Springer number theory b (springer)

Similarly rz - p = ind- f, and the for~nula for ind f follows readilỵ It follows that, for quadratic spaces over a field of the type considered in Proposition 10, a subspace is anisokopi

NUMBER THEORY

An Introduction to Mathematics: Part B

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Contents

Part A

Preface

Sets, relations and mappings

Rings and fields

Vector spaces and associative algebras Inner product spaces

Further remarks

12 Selected references

1 Greatest common divisors

2 The Bezout identity

Contents

I11 More on divisibility

1 The law of quadratic reciprocity

1 The continued fraction algorithm

2 Diophantine approximation

3 Periodic continued fractions

4 Quadratic Diophantine equations

5 The modular group

3 The art of weighing

4 Some matrix theory

5 Application to Hadamard's determinant problem

6 Designs

7 Groups and codes

8 Further remarks

9 Selected references

2 The Hilbert symbol

3 The Hasse-Minkowski theorem

4 Supplements

5 Further remarks

6 Selected references

VIII The geometry of numbers

1 Minkowski's lattice point theorem

I X The number of prime numbers

1 Finding the problem

2 Chebyshev's functions

3 Proof of the prime number theorem

4 The Riemann hypothesis

5 Generalizations and analogues

1 Primes in arithmetic progressions

2 Characters of finite abelian groups

3 Proof of the prime number theorem for arithmetic progressions

4 Representations of arbitrary finite groups

5 Characters of arbitrary finite groups

6 Induced representations and examples

XI1 Elliptic functions

1 Elliptic integrals

2 The arithmetic-geometric mean

3 Elliptic functions

4 Theta functions

5 Jacobian elliptic functions

6 The modular function

VII

We have already determined the integers which can be represented as a sum of two squares Similarly, one may ask which integers can be represented in the form x2 + 2y2 or, more generally, in the form ax2 + 2bxy + cy2, where a,b,c are given integers The arithmetic theory

of binary quadratic forms, which had its origins in the work of Fermat, was extensively developed during the 18th century by Euler, Lagrange, Legendre and Gauss The extension to quadratic forms in more than two variables, which was begun by them and is exemplified by Lagrange's theorem that every positive integer is a sum of four squares, was continued during the 19th century by Dirichlet, Hermite, H.J.S Smith, Minkowski and others In the 20th century Hasse and Siege1 made notable contributions With Hasse's work especially it became apparent that the theory is more perspicuous if one allows the variables to be rational numbers, rather than integers This opened the way to the study of quadratic forms over arbitrary fields, with pioneering contributions by Witt (1937) and Pfister (1965-67)

From this vast theory we focus attention on one central result, the Hasse-Minkowski theorem However, we first study quadratic forms over an arbitrary field in the geometric formulation of Witt Then, following an interesting approach due to Frohlich (1967), we study quadratic forms over a Hilbertfield

1 Quadratic spaces

The theory of quadratic spaces is simply another name for the theory of quadratic forms The advantage of the change in terminology lies in its appeal to geometric intuition It has in fact led to new results even at quite an elementary level The new approach had its debut in a paper by Witt (1937) on the arithmetic theory of quadratic forms, but it is appropriate also if one

is interested in quadratic forms over the real field or any other field

For the remainder of this chapter we will restrict attention to fields for which 1 + 1 # 0 Thus the phrase 'an arbitrary field' will mean 'an arbitrary field of characteristic # 2' The

342 VII The arithmetic of quadratic forms

proofs of many results make essential use of this restriction on the characteristic For any field

F , we will denote by F X the multiplicative group of all nonzero elements of F The squares in

F X form a subgroup Fx2 and any coset of this subgroup is called a square class

Let V be a finite-dimensional vector space over such a field F We say that V is a quadratic space if with each ordered pair u,v of elements of V there is associated an element

(u,v) of F such that

(i) (ul + u2,v) = (ul,v) + (u2,v) for all ul7u2,v E V ;

(ii) (au,v) = a(u,v) for every a E F and all u,v E V ;

(iii) (u,v) = (v,u) for all u,v E V

It follows that

(i)' (u,vl + v2) = (u,vI) + (u,v2) for all u,vl,v2 E V;

(ii)' (u,av) = a(u,v) for every a E F and all u,v E V

Let e l , , en be a basis for the vector space V Then any u,v E V can be uniquely expressed in the form

u = C ;,1 cjq, V = C "rlej,

is a quadratic form with coefficients in F The quadratic space is completely determined by the

quadratic form, since

(u,v) = {(u + 1.?,U + v) - (u,u) - (v,v))/2 (1) Conversely, for a given basis el , .,en of V, any nxn symmetric matrix A = (ajk) with

elements from F , or the associated quadratic form f(x) = xlAx, may be used in this way to give

V the structure of a quadvatic space

Let el', , en' be any other basis for V Then

where T = ( ( z i j ) is an invertible 11x1~ matrix with elements from F Conversely, any such matrix

T defines in this way a new basis el ', ,en1 Since

quadratic foims f and g will be denoted byf -F g or simply f - g

It follows from (2) that

det A = (det n2 det B

Thus, although det A is not uniquely determined by the quadratic space, if it is nonzero, its square class is uniquely dete~mined By abuse of language, we will call any representative of

this square class the determinant of the quadratic space V and denote it by det V

Although quadratic spaces are better adapted for proving theorems, quadratic forms and symmetric matrices are useful for computational purposes Thus a familiarity with both languages is desirable However, we do not feel obliged to give two versions of each definition

or result, and a version in one language may later be used in the other without explicit comment

A vector v is said to be orthogonal to a vector u if (u,v) = 0 Then also u is orthogonal to

v The o~~thogonal conzplement U' of a subspace U of V is defined to be the set of all v E V such that (u,v) = 0 for every u E U Evidently U' is again a subspace A subspace U will be

said to be non-singular if U n u - ~ = {O}

The whole space V is itself non-singulas if and only if V' = (0) Thus V is non-singular if and only if some, and hence every, sy~nmetric matrix describing it is non-singulau, i s if and only if det V # 0

We say that a quadratic space V is the orthogonal sum of two subspaces Vl and V , and we write V = V1 I V2, if V = VI + V2, V1 n V2 = ( 0 ) and (v1,v2) = 0 for all vl E V1, v2 E V2

If Al is a coefficient matrix for V1 and A2 a coefficient matrix for V2, then

is a coefficient matrix for V = V1 I V2 Thus det V = (det Vl)(det V2) Evidently V is non- singulau if and only if both V1 and V2 are non-singulas

VII The arithmetic of quudratic

If W is any subspace supplementasy to the orthogonal complement V' of the whole space

V, then V = v - ~ I Wand W is non-singulas Many problems for arbitrary quadsatic spaces may

be reduced in this way to non-singular quadratic spaces

PROPOSITION 1 If a quadratic space V co~ltailzs a vector u such that (u,u) # 0 , then

where U = <u> is the one-dimensional subspace sparzned by u

Proof For any vector v E V, put v' = v - au, where a = (v,zc)/(u,u) Then (vf,u) = 0 and hence v' E u' Since U n U' = { 0 ) , the result follows 17

A vector space basis u l , , u, of a quadratic space V is said to be an orthogonal basis if

(uj,uk) = 0 whenever j # k

PROPOSITION 2 Any quadintic space V has an ortlzogonal basis

Proof If V has dimension 1, there is nothing to prove Suppose V has dimension n > 1 and the result holds for quadsatic spaces of lower dimension If (v,v) = 0 for all v E V, then any basis

is an orthogonal basis, by (1) Hence we may assume that V contains a vector ul such that

(u,,ul) # 0 If U 1 is the 1-dimensional subspace spanned by ul then, by Psoposition 1,

By the induction hypothesis has an orthogonal basis u2, , u,, and u1,u2, ,un is then an orthogonal basis for V

Proposition 2 says that any symnetic matrix A is congruent to a diagonal matsix, or that

the cossesponding quadratic form f is equivalent over F to a diagonal form F1c12 + + 8,5,2

Evidently det f = 6, 8, and f is non-singular if and only if 6j # 0 (1 I j I n) If A # 0 then,

by Propositions 1 and 2, we can take F1 to be any element of FX which is represented by$ Here y E FX is said to be wpreseizted by a quadratic space V over the field F if there exists a vector v E V such that (v,v) = y

As an application of Psoposition 2 we prove

PROPOSITION 3 If U is a non-singular subspace of the quadratic space V, then

v = U I uL

Proof Let u , , , u, be an orthogonal basis for U Then (uJ,u,) # 0 ( 1 I j I m), since U is non-singular For any vector v E V, let u = allcl + + a,,~,,, where aJ = (v,uj)/(uJ,uj) for

I Quadratic spaces 345

each j Then u E U and (u,uj) = (v,uj) ( 1 I j I m) Hence v - u E u' since u n U' = { 0 ) ,

the result follows

It may be noted that if U is a non-singular subspace and V = U I W for some subspace W ,

then necessarily W = U-L For it is obvious that W c U' and dim W = dim V - dim U = dim u',

by Proposition 3

PROPOSITION 4 Let V be a non-singular quadratic space If v l , , v , are linearly independent vectors in V then, for arbitrary q l , ,qm E F , there exists a vector v E V such that (vj,v) = qj (1 I j I m )

Moreover, if U is any subspace of V, then

(i) d i m ~ + d i r n ~ ' = d i m ~ / ;

(ii) UU = U;

(iii) u - ~ is non-singular ifand only if U is non-singular

Proof There exist vectors v,+~ , , v, E V such that v l , , v, form a basis for V If we put

ajk = (vj,vk) then, since V is non-singular, the nxn symmetric matrix A = (ajk) is non-singular Hence, for any q 1 , , q n E F , there exist unique k 1 , , 5, E F such that v = c l v l + + cnv,

satisfies

( ~ 1 3 ~= )q l ? (vn7v) = rln

This proves the first part of the proposition

By taking U = <vl , , v,> and q = = q , = 0, we see that dim u - ~ = n - m Replacing

U by u-', we obtain dim ULL = dim U Since it is obvious that U c UU, this implies U = UU

Since U non-singular means U n U' = { 0 ) , (iii) follows at once from (ii)

We now introduce some further definitions A vector u is said to be isotropic if u # 0 and

(u,u) = 0 A subspace U of V is said to be isotropic if it contains an isotropic vector and

anisotropic otherwise A subspace U of V is said to be totally isotropic if every nonzero vector in U is isotropic, i.e if U L u' According to these definitions, the trivial subspace 10)

is both anisotropic and totally isotropic

A quadratic space V over a field F is said to be universal if it represents every y E FX, i.e

if for each y E FX there is a vector v E V such that (v,v) = y

PROPOSITION 5 If a non-singular quadratic space V is isotropic, then it is universal Proof Since V is isotropic, it contains a vector u # 0 such that (u,u) = 0 Since V is non- singular, it contains a vector w such that (u,w) # 0 Then w is linearly independent of u and by

346 VII The arithmetic of quadratic forms

replacing w by a scalar multiple we may assume (u,w) = 1 If v = a u + w, then (v,v) = y for

a = { y - (w,w)}/2

On the other hand, a non-singular universal quadratic space need not be isotropic As an example, take F to be the finite field with three elements and V the 2-dimensional quadratic space corresponding to the quadratic form k 1 2 + C22

PROPOSITION 6 A non-singular quadratic form f(51, ,5n) with coefficients from a field

F represents y E F X if and only if the quadratic f o m

is isotropic

Proof Obviously if f ( x l , , x,) = y and xo = 1, then g(xo,xl, , x,) = 0 Suppose on the other hand that g(xo,xl, , x,) = 0 for some xj E F, not all zero If xo # 0, then f certainly represents y If xo = 0, then f is isotropic and hence, by Proposition 5, it still represents y 0

PROPOSITION 7 Let V be a non-singular isotropic quadratic space If V = U I W , then there exists y E F X such that, for some u E U and w E W ,

Proof Since V is non-singular, so also are U and W, and since V contains an isotropic vector

v', there exist U ' E U , W' E W , not both zero, such that

If this common value is nonzero, we are finished Otherwise either U or W is isotropic Without loss of generality, suppose U is isotropic Since W is non-singular, it contains a vector

w such that (w,w) # 0, and U contains a vector u such that (u,u) = - (w,w), by Proposition 5

Proof Let Ul be a maximal totally isotropic subspace of the quadratic space V Then U 1 c u~'

and ull\ U1 contains no isotropic vector Since V' c u~', it follows that V' r U1 If V' is a

I Quadratic spaces 347

subspace of V supplementary to v', then V' is non-singular and U 1 = V' + U 1 ' , where

U1' c V' Since Ul' is a maximal totally isotropic subspace of V', this shows that it is sufficient

to establish the result when V itself is non-singular

Let U 2 be another ~naximal totally isotropic subspace of V Put W = U 1 n U 2 and let W1,W2 be subspaces supplementary to Win U1,U2 respectivelỵ We are going to show that

w2 n w,' = 10)

Let v E W 2 n wlị Since W 2 c U2, v is isotropic and v E u2' c w' Hence v E ul'

and actually v E U 1 , since v is isotropic Since W 2 c U 2 this implies v E W, and since

W n W 2 = { 0 ) this implies v = 0

It follows that dim W 2 + dim wI1 5 dim V But, since V is now assumed non-singular,

dim W1 = dim V - dim w I 1 , by Proposition 4 Hence dim W 2 I dim W 1 and, for the same reason, dim W 1 I dim W2 Thus dim W2 = dim W 1 , and hence dim U2 = dim U1

We define the index, ind V , of a quadsatic space V to be the dimension of any maximal totally isotsopic subspacẹ Thus V is anisobopic if and only if ind V = 0

A field F is said to be ordered if it contains a subset P of positive elements, which is closed under ađition and multiplication, such that F is the disjoint union of the sets {0}, P and

- P = {- x: x E P} The rational field Q and the real field IW are ordered fields, with the usual interpretation of 'positivé For quadsatic spaces over an ordered field there are other useful notions of index

A subspace U of a quadratic space V over an ordered field F is said to be positive definite

if (u,u) > O for all nonzero u E U and ~zegative definite if (u,u) < O for all nonzero 14 E U

Evidently positive definite and negative definite subspaces are anisotropic

PROPOSITION 9 All maxinzal positive definite subspaces of a quadratic space V over an ordered.field F have the s u m dimensio/~

Proof Let Ư be a maximal positive definite subspace of the quadratic space V Since Ư is

certainly non-singular, we have V = Ư I W , where W = ứ, and since Ư is maximal,

(w,w) I 0 for all w E W Since Ư c V , we have V' c W If U- is a maximal negative definite

subspace of W , then in the same way W = U- I Uo, where Uo = U-' n W Evidently Uo is

totally isotropic and Uo c v' In fact Uo = v'-, since U- n V' = {O) Since (v,v) 2 O for all

v E Ư I v', it follows that U- is a maximal negative definite subspace of V

If Ứis another maximal positive definite subspace of V , then Ứ n W = (0) and hence

dim Ứ + dim W = dim (U,' + W) I dim V

348 VIỊ The ar.itlmztic of yliudmtic fonns

Thus dim Ứ I dim V - dim W = dim Ự But Ư and Ứ can be interchanged

If V is a quadsatic space over an ordered field F, we define the positive index ind' V to be

the dimension of any maximal positive definite subspacẹ Similarly all maximal negative definite subspaces have the same dimension, which we call the negative index of V and denote

by ind- V The proof of Proposition 9 shows that

ind' V + ind- V + dim V' = dim V

PROPOSITION 10 Let F denote the real field R or, more genemlly, an ordered field in which every positive element is a squaw The11 any nun-singular quaclvatic folm f in n

variables with coe8icients @om F is equivalerzt over F to a quadratic form

where p E {0,1 , , i l l is ~ m i q z d y determined by j: In fact,

Proof By Proposition 2, f is equivalent over F to a diagonal form 61q12 + + 6,q,2, where

6j z 0 (1 I j I n) We may choose the notation so that Fj > 0 for j I p and 6j < 0 for j > p The change of variables tj = 651/2q, ('j S p ) , cj = (- !i)1/2qj ('j > p ) now brings f to the form g Since the corresponding quadratic space has a p-dimensional maximal positive definite subspace, y = ind'f is uniquely determined Similarly rz - p = ind- f, and the for~nula for ind f follows readilỵ

It follows that, for quadratic spaces over a field of the type considered in Proposition 10, a subspace is anisokopic if and only if it is either positive definite or negative definitẹ

Proposition 10 completely solves the problem of equivalence for real quadsatic forms (The uniqueness of p is known as Sylvester's law of inertiạ) It will now be shown that the problem of equivalence for quadsatic forms over a finite field can also be completely solved

LEMMA 11 I f V is a rlorz-sirzgular 2-dit~~erzsiorzul quadratic space over a firlite field E,, of (ođ) cãdinulity q, then V is ũzivelzral

Proof By choosing an orthogonal basis for V we are reduced to showing that if a,P,y E IFqX,

then there exist t,q E [Fq such that at2 + Pq2 = ỵ As 5 runs through 5,, a t 2 takes (y + 1)/2

= 1 + ( q - 1)/2 distinct values Similarly, as q runs through Fq, y - pq2 takes ( q + 1)/2 distinct values Since ( q + 1)/2 + ( q + 1)/2 > q , there exist t,q E IF, for which a t 2 and

y - pq2 take the same valuẹ

1 Quadratic spaces 349

PROPOSITION 12 Any non-singular quadratic form f in n variables over afinitefield IFq

is equivalent over [Fq to the quadratic form

where 6 = det f is the determinant o f f

There are exactly two equivalence classes of non-singular quadratic forms in n variables over F4, one consisting of those forms f whose determinant det f is a square in

LFqX , and the other those for which det f is not a square in FqX

Proof Since the first statement of the proposition is trivial for n = 1, we assume that n > 1 and

it holds for all smaller values of n It follows from Lemma 11 that f represents 1 and hence, by

the remark after the proof of Proposition 2, f is equivalent over IFq to a quadratic form

E 1 2 + g ( c 2 , ,en) Since f and g have the same determinant, the first statement of the proposition now follows from the induction hypothesis

Since FqX contains (q - 1)/2 distinct squares, every element of [FqX is either a square or a square times a fixed non-square The second statement of the proposition now follows from the first

We now return to quadratic spaces over an arbitrary field A 2-dimensional quadratic space is said to be a hyperbolic plane if it is non-singular and isotropic

PROPOSITION 13 For a 2-dimensional quadratic space V , the following statements are equivalent:

(i) V is a hyperbolic plane;

(ii) V has a basis ul,u2 such that (ul,ul) = (u2,u2) = 0 , (u1,u2) = 1;

(iii) V has a basis vl,v2 such that (vl,vl) = 1, (v2,v2) = - 1 , (v1,v2) = 0;

(iv) - det V is a square in FX

Proof Suppose first that V is a hyperbolic plane and let ul be any isotropic vector in V If v is any linearly independent vector, then (ul,v) # 0, since V is non-singular By replacing v by a scalar multiple we may assume (ul,v) = 1 If we put u2 = v + n u l , where a = - (v,v)/2, then

and ul,u2 is a basis for V

If ul,u2 are isotropic vectors in V such that (u1,u2) = I, then the vectors vl = ul + u2/2

and v2 = ul - u2/2 satisfy (iii), and if vl,v2 satisfy (iii) then det V = - 1

VII The arithmetic of quadratic forms

Finally, if (iv) holds then V is certainly non-singular Let wl,w2 be an orthogonal basis for V and put aj = (wj,wj) (j = 1,2) By hypothesis, 8162 = - y2, where y E FX Since

ywl + 8 1 ~ 2 is an isotropic vector, this proves that (iv) implies (i)

PROPOSITION 14 Let V be a non-singular quadratic space If U is a totally isotropic subspace with basis ul, , urn, then there exists a totally isotropic subspace U' with basis u1 ', ,urnr such that

(uj,uk') = 1 or 0 according as j = k or j + k

Hence U n U' = { 0 ) and

U + U' = H1 I I H,, where Hj is the hyperbolic plane with basis uj,ujl (1 I j I m)

Proof Suppose first that m = 1 Since V is non-singular, there exists a vector v E V such that

(ul,v) # 0 The subspace H1 spanned by ul,v is a hyperbolic plane and hence, by Proposition

13, it contains a vector ullsuch that (ul',ul'j = 0, (ul,ul') = 1 This proves the proposition for

PROPOSITION 15 Any quadratic space V can be represented as an orthogonal sum

where H I , .,Hm are hyperbolic planes and the subspace Vo is anisotropic

Proof Let V l be any subspace supplementary to v' Then V 1 is non-singular, by the definition

of V-L If V l is anisotropic, we can take m = 0 and Vo = V1 Otherwise V 1 contains an isotropic vector and hence also a hyperbolic plane H1, by Proposition 14 By Proposition 3,

where V 2 = H~' n V 1 is non-singular If V 2 is anisotropic, we can take Vo = V, Otherwise

we repeat the process After finitely many steps we must obtain a representation of the required form, possibly with Vo = {O}

1 Qundiatic spaces

Let V and V' be quadsatic spaces over the same field F The quadxatic spaces V,V' are said

to be isometric if there exists a linear map cp: V + V' which is an isometry, i.e it is bijective and

(qv,cpv) = ( v , ~ ) for all v E V

By (I), this implies

((~u,cpv) = (u,v) for all u,v E V

The concept of isometry is only another way of looking at equivalence For if cp: V -+ V'

is an isometry, then V and V' have the same dimension If u l , , u, is a basis for V and

ul', , u,' a basis for V' then, since (u& = (cpuJ,cpuk), the isometry is completely determined

by the change of basis in V' from qul , , (pu,, to u L 1 , , u,'

A particularly simple type of isometry is defined in the following way Let V be a quadratic space and M J a vector such that ( w , ~ ) # O The map z: V -+ V defined by

is obviously linear If W is the non-singular one-dimensional subspace spanned by w, then

V = W I w' Since zv = v if v E W' and zv = - v if v E W, it follows that z is bijective Writing a = - 2(v,w)/(w,w), we have

Tlws z is an isometry Geometrically, z is a reflectioiz in the hypeiplane orthogonal to w We will refer to z = z,, as the reflection cor~esponding to the non-isotropic vector w

PROPOSITION 16 If u,u' a m vectou of a quadratic space V such that (u,u) = (u',u') z 0, then there exists a12 isometry cp: V + V sucl? that cpu = u'

Proof Since

( U + u,',u + u') + (14 - ul,u - u') = 2(u,u) + 2(u',u') = 4(u,u),

at least one of the vectors u + ul,u - u' is not isotropic If u - u' is not isotropic, the reflection z

corresponding to MI = u - u' has the property zu = u', since ( u - u',u - u') = 2(u,u - u') If

u + u' is not isotropic, the reflection z con-esponding to M J = u + u' has the propesty zu = - u'

Since u' is not isotropic, the corresponding reflection cs maps u' onto - u', and hence the isomehy csz maps u onto u'

The proof of Proposition 16 has the following interesting consequence:

352 VII The arithmetic of quadratic forms

PROPOSITION 17 Any isometry cp: V + V of a non-singular quadratic space V is a product

of reflections

Proof Let u l , , u, be an orthogonal basis for V By Proposition 16 and its proof, there exists

an isometry yf, which is either a reflection or a product of two reflections, such that v u l = cpul

If U is the subspace with basis ul and W the subspace with basis uz, , u,, then V = U I W and

W = u - ~ is non-singular Since the isometry cpl = yrlcp fixes u l , we have also cplW = W But

if o: W -+ W is a reflection, the extension z: V -+ V defined by zu = u if u E U , zw = o w if

w E W, is also a reflection By using induction on the dimension n , it follows that cpl is a product of reflections, and hence so also is cp

By a more elaborate argument E Cartan (1938) showed that any isometry of an n-

dimensional non-singular quadratic space is a product of at most n reflections

PROPOSITION 18 Let V be a quadratic space with two orthogonal sum representations

v = U I W = U ' I W'

If there exists an isometry cp: U -+ U', then there exists an isometry v: V -+ V such that

v u = cpu for all u E U and yfW = W' Thus if U is isometric to U: then W is isometric to W' Proof Let u l , ,urn and urn+l , , u, be bases for U and W respectively If uj' = cpuj (1 I j I m ) , then ul' , , u,' is a basis for U' Let urn+l' , , u,' be a basis for W ' The symmetric matrices associated with the bases u l , , u, and ul', , u,' of V have the form

which we will write as A 0 B , A 0 C Thus the two matrices A 0 B , A 0 C are congruent

is enough to show that this implies that B and C are congruent For suppose C = StBS for some invertible matrix S = (oii) If we define ", ., u," by

ui' = C J=rn+loji~j" (m + 1 I i n),

then (uj1',uk") = (uj,uk) (m + 1 I j,k I n) and the linear map y: V -+ V defined by

is the required isometry

By taking the bases for U,W,W' to be orthogonal bases we are reduced to the case in which

1 Quadratic spaces 353

A,B,C are diagonal matrices We may choose the notation so that A = diag [ a l , ,a,], where

a j # O f o r j I r and aj = O f o r j > r If al #O, i.e if r > 0, andif we wliteA'= diag [a2 , , a,],

then it follows from Propositions 1 and 16 that the matsices A'@ B and A' O C are congruent Proceeding in this way, we are reduced to the case A = 0

Thus we now suppose A = 0 We may assume B # 0, C # 0, since otherwise the result is obvious We may choose the notation also so that B = 0, O B' and C = 0, O C', where B' is non-singulas and 0 I s < lz - nz If Tt(O,,+, O C7T = Om+, O B', where

then T4CtT4 = B' Since B' is non-singular, so also is T4 and thus B' and C' are congruent It follows that B and C are also congruent

COROLLARY 19 If a non-singular subspace U of a quadratic space V is isonzetric to another subspace U', then u-' is isometric to u"

Proof This follows at once from Proposition 18, since U'is also non-singular and

The first statement of Proposition 18 is known as Witt's extension theorem and the second statement as Witt's cancellation theorem It was Corollary 19 which was actually proved by Witt (1937)

There is also another version of the extension theorem, which says that if cp: U + U' is an isometry between two subspaces U,U' of a non-singular quadratic space V , then there exists an

isometry y: V -+ V such that yfu = cpu for all u E U For non-singular U this has just been proved, and the singular case can be reduced to the non-singular by applying (several times, if necessary) the following lemma

LEMMA 20 Let V be a non-singular quadratic space If U,U' are singulur subspuces of V

- -

and if there exists an isometry cp: U -+ U ' , the11 there exist subspaces U , U ' properly containing U,U' respectively and an isometry : 8 -+ 8' such that <p u = cpu for all u E U Proof B y hypothesis there exists a nonzero vector ul E U n u' Then U has a basis u l , , u,

with ul as fkst vector By Psoposition 4, there exists a vector w E V such that

(ul,w) = 1, (uj,w) = O for 1 < j I m

354 VII The m-itlzmetic of quadratic forms

Moreover we may assume (W,MJ) = 0, by replacing w by w - a u l , where a = (w,w)/2 If W is the 1-dimensional subspace spanned by M J , then U n W = ( 0 ) and V = U + W contains U

properly

The same construction can be applied to U ' , with the basis q u l , , cpu,, to obtain an isotropic vector w' and a subspace U' = U' + W' The lineas map : V + 0' defined by

is easily seen to have the required propesties

As an application of Proposition 18, we will consider the uniqueness of the representation obtained in Proposition 15

PROPOSITION 21 Suppose the quadratic space V can be represerzted as apz orthogonal sun?

where U is totally isotropic, H is the orthogoml sun1 of nz hyperbolic planes, and the

subspace Vo is anisotropic

Then U = v', nz = ind V - dim v', arid Vo is uniquely detern~ined up to an isometry Proof Since H and Vo are non-singulas, so also is W = H I Vo Hence, by the remark after the proof of Proposition 3, U = w'- Since U c u', it follows that U c v' In fact U = v'-, since w n V' = {O)

The subspace H has two nz-dimensional totally isotropic subspaces U1,Ulf such that

Evidently VI: = V' + U 1 is a totally isotropic subspace of V In fact VI is maximal, since any isotropic vector in U1' I Vo is already contained in U1' Thus nz = ind V - dim v-' is uniquely determined and H is uniquely determined up to an isomet~y If also

where H' is the orthogonal sum of m hyperbolic planes and Vo' is anisotropic then, by Psoposition 18, Vo is isometric to Vat

Proposition 21 reduces the problem of equivalence for quadratic forms over an arbitrary field to the case of anisotropic forms As we will see, this can still be a difficult problem, even for the field of rational numbers

I Quadratic spaces 355

Two quadratic spaces V, V' over the same field F may be said to be Witt-equivalent, in symbols V = V', if their anisotropic components Vo, VO1 are isometric This is certainly an equivalence relation The cancellation law makes it possible to define various algebraic operations on the set W(F) of all quadratic spaces over the field F , with equality replaced by Witt-equivalence If we define - V to be the quadratic space with the same underlying vector space as V but with (vl,v2) replaced by - (vl,v2), then

VL(-V)= { O }

If we define the sum of two quadratic spaces V and W to be V I W, then

Similarly, if we define the product of V and W to be the tensor product V €9 W of the underlying vector spaces with the quadratic space structure defined by

then

It is readily seen that in this way W ( F ) acquires the structure of a commutative ring, the Witt ring of the field F

2 The Hilbert symbol

Again let F be any field of characteristic + 2 and F X the multiplicative group of all nonzero elements of F We define the Hilbert symbol where a,b E F X , by

( ~ , b ) ~ = 1 if there exist x,y E F such that ax2 + by2 = 1 ,

356 VII The arithmetic of quadratic forms

LEMMA 22 F o r anyfield F and any a,b E F X , ( ~ , b ) ~ = 1 if and only if the binuty quadratic form f, = 5 2 - aq2 represents b Morover, for any a E F X , the set G, of all b E FX which

a r e represented by f, is a subgroup of F X

Proof Suppose first that ax2 + by2 = 1 for some x,y E F If a is a square, the quadratic form

f, is isotropic and hence universal If a is not a square, then y # 0 and (j-l)2 - a(xylj2 = b Suppose next that u2 - av2 = b for some u,v E F If - ba-1 is a square, the quadratic form at2 + b q 2 is isotropic and hence universal If - ba-l is not a square, then u # 0 and a(vu-1)2 + b(u-l)2 = 1

It is obvious that if b E G,, then also b-1 E G,, and it is easily ve~lfied that if

(iv) ( 4 - ablF = (a,b)F,

(v) if(a,bIF = 1, then ( ~ , b c ) ~ = (a,cjF for any c E Ex

Proof The first three properties follow immediately from the definition The fourth property follows from Lemma 22 For, since G, is a group and f, represents - a,,f, represents - a b if and only if it represents b The proof of (v) is similar: if fa represents b, then it represents bc if and only if it represents c 0

The Hilbert symbol will now be evaluated for the real field [W = Q, and the y-adic fields

Q, studied in Chapter VI In these cases it will be denoted simply by (a,b),, resp (a,b), For the real field, we obtain at once from the definition of the Hilbert symbol

PROPOSITION 24 Let a,b E RX Then (a,b), = - 1 if and only if both u < 0 and b < 0 El

To evaluate (a,bjp, we first note that we can write a = p W , b = ppb', where a,p E Z and la'/, = lb'lp = 1 It follows from (i),(iij of Proposition 23 that we may assume a$ E { O , l }

2 The Hilbert symbol 357

Furthermore, by (ii),(iv) of Proposition 23 we may assume that a and P are not both 1 Thus

we are reduced to the case where a is ap-adic unit and either b is ap-adic unit or b =pb: where

b' is a p-adic unit To evaluate (a,b)p under these assumptions we will use the conditions for a p-adic unit to be a square which were derived in Chapter VI It is convenient to treat the case

p = 2 separately

PROPOSITION 25 Let p be an odd prime and a,b E Qp with lab = lbip = 1 Then

(9 (a,b), = 1,

(ii) (a,pb), = 1 if and only i f a = c2 for some c E Qgp

In particular, for any integers a,b not divisible by p, (a,b)p = 1 and ( a , ~ b ) ~ = ( a l p ) , where (alp) is the Legendre symbol

Proof Let S c Z p be a set of representatives, with 0 E S , of the finite residue field

Fp = Zp/pZp There exist non-zero ao,bo E S such that

But Lemma 1 1 implies that there exist xo,yo E S such that

Since Ixolp 2 1, lyolp 4 1, it follows that

Hence, by Proposition V1.16, axo2 + byo2 = z2 for some z E Qp Since z f 0, this implies

(a,b& = 1 This proves (i)

If a = c2 for some c E Q,, then (a,pb), = 1, by Proposition 23 Conversely, suppose there exist x,y E Qp such that ax2 + pby2 = 1 Then lax2Ip # bby2Ip, since lalp = lblp = 1 It follows that IxL = 1, lylp I 1 Thus lax2 - 11, < 1 and hence ax2 = z2 for some z E Qpx This proves (ii)

The special case where a and b are integers now follows from Corollary VI.17 0

COROLLARY 26 If p is an odd prime and if a,b,c E Q p are p-adic units, then the quadratic form at2 + bq2 + cC2 is isotropic

Proof The quadratic form - c-lac2 - c-lbq2 - c2 is isotropic, since (- c-la, - ~ l b ) ~ = 1 , by Proposition 25 0

358 VII The arithmetic of quadratic forms

PROPOSITION 27 Let a,b E Q2 with la12 = lb12 = 1 Then

(i) (a,b)2 = 1 if and only i f a t least one of a,b,a - 4,b - 4 is a square in Q2;

(ii) (a,2b)2 = 1 if and only ifeither a or a + 2b is a square in Q2

In particular, for any odd integers a,b, ( ~ , b ) ~ = 1 if and only i f a = 1 or b = 1 mod 4, and ( ~ , 2 b ) ~ = 1 if and only if a = 1 or a + 2b = 1 mod 8

Proof Suppose there exist x,y E Q2 such that ax2 + by2 = 1 and assume, for definiteness, that

lx12 2 ly12 Then ]xi2 2 1 and Ix12 = 2a, where a 2 0 By Corollary VI.14,

where xo E { 1 , 3 } , yo E {0,1,2,3} and x ' , y ' ~ Z2 If a and b are not squares in Q2 then, by Proposition V I 16, la - 1 l2 > 2-3 and Ib - 1 I2 > 2-3 Thus

where ao,bo E {3,5,7} and a',bl E Z2 Hence

where z' E Z2 Since ao,bo are odd and yo2 = 0,l or 4 mod 8, we must have a = 0 , yo2 r 1

mod 8 and a = 5 Thus, by Proposition V I 16 again, a - 4 is a square in Qz This proves that the condition in (i) is necessary

If a is a square in Q 2 , then certainly ( ~ , b ) ~ = 1 If a - 4 is a square, then a = 5 + 8a',

where a' E Z 2 , and a + 4b = 1 + 8c1, where c' E Z2 Hence a + 4b is a square in Q2 and the quadratic form a52 + bq2 represents 1 This proves that the condition in (i) is sufficient

Suppose next that there exist x,y E Q 2 such that ax2 + +by2 = 1 By the same argument as

for odd p in Proposition 25, we must have 1x12 = 1, Iy12 I 1 Thus x = xo + 4x1, y = yo + 4y',

where xo E { 1 , 3 } , yo E {0,1,2,3} and x',yl E Z2 Writing a = a + 8a1, b = bo + 8bf, where

ao,bo E { 1,3,5,7} and a',bf E Z 2 , we obtain adco2 + 2boyo2 = 1 mod 8 Since 2y02 = 0 or 2 mod 8, this implies either a = 1 or a + 2bo = 1 mod 8 Hence either a or a + 2b is a square in

Q2 It is obvious that, conversely, ( ~ , 2 b ) ~ = 1 if either a or a + 2b is a square in Q2

The special case where a and b are integers again follows from Corollary VI.17

For F = [W, the factor group Fx/Fx2 is of order 2, with 1 and - 1 as representatives of the two square classes For F = Qp, with p odd, it follows from Corollary VI.17 that the factor group ~ ~ 1is of order ~ x 2 4 Moreover, if r is an integer such that (r/p) = - 1, then l,r,p,rp are representatives of the four square classes Similarly for F = Q2, the factor group ~ ~ 1is of ~ x 2

2 The Hilbert symbol 359

order 8 and 1,3,5,7,2,6,10,14 are representatives of the eight square classes The Hilbert symbol ( ~ , b ) ~ for these representatives, and hence for all a,b E F X , may be determined directly from Propositions 24,25 and 27 The values obtained are listed in Table 1, where E = (-llp)

and thus

E = + 1 according as p = + 1 mod 4

Table 1: Values of the Hilbert symbol (a,b), for F = Q,

It will be observed that each of the three symmetric matrices in Table 1 is a Hadamard matrix! In particular, in each row after the first row of +'s there are equally many + and -

signs This property turns out to be of basic importance and prompts the following definition:

A field F is a Hilbert field if some a E FX is not a square and if, for every such a, the subgroup G, has index 2 in F X

Thus the real field R = Q, and the p-adic fields Qp are all Hilbert fields We now show that in Hilbert fields further properties of the Hilbert symbol may be derived

360 VII The arithmetic of quadratic forms

PROPOSITION 28 In any Hilbertfield F, the Hilbert symbol has the following additional properties:

(i) i f ( ~ , b ) ~ = 1 for every b E FX, then a is a square in F X ;

(ii) (a,bc), = (a,b)F (U,C)F for all a,b,c E FX

Proof The first property is immediate, since G , # FX if a is not a square If ( ~ , b ) ~ = 1 or

( a , ~ ) ~ = 1, then (ii) follows from Proposition 23(v) Suppose now that ( a , b ) ~ = ( a , ~ ) ~ = - 1

Then a is not a square and fa does not represent b or c Since F is a Hilbert field and b,c E G,,

it follows that bc E G, Thus ( ~ , b c ) ~ = 1

The definition of a Hilbert field can be reformulated in terms of quadratic forms I f f is an anisotropic binary quadratic form with determinant d, then - d is not a square and f is equivalent

to a diagonal form a(52 + dq2) It follows that F is a Hilbert field if and only if there exists an anisotropic binary quadratic form and for each such form there is, apart from equivalent forms, exactly one other whose determinant is in the same square class We are going to show that Hilbert fields can also be characterized by means of quadratic forms in 4 variables

L E M M A 29 Let F be an arbitraryfield and a,b elements of FX with ( a , b ) ~ = - 1 Then the quadratic form

f , a b = 512-at22-b(532-a542)

is anisotropic Morover, the set Ga,b of all elements of FX which are represented by fa,b is

a subgroup of FX

Proof Since (a,b)F = - 1 , a is not a square and hence the binary form fa is anisotropic If forb

were isotropic, some c E FX would be represented by both fa and bf, But then ( u , c ) ~ = 1 and

( ~ , b c ) ~ = 1 Since ( ~ , b ) ~ = - 1, this contradicts Proposition 23

Clearly if c E Ga,b, then also c1 E Gash, and it is easily verified that if

then

It follows that Ga,b is a subgroup of FX

2 The Hilbert symbol 36 1

PROPOSITION 30 A field F is a Hilbert field if and only if one of the following mutually exclusive conditions is satisfied:

(A) F is an ordered jield and every positive element of F is a square;

( B ) there exists, up to equivalence, one and only one anisotropic quaternary quadratic form over F

Proof Suppose first that the field F is of type (A) Then - 1 is not a square, since - 1 + 1 = 0

and any nonzero square is positive By Proposition 10, any anisotropic binary quadratic form

is equivalent over F to exactly one of the forms k2 + q2, - k2 - q 2 and thus F is a Hilbert field Since the quadratic forms c 1 2 + c22 + t 3 2 + e 4 2 and - t12 - 522 - 532 - 542 are anisotropic and inequivalent, the field F is not of type (B)

Suppose next that the field F is of type (B) The anisotropic quaternary quadratic form must be universal, since it is equivalent to any nonzero scalar multiple Hence, for any a E F X

there exists an anisotropic diagonal form

where b',c',d' E F X In particular, for a = - 1, this shows that not every element of F X is a square The ternary quadratic form h = - b'522 - ~ '- d'542 5 ~is certainly anisotropic If ~ h

does not represent 1, the quaternary quadratic form - el2 + h is also anisotropic and hence, by Witt's cancellation theorem, a must be a square Consequently, if a G FX is not a square, then there exists an anisotropic form

- a t I 2 + 522 - b532 - ~ 5 4 ~

Thus for any a E FX which is not a square, there exists b E F X such that (a,b)F = - 1 If

(a,b)F = (a,b')F = - 1 then, by Lemma 29, the forms

are anisotropic and thus equivalent It follows from Witt's cancellation theorem that the binary forms b(t32 - at42) and b1(t32 - aS42) are equivalent Consequently 532 - a542 represents bb'

and (a,bb')F = 1 Thus G, has index 2 in F X for any a E F X which is not a square, and F is a Hilbert field

Suppose now that F is a Hilbert field Then there exists a E FX which is not a square and, for any such a, there exists b E FX such that (a,b)F = - 1 Consequently, by Lemma 29, the quaternary quadratic form fa,b is anisotropic and represents 1 Conversely, any anisotropic quaternary quadratic form which represents 1 is equivalent to some form

VII The arithmetic of quadratic forms

with a,b,c E F X Evidently a and c are not squares, and if d is represented by 532 - ~ 5 ~ 2 , then

bd is not represented by c 1 2 - aa522 Thus ( ~ , d ) ~ = 1 implies (a,bd)F = - 1 In particular,

(a,b)F = - 1 and hence (c,QF = 1 implies (a,djF = 1 Interchanging the roles of Cl2 - ac22 and

t 3 2 - ~ 5 ~ 2 , we see that ( & a F = 1 also implies (c,d)F = 1 Hence ( a ~ , d ) ~ = 1 for all d E FX Thus ac is a square and g is equivalent to

We now show that fa,b and fa,,b are equivalent if (a,b)F = (a',b')F = - 1 Suppose first that

(a,b')F = - 1 Then (a,bb?F = 1 and there exist x 3 j 4 E F such that b' = b (- ax42) ~ ~Since ~

where q3 = x3C3 + ax4k4, q4 = x 4 t 3 + x3t4, it follows that fa,b, is equivalent to fa,b For the same reason fa,b, is equivalent to and thus fa,b is equivalent to fa,,bt By symmetry, the same conclusion holds if (a',b)F = - 1 Thus we now suppose

But then (a,bb?F = (a:bb?F = - 1 and so, by what we have already proved,

Together, the last two paragraphs show that if F is a Hilbert field, then all anisotropic quaternary quadratic forms which represent 1 are equivalent Hence the Hilbert field F is of type (B) if every anisotropic quaternary quadratic form represents 1

Suppose now that some anisotropic quaternary quadratic form does not represent 1 Then some scalar multiple of this form represents 1, but is not universal Thus faPb is not universal for some a,b E FX with (a,b)F = - 1 By Lemma 29, the set Ga,b of all c E FX which are represented by faPb is a subgroup of FX In fact Ga,b = Ga, since Ga c Ga,b, Ga,b f FX and Ga

has index 2 in FX Since fa,b - fb,a, we have also Ga,b = Gb Thus ( a , ~ ) ~ = ( b , ~ ) ~ for all

c E FX, and hence ( a b , ~ ) ~ = 1 for all c E FX Thus ab is a square and ( ~ , a ) ~ = ( ~ , b ) ~ = - 1 Since (a,- a ) F = 1, it follows that (a,- = - 1 Hence fajb -fa,, - fa,-l Replacing a,b by

- 1,a we now obtain (- 1,- = - 1 and fa,-l - fPljpl

Thus the form

f = + b2 + h2 + 542

2 Tlie Hilbert symbol 363

is not universal and the subgroup P of all elements of F X represented by f coincides with the set

of all elements of F X represented by 5 2 + 72, Hence P + P c P and P is the set of all c E FX

such that (- 1 , ~ ) ~ = 1 Consequently - 1 6 P and F is the disjoint union of the sets (01, P and

- P Thus F is an ordered field with P as the set of positive elements

For any c E F X , c2 E P It follows that if a,b E P then (- a,- b)F = - 1 , since a t 2 + l!q2

does not represent - 1 Consequently, if a,b E P, then (- a,- b ) F = - 1 = (- 1,- b ) F and

(- ~ , b ) ~ = 1 = (-l,b)F Thus, for all r E F X , (- 0 , ~ ) ~ = ( - 1 , ~ ) ~ and hence ( a , c ) , ~ = 1 Therefore

a is a square and the Hilbert field F is of type (A)

PROPOSITION 31 I f F is a Hilbert field of type ( B ) , the11 any quadratic form f in more thau 4 variables is isotropic

For any prime p, the field Qi, ofp-adir rzumbe~.s is a Hilbutfield of type ( B )

Proof The quadsatic form f is equivalent to a diagonal form a1512 + + a,tn2, where rz > 4

If g = a 1 t 1 2 + + a4k42 is isotropic, then so also is f If g is anisotropic then, since F is of

type ( B ) , it is universal and represents - U S This proves the first part of the proposition

We already know that Q p is a Hilbert field and we have already shown, after the proof of

Corollaly VI.17, that Qp is not an ordered field Hence Qp is a Hilbert field of type (B)

Proposition 10 shows that two non-singular quadratic forms in n variables, with coefficients from a Hilbert field of type (A), are equivalent over F if and only if they have the same positive index We consider next the equivalence of quadratic forms with coefficients from a Hilbert field of type (B) We will show that they are classified by their determinant and their Hasse invariant

If a non-singular quadsatic form f , with coefficients from a Hilbei-t field F , is equivalent to

a diagonal form a1512 + + a,52, then its Hasse invariant is defined to be the product of

Hilbert symbols

~ l ; ( f ) = nl< j < k < n ( u j ? a k ) ~

We write sJ,(f) for S& when F = Qp (It should be noted that some authors define the Hasse invariant with njSk in place of lI+,) It must first be shown that this is indeed an invariant of

f, and for this we make use of Witt's clzai~z equivalence theo~xwz:

L E M M A 32 Let V be a 17011-singular quadratic space over an arbitrary field F I f 9 =

{ u l , , 11,) and $33 ' = { u , ' , , u,') m e both orthogorzal bases of V , the12 there exists a chairz

of ortlzogo~zul bases $!80,%1, ,!%,,, wit11 aO = 3 and % m = a ' , such that %j-l a d %j differ

by at most 2 vectors for each j E ( 1 , , m }

364 VZI The arithmetic of quadratic forms

Proof Since there is nothing to prove if dim V = n 1 2 , we assume that n 2 3 and the result holds for all smaller values of n Let p = p(%) be the number of nonzero coefficients in the representation of ul' as a linear combination of ul, , u, Without loss of generality we may suppose

ul' = C7=lajuj,

where a j # 0 (1 I j I p ) If p = 1, we may replace ul by ul' and the result now follows by applying the induction hypothesis to the subspace of all vectors orthogonal to ulf Thus we now assume p 2 2 We have

and each summand on the left is nonzero If the sum of the first two terms is zero, then p > 2 and either the sum of the first and third terms is nonzero or the sum of the second and third terms is nonzero Hence we may suppose without loss of generality that

a12(u1,u1) + ~ 2 ~ ( ~ 2 , ~ 2 ) # 0

If we put

v1 = a l u l + a2u2, v2 = u1 + bu2, vj = uj for 3 I j I n,

where b = - al(ul,ul)/a2(u2,u2), then % = {v l, ,vn} is an orthogonal basis and u l ' = v1 + a3v3 + + a,vp T h u s p ( a 1 ) <p(93) By replacing 93 by PA1 and repeating the procedure, we must arrive after s < n steps at an orthogonal basis 93, for which p(93,) = 1 The induction hypothesis can now be applied to 9, in the same way as for 93

PROPOSITION 33 Let F be a Hilhert field If the non-singular diagonal forms a1t12 + + a&,2 and b1t12 + + bnCn2 are equivalent over F, then

Proof Suppose first that n = 2 Since a1t12 + a2tZ2 represents bl, Cl2 + al-1a2522 represents

a l - l b l and hence (- a l - l a 2 , a l - l b l ) F = 1 Thus (albl,- ala2b12)F = 1 and hence (albl,a2bl)F = 1 But (Proposition 28 (ii)) the Hilbert symbol is multiplicative, since F is a Hilbert field It follows that ( b l , ~ l ~ 2 b l ) F = 1 Since the determinants ala2 and blb2 are in the same square class, this implies (al,a2)F = (bl,b2)F, as we wished to prove

Suppose now that n > 2 Since the Hilbert symbol is symmetric, the product JJlgj,ksn(aj,ak)F is independent of the ordering of a l , , a, It follows from Lemma 32 that

we may restrict attention to the case where a1t12 + a2522 is equivalent to b1t12 + b2t22 and

2 The Hilbert symbol 365

But this follows from the multiplicativity of the Hilbert symbol and the fact that ala2 and blb2

are in the same square class

Proposition 33 shows that the Hasse invariant is well-defined

PROPOSITION 34 Two non-singular quadratic forms in n variables, with coefficients from

a Hilbertfield F of type ( B ) , are equivalent over F if and only if their determinants are in the same square class and they have the same Hasse invariant

Proof Only the sufficiency of the conditions needs to be proved Since this is trivial for n = 1,

we suppose first that n = 2 It is enough to show that if

where (a,adjF = (b,bQF, then f is equivalent to g The hypothesis implies (- d , ~ ) ~ = (- d,b)F

and hence (- d,ab)F = 1 Thus El2 + dt22 represents ab and f represents b Since det f and det g are in the same square class, it follows that f is equivalent to g

Suppose next that n 2 3 and the result holds for all smaller values of n Let f ( t l , .,tn)

and g ( q l , .,qn) be non-singular quadratic forms with det f = det g = d and sF(f) = sF(g) By Proposition 3 1, the quadratic form

is isotropic and hence, by Proposition 7, there exists some al E FX which is represented by both f and g Thus

366 VII The arithmetic of padratic f o r m

3 The Hasse-Minkowski theorem

Let a,b,c be nonzero squasefree integers which are relatively prime in paiss It was proved

by Legendse (1785) that the equation

has a nontrivial solution in integers x,y,z if and only if a,b,c are not all of the same sign and the congruences

u 2 = - b c m o d u , v 2 = - c u m o d b , w 2 = - a b m o d c

are all soluble

It was first completely proved by Gauss (1801) that evely positive integer which is not of the form 4l1(8k + 7) can be represented as a sum of three squares Legendre had given a proof, based on the assumption that if a and n7 are relatively prime positive integers, then the arithmetic progression

a, u + nz, a + 2nz,

contains infinitely inany primes Although his proof of this assumption was faulty, his intuition that it had a role to play in the arithmetic theory of quadratic forms was inspired The assumption was first proved by Dirichlet (1837) and will be referred to here as 'Dirichlet's theorem on primes in an arithmetic progression' 111 the present chapter Dirichlet's theorem will simply be assumed, but it will be proved (in a quantitative form) in Chapter X

It was shown by Meyer (1884), although the published proof was incomplete, that a quadsatic form in five or more variables with integer coefficients is isotropic if it is neither positive definite nor negative definite

The preceding results are all special cases of the Husse-Minkowski theorem, which is the subject of this section Let Q denote the field of rational numbers By Ostrowski's theorem (Proposition VI.4), the coinpletions Q, of Q with respect to an arbitray absolute value I 1, are the field Q, = R of real numbers and the fields Qp of p-adic numbers, where p is an arbitrary prime The Hasse-Minkowski theorem has the following statement:

A rzon-singular quadratic form fTcl, ,c,) vt~itlz coefficierzts from Q is isotropic in Q if

a11d only if it is isotropic ill every conlpletion of Q

This concise statement contains, and to some extent conceals, a remarkable amount of information (Its equivalence to Legendre's theorem when n = 3 may be established by

3 The Hasse-Minkowski theorem 367

elementary arguments.) The theorem was first stated and proved by Hasse (1923) Minkowski (1890) had derived necessary and sufficient conditions for the equivalence over Q of two non- singular quadratic forms with rational coefficients by using known results on quadratic forms with integer coefficients The role of p-adic numbers was taken by congruences modulo prime powers Hasse drew attention to the simplifications obtained by studying from the outset quadratic forms over the field Q, rather than the ring H, and soon afterwards (1924) he showed that the theorem continues to hold if the rational field 69 is replaced by an arbitrary algebraic number field (with its corresponding completions)

The condition in the statement of the theorem is obviously necessary and it is only its sufficiency which requires proof Before embarking on this we establish one more property of the Hilbert symbol for the field 69 of rational numbers

PROPOSITION 35 For any a,b E Q X , the number of conzpletions Q , for which (a,b), = - 1

(where v denotes either or an arbitrary prime p) is finite and even

Proof By Proposition 23, it is sufficient to establish the result when a and b are square-free integers such that ab is also square-free Then (a,b), = 1 for any odd prime r which does not divide ab, by Proposition 25 We wish to show that n,(a,b), = 1 Since the Hilbert symbol is multiplicative, it is sufficient to establish this in the following special cases: for a = - 1 and b =

- 1,2,p; for a = 2 and b = p ; for a = p and b = q, where p and q are distinct odd primes But it follows from Propositions 24, 25 and 27 that

Hence the proposition holds if and only if

Thus it is actually equivalent to the law of quadratic reciprocity and its two 'supplements'

We are now ready to prove the Hasse-Minkowski theorem:

THEOREM 36 A non-singular quadratic form f ( t l , ,cn) with rational coefficients is isotropic in Q if and only if it is isotropic in every completion Q,

368 VII The arithmetic of quadratic forms

Proof We may assume that the quadratic form is diagonal:

where ak E Q X (k = 1, , n) Moreover, by replacing kk by rkck, we may assume that each coefficient ak is a square-free integer

The proof will be broken into three parts, according as n = 2, n = 3 or n 2 4 The proofs for n = 2 and n = 3 are quite independent The more difficult proof for n 2 4 uses induction on

n and Dirichlet's theorem on primes in an arithmetic progression

(i) n = 2: We show first that if a E QX is a squase in QVx for all v, then a is already a square in

QX Since a is a square in QmX, we have a > 0 Let a = n p p a p be the factorization of a into powers of distinct primes, where ap E Z and ap # 0 for at most finitely many pi-imes p Since

[alp =P-"P and a is a square in Qp, % must be even But if al, = 2P,, then a = b2, where

b = nppPp

Suppose now that f = a1C12 + a 2 t z 2 is isotropic in Q, for all v Then u : = - a l a 2 is a square in Q, for all v and hence, by what we have just proved, a is a square in Q But if

a = b2, then ala22 + a2b2 = 0 and thus f is isotropic in Q

(ii) n = 3: By replacing f by - a3fand t3 by a3k3, we see that it is sufficient to prove the theorem for

We are going to show that there exists an integer c such that c2 a mod b Since + b is a product of distinct primes, it is enough to show that the congruence x2 z a mod p is soluble for each grime p which divides b (by Corollary 11.38) Since this is obvious if a = 0 or 1 mod p,

we may assume that p is odd and a not divisible by p Then, since f is isotsopic in Qp, ( ~ , b ) ~ = 1 Hence a is a squase mod y by Proposition 25

Consequently there exist integers c,d such that a = c2 - bd Moreover, by adding to c a suitable multiple of b we may assume that Icl I 1b1/2 Then

3 The Hasse-Minkowski theorem

and d z 0, since a is square-free and a z 1 We have

Q2, by Proposition 27 In fact for n = 3 it need only be assumed that f is isotropic in Q, for all

v with at most one exception since, by Proposition 35, the number of exceptions must be even

(iii) n 2 4: We have

f = a1t12 + + ankn2,

where a l , , a, are square-free integers We write f = g - h, where

Let S be the finite set consisting of rn and all primes p which divide 2al a, By Proposition 7 ,

for each v E S there exists c, E Q,X which is represented in Q, by both g and h We will show that we can take c, to be the same nonzero integer c for every v E S

Let v = p be a prime in S By multiplying by a square in Q p X we may assume that

cp =pEp cpt, where ~p = 0 or 1 and lcpllp = 1 If p is odd and if bp is an integer such that

Ic P - b P P - I <p-'p-l, then b p L = Ic,L and lbpcp-1 - l b 5p-1 Hence bpcp-1 is a square in QpX,

by Proposition VI.16, and we can replace cp by hp Similarly if p = 2 and if b2 is an integer such that Ic2 - b2I2 1 2-E2-3, then lb2I2 = lc2I2 and lb2c2y1 - 112 12-3 Hence b2c2-I is a square in Q2X and we can replace c2 by b2

By the Chinese remainder theorem (Corollary 11-38), the simultaneous congruences

c r b2 mod 2E2+3, c r bp mod pEp+l for every odd p E S,

370 VII The arithmetic of quadratic forms

have a solution c E 12, which is uniquely determined mod m , where m = 411pESpE~+1 In exactly the same way as before we can replace bp by c for all primes p E S By choosing c to have the same sign as c,, we can take c, = c for all v E S

If d = llpEspEp is the greatest common divisor of c and m then, by Dirichlet's theorem on primes in an arithmetic progression, there exists an integer k with the same sign as c such that

cld + kmfd = k q,

where q is a prime If we put

a = c + k m = +dq,

then q is the only prime divisor of a which is not in S and the quadratic forms

are isotropic in Q, for every v E S, since c-la is a square in QVX

For all primes p not in S, except p = q, a is not divisible byp Hence, by the definition of

S and Corollary 26, g* is isotropic in Q, for all v, except possibly v = q Consequently, by the final remark of part (ii) of the proof, g* is isotropic in Q

Suppose first that n = 4 In this case, by the same argument, h* = a3532 + a4542 + is also isotropic in Q Hence, by Proposition 6, there exist y l , .,y4 E Q such that

aly12 + a2y22 = a = - a3y32 - a4y42

Thus f is isotropic in Q

Suppose next that n 2 5 and the result holds for all smaller values of n Then the quadratic form h* is isotropic in Q,, not only for v E S, but for all v For if p is a prime which is not in

S, then a3,a4,a5 are not divisible by p It follows from Corollary 26 that the quadratic form

a3t32 + a4t42 + a5t52 is isotropic in Q p , and hence h* is also Since h* is a non-singular quadratic form in n - 1 variables, it follows from the induction hypothesis that h* is isotropic in

Q The proof can now be completed in the same way as for n = 4

COROLLARY 37 A non-singular rational quadratic form in n 2 5 variables is isotropic in

Q i f and only if it is neither positive definite nor negative definite

Proof This follows at once from Theorem 36, on account of Propositions 10 and 31

COROLLARY 38 A non-singular quadratic form over the rational field Q represents a

nonzero rational number c in Q if and only if it represents c in every completion Q,

3 The Hnsse-Minkowski theorem 37 1

Proof Only the sufficiency of the condition requires proof But if the rational quadratic form f(kl, &) represents c in Q,, for all v then, by Theorem 36, the quadratic form

is isotropic in Q Hence f represents c in Q, by Proposition 6

PROPOSITION 39 Two non-singular quadi.atic forms with rational coeflicients are equivalent over Q if and only i f they are equivalevlt over all conzpletions Q,

Proof Again only the sufficiency of the condition requires proof Let f and g be non-singular rational quadratic f o l ~ n s in iz variables which ase equivalent over Q, for all v

Suppose first that n = 1 and that f = at2, g = bq2 By hypothesis, for every v there exists

t, E Qvx such that b = at,,2 Thus ba-I is a square in Q V X for every v , and hence bu-l is a square in QX, by part (i) of the proof of Theorem 36 Therefore f is equivalent to g over Q Suppose now that n > 1 and the result holds for all smaller values of n Choose some

c E Q X which is represented by f in Q Then f' certainly represents c in Q, and hence g represents c in Q,,, since g is equivalent to f over Q, Since this holds for all v, it follows from Corollary 38 that g represents c in Q

Thus, by the remark after the proof of Proposition 2, f is equivalent over Q to a quadratic form cC12 +p(52, ,5,) and g is equivalent over Q to a q~~adratic form c t l 2 + g*(52, ,5,)

Since f is equivalent to g over Q,, it follows from Witt's cancellation theorem t11at~(~~, ,4,>

is equivalent to g*(52, ,5,1) over Q , Since this holds for every v , it follows from the induction hypothesis t h a t p is equivalent to g* over Q, and so f is equivalent to g over Q

COROLLARY 40 Two non-singular quadratic forms f and g in n variables with rational coeflicients are equivalent over the rational field Q i f and only if

(i) (det,fl/(det g) is a square in QX,

(ii) indf f = ind' g,

(iii) sp(R = s J g ) for every prime y

Pmof This follows at once from Proposition 39, on account of Propositions 10 and 34

The strong Hasse principle (Theorem 36) says that a quadratic form is isotropic over the global field Q if (and only if) it is isotropic over all its local completions Q, The weak Hasse principle (Proposition 39) says that two quabatic forms are equivalent over Q if (and only if) they are equivalent over all Q, These local-global pi.inciples have proved remarkably fruitful

372 VII The arithmetic of quadratic forms

They organize the subject, they can be extended to other situations and, even when they fail, they are still a useful guide We describe some results which illustrate these remarks

As mentioned at the beginning of this section, the strong Hasse principle continues to hold when the rational field is replaced by any algebraic number field Waterhouse (1976) has established the weak Hasse principle for pairs of quadratic forms: if over every completion Q,

there is a change of variables taking both fi to gl andf2 to g2, then there is also such a change of variables over Q For quadratic forms over the field F = K(t) of rational functions in one variable with coefficients from a field K, the weak Hasse principle always holds, and the strong Hasse principle holds for K = R, but not for all fields K

The strong Hasse principle also fails for polynomial forms over Q of degree > 2 For example, Selmer (1951) has shown that the cubic equation 3x3 + 4y3 + 5z3 = 0 has no nontrivial solutions in Q, although it has nontrivial solutions in every completion Q,

However, Gusic' (1995) has proved the weak Hasse principle for non-singular ternary cubic forms

Finally, we draw attention to a remarkable local-global principle of Rumely (1986) for algebraic integer solutions of arbitrary systems of polynomial equations

with rational coefficients

We now give some applications of the results which have been established

PROPOSITION 41 A positive integer can be represented as the sum of the squares of

three integers if and only if it is not of the form 4nb, where n 2 0 and b = 7 mod 8

Proof The necessity of the condition is easily established Since the square of any integer is congruent to 0,l or 4 mod 8, the sum of three squares cannot be congruent to 7 For the same reason, if there exist integers x,y,z such that x2 + y2 + z2 = 4nb, where n 2 1 and b is odd, then

x,y,z must all be even and thus ( ~ 1 2 ) ~ + ( ~ 1 2 ) ~ + ( ~ 1 2 ) ~ = 4n-1b By repeating the argument n

times, we see that there is no such representation if b = 7 mod 8

We show next that any positive integer which satisfies this necessary condition is the sum

of three squares of rational numbers We need only show that any positive integer a f 7

mod 8, which is not divisible by 4, is represented in Q by the quadratic form