V is isotropic if and only if S V (-1,-1). Proof. V is isotropic if and only if it is split by a hyperbolic plane, j. e. if and only if V <1> j. <-1> I. < dV > . 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 ± <-1> 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 <—1>. V <1> I ••• I <1>

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•••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) .

Part Three. Arithmetic Theory of Quadratic Forms over Fields

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>1•••±<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_ <6> 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>1 <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 <1> _I_ <1>. q. e. d. 62: la. A regular quadratic space V over a finite field F has a splitting

V .7'f <1>

<1>

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

158

Part Three. Arithmetic Theory of Quadratic Forms over Fields

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.

Chapter VI. The Equivalence of Quadratic Forms

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

Part Three. Arithmetic Theory of Quadratic Forms over Fields

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

Chapter VI. The Equivalence of Quadratic Forms

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

Part Three. Arithmetic Theory of Quadratic Forms over Fields

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 <1 and orda odd. Then b(e) = ao. Proof. Put ow = pd. Clearly b (s) S pd. Suppose if possible that b (e) ç p4 + 1. Then there is a y Eo such that 1

oc (1 ± y) 2 modad+ 1 .

So 1 y (y 2)1 = loci by the Principle of Domination since Iciel = 1701. 141, contrary to the assumpIf we had I Y I 5- 121 we would have loci tions; if we had ly1 > 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

Chapter VI. The Equivalence of Quadratic Forms

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

Part Three. Arithmetic Theory of Quadratic Forms over Fields

§ 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 <671> 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 16> 1 <67c> 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

Chapter VI. The Equivalence of Quadratic Forms

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 <87c> 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 <7i> 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

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

Part Three. Arithmetic Theory of Quadratic Forms over Fields

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

Chapter VI. The Equivalence of Quadratic Forms

take a =

7C. If

167

40C yo Co take a = — p; then

± --A _L <4 a /I> ; 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 <4. SU., SV.

From Remark 58:3 we get the formulas Sp V = fi ((xi, ezi) 1ZiZiZn

Sp

(

Sp ITŒ

Sp - (dU, dW) • S p W ,

(a, (— 1)" ( 1 ÷ 1)12 (d V)n + S p V

63:14. Example. Let V be a quadratic space over a non-dyadic local field with I

•••±

<en >

where all ei are units and n 3. Then V is isotropic. For we can assume

168

Part Three. Arithmetic Theory of Quadratic Forms over Fields

that n = 3. Then SP V = 1 —

(-1, —1 )

Now apply Proposition 58:6. § 63C. Quadratic forms over local fields We are ready to characterize quadratic forms over local fields. The invariants will be the dimension, the discriminant, and the Hasse symbol. The representation problem will be solved by reducing it to the problem

of isometry. By Proposition 63:4 there is a unit of quadratic defect 4o; we fix one such unit and write it A. Let V denote a regular n-ary quadratic space over F. We start the discussion by gathering together some important information about binary spaces. 63: 15. Examplel. In this example V will be a regular binary quadratic space over the local field F. (i) If d V then Q (V) is either uFs or Dt uF.2. For we can assume by scaling that 1 E Q (V). Then V -s-_-. <1> ± < —A >. So for any unit e the quadratic space < e>± <1>j< d> is isotropic by the criterion of Proposition 58:6. Hence V represents E. The same sort of argument will show that V represents no prime elements at all. Hence Q (V) is precisely uFs. (ii) If V is not isotropic and 1 EQ (V), then Q (V) is a subgroup of F with (F Q (V)) = 2 . To prove this we write V <1> ± < a> and construct the group —

—

—

homomorphism Cie,

Y ->-

y

)

of F into {±1}. It is easily seen that Q (V) is the kernel of this homomorphism. And the scalar a is a non square since V is not isotropic, so the homomorphism is surjective. Hence (F: Q ( .1)) = 2 as asserted. (iii) If W is another regular binary space over F, then -

V W if and only if Q (V) =Q (W) . We have to show that V W if Q (V) = Q (W). We can assume that neither V nor W is isotropic. By scaling we can assume that V and W both represent 1. Then V <1> ± < — a> , W <1> ± <— f9>. 1 The results of § 63, particularly of this example, are closely related to the local class field theory. For a discussion of local class field theory we refer the reader to O. F. G. SCHILLING, The theory of valuations (New York, 1950).

Chapter VI. The Equivalence of Quadratic Forms

Now v is in Q (f7) if and only if

169

(7 — 1, and the same with W, hence VYE

(iLl) = (7)

hence a fl is a square by Proposition 63:13, therefore V W. (iv) Suppose neither V nor W is isotropic. Then Q(W) VW• For we can assume by scaling that 1 E Q (V) Ç Q (W). Hence (fi:Q(T))=--- 2 and (E: Q (W)) = 2. But Q (V) Q (TV), so that Q(J2) Q (W). Hence V is isometric to W. 63:16. Example. Let E be a quadratic unramified extension of the local field F. We claim that NE/FE uF2 . Since E IF is a quadratic extension of a field of characteristic not 2 it can be obtained by the adjunction of a square root; since E IF is unramified there is actually a unit LI in F such that E = F 21) and this unit must have quadratic defect 4e by Proposition 63:3. Now NE/F E is the set of non-zero numbers represented by the quadratic space <1> _L <—A>. This set is equal to ufr2 by Example 63:15, so we have proved our claim. Note that by Theorem 14:1 the norm NEIroc of an element a of E is a unit of F if and only if a is a unit of E. Hence Q(V)

= u,

where U is the group of units of E. 63: 17. Let V be an anisotropic quaternary space over a local field. Then V-2= <1> ± <—.%1> 1 1<-7cd> where A is any given unit of quadratic defect 4e. Proof. 1) Suppose we can find a binary subspace U of V with Q (U) S uF 2 ; then Q (U) = uF2 by Example 63:15, hence Q (U*) S nuF2 since V is anisotropic, hence V has the desired form by Example 63:15 again. The same holds if we can find a U with Q (U) S uP'2. 2) Take an arbitrary splitting V = W I W* into binary subspaces. We can assume that W represents some units and some prime elements, else we are through by step 1). The same with W*. From this we can conclude that V has a splitting of the form V

<el > 1 <E2 > 1 <6> 1 • • •

with el, e2, e3 in u. A double application of Proposition 63:11 gives a splitting V 1 < -4 2> ±" •

170

Part Three. Arithmetic Theory of Quadratic Forms over Fields

in which LI, is one of the elements LI and 1, and the same with LI 2. This gives us a binary subspace U with discriminant —LI, hence with Q (el) = uPa by Example 63:15; we have therefore found a binary subspace of the type mentioned in step 1) and we are through. q. e. d. 63:18. Remark. thus we have essentially one anisotropic quaternary space over a local field, its discriminant is 1, it represents all units and all prime elements, and so it is universal. Hence all regular quaternary spaces over a local field are universal. There are two regular quaternary spaces with discriminant 1, the hyperbolic space and the anisotropic space. 63:19. A quadratic space V over a local field is isotropic if dimV 5. 1 Proof. We can assume that V is regular with dim V = 5. Take a splitting V U j W in which U is a line and W is quaternary. Then —Q (U) S Q (W) since W is universal by Remark 63:18. Hence V is isotropic. q. e. d. 63:20. Theorem. Let F be a local field at the prime spot p. Then two regular quadratic spaces U and V over F are isometric if and only if dim U = dim V, dU dV, S p U = Sp V Proof. The equality of the Hasse symbols implies the similarity of the Hasse algebras. Now every regular quinary space over F is isotropic by Proposition 63:19. Hence Theorem 58:8 applies. Hence U is isometric to V. q. e. d. 63:21. Theorem. Let U and V be regular quadratic spaces over a local field with v= dim V — dim U 0. Then U is represented by V if and only if

U z- V V U H V

when v = 0 , when v. 1 , when v = 2 ,

dU = —dV.

Here H denotes a hyperbolic plane.

Proof. The necessity is obvious. So is the sufficiency when y = 0 or v = 1. Next consider v 3. Take splittings U

with cci,

< cci > , 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

Chapter VI. The Equivalence of Quadratic Forms

171

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

• ± < oc>

<

1 • •

Hence U -3-- V. To conclude we must consider y = 2. If dU = —dV we are through by hypothesis. Hence assume that dU+ —dV. In virtue of the case y 3 we have U V H where H is a hyperbolic plane. Hence for some quaternary space W. The assumption about the discriminants gives dW 1. Hence W is isotropic by Proposition 63:17. Hence we .^- H. But this have a splitting W = W1 1 H1 with W1 binary and H1 -.4 implies that U j W 1= V by Witt's theorem. Hence U-).-Vas asserted. q. e. d. 63:22. Theorem. Let V be a regular quadratic space over a local field F. Then there exists a quadratic space V' over F with

dim V' = dim V , dV'. dV , S p r — S p V if and only if V is neither a line nor a hyperbolic plane.

Proof. Clearly all lines with given discriminant are isometric and hence have the same Hasse symbol. This also applies to hyperbolic planes. So we must prove the converse: given that V is neither a line nor a hyperbolic plane, try to change the Hasse symbol without changing the dimension or the discriminant. Define ,

I <7;ZI>

where 4 is, a unit with quadratic defect 4o. Then there is a representation U' U V

by Theorem 63 : 21. Hence there is a quadratic space V' with V' U IV. Clearly V' has the same discriminant and dimension as V. An easy computation with Hasse symbols shows that Sp V' — S, V. q. e. d. 63:23. Theorem. The necessary and sufficient condition that there

exist a regular quadratic space V having invariants

no = dim V , do= d V ,

so= Sp V

over a local field F, is that these quantities satisfy the relation sSn—II d '"

P

\

)

when no = 1, and when n o = 2 with do E —frA. (Here it is assumed that

no ENido Eg, so = ±1).

172

Part Three. Arithmetic Theory of Quadratic Forms over Fields

Proof. The necessity follows from the definition of the Hasse symbol. Let us do the sufficiency. The spaces and <1> 1 <— 1> will work in the exceptional cases. Otherwise, we take any space V with the required dimension and discriminant, say V<1>±•••1<1>i.

Now by Theorem 63:22 we have a space V' with the same dimension and discriminant, but with opposite Hasse symbol to V. Then either V or V' q. e. d. does the job. 63:24. Example. The following result will be used later in the global theory: let V be a regular quaternary space over a non-dyadic local field with dV a unit and .54„ V = 1, and let U be any regular ternary subspace of V, then Q(U) D u. Scaling V by a unit shows that it is enough to prove that 1 E Q(U). By Theorem 63:21 we are sure that U will represent 1 whenever dU+ 1. In the exceptional case we can write V <— 6> I. U where 6 is a unit for which d V .= 6; a computation of Hasse symbols the gives Sp =( —1' —1 ) • hence U is isotropic, hence P it is universal, hence it represents 1. —

5 64. Global notation More definitions are needed before we can proceed with the global theory. Let F be a global field and let D = Di, be its set of non-trivial spots. Recall from § 33D that hi, denotes the group of idèles of the field F. In § 331 we introduced the subgroup of principal idèles Pp, the subgroup of S-idèles J„ and the subgroup P1,= j; n Pp. Here S can be any subset of D but in practice it is usually taken to be a Dedekind set consisting of almost all spots in D. We define IC..; to be the set of idèles i such that lip lp .1 VP ES lip =1 VpED—S.

LI.

Clearly Kis, is a group, in fact a subgroup of (Here I I is the normalized valuation at p, as usual.) Given a spot p E D we define to be the set of idèles i such that

is

iq ---- 1

VcrED—p.

Again .4, is a group. It is a subgroup of J, if and only if p is a natural isomorphism Fi,

S. There

obtained by sending a typical a E Pp to the idèle in I with p-coordinate equal to a and all other coordinates equal to 1.

Chapter VI. The Equivalence of Quadratic Forms

173

Now let EIF be a finite separable extension. For each pair of spots 93 p on EIF we have the norm map Nvit, : E,T3

Fp

defined in § 15B. Consider a typical idèle g in JE. Then by § 33A we have Hence

INVIP B93 ip

Ig43143 •

NTIP B93)pEsap

is an idèle in J. We call this idèle the norm of g and write it This gives a group homomorphism NEIF: JE > JE

NE/Fa.

•

Similarly we obtain the following: let S be a set of spots on F and let the same letter S also denote the set of spots on E for which SIIS; then It is an immediate consequence of the definition of the norm NE1F and of Theorem 15:3 that the diagram E

t JE NEI,"

NE!?

F

is commutative. In other words, the norm of a field element is essentially the same as the norm of the principal idèle which it determines. In particular, NE1FPE P7. For a tower of finite separable extensions FSEç 1-1 of the global field F we find (after looking at a suitably elaborate commutative diagram) that Nrup g = NEy(NH/Eg) for all g hence NHIFIFI NEIFJE•

§ 65. Squares and norms in global fields The purpose of this paragraph is to obtain information about squares and 'norms in global fields. This will be done by specializing the methods and results of global class field theory to quadratic extensions of global fields'. The proofs are of a technical nature involving several auxiliary computations and it would be too cumbersome to repeat our assumptions In the development of the global class field theory one obtains the main results of § 65 (with considerably more effort) for certain extensions of degree n 2. See C. CHIEVALLEY, Ann. Math. 41 (1940), pp. 394-418.

174

Part Three. Arithmetic Theory of Quadratic Forms over Fields

in the enunciation of each proposition. Instead we shall list in the following introductory subparagraph some assumptions and conventions that will remain in force throughout § 65. § 65 A. Notation and conventions F is a global field, D = DE is its set of non-trivial spots. Completions Fp with normalized valuations I lp are taken and fixed at each p ED. S will denote a set of spots on F which satisfies the following properties: (i) every p in S is discrete and non-dyadic, (ii) S consists of almost all spots on F, (iii) S is not all of D, (iv)

JF = PEP; •

Every global field possesses a set S with these four properties in virtue of Corollary 33:14a. The number of spots in Q — S will be written s; so 1

s < 00 .

The group Jig, is generated by the following subgroups in the following way: K;1( 17 IP JF wED-s u will denote the group of S-units (S) defined in § 33F, in other words u {cc EP I kip= 1 for all p E S} . For each discrete spot p E D we let op be the ring of integers and up of units of F,, at p (see § 13B). An element a of F (or ofFr) bethgroup is said to be a square, or an integer, or a unit, etc., at p if it is a square, integer, unit, etc., in Fr . An idèle i is said to be a square, or an integer, or a unit, etc., at a spot p if ip is a square, integer, unit, etc., at p. For example 4, consists of those elements of J7 which are units at all p in S. lip is defined to be the set of S-idèles which are squares at all p E Q — S. This is a subgroup of J, and in fact JP 2 = KI‘(.* 2 . We shall often use the fact that F is an open subset of Fr. At an archimedean p this follows from the ordinary properties of the real and complex numbers, at a discrete p it is Corollary 63: lb. 0 will denote an arbitrary non-square in F and E will be the quadratic extension E =F(Vr)) of F. This extension is separable since the characteristic is not 2, so the theory of § 15 can be applied to EfF. Bar will stand for conjugation in E: cc+ fill0 cc— fillY Vce,flEF• The mapping defined by 11 -4— r is an automorphism of E, and this automorphism plus the identity automorphism of E are the two elements ,,

175

Chapter VI. The Equivalence of Quadratic Forms

of the galois group of Ell'. Thus

FP forai! P EE . E is a global field with its own set of spots DE. The valuations! I v will be normalized valuations on the fixed completions E93 at each 93 E QE. And S' will be the set of spots on E such that Sill S. We shall also use S to denote the set S'. 11 denotes the group of S-units on E, j. c. U = u (S'). For each discrete spot 93 E s2E we let 093 be the ring of integers and 2,1cp be the group of units of Ev at 93 • For any a in /I we have INE/Falp= HINTIP x NE/FF

93 1P

TIP

for each p in .5, hence NER,111.C11C

11

(compare Example 15:6). If we consider the commutative diagram of § 64 we find AVE/F

L

E

hence we have the commutative diagram ISTEIF

NEIF

u

Consider the spot p on F. The extension EIF is certainly abelian, hence by § 15C the local degree n (931p) is the same for all 93 on E which divide p, and the common value of these local degrees is written np . If 0 is a square at p, then there are exactly two distinct spots 931 and 932 on E which divide p and we have n (93ii P)= n(321p)

np = 1

If 0 is a non-square at p, then there is exactly one spot 93 on E which divides p and we have n (TIP) = n,= 2 . Similarly if p is discrete we know that the ramification index e (93 I p) depends only on p and not on 93, and it is written ep. Of course ep is also the ramification index of the extension E93/1793 . Similar remarks apply to the degree of inertia 4 f(3 Fr). So if np = 1 we have ep = fp = np = 1 ,

while if n= we have cp —. 1, fp np = 2 ,

or

P

P

4-- 1

-

176

Part Three. Arithmetic Theory of Quadratic Forms over Fields

We say that EIF is unramified at the discrete spot p if

ip

et,. 1,

.

Thus Ell,* is unramifiea at p if and only if ET/FT is an unramified extension of local fields at each 93 dividing p. Note that EIF is unramified at p whenever the local degree n p is 1. If np . 2 there is a single spot 93 dividing p, so for any P E E we have FIT — 'NovPI= 1NE/F ri p , hence

TIT vrEE

1/113

On the other hand if n p = 1 there will be two spots and here.it can be verified that

q3,,(1.3 2 dividing p,

1/1931 , 1r1,2 VF EE. The set of spots 12 — S on F is broken up into two disjoint subsets A and B as follows: A ={4)(12—Sin t,— 1}

B

E — S np = .

Then S2 is the disjoint union S2=Sv A i B. We let a denote the number of spots in A, and b the number in B. We have aa 0,

b_1:),

We say that an element a of F (or of Fp) is a local norm of the extension ELF at the spot p if ŒENT13;pE13 VTIP• Thus every element of F p is a local norm at p when n p = I. In general we have the formula if n_ 1 (Pp N1 if np — 2 {

for any (P I p, in virtue of Corollary 63:13a. We easily see that a E Fr is a local norm at p if and only if the Hilbert symbol on F 0)

k

- -1.

We say that the idèle i is a local norm at p if i p is a local norm at p. Thus i is a local norm at p if and only if 1

And we have the formula

'p '

n NElp

{1 2

if np . 1 if np — 2.

177

Chapter VI. The Equivalence of Quadratic Forms

65:1. Example. Let p be a discrete non-dyadic spot at which 0 is a unit. Then EfF is unramified at p by Proposition 32:6. And by Example 63 : 16, N93 1 p

p

65:2. Example. Let t be a given idèle in Jp. We claim that i is in NEnrjE if and only if it is a local norm at all p for which tap = 2. The condition is clearly necessary by the definition of the norm of an idèle. Conversely let it be given that i is a local norm at all p for which np = 2. At each of these p take 93 I p on E and 393 E ET such that Ngp 1p 39? ip ; then 1393 193 = IN,431 „ 393 1p = Iip lp, so that 343 193 = 1 for almost all these 93. The remaining p are the ones at which np = 1; here there will be exactly two spots 93,93' dividing p; take 393 E Ev such that Nvip j = ip and B93, 1; again we have 1393 h3 . 1 for almost all these 91 Hence 3= is an idèle of E, and NEIF 3 = I. This proves our assertion.

(a,), EQ,

Incidentally note that the 3 just constructed has the following property: 1343 193 is either iiplp or 1 whenever 93 I p. 65:3. Example. Next observe that NE,,, J..; is the set of idèles i in jf, such that I is a local norm wherever np = 2. For every element of NE/ilig3 is an element of this sort by § 64 and Example 65:2. Conversely if i is an idèle of this sort, then the idèle 3 constructed in Example 65:2 is in and has NEIF I. 65:4. Example. Suppose 0 is a unit at all p in S. In this event NE/FJI is the set of idèles I in jg, such that i is a local norm at all p E 12 S for which np . 2. (Use Examples 65: 1 and 65:3.) § 65B. Index computations

a

6 5 : 5. Let G be a subgroup of P. Suppose that G is a finitely generated abelian group of rank r. Then (G: G 2) is 2' +1 when —1 is in G, otherwise it is 2.

Proof. By the Fundamental Theorem of Abelian Groups and one of the isomorphism theorems of group theory we find (G : G2 ) = ( Go : GS) (G1 : G?) . . . (G7. : G!,.) where Go is a finite group consisting of the elements of finite order in G and where the remaining Gi are infinite cyclic groups. Hence (G: G 2) = , 2'. (Go : G). It remains for us to compute (Go : Gg). In order to do this we consider the surjective homomorphism 9): F F2 defined by 9)x = x2. This has kernel (± 1) since F is a field, and these two elements are distinct since the characteristic is not 2. Hence the kernel of the restriction (p: Go, Gg is ( j- 1) if — 1 is in G, otherwise it is 1. In the first instance we obtain (Go : 1) = 2(G0 : (± 1)) = 2(GS : 1) ,. O'Meara, Introduction to quadratic forms

12

178

Part Three. Arithmetic Theory of Quadratic Forms over Fields

and so (Go : Gg) = 2; in the second instance this argument gives (G0 :G(i)=1. Hence (G: G2) has the value stated. q. e. d. 65:6.

:

(n) 2) = (u : u 2) = 23.

The first equation is simply the fact that the natural isoPF carries u to P. The second equation follows from morphism F q. e. d. the Dirichlet Unit Theorem and Proposition 65:5. Proof.

6 5 : 7. (n, : 2) 4 a. Proof. 1) Consider the surjective homomorphism 7S JF

F PED—S

defined by suppressing the coordinates of an idèle at all spots in S. The kernel of this homomorphism is the group K, and K:;`, Ç 4. 2 Hence by the isomorphism theorems of group theory (j.:, : 4' 2) is the product II (Fp : Fi). We must compute this product. pED-8

2) In the function theoretic case all spots pED—S are discrete and 4 by Proposition 63:9, hence the above non-dyadic, hence (Fr : product is 41 as required. 3) In the number theoretic case if we let d, n, r, c denote the number of spots in Q — S that are respectively dyadic, non-dyadic, real, complex, we obtain p€D-S

(F :F2) = 4a4n2r P P

17

(Np)° P 2

dyadic

= 4 4 ±n 2r 17 12LT1 dyadic

4a -En2r 17 121r arch

---= 4d + n2r 2r 22c ___ 4d +n-Eri-C _48

q. e. d. 65 : 8. 1/ 0 is a unit at all spots of S, then (4: N Ban) =2b Proof. Again consider the surjective homomorphism

Fp pED—S

with kernel K. By Example 65:4 we have K.", c NE pl c J . Again by Example 65:4 we find that the image of NEIF Pi under the above homomorphism is ,F

H P,, PEA

X

H

pE B

I p :E4T3

Chapter VI. The Equivalence of Quadratic Forms

179

So by the isomorphism theorems of group theory we obtain : NEar

= (F p : N vi r 4) = 2b. pEB

q. e. d. 65:9. Herbrand's lemma. Let 0 be a subgroup of finite index in the commutative group G. Consider homomorphisms N, T of G into G such that NTG=TNG=1,

NO SO , TO SO.

Then (G T :NG) (GE : TG)

(O r :NO) (45E:T)

provided the indices on the right are finite. Here TG is the image of G, and GT is the kernel of T. Proof. Clearly NG S GT and NO S OT. By Lemma 63:7 we find

(G: 0) (T: N) = (T G : T 0) (G T : OT) (OT : N) = (TG: TO) (G T : N 0) (T G : T 0) (G T : N G) (N G : N 0) . Interchange N, T to obtain another such equation. Divide. 65:10.

Suppose O

q. e. d.

is a unit at all p in S. Then b land :

NEIF

= (u : NEIF U) = 2b-1.

Proof. 1) The first equation follows from the commutative diagram

in § 65A. We therefore have to prove the second equation and the fact that b 1. Let T : É E be the homomorphism defined by T = Fir V FEt. Throughout this proof we use N for NE/F. Then N : E E is a homomorphism with NI-7 =Fr V FEÈ. We have Nx= x2 , Tx=1 V x Ei hence

I

Tx =x2 , Nx=1 V xETt , NTÉ= TNÉ=--1.

Also TUçU,

NUSu.

2) By the Dirichlet Unit Theorem we can express u as a direct product of a finite group uo with a free commutative group u, of rank 12*

Part Three. Arithmetic Theory of Quadratic Forms over Fields

180

s

I. We define the subgroup 0— th(TU)

of it. It is easily verified that u1 n T U = 1, hence 0 is the direct product of its subgroups th and T U. For any F E it we have f2= (N P) (TE), hence 112 (N

(T

;_ u (T it) = u0 0.

Then

(it :

— (U: NO) (NO : 0)

(U: /12) (u 0 : u 0 r 0) .

Hence by Proposition 65:5 and the fact that tt o is finite we obtain (11 : 0) < 00 .

3) So 0 is a finitely generated abelian group of finite index in the finitely generated abelian group it; from this it follows easily (using the fact that /V it for some i) that rank 0 = rank it. But it has rank a + s 1 by the Dirichlet Unit Theorem. Hence rank 0 = a + s 1. But rank th= s 1. Hence rank T U = a. We are going to apply the Herbrand Lemma to the group it and its subgroup of finite index P. All but the finiteness conditions are evident; so we have (UT NIA) N 0) (16 ni) (ON : T —

—

—

:

:

:

provided the indices on the right turn out to be finite. Now Tu/ --- 1; and if x is a typical element of TU we have Tx x2, so Tx = 1 if and only if x = +.1; but — 1 T(VO ) E T U; hence T ( +1) th. Clearly N 0 Nu l u = Hence by Proposition 65:5 we have —

—

((PT N = 28 .

A similar argument gives ON = T U, TO — (T it) 2, and then (ON : T = 2° + 1

.

Hence we do indeed have

(UT : N = 25-1 (UN : TU)

.

But 11T is clearly equal to u. So if we can prove that (U N T it) = 1 we will have (u : N 21) = 2b -1 and hence b 1. 4) It remains for us to prove that UN Ç T U. So we must show that a typical H in UN is of the form II = T Z for some Z in U. If H = 1 we have 1+H + T(1 + H) , H • ± -- H + —

Chapter VI. The Equivalence of Quadratic Forms

181

with 1 + H E É, since HR = NH 1; if H = —1 we have T( ) --1 H. So we do indeed have Z E E with H=TZ, but this Z need not be in U and a further adjustment is necessary. For each p in S we let93' denote one of the spots on E which divide p; if there is another spot on E which divides p, let it be written 93"; otherwise let93" be93 1 . Now EIF is unramified at each p E S by Example 65:1, hence IF143,, hence there is an ap in F such that laphy = Now Jr = Ppjf, by the general assumption being made throughout this paragraph, from which it follows easily that there is an a in F such that Mr = Iccplp for all p in S, and so I 93' = av = 143, for all 93'. Put Z/Œ. Then IZŒ193,= 1 for all 93'. For each 93" we have

12 1v/10493' ; but TZ=HsoZ.HZ, so IZI93, -= IZIvi since H E 21; hence 1493 , 1. Hence Z. E U and TZŒ =TZ H. q. e. d. 65:111. Let a EF be a square at all spots in Q — S and a unit at all Izal93-=

spots in S. Then a is a square in F. Proof. Let 0 denote any S-unit which is a non-square in F. Then our theory applies to the extension E F(1). So by Proposition 65:10 the set of spots p in D S at which np = 2 is non-empty. In other words, every S-unit 0 which is a non-square in F is a non-square at one of the q. e. d. spots of D — S. Hence the given a must be a square in F. 65: 1 1 a. P79

,-, 4,2 = (

n) . 2

6 5 : 12. (J7 : P7JP 2) = ( J;T: PW 2) = 22 . ProofP7JP 2) = (PFLI P7JP 2) =

(PFJP 2) PFJP 2)

JA;

PFJP2)

PPP 2)

- (n : jp2)

s

(pi jp2 4,2)

J 2) - 48 -I- (n: (4 2) - 28. = 4a ÷ (11, : Pr

q. e. d.

65:13. Suppose that 0 is a unit at all spots in S, and that JE= PEAThen

1

(Jp: PFNEIFJE) = 2 ( P

NE/A:

)

NEIFPI < 00 .

This result can be obtained from Corollary 65:18a.

Part Three. Arithmetic Theory of Quadratic Forms over Fields

182

Proof.Let N denote the norm NE/F. We know from Proposition 65:8 that (Jf, NJ) = 2b< 00. Hence :

(JP' PFN JE)=(Prn: PEN (PEJ)) = (Pr 4: PFN.1g) UPPFN IP: =

PFN

:

= (J17,: NJ) -I- (PP =

2b 4- ( p;

:

NJ)

N J2s3) .

But 2b1 = (11: N P)=

:

n

r\ NJ) (Pig, r\ N J;:N

Hence (J7 : PE N JE) has the desired value ,

q. e. d.

s_

(J7 :PFNEIFJE)<00 . Proof. Take a small enough set S satisfying the general conditions of this paragraph and the special conditions of Proposition 65:13. q. e. d. 65:14. 2

§65C. Squares 65:15. Global Square Theorem. If an element of a global field F is a square at almost all spots on F, then it is a square in F.

Proof. Suppose if possible that there is a non-square 0 in F which is a square at almost all spots on F. Let S be a set of spots which satisfies all the conditions of this paragraph and is so small that 0 is both a unit and a square at each of its spots. Let E be the quadratic extension E = F (11-19-).W We shall deduce from these suppositions that J,ç PFNEIFJE. 2 by ProposiThis is of course impossible since (he: PFNE/FJE) tion 65:14. So consider a typical I in J.F. By the Weak Approximation Theorem there is an a in F such that a is close to 1 at all p E Q - S, and if the approximations are good enough we obtain a I E 11 for all pEQ—S since F is an open subset of F. Then the idèle (cc ) li is a local square, hence a local norm, at all p E Q — S. But np = 1 for all p in S by choice of S. Hence (a) -il is in NBIFJE by Example 65:2. Hence hICPFNEIFJE. This is the contradiction we are after. q. e. d. We shall call the S-units Cl, , ei. a set of generators of u mod u2 if every element of u can be expressed in the form E

62 (-v E Z, 6

E u) ; clearly the condition vi E Z of this definition can be replaced by the is a set of generators of u modu2 we must condition vi — 0,1. If El' • • • 11 s since (u : 2) = 2s. And by the Dirichlet Unit Theorem there have r

Chapter VI. The Equivalence of Quadratic Forms

183

is always a set of generators consisting of exactly s elements. Suppose . . . , Es is such a set. Then every element E of u can be expressed in the form 6 E u) . E . E8P' (5 2 (yi — 0,1; In fact this representation is unique (when the set of generators of u mod u2 consists of s elements). For if not we would have a representation Es C u2, say, and this would imply that the s — 1 elements formed a set of generators of u mod u2. , Es is a set of generators of Consider a non-square e in u. If u mod u2, then after a slight adjustment to the si we can write E =ElE2...Ej for some j 1. So E, E2, . . . , E s is also a set of generators of u mod u2. In other words, every non-square S-unit can be extended to a set of generators of u mod u2 which consists of exactly s elements. We shall use F (ilu) to denote the field obtained by adjoining the square roots of all S-units to F. Clearly F (11u) F We i , • • . , Ves ) whenever si, . . . , S s is a set of generators of u mod u2 . 65:16. [F (47): F] , 22 . Proof. The proposition is a special case of the following abstract result : let K be any field of characteristic not 2, and let /41, . . . , fit be non-squares in K such that any product of distinct fit is again a nonsquare in K; then

V flt) Ki

[K(11131 , - -

2t .

For t = 1 this result is trivial. Let us prove it by induction to t. If we can i < t) is a non-square in prove that any product of distinct flt (2 K(11 131 ) we shall have

[K (01 , • • , fli) K(01)1= 2t-1 and we shall be through. So consider an element x of the form x

. . flevi

(v,_-= 0,1)

which is a square in K (0 1 ). Then

+ fl V fli) 2 = oc2 + fi2 fl1+ 2 a 13 V for some a, 14 in K. Then a i3 — 0 since x is in K, hence either x is in K2 or x fi is in K2 . In either event we have fltvg Kz x

with pi — 0 or 1. Then v2 — • • — v t = 0 by hypothesis. Hence any product of distinct i3t (2 i t) is a non-square in K (IIfl ) , as required. q. e. d. 65:17. Let e 1 , . . . , es be a set of generators 0/ ti modu 2 . Then there are infinitely many spots p In S such that El F; 2 , E i E F;;

(2 _‹ i

s) .

Part Three. Arithmetic Theory of Quadratic Forms over Fields

184

Proof. If s = 1 apply the Global Square Theorem. Therefore assume that s> 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.

Chapter VI. The Equivalence of Quadratic Forms

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

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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
where (

at 16 ,

) denotes the Hilbert symbol over F.

See N. J ACOBSON, Bull. Am. Math. Soc. (1940), pp, 264-268, for a reduction of the theory cif hermitian forms to that of quadratic forms.

Chapter VI. The Equivalence of Quadratic Forms

187

66 : 1. Theorem. A regular quadratic space over a global field is isotropic if and only if it is isotropic at all spots on F. Proof. 1) We need only do the sufficiency. So consider the regular n-ary quadratic space. V over the global field F. We are supposing that n 2 and that Vp is isotropic at all spots p on F. Fix an orthogonal basis x1,.. . ; xn in which V has the splitting

V= j. • • • I (ai EF) . 2) The binary case. Here dV p -- 1, so — al a2 is a square at all p, hence — al a2 is a square in F by the Global Square Theorem, hence dV 1, hence V is a hyperbolic plane, so it is isotropic. 3) The ternary case. By scaling we can assume that V has a splitting V Li P in which < oc> , P <1> 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

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Part Three. Arithmetic Theory of Quadratic Forms over Fields

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. <,u> 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.

Chapter VI. The Equivalence of Quadratic Forms

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--.'. By the Product Formula we can find a finite set of spots T on F which contains all archimedean and dyadic spots and is such that each ai is a unit at all p E D — T. What is the Hasse symbol Sp V at a spot p in ai — 1, hence Sp V = 1, at each T? By Example 63:12 we have ( p in D — T. So in practice one has to check the Hasse invariant at only a finite number of spots. This example also shows that Vp is isotropic at almost all p when n 3.

190

Part Three. Arithmetic Theory of Quadratic Forms over Fields

Chapter VII

Hilbert's Reciprocity Law The Hilbert Reciprocity Law states that

The major portion of this chapter is devoted to the proof of this formula for algebraic number fields. The formula is actually true over any global field, but we shall not go into the function theoretic case here. The Hilbert Reciprocity Law gives a reciprocity law for Hasse symbols, namely

fisp v

I,

and this can be regarded as a dependence relation among the invariants of the quadratic space V. We shall investigate the full extent of this dependence in § 72. § 71. Proof of the reciprocity law Our sole purpose in this paragraph is to prove Hilbert's reciprocity law. Here, as in § 65, we proceed by specializing the methods and results of global class field theory', which concerns itself with abelian extensions EIF, to the case of quadratic extensions EIF. Throughout § 71 we shall assume that (it) F is an algebraic number field, D is the set of non-trivial spots on F, [ lp is the normalized valuation at p, (ii) O is a given non-square in F, (iii) E is the quadratic extension E = F(14) of F. Thus the assumptions about EIF are those of § 65, and in addition F is now an algebraic number field and not just an arbitrary global field. The idèle notation is the same as before. § 71A. Cyclotomic preliminaries Let us recall some facts about roots of unity. Here we limit ourselves to the special case of the algebraic number field F under discussion; so the group of roots of unity is a finite cyclic subgroup of F. We know that is called an m-th root of unity if Cm--= 1. We call a primitive m-th root of unity if it has period m. In this event 1 , C, • • •, For a discussion of the general theory and of Artin's general law of reciprocity we refer the reader to H. FLANDERS, Unification of class field theory, (University of Chicago thesis, 1949). Also see E. ARTIN and J. TATE, Class field theory (Harvard University, 1960).

Chapter VII. Hilbert's Reciprocity Law

191

is a cyclic group of ni distinct ni-th roots of unity; but the number of distinct m-th roots of unity is at most ni since xm— 1 0 has at most ni solutions in a field. Hence a primitive m-th root of unity generates all the m-th roots of unity in a field. If F contains m distinct in-th roots of unity it contains a primitive m-th root of unity. 71:1. Definition. We call an extension KIF an absolute ni-cyclotomic extension if K = FM for some primitive m-th root of unity 71:2. Definition. We call an extension KIF an m-cyclotomic extension if it is a subextension of an absolute m-cyclotomic extension. 71:3. Example. The extension KIF is an absolute m-cyclotomic extension if and only if K is a splitting field of the polynomial en— 1 over F. In particular, F has an absolute m-cyclotomic extension for any natural number in. 71:4. Example. Let H be an extension field of F which contains a and a primitive n-th root of unity ",.„ primitive m-th root of unity with m relatively prime to n. Then Cm C is a primitive mn-th root of unity. And F (Cm, — F (Cm Cn) • 71:5. Example. A compositum of an absolute ni-cyclotomic extension with an absolute n-cyclotomic extension is an absolute mn-cyclotomic extension when m and n are relatively prime. 71:6. Example. Let C be an algebraically closed field containing F. Then C contains exactly one absolute m-cyclotomic extension of F, namely the splitting field H in C of en— 1 over F. If KIT; is any m-cyclotomic subextension of CIF, then F c K c H. So a compositum of cyclotomic extensions is always cyclotomic. And if L is any subfield of C, then KLIFL is m-cyclotomic. The absolute m-cyclotomic extension K F(") with a primitive nt-th root is a splitting field of the polynomial xm--.1 over F, hence KIT . is a galois extension. A typical element e of the galois group 6 (KIF) is completely determined by its action on C. But 9, C is an m-th root of c t. But then unity, hence for some r. Similarly r ,

eT c,

VT_

.re.

T. Hence the galois group of an absolute cyclotomic extension is abelian. Hence every cyclotomic extension is galois and abeli an. Consider an arbitrary finite extension HIF of F with discrete spots (131p. We say that HIF is unramified at 93 (or at 931p) if 11(4-441p is an unramified extension of local fields at 931 p, i.e. if Hence e t.

,

(93 I P) 1 (T(P) (Of course e (93 1 p) (93 1 p) -= n (93 1 p) holds whether or not HIF is unramified at 93.) We say that HIF is unramified at p if it is unramified at e (93 I P)

192

Part Three. Arithmetic Theory of Quadratic Forms over Fields

al193 dividing p. If HIF is abelian, then by § 15C the local degree n (93 I p), the ramification index e(93 p), and the degree of inertia f(3 p), depend only on p and are written ttp ep and 1, respectively. So an abelian extension HIF is unramified at p if and only if

4= ni ,

ep = 1 ,

,

i. e. if and only if it is unramified at some 93 dividing p. (Of course ep fp = np holds whether or not HIF is unramified at p.) If HIF is quadratic we recover the special definitions given in § 65A. All this applies to cyclotomic extensions since they are abelian. For example we see by Corollary 32:6a that an m-cyclotomic extension is unramified at all discrete spots at which m is a unit. 71:7. Let H be a finite extension of an algebraic number field F. Then HIF is unramified at almost all discrete spots 93 on H. Proof. Take H = F (6) and let 1 (x) = irr (x, 6, F). Then by the Product Formula we know that f (x) is integral and f' (6) is a unit at almost all discrete spots 93 on H. For any one of these 93 the extension 1/13/F93 is unramified by Proposition 32:6. q. e• d• 71:8. The absolute p-cyclotomic extensions of F are of degree p - I over F for almost all prime numbers p. Proof. Let Q denote the prime field of F. We shall consider the set of discrete spots p on F for which FpIQp is unramified; this set consists of almost all the discrete spots on F by Proposition 71:7. Each of these spots p induces a p-adic spot on Q, and almost all p-adic spots turn up

in this way. Consider any one such p-adic spot. Let it be induced by p on F. We claim that K = F (C) with a primitive p-th root of unity has degree p - 1 over F. To prove this we take a spot 93 on K which induces the spot p on F. We form •293 S F93 S Ksp. Then — 1 is a root in Ki3 of the polynomial (y + 1)2 — 1 Y

over Q13 ; and this is easily seen to be an Eisenstein polynomial in the sense of Proposition 32:15; hence Q13 ( — 1) has ramification index p - 1 over Q. So the ramification index of Kasp is at least p - 1. But F93/Q93 is unramified by choice of p. Hence the ramification index of Ki3/F93 is at least p 1. So [K:F] [K ip :F931

Hence [K:F]

p - 1.

p-.

q. e. d. 71:9. H is a finite extension of F, and C is an algebraic closure of H. Then for almost all prime numbers p, every p-cyclotomic subextension

Chapter VII. Hilbert's Reciprocity Law

193

KIF of CIF satisfies [KH:H] = [K:F] . Proof. The absolute p-cyclotomic extensions of H are of degree p — 1 over H for almost all prime numbers p. We claim that a p-cyclotomic extension K corresponding_ to any such p satisfies the given equation. Take a primitive p-th root of unity ç in C. Then K Ç F(C) S C. We have [H( ) : H] = p — 1 by choice of p. Hence [F (C) : F] = p — 1. Then

p— 1= [if

(c)

II]

= [H(C): HK] [HK: H] [F (C): K] [HK : H] [F (C) : K] [K :

=p — 1 Hence we must have [KH : H] — [K :F], as required.

q. e. d.

§ 71B. Two cyclotomic special cases Here we prove the Hilbert Reciprocity Law for two special cases (Propositions 71: 12 and 71: 13) that will be used in the general proof. We consider an absolute m-cydotomic extension K = F (C) with ç a primitive m-th root of unity. Assume m> 1. Let S denote the set of all discrete spots on F at which m is a unit. So KIF is unramified at all p in S. We wish to define a Frobenius automorphism al, in G (KIT') at each p in S. We take a spot 93 on K which divides p. Then Kv/F93 is unramified since p is in S, hence by Proposition 32:12 and its corollary there is an automorphism ar E G (K931F93) called the Frobenius automorphism such that o-p = V'TP. If we now use ar to denote the restriction of ar to K we see that at each p in S we have found an element ar E G(KIF) with the property that ar ç Nr • There is clearly just one such element ar. And it is independent of the choice of the primitive m-th root of unity ç in K. We call this ap the Frobenius automorphism of KIF at P (P E For each a E F which is integral at all p in S we define H cry

Es This product makes sense since ordr oc is almost always 0 and since G (KIF) is commutative. 71: 10. Formula. Let 13 be any element of fr with V in K. Then ace (a) O'Meara, Introduction to quadratic forms

irgplEls cc,p 16 ) • 13

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Part Three. Arithmetic Theory of Quadratic Forms over Fields

Proof. It is enough to prove

cIrd"(Vi4) a'rP Vff for each p in S. Take a3 .on K dividing the p under consideration and consider the Frobenius automorphism ap on KI3/Fv. The restriction of dip to K gives the Frobenius automorphism al, on KIF. If 13 is a square at p we have 14-3- EFq3 and so

cr;rdt' a (ifff) = P ) ifg as required. So let us assume that 13 is not a square at p. Then I1T3- is not in the fixed field F93 of ay , hence ap (13) 0 • Hence arp cc (vg) = 6 rg with 6 =4 ± 1, the sign depending upon whether ordp cc is even or odd. a, Furthermore for this same number 6 we have (---P)- e by Examples 63:10 and 63:16 since F93 (1/-43--)/F13 is unramified. Hence our formula is true. q. e. d. 71:11. Formula. Every a in F which is integral at all spots in S and is sufficiently close to 1 at all discrete spots in Q S will satisfy where r is the number of real spots at which a is negative. Proof. Let Q denote the prime field of F, let Z denote the rational integers of Q, let T o denote the set of discrete spots p on Q at which 1m1B,< 1, let T denote the set of discrete spots in 12 — S. Thus TI I To. By Proposition 15:4 and its corollary we know that any a in F which is sufficiently close to 1 at all p in T will be a unit at these spots and will satisfy INF/t2 cc 11„ < 1ml, for all p E To .

So consider any a which is this close to 1 at all p in T, and suppose that a is also an integer at all p in S. Then N.F1Q EZ by Example 15:6. And so ATF/Q cc = 1 modmZ . Hence by Proposition 15:5, (— 1)' N IQ a is a natural number in Z which satisfies (— 1)r N,/QG (— 1) ? modmZ Hence INFN (— 1) r modm Z , where 1 1 0o is the ordinary absolute value on Q. Then riardp cc) (0 kpEs

Chapter VII. Hilbert's Reciprocity Law

195

where

H prio:

i

pEs

= 11 liiccip pEs H kip

arch

ll lNpt 00 0cl oo

arch

INF/Q °C100

(— 1)* modmZ

So o.„C = C( -1Y since is an m-th root of unity.

q. e. d. 71:12. Suppose that the quadratic extension EIF is actually m-cyclotomic. Let q be a discrete spot at which m is a unit. Then .1.`ir„S PFNEIFIE if and only if Elf' has local degree 1 at q.

Proof. 1) If EIF has local degree 1 at q we apply Example 65:2 and find 4, c p,N„,,JE. Conversely we must deduce from this inclusion relation that EIF has local degree 1 at q. We suppose not, i. e. we suppose that the local degree at q is equal to 2, and we use this to get a contradiction. We can take F Ç E SK where K = F (C) with a primitive m-th root of unity. We let S denote the set of discrete spots on F at which m is a unit. The preceding discussion now applies to this situation. Note that q E S. NE/pIE c I by § 65A, Since the local degree is 2 at q We have hence there is an idèle I in I which is not a local norm at q. But we are assuming that I c pi,N,,,JE, hence there is an a in F which is a local norm at all p E Q q, but not a local norm at q. 2) So by § 65A again we have an a in F such that 5+1

(o:,0

P

)

1 —1

ifpEQ—ci if p q.

We must refine this a. At each pED—S we can write a=a

—

O bl; (ap , bp EFp ) .

Take approximations a, b EF to the ap, bp E Fp at all p in D — S. If these approximations are good enough we obtain an element ao . a2 — OP of F which is arbitrarily close to a at all p in Q — S. Hence we can make ao a-1 arbitrarily close to 1 at all p E Q — S. And P 13*

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Part Three. Arithmetic Theory of Quadratic Forms over Fields

for all p E Q. So if we replace a by ao a-4, our new a will have the same property as the original one, and furthermore the new a will be arbitrarily close to 1 at all discrete p in Q S, and will be a square at all real p in Q — S. We claim that actually a can be chosen with the additional property that it is integral at all p in S. For by the StrongApproxirnation Theorem we can find a y in F which is arbitrarily close to 1 at all discrete p in Q — S, which is integral at all p in S, and which makes ay integral at all p in S. Then ay2 satisfies all the properties of the last a, and it is also integral at all p in S. We therefore assume that our a has this additional property. 3) Formula 71:10 applies to this situation with. 13 = O. Hence crOE (K) = 'K. And Formula 71: 11 also applies if a is chosen sufficiently close to 1 at all discrete spots in Q — S. Now our present a is a square at all real spots, hence the r of the formula is O. So o-OEC = C. But this means that ac, Hence EIF is the identity. This contradicts the fact that crOE (K) q. e. d. has to have local degree 1 at q. a —1 —1 71:13. Let a be an element of the number field F such that ( '

- yrr .

for all discrete p on F. Then ot, —1)

PED

—1.

P

Proof. Let K F(C) where C is a primitive 4-th root of unity, and let S be the set of all discrete non-dyadic spots on F. This gives us the situation discussed at the beginning of § 71B with m = 4. An argument similar to the argument used in step 2) of Proposition 71:12 will give a new a which is an integer at all p in S and arbitrarily close to 1 at all dyadic p, and such that the new Hilbert symbols ( a3-1 ) are equal to the original ones at all p in Q. Now C2 = — 1, so by Formula 71:10 we have cfOE C == C. By Formula 71:11 we have cf. = CC-1)? where r is the number of real spots on F at which a is negative. Here r is even since C cf.0 C(-1)'. On the other hand a is negative at a real spot on F if and only if ( mi —1 ) — — 1. Hence ce, —1)

p ED

P

reap

\

(— 1)* ,--- 1 .

q. e. d. § 71C. The ten-field construction Now we have reciprocity in certain special cases involving cyclotomic extensions. The next step in the proof consists in surrounding the quadratic extension EIF with cydotomic extensions and then carrying the reciprocity laws associated with these extensions back to F.

Chapter VII. Hilbert's Reciprocity Law

197

71:14. Lemma. Let a 2 be a natural number. Then there are infinitely many prime numbers p such that a has even period modp, i. e. such that an 1 modp n even.

Proof. Define a sequence A l, A2,. A„, . . . by putting Al = a2, A,, 1 = A. Suppose if possible that A,,+ 1 E 4Z. Then A, is odd, hence a is odd, hence A l — 1 = a2 — 1 ( 4 Z . So Al = 1 + 4), with in Z. But A, is a power of Al. Hence by the binomial theorem we obtain A,— 1 E 4 Z. But we are supposing that A y + 1 E 4 Z. This is therefore impossible. Hence 2'4+ 1 4 Z for all 1. Now A,+ 1 > 2; and we have just shown that A y + 1 is not divisible by 4; hence there is a prime number p,> 2 dividing A y + 1. This p, of courge cannot divide A„— 1; but it does divide A„ + ,— 1 = 4 — 1 (A,,— 1) (A

1) .

Hence we can find prime numbers Pi, P2, "

pv , • • • >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

Part Three. Arithmetic Theory of Quadratic Forms over Fields

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

Part Three. Arithmetic Theory of Quadratic Forms over Fields

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.

Chapter VII. Hilbert's Reciprocity Law

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

Part Three. Arithmetic Theory of Quadratic Forms over Fields

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

± <—cc 13> 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

Chapter VII. Hilbert's Reciprocity Law

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

Part Three. Arithmetic Theory of Quadratic Forms over Fields

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. <

(oei E Pp)

at each p in T. Use the Weak Approximation Theorem to find an ai in F such that loci — ce:l p is small for all p in T. Since .F';; is an open subset of F1, we can obtain

YpET,

arEcF

provided our approximations are good enough. Do all this for 1 I Then take a quadratic space W over F with < oci >

•

n-1.

< oin—i> •

Clearly Wp YpET.

Hence Sp Wp = Sp Up for all p in T. Let R denote the set of spots which Sp Wp and Sp Up are different. Of course R is a finite subset D — T. And Wp -- Up for all p in D — R by Theorem 63:20. If R T empty we are through. Otherwise R consists of those spots in D which S4, W1,= 1; now

./iS p W1, 11 sr wp

at

of is at

•H sp up = Hsp wp = 1 • 12-R

Hence R consists of an even number of spots in D — T. 3) We claim that there are quadratic planes P and P' with the same discriminant over F, with P PI; for all p in D R, and with Pp for all p in R. In fact, take any 0 in F which is a square at all archimedean spots and a non-square at all spots in R. By Theorem 71:19 and its corollary we can find an a in F with f —1 11 1

if PER — R. if p

Define

P <1> < 13> , P' ± <— cc 18> .

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

Part Three. Arithmetic Theory of Quadratic Forms over Fields

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

Chapter VII. Hilbert's Reciprocity Law

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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*

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

Chapter VIII. Quadratic Forms over Dedekind Domains

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)

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

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.

Chapter VIII. Quadratic Forms over Dedekind Domains

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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) <0°; and the number of lattices between K and L is finite. § 81E. Localization

Let us give some attention to the completions Fr of F at the spots p in S. We let I Ir stand for any fixed valuation on Fr at p. Of course the Dedekind theory of ideals applies to Fp at p, and the theory of lattices applies to vector spaces over Fp . We let 1-r denote the group of fractional ideals of Fp at p. Recall that or, up , mr are used instead of o(p), u (p), m (p) on Fr at p. Here the same symbol p denotes two different spots, the one on F and the other on Fr. And p also denotes two different prime ideals, the prime ideal)) n mr of 0 and the prime ideal mr of or. A typical fractional ideal of F at S has the form pvp

E

Chapter VIII. Quadratic Forms over Dedekind Domains

217

with all vp in Z and almost all of them 0; and a typical fractional ideal of Fp at p has the form PAP Cup E Z) • We introduce the surjective homomorphism defined by the mapping 17 pvp

pvp

p ES

The image of a E / (S) under this mapping will be written ap. We call ap the p-ification or localization at p of the ideal a. The following laws follow easily from the results of §§ 22B and 22C: (Œa) p = cop , (a (a + b) p = ap bp , laplp lalp • This last equation shows that a is always contained in ap, and in fact

a= We have

p ES

(F n ap) .

a Sb

al,Sbp VpES,

a=

ap = bp VP ES.

and in particular It is easily seen, again using' the equation laplp op-ideal generated by a in Fp. 81:13. a p is the closure of a in Fp. Proof. We know that

la lp, that a;, is. the

ap = {cc E Fp lialp

ialp} ; but the map a –>– 14 of Fv, into R is continuous; hence ap is closed in Fp. Why is ap the closure of a? We must consider a typical a E ap and an s > 0, and we must find an a E a such that la — 4 s. We can assume that 0 < s < 14) By the Strong Approximation Theorem we can find an a in F such that la — alp < 6

Vci E S 5- lalq The Principle of Domination then insists that lai r = lair Hence lalq 5_ lak q. e. d. for all q in S. So a is in a. Now consider an n-dimensional vector space V over F. As in § 66 we use V,, for the localization Fp V of .V at p. Let L be a lattice in V (with respect to the set of spots S). By the p-ification or localization Lp laic'

Part Four. Arithmetic Theory of Quadratic Forms over Rings

218

of L in Vv at the spot p in S we mean the or-module generated by L in Vv. By the definition of a lattice there is a base xi., . . xn for V such that L

0 + - • • + 0 xt, ,

hence

Lr

+ • • + o v x„ ,

so Li., is a lattice in Vv (with respect to o r, or with respect to the spot p on F).

If L and K are lattices in V, then it is easily seen that VrES. (L K)4, = ±Kr Now (a x) = av x for any a E I (S) and any x in V, hence (not + • • • + ar ti.) v alv z, + • • • + arv zr for any ai E (S) and any zi E V. If we take a base x1 .. x n for V such that L= -F • • • + ar x, , then ,

Li)

p ± • • • +

ar

p .

Hence V n L = (F r a1p)x1 + • • • + (F r arp)xr ,

so

n

(V r\ Lr) L .

pES

We therefore have LSK

Lr Kr

V p ES,

L=K

Lr = Kr

V P ES.

and in particular, If FL = FK, then the Invariant Factor Theorem of § 81D shows that for almost all p in S. 81:14. An or-lattice hp) is given on VI, at each p in S. Suppose there is an o-lattice L on V with Lr = h o for almost all p. Then there is an 0-lattice K on V with Kr = he) for all p in S. Proof. 1) First we prove the following contention: given a spot p in S, there is a base . yn for V such that hp) = °ph ± • ' • + Op Ytt • To prove this we fix a base x1,. , xn for V; multiplying each of these basis vectors by a scalar which is sufficiently small at p allows us to assume that each xi is in hp) . Take a base for hp) , and write each

J4, ) = opih

• • + o plin

+ • + ani x„

(7/27 E 17,Y, (ccii E Fr)

.

Chapter VIII. Quadratic Forms over Dedekind Domains

219

Now -F,p, is the closure of F at p, hence we can find, for each i and j, an element aij CF such that 1

is as small as we wish. If the approximations are good enough we will obtain, in virtue of the continuity of multiplication and addition in the topological field Fp , the inequality det (ai — det(oc ii)1 1, < det (oc ii)i p .

Hence idet

= idet(oc ii)l p by the Principle of Domination. Put

y = a1 5 x 1 lor i

••

ani xn

n. Then 51; — m E op

by choice of the

(t ip

• • + opxn c= hp)

hence

°ph + • • • + opYn hp) • If we write y; f yij n i with all yi; Fp we have (7ii)

(Iii) -1 (a15)

hence det (y o) is a unit in vp , hence opYt ± • " opYrt

by Proposition 81: 8 . This proves our contention. 2) It is enough to prove the following: given a single spot p E S there is an D-lattice K on V such that L K —{ J (qr)

f f

CS —p = P•

The final result will follow by successive applications of this special case. In virtue of step 1) we can find an o-lattice J on V with J = hp) . Use the Invariant Factor Theorem of §81D to find a base z1, zn, for T7 such that Ç L = arzi + • • • + anz. + • • -1-- b.„z„ j=

Construct fractional ideals c i (1 c"

i

n) with

I Dig if q(-S — p bip if q = p

Then K

has the required property.

+ • - • - I- enzn

q. e. d.

220

Part Four. Arithmetic Theory of Quadratic Forms over Rings

§ 82. Lattices in quadratic spaces

An additional structure is imposed on the situation under discussion: the vector space V is made into a quadratic space by giving it a symmetric bilinear form B with associated quadratic form Q. We shall call a lattice L in V binary, ternary, quaternary, quinary, , n-ary, according as its rank is 2, 3, 4, 5, . . . , n. § 82A. Statement of the problem Consider a lattice L in V. Let Ube some other quadratic space over F and consider a lattice K in U. We say that K is represented by L, and write K —›— L, if there is a representation o: FK —>— FL such that aK S L. We say that there is an isometry of K into L, and write L, if there is an isometry : F K >-- FL such that aK Ç L. We say that K and L are isometric, and write or K , if there is an isometry ci: FK >—). FL such that aK L: We pose the following fundamental question: can we determine, say by means of invariants, whether or not two given lattices L and K are isometric. This should be regarded as the integral analogue of the earlier work on quadratic spaces. And just as the question for spaces is a geometric interpretation of the classical problem on the fractional equivalence of quadratic forms, so the question for lattices can be regarded as a geometric interpretation of a classical question on the integral equivalence of quadratic forms. In fact it will be shown in § 82B that finding the lattices isometric to a given free lattice is the same as finding the integral equivalence class of a quadratic form. This poses the question. What about its solution ? It is certainly too much to expect an answer over an arbitrary Dedekind domain at the present time. However, a complete solution is known when o is the ring of integers of a local field, and considerable work has been done over the integers of a global field. These theories will be presented in the remaining chapters of this book. If a solution to the space problem is known over the field F (for instance if F is either a local field or a global field) then there is no loss of generality in assuming that FK and FL are identical. Under these circumstances we can ask our question in the following form: given lattices L and K on the quadratic space V, does there exist an element a of the orthogonal group 0(V) such that aK L? § 82B. The free case Let M be an m x m symmetric matrix and . /sI an n x n symmetric matrix, both over the field F. We write M —>— N (over c)

Chapter VIII. Quadratic Forms over Dedekind Domains

221

and say that M is integrally represented by N if there is an n X m matrix T with coefficients in o such that

M , tTNT , where tT stands for the transpose of T. If this can be done with a unimodular matrix T we say that M is integrally equivalent to N and write

M N

(over o) .

Integral equivalence of matrices is clearly an equivalence relation on the set of all n x n symmetric matrices over F. This equivalence relation depends, of course, on the set of spots S used in defining the ring o = o (S). We let clss N or clsN denote the set of n x n matrices which are integrally equivalent to the n x n symmetric matrix N over F, and we call this set the class of N at S. These classes partition the set of n x n symmetric matrices overF at S. If L is a free lattice on V there is a base x1 . . x. for V such that ,

L =oxi + • • • + oxn . By the matrix of L in the base x1 ,. N (B (xi , xi)), i.e. the matrix of V in

x„ we mean the matrix xn ; we write

LN If there is at least one base x. for which this holds, then we say that L has the matrix N and we write

LN. If

. • . , x is another base for L with x;

2: tl.ixA

(tau ( 0)

then T = (tu) is unimodular by Proposition 81:9, and the matrix N' of L in xi, . . x:, is equal to

N' ITNT In other words a change of base leads from N to a matrix N' in clss N, and every matrix of clss N can be obtained in this way. We can therefore associate one entire class of integrally equivalent matrices with a free lattice on a given quadratic space. And every class of integrally equivalent symmetric matrices can be obtained from a suitable free lattice on a suitable quadratic space. Given any symmetric matrix N we have agreed to let stand for a quadratic space having the matrix N. We shall also use the symbol to denote a free lattice with the matrix N (in a suitable quadratic space).

222

Part Four. Arithmetic Theory of Quadratic Forms over Rings

82: 1. Let K and L be free lattices with matrices

M and N on the

quadratic spaces U and V, respectively. Then (1) 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

Chapter VIII. Quadratic Forms over Dedekind Domains

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

<

r

We then say that L is the orthogonal sum of the sublattices L1, . . or that L has the orthogonal splitting L=4±•••±L,.. We call the L i the components of the splitting. We also use the notation

± L i and

ED Li 1

1

for orthogonal sums and direct sums respectively. We formally define j_ L i = 0 and ED L i = 0 . 0 0 We say that a sublattice K splits L, or that K is a component of L, if there is a sublattice J of L with L=KJ

If X1 . . X,. are sublattices of different quadratic spaces, not necessarily ,

contained in V, we write L

Xi 1 • • • 1 X,.

to signify that L has a splitting L=

1 • • •

± Lr

in which each sublattice L i is isometric to Xi. Suppose we are given quadratic spaces Vi (1 I r) over F and lattices L i in Vi. Then we know that there exists a quadratic space V over F such that V 24 Vi 1 • .j V,. . Hence there always exists a quadratic space V which contains a lattice L such that L ± • • • ± Lt .

.

Chapter VIII. Quadratic Forms over Dedekind Domains

225

There is a slight modification of the preceding construction which we shall call adjunction. Consider quadratic spaces U and V over F. Then there is a quadratic space W which contains the quadratic space V and is such that W = U' ± V with U' U .

We say that W is constructed from V by adjunction of U. Now let K and L be lattices in U and V respectively. Then there is a lattice J in W such that J = K' L with K' K .

We say that J has been constructed from L by an adjunction of K. As an example of the notation consider the equation L

± ,

with M and N symmetric matrices over F. This means that L is a free lattice in a quadratic space with

L=—( Similarly L

(cci >

11/1 0 1— O/sT) . • • 1.
with all cci in F means that L has a base x11 . . x„ in which Q (xi) — cci for 1 I n and B(xi, xj) = 0 for 1 I <j Sn. A base with this property is called an orthogonal base for L. It is easily seen that the direct sum of lattices L= Ll e-••eL,. implies the direct sum of spaces FL = FL, e-••e FL, . Similarly with orthogonal sums. These equations show that rankL = rankLi + • •

rankL,..

We define the radical of a lattice L in a quadratic space V to be the sublattice radL =

EL IB (x, L) = 0} .

We call L regular if radL = O. It is easily seen that radFL = F radL

and radL

L

radFL

In particular, L is regular if and only if FL is regular. We have rad(L i K) = (radL) j (radK) . O'Meara, Introduction to quadratic forms

15

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

In particular, L K is regular if and only if L and K are regular. Suppose J—LIK with L regular; then K is uniquely determined by and L, and in fact K {x J B (x, L) 0} . Let us show that L has a radical splitting, i.e. that there is a lattice K such that L = K j radL. If L is regular just take K L. Now assume that L is not regular and take a base x11 . . xn for FL in which x1,. . x,. span the radical of FL. By Theorem 81:3 there is a base yi, . y,z for FL in which ' • • + arYr + • • • + anYn and radFL == Fyi • • + F y,. . Then radL L radFL = a1 y1 + • • a,.y,.. So taking K = ar_Fo r+i + • • • + anYn gives us the desired radical splitting L = K radL. Incidentally, the lattice K in any radical splitting L = K j radL is always regular since the equations radL = radK j rad (radL) radK j radL

imply that radK is O. If L = K j radL and L1 = K1 I radLI are two radical splittings, then it can be shown (using .a proof somewhat similar to the proof of Proposition 42:8) that L is isometric to L1 if and only if K K 1 with

radL212- radLi •

82 : 7. Let L be a lattice in the quadratic space V and let K be a regular sublattice of L. Suppose K has a splitting K=

_1_ • • • ± K,. .

Then K is a component of L if and only if each K i is. Proof. If K splits L, then it is clear that each K i does too. Conversely suppose L is split by each K. Since FK is regular we have a splitting FL = FK in which U is the orthogonal complement of FK in FL, by Proposition 42:4. We claim that L K (U n . In order to prove this it is enough to show that L = K (U L). So consider a typical x in L. For r we can write x = xA 3/2 with xA E KA and B (y A, K A) = 0, since KA splits L. Then B (x—E xA , K i) = B(x— xi, K i) B (y i , K i) = 0 2

Chapter VIII. Quadratic Forms over Dedekind Domains

for 1

227

i < r. Hence B (x — E xA,F K) . 0. a ‘

So x— E xA E U n L . a

Hence -47 XA) E K + (U n L) . X = (I ' XA)± (X q. e. d.

§ 82E. Scale, norm and volume Consider a lattice L in the quadratic space V. By the scale sL

of L we mean the o-module generated by the subset B (L, L) of F. Clearly sL = 1 E B (x, y)jx,y E

9.

I fin

Since L is finitely generated we can write L = o z1 + • • • + o zr, hence sL C E B(z i, zj)o , ci

so sL is either a fractional ideal or 0. We define the norm nL

to be the o-module generated by the subset Q (L) of F. Since Q (L) s B (L , L) it follows that nL is also either a fractional ideal or 0. Now for any x, y in L we have 2B (x, y) = Nx + A - 12 (x) - Q(y) E nL . Hence 25L ç la C sL . We also have Q (F L) = F2 Q (L) .

So all the sets oL, nL, B(FL,FL), Q(FL)

are 0 if one of them is. If L has the splitting L = J ± K, then it is easily verified that sL = s, f + 6K,

nL—ni-l-nK.

82:8. Let the lattice L in the quadratic space V have the form L = arzi -I- • • • + ar zt with the a i in I and the zi in V. Then (1) sL = E a i ai B(z i,zi), iI

(2) nL -- E 4Q(z) +26L. i

15*

Part Four. Arithmetic Theory of Quadratic Forms over Rings

228

Proof. 1) First we do the case r — 1. Clearly n(a01) c (a01) Ç a7+2' (z1) .

Let us prove that aTQ (xi) C n(a01). Write al — ao + fib with a, /3 in al. Then cc2 Q (xi) = 02 (azi) E 02 (al zi) ç n (al zi) , and similarly 132 Q (xi) E n(aiz,L). Hence by Example 22:3

alQ (zi) —

(x20 + 1320)

Q(z) ç n (alz1) .

So n (a1 z1) . s (aizi) . alo2 (z1) .

This proves the proposition in the case r — 1. 2) Next we find the scale for any r> 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).

Chapter VIII. Quadratic Forms over Dedekind Domains

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.

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

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,

Chapter VIII. Quadratic Forms over Dedekind Domains

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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 .

Chapter VIII. Quadratic Forms over Dedekind Domains

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.

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

§ 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 -

Chapter VIII. Quadratic Forms over Dedekind Domains

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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.

Chapter VIII. Quadratic Forms over Dedekind Domains

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 <4. 6L S a . It is easily seen that La = aL#n L. In particular this shows that FLa=FL. For any splitting L=J IK we have ± Kr= PI Ka. Now consider another fractional ideal b. Then ab-1 Lb S LaÇ Lb if a S b follows directly from the definitions. If L is b-modular, then IL if b S a La = ab -'L if b2a, since La=aL4*nL—ab-lLnL.

Part Four. Arithmetic Theory of Quadratic Forms over Rings

238

§ 82J. Scaling

Consider the lattice L in the quadratic space V. Let a be a non-zero scalar. Recall that 'Pc denotes the vector space V provided with a new bilinear form BOE (x, y) = a B (x, y). We shall use LOE to denote the lattice L when it is regarded as a lattice in Va. Thus the properties of LI as a lattice are identical to those of L; the superscript a merely emphasizes that our interest has shifted from the properties of B on L to those of B1 on L. We easily see that SLOE

(6 L) , n LOE = cc (n L) , t, LOE ---

(b L)

where r denotes the rank of L. If L is a-modular in V, then La is a amodular in Va. If L is a-maximal on V, then LΠis cx a-maximal on Vcc. If Ka. So for a lattice L L K is a representation, then so is cp: LOE on the regular non-zero space V we have (L) ---- 0 (La) , 0+ (L) = 0+ (L a ),

and cis L cis

cls+L = cls+LŒ.

If L is a free lattice with matrix M, then LI is free with matrix cc M; so here d La = ccr (dL) .

Suppose L is regular and let a be a fractional ideal. Then the notation Lcc should not be confused with the notation L. For instance, Lao denotes a certain sublattice of L in the quadratic space V, while LI denotes the original lattice L in the quadratic space VI. It is easily verified that (LŒ) a = (Lc‘ - ' a)a . § 82K. Localization

Consider the lattice L in the quadratic space V. Let p be any spot in the Dedekind set of spots S on which our ideal theory is based. Consider the quadratic space VI, (i. e. the F;, -ification Fp V of the quadratic space V) and the localization Lp of L in Vp . Take a base x1 .. , x,. for FL in which L has the form ,

L a i xi + • • • + ar xr .

Then by § 81E Lp aip + • • • ± afp x,.

so by Proposition 82:8 we have sL p

E ai;, a1 , B (xi, xj) = (E aj ai B (xi, xj)) =

Similarly with norm and volume. In other words, sL p = (s

, n L p = (n L) ;, ,

= (»L)p

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

239

holds for all p in S. In particular two lattices L and K have the same scale (or norm or volume) if and only if Lp and Kp have the same scale (or norm or volume) at all p in S. From this it follows, using the definition of a modular lattice, that L ay-modular for all p in S. is a-modular if and only if L It is also true that L is a-maximal on V if and only if Lp is ap-maximal a y-maximal for all p in S. on Vp for all p in S. First suppose that L If L is not a-maximal, there is a lattice K on V such that KDL and rtK ç a. Then by § 81E there is a of in S at which K g D L g . But nKg ç a q . This denies the maximality of L g . Conversely let us be given an a-maximal lattice L on V, and let us suppose that L g is not aq-maximal at some spot of in S. Then there is a lattice J(q) on Vq which strictly contains L g and has n j (q) C aq . By Proposition 81: 14 there is a lattice K on V with if if

p C S—q p -- q .

Here we have KD L. But nK), C. ap for all p in S. Hence nK .Ç. a. This denies the maximality of L. Chapter IX

Integral Theory of Quadratic Forms over Local Fields This chapter classifies quadratic forms under integral equivalence over local fields'. We continue using the notation of the last chapter, except that the field F is now taken to be a local field, S is the single spot p, o is the ring of integers o (p), p is the maximal ideal m (p), and u is the group of units u (p). We let i 1-= I l y be a valuation, say the normalized valuation, in the spot p. The letter a denotes a prime element of F at p. Ideal theory and lattice theory are with respect to the single spot p. The quadratic space V provided with its symmetric bilinear form B and its associated quadratic form Q is assumed to be regular and nonzero. We shall consider non-zero regular lattices L, K, J, . . . in V. Note that every line Fx in FL contains a maximal vector of L since the class number of F at p is 1. And every non-zero lattice is free. In particular the discriminant dL of § 82B is available for use.

5 91.

Generalities

Here F can be dyadic or non-dyadic. Note that L has the following property: if we take any two vectors x, y in L for which 1B (x, 3))1 is ' For the local integral theory of hermitian forms see R.

Math. (1962), pp. 441--465.

j ACODOWITZ,

Am. J.

240

Part Four. Arithmetic Theory of Quadratic Forms over Rings

largest, then B (L, L) = B(x y) = L

Similarly if we take any vector x E L for which 1Q (x)I is largest, then Q (x) 0 = n L § 91A. Maximal lattices 91:1. Theorem. V is an anisotropic quadratic space over a local field, and L is an a-maximal lattice on V. Then L ={x E VIQ (x) Ea}. In particular, all a-maximal lattices on V are equal. Proof First let us show that the set X = {x E V IQ(x) E a} is closed under addition. Consider typical vectors x, y in X. Thus Q (x), Q (y) E a and we must show that Q (x + y) c a. Suppose not. Then 2 B (x, Y) = Q (x + y) - Q (x) - Q (y) a, hence 12 B (x, y)I > [al and so Q (x)Q (Y) B

I -I •

Now the discriminant dB (x, y) is equal to

—B (x, y) 2 11

Q (x) Q (Y)1 B(r y) 2

so —dB (x, y) is a square in F by the Local Square Theorem. So Fx Fy is a hyperbolic plane. This is impossible since V is given anisotropic. Hence X is indeed closed under addition. Hence it is an o-module. Suppose X is not equal to L. Then L C X by dëfinition of X, so we can pick z E X — L. Then L oz is a lattice contained in the o-module X, so Q (L o z) Q (X) Ç a, q. e. d. z) S a. This contradicts the maximality of L. hence n (L 91:2. Theorem. Let K anti L be a-maximal lattices on the regular quadratic space V over the local field F. Then cis K = cis L. Proof. We can suppose that V is isotropic by Theorem 91: 1. By Proposition 82:23 we have a splitting V Hi 1 • • • L. H,. ± Ho in which H1, , H,. are hyperbolic planes and 1/0 is either 0 or an anisotropic space, with L (L _L - - (L Hp) (L n Ho) , 1K (Kñ H 1) 1 • • • ± (Kñ H,.) j (K H 0) .

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

241

Now L r Ho and K r Ho are either both 0 or both a-maximal on Ho, hence they are equal by Theorem 91:1. And L r Hi and K r Hi are isometric for 1 i r by Example 82:22. So for 0 I r we have E 0 (II1 ) with a (L r H1) K r H1. Then —

a=a1±*••Ictriao

is an element of 0(V) vvith L = K. Hence K E cis L. Hence cis K = cIsL. q. e. d. 91:3. Example. If L is an a-maximal lattice on a regular quadratic space V over a local field F, then Q (V) r\ a Q (L). For clearly Q (V) n a Q (L). On the other hand if oc is a non-zero element of Q (V) n a, there is a vector x in V with Q (x) = cc E a. This vector x is contained in an a-maximal lattice M on V by Propositions 82:9 and 82:18. But cis L = cis M by Theorem 91:2. Hence a E Q (L) as required. In particular if dim V 4 we must have Q (L) = a since every regular quadratic space V with dim V 4 over a local field is universal by Remark 63:18. If dim V = 3 we can use Remark 63:18 to show that V represents either all units or all prime elements, hence for any # in a we have either Q (L) 9 u or Q (L) 2 flu. —

§ 91B. The group of units of a lattice

Consider the lattice L on the regular, non-zero quadratic space V over the local field F. Let u be a maximal anisotropic vector of L. We claim that the symmetry T.. of V is in 0(L) if and only if 2 B (u , L)

Q (u)

S0.

First suppose Zu is a unit of L. Then 1-ti L = L and so 2 B (u,

Q (u)

u

x—

xEL

for all x in L. But the coefficient of u is o since u is maximal in L, hence 2E (u, x)112 (u) E o as required. Conversely if this condition is satisfied it is dear that .ru L Ç L and so ru L = L by Example 82 :12, i. e. E 0 (L) as required. 91:4. Let L be a lattice on the regular non-zero quadratic space V over the local field F. Then 0 (L) contains a symmetry of V. Proof. Take u E L with Q (u) 0 = nL. Then u is clearly a maximal vector in L; and 2 B (u, L) C28L SnL = Q(u) o , so that 2B (u, L)/Q(u) Ç o; hence x„, is a unit of L. 91:4 a. (0(L): 0-E(L)) = 2 and so cis L = cls-FL. O'Meara, Inirodlielion to quadratic forms

q. e. d. 16

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

91:5. Let L be a maximal lattice on an anisotropic quadratic space V over a local field. Then 0(L) = 0(V). Proof. We must show that a typical a E 0(V) is in 0(L). Now by Theorem 91: 1 there is a fractional ideal a such that

L {x E V IQ (x) E a} .

For any x in L we have Q (a x) = Q (x) E a since a is an isometry, hence aL CL, hence a E 0(L). q. e. d. 91:6. Let V be a regular quadratic space overa local field with dimV 3. Then 0 (0 -E(V)) = F. Proof. If dim V

4 the space is universal by Remark 63:18, hence there is a symmetry with any preassigned spinor norm, hence there is a rotation with any preassigned spinor norm. We are left with the ternary case. By Remark 63:18 the space V will represent all a in F for

which VI <--oc>

has discriminant not equal to 1. So if the discriminant of V is a unit, V will represent all prime elements and at least one unit; now every element of F is a product of exactly two such elements times a square in F; hence 0(0-F(V)) F. A similar argument applies when the discriminant of V is a prime element. q. e. d. 91:7. Let V be a regular quadratic space over a local field with dimV 3. Then there is a lattice L on V with O (0 -E(L)) . F. Proof. If V is anisotropic we take any maximal lattice L on V. Then by Propositions 91:5 and 91:6 we have 0(04-(L)) = 0(0 4- (V))

.

Hence we may assume that V is isotropic. Take a base x, y, z, . . . for V in which 1

1 • " . V< — Î > 0 By suitably scaling the bilinear form defining V we may assume that 27c where v is a prime element. Let L be the lattice L

(ox

oy)

j_ (0.Z) j_ • • • .

Then for any unit e we have ey, L)20 C0 ' Q (x ey) = 2e

2B

hence the symmetry rx+81/ is a unit of L. Now this symmetry has spinor norm 2 E F2. Hence 0(0+ (L)) D uF2. Using the same argument with z instead of x ey we obtain a symmetry ; which is a unit of L and has

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

243

spinor norm 27t 1 ; then O(ux .i.„) = Emil; hence 0(0+(L)) D nub:2 and we are through. q. e. d. 91:8. L is a maximal lattice on a regular quadratic space V over a local field with dim V 3. Then 0(0+ (L)) D u. Proof. By scaling V we can assume that L is 20-maximal. If V is anisotropic, then 0+ (L) = 0+(V) by Proposition 91:5 and so 0(0±(L)) F D u by Proposition 91:6. Now suppose that V is isotropic. By the results of § 82H there is a splitting L = U I. • • • in which U has a base U = ox + ey with matrix CI 01) . Consider any e in u. Then ey,

2 B (.7;

20

2e

+

0

hence the symmetry Toe+ , is a unit of L. Now this symmetry has spinor norm 2eP2. Hence 0 (0+ (L)) D UF2 q. e. d. § 91C. Jordan splittings

Consider a non-zero regular lattice L in the quadratic space V. We claim that L splits into 1- and 2-dimensional modular lattices. If there is an x in L with Q(x) o = L. then J = ox is an 6L-modular sublattice of L. Otherwise Q(x) o ( 8L for all x in L; in this event we can find a binary sL-modular sublattice j of L: we pick x, y in L with B(x,y)o=sL; then the vectors x, y have discriminant d (x, y) = Q (x) Q (y) — (z, y) 2 and this is not zero by the Principle of Domination; hence J = o z + 0y is a binary lattice; a direct computation shows that J

B (x, y)

sL , J= B (x, y) 2o = (54 2 , so that J is actually a binary 5L-modular sublattice of L. Hence L always contains a 1- or 2-dimensional 6L-modular sublattice J. Then L has a splitting L=JIK by Corollary 82: 15a. Now repeat the argument on K, etc. Ultimately we obtain a splitting of L into 1- or 2-dimensional modular components. This establishes our claim.

If we group the modular components of the above splitting in a suitable way we find that L has a splitting L

_L • • • _L L i

in which each component is modular and 64. • • • )54 Any such splitting is called a Jordan splitting of L. We have therefore proved that every non-zero regular lattice L in a quadtatic space V over

a local field F has at least one Jordan splitting. The rest of this chapter is really a study of the extent of the uniqueness of the Jordan splittings of L. 16*

244

Part Four. Arithmetic Theory of Quadratic Forms over Rings

We shall need the lattice La of § 821. First suppose that L is b-modular. If b = a, then La= L and so L a is a-modular. If b C a, then La= L is b-modular with b C a. And if b D a, then La = a b-1 L is a2 b-4-modular with a2b--' c a. So for a b-modular lattice L we have the following result: La is a-modular if and only if a = b, otherwise it is c-modular with c C a. What does this mean in general? Let us consider any non-zero regular lattice L in the Jordan splitting

L = L1 I • • • 1 L i .

We know that sLa S a. And we have a splitting L a DILL • • • ± Ltt' into modular components. If L i is not a-modular, then L will be cmodular with c c a. And if L i is a-modular, then Lf will be equal to L. Hence we find that s La = a if and only if there is an a-modular component in the given Jordan splitting. Otherwise &La C a. Now consider L 1 in the given splitting, and suppose it is a-modular. If we group the components of the splitting LI ± • • ± .14` we obtain a Jordan splitting of La in which the first component is Li. So we have proved that an a-modular component in a given Jordan splitting of the lattice L is the first component in some Jordan splitting of the lattice La. 91:9. Theorem. Let L be a lattice in the quadratic space V over the local field F, and let L = L1 1,•• • ± L i , L = ± • • • ± Kr

t we have be two Jordan splittings of L. Then t = T. And /or 1 A (1) 5 LA = 8K2, dim LA = dim Ka, (2) nLa sLa if and only if nKA &KA . Proof. We shall use the results of the preceding discussion in the proof. 1) Suppose there is an a-modular component in the first Jordan splitting. Then sLa a. Hence there is an a-modular component in the second splitting. Hence, on grounds of symmetry, we have t = T and 2, t. 8L 2 = sICA for 1 Consider a typical A with 1 A I. We must prove that dim LA = dim KA and that n LA=- 841 if and only if nlça = 6 K2. Now LA and KA are first components in Jordan splittings of La where a = sLA = sKA . Hence we can assume that A . 1. And by suitably scaling the bilinear form B on V we can assume that sL o. On grounds of symmetry we are therefore reduced to proving the following: given 54— 6 K1 — s L=0 , prove that dim././ dimKi , and also that n.Li = n implies niC, = 0. This we now do. 2) We shall need the projection map 99: FL„--›-FK„. This is defined as follows: each x in FL, is an element of FL and hence has a unique -

-

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

245

expression x—y+z with y C FKI and z C F(K2 1_ • _L Kg). Put ipx y. Then Tx is a well-defined element of FK1 , and it is easily seen FK1 determined by x --->— 99X is F-linear. that the map T: FL, Now when x in L 1 is expressed as z y + z in the above way we have y E K1 and z(K 2 I''• L K t . Hence 99L1 c K 1 . And for all )c, x' in L1 we have B (Tx, T xi) = B (x — z, z') B (x, x') modp since B (K2 • • • K t , L) p In other words B (99 x Tx') — B (x, z'), Q( T x) Q (x) modulo p for all x, x' in L1 . Suppose we had Tx = 0 for a non-zero vector x of FL,. Then we would have cp-x — 0 for a maximal vector x of the lattice L1 . Since L 1 is unimodular there is a vector y in L1 with B(x, y) 1 by Proposition

82:17. Then 1

B(x, y)

B (Tx, Ty) — 0 mod p

,

and this is absurd. Hence 99 is an isomorphism of FL, into FKi . Hence

dim L1 = dimFLI dimFK1 — dimKi , as required. 3) Finally we have to prove that nK/ — if nL1 = v. Since nil find a vector x in L 1 with Q (x) s where e is in u. Then wecan

Q( T x) Q (x) e modp Hence Q( T x) C u. Hence nKI D o. Hence nKi q. e. d. Consider non-zero regular lattices L and K in quadratic spaces V and U over the same field F. Let

L=

_L • • L t , K =

• - • 1. KT

be Jordan splittings of L and K. We say that these _Jordan splittings are of the same type if t = T and, whenever 1 t, we have

eLA 6K2 ,

dimL2 = dim KA

and !IL A --= 6L 2

if and only if

n KA

= 6 Kz

.

We know from Theorem 91:9 that any two Jordan splittings of L are of the same type. And the same with K. We say that the lattices L and K are of the same Jordan type if their Jordan splittings are of the same type. Isometric lattices are of the same Jordan type. 91:10. Notation. Given a lattice L and a Jordan splitting L -•• L I , we put

L (i)

L1

--

Li

246

Part Four. Arithmetic Theory of Quadratic Forms over Rings

and we call L(1) C L(2) C • • • CL()

the Jordan chain associated with the given splitting. We put = L.

•••_L

and we call LA ) D L r2) D • • • D / (t )

the inverse Jordan chain associated with the given splitting. Clearly L = L( 1) L 4 4. 1) .

A Jordan chain is determined by one and only one Jordan splitting of L. 91:11. Example. Let L(1) C • L. (t) and K(1) C» • C K(0 be Jordan chains of lattices L and K of the same Jordan type. Then L (i) and Km are lattices of the same Jordan type. Also )3 L(0 = K. Hence there is a unit s such that d LtOld K = S.

The same applies to inverse Jordan chains of L and K. § 92. Classification of lattices over non - dyadic fields Throughout this paragraph we assume that the local field under discussion is non-dyadic. We consider a non-zero regular lattice L in the quadratic space V. We know from § 82E that 2 el. Ç nL S EL. But 2 is a unit in o since F is non-dyadic. Hence nL = eL, i. e. norm and scale are equal over non-dyadic fields. We can therefore pick x E L with Q(x) o = ttL = 5 L . Then J = ox is an EL-modular sublattice of L, hence L has a splitting L =J _L K. If we repeat on K, etc. we ultimately find a splitting L = oxi _L • • • 1, o x, . In other words, in the non-dyadic case every non-zero regular lattice has an orthogonal base. 92: 1. Let L be a unimodular lattice on the quadratic space V over the non-dyadic local field F. Then there is a unit 8 such that L.r. <1> ••• ± <1> _L <E> .

Proof. Since L has an orthogonal base we can write L 2-4 <8 1 > _L • • • I <811> (Si E u) . Put s = e en . Then <1> • • • <1> ± <8> by the criterion of Theorem 63:20 in virtue of the fact that the Hilbert symbol `3' ) is 1 whenever 6, 6' are units in a non-dyadic local field. FL

(-!—

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

247

So there is a lattice K over V with K <1> 1 • • • 1 <1> J.. <e>. Now K and L are 0-maximal on V by Proposition 82:19, hence they are isometric by Theorem 91:2. q. e. d. 92 : 1 a. There are essentially two unim,odular lattices of given dimension over a given non-dyadic local field. 92: lb. Q (L) u if dimL = 2. And Q (L) = o if dimL 3. Proof. If dimL = 2 we apply the proposition twice to obtain

L <1> I <e> I. <de> for any (5 in u. Hence ô E Q (L) as required. If dimL 3 we must prove that a E Q (L) for any a in p. But by the case dimL = 2 we have L <— 1> I_ <1 ± a> - • • . Hence a E Q(L). q. e. d. 92:2. Theorem. 1 Let L and K be lattices of the same Jordan type on the regular quadratic space V over the non-dyadic local field F. Consider Jordan splittings L ± • • • ± Li K = ± • • ± Kt . Then clsL clsK if and only if dL i = dK i for 1 Proof. 1) First suppose that clsL = clsK. Then K aL for some a in 0 (V), so we can actually suppose that K = L. Consider L i and K i. By suitably scaling the bilinear form B on V we can assume that Li and K i are unimodular. Then Li and K i are the first components of Jordan splittings of L°. We may therefore assume that i 1. We saw in step 2) of the proof of Theorem 91:9 that there is an F-linear isomorphism gi: FKIL with ç)L1 S K and such that B (Tx, Ty) B (x, y) modp for all x, y in LI. Take a base L1 — • • • + 0x,.. Then 994 is a sublattice of K1 which is on FKi and has a base 91-1 = 0 (Tx:0 + • • • + (Tx?) Write dB (x1, . , x,.) = e. Then e is a unit since Li is unimodular. Now dB (px,.) -=-- e modp.

Therefore d B (47) . . . , x,.) is a unit. Hence D (974) 9)L1 = K1 by Corollary 82:11a. So K1 has the base

o

—» Kr So

0 (ço xi) + • • • + 0 (99 x,.) . 1

O. T.

A similar theorem holds for representations in the non-dyadic case. See O'MEARA, Am. J. Math. (1958), p. 850.

248

Part Four. Arithmetic Theory of Quadratic Forms over Rings

Hence dK, s (1 + a) with cc in p. Hence dKi s by the Local Square Theorem. Hence dK, = 2) Conversely suppose that dL i = dfCi for 1 i t. Then it follows easily from Proposition 92: 1 that there is an isometry ai of FLi onto FICI such that o•i L i = Ki . Put = • • • I at . Then a is an element of 0 (V) such that cf. = K. Hence K E cIsL. Hence q. e. d. clsK = clsL t. 92:2a. Corollary. cIsL = clsK if and only if FLi F Ki for 1 92: 2b. Corollary. Let L(, ) C» • • C L(0 and K(1) C • • • C Kw be the Jordan chains associated with the given splittings of L and K. Then clsL clsK if and only F K(i) for 1 I t. 92:3. Theorem. Let L and K be isometric lattices on the regular quadratic spaces V and U over the non-dyadic local field F. Suppose there are splittings L = L' I. L" and K = K' J. K" with L' isometric to K'. Then L" is

isometric to K". Proof. This is an easy application of Theorem 92:2 and is left as an q. e. d. exercise to the reader,

92:4. Theorem. Let L be a lattice on a regular n-ary quadratic space over a non-dyadic local field. Then every element of 0 (L) is a product of at most 2n —1 symmetries in 0 (L). Proof. Let V be the quadratic space in question. The proof is by induction to n. The case n = 1 is trivial', so we assume that n> 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=-,- <1> ± • ± <1> .

Hence there is a lattice L on V with Lc <1> j. <2>

...

< 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.

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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 ,

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

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 .

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

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.

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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'.'

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

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
(0, 0))

E -fr)

If = a for each i, then J is a-modular and we call the adjunction of J an a-modular hyperbolic adjunction. 93: 14 a. Corollary. A lattice L on a regular quadratic space over a dyadic local field has splittings L = J J K and L J j K1 with J isometric to J1. Suppose that J is a-modular with ..

giSg Ka and gJ1 ÇgK. Then K is isometric to

Proof. By scaling we can assume that a --- 0. Adjoining j-1 to L shows that we may assume, without loss of generality, that J and L. are isometric unimodular lattices on hyperbolic spaces with gjSgK° and g S gig'. Now by Proposition 82:17 a unimodular lattice on a hyperbolic space has the form (oci, 0)) ± • - • 1
17

258

Part Four. Arithmetic Theory of Quadratic Forms over Rings

hence we have a splitting L

(A (cri, 0)) ± • • • ± (A (a,.,0)) j_ K

in which the ai are in g f and hence in g K°. Successive application of Example 93:13 now gives L

(21(0, 0)) ± • • • ± (A (0, 0)) K

Similarly (A (0, 0)) j • • • ± (A (0, 0) > 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)) ± <1>

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 <4> 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)

<Œ>

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.

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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


(iii) Next we consider dimL = 4 with ordp a ordp b odd. Here we suppose that d has been expressed in the form d = 1 + cc with b (d) = oco. It follows from Example 93:10 that we must have a Eab 0. Take lattices J, K on quadratic spaces over F with .1 (A (b, 0)) K (A (a, — (oc — 4e) a-4)) 1 (A (b, 4e b-')>. Let us prove that L is isometric to J or to K, but not to both. First we note that every number represented by J is in ao2 bo g L, hence g/ C gL; but a is a norm generator of J and J represents the number b with ordp a ordp b odd, so b E wJ , hence g j = ao 2 to f 2 g L . Similarly with K. Hence we have proved that g/=gL.gK.

It is easily checked that di = dL dK

If we compute Hasse symbols using the rules of § 63B we obtain Sp (F j) — S (F . This shows first of all that L cannot be isometric to both J and K.

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

261

Secondly, it shows that FL is isometric to one of F j or F K, by Theorem 63:20, hence L is isometric to j or K by Theorem 93:16. (iv) Now let dimL = 3 with ord p a 4- ord p b odd. We claim tht

L

(A (b, 0)) ± (— d>

or

L

(A (b , 4 b-'))

(— d (1 — 4 ) > ,

but not both. This can be derived from the 4-dimensional case. Take a unit e in Q (L) and adjoin <e> to L to obtain a lattice K with (e) j. L. Then gL = gK and mi. tr•K so that

K

(A (e, . .

± (A (b, 0) >

or

K

(A (e, . . ) > _L (A (b, 4 e

Therefore

(e)

j

L

<e> J (. . . >1 (A (b, 0))

or

<e> 1 L

<e>

(. . . >


>

By Corollary 93:14a we can cancel <e>, so L has the desired form. A computation with Hasse symbols shows that L cannot have both the given forms. (v) If dimL 5 we claim that

L

(A (0 , 0) > 1 . • • , and so gL = Q (L). We can assume that ordp a 4- ordp b is odd. First we take a quaternary unimodular sublattice of L which has norm a o. If this quaternary lattice has weight b p 2 r + 1 (r 0) it will have the form (A (0, 0)) I . . . by (ii) and we will be through. Otherwise it will have weight b p2r (r 0) in which case we obtain L (A (b 0.1 2r 2 co > K for some a in o by (iii). Here nK is still ao. If we repeat the above procedure on a suitable ternary or quaternary sublattice of K we either obtain IC (A (0, 0) > ± • - • , or else L (A (b n2r , 2x)) J (A (1)7E2' , 2 /3)) I • • for some I 0 and some f3 in o. In the first event we are through. Otherwise we can assume that r t, say. Then L (A (2 b ;Or, 2 a) ) _L - • • L". (A (0 , 0) > 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

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

by (ii) and Example 93:9. If ordp a + ordp b is odd, then L (A (a, O)) J
(A (a, 4a-le)) j_

(b, 4 b-i e) .

The first of these is clearly isotropic, and the second is not by Example 1, then 63 :15 (i). So we have the following result: if dimL = 4 and dL L

(A (a, 4a-1 2)) ±
with A = 0 when FL is isotropic (in particular when ordt, a ± ordr b is even) and with A = e when FL is not isotropic. 93: 19. Let L be a modular lattice in a quadratic space over a dyadic 3. Consider any 2 E g L. Then local field with sL S p and dim L (A (a, 13)) 1 L (21(a + 2, ()) K with gL c gK. Proof. Put sL = o with E F. Then by Examples 93:13 and 93:18 we have (v) (A (a, fl)) 1 aA(0,0)>IL L-4 (A (a, p)) 1 (EA ( -1. 2, 0) )1. L .)) (A (a + 2, 13))

We have gL S

(A (a + 2, )> ± ( 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

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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
with a E gi , E gi+1 when ui ui+1 is odd, and we put

cui+ tii+0+ 8i ±2p 2 5 ffi = b (cc fi) with a C gi 13 C g11 when u i ui±i is even. These ideals fi also depend ,

just on the fundamental type. It is clear that in all cases fi

The modular lattice L i is the first component in some Jordan splitting = Li ± • • •

of L 6i. Hence we have

g L i gL6'= g i . We shall call the given Jordan splitting L = L1 I • • • I L i saturated if

gLi gi for

1

t

If L = L1 ± • • ± L i is saturated, then ai is a norm generator of L i and

wi —wL i , since L i and L 6i have the same norm group and the same scale.

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

265

93:21. Suppose L has a Jordan splitting L = L 1 1 • • • I L t in which dim L1 3 /or 1 i t. Then L has a saturated Jordan splitting. Proof. 1) First let us prove the result in the case where all dim Li 7. Take a set of fundamental norm generators a t and a set of fundamental weight generators b1, . . bt. So

for 1

g i = aio2 bi o

Consider a single index i and put 93:18 L i has a splitting L.

o

a A (0, 0)) ±

with in F. Then by Example

aA (0, O)> ± • - •

and this induces a splitting L

<EA (0, O)> ±

Now

aA (0, O)> j. J .

hence ai, bi are in gr°. If we now apply Example 93:13 twice we find L

aA(

-3. at , 0)) ± a A (t bi, 0)) I

I•

Hence we have the following: if we start with a Jordan splitting L = L 1 ± • • • ± Lt there is another Jordan splitting L = K1 1 • ± Kt in which ai, bi E gKi and K 5 n-2 L i for j +1. It then follows from the properties of the weight (§ 93A) that bi o ( gK i. Hence gi = ai 02 b io gK i R gi ,

and so gi = g Ki . Do all this for î = 1, then for i = 2, etc. Ultimately we obtain a saturated Jordan splitting for L. 2) Now assume that dim L 1 3 for 1 i t. Make a 4-dimensional s -modular hyperbolic adjunction of a lattice H1 to L. Then repeat with 62, etc. We obtain a lattice L with the same fundamental norm groups as L and a Jordan splitting L =(H1 1_ Li) ± • • • 1 (Ht Lt)

with dim (Hi ± L i) splitting

7 for 1 < i t. But L has a saturated Jordan j_ • • •

Kt

by step 1). Now K1 H. ± ji by Example 93:18 since dimHi =4 and dim K1 7. Hence we have a splitting L (H1

JD ± • • I_ (He ± Jt)

with g ji == gK1 g i. By Theorem 93:14 L-=' A

Hence L has a saturated splitting.

•

-LIt •

q. e. d.

266

Part Four. Arithmetic Theory of Quadratic Forms over Rings

9 3 :22. Example. Suppose the Jordan splitting L = L1 ± • • • I L t has dim L1 3 for 1 S t. Then it follows from the definition of a saturated splitting and from Proposition 93:21 that the given Jordan splitting •• • ±L't is saturated if and only if it has the following property: if L. is any Jordan splitting of L, then gL t g./.4 for 1 t. So in this case a saturated splitting is one that maximizes all the groups g L i. 9 3 :23. Example. What happens to the fundamental invariants under scaling? If L has the Jordan splitting L • • • ± L t , then LŒ has the Jordan splitting Lee = L J • • L. Hence the new fundamental ...

-

scales are ael. )•••Dast .

We have g (Lec ) e4 = g (L)Œ = CC gLel = a gi , hence the new fundamental norm groups are otgi-•••2 0cgt-

Similarly with weights. So we obtain a new set of fundamental invariants t, dimLi, cc si, cc w., cc cti for La. It follows from the definition that the ideals fi'.. , fi _l are the same for La as for L. A Jordan splitting is saturated for L if and only if it is saturated for L. 93:24. Example. What happens to the fundamental invariants as we pass from L to L#? Consider a Jordan splitting L =-- L1 1 • • 1 Lt. Then by § 82F and Proposition 82:14 we get a Jordan splitting L4L--

••-

Lr=

x-sgL t i • • • ± x-8.4

in which the fundamental scales of L# are The fundamental norm groups and weights of L and L# are not altered by making si-modulas hyperbolic adjunctions to L. Let us therefore assume that dim L i 3 for 1 I t. Suppose the given Jordan splitting of L is saturated; then it follows easily from Example 93:22 and the fact that L## = L that the Jordan splitting 7r-8, L t ± • • • I

is saturated. Hence by Example 93:7 we have —

i+1

or

-2 9 _, wt-i+i 2

alt=

where the it symbol refers to the invariants of LA. Hence for 1 i we have f *- f 1— • .

t-1

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

267

It follows from Example 93:24 that the fundamental norm groups and weights satisfy 93:25. Example.

n -28: gt

•

a -28, 111.

71 --2 8tm t

Hence

6r 2 at

61 2

ideals fi can be expressed in terms of the fundamental invariants. Consider the lattice L with fundamental 93:26. Example. The

invariants

t, dim L 1 ,6i ,w 1 , ai .

Consider any i with 1

i < t— 1. Then it is clear by inspection that = ai ai+l o

when u • u i 41 is odd. And a straightforward computation using the fact that ai ci-2 ç a11 si-±21 will show that (Ug+ Ui+i) 8i

5F fi (aiai+i) + + ai+1w2: + 2 p" when ui ui +1 is even. 93:27. Example. Let us continue with the preceding example and compute sff i in certain special cases that will be needed later. We know that u, 1 ui since g 1+1 Ç g i. We claim that { ai±troi 6i2 fi .

ai ai +1 o

a1 w 11

if if if

U i +1

U1+1

= u. Ui

± 1

Ui

+ 2 si+1 = si +

1.

The first two computations are direct. Let us do the third. We know from Example 93:25 that n-28€-Fxg 1+1 D n-28i g i , so in the present situation we have n2 gi gi +1 cg1, hence n2 a1 may be taken as an (i V h fundamental norm generator, in other words we may assume that ci1+1 = 7r2 a1 . We also have n2 W 1

w i +1

Now substitute these values in the second formula of Example 93:26. We find that sf fi = ai w i 44 as required. § 93F. Determination of the class 93:28. Theoreml. Let L and K be lattices on a regular quadratic space over a dyadic local field F, suppose that L and K have the same fundamental I A different classification involving the so-called "Gauss sums" can be found in O. T. O'MEARA, Am. J. Math. (1957), pp. 687-709.

268

Part Four. Arithmetic Theory of Quadratic Forms over Rings

invariants, and let

Lq) ( • • • C L( i) and K( .1) C- • • (K be Jordan chains for L and K. Then clsL = clsK if and only if the following conditions hold for 1 I t 1:

(i) dL( i)IdK( i) 1 modfi (ii) FL(i) FK(i) (iii) FL(1) F K( i) j..
when when

fic 4 ai +1 tv-ai

fi C 4ai wi71 Proof. Let V denote the quadratic space in question. Let L=

-•-

Li , K

1. • • • 1. K t

denote Jordan splittings which determine the given Jordan chains. The fuhdamental invariants t, dim Li, ai refer to both L and K since it is given that L and K have the same fundamental invariants. Recall the auxiliarly notation si = ordp 6i and ui = ordp ai. Let b, .. bt denote a set of fundamental weight generators. Proof of the necessity. Here we are given that clsL clsK. Hence there is an isometry a E O (V) such that K = a L. Now a preserves the fundamental invariants, and if (i), (ii), (iii) are established for K and a L they can be carried back to K and L by a-1, hence we can assume that K = L. The above Jordan chains and splittings now refer to the same lattice L. 1) The first step is to prove (i) and (iii) for the following special case: — 1, FL 4-dimensional hyperbolic, LI unimodular, gLi gK, = If u1 = u2 we have h = a w1 by Example 93:27; but dL i. = 1 and dK, c-2. 1 moda , eh by Example 93:10; hence dL/friKi 1 modh and we have (i); condition (iii) is vacuous since w1 D 261 = 2e implies that ,

,

,

4 ai

Hence the special case is proved when u1 = u2. We therefore assume that u1 < u2. We now choose a base (exi ex2) j. (0x3 + in which <2,1(a 0)> 1, where a; and ai' are norm generators of PI; the existence of such a base follows easily from the fact that FL" is hyperbolic and that LI has norm aio. Now each of the vectors xi (1 i s-7. 4) is in L = K, j . Kr2) , hence we can find vectors yEK,, E Kr2)

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

for 1

i

269

4 such that xi = Y

Then Yi E Lel and so f B (xi, xi) = B(31 , 31,) mod 5 2 Q (xi) Q (y e) mod g2

for 1 S I

4. In particular we have

4 and 1 j

B (xi , xi) = B (yi , 5/I ) modp so dn(y1 , • y4) is a unit, hence oh + • • • + oh is a 4-dimensional sublattice of K1 with volume equal to o, hence = 0y1 + • • • -}-0y4.

Now solve the pair of equations

B 3/ 2) + B (y2, y 2) + for E3, 773. We have {

Q(y 1 ) + 773 B (yi, y2) = 0 B Y) + 273 Q (Y =

(311) Q(y2) — B (y1, Y2) 2 EU,

and so the solutions 3, 713 are in s2. Then 34 = Y3 + Eah.

2702

is a vector of K1 which is orthogonal to both Yi and y2. Similarly obtain 3/4 = Y4 + .1)11

orthogonal to y 1 and y2 with = ( 11

174Y2

and 774 in 62. Then 0 Y2) -I- (0 )4 + *A) •

We find that we still have {

for 3

B (xi, xi) B yi) mod6 2 Q (xi) Q (y;) mod g2

4 and 3 j of f we find that

4. By a direct calculation using the definition

dB (yi., Y2)1 modh ,

dB

—1 modh ,

hence &K.,. 1 mod f1. Hence dL ifeiK i. , 1 mod h. This proves (i) in the special case under discussion at the moment. We must prove (Hi). We assume that h C 4 al toTl. We have Q(y i )o = = n.K i since Q Q(x 1) modg2, and since g 2 Ç al p by the assumption that u1 < u2. Similarly with A. So we have just found a splitting = J j_ J'

270

Part Four. Arithmetic Theory of Quadratic Forms over Rings

in which J and J' have norm ale and

dJ

—1 mod

, d

—1 mod fi .

Then by Example 93:17,

.T=. 21 (aicc2 + fl, 71) with a E v, flEw1, vE aT'f1 c 4wr'p. Hence =.
A> I r 1_
Now

J'J .

(2% 0C2 + fl) y E (w1) (41V/71) hence is isotropic by the Local Square Theorem and Proposition 42:9, hence

<%) =. J r L instead of J • This gives ±


.)>



<•••>.

Hence F.Kl

I. < • • > .

Hence FL/ ± , as required. 2) Now we prove (i) and (ii) in general. By making suitable hyperbolic adjunctions we can increase all the fundamental dimensions without changing the remaining fundamental invariants. If we can prove conditions (i) and (ii) for the enlarged lattice, then we shall have them for the original lattice L by cancellation. In effect, this allows us to assume that all fundamental dimensions are 3. Next we take a saturated splitting ji. ± • • • IL and make the adjunction • • • _Lit to L. The only fundamental invariants that are changed by this adjunction are the fundamental dimensions, and it suffices to prove conditions (i) and (ii) for the enlarged lattice. In effect this allows us to assume that the given Jordan splittings are saturated. A similar argument involving the adjunction of the lattice

. Li-1 ± • • 1. Li4 allows us to assume that each space FL i (but not necessarily each TWO is hyperbolic. Suppose we wish to verify (i) and (ii) for a specific value of i. By suitably scaling V we can assume that Li+, and K i +1 are unimodular. Now Lti 1) is a hyperbolic space. Hence by Proposition 93:3 there is a

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

271

unimodular lattice J on FLit + » with Lri+i) _Ç J FLA+ » and g J =g Put L' = L + J. By considering the Jordan splitting L' = L1 ± • • ± L j. J it is easily verified that the first i -I- 1 fundamental invariants other than the fundamental dimensions are the same for L and L'. Obviously the first i fundamental dimensions also agree. So the above splitting for L' is saturated. If 1 A j the lattice KA is 62-modular with $2 D o, and B (KA, L') cC B (KA, L) B (KA , J) ç B (KA, L) B (L', sa Hence KA splits L' by Proposition 82:15. Hence by Proposition 82:7

there is a splitting L'

± • • • 1.

±I.

It is easily seen that this splitting is saturated. If we can prove (i) and (ii) for the lattice L' at i, then we shall have these conditions for L at i. In effect this means that we can make the further assumption that i = t —1 with L t unimodular. By Example 93:18 we have a splitting L e = L j . Lr with dim/4 = 4 and Lr
.

ft

_,

dL eldK t L, .-- 1 modf? .

by Example 93:24, and dL( t _i) • dL t dK( t _i) dK t, 1 modft _i .

We have therefore completely proved (i). Now (ii). We suppose that h-1C 4 at WTI. By Example 93:24 this condition translates into C 4 ar(w4t) -1 . So by step 1), FL FK j <—ar>. Hence FL<•••> KTI I since FL is hyperbolic. Adjoining V to both sides gives F.L(t _i) j. FL t i,FLt _L<- • •> FK( t _D _LFIfe lFifT1 ± . .

272

Part Four. Arithmetic Theory of Quadratic Forms over Rings

Hence FL(t _1) F 4_0 ± by Witt's theorem. 3) Finally (iii). Consider any i for which fi C 4 ai zio.V.. Then IA a?-i+1 (tv?-i+1) -1 by Example 93:24. But this is condition (ii) for the lattice L# at t And (ii) was completely established in step 2) of this proof. Hence

F.Kil+1)

4)

J
.

Adjoining V to each side gives FL(i)

j

FL;+1) FK'4 ÷/) J < • • >

FK(i) j FKri+i) F ..

.

Hence FL(i) FK( i) J by Witt's theorem. This completes the proof of the necessity. Proof of the sufficiency. 1) We shall prove the sufficiency by induction on the quantity st — s1 O. We start with st = sr Then L and K are modular of the same scale. And gL = gK since L and K have the same fundamental invariants. Hence clsL = cliK by Theorem 93:16. So we are through with the case st = s1. We therefore assume that si > 0 (and so t> 1) and we proceed with the induction. 2) By making suitable hyperbolic adjunctions we can increase all the fundamental dimensions without altering the other fundamental invariants or the given conditions. By Theorem 93:14 it is enough to prove the isometry of the enlarged lattices. In effect this allows us to assume that all fundamental dimensions are 3. Now take a saturated splitting L =J1 .1 • • • 1_ t and make the adjunction of ji ± • • • ± it to L and to K. The only fundamental invariants that are changed by this adjunction are the fundamental dimensions; and if the given conditions hold for L and K they will also hold for L j . L and L j K; if we can prove that L L and L j.. K are isometric, then we can obtain an isometry of L onto K by using Corollary 93: I4a to cancel off one J. at a time. In effect this allows us to assume that the given splittings of L and K axe saturated.

A similar argument involving the adjunction of the lattice L -1 Lr 1 ± • • • I LT1

allows us to assume that each space FLi (but not necessarily each FKi) is hyperbolic. Now by Example 93:18 we know that each Li (or Ki) is of the form L1 A (O, --A (O, with dimg = 4. By making suitable cancellations to the L i and K i we see that we can assume that dimLi dimKi 4 for 1 I t. Finally we scale V to make L1 and K1 unimodular.

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

273

So we have reduced things to the following: we are given lattices L and K with the same fundamental invariants in saturated Jordan splittings

L=

K=

••• I K t

which satisfy conditions (i)—(iii), the first components L1 and K1 are unimodular, and all spaces FL are 4-dimensional hyperbolic, and we must prove, using the inductive hypothesis, that L and K are isometric. Note that Example 93: 18 (vi) asserts that L1 has the form L1 ,-_1-4, (A (al, C)> I . 3) The splitting L = L1 I•••IL t will be left fixed throughout. The computational part of the proof consists in successively changing the Jordan splitting K1. I K2 until a new splitting is obtained in which K1 is isometric to L1 . It is understood that each splitting employed in this procedure induces a saturated splitting on K. We shall use (without further referênce) the rules of Example 93:13 and Proposition 93:19 to effect these changes to K1 I K2 . In connection with this matter we must mention that conditions (i)—(iii) will continue to hold if the given splitting K1 I • • • I K t is replaced by any other Jordan splitting K I • I K. For dL( i)/dK( i) ,- 1 mod fi by hypothesis, and dK (i)/c/Kr(i) 1 mod fi by the necessity of this theorem, hence dL (i)/c/K i) 1 modfi and so (i) holds in the new splitting. Let us do (ii). Assume we have ft c 4 a t+1 tv,71 11 . Then FL (t) FK(i) I by hypothesis; now the discriminant shows that we can take ai+ ai +1 mod ai +1 ; C tto i +1 ; so a but ai +1 fi c 44+1 is a fundamental norm generator; hence we have an isometry FK(i) I < • • > FK ) I by the necessity. Then FL() I (a:: + 1 > I FIC(i) I <• • -> F K(i) I <6 ) I F K'(i) 1 as required. We establish (iii) in the and so FL (i) same way. This shows that we will be through once we obtain a new saturated Jordan splitting K = K1 ± • - • I K, in which L1 K1 . For the new splitting will still satisfy conditions (i)—(iii). And we can assume that — K1 by Witt's theorem, so LN and Kr2) become lattices on the same quadratic space FL ) . But the first , . . , (t — 1) th fundamental invariants of Lr2) and Kr) ) are equal to the second , , i invariants of L and K since the splittings L1 I•••IL t and K1 I•••IKt are O'Meara, introcluction to quadratic forms

18

274

Part Four. Arithmetic Theory of Quadratic Forms over Rings

saturated. Hence Lr2) and KA) satisfy conditions (i)—(iii), hence they are isometric by the inductive hypothesis. Of course this implies the isometry of L and K. 4) Now we start the computational part of the proof. First suppose that u2 = u1 . So a2 o = ai o. In this event we have fi a2tv/ by Example 93:27. We know that dLi. = 1. Let us arrange dif/ 1. By condition (i) we have c/K/ 1 modfl, hence d.K 1 = 1 + ad. for some A E w1 . Then


.

K2' 1.

± K 1 ± K _L K _L K.

Hence K1 ± K2 has a saturated splitting K1 _L .1q in which c/K1 = 1. In other words we can assume that difi = 1. By Example 93:18 (vi) there is a splitting

I. in which A = 0 when FK/ is isotropic and A. = e when FK/ is not isotropic. If w2 c 4tr•i-I, then fi = a2w1 (4 a 2 w' and condition (ii) holds. Hence in this case we have FL1 FK/, so FK/ is isotropic, therefore = O. Otherwise tv2 4w11 = 4 WI). Then


_L

.

So we can take A. = 0 in all cases. Then K1 LI. Hence K L by the observation at the end of step 3). 5) Next suppose aio = 2o. By step 4) we can assume that a2o S 2p. Hence f/ c. 4p. So d.KI 1 mod4 p, hence dKI = 1 by the Local Square Theorem. Hence by Example 93:18 K i has the form

. But Li has the same form. Therefore L1 K1 and L K by the observation at the end of step 3). 6) Now let u2 = U l + 1. We can assume that ao D 2o by step 5). Then a2 E w1 since a2 E g2 S g/ with ordp al ordp a2 odd, hence al p = a2o is contained in g hence ai o is the maximal ideal in gi , hence g/ = And bl o tv1 a2o since ai o D 2o. Now f/ al a2 c' by Example 93:27, so we can proceed as in step 4) to get a new splitting K1 ± K2 in which //K t - 1. By Example 93:18 there is therefore a splitting ,

f-e. _L
Chapter IX. Integral Theory of Quadratic Forms over Local Fields

275

in which A = 0 if FK„ is isotropic and A = e if FK„ is not isotropic. If i 1 0 we have fi = a1 a2 o (4a2 toy' and condition (ii) holds. This tv2 C 4a— irnplies that FL, FK„ and so A = O. If tvi D4ar1P we can proceed as in step 4) to get a new splitting K11 K2 with A = O. So in any case we can arrange A — O. Then Li K1 . Hence L K. So the case u2 = u, 1 is proved. 7) Next we settle all cases in which .S2 = 1, j. e. in which the second component is p-modular. By steps 4), 5), 6) we can assume that a 2 o Ça1 p 2 and al o 20. Then in fact a 2 0 = a1 p 2 since a2 g 1 c g2 c g„. We have a1 tv2 by Example 93:27, so we can arrange the splitting K1 I K2 to be such that dK„ = 1 as in step 4). Continuing as in step 4) we obtain a new splitting ± K2 (A (a1 , C)> ± (A (b1, OD 1 q and then K L. (There are two variations in this part of the proof: first, one uses (iii) instead of (ii), secondly one needs the inclusion a2 o a 1 p2 4aTio .) 8) Finally we consider the following situation: s2 > 1, a2 oÇ a1 p 2 , al p D 2o. Perform the adjunction of the same lattice J to L and K, where <7. cA (0, 0) > . Put L' = L I J and K' K I J. Suppose we can show that L' and K' satisfy the conditions of the theorem. Then s; si st s, and 4 1, so L' and K' will be isometric by steps 2) and 7). Hence L and K will be isometric by Theorem 93:14. The rest of the proof therefore concerns itself with showing that L' and K' satisfy the conditions of the theorem. We have Jordan splittings —

L'— L 1 1 L i t,

L2 1 • • • 1

K1 1 K 1 1 12 1 K 2 1 • • • 1 K t

Lt ,

in which L 1 1 1 J = K 1 1 1 . It is easily seen that L' and K' have the same fundamental invariants. And it is clear what happens to the invariants t, dimLi , 5 i as we pass from L to L'. What about the new fundamental norm groups, weights, and norm generators ? We find that we can take

(i = 1, 2, . .

t) ,

hence from the definition

f;

fi

(i

-

2,

•

• , t) •

We have (L')P =

j

L2 1 ••• ILt,

hence n (L')P — :7 2 ai ø and we can take ali 12 = a2 a1 as a fundamental norm generator. By Example 93:25, :72 gi C g 11 ç g1

gr2 w1 18*

276

Part Four. Arithmetic Theory of Quadratic Forms over Rings

and g2

g 1 1 /2 P W 2 -C W1 11 2 •

Hence from the definition

-

fi

Ti fi ,/ It is now obvious that L' and K' satisfy (i) for all i. Also that they satisfy (ii) and (iii) for i — 2, . . . , t. Also that they satisfy (ii) for i = 1'/ 2 . We see that (ii) is vacuous for i 1 since then 2

wit h fi = a1 wIt h with 7 4 al p 2 = 4 ait h . We leave the verification of (iii) for 1 — 1 and i= 1'12 to the reader. q. e. d. § 93G. The 2-adic case The ideal 21) in a dyadic local field F is always contained in the maximal ideal p since the residue class field has characteristic 2. Let us suppose for the rest of this subparagraph that we actually have 20 = p, or equivalently that 2 is a prime element in F. We shall call such a field a 2-adic local field in order to signify that 2 is one of its prime elements. (The field of 2-adic numbers Q2 is an example of a 2-adic local field.) Otherwise let the situation be the one already under discussion in this paragraph. In particular let L be a lattice in a regular quadratic space V over F with the Jordan splitting L

j_ • • •

Lt

If a is any fractional ideal in F, then there are no fractional ideals properly situated between 2a and a. Hence n L is either 5L or 2(L). So by Theorem 91:9 the ideal tt L i is independent of the Jordan splitting L 1- • "L t. We define

t ni n L i for 1 (This definition applies only in the 2-adic case since otherwise the ni not uniquely determined by the Jordan splitting.) We call the are quantities t, dim L1 , s i , ni the Jordan invariants of L. It is clear that two lattices have the same Jordan invariants if and only if they are of the same Jordan type. We put

ui ordp ni We have

nL1+1 ç sL i ç 2(sL 1) ç nL i and so th 7 n2

•••

.

What is the connection between the Jordan invariants and the fundamental invariants of L? The maximal ideal in L in the norm group

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

277

gL must be all of nL since 2 (BL) ( na SnL with ordpmL + ordp nL even. Hence gL =- nL and toL = 2(sL) . Then gLei = nLei = nL i and so gi — ni , tvi = 2si for l

it.

In particular this shows that the quantity ui = ordp ni is equal to the quantity ui = ordp ai defined in § 93E. We can take any element of order ui as a fundamental norm generator, in particular we can take ai . Th. The lattice L now has the following set of fundamental invariants: t, dimL i, si , 2,

This shows that any two lattices with the same Jordan invariants (over a 2-adic local field) have the same fundamental invariants. 93:29. Theorem.' Let L and K be lattices on a regular quadratic space over a 2-adic local field F, suppose that L and K have the same Jordan invariants, and let L6.) C • • • C L( t) and Kb) C • - • (K be Jordan chains for L and K. Then clsL = clsK if and only if the following t — 1: conditions hold for 1 i

(i)

dL( i)IdK( i) e. 1 modni ni _ols., (ii) FL(i) -4— FIC(i) i (2ui> when n1+1 (4n. Proof. 1) We obtain the necessity by suitably interpreting Theorem 93:28. Take a i = 2" for 1 ._ i_.< t. Then it follows by direct calculation from Example 93:26 that s1 fi D ni ni+„ and that fi ç 80. 61 fi D ni ni+1 Hence (i) is a consequence of condition (i) of Theorem 93:28 and the Local Square Theorem. We find that n i +1 S 4ni implies that fi c 4ni triTi, therefore (ii) is a consequence of condition (iii) of Theorem 93:28. 2) The sufficiency is proved in essentially the same way as the sufficiency of Theorem 93:28. Let V denote the quadratic space in question. Let L = LI _L • • - _L L t , K . K i _L - - • _L K t denote Jordan splittings which determine the given Jordan chains. We shall prove the sufficiency by induction on the quantity t. If t = 1 we See O. T. O'MEARA, Am. J. Math. (1958), pp. 843-878, for the solution of the representation problem over 2-adic fields. A general solution over arbitrary dyadic fields is not known at present, but much progress has been made on this difficult problem by C. RIEHM, On the integral representations of quadratic forms over local fields (Princeton University thesis, 1961). 1

278

Part Four. Arithmetic Theory of Quadratic Forms over Rings

have gL1 n1 gKi and so clsL = clsK by Theorem 93:16. We therefore assume that t> 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


2(70> ± (A (2, 2 i3)> with a, 9 E 0, hence by Proposition 93:19 K1 ± K2 (.11 (2u1, j_ with 2. equal to 0 or e and the same with #. But n2 = 4o Ç 4n1, hence ,

FL,

,

<s)

± <1>

where. e is the unit (1-42.) (1 - 4 It). A computation of Hasse symbols, say, will show that # = 0. Then

± K2

j_ j_ .1q

since g K2 = n2 4o. But


j_ .

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

279

Finally let n2 S 80. Then dL I/dK i ._—_.t 1 mod8o, hence difi = dLi = 1. But n2 S 8o S 4n1, so that FL, ± (1>FK1 ± <1>, hence F Ki is isotropic. Hence _L q. e. d. § 94. Effective determination of the invariants Let us show how to compute the invariants of the local integral theory. We suppose that L is a regular lattice given in the base L = o - - • + ox, and that the symmetric bilinear form B is determined by specifying the values B (xi , xj) = aij . (In practice it is just the symmetric matrix (aii) that is given.) To find the scale we simply take the largest of the ideals B (xi, xj ) o, i. e. 5L = f (a • • o) •

To find the norm we take the largest of the ideals Q (xi) o and 2 (L), i. e. (a i o) 2 (5 L) . nL These rules are justified by Proposition 82:8. Next we show how to obtain a Jordan splitting from the given base. First the non-dyadic case. If there is no xi with Q (xi) o = sL, then there will be two distinct basis vectors, x1 and x2 say, with B (xi, x2) o applying the Principle of Domination to the equation Q (x1 + x2) =Q (xi) Q (x2) + 2B (xi, x2) shows that Q (xi + x2) o = 5 L. (Note that this will not work in -the dyadic case.) Hence a minor adjustment to the given base allows us to assume that Q (x1) o = BL. All the coefficients B (xi, xi)fQ (xi) are now in o, hence the vectors xi, y2, . . y. with Yi = xi

B (xi, x1)

form a new base for L and in fact we have a splitting L= ox1 ± (oy2 + • • • + oyn) . Repeat all this on o y2 + - • + oyn, etc. Ultimately we arrive at an orthogonal base for L, and a Jordan splitting is formed by suitably grouping these basis vectors. The dyadic procedure is somewhat longer. If there is a vector, say xi, with Q(x1) o = sL, use the non-dyadic method to split off o xi. Otherwise Q (xi) oc sL for 1 i n. Then B(x1 x2) o=sL, Q (xi) oCeL ,

280

Part Four. Arithmetic Theory of Quadratic Forms over Rings

say. This implies that the equations

B (x1, xi) + Ei Q (x1) + B (x1, x2) = 0 B (x2, xi) + Et B (xi, x2) + 1h Q (x2) = 0 have integral solutions X1, x2, y 3 , • • • , yn with

Ei ni for ,

3 I

n. Hence the vectors

yi = xi + ni x2 form a base for L, and in fact there is a splitting

L (o + o x 2) ± (0y3 + • • • + oy.„) with O X1 + O x2 modular. Repeat on oy3 + • • • + oyn, etc. Ultimately we obtain a Jordan splitting for L. The Jordan splitting is all that is needed in the non-dyadic theory. The Jordan invariants are also needed in the 2-adic theory, but these can be read off from the Jordan splitting. So Theorems 92:2 and 93:29 can be applied in practice. The general dyadic.theory requires a knowledge of the fundamental invariants ai and wi (the invariants fi are obtained from the fundamental invariants using Example 93:26). First let us find a norm generator a and the weight t0 of L. The norm generator is easy: if nL = 2 (s L) take any a with ao ---- 2 (s L) , otherwise take a = Q (xi) for any value of I which makes Q (xi) o largest. The weight is then given by the formula w=f ab(Q (xj)Ia) + 2 (s L) .

We leave the verification of this formula to the reader. In order to find , w t we apply this the fundamental invariants a1,.. ai and w1, Pg. These lattices ' can be procedure to each of the lattices Lei, expressed in terms of the Jordan splitting L L 1 I • • • ± L t by the formula (grgi--814) -••± Li _i) ± L i ± • • .

§ 95. Special subgroups of 0.(V) We conclude this chapter by giving the structure of the groups

n Z,,, On'

, 0„410„'

of a regular n-ary quadratic space V over an arbitrary local field. Recall that we first raised this question over a general field in § 56, and that we have already described these groups over complete archimedean fields and over finite fields in §§ 61 and 62. 95:1. Let V be a regular n-ary quadratic spq.ce over a local field F. Then 0;, .Q.„ with the following exception: V is a quaternary anisotropie space over a non-dyadic field. In the exceptional case we have (0'4 : D.4) - -- 2.

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

281

Proof. 1) If V is isotropic, in particular if n 5, then On = 12„ by Corollary 55:6a. If 1 3 we have On,' = Q Proposition 55:5. So the quaternary anisotropic case remains, and we must prove that 04 = Q4 if F is dyadic and that (OI: Q4) — 2 if F is non-dyadic. Note that d V = 1 by Proposition 63:17. If cc is a non-zero scalar in Q (V) we let .r. „› stand for a symmetry with respect to a vector x with Q (x) = oc. This is not a precise notation since -Q,c> is not uniquely determined by oc. However the coset -r <,(> Q4 is well-defined in virtue of Proposition 55:3. In fact we can write TC1x) - • • r
( ai

E 17)

without fear of ambiguity, and this coset is even independent of the order in which the cci appear. 2) We are ready to do the dyadic case. We must prove that a typical element a of 04 is actually in Q4. Express a as a product of four symmetries, say T(Œ1)..

• T OO

(az

C

Here oc1oc2oc3oc4 is in F2 since a is in 0,'4 .Clearly oc1 oc2 , oc3, cc4 fall in at . . most four distinct cosets of I; modulo F2 . On the other hand (F:F2 ) 8 by Proposition 63:9. Hence there is a e in F such that eocl , eoc2 , eoc3 , ecc4 are non-squares in F. Now V represents —e since V is universal, hence there is a ternary subspace U of V such that V (—e) I U. Here we have dU = —e since d V = 1, hence the discriminant of the space U is a non-square, hence this space is isotropic, hence U represents ocl, oc2 , oc3 , oc4. We can therefore write ,

cr•Q4 —

T(cc i> • • • 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 • • a Here again oc1 cc2oc3oc4 E F2 . If two of the ai are in the same coset of F modulo F2, then so are the remaining two, hence 012 4 — r

> r<> r r(ccj> Q4 — Q4 -

Otherwise we can write Cr 124

TO)

r Q4 •

282

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

<1> I (--/1>

and

L2

(n> 1 <----r/1).

Then L1 I L 2 Ç L. But L cannot be unimodular, for if it were the space V would be isotropic by Example 63:14. Hence by a volume argument

L

L i j_ L 2 .

We shall need the lattice LP defined in § 821. Note that LP

p

I L 2 pL

Since D' is an additive subgroup of L we can form the factor group L/LP. Let l denote this factor group and let bar denote the natural group homomorphism

L --> = LILP Also let bar denote the natural ring homomorphism of o onto the residue class field F of F. We put a scalar multiplication on V by defining

ax

(c

, x E L) .

This is well-defined since p L Ç I)', and it makes V into a vector space over F. It is easily seen that any base for the lattice L1 becomes a base for the vector space F , hence dim V - 2. Put a symmetric bilinear form on V by defining B(, )7) B (x , y) V x, y E L This is well-defined and it makes V into a regular binary space with discriminant —,71. Each a in 04 is a unit of L by Proposition 91:5; and aL = D' by definition of D'; so a induces a mapping 5: V -->- 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)

Chapter IX. Integral Theory of Quadratic Forms over Local Fields

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 <1> ± <1>

± ,

therefore , iv =. TO> TO> .r<x> 2.<,1 > E and so 124 n Z4 = { ± l v}. Conversely suppose that —1 v E124. If-1 were —

284

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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 .

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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,

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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.

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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 <

at all p in T, in virtue of the continuity of multiplication in LFp (Vp). Put a = rui . Tur E 0 ± (V). Then !lap — Ta p < for all p in T or, in the relaxed notation, o— cppil p < e . q. e. d. 101:8. Let V be a regular quadratic space over a global field F with dim V 3. Then 0(0+ (V)) consists of the set of elements of F which are positive at all real spots p at which V I, is anisotropic. Proof. Let R denote the set of real spots p of F at which Vp, is anisotropic. (Needless to say, R may be empty.) Every element of 0(0+(V)) is a product 6f elements of the form Q 1) • - Q

&)

where the yi are anisotropic vectors of V. Each Q (y i) is negative at those p in R at which Vp is negative definite, hence the above product is positive at these p; the same applies at the spots p of R at which VI, is positive definite; hence the above product is positive at all p in R. Conversely, let cc be any element of F which is positive at all p in R. First let n 4. Take any fixed non-zero element 13 in Q (V). Then a 13 is represented by V p at all p in R; and Vp is universal at all remaining spots on F; hence a 13 is represented by Vp at all p on F; hence a /3 is represented by V, by Theorem 66:3. Hence a ,82 = (a 13) fl is in 0(0 1- (V)), hence a is. Now let n , 3. Scaling V by its discriminant shows that we can assume that the discriminant dV is actually equal to 1. Then VI, is positive definite at all p in R. Let T denote the set of discrete spots p on F at which V p is anisotropic. So R i T consists of all the spots p on F at which VI, is anisotropic. Use the Weak Approximation Theorem of § 11E to find an element ,8 of F which is positive at all p in R and which has the following property: both and —a ,8 are non-squares at each p in T. Then by Proposition 63: 17 we see that <—,8> ± V is isotropic at all p in T, it is also isotropic at all p in R, hence it is isotropic at all p, hence it is isotropic, hence V represents #. Similarly V represents a 13. Then a#2 — (a #) 13 is in 0(0+ (V)), hence a is. q. e. d. 19*

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101 : 8 a. If F is a function field, then

101:9. Let V be a regular quaternary quadratic space over a global field F. 'Consider a set of spots T on F such that D4 (Vp) c 0;(V p) for all p in T. Then there is a a in 0;(V) such that ar (24 (V p) for all p in T and ap E .Q4 (V p) at all remaining p on F. Proof. By scaling V we can assume that 1 E Q(V). The set T is a subset of the set of non-dyadic spots p at which Vp is anisotropic. Hence T is a finite set. Let W denote the remaining set of spots p (archimedean or discrete) at which VI, is anisotropic. Again W is a finite set, possibly

empty. Using the Weak Approximation Theorem and the fact that IT, is open in Fp we can find an element 4 in F which is a non-square unit at all p in T and a square at all p in W. Similarly we can find an element in F which is a prime element at all p in T and a square at all p in W. Then z1 is represented by Vp at each p in T since a regular quaternary space over a local field is universal; and z1 is represented by Vp at each p in W since A is a square there and 1 is in Q (V); and Z1 is represented by Vp at all remaining spots p on F since Vp is then isotropic by definition of T and W. Hence Z1 E Q (V) by Theorem 66:3. Similarly E Q (V). Similarly A E Q (V). Hence 1, Li, a, nZI E (V) • There must therefore exists a a in 04 (V) of the form =- <1) (A> (x> r Os A> •

(Here, as in § 95, t(Œ > denotes a symmetry with respect to a vector x with Q (x) a.) Then a is in Qi (V) by definition of 0'4 (V). And ap Q4 (V) at each p in T by Corollary 95: la. And ap E X2 (Vp) at each p in W since we then have =- To> 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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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.

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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.

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

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.

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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.

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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 <1> ± <11>. 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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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

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Part Four. Arithmetic Theory of Quadratic Forms over Rings

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)

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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).

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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. Since S is indefinite for V there is a spot go in D S at which Vq , is isotropic. We fix this spot go. By adjoining all discrete spots in (S u go) to T (and hence to S) we see that we can make the following assumption: S consists of all discrete spots when go is archimedean, S u go consists of all discrete spots when go is discrete. Let x11 . . xn, be a base for V which determines all the given norms It It p. Suppose this base is replaced by the base 6 z1 ,. , ô xn where 6 denote norms with respect to this new is any (S — T)-unit, and let base. Clearly liii p = liii for all p in S — T. Now Proposition 33:8 provides us with an (S T)-unit ô which is arbitrarily large at all p in T. In fact 6 can be chosen in such a way that lizpli p 1 for all p in T, and hence for all p in S. In effect this allows us to assume that the given norms satisfy lizpli p 1 for all p in S. We define the lattice L with respect to S on V by the equation L = o + • • + o x,, .

If p is in S and x is in Vp, then x E L p if and only if 11.x1I p 1. So each of the given vectors xi, is assumed to be in L. The required vector z will actually be found in L, i. c. it will satisfy lIzil p 1 for all p in S. We adjoin to T all those spots p in S which are either dyadic or such that L not unimodular. The enlarged set T is still finite. It suffices to prove the theorem for the new T (since s < 1). We therefore assume for the rest of the proof that qo is either discrete or archimedea.n, that S u go contains all discrete spots on F, that every spot p in S T is non-dyadic with L p unimodular, and that xi, E Lp for all p in S. We illustrate these facts in the spot diagram

apoikiedeen

discrete

2) Pick x E V with Q (x) = 1. By Witt's theorem there is a rotation (pp in 19+ (Vp) at each p in T such that zp =(pp•x. By the Weak Approximation Theorem for Rotations there is a rotation 41 in 0-F (V) with (ppil p arbitrarily small at each p in T. Hence [lax — zpil p

!la — (pplir

Chapter X. Integral Theory of Quadratic Forms over Global Fields

313

is arbitrarily small at all p in T. So there exists a vector y in V with Q (y) = 1 and lb/ — zpl! p <s Vp E T. This implies that IlYllp 1 for all p in T. By the Product Formula, = 1 holds at almost all p; hence by the Strong Approximation Theorem there is a A in 0 such that IIYII:+ I

{

V PE T V p ES—T.

1 Put y = Ay. Then Ily — zp li p < e holds for all p in T. The rest of the proof now consists in using this A and this y to find a vector z in L with Q(z) = 1 and such that liz — O p < s for all p in T. Once this is done we shall have our vector z and we shall be through. It is easily seen that A and y have the following properties: we have a scalar A and a vector y such that A E 0, v E L, Q (v) = A2 with 1 2—lj,<e VpET. Only these properties of A and y will be used in the construction of z. 3) Obviously A is a unit at all spots in T. Let TA denote the set of spots p in S at which A is not a unit, i. e. at which I Al p < 1. Then TA is a finite, possibly empty, subset of S — T. Take the splitting V = Fv _L U. Then L r U is a lattice on U with respect to S, and L p n Up is a lattice on Up with respect to p. (All localizations are taken in Vp .) It is easily seen, say by Theorem 81:3, that (L n U) p = L p n Up VrES. So by Proposition 81:14 there is a lattice K on U such that Kp = L p n Up Vp ES— T and K p Lp n Up with liKpil p ‹s Vp ET. Thus K Ç L. This lattice K has the following properties at the different spots in S: at each p in S — (T u T A) the lattice K p is unimodular since the unimodular lattice L p is split by the unimodular sublattice o p v, hence by Corollary 92: lb Q (K p) =op V p ES— (T T a) ; at each p in TA the lattice L p is unimodular, hence one can use Proposition 82:17 to find a binary unimodular sublattice of Lp that contains y, this sublattice splits L p by Proposition 82:15, hence Kip contains a binary unimodular sublattice, hence by Corollary 92: lb Q (Kp ) up V p TA; illYiip

Part Four. Arithmetic Theory of Quadratic Forms over Rings

314

at each p in T the lattice Kp will contain a maximal lattice of some norm, hence by the last part of Example 91:3 there is an element ap in Fp with < lapip < 14Ip such that Q (K ) 2: ap up V p T • By Lemma 104:1 there is a scalar ts in o which is a unit at all spots in TAU T and which has the following property: every element of tto which is represented by U and also by Kp at each p in S is represented by K. We let To denote the set of spots p in S at which IA is not a unit,

e. at which Lui p < 1. Clearly To is a finite, possibly empty, subset of S — (Ta v T). We now have the spot diagram j.

Mr=

NV.k■

N'kk. dihrete

I

archimedeall

4) We claim that there exists a scalar /3 in F such that 11 — filp 5- liuip 5- 1 , iflip 1 , 1 — 2,2 /32 E Q(K p) Vp ES and flip < 8 V P ET and 1_12 132 E Q(up )

v p 0-2—s

In order to prove this we use the Very Strong Approximation Theorem and its corollary (§ 33G) to find a 13 in F in which approximations are made in the following way at the various spots on F: 1, 1. if p ES — (T TAU T) make in, /32I p !duly, 2. if p E To make 1 )1— 2.-4 1p so small that 11 3. if p E TA make Ply 1, 4. if p ET make /3— 2-4 (1 + exp)lp so small that 11 — 131p < e and

1 1 — #21r 5. if

(To make AN so large that

-(1 ___ ) 2 (l2) = 2. 2 /32 (1

1/ 2.2 (l 2) E

6. if p is a real spot in D— (S v q0) at which Up represents 1 make fi2 E 1/31p so small that 1 7. if p is one of the remaining real spots in D— (S v q0) make I p1p so large that —(1 — 2.2 (l 2) E 8. if p is a complex spot in D— (S v cro) nothing special is needed. Using the results of step 3) it follows easily by direct calculation at each stage of the above approximation that the element /3 chosen in the above way satisfies the conditions stated at the beginning of step 4).

Chapter X. Integral Theory of Quadratic Forms over Global Fields

315

5) Consider the element fi with the properties stated at the beginning of step 4). Then 1— 2.2 P2 E Q (U at all p in D, hence 1 _22 9 2 E Q (u) by Theorem 66:3. Now 1— 2,2 P2 E ,u 0. And 1 _ a2 P2 is represented by Kv at all p in S. Hence 1 --- 2,2 EQ(K) by the choice of ,u in step 3). Take y in K with Q (y) = 1— 2 p 2 and put z = 13 y ± y. Then z is an element of L with Q (z) = 1. Also llYllp < e for all p in T by the choice of K in step 3), hence we have )

vilp max — v1lp , 11Y11p) = max — #1p fivilpi 1IY!!0 < for all p in T. Therefore z satisfies the conditions mentioned in step 2). So the theorem is proved, q. e. d. 104:4. Strong Approximation Theorem for Rotations. V is a regular quadratic space over the global field F with dim V 3, S is an indefinite set of spots for V, and T is a finite subset of S. A rotation Tv is given in 0' (Vp) at each p in T. Then for each s > p there is a rotation o. in 0' (V) ilz

such that 92 pli p< 8 VPET

and

11016 1 VpES—T. Proof. The proof consists of two major steps, the first for n 4 and the second for n = 3. We let . , xn denote a base for V which determines all the norms pp. We can assume that 0 <e < 1. la) We start with n. 4. First suppose that each cpv is a "short commutator", i. e. that each rpv has the form = TUp TVp rliprVp

where ruv and reyv are symmetries of Vv with respect to the anisotropic vectors up and yv of V. By using the Weak Approximation Theorem on the coordinates of the up in the base x11 . , x„ we can find a vector u in V which is arbitrarily close to up at each p in T. So by the continuity of the map x Tx (see Example 101:2) we can assume that Tu is arbitrarily close to ruv at each p in T. Now do the same thing with the vv to obtain a vector y in V such that the symmetry Tv is arbitrarily close to Tvp at each p in T. Then by the continuity of multiplication in L,Fr (Vv) we can assume that we have found vectors u and y in V such that Tu Tv ru — TUp TVp TUp TVP < 8 p E T. Put w = ;u. Then Tw = Tv Tu Tv. Hence we have a pair of vectors u and w in V with Q (u) = Q (w) 0 such that llrurto — (Pp I P < 8 VPET• Therefore it is enough to find o E (V) with llolip = IL for all p in S — T

such that

— ru'rwiip ‹E

Vp E T.

316

Part Four. Arithmetic Theory of Quadratic Forms over Rings

Let L be the lattice L = o + • • ox, on V. We can assume that u and w are in L (replace them by Au and Aw with a suitable A in o if necessary). Let T1 be a finite subset of S T such that L is unimodular with Q (u) = Q (w) a unit at each p in S (T1 u T). By Theorem 104:3 we can find a vector z in L with Q (z) Q (u) = Q (w) such that z is arbitrarily close to u at all p in T1 and arbitrarily close to w at all p in T. So by the continuity of the map x rz we can assume that the symmetry rz with respect to this vector z is arbitrarily close to ru at all p in T1 and arbitrarily close to -r at all p in T. Put a ru t... Then a is in 0' (V). By the continuity of multiplication we can suppose that we have found a a which is arbitrarily close to 1vp at all p in T1 and arbitrarily close to ru Tto at all p in T, in other words such that

1110.11p = 1 11 Icl — wlIp<S

vP E VPET•

It remains for us to prove that --- 1 for all p in S — (T1 u T). Now at each such p both u and z are elements of Lp with Q (u) =Q (z) a unit at p, hence ru and; are units of Lp by § 91B, hence 0.4, is a unit of L p , hence Doll y = 1. 1b) We continue with the case n 4, but now we let the gap be arbitrary elements of the commutator subgroups Du (Vp) at the spots p in T. We can express each Tr in the form 99p =

- - • V;

where /pi'? is a short commutator in On (Vp) (see Proposition 43:6). We can assume that the saine r applies at all p in T, by adjoining trivial short commutators wherever necessary. By step 1 a) we can find vi in 0' (V) with Vi arbitrarily close to 4, at each p in T and My = 1 at each p in S T. Do this for i = 1, ..., r. Then 1 = tp* is the required element. 1c)We conclude the case n 4 by considering arbitrary elements Tr in O' (Y u) at each p in T. We can assume that n — 4, for otherwise 0;,(Vp) = Q (Vp) for all p in T and this is covered by step 1b). We enlarge T by adjoining to it all spots p in S T at which Vp is anisetropic, and we define tpp to be 1 -vp at each of the new spots in T. In effect this allows us to assume that VI, is isotropic, hence that (4 (Vp) =f24(Vp), at all p in S — T. By Propositions 95:1 and 101:9 we can find e in Q't (V) such that gp epp is in D4 (Vp) at all p in T. Let T1 be the set of spots in S - T at which > 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) ,,

,,

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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


with al, a2 in up at each p in S — T. 2b) Suppose the localization Vp of V at p is replaced by some other localization V10, of V at p. Then there is a unique isometry Vp >--> 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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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-

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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

<1> I <-1> 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>

• <1> <

-

1>

<

-

1> .

Deduce that

L <1>

<1 > <- 1 > ± • ±

• •

(- 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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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> ± (2> j_ <1> ± (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

Part Four. Arithmetic Theory of Quadratic Forms over Rings

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

<1> ± • • • ± <1> .

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

Chapter X. Integral Theory of Quadratic Forms over Global Fields

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..
with n 0 mod 4. A computation of Hasse symbols over Q2 shows that n 4 mod 8 is impossible. Hence n 0 mod 8. 3) Conversely let us assume that n 0 mod 8. Then the criterion of Theorem 63:20 informs us that QI (0, 0)) ± • - i Hence by Proposition 81:14 there is a lattice L on V with L ,, In when V2

p = 3, 5, 7, . . .

and L2

I • • • 1

This L is clearly unimodular and even,

q. e. d.

§ 106B. Adjacent lattices

We continue our investigation of unimodular lattices with respect to Z on the quadratic space V with matrix I , over Q. For any such lattice L we define a (L) to be the set of all unimodular lattices K with respect to Z on V such that K,

L a, for

a

p

3, 5,.7,

.

Note that aa (L) = (aL) for any 0- in On (V). 106:2. Suppose n 5 and let K and L be any two unimodular lattices on the space V under discussion. Then there is a lattice J in cls+K such that J E (L). Proof. Let S' denote the set of all non-dyadic spots on Q and put Z' = o (S'). Then Z Z' S Z„ holds for each odd p. The Z'-module L' generated by L in V is clearly a lattice on V with respect to Z', and L S L' Ç L„, holds for each odd p. Hence L' has the property that Li,' =4 for each odd p. Now do the same with K to obtain a lattice K' with respect to Z' with analogous properties. The set of spots S' is indefinite for V since the localization V2, having dimension 5, is isotropic. But L' and K' are unimodular of norm Z'. Hence cls+K' by Theorem 104:10. So we can pick a E (V) such that a K' = L'. Put J = a K. So J E cls+ K . And for each odd p we have

L„ . J = (a K)„ = a„K„ = a.K;„ = (a K'),, q. e. d. Therefore j E (L). If K and L are unimodular lattices on V, then K is in (L) if and only if the invariant factors of K in L are of the form

a

a

Z,

. , 2r* Z 21a

326

Part Four. Arithmetic Theory of Quadratic Forms over Rings

(with r 0 and r„ 0). We say that K is adjacent to L if the invariant factors of K in L are equal to 1 -2- Z, Z, . . ., Z, 2Z (assume n 2 for this definition). Suppose K is adjacent to L. Then it follows immediately from the definitions that L is adjacent to K, that K is in (L), that there is a base xi, . . . , xn, for V in which

a

{ L . Zx l. + • • • + Zxn 1 K . Z (-T xi) + • • • + Z (2 xn) , 1

that 2 L C K -2-L, and finally that aK is adjacent to aL for any a in 0„(V). The number of lattices adjacent to L is finite since the index (--L : 2 L) is finite. 2 106:3. Suppose n-=--- 0 tnod8 and let L be any unimodular lattice on the space V under discussion. Then there are even and odd unimodular lattices on V which are adjacent to L. Proof. By the criterion of Theorem 63:20 we know that V2 is the orthogonal sum of hyperbolic planes, hence by Example 93:18 we have

either L2 2f—


L2-2-'--

± • • • 1 (A (1, 0» .

or

There is clearly a unimodular lattice K2 on V2 which is either odd or even as desired, and whose invariant factors in L2 are equal to 1

ay

6

. .7

2 dr...2, i...2, . . • , Z2,

2 Z2 .

Take a lattice j on V with if p . 2 if p . 3, 5, 7, . . This j is the required lattice on V. q. e. d. 106:4. Suppose n 2 and let K and L be any two unimodular lattices (L) and K+ L. Then there on the space V under discussion with K is a chain of unimodular lattices

Ea

L — L., .12, • • • , Tt - K on V with j i+1 adjacent to L. Proof. It is enough to find a chain of unimodular lattices

P, J2, . . . , »

(over Z2)

Chapter X. Integral Theory of Quadratic Forms over Global Fields 327

on the localization V2 with P = L2 and ft = invariant factors of ja+ 1 in ji are equal to I -

y

K2,

and such that the

Z2; Z 2; • • • 3 Z fir 2Z 2 •

For then we can define lattices fi (1 i t) with respect to Z by the equations T {If if p . 2 .1 fp= L„ if p . 3, 5, 7, . . and L = L., j2 , .. ., j t = K will provide the desired chain from L to K. If n = 2 we have

I

L2 =---- Z2X + Z2y K2 = Z2 (-2X7)

+ Z2(2r y)

and the required local chain is obvious. We may therefore assume that n 3.

By Proposition 106:3 we can assume that both L and K are odd, hence that the localizations L2 and K2 are odd. The local theory (Theorem 93:29) then gives an isometry a in 0„(V2) such that K2 = CIL2. By expressing a as a product of symmetries on V2 we see that it is enough if we assume that a is itself a symmetry such that aL 2 + L2. Thus a = *ru with u a maximal anisotropic vector in L2, say. It follows easily from the local theory that there is either a 1- or a 2-dimensional unimodular sublattice M of L2 which contains u. In the first event we would have ;42,2 = L2. Hence M is actually binary. Take the splitting L2 = M IN. Then N = -44 N C K2 and we therefore have a splitting K2 = M' IN. Write M — Z 2 x + Z2 y M' . Z2 (j-) + Z2(2r y) . {

The required local chain from M to M', hence from L2 to K2, is now q. e. d. obvious, We are beginning to see how the theory of adjacent lattices can be used in determining the unimodular classes on V. We start with a fixed lattice D of the form D

<1> j_ - • • ± <1>

on V. By the finiteness of class number (Theorem 103:4) and by Proposition 106:2 there are lattices K,,. . . , K t in (D) such that

a

cls+Ki, . . ., cls+Kt

are all the distinct proper unimodular classes on V (at least for n 5). So we shall certainly achieve our purpose if we can find the lattices

Part Four. Arithmetic Theory of Quadratic Forms over Rings

328

, K. How is this to be done ? By a step-by-step construction of adjacent lattices starting with the fixed lattice D. First we construct by a certain procedure all the lattices adjacent to D (we have already seen that these are finite in number); then all lattices adjacent to these; and so on. By Proposition 106:4 we can obtain K1,. , K t by steps of the above type. The essential technical features of this procedure are the following: the method of construction, deciding when all Ki have been found, eliminating duplications. The rest of the chapter is devoted to these matters and their application up to n — 9.

§ 106C. Rules of construction Here we give rules for operating with adjacent lattices. Throughout this subparagraph we assume without further reference that K and L are two unimodular lattices with respect to Z on the quadratic space V with matrix I„ over Q. 106:5. The following assertions are equivalent: (1) K is adjacent to L

L) = (3) 13 (K n L) = 4Z. Proof. That (1) implies (2) is immediate from the definition of adjacent lattices and the definition of the volume b. To prove that (2) implies (3) we use the fact that LA = L holds for any unimodular lattice L; then (2)

b (K

(K L) =

(IC4* n .0) = » (K+L)*== 4Z.

That (3) implies (1) follows easily from the Invariant Factor Theorem. q . e. (1. 10 6 : 6. Suppose K is adjacent to L and let y be a vector in L — K. Then (1) L K = Zy K (2) L Zy (L K) (3) L K = (Zy K)#. {w EKIB (w , y) E Z} Proof. (1) We have KcZy+K(L+K. But there are no lattices properly between K and L K by Proposition 106:5. Hence Zy K = L K. (2) We have L r KCZy (L r K) L. But there are no lattices properly between L r. K and L. Hence Zy (L n K) = L. (3) Since L and K are assumed to be unimodular we have .

L nK=L4 r

= (L

(Zy K)#

The second equation follows from the definition of the dual (Zy

K) 4t. q. e. d.

106:7. Let y be any vector in (21- K) K with Q (y) in Z. Then there is exactly one unimodular lattice J . on V that contains y and is adjacent

Chapter X. Integral Theory of Quadratic Forms over Global Fields

329

to K. This lattice can be constructed by forming

Zy + (Zy + K)* .

Proof. The fact that J, if it exists, will have the above form is an immediate consequence of Proposition 106:6. Incidentally, this also proves the uniqueness. We now define a lattice J. on V by the above equation and we prove that the J so defined is first of all unimodular, and secondly that it is adjacent to K. A direct computation shows that B (J, .7) Ç Z, hence 6 J Ç Z. And if we write K =Zw + Zx 2 + + Zxy, with 2y = mw (m E Z), then Zy + K ç_ Z

+ Z; + • • • + Zx n

so that Z b

(Z y

K)

44- Z ,

hence 1) (Zy + K) = Z, hence b (Zy + K) 4* = 4Z. Now y is not in K# = K, hence it is certainly not in (Zy + K)#, hence 4Zc jS Z. Therefore »j = Z. SO J is unimodular. Since (Zy + n K we must have 4 Z b(J nK)CZ, hence K)= 4Z, hence J is adjacent to K. q. e. d. 106:8. Suppose K is adjacent to L, let y be a vector in L K, let z be any vector in V. Then y + z is in L — K if and only if z is in LnK—{wEKIB(w,Y)EZ}•

Proof. If y + z is in L — K, then by the definition of adjacent lattices we have (L + K : K) = ( 2-1Z: Z) --- 2, hence z = (y z) — y is in K, hence z is in L n K. The converse is obvious, q. e. d. 106:8a. Let L and L' be unimodular lattices adjacent to K, let y be an element of L — K, let y' be an element of L' — K. Then L = L' if and only if

y— y' E K

with

B(y, y') EZ Z.

We now have a procedure for constructing all unimodular lattices adjacent to K. Start with a complete set of representatives of T K modulo 2K. Eliminate all those on which the quadratic form Q is not integral and also all those that fall in K. Let y. . . , y,. be the remaining representatives. Form the lattices Zyi + (Zyi + K) 4t for 1 i r. This gives every unimodular lattice adjacent to K at least once. And when do we have duplication, say L = L i ? If and only if yi yi E K with B (yi, yi) E Z.

330

Part Four. Arithmetic Theory of Quadratic Forms over Rings

106:9. Suppose K is adjacent to L and let y be a vector in L — K which has the form y = w ± z with Q (w) .1,

zEK,

B (y, w) E(--i i Z) — z.

Then cla = clsK. Proof. In fact we shall prove that K .--. r L where T. is the symmetry Tw. Here TX = X— 2 B(x, w) w V x EV . Note that 2w = 2y — 2z is in L n K. If we let m denote the odd integer 2 B (y, w) we have Ty

= y — rnw = (y— w) — (m — 1) w = z— (m — 1) w ,

hence Ty is an element of K. And for each x in L n K we have B (x , w) = B (x , y) — B(x, z) E Z

so that Tx is an element of L n K. Hence r L = r (Zy + (L n K)) C K . q. e. d. § 106 1) . 1...< is _‹ 7 1 06: 1 O. Let L be a unimodular lattice on the space V under discussion, and sup ose that 1 n S 7. Then L'-'---- In' Proof. By adjoining a lattice with matrix 1-7 _n to L we can assume, in virtue of Theorem 105:1, that we have n = 7. Let D be a completely decomposable unimodular lattice on V. By Proposition 106:2 we can assume that D E(L). By Proposition 106:4 it will be enough if we prove that clsD = cls.L under the assumption that D is adjacent to L. Take an orthogonal base for D, say D= Zxj . .1. • • - 1 Zx7. Let y be any vector in L — D. By reordering the above base for D if necessary,

we can write 1

y ---- T (a1x1 + • • • ± ax) + ar+1 x,÷1 + • • • + a7x7 with all ai in Z and ai, . . . , a, odd. Here we must have r= 4 since i (xi + x2 + x3 ± x4) and Q (Y) E Z and since y is not in D. Put w — -y then define z by the equation y . w + z. We have Q (w) = 1 ,

1 zED, B(y,w)E(- 2-Z)—Z,

provided the sign ± in the definition of w is correctly chosen. Then cis /. . cis D by Proposition 106:9. q. e. d.

Chapter X. Integral Theory of Quadratic Forms over Global Fields

§ 106E. The matrix On(/'

331

8, 12, 16, . . .)

Throughout this subparagraph we assume that the dimension n of the space V under discussion is one of the numbers 8, 12, 16, . . . We take a completely decomposable unimodular lattice D on V and we fix it. We take a base for D in which D = Zxl ± • • ± Z x7, and we fix it. Our first purpose is to show that there is always an indecomposable unimodular lattice adjacent to D, that all such lattices are in the same proper class, that all have the matrix

where P,,, is the matrix

in particular that the matrix On just defined is unimodular for the specified values of n. Needless to say, every lattice on V with matrix On can be obtained from a suitable completely decomposable lattice in the above way. Put y = (x1 + • • + x,). Then y is a vector in G1 D) — D, Q (y) E Z since n O mod4, hence by Proposition 106:7 there is exactly one unimodular lattice E on V which contains y and is adjacent to D. (This E will be a lattice with the desired properties.) Now D = Z (2y) Z x 2 Z (x3 — x2) + • • • + Z (xn xn--1) • And Zy + Z (24 Z (x3 — x2) + • • • + Z (xn xn_1) is a unimodular lattice adjacent to D which contains y, hence E = Zy Z (2x2) + Z (x3 — x2) + • • • + Z (x„— . By inspection we see that the matrix of E in the base y, 2x2, — (x3 — x2), (x4 — x3), . . . , (xn — is equal to On . Before we prove that E is indecomposable we must give a coordinate description of the vectors in E.

332

Part Four. Arithmetic Theory of Qua.dratie Forms over

106: 11.

E A,x, is in E if and only if A i — Aj EZ, E A i E2Z

Ai E hold for 1

Rings

I

n and I j

n.

Proof. Put x= E A.;. First suppose that x is in E. Then x is in -2- D and so all A i E-21 Z. And xi — x E (Zy D)it CE, so B(x,xi—x) E Z, 1 LT A i E Z. hence A i — Ai E Z. And B(x, y) E Z so that T Now the converse. If one Ai is in (1 1 Z) — Z, then so are they all since A i — Ai E Z for all j. In this event replace x by x + y. We may therefore assume that all A i are in Z. We still have EA E 2Z. Now x = E A is an element of D and it has the property B(x, y) E Z, hence xEDnEçEby Proposition 106:6. q. e. d. The last proposition can be used to show that E does not represent 1. Suppose we have E A = 1 with E A i xi in E. If one A i is in Z, then so are they all, and in this event Al = ± 1, A 2 = A 3 = • • • = A n = 0, say. But then E A i is not in 2 Z, and this contradicts the fact that EA; is in E. So E cannot represent 1 in the above way. On the other hand, if each A i is in (--ï l Z) — Z we have Ai+ • • • + 2,1!

2.

So E does not represent 1 in any way, as we asserted. Why is E indecomposable ? Consider the vectors x2 — xi, x3 — x2, . .

xn,

+ xn in E.

Call them yl, , y„. These vectors obviously span V. Now Q (yi) - 2, and E does not -represent 1, hence each yi must fall in exactly one component of the indecomposable splitting of E. But B(y i, yi+i) =f 0. Hence all yi fall in the same component of the indecomposable splitting. Hence this component has dimension n, so it is equal to E. Hence E is indecomposable.

Finally we must prove that any other indecomposable unimodular lattice E' adjacent to D is in the same proper class as E. It is enough to prove that E' is in the same class as E since the symmetry rxi_ z, is a unit of E. Let Y

, 2

(al

+

+ an xn)

(ai E Z)

Chapter X. Integral Theory of Quadratic Forms over Global Fields

333

be an element of E'— D. If one of the a i were even we would have xi E E' n D S E' by Proposition 106:6 and E' would then be split by Zxi. So all ai are odd. By making suitable choices of sign we can write a= 1 — 4 b i (b i EZ) bn xn) E2D S E' r D. Then n. Put z = 2 (bi + • for 1 ± • • • ± x.)

is also in E'—D. Define an isometry o. E O, (V) by the equations axi = ±xi for 1 .< I n. Then a D=D and ay = y' z. Hence o.E is a unimodular lattice adjacent to D —crD which contains y'd- z. Therefore o..E = E'. So E' is in the same class as E. 1 06:1 2. Suppose n = 8. If L is a unimodular lattice adjacent to E, then L is either in the class of D or in the class of E. Proof. Take a vector z in L — E. Then 2z is in E and so we can write z

1

(Ai xi ± • • • + A 8 x8)

with all

8 A .- AiEZ,

Ai

A i E2Z.

1) First suppose that all A. are even integers. Then E (T A.) is an odd integer since z is not in E. Put z' =x1. Then z z' is in E and B (z, z') E Z. Now L is a unimodular lattice adjacent to E with zEL—E, and D is a unimodular lattice adjacent to E with z' E D— E. Hence L = D by Corollary 106:8a. 2) Next suppose that all A i are odd integers. Then E (--A) is again an odd integer since z is not in E. Put , z =

(—x1+ x2 + • • • + x8) •

There is a unimodular lattice adjacent to E that contains z', and by Corollary 106:8a this lattice must be equal to L, hence z' is in L — E. Then z' = —

+ (x1 + • • • ± x8) .

So clsL = clsE by Proposition 106:9. 3) Let us complete the case of integral A i. Since 12(z) E Z we can suppose that exactly four A i are odd, say A 1, . . , A4 odd and A 8,...,A 8 even. By successively adding and subtracting suitable vectors of the form xi ± xi to z we can find a new z in L E which hag one of the four forms z

2(

+ X2 ± x3 ± x4),

Z

= 2(

+ X2 + X3 ± X4) + X5

334

Part Four. Arithmetic Theory of Quadratic Forms over Rings

(apply Proposition 106:8). In the first two cases L contains a vector z with Q (z).. 1, hence it is decomposable, hence L c4. 18 by Proposition 106:10, hence clsL = clsD. In the second two cases write 1

z = -j-- (T xl + x2 + x3 + x4) + ( ± xl + x5) , and apply Proposition 106:9 to find cIsL = clsE. 4) Finally assume that all A i are in (-- Z) — Z. All the vectors 4x1, ± x1 +

X2 , . • . , ± Xi ±

x8

are in E; using Proposition 106:8 we can successively add and subtract certain of the above vectors (with correct signs attached) to obtain a new zinL—E of the form 1 z .— 4 ( ± a xl ± x2 ± • • • ± x8)

where a is one of the numbers 1, 3, 5, 7. Now a cannot be 1 or 7 since Q (z) E Z. If a — 3 we have Q (z) . 1, hence L splits, hence L-.:-41. 1 8 by Proposition 106:10, hence cIsL = clsD. If a . 5 we write 1, „ z = -71- (T a xi. ± x2 ± • • - ± x2) ± 2x1

and apply Proposition 106:9 to find clsL = clsE.

q. e. d.

§ 106 F. Summary 106:13. Theorem. V is an n-ary quadratic space over the field of rational numbers Q with positive definite localization V.., and L is a unimodular lattice with respect to Z on V. Suppose 1 _< n __< 9. Then L has

exactly one of the forms /n , 08, Os I Ii.

Proof. The fact that L cannot have more than one the above forms is clear from Theorem 105:1 and the fact that any lattice with matrix 08 is indecomposable. If 1 s n s 7 we have L .'_". in by Proposition 106:10. Next let n = 8. Consider adjacent lattices I and K on V. If JL.-_-- /8, then K-...- 4 1f K splits and K .s...A 08 if K does not split. If J L--.._• 08, then K has matrix /8 or 08 by Proposition 106:12. Hence by Propositions 106:2 and 106:4 we have L a= 1.8 or L ae Os. Finally n = 9. Here it is enough to prove that L represents 1, since then L will split. This will be achieved if we can prove the following: if J and K are adjacent unimodular lattices on V such that J represents 1, then K also represents 1. Since J represents 1 we have a splitting J . Zxi i_ J'. Let y be any vector in K — J. If J' ae 08 we write 2y = tnxi + z with in E Z and z E r ; then J' is even and so the fact

Chapter X. Integral Theory of Quadratic Forms over Global Fields

335

that Q (y) E Z implies that m E 2Z; hence E (Zy j)* --JnKSK and K therefore represents 1 as required. Otherwise r /8 ; then we have a base for J in which = Z ± • • - Z xi, ; this time write 2y = m1 x, + m8 x8 (mi E Z) ; then at least one nii must be even since Q (y) E Z; for this xi we have xi E (Zy j) 4* = I n K _S K and K therefore represents I as required. q. e. d. With some perseverance it is possible to extend these results' to higher dimensions using the general principles of this paragraph. For 1 n S 13 the distinct proper classes on V are determined by lattices of the form , 08 I 1 2 112 One obtains new indecomposable lattices in 14, 15, and 16 dimensions. 9

45

See M. KNESER, Arch. Math. (1957), pp. 241-250. For an example of the classical approach using "reduction theory" we refer the reader to B. W. JONES, The arithmetic theory of quadratic forms (Buffalo, 1950).

Bibliography Our original intention was to provide a complete bibliography with full documentation, but we were soon discouraged by the complexity of such a task. Instead we have decided to give the following short list of books and articles' in order to enable the reader to trace individual results to their source and to lead him to other fields of interest in number theory and the theory of quadratic forms. E. Arlin, Algebraic numbers and algebraic functions (Princeton University, 1951). E. Artin, Geometric algebra (New York, 1957). C. Chevalley, The algebraic theory of spinors (New York, 1954). L. E. Dickson, Studies in the theory of numbers (Chicago, 1930). L. E. Dickson, History of Me theory of numbers vol. III (New York, 1952). J. Dieudonné, La géométrie des groupes classiques (Berlin, 1955). M. Eichler, Quadratische Formen und orthogonale GrupPen (Berlin, 1952). H. Hasse, Zahlentheorie (Berlin, 1949). B. W. Jones, The arithmetic theory of quadratic forms (Buffalo, 1950). M. Kneser, Klassenzahlen indefiniter quadratischer Formen in drei oder mehr VerAnderlichen, Arch. Math. (1956), pp. 323-332. H. Minkowski, Gesammelte Abhandlungen (Berlin, 1911). • O. T. O'Meara, Integral equivalence of quadratic forms in ramified local fields, Am. J. Math. (1957), pp. 157-186. C. L. Siegel, -Ober die analytische Theorie der quadratischen Formen, Ann. Math. 36 (1935), pp. 527-606; 37 (1936), pp. 230-263; 38 (1937), pp. 212-291. B. L. van der Waenlen, Die Reduktionstheorie der positiven quadratischen Formen, Acta Math. (1956), pp. 265-309. G. L. Watson, Integral quadratic forms (Cambridge, 1960). A. Weil, Adeles and algebraic groups (Institute for Advanced Study, 1961). E. Witt, Theorie der quadratischen Formen in beliebigen Kiirpern, Crelle's J. 176 (1937), pp. 31-44. 1 Additional references to specific points of interest are given in footnotes in the text.

Index Abelian extension, 35 Absolute units, 73 Adapted base, 212 Adjacent lattices, 326 Adjunction, 225 hyperbolic, 257 Algebraic function field, 58 Algebraic number field, 58 Algebras, 114 central, 119 Clifford, 132 commutative, 115 compatible with quadratic spaces, 131 division, 115 E-ification of, 130 Hasse, 150 homomorphism of, 115 ideals in, 119, 121 quaternion, 142 reciprocal, 128 similar, 128 simple, 119 splitting of, 128 tensor product of, 116 Almost all elements of a set, xi Analytic map, 3 Anisotropic quadratic space, 94 Anisotropic vector, 94 Approximation theorems, strong, 42 strong for rotations, 315 very strong, 77 weak, 8 weak for rotations, 290 ARTIN, E., viii, 52, 54, 78, 79, 141, 190, 336 Automorphs, 101 intepal, 223 proper, 101 proper integral, 223 Axis of a ternary rotation, 105 Bilinear form, alternating, 82 hermitian, 82 symmetric, 82

Bilinear mapping, 113 multiplicative, 116 Brauer group, 128 CASSELS, J. W. S., 187 Cauchy sequence, 9 CHEVALLEY, C., 112, 158, 173, 336 Class number, 48 finiteness of, 79 for quadratic forms, 309 Class of a lattice, 222 of a matrix, 221 Classical ideal theory, 81 Classification of quadratic forms, complex, 156 for finite fields, 158 general, 152 global, 189 global integral, 320, 334 local, 170 local integral, 247, 248, 267, 277 real, 156 Clifford algebra, 132 derived base for a, 135 even element of a, 135 Coefficient of a vector in a lattice, 210 Compatibility of an algebra with a quadratic space, 131 Completeness, 9 Completion, 9 notation for, 11 Components of a splitting, 89, 224 Constant field, 55 Cyelotomic extension, 191 absolute, 191 Decomposable lattice, 321 completely, 324 Decomposition field, 36 Decomposition group, 35 Dedekind domain, 52 Dedekind set of spots, 42 Defining base of a quaternion algebra, 142 Defining matrices, 122

338

Index

Definite quadratic space, 154 Definite set of spots, 311 Degree of inertia, 24, 37 Derived base for a Clifford algebra, 135 DICKSON, L. E., 336 DIEUDONNi, 3., 82, 336 Dimension of a lattice, 213 Direct sum of lattices, 224 Dirichlet unit theorem, 77 Discriminant of a free lattice, 222 of a quadratic space, 87 of a set of vectors, 87 Dual base, 92 Dual of a lattice, 230 Dyadic local field, 158

M., vii, 336 E-ification, 129 EICHLER,

of a quadratic space, 131 of an algebra, 130 Eisenstein polynomial, 64 Elementary transformation, 100 Equivalence of matrices, 85 integral, 221 Even element of a Clifford algebra, 135 Even lattice, 324 Finiteness of class number, 79 for quadratic forms, 309 Fixed space of an isometry, 103 FLANDERS, FL, 190 Forms of degree d, 87 Free lattice, 213 Frobenius automorphism, 64, 193 Function field, 58 Fundamental invariants of a lattice, 263 same set of, 264 Fundamental type of a lattice, 263 Galois extension, 35 General linear group, 84 Genus of a lattice, 297 Geometry, orthogonal, 82 symplectic, 82 unitary, 82 Global field, 58 rational, 55 separable generation of a, 58 Global square theorem, 182 Group of units of a lattice, 223

B ASSE, H., 336 Hasse algebra, 150 Hasse norm theorem, 186 Hasse symbol, 167 Hasse-Minkowski theorem, 189 Ecxil, E., 52 Hensel's lemma, 26 Herbrand's lemma, 179 Hilbert symbol, 164 Hilbert's reciprocity law, 190, 201 Hyperbolic plane, 94 Hyperbolic space, 99 Ideals, 44 class group, 48 class number, 48 decomposable, 48 divisibility of, 44 finiteness of class number, 79 fractional, 44 g.c.d. of, 45 indecomposable, 48 integral, 44 1.c.m. of, 45 localization of, 217 multiples of, 44 order of an ideal, 50 p-ification of, 217 prime, 48 principal, 44 product of, 45 relatively prime, 45 value of, 45 Mee notation, 68-69, 80, 172-174, 176 Identification process, 130 Indecomposable splitting of a lattice, 321 Indefinite quadratic space, 154 Indefinite set of spots, 311 Index of a quadratic space, 99 Integers, algebraic, 81 at a set of spots, 42 at a spot, 20, 174 p-adic, 57 rational, 54-55 Integral equivalence of matrices, 221 Integral matrix, 214 Integral polynomial, 25 Integral representation of matrices, 221 Invariant factor theorem, 214 Invariant factors, 216

Index Involution, 96 Isometry, 84 at a spot, 186 fixed space of an, 103 of lattices, 220 spinor norm of an, 137 Isotropy, 94 at a spot, 186 Isotropy theorems, for finite fields, 158 for function fields, 188 general, 153 global, 187 local, 170 IWASAWA, K., 78

viii, 239 N., 186 JONES, B. W., 335, 336 Jordan chain, 246 Jordan invariants, 276 Jordan splitting, 243 saturated, 264 Jordan type, 245 JACOBOWITZ, R., JACOBSON,

KNESER, M., 293, 296, 335, 336 LANG, S., 170 Lattice, 209 adapted base for a, 212 adjacent, 326 adjunction of a, 225 almost free, 213 base for a, 212 class of a, 222 coefficient of a vector in a, 210 completely decomposable, 324 components of a, 224 decomposable, 321 dimension of a, 213 direct sum of lattices, 224 discriminant of a, 222 dual, 230 even, 324 free, 213 fundamental invariants of a, 263 fundamental type of a, 263 genus of a, 297 group of units of a, 223 in a space, 209 indecomposable splitting of a, 321 isometry of a, 220 localization of a, 217

339

matrix of a, 221 maximal, 234 maximal vector of a, 210 modular, 231 n-ary, 220 norm generator of a, 253 norm group of a, 251 norm of a, 227 odd, 324 on a space, 209 orthogonal base for a, 225 orthogonal splitting of a, 224 orthogonal sum of lattices, 224 p-ification of a, 217 radical of a, 225 rank of a, 212 regular, 225 representations of a, 220 saturated splitting of a, 264 scale of a, 227 scaling a, 238 spinor genus of a, 298 unimodular, 231 volume of a, 229 weight generator of a, 253 weight of a, 253 Legendre symbol, 205 Local degree, 31, 37 Local field, 59 dyadic, 158 non-dyadic, 158 2-adic, 276 Local norm, 32 at a spot, 176 Local square theorem, 159 Local trace, 32 Localization, 186 convention, 290 of a lattice, 217 of a linear transformation, 289 of an ideal, 217 Matrices, class of, 221 equivalence of, 85 integral, 214 integral equivalence of, 221 integral representation of, 221 of a lattice, 221 of a quadratic space, 85 representation of, 85 unimodular, 214 Maximal ideal at a spot, 21

22*

340

Index

Maximal lattice, 234 Maximal vector of a lattice, 210 MrismowsKI, H., 336 Modular lattice, 231 Multiplicative bilinear mapping, 116 Negative definite quadratic space, 154 Negative element at a real spot, 20 Negative index of a quadratic space, 155 Non-dyadic local field, 158 Norm (algebraic), xi of a quaternion, 143 Norm generator, 253 Norm group, 251 Norm of a genus, etc., 299 Norm of a lattice, 227 Norm of a linear transformation, 286 of a matrix, 287 of a vector, 12, 285 Number field, 58 Odd lattice, 324 O'MEARA, O. T., 247, 267, 277, 336 Order of an element, 38 of an ideal, 50 Ordinary absolute value, 14, 19 Orthogonal base, 90, 225 Orthogonal complement, 90 Orthogonal group, 84 center of, 108 commutator subgroup of, 106 generation by symmetries, 102 special subgroups of, 100, 141, 156, 158, 280, 293 Orthogonal sets, 83 Orthogonal splittings, 88, 224 p-adic numbers, 57 integers, 57 spot, 55 p-adic topology, 7 p-ification, 186 convention, 290 of a lattice, 217

of a linear transformation, 289 of an ideal, 217 Pigeon holing principle, 69 Place, 29 POLLAK, B., viii Positive definite quadratic space, 154 Positive element at a real spot, 20

Positive index of a quadratic space, 155 Power series expansion, 39 Prime, 55 Prime element, 38 Prime function, 55 Prime ideal, 48 Prime number, 54 Prime polynomial, 55 Prime spot, see under Spots Principle of domination, 4 Product formula, 66 Quadratic defect, 160 Quadratic form, 87-88 of a quaternion algebra, 145 Quadratic map, 83 Quadratic reciprocity law, 207 Quadratic space, 83 anisotropie, 94 definite, 154 discriminant of a, 87 E-ification of a, 131 hyperbolic, 99 indefinite, 154 index of a, 99 isotropic, 94 matrix of a, 85 n-ary, 83

negative definite, 154 negative index of a, 155 orthogonal base for a, 90 positive definite, 154 positive index of a, 155 radical of a, 91 regular, 91 representations of a, 83, 84 scaling a, 83 totally isotropic, 94 universal, 83 Quaternion, 142 conjugate, 143 norm of a, 143 pure, 142 scalar, 142 trace of a, 143 Quaternion algebra, 142 defining base of a, 142 quadratic form of a, 145

Radical, 91, 225 Radical splitting, 93, 226 Ramification index, 22, 37

Index Ramified extension, 60 fully, 60 Rank of a lattice, 212 Rational function field, 55 constant field of a, 55 integers of a, 55 Rational global field, 55 integers of a, 55 Rational number field, 54 integers of a, 54 Reciprocal algebra, 128 Reducibility criterion, 28 Refle.xion, 100 Regular lattice, 225 Regular quadratic space, 91 Representation, 83 for lattices, 220 for matrices, 85 for quadratic spaces, 84 integral, for matrices, 221 Residue class field, 21 notation for, 21 obtained by natural restriction, 23 representative set of a, 39 RIEHM, C., viii, 277 Root of unity, 190 primitive, 190 Rotation, 100 axis of a ternary rotation, 105

SAM, C. H., viii, 250 SAMUEL, P., 52, 58 Saturated Jordan splitting, 264 Scale of a genus, etc., 299 Scale of a lattice, 227 Scaling a lattice, 238 a quadratic space, 83 a vector, 83 SCHILLING, O. E G., 60, 168 SCHMIDT, F. K., viii Sequence, limit of, 3 Series, sum of, 3 SIEGEL, C. L., 187, 336 Similarity of algebras, 128 Spinor genus of a lattice, 298 Spinor norm of an isometry, 137 notation, 138 Split rotation, 295 principal, 296 Splittings, of algebras, 128 of lattices, 224

341

of quadratic spaces, 88 radical splittings, 93, 226 Spots, 7 archimedean, 7 carrying a spot, 7 complete, 9 complex, 19 Dedekind set of, 42 definite set of, 311 discrete, 37 dividing, 7, 8 fully dividing, 8 indefinite set of, 311 induced, 7 infinite, 55 integer at a spot, 20, 174 isometry at a spot, 186 isotropy at a spot, 186 local norm at a spot, 176 non-archimedean, 7 p-adic, 55 real, 19 square at a spot, 174 trivial, 7 unit at a spot, 20, 174 unramified at a spot, 60, 176, 191 Square at a spot, 174 Strong approximation theorem, 42 for rotations, 315 Symmetric bilinear form, 82 Symmetry, 96 TATE, J., 190 Ten-field construction, 196 Tensor products, 113 notation, 114, 118 of algebras, 116 Topological field, 2 ring, 287 vector space, 286 Triangle law, 1 Unimodular lattice, 231 Unimodular matrix, 214 Unique factorization theorem, 49 Units, absolute, 73 at a set of spots, 42, 73 at a spot, 20, 174 of a lattice, 223 Universal quadratic space, 83 Unramified at a spot, 60, 176, 191

342

Index

Valuated field, 1 Valuations, 1 archimedean, 1 carrying a valuation, 3 complete, 9

discrete, 37 equivalent, 5 general, 29 non-archimedean, 1 normalized, 19, 59, 65 trivial, 2 valuation ring, 20, 29 Value group, 2 Very strong approximation theorem, 77 Volume of a genus, etc., 299

Volume of a lattice, 229 Volume of an idèle, 68 WAERDEN, B. L. VAN DER, 125, 336 WATSON, G. L., 305, 336 Weak approximation theorem, 8 for rotations, 290 Wedderbum's theorem, 125 Weight generator, 253 Weight of a lattice, 253 WEIL, A., 336 WHAPLES, G., viii, 29, 54, 78 WrIT, E., 151, 336 Witt's theorem, 97, 98 ZARISKI, 0., 52, 58 ZASSENHAUS, H., 137

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