Arithmetic of Quadratic Forms

1 Foundation

Throughout this section, F always denotes a field of characteristic different from 2.

1.1 Quadratic Forms and Quadratic Spaces

An (n-ary) quadratic form over F is a polynomial f in n variables x1, . . . , xn over F that ishomogeneous of degree 2. In general, f takes the form

f(x1, . . . , xn) =n∑

i,j=1

bijxixj , bij ∈ F.

To render the coefficients symmetric, it is customary to rewrite f as

f(x1, . . . , xn) =

n∑i,j=1

aijxixj ,

where aij = (bij + bji)/2. In this way, f determines a symmetric matrix (aij), which weshall denote by Af . In terms of matrix multiplication, we have

f(x) = xtAf x

where x = (x1, . . . , xn)t.Let a be an element in F . We say that a is represented by f if the equation

(∗) f(x) = a

has a solution in Fn. The representation problem of quadratic forms is to determine, in aneffective manner, the set of elements of F that are represented by a particular quadraticform over F . We shall discuss the case when F is a field of arithmetic interest, for instance,the field of complex numbers C, the field of real numbers R, a finite field F, and the field ofrational numbers Q. The representation problem for quadratic forms over any one of thesefields has a very satisfactory solution. At the end, we shall discuss the solubility of equation(∗) over a subring R of F , that is, the problem of finding solution of (∗) in Rn. The mostinteresting and difficult case is when R is the ring of integers Z for which there is still a lotof questions left unanswered.

Let f and g be two n-ary quadratic forms. We say that f and g are equivalent, writtenf ∼= g, if there exists an invertible matrix C ∈ GLn(F ) such that f(x) = g(Cx). This is thesame as saying that there is an invertible homogeneous linear substitution of the variablesx1, . . . , xn which takes the form g to the form f . Since

g(Cx) = (Cx)tAg (Cx) = xt(CtAgC)x,

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the condition f(x) = g(Cx) is equivalent to the matrix equality

Af = CtAgC.

Thus equivalence of forms amounts to congruence of the associated symmetric matrices. Asexpected, equivalence of forms is indeed an equivalence relation. It is clear that equivalentforms represent the same set of elements of F .

Example 1.1 Let g(x, y) be the binary form xy. If we make the substitution x 7→ x+y, y 7→x− y, then g changes to

g(x+ y, x− y) = (x+ y)(x− y) = x2 − y2 = f(x, y).

The matrix C in this case is(

1 11 −1

). This can be verified by

Af =

(1 00 −1

)=

(1 11 −1

)(0 1/2

1/2 0

)(1 11 −1

)= CtAgC.

Any quadratic form f gives rise to a map Qf : Fn −→ F defined by Qf (x) = xtAf x. Weshall refer to Qf as the quadratic map defined by f . The notion of equivalence of quadraticforms, f ∼= g, amounts to the existence of a linear automorphism C of Fn (that is, aninvertible matrix in GLn(F )) such that Qf (x) = Qg(Cx) for all x ∈ Fn. Note that thequadratic map Qf determines the quadratic form f uniquely. For, suppose that Qf = Qgas maps from Fn to F . Let e1, . . . , en be the standard basis for Fn. Then for any i, wehave

(Af )ii = Qf (ei) = Qg(ei) = (Ag)ii.

For i 6= j, we haveQf (ei + ej) = Qf (ei) +Qf (ej) + 2(Af )ij ,

and a similar equation for Qg(ei+ej). Therefore, (Af )ij = (Ag)ij ; thus Af = Ag and f = g.The quadratic map Qf satisfies the following properties:

(1) For any a ∈ F and x ∈ Fn, Qf (ax) = a2Qf (x).

(2) The function Bf (x,y) = 12 [Qf (x + y)−Qf (x)−Qf (y)] is a symmetric bilinear form

on Fn. That is, Bf is a function from Fn × Fn to F satisfying

(i) Bf (x,y) = Bf (y,x) for all x,y ∈ Fn, and

(ii) Bf (ax + by, z) = aBf (x, z) + bBf (y, z) for all a, b ∈ F and x,y, z ∈ Fn.

Note that the quadratic map Qf can be recaptured by the symmetric bilinear form Bf ,since

Qf (x) = Bf (x,x), ∀x ∈ Fn.

This motivates the following geometric approach to the notion of a quadratic form. LetV be an n-dimensional vector space over F equipped with a symmetric bilinear form B :

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V ×V −→ F . The pair (V,B) is called a quadratic space, and associate with it is a quadraticmap Q = QB : V −→ F given by Q(v) = B(v, v) for all v ∈ V . As in (1) and (2) above, wehave Q(av) = a2Q(v) for all a ∈ F and v ∈ V , and 2B(u, v) = Q(u+ v)−Q(u)−Q(v) forall u, v ∈ V . Therefore, Q and B determine each other and hence it is legitimate to write(V,Q) to represent the quadratic space (V,B).

Now, suppose that v1, . . . , vn is a basis for V . Then the quadratic space (V,B) givesrise to a quadratic form

f(x1, . . . , xn) =∑i,j

B(vi, vj)xixj ,

withAf = (B(vi, vj)).

If we identify V with Fn via the basis v1, . . . , vn that is, we identify each vector v =x1v1 + . . . xnvn with the column x = (x1, . . . , xn)t, then Q = QB is precisely the quadraticmap Qf associated with the form f . An element a ∈ F is represented by the form f if andonly if a is represented by V , that is, there exists a vector v ∈ V such that Q(v) = a.

Now, let us choose another basis u1, . . . , un for V , and let g be the resulting quadraticform. Suppose that ui =

∑k ckivk, then

(Ag)ij = B(ui, uj)

= B(∑k

ckivk,∑`

c`jv`)

=∑k,`

ckiB(vk, v`)c`j

= (CtAfC)ij

where C = (ck`). Thus the quadratic space (V,B) determines uniquely an equivalence classof quadratic forms, which we shall denote by [fB].

If (V,B) and (V ′, B′) are two quadratic spaces, we say that they are isometric, written(V,B) ∼= (V ′, B′), if there exists an isomorphism σ : V −→ V ′ such that

B′(σ(u), σ(v)) = B(u, v), ∀u, v ∈ V.

Such σ is called an isometry from V to V ′. The set of all isometries from V to V itself forma group O(V ) under the composition of functions. It is clear that

(V,B) ∼= (V ′, B′)⇐⇒ [fB] = [fB′ ].

Thus there is a one-to-one correspondence between the equivalence classes of n-ary quadraticforms and the isometry classes of n-dimensional quadratic spaces. In this set of lecture notes,we shall adopt the geometric language of quadratic spaces.

Let (V,B) be an n-dimensional quadratic space, and v1, . . . , vn be a basis for V . Let Abe the symmetric matrix (B(vi, vj)). We call A a matrix for V and write

V ∼= A.

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So, V ∼= A ∼= CtAC for any C ∈ GLn(F ). We call V nondegenerate if det(A) is nonzero.Otherwise, V is degenerate. For a nondegenerate space V , its discriminant of V , denotedd(V ), is defined to be the square class det(A)F×2 in F×/F×2. For convention, the zerospace is considered to be nondegenerate and its discriminant is defined to be F×2.

If W is a subspace of V , the map BW : W ×W −→ F defined by BW (x, y) = B(x, y)for all x, y ∈ W is a symmetric bilinear form on W . Thus (W,BW ) is also a quadraticspace. We say that W is a nondegenerate subspace of V if (W,BW ) is nondegenerate as aquadratic space.

Let (V,B) be a quadratic space. Let V ∗ be the vector space of all homomorphisms fromV to F . It is called the dual space of V . The function B : V −→ V ∗ defined by

B(x)(y) = B(x, y)

is clearly linear.Suppose that E = v1, . . . , vn is a basis for V . For each i, define v∗i ∈ V ∗ by

v∗i (vj) = δij =

1 if i = j,

0 otherwise.

Then E∗ = v∗1, . . . , v∗n is a basis for V ∗.

Lemma 1.2 The matrix of B with respect to the bases E and E∗ is the symmetric matrixfor V associated with E.

Proof. From B(ei)(ej) = B(ei, ej) follows

B(ei) =

n∑j=1

B(ei, ej)e∗j .

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Corollary 1.3 A quadratic space (V,B) is nondegenerate if and only if the map B : V −→V ∗ is an isomorphism.

1.2 Orthogonal Decomposition

Let (V,B) be a quadratic space. Two vectors x, y ∈ V are orthogonal if B(x, y) = 0. Twosubsets X and Y of V are said to be orthogonal if B(x, y) = 0 for all x ∈ X, y ∈ Y . Witheach subset X of V , the orthogonal complement of X in V is the set

X⊥ = v ∈ V : B(v, x) = 0 for all x ∈ X.

The orthogonal complement of V itself is called the radical of V , denoted by rad(V ) (butnot by V ⊥).

Lemma 1.4 Let X,Y be subsets of V . Then

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(a) X⊥ is a subspace of V .

(b) If X ⊆ Y , then X⊥ ⊇ Y ⊥.

(c) X ⊆ X⊥⊥

(d) V is nondegenerate if and only if rad(V ) = 0.

Proof. Parts (a), (b) and (c) are direct consequences of the definition of orthogonal com-plements. For (d), note that the kernel of the map B is precisely rad(V ). Therefore,dim(V ) = dim(rad(V )) + dim(Im(B)). By Corollary 1.3, V is nondegenerate if and onlyif B is an isomorphism. Thus V is nondegenerate if and only if dim(rad(V )) = 0, that is,rad(V ) = 0.2

Definition 1.5 Let (V1, B1) and (V2, B2) be two quadratic spaces. The orthogonal sumV1 ⊥ V2 is the quadratic space V1 ⊕ V2 with the symmetric bilinear form B defined byB(x1 + x2, y1 + y2) = B1(x1, y1) +B(x2, y2) for all x1, y1 ∈ V1 and x2, y2 ∈ V2.

So if V = V1 ⊥ V2 as above, then (Vi, Bi) ∼= (Vi, B|Vi) for i = 1, 2, and V1 and V2 areorthogonal in V .

Theorem 1.6 Let (V,B) be a quadratic space. Suppose that V = rad(V ) ⊕W for somesubspace W . Then

(a) V = rad(V ) ⊥W .

(b) W is nondegenerate.

(c) (W,B|W ) is determined up to isometry by V .

Proof. Part (a) is clear. For part (b), suppose that x ∈ W is orthogonal to all vectors inW . Then x is orthogonal to all vectors in rad(V ) ⊥ W = V . Therefore, x ∈ rad(V ) ∩Wand hence x = 0.

For part (c), suppose that V = rad(V )⊕W1. Then each x ∈W can be written uniquelyas

x = y + α(x), y ∈ rad(V ), α(x) ∈W1.

One can check that the map α : W −→ W1 defined by x 7−→ α(x) is a vector spacehomomorphism. It is clear that α is injective and hence surjective since dim(W ) = dim(W1).If x = y + α(x) and x′ = y′ + α(x′), then

B(x, x′) = B(y + α(x), y′ + α(x′)) = B(α(x), α(x′)).

Hence α is an isometry.2

The isometry class of (W,B|W ) obtained in the above theorem is called the nondegen-erate component of (V,B). The classification of general quadratic spaces reduces to theclassification of their nondegenerate components.

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Lemma 1.7 Let V,W, V ′,W ′ be quadratic spaces.

(a) V ⊥ V ′ ∼= V ′ ⊥ V .

(b) If V ∼= W and V ′ ∼= W ′, then V ⊥ V ′ ∼= W ⊥W ′.

(c) If A is a matrix for V and A′ is a matrix for V ′, then

V ⊥ V ′ ∼=(A 00 A′

).

(d) V ⊥ V ′ is nondegenerate if and only if both V and V ′ are nondegenerate. In this case,d(V ⊥ V ′) = d(V )d(V ′).

Proof. Everything is obvious.2

Proposition 1.8 If W is a nondegenerate subspace of a quadratic space (V,B) , thenV = W ⊥W⊥.

Proof. It suffices to show that V = W ⊕W⊥. Since W is nondegenerate, 0 = ker(BW ) =W ∩W⊥. If x ∈ V and h = B(x)|W , then because W is nondegenerate there exists y ∈Wwith h = BW (y). Hence for all z ∈W ,

B(x, z) = B(x)(z) = h(z) = BW (y)(z) = B(y, z)

and we can write x = y + (x− y) where y ∈ W and (x− y) ∈ W⊥ by the above equation.Thereby V = W +W⊥ is proven.2

The above proposition implies that dim(W ) + dim(W⊥) = dim(V ) whenever W is anondegenerate subspace of V . Below we show that the same additive property of thedimension holds for every subspace provided V itself is nondegenerate.

Proposition 1.9 Let (V,B) be a nondegenerate quadratic space, and W be a subspace ofV . Then

(i) dim(W ) + dim(W⊥) = dim(V ).

(ii) (W⊥)⊥ = W .

Proof. Part (ii) is a direct consequence of part (i) because W ⊆ (W⊥)⊥. The map B :V −→ V ∗ is an isomorphism. Since W is a subspace of V , the canonical projection V ∗ −→W ∗ is surjective. The kernel of the composition V −→ V ∗ −→ W ∗ is W⊥. Therefore,dim(V ) = dim(W⊥) + dim(W ∗) = dim(W⊥) + dim(W ). 2

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1.3 Witt’s Theorems

Let (V,B) be a quadratic space. A nonzero vector v ∈ V is called isotropic if Q(v) = 0.Otherwise, v is called anisotropic.

Theorem 1.10 Every quadratic space has an orthogonal basis.

Proof. Let (V,B) be a quadratic space. Since rad(V ) is an orthogonal summand of V , wemay assume that V is nondegenerate. Suppose that Q(x) = 0 for all x ∈ V . Then for anyu, v ∈ V ,

B(u, v) =1

2[Q(u+ v)−Q(u)−Q(v)] = 0,

which implies that rad(V ) = V which is impossible.Now, pick an anisotropic vector x ∈ V . Then Fx is a nondegenerate subspace, and

hence we have a decomposition V = Fx ⊥ V ′ for some subspace V ′. The subspace V ′

is also nondegenerate. An induction argument on the dimension of V will complete theproof.2

Corollary 1.11 Every invertible symmetric matrix in GLn(F ) is congruent to a diagonalmatrix.

We use the notation 〈a1, . . . , an〉 to denote a diagonal matrix with a1, . . . , an as thediagonal entries. Then V ∼= 〈a1, . . . , an〉 if V has an orthogonal basis v1, . . . , vn such thatQ(vi) = ai for all i.

Let v be an anisotropic vector in a quadratic space (V,B). Define a map τv : V −→ Vby

τv(x) = x− 2B(v, x)

Q(v)v.

It is easy to verify that τv is linear. It is called the symmetry with respect to v (or areflection with respect to the hyperplane v⊥).

Lemma 1.12 For each anisotropic vector v ∈ V , τv is an isometry of V and det(τv) = −1.

Proof. The first assertion can be checked directly. Since Q(v) is nonzero, the subspace Fvis nondegenerate and hence V = Fv ⊥ V ′ for some subspace V ′. Let v2, . . . , vn be a basisfor V ′. Then v, v2, . . . , vn is a basis for V . The matrix of τv with respect to this basis is thediagonal matrix 〈−1, 1, . . . , 〉. Therefore, det(τv) = −1.2

Lemma 1.13 Let (V,B) be a quadratic space. If x and y are anisotropic vectors of V withQ(x) = Q(y), then there is an isometry σ of V such that σ(x) = y.

Proof. Let u = (x + y)/2 and v = (x − y)/2. Then B(u, v) = 0 and Q(x) = Q(u) + Q(v).Either Q(u) or Q(v) is nonzero. In the first case, −τu(x) = y and in the second caseτv(x) = y.2

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Lemma 1.14 Let (V,B) and (V ′, B′) be two quadratic spaces. If σ : V −→ V ′ is anisometry and W is a subspace of V , then σ(W⊥) = σ(W )⊥.

Proof. Let x ∈W⊥. For any w ∈W ,

B′(σ(x), σ(w)) = B(x,w) = 0.

Therefore, σ(W⊥) ⊆ σ(W )⊥. The reverse inclusion can be proved similarly.2

Theorem 1.15 (Witt’s Cancellation Theorem) If V, V1, V2 are nondegenerate quadraticspaces such that

V1 ⊥ V ∼= V2 ⊥ V,

then V1∼= V2.

Proof. Since V is the orthogonal sum of 1-dimensional subspaces, it suffices to consider thecase where dim(V ) = 1; thus V = Fx. It follows from the hypothesis and Lemma 1.13 thatthere exists an isometry

Σ : V1 ⊥ Fx −→ V2 ⊥ Fx

such that Σ(x) = x. By Lemma 1.14, Σ(V1) = V2.2

Corollary 1.16 (Witt’s Extension Theorem) Let V and V ′ be isometric nondegeneratequadratic spaces. Suppose that W and W ′ are nondegenerate subspaces of V and V ′, respec-tively, and that σ : W −→ W ′ is an isometry. Then there exists an isometry Σ : V −→ V ′

which extends σ, that is, Σ|W = σ.

Proof. Let τ : V ′ −→ V be an isometry. Then by Proposition 1.8,

W ⊥W⊥ = V = τ(W ′) ⊥ τ(W ′)⊥.

Since W ∼= τ(W ′), it follows from Theorem 1.15 that there exists an isometry σ′ from W⊥

to τ(W ′)⊥. Then Σ = σ ⊥ (τ−1σ′) is the required isometry.2

1.4 Witt Index

Definition 1.17 A quadratic space V is said to be isotropic if it contains an isotropicvector. Otherwise, it is anisotropic. The space V is said to be totally isotropic if everynonzero vector in V is isotropic.

Theorem 1.18 Let (V,B) be a 2-dimensional nondegenerate quadratic space. The follow-ing conditions are equivalent:

(a) V is isotropic.

(b) V ∼=(

0 11 0

).

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(c) V ∼= 〈1,−1〉 ∼= 〈α,−α〉 for any α ∈ F×.

(d) For any α ∈ F× and any γ ∈ F , V ∼=(

0 αα γ

).

(e) −d(V ) = F×2.

Proof. Suppose that V is isotropic. Then V has an isotropic vector x, and V = Fx ⊕ Fyfor some y ∈ V . Let α = B(x, y) and γ = Q(y). Since V is nondegenerate, α ∈ F×. If we

let e = 1αx and f = − γ

2αx + y, then the symmetric matrix associated to e, f is

(0 11 0

).

This proves (a) =⇒ (b).Suppose (b) holds and the basis which yields the matrix is x, y. For any α ∈ F×,

Q(12x + αy) = α. So, the subspace spanned by u = 1

2x + αy is nondegenerate and henceV = Fu ⊥ Fv for some v ∈ V . Comparing the discriminants of both sides shows that vcan be chosen so that Q(v) = −α. This proves (b) =⇒ (c).

If (c) holds, then V has an isotropic vector, say x, which can be extended to a basisx, y of V . Note that B(x, y) must be nonzero because V is nondegenerate. Now, let

e =α

B(x, y)x, f =

γ −Q(y)

2B(x, y)x+ y.

Then e, f is a basis for M whose associated symmetric matrix is the one stated in (d).The implication (d) =⇒ (e) is trivial. We now show that (e) =⇒ (a). If x, y is an

orthogonal basis for V and 〈α1, α2〉 is the associated symmetric matrix, then α1α2 = −γ2

for some γ ∈ F×. One easily checks that the vector x+ γβ−12 y is isotropic.2

The preceding theorem implies that there is only one isometry class of nondegenerateisotropic 2-dimensional quadratic space. Any one of such space is called a hyperbolic planeand is denoted by H.

Definition 1.19 A quadratic space V is said to be universal if V represents all elementsin F .

Corollary 1.20 Let (V,B) be a nondegenerate isotropic quadratic space. Then V ∼= H ⊥W for some subspace W . In particular, every nondegenerate isotropic quadratic space isuniversal.

Proof. The second assertion is a consequence of Theorem 1.18. Let x be an isotropic vectorin V . Since V is nondegenerate, there exists y ∈ V such that B(x, y) = 1. The subspaceFx⊕ Fy is isometric to H.2

By the preceding corollary, every nondegenerate quadratic space V has an orthogonaldecomposition of the form

V ∼= H ⊥ · · · ⊥ H ⊥ V0 = Hm ⊥ V0

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where V0 is nondegenerate and anisotropic. The number m is well-defined, that is, it isindependent of the way we obtain the above orthogonal decomposition. For, suppose that

V ∼= Hk ⊥ V1

where V1 is nondegenerate and anisotropic. If k > m, by Witt’s Cancellation Theorem itfollows that

Hk−m ⊥ V1∼= V0,

which contradicts that V0 is anisotropic. Therefore, k = m, and V0∼= V1 by Witt’s Cancel-

lation Theorem once again. In summary, we have

Corollary 1.21 (Witt’s Decomposition) Every nondegenerate quadratic space V has anorthogonal decomposition

V = Hm ⊥ V0,

where V0 is nondegenerate and anisotropic. The integer m and the isometry class of V0 areuniquely determined by the isometry class of V .

The number m, which is the number of copies of H in the above decomposition, is calledthe Witt Index of V and is usually denoted by Ind(V ).

Definition 1.22 A quadratic space is called hyperbolic if it is an orthogonal sum of copiesof H.

Proposition 1.23 Let (V,B) be a nondegenerate quadratic space, and W be a totallyisotropic subspace of dimension k. Then W is contained in a hyperbolic subspace of Vof Witt index k.

Proof. Let e1, . . . , ek be a basis for W . Since V is nondegenerate, there exists f1 ∈ Vsuch that B(e1, f1) = 1 and B(f1, ei) = 0 for i = 2, . . . , k. The subspace H1 = Fe1 ⊕ Ff1

is isometric to H; hence V = H1 ⊥ V1 for some nondegenerate subspace V1. Moreover,e2, . . . , ek are vectors in V1 which span a totally isotropic subspace. An induction on thedimension of W will complete the proof of the proposition.2

Corollary 1.24 The dimension of a maximal totally isotropic subspace of a nondegeneratequadratic space V is equal to the Witt index of V .

At last, the following lemma is useful when deciding which element in F is representedby V .

Lemma 1.25 Let a be a nonzero element in F . If V is nondegenerate, then a is representedby V if and only if V ⊥ 〈−a〉 is isotropic.

Proof. Let Fe be a 1-dimensional quadratic space over F with Q(e) = −a. Suppose thata is represented by V . Then there exists x 6= 0 in V such that Q(x) = a. Therefore,Q(x+ e) = 0 which implies that V ⊥ 〈−a〉 is isotropic.

Conversely, suppose that V ⊥ Fe is isotropic. If V is isotropic, then we are done.Otherwise, there exist x ∈ V \ 0 and t ∈ F× such that Q(x + te) = 0. Then Q(t−1x) =−Q(e) = a. 2

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1.5 Quadratic spaces over C

The field of complex numbers C is algebraically closed. Therefore, every nonzero complexnumber is a square. If a is a nonzero complex number, then 〈a〉 ∼= 〈1〉 ∼= 〈−1〉 over C.

Let (V,B) be an n-dimensional nondegenerate quadratic space over C. Let r be thegreatest integer smaller than or equal to n/2. There are 2r pairwise orthogonal vectorsx1, . . . , xr, y1, . . . , yr of V such that

Q(xi) = 1, Q(yj) = −1, 1 ≤ i, j ≤ r.

By Theorem 1.18, the binary subspaces Cxi +Cyi are all isometric to H. This implies thatV has an orthogonal decomposition isometric to Hr ⊥ V0, where V0

∼= 〈1〉 or 0. There areseveral implications of this observation:

(i) All nondegenerate quadratic spaces over C of dimension ≥ 2 are isotropic.

(ii) Every nondegenerate quadratic space V has maximal Witt index, that is, Ind(V ) isalways bn/2c.

(iii) The dimension of V determines the isometry class of V .

1.6 Quadratic Spaces over R

The quotient group R×/R×2 has two elements which are represented by 1 and −1, respec-tively. Therefore, if V is a nondegenerate quadratic space, V can be decomposed as

V ∼= 〈1〉p ⊥ 〈−1〉q.

The integer p is called the positive index of V and is denoted by Ind+(V ). Similarly,q = Ind−(V ) is the negative index of V . The difference Ind+(V ) − Ind−(V ) is called thesignature of V . It is a consequence of the Witt’s Cancellation Theorem that both Ind+(V )and Ind−(V ), and hence the signature, depend only on the isometry class of V .

Theorem 1.26 (Sylvester’s Law of Inertia) Let V and W be two nondegenerate quadraticspaces over R. Then V ∼= W if and only if Ind+(W ) = Ind+(V ) and Ind−(W ) = Ind−(V ).

A quadratic space (V,Q) over R is said to be positive definite if Q(x) > 0 for allx 6= 0. A negative definite quadratic space over R is defined analogously. A nondegeneratequadratic space over R is called indefinite if it is neither positive definite nor negativedefinite. In particular, an nondegenerate indefinite quadratic space over R must be isotropicand universal.

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1.7 Quadratic Spaces over Finite Fields

Let F be a finite field of q elements. We always assume that q is odd. The number of squareclasses in F× is 2. Let ∆ be a fixed nonsquare element in F×.

Proposition 1.27 For n ≥ 2, every nondegenerate n-dimensional quadratic space over Fis universal.

Proof. It suffices to prove the proposition for a binary quadratic space (V,B) over F. Wemay assume that V ∼= 〈δ, ε〉 where δ, ε ∈ 1,∆. If V ∼= 〈1,∆〉, then we are done. IfV ∼= 〈∆,∆〉, the setQ(V ) is equal to ∆·Q(〈1, 1〉). Therefore, we may assume that V ∼= 〈1, 1〉.Our goal is to show that V represents ∆. If −1 is a square, then 〈1, 1〉 ∼= 〈1,−1〉 ∼= H whichis universal. Therefore, We may further assume that −1 is a nonsquare.

The sets F×2 and 1 + F×2 have the same number of elements. They are not equal since1 is not inside 1 + F×2. Therefore, there exists α ∈ F× such that 1 + α2 is not in F×2.This element 1 + α2 cannot be zero because −1 is not a square. Hence V represents anonsquare.2

Corollary 1.28 Let V be a nondegenerate quadratic space over F. Then

V ∼= 〈1〉 ⊥ · · · ⊥ 〈1〉 ⊥ 〈d(V )〉.

Proof. We have a decomposition V ∼= 〈1〉 ⊥ V0 whenever V is universal.2

Corollary 1.29 Every nondegenerate quadratic space of dimension ≥ 3 over F is isotropic.

Proof. Let (V,B) be a quadratic space of dimension ≥ 3 over F. We may assume thatdim(V ) = 3; hence V has a decomposition V ∼= 〈1, 1, d(V )〉. Since 〈1, d(V )〉 is universal, itrepresents −1. As a result, V contains a binary subspace which is isometric to 〈1,−1〉 ∼= H.Thus V is isotropic.2

Theorem 1.30 Let V and W be nondegenerate quadratic spaces over F. Then V ∼= W ifand only if d(V ) = d(W ) and dim(V ) = dim(W ).

Proof. If V ∼= W , then of course d(V ) is equal to d(W ) and dim(V ) = dim(W ). Theconverse is a consequence of Corollary 1.28.2

2 The p-adic Numbers

2.1 Valuations

Let F be a field whose characteristic is different from 2. A valuation on F is a function | |of F into R which satisfies:

V1 |a| > 0 if a 6= 0, |0| = 0;

12

V2 |ab| = |a||b|;

V3 |a+ b| ≤ |a|+ |b|,

for all a, b ∈ F . A function which satisfies V1, V2 and

V3’ |a+ b| ≤ max|a|, |b|

will satisfy V3 and therefore is a valuation. Axiom V3 is called the triangle inequality andV3’ is called the ultra triangle inequality. A valuation is called nonarchimedean if it satisfiesV3’; otherwise it is called archimedean.

There is always a valuation on F , namely the trivial valuation obtained by putting|a| = 1 for all a ∈ F×. Since this valuation has no significant interest, therefore we assumethat every valuation in the subsequent discussion is nontrivial.

Given a valuation | | on F , the distance function d(a, b) = |a− b| makes F into a metricspace. By a completion of F with respect to d (or | |) we mean a field F together with avaluation which satisfies

(i) F is complete with respect to the given valuation;

(ii) F is a subfield of F and the given valuation on F extends | |;

(iii) F is dense in F .

Theorem 2.1 A completion of F with respect to a valuation exists

Proof. We know from topology that as a metric space F has a completion, that is, there isa metric space F which is complete and contains F as a dense subset. Moreover, the metricon F induces the metric d on F . Without causing any confusion, we denote the metric onF also by d. We have to define addition and multiplication on F to make it become a field.Let α and β be two elements in F . They are the limits of two Cauchy sequences an andbn in F , respectively. It is obvious to see that both an+ bn and anbn are Cauchy andhence they converge to some elements in F . Define

α+ β = limn

(an + bn), αβ = limn

(anbn).

One can check that these definition are independent of the choices of an and bn. Takethe original 0 and 1 of F as the 0 and 1 of F . These all together make F into a field.

Finally define ‖α‖ = d(α, 0) for all α ∈ F . It is clear that ‖ ‖ extends the originalvaluation on F . Suppose that α is the limit of a Cauchy sequence an of F . By thetriangle inequality,

−d(an, α) ≤ d(an, 0)− d(α, 0) ≤ d(an, α).

Since |an| − ‖α‖ = d(an, 0) + d(α, 0), therefore

limn|an| = ‖α‖.

13

Hence, if bn −→ β, then

‖αβ‖ = limn|anbn| = lim

n|an| lim

n|bn| = ‖α‖‖β‖.

Similarly ‖α+β‖ ≤ ‖α‖+‖β‖. Therefore, ‖ ‖ is a valuation on F , and the metric associatedwith ‖ ‖ is d because

‖α− β‖ = limn|an − bn| = lim

nd(an, bn) = d(α, β).

2

Remark 2.2 Note that ‖ ‖ is nonarchimedean if | | is nonarchimedean.

From now on, by abuse of notation, the valuation on F that extends | | on F is alsodenoted by | |.

2.2 Nonarchimedean Valuations

Let | | be a nonarchimedean valuation on a field F . By Remark 2.2, the extension of | | toF is also nonarchimedean. Here are some consequences of the ultra triangle inequality V3’:

(1) (Principle of Domination) Let α1, . . . , αn ∈ F . If |αi| < |α1| for all i, then

|α1 + · · ·+ αn| = |α1|.

(2) Let αn be a sequence of elements of F . If∑αn converges, then of course αn −→ 0.

Conversely, suppose that αn −→ 0. Then for all M ≥ N ,

|αN + · · ·+ αM | ≤ max|αj | : N ≤ j ≤M.

Therefore, the partial sums of∑αn form a Cauchy sequence. Thus

∑αn converges

to some element in F .

(3) Let α ∈ F×. There exists a sequence an of elements in F which converges to α.Therefore, for all sufficiently large n, |an − α| < |α|. This shows that |an| = |α| forall sufficiently large n. This in particular shows that the two sets |F | and |F | are thesame.

(4) Let o be the subset of F containing all elements of F with valuation ≤ 1. The principleof domination implies that o is a subring of F . Let o be the closure of o in F . Itfollows from (3) o is a subring of F . It is called the valuation ring of F . Since F iscomplete and o is closed, o itself is a complete metric space.

(5) Let α ∈ o such that |α| = 1. Then α 6= 0 and |α−1| = 1 also. This means that α−1 isin o and hence α is a unit of o. This shows that o has only one maximal ideal

p = x ∈ o : |x| < 1.

Note that the group of units of o is precisely the set o \ p.

14

The valuation | | is called a discrete valuation if the image the composition

F× −→ R+ log−→ R

is an infinite cyclic subgroup of R. Let π ∈ o be a pull back of a generator of this cyclicgroup. Such an element is called a prime element of F . The ideal p is generated by π ando is a principal ideal domain. Every element in F× is of the form πnu for some n ∈ Z anda unit u of o. Let C be a complete set of representatives of cosets of p in o. We always pick0 to represent p.

Proposition 2.3 Suppose that | | is a discrete valuation on F . Let C be a complete set ofrepresentatives of p in o as described above. Then every element α ∈ F can be expresseduniquely by a Laurent series ∑

j≥ncjπ

j

where cj ∈ C for all j, |α| = |π|n and cn 6= 0.

Proof. We may assume that α 6= 0. Suppose that α = πnu for some unit u of o. Choosecn ∈ C such that u ≡ cn mod p. Then cn 6= 0 and

α = cnπn + α1

with |α1| ≤ |π|n+1. Next apply this procedure to α1 to obtain an α2, then to α2, and so on.From this procedure we obtain a sequence cn, cn+1, . . . of elements of C such that for eachm ≥ n, there exists βm+1 ∈ pm+1 and

α = cnπn + cn+1π

n+1 + · · ·+ cn+mπn+m + βm+1.

The partial sumcnπ

n + · · ·+ cn+mπn+m

clearly converges to α.Suppose that there are two Laurent series∑

j≥ncjπ

j =∑j≥n

djπj

with cj , dj ∈ C for all j but ci 6= di for some i ≥ j. Let i be the smallest index with thisproperty. Then ci − di 6∈ p which means that ci − di is a unit of o. Then

0 =

∣∣∣∣∣∣∑j≥n

cjπj −

∑j≥n

djπj

∣∣∣∣∣∣ = |π|i,

which is impossible.2

15

2.3 Valuations on Q

Our primary objects of investigation here are the valuations of Q. The usual absolute valueis an archimedean valuation on Q. We denote it by | |∞. The completion of Q with respectto | |∞ is the field of real numbers R.

Beside | |∞, Q has other valuations. Let p be a prime number. Any α ∈ Q× can bewritten as

α = pia

b

where a and b are integers prime to p. Put

|α|p =1

pi.

It is easy to show that this defines a discrete nonarchimedean valuation on Q. The comple-tion of Q with respect to | |p is the field of p-adic numbers Qp. Its valuation ring is the ringof p-adic integers Zp. The topology on Q or Qp induced by | |p is called the p-adic topology.In the p-adic topology, every pnZp is both open and closed.

Lemma 2.4 The maximal ideal of Zp is generated by p.

Let α be a p-adic integer. Since Zp is the closure of Z under the p-adic topology, thereforethere exists a ∈ Z such that α ≡ a mod pZp. Hence we can choose C = 0, 1, . . . , p − 1to be a complete set of representatives of Zp/pZp. In this way, every p-adic number can berepresented uniquely as a Laurent series in p

∞∑j=n

cjpj

with cj ∈ C.

Corollary 2.5 Zp/pZp is a finite field of p elements.

Proof. Let α ∈ Zp. Then α can be represented by a Taylor series in p

α =

∞∑j=0

cjpj

with cj ∈ C. The function α 7→ c0 mod p is clearly a surjective ring homomorphism fromZp onto Z/pZ with kernel pZp.2

Corollary 2.6 Zp is a compact topological space.

Proof. Let Oλ : λ ∈ Λ be an opening covering of Zp. Suppose that it has no finitesubcovering. Note that

Zp =

p−1⋃i=1

(i+ pZp).

16

Therefore, there exists c0 ∈ C = 0, 1, . . . , p−1 such that c0 +pZp is not covered by finitelymany of the Oλ. Similarly, there exists c1 ∈ C such that c0 + c1p + p2Zp is not finitelycovered. By continuing this process, we can construct a sequence c0, c1 . . . of elements of Csuch that for each j ≥ 0,

c0 + c1p+ · · ·+ cjpj + pj+1Zp

is not finitely covered. Let

α =

∞∑j=0

cjpj .

Then α ∈ Zp, and α must be in Oλ0 for some λ0 ∈ Λ. Since Oλ0 is open, there exists m ≥ 1such that α+ pmZp ⊆ Oλ0 . But then

c0 + c1p+ · · ·+ cm−1pm−1 + pmZp ⊆ Oλ0

which is a contradiction.2

Definition 2.7 Two valuations | | and ‖ ‖ on a field F are said to be equivalent if thereexists a nonzero real number k such that | |k = ‖ ‖.

Equivalent valuations on a field F define the same topology on F . It is not hard to seethat the absolute value | |∞ and the p-adic valuations | |p are inequivalent valuations on Q.

Theorem 2.8 (Ostrowski) Every nontrivial valuation of Q is equivalent to | |∞ or | |p forsome prime p.

Before giving the proof of Ostrowski’s theorem, we need a lemma to characterize thenonarchimedean valuations on Q.

Lemma 2.9 A valuation ‖ ‖ on Q is nonarchimedean if and only if it is bounded on Z.

Proof. Suppose that ‖ ‖ is nonarchimedean. Then for any positive integer n,

‖n‖ = ‖1 + · · ·+ 1‖ ≤ ‖1‖ = 1.

Therefore, ‖ ‖ is bounded on Z.Conversely, suppose that ‖m‖ < K for all m ∈ Z. Then for any positive integer k and

any rational numbers x, y,

‖x+ y‖k ≤ K(|x‖k + ‖x‖k−1‖y‖+ · · ·+ ‖y‖k) ≤ K(k + 1)max‖x‖, ‖y‖k.

Taking the k-th root on both sides and letting k −→∞, we see that ‖ ‖ is nonarchimedean.2

Proof of Ostrowski’s theorem. Let ‖ ‖ be a nonarchimedean valuation on Q. Then ‖n‖ ≤ 1for all n ∈ Z. So, there must be a prime number p such that ‖p‖ < 1 because, if not, theFundamental Theorem of Arithmetic would imply ‖x‖ = 1 for all x ∈ Q×.

17

The set a = a ∈ Z : ‖a‖ < 1 is an ideal of Z satisfying pZ ⊆ a 6= Z. Thus pZ = a. Ifa ∈ Z and a = pmb with gcd(p, b) = 1, then ‖b‖ = 1 and hence

‖a‖ = ‖p‖m = |a|sp

where s = − log ‖p‖/ log p. Consequently ‖ ‖ is equivalent to | |p.Now, suppose that ‖ ‖ is archimedean. Let m > 1 and n > 1 be two natural numbers.

Thenm = a0 + a1n+ · · ·+ arn

r

where ai ∈ 0, 1, . . . , n− 1 and nr ≤ m. Observe that r ≤ logm/ log n and ‖ai‖ ≤ ai‖1‖ ≤n. So, ‖m‖ < n(1 + ‖n‖+ · · ·+ ‖n‖r). If ‖n‖ < 1, then ‖m‖ < n/(1−‖n‖) and this is truefor all m ∈ Z. This contradicts that ‖ ‖ is archimedean. Therefore, ‖n‖ ≥ 1 and we obtainthe inequality

‖m‖ ≤n∑i=1

‖ai‖ · ‖n‖r ≤(

1 +logm

log n

)n · ‖n‖logm/ logn.

Substituting here mk for m, taking k-th roots on both sides, and letting k tend to ∞, onefinally obtains

‖m‖ ≤ ‖n‖logm/ logn, or ‖m‖1/ logm ≤ ‖n‖1/ logn.

Interchanging the roles of m and n gives

‖m‖1/ logm = ‖n‖1/ logn.

Putting c = ‖n‖1/ logn and pick s so that c = es. Then for any positive rational number x,

‖x‖ = es log x = |x|s.

Therefore ‖ ‖ is equivalent to | |∞.2

Proposition 2.10 (Product Formula) For any a ∈ Q×, we have

|a|∞∏p

|a|p = 1.

Proof. The proof is obvious. Note that |a|p = 1 for almost all p.2

2.4 Square Classes of Qp

Let p be a prime number. It is clear that every square class of Q×p must contain either aunit or a prime element, but not both.

18

Lemma 2.11 (Local Square Theorem)Let α ∈ Q×p and suppose that

|α− ε2|p ≤ |4p|pfor some unit ε in Zp. Then α is a square in Z×p .

Proof. The hypothesis implies that α is in Z×p . Write ε as ε0 and suppose that for anynon-negative integer k ≤ n, there is εk ∈ Z×p such that

α ≡ ε2k mod 4pk+1.

Let b ∈ Z such thatα− ε2n4pn+1

≡ bεn mod p.

Putting εn+1 = εn + 2bpn+1 gives

α− ε2n+1 = 4pn+1

(α− ε2n4pn+1

− bεn)− 4b2p2(n+1).

The choice of b implies α ≡ ε2n+1 mod 4pn+2. Thus we have proved the existence of asequence εn in Z×p such that α ≡ ε2n mod 4pn+1 and εn+1 ≡ εn mod pn+1 for every n ≥ 0.

Now, for any m > n,

|εm − εn|p = |(εm − εm−1) + · · ·+ (εn+1 − εn)|p≤ p−(n+1)

Therefore εn is a Cauchy sequence and it must converge to some x in Zp. As α ≡ ε2n+1

mod 4pn+2 for all n, we see that α = x2.2

Corollary 2.12 For any α ∈ Q×p , the set αQ×2p is an open subset of Q×p .

Proof. Let αy2 ∈ αQ×2p . For any x ∈ Q×p such that |x− αy2|p ≤ |4pαy2|p, we have∣∣∣∣ xαy2

− 1

∣∣∣∣p

≤ |4p|p.

Therefore x(αy2)−1 is a square and hence x ∈ αQ×2p .2

Corollary 2.13 Q×2 /Q×22 is of order 8 and generated by 2,−1 and 5. For p 6= 2, let ∆ be

a non-square unit in Zp. Then Q×p /Q×2p is of order 4 and generated by p and ∆.

Proof. The assertion for p 6= 2 is clear since (Z/pZ)× has only 2 square classes representedby 1 and ∆ respectively. Suppose α = x2 in Z×2 . Let x =

∑∞j=0 cj2

j with cj ∈ 0, 1 andc0 = 1. Then

α = (c0 + 2c1)2 + 2(c0 + 2c1)∞∑j=2

cj2j +

∞∑j=2

cj2j

2

≡ (c0 + 2c1)2 mod 8

≡ 1 mod 8.

19

Therefore, α ∈ Z×22 if and only if α ≡ 1 mod 8. Thus the square classes represented by the

units are generated by −1 and 5.2

In the following proposition, δ ∈ ∆, p∆, p if p > 2, and δ ∈ −1, 3, 5, 2,−2, 6, 10 ifp = 2.

Proposition 2.14 (a) If p > 2, then x2 − δy2 ∈ Zp if and only if x and y are in Zp.(b) If p = 2 and δ 6= 5, then x2 − δy2 ∈ Z2 if and only if x, y ∈ Z2.(c) If p = 2 and δ = 5, then x2 − 5y2 ∈ Z2 if and only if x = s/2, y = t/2, s, t ∈ Z2 ands− t ∈ 2Z2.

Proof. (a) We assume that x2−δy2 ∈ Zp. The statement is clear if either x or y is zero. Thuswe further assume that xy 6= 0. Consider the case when δ = ∆ first. Suppose |x|p ≥ |y|p.Since the space 〈1,−∆〉 is anisotropic over Zp/pZp ∼= Z/pZ, therefore 1− (y/x)2∆ is alwaysa unit in Zp. In other words, |x|p ≤ 1 and hence |y|p ≤ 1 also. Similar argument applies tothe case when |x|p ≤ |y|p. Now suppose δ = pε with ε = 1 or ∆. If |x|p ≥ |y|p, then

|x2 − y2pε|p = |x|2p∣∣∣1− (

y

x)2pε

∣∣∣p

= |x|2p.

Thus both x and y are in Zp. If, on the other hand, |y|p ≥ |x|p, then we consider

x2 − y2pε = y2

((x

y

)2

− pε

)and deduce that |(x/y)2 − pε|p is either 1 or |p|p. Therefore, y is in Zp and hence so is x.

For (b), we assume that x2 − y2δ ∈ Z2 and xy 6= 0. If 2 | δ, then the above argumentfor p > 2 can apply here. Suppose δ = −1. If y/x ∈ Z2, then (y/x)2 ≡ 0 mod 4 or 1 mod 8.So, 1 + (y/x)2 is a unit or it is congruent to 2 mod 8. The former implies that |x|22 ≤ 1 andthus x ∈ Z2. This implies y ∈ Z2 as well. If 1 + (y/x)2 ≡ 2 mod 8, then 2x2 ∈ Z2 whichimplies x ∈ Z2. Thus y ∈ Z2 also. Same argument works when x/y ∈ Z2. The case δ = 3can be done in a similar manner.

For (c), it is clear that if x = s/2 and y = t/2 with s, t ∈ Z×2 , then x2 − 5y2 ∈ Z2 andxy 6= 0. Conversely, suppose x2 − 5y2 ∈ Z2. If y/x ∈ Z2, then 1 − 5(y/x)2 is either a unitor ≡ 4 mod 8. The first option implies that x ∈ Z2 and hence y ∈ Z2 as well. The secondhappens only if y/x is a unit and x = s/2 with s ∈ Z2. So, y = t/2 with t ∈ Z2. Note thats− t ∈ 2Z2 whenever |s|2 = |t|2. The argument is similar when x/y ∈ Z2.2

2.5 Quadratic Reciprocity

Let p be an odd prime and Fp be a finite field of p elements. For any x ∈ F×p , define(x

p

):=

1 if x is a square,−1 otherwise.

It is called the Legendre symbol mod p. Its definition is often extended to integers that arerelatively prime to p. It is clear that the Legendre symbol is multiplicative in x.

20

Lemma 2.15 For any a ∈ F×p ,

ap−12 =

(a

p

)in Fp.

Proof. Let ξ be a generator of F×p . For any integer n, the order of ξ(p−1)n

2 is 1 if n is even

and 2 otherwise. Therefore ξ(p−1)n

2 is 1 if n is even and −1 otherwise. The lemma followsfrom the fact that a can be written as ξn for some n and a is a square if and only if n iseven.2

Lemma 2.16 For any odd prime p,(−1

p

)= (−1)

p−12 ,

(2

p

)= (−1)

p2−18 .

Proof. The first equality is clear. For the second one, let ξ be a primitive 8-th root of unityin the algebraic closure of Fp. Then ξ4 = −1 and ξ2 + ξ−2 = 0. Therefore, if y = ξ + ξ−1,then y2 = 2. Hence 2 is a square in Fp if and only if y ∈ Fp, or equivalently, yp = y. Thelast condition is satisfied if and only if ξp+1(ξp−1 − 1) = ξp−1 − 1, which is the same asξp−1 = 1 or ξp+1 = 1. Therefore, 2 is a square in Fp if and only if p ≡ ±1 mod 8. 2

Theorem 2.17 (Quadratic Reciprocity Law) Let q 6= p be two odd primes. Then(q

p

)(p

q

)= (−1)

p−12

q−12 .

Proof. For simplicity, we put χ(a) =(ap

)for any a ∈ F×p and χ(0) = 0. Since half of the

elements in F×p are squares and the other half consists of non-squares,∑

b∈Fp χ(b) = 0. Letx be a primitive p-th root of unity in the algebraic closure of Fq. So it makes sense to talkabout xa for any a ∈ Fp ∼= Z/pZ. Put

g :=∑a∈Fp

χ(a)xa.

Theng2 =

∑a,b∈Fp

χ(ab)xa+b =∑c∈Fp

xc∑a∈Fp

χ(a(c− a)).

21

If c 6= 0, then we have ∑a∈Fp

χ(a(c− a)) = χ(−1)∑a∈F×p

χ(a2 − ac)

= χ(−1)∑a∈F×p

χ(1− a−1c)

= χ(−1)∑a∈F×p

χ(1− a)

= χ(−1)∑b∈Fp

χ(b)− χ(−1)

= −χ(−1).

Thusg2 = −χ(−1)

∑c6=0

xc +∑a∈Fp

χ(−a2) = χ(−1) + (p− 1)χ(−1) = pχ(−1).

On the other hand, if we raise g to the q-th power, we get

gq =∑a∈Fp

χ(a)xqa = χ(q)∑a∈Fp

χ(qa)xqa = χ(q)g

and so gq−1 = χ(q). Therefore, in Fq,

χ(−1)q−12 p

q−12 =

(q

p

).

The left hand side is equal to (−1)q−12

p−12

(pq

)in Fq. Therefore

(−1)p−12

q−12 =

(q

p

)(p

q

)in Fq.

This can be seen as an equality in Z because both sides are ±1.2

2.6 Hilbert Symbols and Hilbert Reciprocity

In this subsection, let F be either R or Qp for some prime p. Let a, b be two nonzeroelements in F . The Hilbert symbol (a, b)F is defined as

(a, b)F =

1 if 〈a, b,−1〉 is isotropic over F ,−1 otherwise.

For simplicity, we will use (a, b) unless confusion arises. It is clear from the definition thatthe Hilbert symbol is a symmetric function defined on F×/F×2 × F×/F×2. Moreover,(a, b) = 1 whenever a or b is a square.

22

Lemma 2.18 For any a, b, c ∈ F×,

(a) (a, b) = (b, a), (a, b2) = 1.

(b) (a,−a) = (a, 1− a) = 1 if a 6= 1.

(c) (a, a) = (a,−1).

(d) (a, bc) = (a, b)(a, c).

Proof. Part (a) and (b) are obvious. For part (c), since a 6= 0, 〈a, a,−1〉 is isotropic if andonly if 〈−1,−1, a〉 is isotropic.

For part (d), we may assume that a is not a square. The case F = R can be verifieddirectly: (−1, d) = 1 if and only if d > 0.

When F = Qp, letNa = x2 − ay2 6= 0 : x, y ∈ F.

Note that Na is in fact a subgroup of F×. Observe that (d, a) = 1 if and only if d ∈ Na.Therefore, part (d) is valid if either b or c is in Na. It remains to show that bc ∈ Na ifb, c 6∈ Na. It suffices to show that [F× : Na] ≤ 2.

Suppose p > 2. Then we may assume that a ∈ ∆, p, p∆, where ∆ is a nonsquareunit in Z×p . If a = p or p∆, then −a ∈ Na, which implies [Na : F×2] ≥ 2 and thus[F× : Na] ≤ 2. If a = ∆, the space 〈∆,−1,−1〉 is isotropic over Zp/pZp but 〈∆,−1〉 isanisotropic. Therefore, there exist z ∈ Z×p and x, y ∈ Zp such that (∆x2 − y2) − z2 ≡ 0mod p. By Lemma 2.11, ∆x2−y2 itself must be a square and hence ∆x2−y2 = t2 for somet ∈ Zp. Therefore, ∆t2 ∈ Na and [Na : F×2] ≥ 2.

If p = 2, we have to show that [Na : F×2] ≥ 4. It suffices to produce 3 different squareclasses in Na. We may assume that a is coming from 3, 5,−1, 2, 6, 10,−2. If a is divisibleby 2, the numbers 1,−a, 1−a fall into different square classes in Na. When a is not divisibleby 2, then 1, 1− a, 1− 4a will do the job.2.

Note that even when a is a square, the assertion that (d, a) = 1 if and only if d ∈ Na isstill true.

For a ∈ Q×p , we can write a = pmu with m ∈ Z and u ∈ Z×p . The integer m is called theorder of a, denoted ordp(a).

Lemma 2.19 Let ∆ be a nonsquare unit in Zp if p > 2 and ∆ = 5 if p = 2. Then,

(∆, a)Qp =

1 if ordp(a) is even,−1 otherwise.

Proof. Obviously, we may assume that ordp(a) is 0 or 1. We first treat the case whenp > 2. Since the binary space 〈1,−∆〉 is anisotropic over Zp/pZp, therefore if x2−∆y2 ≡ 0mod p, both x and y are ≡ 0 mod p and hence ordp(x

2 −∆y2) ≥ 2. In this case, a 6∈ N∆

and so (a,∆)Qp = −1. On the other hand, suppose ordp(a) = 0. Since the space 〈1,−∆〉

23

is universal over Zp/pZp, there exist x, y ∈ Z×p such that x2 − ∆y2 ≡ a mod p. By theLocal Square Theorem, there exist z ∈ Z×p such that x2 − ∆y2 = az2, which implies that(a,∆)Qp = 1.

Let us consider the case p = 2. Suppose ord2(x2 − 5y2) = 1. If x, y ∈ Z2, then both ofthem must be units in Z2. So x2 − 5y2 ≡ 4 mod 8 which is a contradiction. This impliesthat x = s/2 and y = t/2 for some units s, t and s2− 5t2 ≡ 0 mod 8. But this is impossiblebecause s2 − 5t2 ≡ 4 mod 8 for any units x, y. Therefore, 2 6∈ N5 and so (5, 2)Q2 = −1.

If a ∈ Z×2 , we can always find x, y ∈ Z2 such that x2 − 5y2 ≡ a mod 8. By the LocalSquare Theorem again, we see that a ∈ N5 and so (5, a)Q2 = 1. 2

Corollary 2.20 If p > 2 and ε, δ ∈ Z×p , then (ε, δ)Qp = 1.

Corollary 2.21 If p > 2 and ε ∈ Z×p , then

(ε, p)Qp =

1 if ε is a square,−1 otherwise.

Consequently, (ε, p)Qp is the Legendre symbol(εp

), where ε is the class of ε mod pZp.

Theorem 2.22 (Hilbert Reciprocity Law) For any a, b ∈ Q×, we have

(a, b)R∏p

(a, b)Qp = 1.

Proof. Put

Fa(x) := (a, x)R∏p

(a, x)Qp .

Since (a, x)Qp = 1 for almost all p, therefore Fa(x) is well defined. Moreover, Fa(xy) =Fa(x)Fa(y), Fa(x) = Fx(a) and Fa(a) = Fa(−1). So it suffices to show that F−1(−1),F−1(2), F−1(p), F2(p), and Fp(q) are all equal to 1, where p, q are odd prime numbers.

Note that (a, b)R = −1 if and only if both a and b are negative. If a is a unit in Z2, thenx2 + y2 = a is solvable in Q2 if and only if a ≡ 1 mod 4. Therefore, (−1,−1)Q2 = −1 and

(−1, p)Q2 = (−1)p−12 . The equation x2−2y2 = p is solvable in Q2 if and only if p ≡ ±1 mod

8. Therefore (p, 2)Q2 = (−1)p2−1

8 and in particular (−1, 2)Q2 = 1. Together with Lemma2.16 and Corollary 2.20, these show that

F−1(−1) = F−1(2) = F−1(p) = F2(p) = 1.

For Fp(q) = 1, we have to show that (p, q)Q2 = 1 if and only if p or q is congruent to 1 mod4. We first assume that either p or q is congruent to 1 mod 4. Without loss of generality,suppose p ≡ 1 mod 4. We may further assume that p is either 1 or 5. Clearly, (1, q)Q2 = 1for all q and Lemma 2.19 shows that (5, q)Q2 = 1 as well.

24

Now suppose that (p, q)Q2 = 1. So there exist x, y ∈ Q2 such that x2−py2 = q. Assumethat p and q are ≡ 3 mod 4. Then x and y are in Z2. Modulo 4, we have

3 ≡ q ≡ x2 − py2 ≡ 1 mod 4

which is a contradiction. Thus either p or q is ≡ 1 mod 4.2

For our convenience, we extract the following from the proof above.

Corollary 2.23 If a and b are odd integers, then(a, b)Q2 = (−1)(a−1)

2(b−1)

2 ,

(a, 2)Q2 = (−1)a2−1

8 .

Our last remark is that if a ∈ Q×p which is not a square, then ( , a) : Q×p → ±1 is asurjective group homomorphism with kernel Na.

3 Quadratic Spaces over Qp

3.1 Hasse Invariants

Within this subsection, F is either R or Qp for some prime p. Let (V,B) be a nondegeneratequadratic space over F . Take an orthogonal basis for V and suppose

V ∼= 〈α1, · · · , αn〉

in this basis. We define the Hasse invariant of V , written as SF (V ) or simply S(V ) if F isclear from the context, to be ∏

1≤i≤j≤n(αi, αj),

where all the Hilbert symbols are over F .

Theorem 3.1 S(V ) is independent of the choice of the orthogonal basis.

Proof. If n = 1 and V ∼= 〈α〉, then (α, α) = (α,−1) = (d(V ),−1) which depends only on V .Suppose n = 2. A direct computation shows that∏

i≤j(αi, αj) = (d(V ),−1)(α1, α2).

Now, (α1, α2) = 1 if and only if 〈α1, α2,−1〉 ∼= V ⊥ 〈−1〉 is isotropic. This is certainlydependent on V , not on the basis chosen.

25

When n ≥ 3, we need a couple of definitions. Let vi and ui be two orthogonal basesfor V . We write vi ∼ ui if Fvi = Fuj for some i, j. We say that vi and ui are

linked if there exist orthogonal bases zji of V such that

vi = z1i ∼ z2

i ∼ · · · ∼ zmi = ui.

Claim: If dim(V ) ≥ 3, then every pair of orthogonal bases for V are linked.

Let us show how the theorem follows from the claim. Using the claim, we may assumethat

V ∼= 〈α1, . . . , αn〉 ∼= 〈β1, . . . , βn〉

and αh = βk for some h, k. It suffices to show that S(V ) is unchanged in the following twosituations:

(i) αk = β1 for some k 6= 1, and αi = βi for all i 6= 1, k. Note that we could then assumethat α1 = βk since α1 · · ·αn and β1 · · ·βn are in the same square class.

(ii) α1 = β1.

For a fixed h, ∏i≤j

(αi, αj) =∏h≤j

(αh, αj)∏i<h

(αi, αh)∏

i≤j;i,j 6=h(αi, αj)

= (αh, d(V ))∏

i≤j;i,j 6=h(αi, αj).

In case (i), we have∏i≤j

(αi, αj) = (α1, d(V ))∏

1<i≤j(αi, αj)

= (α1, d(V ))(αk, α1d(V ))∏

i≤j;i,j 6=1,k

(αi, αj)

= (α1αk, d(V ))(αk, α1)∏

i≤j;i,j 6=1,k

(αi, αj)

= (β1βk, d(V ))(βk, β1)∏

i≤j;i,j 6=1,k

(βi, βj)

=∏i≤j

(βi, βj)

In case (ii), if α1 = β1, then∏i≤j

(αi, αj) = (α1, d(V ))∏

1<i≤j(αi, αj) = (α1, d(V ))S(〈α2, . . . , αn〉).

26

Witt’s Cancellation Theorem implies that 〈α2, . . . , αn〉 is isometric to 〈β2, . . . , βn〉. There-fore, case (ii) can be completed by an induction on dim(V ).

Proof of the claim Write v1 = a1u1 + · · · + amum, where ai ∈ F and am 6= 0, and defineν(ui) = m. Observe that m = 1 means that vi and ui are linked. Therefore, all weneed to show is that if m > 1, then there exists another orthogonal basis u′i such thatu′i ∼ ui and ν(u′i) < m.

Suppose aj = 0 for some j < m. Define u′i as follows:

u′i =

ui if i < j,

ui+1 if j ≤ i < n,

uj if i = n.

Then u′i and ui are linked and ν(u′i) = m− 1. Thus we may assume that aj 6= 0 forall j < m. If m ≥ 3, then we may suppose Q(a1u1 + a2u2) = a2

1Q(u1) + a22Q(u2) 6= 0 since

one ofQ(a1u1 + a2u2), Q(a2u2 + a3u3), Q(a1u1 + a3u3)

is not equal to 0 (if they were all zero, thenQ(u1) = Q(u2) = Q(u3) = 0 which is impossible).Then we can extend a1u1 + a2u2, u3, . . . , um to an orthogonal basis for V that is linkedto ui, and the value of ν for this basis is m− 1.

Suppose m = 2. Then a1u1 + a2u2, a2Q(u2)u1 − a1Q(u1)u2, u3 . . . , is an orthogonalbasis whose value of ν is equal to 1, and hence it is linked to vi. But it is obvious thatthis basis is also linked to ui. 2.

Suppose thatV ∼= 〈α1, . . . , αn〉.

Lemma 2.18 implies that

S(V ) =∏

1≤i≤n(αi, α1 · · ·αi).

For a nontrivial decomposition V = U ⊥W , we obtain

S(V ) = S(U)S(W )(d(U), d(W )).

For any α ∈ F×, we let V α be the space (V,Qα) where Qα(x) = αQ(x) for all x ∈ V . Wecall V α the scaling of V by α. The discriminant of V α is αnd(V ). Also,

S(V α) = (α, (−1)n(n+1)/2d(V )n+1)S(V ).

3.2 Classification

In this subsection, let p be a prime. The Hilbert symbol over Qp is denoted by ( , ). If Vis a nondegenerate quadratic space over Qp, then the Hasse invariant of V is denoted byS(V ).

27

Proposition 3.2 A nondegenerate ternary space V over Qp is isotropic if and only ifS(V ) = (−1,−1).

Proof. After a suitable scaling, we may assume that V ∼= 〈a, b,−1〉. Therefore V is isotropicif and only if (a, b) = 1. However, S(V ) is equal to (a, b)(−1,−1). As a result, V is isotropicif and only if S(V ) = (−1,−1).2.

Corollary 3.3 Let a, b, c be units in Z×p with p > 2. Then 〈a, b, c〉 is isotropic.

Proof. The Hasse invariant of 〈a, b, c〉 is equal to 1 = (−1,−1).2

The above corollary is false if p = 2. For example, the ternary space 〈1, 1, 1〉 is anisotropicover Q2.

Proposition 3.4 A nondegenerate quaternary space V over Qp is isotropic if and only ifone of the following holds:

(a) d(V ) = 1 and S(V ) = (−1,−1);

(b) d(V ) 6= 1.

Proof. If V is isotropic, then V ∼= H ⊥ W . Suppose d(V ) = 1. Then d(W ) = −1 and thusW is also a hyperbolic plane. A computation shows that S(V ) = (−1,−1) in this case.

For the converse, we have to show that for an anisotropic quaternary space V , bothd(V ) = 1 and S(V ) = −(−1,−1) hold. Suppose V ∼= 〈a, b, c, d〉. Then both 〈1, ab, ac, ad〉and 〈1, cd, ad, bd〉 are anisotropic. Suppose that (x,−ab) = 1 for some x ∈ Q×p . Then〈x,−ab,−1〉 is isotropic or x is represented by the space 〈ab, 1〉. Hence x is not representedby 〈−ac,−ad〉 and so 〈x, ac, ad〉 is anisotropic. By Proposition 3.2, we see that

(x, xcd)(ac, cd)(ad, ad) = −(−1,−1).

The ternary space 〈1, ac, ad〉 is also anisotropic. Using Proposition 3.2 again we have(ac, cd)(ad, ad) = −(−1,−1) and hence (x,−cd) = 1. We thus show:

(x,−ab) = 1 =⇒ (x,−cd) = 1,(x,−cd) = 1 =⇒ (x,−ab) = 1.

Hence (x,−ab) = (x,−cd) for all x ∈ Q×p . This implies that d(V ) = abcd must be a square.Another consequence is that S(V α) = S(V ) for all α ∈ Q×p . Therefore

S(V ) = S(〈1, ab, ac, ad〉)= S(〈ab, ac, ad〉)= −(−1,−1)

since 〈ab, ac, ad〉 is anisotropic.2

28

Theorem 3.5 If the dimension of a quadratic space V over Qp is at least 5, then V isisotropic.

Proof. We may assume that the dimension of V is 5 and that V is nondegenerate. SupposeV ∼= 〈a, b, c, d, e〉, and yet V is anisotropic. Proposition 3.4 implies that the product ofany four different elements chosen from a, b, c, d, e is a square. This implies that V ∼=〈a, a, a, a, a〉 which is a scaling of 〈1, 1, 1, 1, 1〉. So we may assume that V ∼= 〈1, 1, 1, 1, 1〉. Ifp > 2, then V contains an isotropic subspace isometric to 〈1, 1, 1〉. If p = 2, then

12 + 12 + 12 + 22 +√−7

2= 0.

Therefore, V is isotropic. 2

Corollary 3.6 Every nondegenerate quaternary space over Qp is universal.

If two nondegenerate quadratic spaces over Qp are isometric, then their dimensions,discriminants, and Hasse invariants are the same. The main theorem in this subsection isthat these three invariants determine the isometry class of a quadratic space over Qp.

Theorem 3.7 Let V and W be two nondegenerate quadratic spaces over Qp. Then V ∼= Wif and only if dim(V ) = dim(W ), d(V ) = d(W ), and S(V ) = S(W ).

Proof. Let the dimension of V be n. If n = 1, then the theorem is trivial. If n = 2, we mayassume that d(V ) = d(W ) 6= −1; otherwise both V and W are isometric to the hyperbolicplane and hence V ∼= W . Let U be the quaternary space W ⊥ V −1. Then d(U) = 1 andS(U) = (−1,−1) by a routine calculation. Therefore, U is isotropic by Proposition 3.4.Since both V and W are anisotropic, we can find nonzero vectors v ∈ V , w ∈W such thatQ(v) = Q(w) = a. Thus

V ∼= 〈a, ad(V )〉 ∼= W.

Suppose now that n ≥ 3. Let U be the space W ⊥ V −1 whose dimension is at least6. Therefore, it is isotropic which implies that there is an a ∈ Q×ν which is represented byboth V and W . Hence

V = 〈a〉 ⊥ V1, W = 〈a〉 ⊥W1.

Direct computations show that d(V1) = d(W1) and S(V1) = S(W1). So, an inductionargument (on the dimension) would show that V1 and W1 are isometric. Therefore, V andW are isometric.2

Corollary 3.8 If V is an anisotropic quaternary space over Qp, then

V ∼= 〈1,−∆, p,−∆p〉.

Proof. From Proposition 3.4, we have d(V ) = 1 and S(V ) = −(−1,−1). These two con-ditions are satisfied by 〈1,−∆, p,−∆p〉. Theorem 3.7 implies that such a space must beunique up to isometry.2

29

4 Quadratic Spaces over Q

4.1 Strong Hasse Principle

Let Ω be the set of all prime numbers and the formal symbol ∞. We call an element in Ωa place of Q; a prime number is called a finite place and ∞ is the infinite place. Let V bea quadratic space. For each ν ∈ Ω, let Vν be the vector space Qν ⊗Q V over Qν , where weput Q∞ = R. We call Vν the local completion of V at ν. We identify V as a subset of Vνvia the embedding v 7→ 1⊗ v.

Let v1, . . . , vn be a basis for V . Define a quadratic map Qν on Vν by

Qν(n∑i=1

xivi) :=n∑

i,j=1

xixjB(vi, vj), xi ∈ Qν for all i,

where B is the bilinear form on V . It is clear that (Vν , Qν) becomes a quadratic space overQν . Since Q and Qν are essentially the same, we will drop the subscript and use the sameQ to denote the quadratic form on Vν . Note that if A is a symmetric matrix for V , then Ais also a symmetric matrix for Vν . We say that V is isotropic at ν if Vν is isotropic. TheHasse invariant of Vν over Qν will be written as Sν(V ). In particular, if V ∼= 〈a1, . . . , an〉,then

Sν(V ) =∏

1≤i≤j≤n(ai, aj)Qν .

Lemma 4.1 Let V be a nondegenerate quadratic space over Q. Then∏v∈Ω

Sν(V ) = 1.

Proof. This follows from the Hilbert Reciprocity Law.2

Theorem 4.2 (Strong Hasse Principle) A nondegenerate quadratic space V over Q isisotropic if and only Vν is isotropic for every ν ∈ Ω.

Proof. The only if part is clear. We prove the converse by an induction on n, the dimensionof V . There is nothing to verify when n = 1 since all one dimensional nondegenerate spaceare anisotropic.

Suppose that n = 2 and that V ∼= 〈a, b〉. Since Vν is isotropic, −ab is a square in Qν forevery ν ∈ Ω. Hence −ab is a square in Q and V is isotropic.

When n = 3, we scale V so that V represents 1. Hence we can suppose

V ∼= 〈1,−a,−b〉.

Moreover, by multiplying a and b by squares, we may assume that a and b are integers. Wehandle the problem by induction on |a| + |b|. Here | | is the usual absolute value on Q. If

30

|a|+ |b| = 2, then |a| = |b| = 1. Since V∞ is isotropic, therefore either a or b is positive. Asa result, V ∼= 〈1, 1,−1〉 or 〈1,−1,−1〉 and V is isotropic.

Suppose |a| + |b| > 2 and |a| ≤ |b|. Then we must have |b| ≥ 2. Moreover, by dividingsquares if necessary, we may assume that both a and b are squarefree integers. If a = 1,then V is obviously isotropic. So we can assume that a 6= 1. We claim that the congruencex2 ≡ a mod b is solvable with x ∈ Z. Up to ±, b is a product of distinct primes. Thereforeit suffices to prove the claim for every prime divisor p of b. We just need to look at those pwhich do not divide a. Moreover, we may solve the congruence over Zp instead of Z. SinceVp is isotropic, there exist x, y, z ∈ Zp, not all zero, such that x2 − ay2 − bz2 = 0. We mayassume that one of them is a unit in Zp. Since b 6≡ 0 mod p2, y must be in Z×p . Thena ≡ (xy−1)2 mod p and xy−1 ∈ Zp.

Now, there exists an integer c such that c2 = a+ bd for some integer d. We can furtherassume that |c| ≤ |b|/2. Note that d 6= 0 because a is not a square. Then

|d| = |c2 − a||b|

≤ |b|4

+ 1 < |b|

by virtue of |b| ≥ 2. Note that bd = c2 − a 12 which is a norm of the extension Q(√a)/Q.

Therefore, bd is also a norm of F (√a)/F whenever F is a field extension of Q. This means

that b is a norm of F (√a)/F if and only if d is.

By the assumption, 〈1,−a,−b〉 is isotropic over Qν and it is equivalent to saying thatb is a norm of Qν(

√a)/Qν , and hence d is a norm of Qν(

√a)/Qν . Apply the induction

hypothesis to 〈1,−a,−d〉, we find that d is a norm of Q(√a)/Q and hence b is also a norm

of Q(√a)/Q. This means that V is isotropic.

When n ≥ 4, putV = U ⊥W

where U = 〈a1, a2〉 and W = 〈a3, . . . , an〉. All the ai can be assumed to be integers. Forevery ν ∈ Ω, Vν is isotropic. Therefore, there exists bν ∈ Q×ν such that

(∗) bν ∈ Q(Uν) ∩ −Q(Wν).

Scaling V by −1 if necessary, we can assume that b∞ > 0. Write

bp = cppe(p), cp ∈ Z×p

at every finite place p. We need to invoke an analytic result in number theory:

Theorem 4.3 (Dirichlet’s Theorem on Primes in an Arithmetic Progression) Let m > 0and a be relatively prime integers. Then there exist infinite many primes p ≡ a mod m.

Let d be the product 2a1 · · · an and let

c =∏p|d

pe(p).

31

For every p | d, cp(cp−e(p))−1 is in Z×p . Therefore, there is a sufficiently large prime q such

thatq ≡ cp(cp−e(p))−1 mod p3,

for all p | d. If we set b = qc > 0, then b/bp ≡ 1 mod p3 for p | d. This implies thatb/bp ∈ Z×2

p for all p | d. By (∗),

(∗∗) b ∈ Q(Uν) ∩ −Q(Wν)

for all ν | d and for ν = ∞. If p is a prime which does not divide qd, then b and all the aiare in Z×p . Therefore, both U ⊥ 〈−b〉 and W ⊥ 〈b〉 are isotropic at that p. So (∗∗) holds atall p 6= q. We claim that (∗∗) also holds at q. The ternary space U ⊥ 〈−b〉 is isotropic atall p 6= q. Hence Sp(U ⊥ 〈−b〉) = (−1,−1)Qp at all p 6= q. It follows from Lemma 4.1 that

Sq(U ⊥ 〈−b〉) =∏v 6=q

(−1,−1)Qν = (−1,−1)Qq .

Thus U ⊥ 〈−b〉 is isotropic at q also. When n = 4, W ⊥ 〈b〉 is isotropic at q by the samereason. When n > 4, then W is already isotropic at q because the dimension of W is atleast 3 and each ai is in Z×q (note that q is odd by choice). Hence W ⊥ 〈b〉 is isotropic at q.

Since the Strong Hasse Principle is proved for ternary spaces, therefore U ⊥ 〈−b〉 isisotropic. By induction hypothesis, W ⊥ 〈b〉 is also isotropic. Therefore, b ∈ Q(U)∩−Q(W )and thus V = U ⊥W is isotropic.2

4.2 Weak Hasse Principle

Let a be any number in Q. Since V represents a if and only if V ⊥ 〈−a〉 is isotropic,therefore we have

Proposition 4.4 A nondegenerate quadratic space V over Q represents a ∈ Q if and onlyif Vν represents a for every ν ∈ Ω.

Suppose that U is another nondegenerate quadratic space over Q. A necessary conditionfor U ∼= V is that Uν ∼= Vν for every ν ∈ Ω. The converse also holds and we are going toprove it by induction on the dimension of U . The case where dim(U) = 1 can be deducedfrom the last proposition. Suppose now that dim(U) > 1. Let a be an nonzero element inQ(U). Then a ∈ Q(Uν) = Q(Vν) for all ν ∈ Ω; hence a is represented by V . So we have adecomposition

U = 〈a〉 ⊥ U ′, V = 〈a〉 ⊥ V ′.

It follows easily from Witt’s Theorem that U ′ν∼= V ′ν for all ν ∈ Ω, and V ′ ∼= U ′ by the

induction assumption. Thus we have shown the following:

Theorem 4.5 (Weak Hasse Principle) Let U and V be nondegenerate quadratic spaces overQ. Then U and V are isometric if and only if Uν and Vν are isometric for all places ν ofQ.

32

We have shown that the arithmetic of a quadratic space over Q (or its equivalent, aquadratic form over Q) is completely determined by the local behavior of the space at everyplace of Q. Using the earlier results proven for quadratic spaces over Qp and R, we obtainthe following complete set of isometry class invariants for a nondegenerate quadratic spaceV over Q:

(i) the dimension dim(V ),

(ii) the discriminant d(V ),

(iii) the Hasse invariants Sp(V ) at all primes p,

(iv) the positive index Ind+(V∞).

If V ∼= 〈a1, . . . , an〉 with a1, . . . , an ∈ Q×, then there exists a finite set of places S, whichcontains ∞ and 2, such that each ai is a unit in Zp for any p 6∈ S. Therefore, Sp(V ) = 1for each p 6∈ S. So in practice one has to check the Hasse invariants for only finitely manyplaces. This also shows that Vν is isotropic for almost all ν if dim(V ) ≥ 3.

There is a version of Theorem 4.5, which is often referred as the Hasse Principle, whichconcerns general representations of quadratic spaces. Let W and V be nondegeneratequadratic spaces over a field F . We say that W is represented by V if there is an linearmap σ : W → V such that Q(σ(w)) = Q(w) for all w ∈W . If dim(W ) = 1 and w is a basisvector for W with Q(w) = a , then W is represented by V if and only if a is represented byV .

Theorem 4.6 (Hasse Principle) Let W and V be nondegenerate quadratic spaces over Q.Then W is represented by V if and only if Wν is represented by Vν for all ν ∈ Ω.

Proof. We proceed by an induction on dim(W ). If dim(W ) = 1, this is just Proposition4.4. Suppose that dim(W ) ≥ 2. Let a be a nonzero rational number represented by W .Then a is represented by Vν for all ν ∈ Ω. Therefore, a is represented by V . Hence

W = 〈a〉 ⊥W ′, V = 〈a〉 ⊥ V ′.

By Witt’s Cancellation Theorem, W ′ν is represented by V ′ν for all ν ∈ Ω. By the inductionassumption, W ′ is represented by V ′ and we are done.2

5 Quadratic Forms over PID

In this section, R denotes a principal ideal domain (PID) and F is the field of fractions ofR. We keep the general assumption that the characteristic of F is not 2. In the subsequentsections, R is usually taken to be Z or Zp for some prime p. Unless stated otherwise, (V,Q)is a quadratic space over F .

33

5.1 Lattices on Quadratic Spaces

Definition 5.1 A subset L of V is called an R-lattice (or simply a lattice) in V if L isa finitely generated R-module. We say that L is a lattice on V if it is a lattice in V andFL = V .

Let L be a lattice on V . An element a ∈ F is said to be represented by L if there existsa vector v ∈ L such that Q(v) = a. To see the connection with number theory, let us fix abasis v1, . . . , vn for L. It is also a basis for V over F . Let f(x1, . . . , xn) be the quadraticform associated with this basis. Then a is represented by L if and only if the diophantineequation

f(x1, . . . , xn) = a

has a solution (x1, . . . , xn) ∈ Rn, not only in Fn. The fundamental question is still therepresentation problem which asks for an effective determination of Q(L), the set of elementsof F represented by L.

Two lattices M and L are isometric, written M ∼= L, if there exists an isometry σ :FM −→ FL such that σ(M) = L. It is clear that isometric lattices represent the same setof elements of F . Therefore, it is also important to determine, say by means of computableinvariants, whether or not two given lattices are isometric. For any lattice L on V , theisometry group of L is the set

O(L) = σ ∈ O(V ) : σ(L) = L.

Definition 5.2 Let L be a lattice on V . A sublattice N of L is called primitive if N is adirect summand of L.

So a sublattice N of a lattice L is primitive if and only if L/N is free. A vector x ∈ Lis said to be primitive in L if the rank one lattice Rx is a primitive sublattice of L.

Lemma 5.3 Let L be a lattice on V and x1, . . . , xn be a basis for L. Suppose that e is anonzero vector in L with e = a1x1 + · · · + anxn. Then e is primitive in L if and only ifgcd(a1, . . . , an) = 1.

Proof. If d := gcd(a1, . . . , an) is not 1, then d−1e is a vector in L and so L/Re is not free.In other words, e is not primitive in L.

Conversely, suppose that e is not primitive in L. Then L/Re is not free. So there existsx ∈ L, but not in Re, and n 6= 0 such that nx ∈ Re. Note that n is not a unit in R.Therefore nx = me for some m ∈ R and we may assume that gcd(m,n) = 1. Then

x =m

n(a1x1 + · · ·+ anxn)

which implies that n divides gcd(a1, . . . , an). In particular, gcd(a1, . . . , an) 6= 1.2

Definition 5.4 A fractional ideal of F is a subset of F of the form (a) = ar : r ∈ R forsome a ∈ F×.

34

Lemma 5.5 A subset I of F is a fractional ideal of F if and only if I is a nonzero R-moduleand λI ⊆ R for some nonzero λ ∈ F .

Proof. It is clear that every fractional ideal is a nonzero R-module. If I is the fractionalideal (a), a ∈ F×, then a−1I = R. Conversely, suppose that I ⊆ F is a nonzero R-module,and that λI ⊆ R for some λ ∈ F×. Since λI is also a nonzero R-module and λI ⊆ R, it isjust an ideal of R. Therefore, λI = (α) from some nonzero α ∈ R. Then I = (λ−1α).2

Let L be a lattice on V . For any x ∈ V , define the coefficient of x with respect to L tobe the set

ax = t ∈ F : tx ∈ L.

We claim that ax is a fractional ideal of F if x 6= 0. It is clear that ax is an R-module in F .Let u1, . . . , un be a basis for L and write x = a1u1 + · · ·+anun where ai ∈ F for all i. Thereexists a nonzero r ∈ R such that rai ∈ R for all i. Therefore, r ∈ ax which shows that ax isnonzero. For each i, write ai = αiβ

−1i where αi and βi are in R such that gcd(αi, βi) = 1.

Let g be the gcd of the αi and ` be the lcm of the βi. Suppose that t = αβ−1 ∈ ax withα, β ∈ R. So, tαiβ

−1i ∈ R for all i. Therefore, we must have

β | αi and βi | α for all i

and hence β | g and ` | α. If we set λ = g`−1, then λt ∈ R, that is, λax ⊆ R.

Lemma 5.6 Let L be a lattice on V and x be a nonzero vector in V . Then there existsa ∈ F× such that ax is primitive in L.

Proof. We know that ax = (a) for some a ∈ F×. In particular, ax ∈ L. Let y ∈ L andsuppose that there exists an nonzero t ∈ R such that ty ∈ Rax. Then ty = rax for somer ∈ R, and hence t−1rax ∈ L. This means that t−1ra ∈ ax and therefore t−1r ∈ R. As aresult, y = t−1rax ∈ Rax and so L/Rax is torsion free.2

We define the radical of a lattice L in V to be the sublattice

rad(L) = x ∈ L : B(x, L) = 0.

We call L nondegenerate if rad(L) = 0. It is easy to see that

F rad(L) = rad(FL)

andrad(L) = L ∩ rad(FL).

In particular, L is nondegenerate if and only if FL is nondegenerate.

Proposition 5.7 Let L be a lattice in V . Then there exists a nondegenerate lattice K suchthat L = K ⊥ rad(L).

Proof. It is clear that if K exists then it must be nondegenerate. It remains to demonstratethe existence of K. It suffices to show that rad(L) is a primitive sublattice of L, equivalently,

35

L/rad(L) is a torsion free R-module. Suppose that there exists v ∈ L and r ∈ R such thatrv ∈ rad(L). Then B(rv, L) = 0 which implies that r = 0 or B(v, L) = 0, that is,v ∈ rad(L).2

Remark 5.8 In the decomposition L = K ⊥ rad(L), the lattice K is not unique, but itsisometry class is uniquely determine by that of L. As a consequence, if L′ = K ′ ⊥ rad(L′)is another lattice, then L ∼= L′ if and only if K ∼= K ′ and rank(rad(L)) = rank(rad(L′)).

Let L be a nondegenerate lattice and v1, . . . , vk be a basis for L. If A is the symmetricmatrix (B(vi, vj)), we shall write

L ∼= A.

We call such an A a matrix for L. Since FL is a nondegenerate quadratic space, thedeterminant of the matrix A is nonzero. The discriminant of L, denoted d(L), is thecanonical image of det(A) in F×/R×2. Note that d(L) is well-defined because if A′ is thesymmetric matrix associated to another basis for L, then

A′ = T tAT

for some matrix T ∈ GLk(R); thus det(A′) = det(A) det(T )2 ∈ det(A)R×2. Quite often wealso consider d(L) as an element of F×. In this case, d(L) is determined by L up to themultiplication of an element in R×2.

Let M be a sublattice of L of the same rank. Let [L/M ] be the product of the invariantfactors of the quotient R-module L/M . It is an ideal of R. By abusing the notation, wealso use [L/M ] to denote one of its generators.

Proposition 5.9 Let L be a nondegenerate lattice and M be a sublattice of L of the samerank. Then d(M) = [L/M ]2d(L).

Proof. There exist a basis e1, . . . , ek for L and nonzero elements a1, . . . , ak of R such thata1e1, . . . , akek is a basis for M . Then [L/M ] = a1 · · · ak, and

d(M) = det(B(aiei, ajej))

= (a1 · · · ak)2 det(B(ei, ej))

= [L/M ]2d(L)

2

Remark 5.10 If R = Z or Zp, the group index [L : M ] is a generator of [L/M ]. In eithercase, we have d(M) = [L : M ]2d(L).

Consider a nondegenerate lattice L. The dual of L is the set

L# = x ∈ FL : B(x, L) ⊆ R.

Let v1, . . . , vn be a basis for L. As FL is nondegenerate, there exists a basis u1, . . . , unfor FL such that

(B(vi, uj)) = In.

36

It is easy to see that u1, . . . , un is a basis for L#; hence L# is a lattice on the space FLand (L#)# = L. Moreover, L# is isomorphic to Hom(L,R) through the map x 7→ B(x, ).Let T be the matrix defined by

(v1, . . . , vn) = (u1, . . . , un)T.

Then In = (B(ui, uj))T and (B(vi, vj)) = T . Therefore,

1 = d(L#) det(T ), det(T ) = d(L).

Thus we haved(L#) = d(L)−1.

If L has a decomposition L = J ⊥ K, then

L# = J# ⊥ K#.

Finally, for any nonzero a ∈ F ,(aL)# = a−1L#.

5.2 Modular Lattices

Consider a lattice L in V . The scale of L, denoted s(L), is the R-module generated by thesubset of B(L,L) of F . If e1, . . . , ek is a basis for L, then

s(L) ⊆∑i,j

B(ei, ej)R

So, either s(L) is either a fractional ideal or 0. We define the norm n(L) to be the R-modulegenerated by the subset Q(L) of F . Since Q(L) ⊆ B(L,L), it follows that n(L) is also afractional ideal or 0. Now, for all x, y ∈ L we have

2B(x, y) = Q(x+ y)−Q(x)−Q(y) ∈ n(L).

Hence2s(L) ⊆ n(L) ⊆ s(L).

If L = J ⊥ K, then it is easily verified that

s(L) = s(J) + s(K), n(L) = n(J) + n(K).

Let a be a nonzero element of F . Recall that V a denotes the vector space V providedwith a new quadratic map Qa such that Qa(x) = aQ(x) for all x ∈ V . We shall use La todenote the lattice L when it is regarded as a lattice in V a. It is easy to see that

s(La) = as(L), n(La) = an(L), d(La) = akd(L)

where k = rank(L).

37

Definition 5.11 A lattice L is called unimodular if s(L) ⊆ R and d(L) is a unit. It iscalled (a)-modular if La

−1is unimodular.

Lemma 5.12 Let L be a nondegenerate lattice in V . Then L is (a)-modular if and only ifaL# = L. In particular, L is unimodular if and only if L# = L.

Proof. First suppose that aL# = L. Then

B(L,L) = B(L, aL#) ⊆ (a),

and so s(L) ⊆ (a). Therefore, s(La−1

) ⊆ R. Furthermore, if rank(L) = k, then

d(L) = d(aL#) = a2kd(L#) = a2kd(L)−1.

Therefore, d(La−1

) = a−kd(L) is a unit of R.Now, assume that L is (a)-modular. Then La

−1is unimodular. Therefore, s(La

−1) ⊆ R

and d(La−1

) = a−kd(L) is a unit. For any v ∈ L, we have

a−1B(v, x) ⊆ R, for all x ∈ L.

So a−1v ∈ L# or v ∈ aL#; hence L ⊆ aL#. Now,

d(aL#) = a2kd(L#) = a2kd(L)−1 = d(L)2d(L)−1 = d(L).

Thus L = aL#.2

Corollary 5.13 Suppose that L is an (a)-modular lattice in V . Then

L = x ∈ FL : B(x, L) ⊆ (a).

Proof. This is clear because x ∈ aL# if and only if B(a−1x, L) ⊆ R, which is equivalent toB(x, L) ⊆ (a).2

Proposition 5.14 Let L be a nondegenerate lattice in V . If K is an (a)-modular sublatticeof L with s(L) = (a), then K is an orthogonal summand of L.

Proof. Since K is (a)-modular, it is nondegenerate. Therefore, there is an orthogonaldecomposition FL = FK ⊥W for some nondegenerate subspace W of FL. We claim that

L = K ⊥ (L ∩W ).

It suffices to show that L = K + (L ∩W ). Let x ∈ L. Write x = y + z, where y ∈ FK andz ∈W . Then

B(y,K) = B(x,K) ⊆ B(L,K) ⊆ s(L) = (a).

But y is in FK; hence y ∈ K.2

38

6 Quadratic Forms over ZpIn this section, (V,Q) is always a quadratic space over Qp for some prime p. Every latticeconsidered here is a Zp-lattice.

6.1 Classification in Terms of Jordan Decompositions

Lemma 6.1 (a) If p > 2, every nondegenerate Zp-lattice has an orthogonal basis.(b) Every nondegenerate Z2-lattice is an orthogonal sum of modular sublattices of rank 1 or2.

Proof. Let L be a nondegenerate lattice on V . If s(L) = n(L), then there exists a vectorv ∈ L such that (Q(v)) = s(L) and by Proposition 5.14 it follows that Zpv is an orthogonalsummand of L. This proves (a).

If p = 2 and n(L) = 2s(L), then there exist u, v ∈ L such that Q(v), Q(u) ∈ 2s(L)and (B(u, v)) = s(L). The binary submodule Z2u + Z2v is s(L)-modular and therefore isan orthogonal summand of L. Thus repeating this argument, L is an orthogonal sum ofmodular sublattices of rank 1 or 2.2

Let L be a nondegenerate lattice on V . We can write

L = L1 ⊥ · · · ⊥ Lt,

where Li is s(Li)-modular and s(L1) ) · · · ) s(Lt). Such an orthogonal decomposition iscalled a Jordan decomposition of L, and each Li is a Jordan component. The first componentL1 is called the leading component of that Jordan decomposition. In general, a Jordandecomposition is not unique.

For any fractional ideal a of Qp, let

La = x ∈ L : B(x, L) ⊆ a.

It is clear that La is a lattice in V .

Lemma 6.2 Suppose that M is a b-modular lattice and a 6= b. Then s(Ma) a.

Proof. It suffices to prove the case when a b. By scaling the bilinear form, we may assumethat M is unimodular. Consequently, a Zp. In particular, a = (pa) with a ≥ 1. Let

M = M1 ⊥ · · · ⊥M` ⊥M`+1 ⊥ · · · ⊥Mk

be an orthogonal decomposition of M such that rank(Mj) is 1 for all j ≤ ` and 2 for allj > `. Note that ` < n happens only when p = 2. Let x ∈Ma and write x = x1 + · · ·+ xkwhere xj ∈Mj for all j.

Suppose that j ≤ `. Let zj be a basis vector for Mj . Then Q(zj) ∈ Z×p . FromB(xj , zj) ⊆ B(x,Mj) ⊆ B(x,M) ⊆ a we see that xj ∈ paMj .

39

If p = 2 and j > `, let ej , fj be a basis for Mj such that Q(ej), Q(fj) ∈ 2Z2 andB(ej , fj) = γ ∈ Z×2 . Write xj = αej + βfj where α, β ∈ Z2. Then

B(xj , ej) = αQ(ej) + βγ ∈ a (1)B(xj , fj) = βQ(fj) + αγ ∈ a (2)

If α 6∈ a, then from (2) |α|2 = |βQ(fi)|2. This implies that |αQ(ej)|2 = |βQ(fj)Q(ej)|2 <|β|2, and by (1) we have β ∈ a. But then (2) says that αmust be in a which is a contradiction.Therefore, α ∈ a and from (1) it follows that β is also in a.

Finally, the above shows that Ma ⊆ paM . Therefore, s(Ma) ⊆ p2as(M) a.2

Lemma 6.3 If L = L1 ⊥ · · · ⊥ Li ⊥ · · · ⊥ Lt is a Jordan decomposition and s(Li) = a,then Li is the leading component of a Jordan decomposition of La.

Proof. For any j, Laj is a subset of La. If x ∈ La and x = x1 + · · ·+ xt with xj ∈ Lj for all

j, thenB(xj , Lj) = B(x, Lj) ⊆ a.

Therefore, xj ∈ Laj and hence La = La

1 ⊥ · · · ⊥ Lat . Lemma 6.2 implies that s(La

j) a forall j 6= i. Therefore, La = Li ⊥M where s(M) s(Li). This proves the lemma.2

Theorem 6.4 Suppose p > 2. If L = L1 ⊥ · · · ⊥ Lt = K1 ⊥ · · · ⊥ Ks are two Jordandecompositions of L, then t = s and Li ∼= Ki for all i.

Proof. Suppose s(Li) = a. Then La is not empty and hence there must be one and onlyone j such that s(Kj) = a. Similarly, for any Ki, there is a unique L` with s(L`) = s(Ki).Therefore, t = s and s(Li) = s(Ki) for 1 ≤ i ≤ t.

Since Li and Ki are the leading components of two Jordan decompositions of the latticeLa, we may assume that i = 1 and s(L1) = s(K1) = Zp by scaling the quadratic form.The quotient L/pL becomes a quadratic space over the finite field F = Zp/pZp. While L1

and K1 induce nondegenerate subspaces of L/pL, both L2 ⊥ · · · ⊥ Lt and K2 ⊥ · · · ⊥ Kt

become the radical of L/pL. Thus L1/pL1∼= K1/pK1 as quadratic spaces over F, and

therefore rank(L1) = rank(K1) and d(L1) = d(K1) in F×/F×2. Since p > 2, d(L1) = d(K1)in Z×p /Z×2

p . Then theorem follows from Corollary 6.20.2

The above theorem does not hold when p = 2. For example, consider the Z2-lattice Lwith Jordan decomposition

L ∼=(

2 11 2

)⊥ 〈−2〉.

Suppose that x, y, z is a basis which gives the symmetry matrix on the right. Then thevectors x + z, y + z span a sublattices which is isometric to H. Therefore, L has anotherJordan decomposition

L ∼=(

0 11 0

)⊥ 〈6〉.

Obviously, the leading components of the two Jordan decompositions are not isometric.

40

Theorem 6.5 Let L = L1 ⊥ · · · ⊥ Lt = K1 ⊥ · · · ⊥ Ks be two Jordan decompositions of aZ2-lattice L. Then

(a) t = s;

(b) s(Li) = s(Ki), rank(Li) = rank(Ki) for 1 ≤ i ≤ t;

(c) n(Li) = n(Ki) for 1 ≤ i ≤ t;

(d) n(L1) = n(L) and s(L1) = s(L).

Suppose that L = L1 ⊥ · · · ⊥ Lt is a Jordan decomposition of a Z2-lattice L. For1 ≤ i ≤ t, write si = s(Li) and ni = n(Li), and set ui = ord2(n(Li)). The lattice L now hasthe following set of invariants:

t, rank(Li), si, ui, 1 ≤ i ≤ t.

They are called the Jordan invariants of L. For each i, let L(i) = L1 ⊥ · · · ⊥ Li.

Theorem 6.6 Let L and K be lattices on a nondegenerate quadratic space over Q2. Sup-pose that L and K have the same Jordan invariants. Let t be the number of Jordan compo-nents in a Jordan decomposition of L. Then L ∼= K if and only if the following conditionshold for 1 ≤ i ≤ t− 1:

(i) d(L(i))/d(K(i)) ≡ 1 mod nini+1/s2i ,

(ii) Q2K(i) ⊥ 〈2ui〉 has a subspace which is isometric to Q2L(i) when ni+1 ⊆ 4ni.

6.2 Maximal Lattices

Definition 6.7 Let a be a nonzero element in Qp. A lattice L on a nondegenerate spaceV over Qp is called (a)-maximal if n(L) ⊆ (a) and for any lattice M containing L properly,n(M) * (a) holds.

Suppose that L is (a)-maximal but n(L) 6= (a). If n(L) ⊆ (p2a), then p−1L properlycontains L and n(p−1L) ⊆ (a). This contradicts the maximality of L. Therefore, for any(a)-maximal lattice L, n(L) is either (a) or (pa).

Lemma 6.8 Let L be a lattice on a nondegenerate quadratic space V of dimension n overQp. If n(L) ⊆ (a) and ((2a−1)nd(L)) = Zp or pZp, then L is (a)-maximal.

Proof. Let M be a lattice on V which contains L and n(M) ⊆ (a). Then

(d(M)) ⊆ s(M)n ⊆ (2−1n(M))n ⊆ (a/2)n.

However, d(L) = [M : L]2d(M) which implies that

((2a−1)nd(L)) = ([M : L]2(2a−1)nd(M)) ⊆ [M : L]2Zp.

Therefore, [M : L] = 1 and hence M = L.2

41

Corollary 6.9 If L is (a)-modular with n(L) = (2a), then L is (2a)-maximal. In particular,if p > 2 then every (a)-modular lattice is (a)-maximal.

Over Z2, there are unimodular lattices which are not Z2-maximal. For example, theZ2-lattice N ∼= 〈1, 3〉, with respect to a basis e, f, is unimodular. However it is notZ2-maximal because it is a sublattice of M = Z2

12(e+ f) + Z2

12(e− f) and n(M) = Z2.

Proposition 6.10 Let V be an anisotropic quadratic space over Qp. A lattice L on V is(a)-maximal if and only if

L = x ∈ V : Q(x) ∈ (a).

In particular, there is one and only one (a)-maximal lattice on V .

Proof. We first claim that the set M = x ∈ V : Q(x) ∈ (a) is a lattice. It is clear that Mis closed under multiplication by Zp. Let x, y ∈M . Assume that 2B(x, y) 6∈ (a). Let t ≥ 1be an integer such that (2ptB(x, y)) = (a). Then

Q(x)Q(y)B(x, y)−2 ∈ (4p2t)

and, since we also have Q(x + y) = Q(x) + Q(y) + 2B(x, y), 1 − Q(x)Q(y)B(x, y)−2 is asquare of some element c ∈ Z×p . This implies that

Q(x)Q(y)−B(x, y)2 = −(cB(x, y))2 6= 0,

and, as a result, Qpx⊕Qpy is a hyperbolic plane which contradicts the assumption on V .Next we show that M is finitely generated. Fix a lattice K on V with n(K) ⊆ (a). Then

K must be a subset of M . Let N be a finitely generated Zp-module such that K ⊆ N ⊆M ,and let x1, . . . , xn be a basis for N . Since xi ∈ M for all i, the diagonal entries and theoff diagonal entries of the matrix (B(xi, xj)) are in (a) and (a2 ) respectively. In particular,det(B(xi, xj)) ∈ (a2 )n and hence

Zp ⊇ ([N : K]2) = (d(K) det(B(xi, xj))−1) ⊇ (d(K))

(a2

)−n.

This implies that [N : K] is bounded, and M must be finitely generated.It is clear by now that M is an (a)-maximal lattice. If L is an (a)-maximal lattice on

V , then L ⊆M . But n(M) is clearly contained in (a). Therefore, L = M . 2

Lemma 6.11 Let L be a lattice on the hyperbolic plane. The following three conditions areequivalent:

(i) L ∼=(

0 cc 0

);

(ii) L is (c)-modular and n(L) ⊆ (2c);

(iii) L is (2c)-maximal.

42

Proof. (i) =⇒ (ii) is obvious, and (ii) =⇒ (iii) follows from Corollary 6.9 by putting a = 2c.For (iii) =⇒ (i), let x, y be a basis for L such that Q(x) = 0. Write B(x, y) = a andQ(y) = b for some a, b ∈ Qp. Then

a ∈ s(L) ⊆ 2−1n(L) ⊆ (c), and b ∈ n(L).

We claim that (a) = (c). If not, then a ∈ (cp) and for any α, β ∈ Zp,

Q(αp−1x+ βy) = 2αβp−1a+ β2b ∈ (2c).

Therefore, the lattice Zpp−1x+ Zpy contains L properly and its norm is contained in (2c).This contradicts the maximality of L.

Now, write c = aε for some ε ∈ Z×p . Then εx,−b(2a)−1x + y is a basis for L whose

associated symmetric matrix is

(0 cc 0

).2

Theorem 6.12 Let V be a nondegenerate isotropic quadratic space over Qp, and let L andK be (a)-maximal and (b)-maximal, respectively, lattices on V . Then there exists a splittingV = U ⊥W in which U is a hyperbolic plane and

L = (L ∩ U) ⊥ (L ∩W ), K = (K ∩ U) ⊥ (K ∩W ).

Proof. For any isotropic vector x of V , let ax be the coefficient of x with respect to L, thatis,

ax = t ∈ Qp : tx ∈ L.

It has been shown that ax is a fractional ideal. We let bx be the coefficient of x with respectto K. Now αK ⊆ L for some nonzero α in Qp. Therefore, rx := bx/ax is inside (α−1); hencewe can pick an isotropic vector x in V for which rx is maximal among all the coefficients ofthe isotropic vectors in V . For our convenience, we let ` and k be generators of ax and bxrespectively. In particular, `x and kx are basis vectors of L and K respectively.

Note thatB(`x, L) ⊆ s(L) ⊆ 2−1n(L) ⊆ (a/2).

We claim that B(`x, L) = (a/2). If not, then B(`x, L) ⊆ (pa/2). So, for any v ∈ L,

Q(v + p−1`x) = Q(v) + 2p−1B(v, `x) ⊆ (a).

This implies that the norm of the lattice L + Zpp−1`x is contained in (a). But L $ L +Zpp−1`x and this contradicts the maximality of L.

Now, let w ∈ L such that B(`x, w) = a/2. Put y = w − a−1Q(w)`x ∈ L. Then

Zp`x+ Zpy ∼=(

0 a/2a/2 0

).

43

So, Zp`x + Zpy is an (a/2)-modular lattice on a hyperbolic plane. Since s(L) ⊆ (a/2),Zp`x+ Zpy must be an orthogonal summand of L.

Let k′ be a generator of by. Then Zpkx+ Zpk′y ⊆ K and so kk′/` ∈ (b/a), that is,

rxry ⊆ (b/a).

As is done in the last paragraph, there exists isotropic vector z ∈ K such that Zpz + Zpk′ysplits K and B(z, k′y) = b/2. Let `′ be a generator of az. Then Zp`′z + Zpy ⊆ L and so`′b/k′ ∈ (a), that is,

(b/a) ⊆ rzry.

But rx is chosen to be maximal. Therefore,

rxry = (b/a)

which implies that the scale of Zpkx+ Zpk′y is precisely (b/2). Hence Zpkx+ Zpk′y splitsK. Thus U = Qpx+Qpy gives the desired splitting of V .2

Corollary 6.13 Let L be an (a)-maximal lattice on a nondegenerate quadratic space Vover Qp. Then

L ∼=(

0 a/2a/2 0

)m⊥ L0

where m is the Witt index of V and L0 is anisotropic. Moreover, the isometry class of L0

is determined by that of V . As a result, all (a)-maximal lattices on V are isometric.

Proof. By Theorem 6.12, there is a splitting V = U1 ⊥ · · · ⊥ Um ⊥ V0, in which U1, . . . , Umare hyperbolic planes and V0 is anisotropic, such that

L = (L ∩ U1) ⊥ · · · ⊥ (L ∩ Um) ⊥ (L ∩ V0).

It is clear that each L ∩ Ui is isometric to(

0 a/2a/2 0

). The isometry class of V0 is uniquely

determined by that of V and L∩V0 is the only (a)-maximal lattice on V0. Thus the isometryclass of L ∩ V0 is uniquely determined by that of V .2

Corollary 6.14 Let L and K be (a)-maximal lattices on a nondegenerate quadratic spaceover Qp. Then [L : L ∩K] = [K : L ∩K].

Proof. By Theorem 6.12, there are hyperbolic pairs xi, yi, 1 ≤ i ≤ m, and anisotropic latticeN such that

L =

(m∑i=1

Zpxi + Zpyi

)⊥ N, K =

(m∑i=1

Zpaixi + Zpa−1i yi

)⊥ N,

where a1, . . . , am are nonzero elements in Zp. Then L ∩ K = (∑m

i=1 Zpaixi + Zpyi) ⊥ Nand the corollary now follows immediately.2

44

Lemma 6.15 Let N be a lattice in a nondegenerate quadratic space V over Qp . If n(N) ⊆(a), then there exists an (a)-maximal lattice on V which contains N .

Proof. Let v1, . . . , vm be a basis for N . Extend it to a basis v1, . . . , vn for V (m ≤ n).Let

M = N ⊕n∑

i=m+1

Zpptvi

for some integer t. If t is large enough, then n(M) ⊆ (a). Let L be a lattice on V whichcontains M and n(L) ⊆ (a). Then d(M) = [L : M ]2d(L) and d(L) ∈ s(L)n ⊆ (2−1a)n.Since [L : M ] is an integer, it must be bounded and hence there exists a lattice K which ismaximal with respect to inclusion such that K ⊇ M and n(K) ⊆ (a). Such K must be an(a)-maximal lattice on V .2

Theorem 6.16 Let L be an (a)-maximal lattice on a nondegenerate quadratic space V overQp. For any b ∈ Q×p , b is represented by L if b is represented by V and b ∈ (a).

Proof. Let v be a vector in V such that Q(v) = b. Then the norm of the rank 1 latticeZpv is (b) ⊆ (a). Using the previous lemma we find an (a)-maximal lattice K on V whichcontains Zpv. By Theorem 6.12, K is isometric to L and therefore there exists w ∈ L withQ(w) = b.2

Corollary 6.17 Let L be an (a)-maximal lattice on a nondegenerate quadratic space V ofdimension ≥ 4 over Qp. Then L represents every element in (a).

Proof. Since dim(V ) ≥ 4, V is universal and hence it represents all elements in (a). ByTheorem 6.16, L represents every element in (a).2

6.3 Modular Lattices over Zp, p > 2

In this subsection we shall present a solution to the classification problem of modular latticesover Zp when p > 2. Together with Theorem 6.4 they provide a solution to the classificationof arbitrary Zp-lattices when p is odd. By Corollary 6.9, all (a)-modular lattices are (a)-maximal. Therefore, if L is an (a)-modular lattice on a nondegenerate quadratic space overQp, then

L ∼=(

0 aa 0

)m⊥ L0

where m is the Witt index of the ambient space and L0 is anisotropic.

Corollary 6.18 If p > 2 and L is an (a)-modular lattice of rank ≥ 3, then

(0 aa 0

)is an

orthogonal summand of L. In particular, L represents all elements in (a).

Proof. Under the assumption on the rank of L, the ambient quadratic space is isotropic.2

45

Lemma 6.19 If p > 2 and L is a unimodular lattice of rank 2, then L represents everyunit of Zp.

Proof. Let ∆ be a nonsquare unit of Zp. By Lemma 6.1, L has an orthogonal basis; henceL is isometric to one of the following binary lattices:

〈1, 1〉, 〈∆,∆〉, 〈1,∆〉.

It is clear that the last lattice represents all units in Zp. The first two are isometric becausethey are Zp-maximal lattices on isometric quadratic spaces over Qp; see Theorem 6.12.Therefore, it is enough to consider the case where L ∼= 〈1, 1〉. Then L is a Zp-maximallattice on the space 〈1, 1〉, and this space represents all units of Zp. By Theorem 6.16, Lrepresents all units of Zp.2

Corollary 6.20 If p > 2, the isometry class of an (a)-modular lattice L is determined byrank(L) and d(L). In particular, if L is unimodular, then

L ∼= 〈1〉 ⊥ · · · ⊥ 〈1〉 ⊥ 〈d(L)〉.

Proof. We may assume that L is unimodular. By Corollary 6.18 and Lemma 6.19, Lrepresents 1 if rank(L) ≥ 2. Therefore, L ∼= 〈1〉 ⊥ L′ for some unimodular lattice L′. Aninduction on the rank of L will complete the proof. 2

6.4 Modular Lattices over Z2

If L is an (a)-modular Z2-lattice, then s(L) = (a) and n(L) is either (a) or (2a). We saythat L is improper if n(L) = (2a); proper otherwise. If L is an improper Z2-modular lattice,then n(L) = (2) and we call L even unimodular. On the hyperbolic plane H, there is a

unique unimodular lattice isometric to

(0 11 0

)which we also denote by H.

Lemma 6.21 Let L be a binary even unimodular Z2-lattice on a space V .

(i) If L is isotropic, then L ∼= H, d(L) = −1, S(V ) = −1 and Q(L) = (2).

(ii) If L is anisotropic, then L ∼=(

2 11 2

), d(L) = 3, S(V ) = 1 and Q(L) = x ∈ Z2 :

ord2(x) ≡ 1 mod 2 ∪ 0.

Proof. The lattice L is (2)-maximal by virtue of Corollary 6.9. The case where L is isotropiccan be deduced from Lemma 6.11. Suppose that L is anisotropic. Let x be a basis vectorof L. Since L is unimodular, there exists y ∈ L = L# such that B(x, y) = 1. Then x, ymust be a basis for L and the corresponding symmetric matrix for L is(

2a 11 2b

).

46

If a or b is divisible by 2, then d(L) = 4ab − 1 ≡ −1 mod 8. So L is isotropic which is acontradiction. As a result, both a and b are units and d(L) = 3. Moreover, V ∼= 〈2a, 6a〉and a direct computation shows that S(V ) = 1. By the classification of quadratic spacesover Q2, we see that

V ∼=(

2 11 2

),

which has a unique (2)-maximal lattice isometric to

(2 11 2

). Therefore, L ∼=

(2 11 2

).

For any x ∈ Q×2 , x ∈ Q(L) if and only if x ∈ Q(V ) and x ∈ (2) by Theorem 6.16.However, x ∈ Q(V ) if and only if V ⊥ 〈−x〉 is isotropic, which is equivalent to requiringthat its Hasse invariant is equal to (−1,−1) = −1. A direct computation shows that theHasse invariant of V ⊥ 〈−x〉 ∼= 〈2, 6,−x〉 is (5,−x) = (−1)ord2(x). This completes the proofof the lemma.2

The last lemma says that there is only one isometry class of anisotropic binary evenunimodular Z2-lattice. Any one of such lattices is denoted by A.

Lemma 6.22 Suppose that L ∼= 〈ε1, ε2, ε3〉 where εi ∈ Z×2 for all i. Then L ∼= P ⊥ 〈ε〉for some ε ∈ Z×2 and an even unimodular lattice P. Moreover P = H if and only if L isisotropic.

Proof. Let v1, v2, v3 be an orthogonal basis for L in which Q(vi) = εi for all i. Let P bethe binary lattice spanned by v1 + v2, v2 + v3. Then

P ∼=(ε1 + ε2 ε2ε2 ε2 + ε3

).

Since the sum of any two units of Z2 is divisible by 2, P is a binary even unimodular latticeand hence P = H or A. Moreover, L ∼= P ⊥ 〈ε〉 for some ε ∈ Z×2 . Lemma 6.21 implies thatL is isotropic if and only if P = H.2

Theorem 6.23 Let L be a unimodular Z2-lattice. If L is even, then L ∼= H or H ⊥ A,where H is an orthogonal sum of some copies of H. If n(L) = Z2, then L has an orthogonalbasis.

Proof. Suppose that L is even unimodular. Then L is (2)-maximal, and by Theorem 6.12L is an orthogonal sum of binary even unimodular lattices. To complete the proof of thiscase, it suffices to show that

H ⊥ H ∼= A ⊥ A.

Both lattices are (2)-maximal, and hence we only need to show that their ambient spacesare isometric. This can be done by comparing the Hasse invariants and the discriminants.

Suppose n(L) = Z2. Then L ∼= 〈ε〉 ⊥ L′ for some ε ∈ Z×2 and a unimodular lattice L′.Therefore it suffices to show that for any binary even unimodular lattice P, 〈ε〉 ⊥ P has anorthogonal basis. This is done by Lemma 6.22.2

47

Corollary 6.24 Let L be a unimodular Z2-lattice with rank ≥ 5. Then H is an orthogonalsummand of L.

Proof. If L is even unimodular, then Theorem 6.23 implies the assertion. So we may supposethat

L ∼= 〈ε1, . . . , εn〉where εi ∈ Z×2 for all i. If 〈ε1, ε2, ε3〉 is isotropic, then we are done by Lemma 6.22. Otherwise〈ε1, ε2, ε3〉 ∼= A ⊥ 〈α〉 for some α ∈ Z×2 . We may then assume that 〈α, ε4, ε5〉 ∼= A ⊥ 〈β〉 forsome β ∈ Z×2 . But then A ⊥ A ∼= H ⊥ H is an orthogonal summand of L.2

7 Quadratic Forms over Z

Unless stated otherwise, (V,Q) is always a quadratic space over Q and all lattices consideredin this section are Z-lattices. Since ±1 are the only units in Z, the discriminant of anondegenerate lattice in V is well-defined as a rational number.

7.1 Preliminaries

A lattice L in V is called integral if B(L,L) ⊆ Z. For an integral lattice L, L# contains Land d(L) = [L# : L]2d(L#). Since d(L#) = d(L)−1, we have

[L# : L] = |d(L)|.

Lemma 7.1 Let L be an integral lattice on a nondegenerate quadratic space V over Q.There are only finitely many integral lattices on V which contain L.

Proof. If M is an integral lattice on V which contains L, then L ⊆ M ⊆ L#. Since L#/Lis a finite abelian group, there are only finitely many such M .2

Lemma 7.2 Let L be an integral lattice on a nondegenerate quadratic space V over Q, andN be a nondegenerate sublattice of L. Then d(N⊥) divides d(L)d(N).

Proof. Since L is integral, therefore for any x ∈ L the linear map z 7→ B(x, z) is a homo-morphism from N to Z. But N is nondegenerate; hence there exists a unique y ∈ N# suchthat B(x, z) = B(y, z) for all z ∈ N . Define a map φ : L −→ N# by φ(x) = y. Sinceker(φ) = N⊥ and φ(x) = x for all x ∈ N , φ−1(N) = N ⊥ N⊥ and this implies

[L : N ⊥ N⊥] = [φ(L) : N ].

The right hand side of the above equation certainly divides [N# : N ] = |d(N)|. Butd(N) d(N⊥) = d(L)[L : N ⊥ N⊥]2. Therefore, d(N⊥) divides d(L)d(N).2

Theorem 7.3 (Hermite) Let L be a lattice on a nondegenerate quadratic space V over Q.Let m(L) = min|Q(x)| : x ∈ L \ 0. Then

m(L) ≤(

4

3

)n−12

|d(L)|1n .

48

Proof. We may assume that L is anisotropic. The proof will be by induction on n = dim(V ).When n = 1, the theorem is obviously true. Assume that the theorem holds for latticeson quadratic spaces of dimension ≤ n− 1. Let e1 ∈ L such that |Q(e1)| = m(L). Then e1

must be primitive in L. If not, there would be an x ∈ L and nonzero integer a such thatax = e1. Then

|Q(x)| = a−2|Q(e1)| < |Q(e1)|

which is impossible.Now, extend e1 to a basis e1, e2, . . . , en for L. Let Φ : V −→ V be the orthogonal

projection of V onto (Qe1)⊥. Then

Φ(x) = x− B(x, e1)

Q(e1)e1, x ∈ V,

and L′ := Φ(L) =∑n

i=2 ZΦ(ei). Let f1 = e1 and fi = Φ(ei) for i ≥ 2. Then

|d(L)| = |det(B(ei, ej))| = | det(B(fi, fj))| = m(L)|d(L′)|.

For any x′ ∈ L′, there exist t ∈ [−12 ,

12 ] and x ∈ L such that x′ = x+te1. Pick x′ ∈ L′ so that

|Q(x′)| = m(L′). Then Q(x) = Q(x′)+t2Q(e1) and hence m(L) ≤ |Q(x)| ≤ m(L′)+ 14m(L).

That is,

3

4m(L) ≤ m(L′)

≤(

4

3

)n−22

|d(L′)|1

n−1 ,

where the last inequality is from the induction assumption. Using |d(L)| = m(L)|d(L′)| weobtain the desired inequality for m(L).2

Theorem 7.4 For any given positive integer n and nonzero rational number d, there areonly finitely many non-isometric lattices of rank n and discriminant d.

Proof. The theorem is obvious if n = 1. Let us assume that the theorem holds for all latticesof rank ≤ n − 1. Let L be a lattice of rank n and discriminant d. By scaling, we couldassume that L is integral. Pick e1 ∈ L so that |Q(e1)| = m(L). Construct a sublattice Nof L as follows:

• If m(L) 6= 0, put N = Ze1.

• Suppose that m(L) = 0. Since L is nondegenerate, the set B(e1, L) is a nonzero idealof Z with a positive generator a. Then a2 | d; so there are only finitely many choicesfor a. Let e2 ∈ L such that B(e1, e2) = a. Note that e1 must be linearly independentfrom e2. Since

Q(te1 + e2) = Q(e2) + 2tB(e1, e2)

49

andB(e1, te1 + e2) = a,

we can choose e2 so that |Q(e2)| ≤ a. We set N = Ze1 + Ze2.

In both cases, there are only finitely many possibilities for the isometry class of N . ByLemma 7.2, d(N⊥) divides d(N) · d. By the induction assumption, there are only finitelymany non-isometric lattices with rank and discriminant the same as those for N⊥. Since Lcontains N ⊥ N⊥, the theorem is now a consequence of Lemma 7.1.2

7.2 Genus and Class

Let L be a lattice on V . If p is a prime, the p-adic completion of L at p is the Zp-latticeLp = Zp⊗L. Let K be another lattice on V . By the Invariant Factor Theorem, there exista basis x1, . . . , xn for L and nonzero rational numbers r1, . . . , rn such that r1x1, . . . , rnxnis a basis for K. Consequently, Lp = Kp for almost all primes p. Since the only rationalnumbers that are p-adic units for every prime p are ±1, therefore L = K if and only ifLp = Kp for all primes p.

Let σ : V → V be a linear map. For every prime p, σ can be extended uniquely to alinear map from Vp to Vp itself which we again denote by σ. It is not hard to check thatσ(Lp) = σ(L)p.

Suppose that M is a lattice in another quadratic space over Q. A necessary conditionfor M ∼= L is that Mp

∼= Lp for all primes p and RM ∼= V∞. This necessary conditionimplies that QνM ∼= Vν for all places ν of Q. It follows from the Weak Hasse Principle thatthere exists an isometry QM −→ V which sends M into V . Therefore, when studying theproblem of deciding when two lattices are isometric, it is enough to consider lattices on thesame quadratic space.

The class of L, written cls(L), is the set of all lattices M on V such that M ∼= L. Thegenus of L, denoted by gen(L), is the set of all lattices M on V such that Mp

∼= Lp for allprimes p. Note that RM ∼= V∞ is automatic since M is also on V . It is obvious that ifM ∈ gen(L), then d(M) = d(L) and rank(M) = rank(L).

Theorem 7.5 The genus of a lattice is partitioned into finitely many classes.

Proof. This follows from Theorem 7.4.2

Definition 7.6 The number of classes in gen(L) is called the class number of L.

Example 7.7 A lattice L is said to be positive definite if Q(x) ≥ 0 for all x ∈ L andQ(x) = 0 if and only if x = 0. Now, suppose that L is a positive definite unimodularlattice of rank ≤ 5. Then Hermite’s theorem implies that L must represent 1. Therefore,L ∼= 〈1, . . . , 1〉. In particular, the class number of L is 1. The same conclusion holds forpositive definite unimodular lattices of rank ≤ 7, but the proof requires a much involvedargument.

50

Theorem 7.8 Let L be a lattice on a quadratic space V over Q. For each prime p, letM(p) be a Zp-lattice on Vp. Then there is a lattice M on V such that Mp = M(p) for all pif and only if Lp = M(p) for almost all p.

Proof. We first assume that there exists M on V such that Mp = M(p) for all p. Since wehave seen that Mp = Lp for almost all p, therefore Lp = M(p) for almost all p.

Conversely, suppose that Lp = M(p) for almost all p. It suffices to show that given asingle prime q, there is a lattice K on V such that

Kp =

Lp if p 6= q;

M(q) if p = q.

The theorem will follow by successive applications of this special case.We first prove the following contention: given a prime q, there is a basis y1, . . . , yn for

V such thatM(q) = Zqy1 + · · ·+ Zqyn.

To prove this we fix a basis x1, . . . , xn for V . By multiplying a suitable power of q toeach xi we may assume that all xi are in M(q). Suppose

M(q) = Zqz1 + · · ·+ Zqzn.

Write each zj as a linear combination of the xi, say

zj = α1jx1 + · · ·+ αnjxn, αij ∈ Qq.

Since Q is dense in Qq, we can find, for each i and j, a rational number aij ∈ Q such that|aij − αij |q is as small as we wish. If the approximations are good enough we will obtain,by virtue of continuity of multiplication and addition in Qq,

|det(aij)− det(αij)|q < |det(αij)|q.

Hence |det(aij)|q = |det(αij)|q. Put yj = a1jx1 + · · ·+ anjxn for 1 ≤ j ≤ n. Then

yj − zj ∈ Zqx1 + · · ·+ Zqxn ⊆M(q)

by our choice the aij , hence

Zqy1 + · · ·+ Zqyn ⊆M(q).

If we write yj =∑

i γijzi with all γij ∈ Zq we have

(γij) = (αij)−1(aij),

hence det(γij) is a unit in Zq. As a result,

Zqy1 + · · ·+ Zqyn = M(q).

51

Now, we can find a lattice J on V such that Jq = M(q). By the Invariant FactorTheorem of finitely generated modules over PID, there exists a basis e1, . . . , en for L andrational numbers r1, . . . , rn such that r1e1, . . . , rnen is a basis for J . For any i, let ti be theq-part of ri. Then K = Zt1e1 + · · ·+ Ztnen has the desired property.2

A rational number a is said to be represented by the genus of L if V∞ represents a andLp represents a for every p. This is a necessary condition for a to be represented by L.

Theorem 7.9 If a is represented by the genus of L, then a is represented by some latticein gen(L).

Proof. We may assume that a 6= 0. From the Strong Hasse Principle it follows that a isrepresented by V . Let v ∈ V such that Q(v) = a. Then for almost all p, v ∈ Lp. If v 6∈ Lpfor a prime p, then Witt’s extension theorem implies that there exists an isometry σp of Vpsuch that σp(v) ∈ Lp. We define a lattice M on V by

Mp =

Lp if v ∈ Lp;σ−1p (Lp) otherwise.

This is well-defined by Theorem 7.8. Then M ∈ gen(L) and v ∈Mp for all p; so v ∈M .2

Corollary 7.10 If L has class number 1, then L represents all rational numbers that arerepresented by its genus.

Example 7.11 Consider the Z-lattice L = 〈1, 11〉. The equation

3 =

(8

5

)2

+ 11

(1

5

)2

shows that V∞ represents 3. Since 5 ∈ Z×p for all p 6= 5, Lp represents 3 for those p; butL5∼= H also represents 3. Therefore 3 is represented by gen(L). But it is obvious that 3 is

not represented by L. Incidentally this shows that there are at least two classes in gen(L).One can show that

M =

(3 11 4

)is a lattice in gen(L) which represents 3, and that M 6∼= L.

7.3 Sum of Squares

For any integer n ≥ 1, let In be the lattice corresponding to the sum of n squares, that is,

In = 〈1, . . . , 1〉.

When n ≤ 7, In has class number 1. Therefore, the set Q(In) is precisely the set ofpositive integers that are represented by the genus of In.

52

Theorem 7.12 (Euler) A positive integer m is a sum of two integer squares if and only ifordp(m) is even for all primes p ≡ 3 mod 4.

Proof. Let p be a prime. We need to determine the set of p-adic integers that is representedby the lattice 〈1, 1〉 over Zp.

(a) p ≡ 1 mod 4: In this case, −1 is a square in Zp. Therefore, 〈1, 1〉 ∼= H over Zp andhence 〈1, 1〉 represents every p-adic integer.

(b) p ≡ 3 mod 4: Since −1 is a nonsquare in Zp, 〈1, 1〉 is a Zp-maximal lattice on ananisotropic quadratic space over Qp. It follows from Theorem 6.16 that a nonzerom ∈ Zp is represented by 〈1, 1〉 if and only if the space 〈1, 1,−m〉 is isotropic, and thelatter is the same as saying that ordp(m) is even.

(c) p = 2: A direct computation shows that a nonzero 2-adic integer m is represented by〈1, 1〉 if and only if m = 2αb, where b ≡ 1 mod 4.

The theorem is now a consequence of (a), (b) and (c) together.2

Theorem 7.13 (Legendre) A positive integer m is a sum of three integer squares if andonly if m is not of the form m = 4a(8b+ 7), a, b ∈ Z,

Proof. Since 〈1, 1, 1〉 is unimodular over Zp for all odd primes p, by Corollary 6.18 it repre-sents all elements of Zp. Therefore, it suffices to show that over Z2, 〈1, 1, 1〉 represents all2-adic integers not of them 4a(8b+ 7).

Over Q2, it is direct to show that the space 〈1, 1, 1,−m〉 is anisotropic if and onlyif m ∈ −Q×2

2 . Therefore, the lattice 〈1, 1, 1〉 does not represent any integer of the form4a(8b+ 7). But it is clear that 〈1, 1, 1〉 represents 1, 3 and 5. Furthermore, since

〈1, 1, 1〉 ∼= A ⊥ 〈3〉

by Lemma 6.22, 〈1, 1, 1〉 represents all 2-adic integers m with ord2(m) ≡ 1 mod 2; seeLemma 6.21. Consequently, 〈1, 1, 1〉 represents all 2-adic integers of the form not of them4a(8b+ 7).2

Theorem 7.14 (Lagrange) Every positive integer is a sum of four integer squares.

Proof. By Corollary 6.18 I4 represents all elements of Zp when p > 2. Over Z2, it can bechecked directly that I4 represents all 2-adic integers. Therefore, the genus of I4 representsall positive integers.2

7.4 Spinor Norms

In this subsection, let (V,Q) be a nondegenerate quadratic space over an arbitrary field Fwhose characteristic is not 2. If W is a nondegenerate subspace of V , then V = W ⊥W⊥.Thus every isometry σ of W can be extended to an isometry of V by defining, σ(x) = x forall x ∈W⊥. So we can identify O(W ) as a subgroup of O(V ).

53

Proposition 7.15 Every isometry in O(V ) is a product of symmetries.

Proof. The proof is by induction on n = dim(V ). The case n = 1 is trivial, so we assumethat n > 1. Take σ ∈ O(V ) and any v ∈ V such that Q(v) 6= 0. Then

Q(v − σ(v)) +Q(v + σ(v)) = 4Q(v) 6= 0.

Hence either Q(v − σ(v)) or Q(v + σ(v)) is nonzero. In the first case we have

τv−σ(v)(v) = σ(v),

while in the second caseτv+σ(v)τv(v) = σ(v).

So in either case there exists ρ ∈ O(V ) which is a product of one or two symmetries suchthat ρ(v) = σ(v). Let W be the orthogonal complement of v. Then ρ−1σ is in O(W )and induction hypothesis implies that ρ−1σ is a product of symmetries τv1 · · · τvk in O(W ).Then

σ = ρτv1 · · · τvnwhich is a product of symmetries in O(V ).2

Remark 7.16 The proof of the above proposition can be modified to show that if p > 2and L is a modular Zp-lattice, then O(L) is also generated by symmetries. For, by scalingthe quadratic form suitably we may assume that L is unimodular. Then L contains a vectorv with Q(v) ∈ Z×p . Hence either Q(v+ σ(v)) or Q(v− σ(v)) is in Z×p . The rest of the proofwill go over in nearly verbatim fashion.

If σ ∈ O(V ) and σ = τv1 · · · τvk for some anisotropic vectors v1, . . . , vk ∈ V , we definethe spinor norm of σ to be

θ(σ) = Q(v1) · · ·Q(vk)F×2 ∈ F×/F×2.

Theorem 7.17 θ : O(V ) −→ F×/F×2 is a well-defined group homomorphism.

We are not going to be too rigid in our use of the θ notation, although it will always beclear from the context just what we have in mind. The equality θ(σ) = a with a ∈ F× willoften appear; this really means θ(σ) is the canonical image of a in F×/F×2. Occasionallywe regard θ(σ) as the full coset aF×2 taken as a subset of F×. More generally, if X is asubset of O(V ), θ(X) is the image of X in F×/F×2 under θ; but we shall also regard θ(X)as the union of the cosets

θ(X) =⋃σ∈X

θ(σ)F×2.

Lemma 7.18 If σ ∈ O(V ), then det(σ) = ±1.

54

Proof. Let B = v1, . . . , vn be a basis for V and A be the symmetric matrix (B(vi, vj)).For 1 ≤ j ≤ n, write

σ(vj) =n∑i=1

tijvi.

Then the matrix T = (tij) is the matrix representation of σ with respect to the basis B. Ifwe identify V with the column space Fn, then for any x, y ∈ V ,

xtAy = B(x, y) = B(σ(x), σ(y)) = xtT tATy,

hence A = T tAT . Therefore det(T ) = ±1.2

The special orthogonal group of V , denoted by O+(V ), is the set

O+(V ) = σ ∈ O(V ) : det(σ) = 1.

The kernel of the homomorphism θ : O+(V ) −→ F×/F×2 is written asO′(V ). It is clear thatO′(V ) is a normal subgroup of O(V ) (not only O+(V )). Note that the groups O′(V ), O+(V ),and O(V ) are unchanged upon scaling the quadratic map on V by any nonzero element ofF .

7.5 Spinor Genus

Suppose that V is a nondegenerate quadratic space over Q. Let L be a lattice on V . Theproper class of L is the set cls+(L) of all lattices K on V such that K = σ(L) for someσ ∈ O+(V ). It is clear that

cls+(L) ⊆ cls(L).

Each class in gen(L) breaks into at most two proper classes, and cls(L) = cls+(L) if andonly if O(L) contains an isometry of determinant −1.

We can define the proper genus of L to be the set gen+(L) of all lattices K on V suchthat Kp = σp(Lp) for some σp ∈ O+(Vp) at each p. However, this turns out to be an obsoletedefinition.

Lemma 7.19 gen+(L) = gen(L).

Proof. It suffices to show that at each p, O(Lp) contains a symmetry. If Lp has an orthogonalbasis, say x1, . . . , xn, then it is easy to verify that τxi is a symmetry in O(Lp). Thereforewe are left with the case when p = 2 and L2 is an orthogonal sum of binary impropermodular lattices. Upon scaling the quadratic form on L suitably, we may assume thatL2 = J ⊥ K with J ∼= H or A. Take any vector v ∈ J with Q(v) = 2. The symmetry τv isreadily an element in O(L2).2

Lemma 7.20 If dim(V ) ≥ 3, then for any prime p, θ(O+(Vp)) = Q×p .

Proof. If dim(V ) ≥ 4, then Vp is universal. Hence there is a symmetry with any prescribedspinor norm and the proposition follows immediately.

55

Suppose that dim(V ) = 3. It is known that Vp represents a ∈ Q×p if the discriminant ofVp ⊥ 〈−a〉 is not 1. So if d(Vp) is a unit, Vp represents all prime elements of Qp and at leastone unit in Zp. Now every element of Q×p is a product of exactly two such elements timesa square in Q×p ; hence θ(O+(Vp)) = Q×p . A similar argument applies when d(Vp) is a primeelement.2

A lattice K is in the proper spinor genus of L if there exist φ ∈ O+(V ) and σp ∈ O′(Vp)at each prime p such that

φ(Kp) = σp(Lp) for all p.

The proper spinor genus of L is written as spn+(L).

Theorem 7.21 If a is represented by gen(L) and rank(L) ≥ 4, then a is represented bysome lattice in spn+(L).

Proof. We may assume that a 6= 0 and, by virtue of the Strong Hasse Principle, that Vrepresents a. Let v be a vector in V such that Q(v) = a. Let S be the set of primes forwhich v 6∈ Lp. Note that S is a finite set. For p ∈ S, since Lp represents a, there existsan isometry σp ∈ O(Vp) such that σp(v) ∈ Lp. Since O(Lp) contains a symmetry, we mayassume that σ is in O+(Vp). Let W be the orthogonal complement of v in V . By Lemma7.20 there exists ρp ∈ O+(Wp) such that θ(ρp) = θ(σp) for each p ∈ S. Define a lattice Kon V by

Kp =

Lp if p 6∈ S;ρpσ

−1p (Lp) if p ∈ S.

Then K ∈ spn+(L) and v ∈ K.2

8 Strong Approximation

8.1 Norms on orthogonal groups

In this subsection, let p be a prime and V be a nondegenerate quadratic space over Qp. Weshall define a norm ‖ ‖p, or simply ‖ ‖ if no confusion arises, on O(V ). Let x1, . . . , xnbe a basis for V . The norm ‖ ‖ which we are about to define is with respect to the basisx1, . . . , xn unless stated otherwise. Let M be the Zp-lattice spanned by this basis.

We first define the norm on V . For any x ∈ V , express it as

x = α1x1 + · · ·+ αnxn, αi ∈ Qp,

and define the norm of x by‖x‖ = max

i|αi|

where | | is the p-adic valuation on Qp. It is not hard to see that ‖ ‖ is a real-valued functionwhich satisfies the following three properties:

(i) ‖x‖ ≥ 0 for all x ∈ V and ‖x‖ = 0⇔ x = 0;

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(ii) ‖αx‖ = |α|‖x‖ for all α ∈ Qp and x ∈ V ;

(iii) ‖x+ y‖ ≤ max‖x‖, ‖y‖, with equality sign holds when ‖x‖ 6= ‖y‖.

Technically speaking, V becomes a normed vector space over Qp. We can make V into ametric space by defining the distance between two vectors x and y to be ‖x− y‖. Additionand scalar multiplication will become continuous operations with respect to the topologyinduced by ‖ ‖.

Let L be the vector space of linear operators on V . Consider a typical element σ ∈ L.For each j, let

σ(xj) =

n∑i=1

αijxi.

Define the norm of σ to be

‖σ‖ = maxi,j|αij | = max

j‖σ(xj)‖.

Then ‖ ‖ makes L into a normed vector space over Qp. Therefore, L is a metric spaceand the distance between σ and τ is ‖σ − τ‖. The addition and scalar multiplication arecontinuous operations on L. However, we also have multiplication in L to consider. Wefind that

‖σ(x)‖ ≤ ‖σ‖‖x‖

for all σ ∈ L and x ∈ V . This shows that

‖στ‖ ≤ ‖σ‖‖τ‖

for any σ and τ in L. Therefore, multiplication is continuous on L.For any σ ∈ O(V ), we have det(σ) = ±1. Therefore ‖σ‖ ≥ 1. Then ‖σ‖ = 1 if and only

if ‖σ‖ ≤ 1, and this is equivalent to σ(M) ⊆ M . Hence σ(M) = M since the discriminantof σ(M) is the same as that of M . Therefore,

O(M) = σ ∈ O(V ) : ‖σ‖ = 1.

Let x′1, . . . , x′n be another basis for V . Let ‖ ‖′ denote the norm on V and L definedby this new basis. Suppose that

x′j =

n∑i=1

aijxi and xj =n∑i=1

bijx′i.

Let A = maxij |aij | and B = maxij |bij |. Then for any x ∈ V and σ ∈ L,

‖x‖A≤ ‖x‖′ ≤ B‖x‖

and‖σ‖AB≤ ‖σ‖′ ≤ AB‖σ‖.

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Therefore, ‖ ‖ and ‖ ‖′ induce equivalent topology on V and L. Now let K be the Zp-latticeZpx′1 + · · · + Zpx′n. If σ ∈ O(V ) is sufficiently close to 1 under the metric induced by ‖ ‖,then ‖σ − 1‖′ < 1. Each such σ satisfies ‖σ‖′ = 1, hence σ(K) = K.

Now consider a third Zp-lattice N on V , and suppose that N = λ(M) for some λ ∈ O(V ).We claim that σ(M) = N for all σ ∈ O(V ) which are sufficiently close to λ. For, by choosingσ sufficiently close to λ we can make

‖λ−1σ − 1‖ ≤ ‖λ−1‖ ‖σ − λ‖

arbitrary small. But all λ−1σ which are sufficiently close to 1 make λ−1σ(M) = M . Henceall σ which are sufficiently close to λ make σ(M) = λ(M) = N .

8.2 Approximation for Isometries

Let V be a nondegenerate quadratic space over Q. Fix a basis for V and use it to definethe norms ‖ ‖p on O(Vp) for all primes p. The following Strong Approximation Theorem isdue to Eichler.

Theorem 8.1 (Strong Approximation for O′(V )) Let V be a nondegenerate quadratic spaceover Q with dim(V ) ≥ 3. Suppose that Vν is isotropic for some place ν (ν could be ∞ ora prime). Let S be a finite set of primes with ν 6∈ S. For any ε > 0 and any given familyσp ∈ O′(Vp) : p ∈ S, there is an isometry σ ∈ O′(V ) such that

(i) ‖σ − σp‖p < ε for all p ∈ S;

(ii) ‖σ‖p = 1 for all p 6∈ S ∪ ν.

A quadratic space (V,Q) over Q is called indefinite if V∞ is an indefinite quadratic spaceover R. This implies that Q takes on both positive and negative rational numbers. Thefollow theorem is a consequence of the Strong Approximation Theorem.

Theorem 8.2 Let L be a lattice on a nondegenerate quadratic space V over Q. If V isindefinite and dim(V ) ≥ 3, then cls+(L) = spn+(L).

Proof. Let K be a lattice in spn+(L). There exist isometries φ ∈ O+(V ) and σp ∈ O′(Vp)at each p such that

φ(Kp) = σp(Lp) for all primes p.

Let T be the set p : φ(Kp) 6= Lp, which is a finite set. Let S be the set of primes p forwhich p 6∈ T and Lp is not the Zp-lattice spanned by the basis for V that defines ‖ ‖p. Notethat S is also a finite set. By the Strong Approximation Theorem, there exists σ ∈ O′(V )such that

(i) for all p 6∈ T ∪ S, ‖σ‖p = 1;

(ii) for all p ∈ T , σ and σp are sufficiently close so that σ(Lp) = σp(Lp);

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(iii) for all p ∈ S, σ and 1Vp are sufficient close so that σ(Lp) = Lp.

For any p 6∈ T , σ(Lp) = Lp = φ(Kp). For any p ∈ T , σ(Lp) = σp(Lp) = φ(Kp). As aresult, σ(Lp) = φ(Kp) for all p and hence K = φ−1σ(L). In particular, K is in cls+(L).2

Corollary 8.3 Let L be a lattice on an indefinite quadratic space of dimension ≥ 4 overQ. If a is represented by gen(L), then a is represented by L.

Proof. This follows from Theorems 7.21 and 8.2.2

Corollary 8.3 does not hold for ternary lattices. For example, let L be the Z-latticecorresponding to the quadratic form −9x2 + 2xy+ 7y2 + 2z2. In terms of symmetric matrix

L ∼=

−9 1 01 7 00 0 2

.

Since d(L) = −27, Lp is unimodular for all primes p ≥ 3. In particular, Lp represents 1for all p ≥ 3. Over Z2, −9 is −1 times a square of a unit. Therefore, L2 also represents 1.This means that gen(L) represents 1. We claim that L does not represent 1. The followingelementary proof is due to D. Zagier.

Assume on the contrary that L represents 1, i.e. there exist integers x, y, z such that

−9x2 + 2xy + 7y2 + 2z2 = 1.

We may rewrite this equation as

2z2 − 1 = (x− y)2 + 8(x− y)(x+ y).

The left hand side is odd; hence (x− y) and hence x+ y as well are odd. This shows thatthe right hand side is congruent to 1 mod 8, and thus z is odd and then 2z2 − 1 ≡ 1 mod16. But 8(x− y)(x+ y) ≡ 8 mod 16. If follows that (x− y)2 ≡ 9 mod 16 and hence

(x− y) ≡ ±3 mod 8.

In particular, (x− y), and 2z2 − 1 as well, has a prime factor p ≡ ±3 mod 8. On the otherhand, if such p divides 2z2 − 1, then 2 is a square mod p and it follows that p ≡ ±1 mod 8which is a contradiction.

Here is another consequence of the Strong Approximation Theorem. Its proof is similarto that of Theorem 8.2

Theorem 8.4 Let L be a Z-lattice on a nondegenerate quadratic space V over Q, and qbe a prime for which Vq is isotropic. If dim(V ) ≥ 3, then Z[1/q]L ∼= Z[1/q]M for anyM ∈ spn+(L).

Suppose that V is an indefinite quadratic space over Q. Let L be a Z-lattice on Vand let M be a Z-lattice in gen(L). Let T be a finite set of primes such that Lp = Mp

59

for all p 6∈ T . For each p ∈ T , let σp ∈ O+(Vp) such that σp(Mp) = Lp. If the StrongApproximation Theorem held for O+(V ), then there would be an isometry σ ∈ O+(V ) suchthat σ(Mp) = σp(Mp) for all p ∈ T and that σ(Mp) = Mp for all p 6∈ T . Then σ(M) = Land this means that gen(L) has only one class. But from the class number formula obtainedin the next section we know that there exists indefinite Z-lattice whose class number is not1. Therefore, the Strong Approximation Theorem does not hold for O+(V ). However, thereis a weak approximation for O+(V ).

Theorem 8.5 (Weak Approximation for O+(V )) Let V be a nondegenerate quadratic spaceover Q and let T be a finite set of primes on Q. Suppose that σp is given in O+(Vp) at eachp ∈ T . Then for every ε > 0, there is a σ ∈ O+(V ) such that

‖σ − σp‖p < ε for all p ∈ T.

Proof. For each p ∈ T , we may write

σp = τx1(p) · · · τxm(p)

where every xi(p) is in Vp. Since m must be even and all symmetries have order 2, we maysuppose that m is independent of p ∈ T . By Chinese Remainder Theorem, we can choosexi ∈ V such that xi and xi(p) are sufficiently close in Vp for p ∈ T . Then τxi and τxi(p) aresufficiently close so that σp and σ := τx1 · · · τxm are, for p ∈ T .2

Corollary 8.6 Let L be a Z-lattice on a nondegenerate quadratic space V over Q, and letT be a finite set of primes. For every Z-lattice M in gen(L), there exists K ∈ cls+(M) suchthat Kp = Lp for all p ∈ T .

Proof. For each p ∈ T , let σp ∈ O+(Vp) such that σp(Mp) = Lp. It follows from the WeakApproximation Theorem for O+(V ) that there is a σ ∈ O+(V ) satisfying σ(Mp) = σp(Mp).Let K = σ(M). Then Kp = Lp for all p ∈ T .2

Finally, we can improve Theorem 8.3 to representations with approximation property.

Theorem 8.7 Let L be a Z-lattice on an indefinite quadratic space V over Q with dim(V ) ≥4 and T be a finite set of primes. Let a be a nonzero rational number which is representedby gen(L), and suppose that for each p ∈ T there is a vp ∈ Lp with Q(vp) = a. Then foreach ε > 0, there exists v ∈ L with Q(v) = a and

‖v − vp‖p < ε ∀ p ∈ T,

where all ‖ ‖p are defined by L

Proof. Since a is represented by gen(L), there exists z ∈ V such that Q(z) = a. We enlargeT so that

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(a) T contains all the prime divisors of 2d(L), and

(b) z ∈ Lp for all p 6∈ T .

It suffices to prove the theorem for this enlarged T .By Witt’s theorem, there is an isometry ρp ∈ O+(Vp) at each p ∈ T such that ρp(z) =

vp. Let W be the orthogonal complement of Qz in V . Since dim(W ) ≥ 3, there existsτp ∈ O+(Wp) at each p ∈ T such that θ(τp) = θ(ρp). Let σp = ρpτp. Then σp ∈ O′(Vp) forall p ∈ T . By the Strong Approximation for O′(V ), there exists σ ∈ O′(V ) such that

‖σ − σp‖p <ε

‖z‖p∀ p ∈ T

and‖σ‖p = 1 ∀ p 6∈ T.

Let v be σ(z). It is clear that ‖v − vp‖p < ε for each p ∈ T . Since vp ∈ Lp and ε < 1,v ∈ Lp as well. For p 6∈ T , σ ∈ O′(Lp) and z ∈ Lp. So v = σ(z) ∈ Lp. Consequently, v ∈ Lpfor all primes p; hence v ∈ L.2

9 Adelic Theory of Spinor Genus

In this section, let V be a nondegenerate quadratic space over Q and L be a Z-lattice onV . We shall provide a formula for the number of proper spinor genera in gen(L). When Vis indefinite, this formula gives the number of proper classes in gen(L). We also describe aprocedure to decide if two given lattices in a given genus are inside the same proper spinorgenus.

9.1 Adeles

Let Ω be the set of all places of Q. Fix a basis B of V and use it to define all the norms‖ ‖p on O(Vp). If Σ is an element in the direct product

∏ν∈ΩO

+(Vν), then Σν will denoteits ν-component. The adelization of O+(V ) is the set

O+A (V ) = Σ ∈

∏ν∈Ω

O+(Vν) : ‖Σp‖p = 1 for almost all primes p .

Lemma 9.1 O+A (V ) is well-defined, that is, it is independent of the choice of the basis B.

Proof. Let L be the lattice spanned by the vectors in B. Then for a prime p, ‖Σp‖p = 1 ifand only if Σp(Lp) = Lp. Now, suppose that B′ is another basis for V . Let L′ be the latticespanned by the vectors in B′. Then for almost all p, Lp = L′p. So, if Σ is in O+

A (V ), then

Σp(L′p) = Σp(Lp) = Lp = L′p

for almost all primes p.2

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An element in O+A (V ) is called an adele of O+(V ). It is clear that O+

A (V ) is a subgroupof the direct product. The set of all elements Σ ∈ O+

A (V ) with the property

Σν ∈ O′(Vν) for all ν ∈ Ω

is clearly a subgroup of O+A (V ). We shall denote this subgroup by O′A(V ). It is evident that

O′A(V ) contains the commutator subgroup of O+A (V ).

Consider a typical element σ ∈ O+(V ). Then σ can be regarded as an element in O+(Vν)for all ν ∈ Ω. Let T be the matrix representation of σ with respect to the basis B. Thenthe entries of T and det(T ) are units in Zp for almost all p. Therefore, ‖σ‖p = 1 for almostall primes p. Hence we can identify O+(V ) as a subgroup of O+

A (V ) through the diagonalembedding.

For any Σ ∈ O+A (V ) and any lattice L on V , define the lattice Σ(L) by specifying its

local completions asΣ(L)p = Σp(Lp) for all p.

This definition is meaningful since Σp(Lp) = Lp for almost all p. Therefore, we have anaction of the group O+

A (V ) on the set of lattices on V .

Proposition 9.2 We have cls+(L) = O+(V )L, spn+(L) = O+(V )O′A(V )L, and gen(L) =O+A (V )L.

Proof. We only demonstrate the equality gen(L) = O+A (V )L; the other two equalities can

be proved in a similar way. It is clear that Σ(L) ∈ gen(L) for any Σ ∈ O+A (V ). Suppose

that M ∈ gen(L). Then for each prime p, there exists Σp ∈ O+(Vp) such that

Σp(Lp) = Mp.

Since Lp = Mp for almost all p, therefore, ‖Σp‖p = 1 for almost all p. Let Σ∞ be anyisometry in O+(V∞) and put Σ = (Σν). Then M = Σ(L).2

The stabilizer of L in O+A (V ) is the set

O+A (L) = Σ ∈ O+

A (V ) : Σν(Lp) = Lp for all primes p.

9.2 Number of Spinor Genera

Lemma 9.3 For any Σ ∈ O+A (V ), we have Σ(spn+(L)) = spn+(Σ(L)).

Proof. Let K be a lattice in Σ(spn+(L)). Then there exist φ ∈ O+(V ) and σp ∈ O′(Vp) ateach p such that

Kp = Σpφσp(Lp) for all p.

Since O′(Vp) contains the commutator subgroup of O+(Vp), there exists ρp ∈ O′(Vp) suchthat

Σpφσp = ρpφσpΣp.

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However, as O′(Vp) is normal in O+(Vp), there exists τp ∈ O′(Vp) such that

ρpφσpΣp = φτpσpΣp.

Therefore, K ∈ spn+(Σ(L)), and hence Σ(spn+(L)) ⊆ spn+(Σ(L)). The reverse inclusioncan be proved in the same fashion.2

If Σ,Λ are elements in O+A (V ), then the normality of O′A(V ) in O+

A (V ) implies thatΣO′A(V ) = O′A(V )Σ, and the fact that O′A(V ) contains the commutator subgroup of O+

A (V )implies that ΣΛO′A(V ) = ΛΣO′A(V ), hence the set O′A(V )ΣΛ is independent of the orderof O′A(V ),Σ,Λ. From this it follows that the set O+(V )O′A(V )O+

A (L) is independent of theorder of O+(V ), O′A(V ), O+

A (L), and that this set is actually equal to the group generatedby O+(V ), O′A(V ), O+

A (L), which is a normal subgroup in O+A (V ).

Proposition 9.4 The number of proper spinor genera in gen(L) is equal to the index

[O+A (V ) : O+(V )O′A(V )O+

A (L)].

Proof. It is clear that the group O+A (V ) acts transitively on the set of proper spinor genera

in gen(L) and O+(V )O′A(V )O+A (L) is contained in the stabilizer of spn+(L). Now, suppose

that Σ(spn+(L)) = spn+(Σ(L)) = spn+(L) for some Σ ∈ O+A (V ). Then Σ(L) ∈ spn+(L)

and this means that there exist φ ∈ O+(V ) and σp ∈ O′(Vp) at each p such that

Σp(Lp) = φσp(Lp) for all p.

Therefore, Σ ∈ O+(V )O′A(V )O+A (L).2

Corollary 9.5 The number of proper spinor genera in a given genus is power of 2.

Proof. At each place ν, the square of any element in O+(Vν) has trivial spinor norm.Therefore, the square of element in O+

A (V ) is in O′A(V ), and hence the exponent of thefinite abelian group O+

A (V )/O+(V )O′A(V )O+A (L) is 2.2

The set of ideles of Q is the set

J = x = (xν) ∈∏ν∈Ω

Q×ν : |xp|p = 1 for almost all p .

It is clear that J is a group.

Lemma 9.6 If p > 2 and Lp is a unimodular Zp-lattice of rank ≥ 2, then θ(O+(Lp)) =Z×p Q×2

p .

Proof. By Lemma 6.19, Lp represents all units of Zp. Therefore, θ(O+(Lp)) contains Z×p Q×2p .

For the reverse inclusion, we first note that O+(Lp) is generated by symmetries (see Remark7.16). Therefore it is enough to show that the spinor norm of any symmetry in O+(Lp) is a

63

unit. Take a typical symmetry τy ∈ O+(Lp). We may assume that y is primitive in Lp andthus p−1y 6∈ Lp. For any x ∈ Lp,

τy(x) = x− 2B(x, y)

Q(y)y.

Therefore B(y, Lp) ⊆ Q(y)Zp. But B(y, Lp) = Zp because Lp is unimodular. Hence θ(τy) =Q(y)Q×2

p ⊆ Z×p Q×2p .2

Corollary 9.7 Let L be a lattice on V . If dim(V ) ≥ 2, then for any Σ ∈ O+A (V ), θ(Σp) ⊆

Z×p Q×2p for almost all p.

Proof. This is because Lp is unimodular for almost all p.2

Let Σ be an element in O+A (V ). If rank(L) ≥ 3, then by Corollary 9.7 there exists an

x ∈ J such that xν ∈ θ(Σν) for all ν ∈ Ω. If y is another element in J which is associatedto Σ in this way, then y ∈ xJ2. Let JL be the subset of J defined by

JL = x = (xν) ∈ J : xν ∈ θ(O+(Lν)) for all ν ∈ Ω .

Then J2 ⊆ JL, and hence the image of x and y in J/Q×JL are equal. We therefore have awell-defined homomorphism

θA : O+A (V ) −→ J/Q×JL.

Theorem 9.8 If rank(L) ≥ 3, then the number of proper spinor genera in gen(L) is equalto the index [J : Q×JL].

Proof. It suffices to show that θA is a surjective homomorphism with O+(V )O′A(V )O+A (L)

as its kernel. Since the definition of O+A (V ) is independent of the choice of the basis B of

V , we may very well assume that B is in fact a basis for L.Take a typical element x ∈ J . By considering −x instead if necessary, we may assume

that x∞ ∈ θ(Σ∞) for some Σ∞ ∈ O+(V∞). For almost all p, xp is a unit. Since Lp isunimodular for almost all p and θ(O+(Lp)) contains Z×p at those p, we can find Σp ∈ O+(Lp)such that xp ∈ θ(Σp) for almost all p. For the remaining finite number of primes q, thereexist Σq ∈ O+(Vp) such that xq ∈ θ(Σq) because dim(Vq) ≥ 3; see Theorem 7.20. ThenΣ = (Σν) is an element in O+

A (V ) and θA(Σ) = x.It is clear that O+(V )O′A(V )O+

A (L) is a part of the kernel of θA. Now suppose thatθA(Σ) ∈ Q×JL. Let us assume the following statement at this moment:

θ(O+(V )) =

a ∈ Q× : a > 0 if V∞ is anisotropic,Q× otherwise.

Then we can take σ ∈ O+(V ) and Λ ∈ O+A (L) so that

θA(ΣσΛ) ⊆ J2.

As a result, ΣσΛ ∈ O′A(V ) and hence the kernel of θA is O+(V )O′A(V )O+A (L).2

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Proposition 9.9 If dim(V ) ≥ 3, then

θ(O+(V )) =

a ∈ Q× : a > 0 if V∞ is anisotropic,Q× otherwise.

Proof. Let a ∈ Q× and suppose that a > 0 if V∞ is anisotropic. If the latter situationarises, then by multiplying −1 to the quadratic form if necessary, we may assume that V∞is negative definite.

Let S be the set of primes for which Vp is anisotropic. Note that S is a finite set. Ifp ∈ S, we take bp ∈ Q×p such that both bpd(V ) and abpd(V ) are not squares. Using theChinese Remainder Theorem we can find 0 < b ∈ Q× such that b and bp are sufficientlyclose for all p ∈ S. If S is empty, then we set b = 1. Then both Vv ⊥ 〈b〉 and Vv ⊥ 〈ab〉 areisotropic for every place v of Q. From the Strong Hasse principle it follows that both −band −ab are represented by V . Let x, y ∈ V such that Q(x) = −b and Q(y) = −ab. Thenθ(τxτy) = Q(x)Q(y) ∈ aQ×2. Thus

θ(O+(V )) ⊇a ∈ Q× : a > 0 if V∞ is anisotropic,Q× otherwise.

The reverse inclusion is obvious.2

Corollary 9.10 Let L be a nondegenerate Z-lattice. If rank(L) ≥ 3 and θ(O+(Lp)) ⊇ Z×pfor all primes p, then gen(L) = spn+(L).

Proof. It suffices to show that the quotient group J/Q×JL is trivial. Take a typical elementx ∈ J . By multiplying −1 if necessary, we could assume that x∞ > 0 and hence x∞ ∈θ(O+(V∞)). Let S be the set of primes p for which xp 6∈ Z×p Q×2

p . Then there exists arational number a > 0 such that ordp(a) = 0 for all p 6∈ S and axp ∈ Z×p . As a result,ax ∈ JL and hence x ∈ Q×JL.2

Example 9.11 Let L be a nondegenerate unimodular Z-lattice of rank ≥ 3. If p > 2, thenθ(O+(Lp)) = Z×p Q×2

p since Lp is unimodular. At p = 2, P is an orthogonal summand of L2,

where P = H or A, and θ(O+(P)) ⊇ Z×2 Q×22 . Therefore, gen(L) = spn+(L). If, in addition,

L is indefinite, then gen(L) = cls+(L) and the class number of L is 1.

9.3 Spinor Equivalence

Let L be a Z-lattice on a nondegenerate space V , and K is another Z-lattice in gen(L).How do we know if K is inside spn+(L)? By definition, K = ΣL for some Σ ∈ O+

A (V ).Since the adelic spinor norm θA induces an isomorphism

O+A (V )/O(V )O′A(V )O+

A (L) −→ J/Q×JL,

therefore K is in spn+(L) if and only if θA(Σ) ∈ Q×JL. The remaining question is: how dowe construct Σ if L and K are given explicit enough?

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It is clear that we can replace K by any lattice in cls+(K). So by virtue of the Weakapproximation Theorem we can assume that L and K only differ at the primes not dividing2d(L), that is Lp = Kp for all primes p | 2d(L). Now take a prime p so that Lp 6= Kp. Thenp > 2 and both Lp and Kp are unimodular Zp-lattices. Thus [Lp : Lp∩Kp] = [Kp : Lp∩Kp]by Corollary 6.14. Consequently, [L : L ∩K] = [K : L ∩K].

Lemma 9.12 Suppose Lp 6= Kp. If σp is an isometry in O+(Vp) which sends Lp to Kp,then θ(σp) ∈ [Lp : Lp ∩Kp]Z×p Q×2

p .

Proof. For simplicity, let e = ordp([Lp : Lp ∩Kp]). Since θ(O+(Lp)) = Z×p Q×2p , it suffices

to exhibit an isometry in O+(Vp) of spinor norm pe which sends Lp to Kp. By Theorem6.12, we may assume that Lp and Kp are unimodular lattices on a hyperbolic plane witha hyperbolic pair x, y such that Lp = Zpx + Zpy and Kp = Zppex + Zpp−ey. Then theisometry τx−yτx−p−ey, which has spinor norm pe, sends Lp to Kp.2

Definition 9.13 The intersecting idele of L and K is the idele j(L,K) ∈ J whose p-component is pordp([L:L∩K]).

Theorem 9.14 Let L and K be Z-lattices in the same genus. If [L : L ∩K] is relativelyprime to 2d(L), then K ∈ spn+(L) if and only if j(L,K) ∈ Q×JL.

Proof. If Σ(L) = K where Σ ∈ O+A (V ), then θA(Σ) ∈ j(L,K)Q×JL by Lemma 9.12. Thus

K ∈ spn+(L) if and only if j(L,K) ∈ Q×JL.2

10 Representations of Spinor Genus

Throughout this section, L is a Z-lattice on a nondegenerate space V of dimension ≥ 3.Suppose that a is a nonzero rational number which is represented by gen(L). Then a isrepresented by some lattice in gen(L). We say that a is represented by a proper spinorgenus S in gen(L) if a is represented by some lattice in S. The main question we would liketo answer is: which proper spinor genera in gen(L) represent a?

Let v ∈ V such that Q(v) = a and let W be the orthogonal complement of the spacespanned by v. For the questions with which we are concerned we may, without loss ofgenerality, assume that v ∈ L.

10.1 Generators

An adele Σ = (Σp) ∈ O+A (V ) is called a generator for L and v if v ∈ Σ(L). Let XA(L, v) be

the set of generators for L and v. It is clear from the definition that

XA(L, v)O+A (L) = O+

A (W )XA(L, v) = XA(L, v).

The importance of XA(L, v) is illustrated in the following lemma.

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Lemma 10.1 Let Σ ∈ O+A (V ). Then a is represented by spn+(Σ(L)) if and only if Σ ∈

O+(V )O′A(V )XA(L, v).

Proof. This is clear from the definitions.2

Note that in general XA(L, v) may not be a group. However, we have

Theorem 10.2 The set O′A(V )XA(L, v) is a group.

Proof. It suffices to prove that θA(XA(L, v)) is a group. Let p be a prime, and let X(Lp, v)be the set of local generators, that is,

X(Lp, v) = σ ∈ O+(Vp) : v ∈ σ(Lp).

We need to show that θ(X(Lp, v)) is a group. Since X(Lp, v)O+(Lp) = X(Lp, v), it is onlynecessary to consider the case when

[Q×p : θ(O+(Lp))] > 2.

This implies that Lp has only one dimensional Jordan components. When p > 2 thisfollows from the fact that the spinor norm of the isometry group of any binary unimodularZp lattices contains all p-adic units (Lemma 9.6). For p = 2, this follows from the explicitcalculation of spinor norms of the isometry groups of modular Z2-lattices by Hsia.

So, we may write Lp = Zpx1 ⊥ · · · ⊥ Zpxn with ordp(Q(x1)) < · · · < ordp(Q(xn)). Wemay assume that ordp(Q(v)) = ordp(Q(x1)); otherwise one can replace Lp by L′p = Zppx1 ⊥· · · ⊥ Zpxn because X(Lp, v) = X(L′p, v) and the argument can be repeated. Also, x1 = vcan be assumed.

Let σ ∈ X(Lp, v). Then σ−1(v) ∈ Lp and hence σ−1(v) = αv + w where α ∈ Z×p andw ∈W . One of 1− α and 1 + α is not in 2pZp. Without loss of generality, we assume that1−α 6∈ 2pZp. A direct verification shows that τσ−1v−v is in O(Lp) and τσ−1v−v(v) = σ−1(v).Therefore,

σ τσ−1v−v τx2(v) = v

which implies that σ ∈ O+(Wp)O+(Lp). However, it is clear from the definitions that

O+(Wp)O+(Lp) ⊆ X(Lp, v). Thus X(Lp, v) = O+(Wp)O

+(Lp) and θ(X(Lp, v)) is a sub-group of J .2

We therefore have the following chain of containment of groups

O+(V )O′A(V )O+A (L) ⊆ O+(V )O′A(V )O+

A (L)O+A (W ) ⊆ O+(V )O′A(V )XA(L, v) ⊆ O+

A (V ).

Lemma 10.3 We have O+A (V )/O+(V )O′A(V )O+

A (L)O+A (W ) ∼= J/Q×JLθA(O+

A (W )) and

O+A (V )/O+(V )O′A(V )XA(L, v) ∼= J/Q×θA(XA(L, v)).

Proof. This follows from the application of the adelic spinor norm map θA.2

Finally, we show that everything discussed thus far is independent of the choice of v andL.

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Lemma 10.4 Suppose that v′ ∈ L′ with Q(v′) = Q(v) and L′ ∈ gen(L). Then thegroups generated by XA(L, v) and XA(L′, v′) are conjugate in O+

A (V ) so that the subgroupO+(V )O′A(V )XA(L, v) is independent of the choice of v and L.

Proof. Let G(L, v) be the group generated by XA(L, v). Let Σ ∈ O+A (V ) such that L′ =

Σ(L). We first suppose that v = v′. Then Σ ∈ XA(L, v) by definition. If g ∈ XA(L′, v), thenv ∈ gΣ(L); thus gΣ ∈ XA(L, v) and hence g ∈ G(L, v). This shows that G(L′, v) ⊆ G(L, v).By reversing the roles of L and L′ we see that G(L, v) = G(L′, v).

In general, let φ ∈ O+(V ) such that φ(v) = v′. Then

φG(L, v)φ−1 = G(φ(L), φ(v)) = G(φ(L), v′) = G(L′, v′).

2

From now on we shall write XA(L, a) for XA(L, v).

10.2 Representations

We know from the previous subsection that the number of proper spinor genera in gen(L)representing a is given by the following indices

[O+(V )O′A(V )XA(L, a) : O+(V )O′A(V )O+A (L)] = [Q×θA(L, a) : Q×JL].

For simplicity, we have used θA(L, a) to denote θA(XA(L, a)).Suppose first that either dim(W ) ≥ 3 or W is a hyperbolic plane. Then θ(O+(Wp)) =

Q×p for all primes p. Therefore Q×θA(O+A (W )) = J , so that a is represented by every proper

spinor genus in gen(L). Note that this recaptures Theorem 7.21.Consider next the case dim(W ) = 2 and W is anisotropic. Let δ = d(W ) and E =

Q(√−δ). Then E is a quadratic extension of Q. Let ΩE be the set of places of E (that

is, the set of equivalence classes of valuations on E). Each place ν of Q has one or twoextensions in ΩE . Let JE be the group of ideles of E defined by

JE = (xω) ∈∏ω∈ΩE

E×ω : |xω|ω = 1 for almost all ω.

Define the norm map NE/Q : JE → J by

NE/Q((xω)) = (∏ω|ν

Nω/ν(xω)),

where Nω/ν is the norm of the extension Eω/Qν .

Remark 10.5 The definitions of JE and the associated norm map NE/Q extend naturallyto all finite extensions of Q.

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Lemma 10.6 We have θA(O+A (W )) = NE/Q(JE).

Proof. It suffices to show that θ(O+(Wν)) =∏ω|ν Nω/ν(E×ω ) for every place ν of Q. This is

obvious when ν splits since both groups are equal to Q×ν .Suppose that ν does not split in E. Then there is only one place ω of E lying above

ν and Eω is a quadratic extension of Qv. Take an anisotropic vector x ∈ W and letα = Q(x). Then W ∼= 〈α〉 ⊥ 〈αδ〉. Thus, after scaling by α, the quadratic space Wν can beidentified with Eω and the quadratic map is simply the norm Nω/ν . Therefore, if t is givenin Nω/ν(E×ν ), then there must be a vector y ∈W such that αQ(y) = t and so

t = α · α−1t = θ(τxτy) ∈ θ(O+(Wν)).

Conversely, let σ ∈ O+(Wν). Then σ is a product of even number of symmetries of Wν .It follows from the above discussion that θ(σ) ∈ Nω/ν(E×ω ). 2

By class field theory, [J : Q×NE/Q(JE)] = [E : Q] = 2. We offer another proof here.

Lemma 10.7 If E/Q is a quadratic extension, then [J : Q×NE/Q(JE)] = 2.

Proof. Define a function R : J → ±1 by

R((xν)) =∏ν

(xν ,−δ)ν ,

where ( , )ν is the Hilbert symbol over Qν . It is well-defined since for almost all odd primesp, xp and −δ are p-adic units and hence (xp,−δ)p = 1. Since −δ 6∈ Q×2, there is a place νat which −δ 6∈ Q×2

ν . Hence there must be xν ∈ Q×ν such that (xν ,−δ)ν = −1. This showsthat R is surjective. We shall show that kerR is precisely Q×NE/Q(JE).

Recall that (xν ,−δ)ν = 1 if and only if xν is a norm of the field extension Qp(√−δ)/Qp.

Thus NE/Q(JE) ⊆ kerR. It follows from Hilbert Reciprocity Law that Q× ⊆ kerR. HenceQ×NE/Q(JE) ⊆ kerR. For the other inclusion, let (xν) ∈ kerR and let ν1, . . . , νm be theplaces for which (xν ,−δ)ν = −1. Note that m is even and nonzero. Let S be a finite set ofprimes which contains the prime 2 and all the finite places among the νi, and such that forany p 6∈ S, xp and δ are p-adic units. Put ep = ordp(xp) and define

h =∏p∈S

pep .

We take a prime q 6∈ S such that

q ≡ ηh−1xp mod pt, for all p ∈ S,

where t is a sufficiently large integer and η = 1 if all the places νi are finite; otherwiseη = −1. We claim that qηh−1(xν) ∈ NE/Q(JE), that is (qηh−1xν ,−δ)ν = 1 for all places ν.This will complete the proof.

If a prime p is not in S and p 6= q, then p is odd and qηh−1xp and −δ are in Z×p ,and thus (qηh−1xp,−δ)p = 1. If p is in S, then qηh−1xp is a square in Q×p (since t is

69

chosen large enough) and hence (qηh−1xp,−δ)p = 1 also. If all the places νi are finite, thenqηh−1 > 0 and then (qηh−1x∞,−δ)∞ = (x∞,−δ) = 1. If νi = ∞ for some i, then η = −1and (x∞,−δ)∞ = −1. This means that x∞ < 0. So, (qηh−1x∞,−δ)∞ = (−x∞,−δ)∞ = 1.Finally at the prime q, since qηh−1(xν) ∈ kerR, we have

(qηh−1xq,−δ)q =∏ν 6=q

(qηh−1xν ,−δ)ν = 1.

2

Since there are inclusions of groups

Q×NE/Q(JE) ⊆ Q×θA(L, a) ⊆ J,

therefore, Q×θA(L, a) 6= J if and only if Q×θA(L, a) = Q×NE/Q(JE).

Lemma 10.8 If Q×θA(L, a) = Q×NE/Q(JE), then θA(L, a) = NE/Q(JE).

Proof. Assume the contrary that there exists (xν) ∈ θA(L, a) which is not in NE/Q(JE).Then there must be a place ` for which (x`,−δ)` = −1. Construct an idele j = (jν) suchthat jν = 1 if ν 6= ` and jν = xν . Then j ∈ θA(L, a) and so we can write j = αn for someα ∈ Q× and n ∈ NE/Q(JE). For all ν 6= `, since j` = 1, we have (α,−δ)ν = 1. By HilbertReciprocity Law, (α,−δ)` = 1 as well. Thus (xν ,−δ)ν = 1 which is a contradiction.2

Therefore, Q×θA(L, a) 6= J if and only θA(L, a) = NE/Q(JE). When this is the situationwe shall call a a spinor exception. In this case, we have

[Q×θA(L, a) : Q×JL] =[J : Q×JL]

[J : Q×θA(L, a)]=

[J : Q×JL]

[J : Q×NE/Q(JE)]=

1

2[J : Q×JL]

so that a is represented by exactly half of the proper spinor genera in gen(L). To summarize:

Theorem 10.9 Suppose that a is represented by the genus of a ternary Z-lattice L and−ad(L) 6∈ Q×2. Then a is a spinor exception if and only if θA(L, a) = NE/Q(JE), where

E = Q(√−ad(L)). Moreover, spn+(ΣL) represents a if and only if θA(Σ) ∈ Q×θA(L, a).

The groups θA(L, a) are explicitly determined by Schulze-Pillot.

10.3 A Class Field Interpretation

Let F be a number field and O be the ring of integers in F . Let ΩF be the set of places ofF . The group of ideles of F is defined by

JF = (xν) ∈∏ν∈ΩF

F×ν : xν ∈ O×ν for almost all ν.

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Let ΩfF and Ω∞F be the set of finite and infinite, respectively, places of F . We impose a

topology on JF as follows. Let

E =∏ν∈Ω∞F

F×ν∏ν∈ΩfF

O×ν .

It is easy to see that E is a subgroup of JF . Each factor of E has its metric topology inducedby the associated valuation. Each F×ν is locally compact while each O×ν is compact. Wegive E the product topology so that it becomes a locally compact topological group. Nextwe consider the following collection of subsets of JF

jA : j ∈ JF and A is open in E.

We declare each subset in this collection open in JF . It turns out that these open subsetsform a basis of a topology on JF which we call the restricted product topology.

An open subgroup H of JF is called principal if F× ⊆ H and [JF : H] <∞. For a finiteabelian extension E/F , F×NE/F (JE) is a principal open subgroup of JF . A main theoremin Class Field Theory says that

E 7→ F×NE/F (JE)

is a bijective inclusion-reversing correspondence between finite abelian extensions of F andprincipal open subgroups of JF . Moreover, Artin Reciprocity Law, which is a generalizationof Hilbert Reciprocity Law, gives an isomorphism

JF /F×NE/F (JE) −→ Gal(E/F ).

Let us go back to our discussion of spinor genera. In our case, the base field is Q (eventhough all the results discussed thus far about spinor genera can be generalized to anynumber field setting). The group Q×JL is clearly a principal open subgroup of J(= JF ).Associated to it by Class Field Theory is an abelian extension Σ/Q. Since JL ⊇ J2, Σ/Fis a multiquadratic extension. We call Σ the spinor class field of gen(L).

It is clear that we can rephrase all the results, especially Theorem 10.9, in terms of thespinor class fields. Here we offer a result that can be deduced easily from this class fieldsetting but is not obvious from the previous discussion. The group Q×θA(L, a) is also aprincipal open subgroup of J and hence one can associate to it an abelian extension ΣL/a

called a relative spinor class field. Since we have a chain of containments of groups

Q×JL ⊆ Q×θA(L, a) ⊆ J,

we have a chain of fieldsQ ⊆ ΣL/a ⊆ Σ.

If a is a spinor exception of gen(L), then θA(L, a) = NE/Q(JE) where E = Q(√−ad(L)).

In particular, ΣL/a is in fact the quadratic extension Q(√−ad(L)). Since Σ can have only

finitely many quadratic subextensions of Q, therefore we have

Proposition 10.10 Let L be a ternary nondegenerate Z-lattice. Then the spinor exceptionsof gen(L) fall into finitely many square classes.

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11 Representations of Definite Quadratic Forms

In this section, we will prove the following theorem.

Theorem 11.1 (Tartakowsky, Kloosterman, Pall and Ross) If L is a positive definite Z-lattice of rank ≥ 5. Then there exists a constant c(L) such that a is represented by L if ais represented by the genus of L and a > c(L).

The above theorem is false if the rank of L is 4. Consider the Z-lattice L = 〈1, 1, 25, 25〉.If p 6= 5, Lp is just the sum of 4 squares. Hence Q(Lp) = Zp for p 6= 5. Since −1 is asquare in Z5, L5 is split by 〈1, 1〉 ∼= H; hence Q(L5) = Z5. Consequently, the genus of Lrepresents all positive integers. We claim that L does not represent all integers of the form3 · 2r. Let v1, v2, v3, v4 be an orthogonal basis for L whose associated symmetric matrixis 〈1, 1, 25, 25〉. Suppose that there exists v ∈ L such that Q(v) = 3 · 2r. If v =

∑aivi, then

3 · 2r = a21 + a2

2 + 25a23 + 25a2

4.

If r ≤ 2, then a3 = a4 = 0. But then the equation

3 · 2r = a21 + a2

2

is not soluble over Z. If r ≥ 3, then all the ai must be even, and hence 3 · 2r−2 is alsorepresented by L. Eventually we arrive at the conclusion that L represents an integer ofthe form 3 · 2m with m ≤ 2, which is impossible as shown earlier.

11.1 Some Local Results

In this subsection, L is a Zp-lattice on a nondegenerate quadratic space V over Qp. Recallthat L is a maximal lattice if L is an (a)-maximal lattice for some a ∈ Q×p .

Proposition 11.2 If L is a maximal lattice on V and dim(V ) ≥ 3, then θ(O+(L)) ⊇ Z×p .

Proof. By scaling the quadratic form on V , we may assume that L is (2)-maximal. If Vis anisotropic, then L is the unique (2)-maximal lattice on V . This means that O+(L) =O+(V ). By Lemma 7.20, θ(O+(L)) = θ(O+(V )) = Q×p .

Suppose that V is isotropic. Then L is split by the hyperbolic plane H; see Theorem6.12. So, θ(O+(L)) ⊇ θ(O+(H)) ⊇ Z×p .2

Lemma 11.3 Suppose that p > 2 and that L is a unimodular lattice. If x and y areprimitive vectors of L with Q(x) = Q(y), then there exists σ ∈ O(L) such that σ(x) = y.

Proof. Let w = y − x. Then Q(w) = −2B(x,w). If Q(w) ∈ Z×p , then τw ∈ O(L) andτw(x) = y. Therefore, we may assume that Q(w) ∈ (p). Suppose that we can find an u ∈ Lsuch that

Q(u), B(u, x), B(u, y) ∈ Z×p . (∗)

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Let v = y − τu(x). Then v ∈ L because τu ∈ O(L). Moreover,

Q(v) = Q(w) + 4B(u, x)B(u, y)Q(u)−1 ∈ Z×p .

So, τv ∈ O(L) and a direct computation shows that τv(y) = τu(x). This implies τvτu(x) = y.Now it remains to show the existence of v satisfying (∗). Let L be the quadratic space

L/pL over the finite field F = Zp/pZp. Since L is unimodular, L is a nondegenerate spaceover F. For any vector t ∈ L, let t be its canonical image in L. By abusing the notation,we also use Q and B to denote the quadratic form and bilinear form, respectively, inducedon L. Condition (∗) is equivalent to

Q(L \ (x⊥ ∪ y⊥)) 6= 0.

Since x is a primitive vector of L, there exists z ∈ L such that B(z, x) = 1. Therefore,dim(x⊥) = dim(L)− 1. Similarly, dim(y⊥) = dim(L)− 1.

Suppose on the contrary that

Q(L \ (x⊥ ∪ y⊥)) = 0.

Let t ∈ x⊥ ∩ y⊥ and s ∈ L \ (x⊥ ∪ y⊥). Then at+ s 6∈ x⊥ ∪ y⊥ for all a ∈ F. Therefore,

Q(at+ s) = a2Q(t) + 2aB(t, s) +Q(s) = 0, for all a ∈ F.

Since p > 2, this implies that

Q(t) = B(t, s) = Q(s) = 0 ∀t ∈ x⊥ ∩ y⊥ and ∀s ∈ L \ (x⊥ ∪ y⊥). (∗∗)

Note that w ∈ x⊥ ∩ y⊥ and the set L \ (x⊥ ∪ y⊥) spans the whole space L. Therefore,B(w, L) = 0 and then x⊥ = y⊥. Together with (∗∗) we deduce that Q(x⊥) = Q(L\ x⊥) = 0,and hence Q(L) = 0 which is impossible.2

Theorem 11.4 Suppose that p > 2 and that L is a unimodular lattice. If M and N areisometric primitive sublattices of L and τ : M → N is an isometry, then there existsσ ∈ O(L) such that σ|M = τ . In particular, σ(M) = N .

Proof. We proceed by an induction on the rank of M . If the rank of M is 1, then thetheorem is a consequence of Lemma 11.3. Suppose now that the rank of M is at least 2. Ifthere exists x ∈M with Q(x) ∈ Z×p , then

M = Zpx ⊥M ′, L = Zpx ⊥ L′,

where L′ is unimodular. By Lemma 11.3, there exists an isometry of L which sends x toτ(x). So, we may assume that x = τ(x), and hence

N = Zpx ⊥ N ′.

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Both M ′ and N ′ are primitive sublattices of L′. Moreover, by Theorem 6.4, M ′ and N ′

are isometric. By the induction assumption, there exists an isometry φ ∈ O(L′) such thatφ|M ′ = τ |M ′ . Then σ = 1 ⊥ φ is an isometry of L and φ|M = τ .

Now, suppose that Q(M) ⊆ (p). Let x1, . . . , xn be an orthogonal basis for M . SinceM is primitive in L, there exists z ∈ L such that B(z, x1) = 1 and B(z, xi) = 0 for all i ≥ 2.Then the binary lattice H = Zpx1 + Zpz is a hyperbolic plane. Therefore,

L = H ⊥ L1,

where L1 is unimodular. Note that the lattice M1 = Zpx2 + · · · + Zpxn is a primitivesublattice of L1. As is in the previous paragraph, we may assume that x1 = τ(x1). ThenN1 := Zpτ(x2) + · · · + Zpτ(xn) is also a primitive sublattice of L1, and N1 is isometric toM1. By Lemma 11.3, there exists an isometric ρ ∈ O(L1) such that ρ|M1 = τ |M1 . Thenσ = 1 ⊥ ρ, which is an isometry of L, is the desired isometry.2

Remark 11.5 When p = 2, Theorem 11.4 holds if L is an even unimodular Z2-lattice.

11.2 Main Result

From now on, let L be a Z-lattice on a positive definite quadratic space V over Q. LetQ(gen(L)) be the set of rational numbers represented by L. For any positive integer r, let

Qr(gen(L)) = n ∈ Q(gen(L)) : ordp(n) ≤ r for all p for which Vp is anisotropic

andQr(L) = Q(L) ∩Qr(gen(L)).

If dim(V ) ≥ 5, then Q(gen(L)) = Qr(gen(L)) for every r, since Vp is isotropic for everyprime p. If dim(V ) = 4, the above definitions make sense because there are only finitelymany p for which Vp is anisotropic. We will prove the following theorem which has Theorem11.1 as a consequence.

Theorem 11.6 Let L be a positive definite Z-lattice of rank ≥ 4. Then Qr(gen(L))\Qr(L)is finite.

Lemma 11.7 Let a be a positive rational number and M be a positive definite Z-lattice ofrank ≥ 3. Suppose that the genus of M contains only one spinor genus. Let q be a primefor which Mq is isotropic. Then there exists s ∈ Z such that

(a) Q(gen(qsM)) ⊆ Q(M);

(b) Qr(gen(qsM ⊥ 〈a〉)) \Q(M ⊥ 〈a〉) is finite.

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Proof. (a) By Theorem 8.4, Z[1/q]M ∼= Z[1/q]K for every K ∈ spn+(M). Therefore, thereexists s ∈ Z such that qsK ⊆M and hence

Q(K) ⊆ 1

q2sQ(M).

This s can be chosen to be independent of K because there are only finitely many classesin spn+(L). Let n ∈ Q(gen(qsM)). Then q−2sn is represented by some K ∈ gen(M) =spn+(M). So n ∈ Q(M).(b) For our convenience, let L′ be the lattice qsM ⊥ 〈a〉. Let S be the set containing all theprime divisors of 2d(L′). Note that L′` is unimodular for all ` 6∈ S. For each prime p, let

Qr(L′p) = t ∈ Q(L′p) : ordp(t) ≤ r if L′p is anisotropic.

Let p ∈ S. Suppose that t ∈ Qr(L′p) and t = Q(v) for some v ∈ qsMp. If t 6= 0, then for

a sufficiently large integer k, t− ap2k = tε2 for some ε ∈ Z×p . So, t = Q(εv) + ap2k. If t = 0and v is an isotropic vector in qsMp, then there exist integer m ≥ 0 and w ∈ qsMp suchthat Q(w) = −ap2m. So, t = Q(w) + ap2m. If t = 0 but qsMp is anisotropic, then L′p mustbe isotropic. So, there exists v′ ∈ qsMp and nonzero u ∈ Zp such that t = Q(v′) + au2.

In summary, for each t ∈ Qr(L′p), there exists u ∈ Zp, u 6= 0, such that t ∈ Q(qsMp) +au2. Since the set Qr(L

′p) is compact and Q(qsMp) is open, there is a finite set Ip of nonzero

p-adic integers such that

Qr(L′p) ⊆

⋃up∈Ip

(Q(qsMp) + au2p).

Therefore, the product ∏p∈S

Q(L′p)

is contained in the union of the sets∏p∈S

(Q(qsMp) + au2p)

where up ∈ Ip for each p ∈ S. By the Chinese Remainder Theorem, we can find a finite setof nonzero integers J such that for each collection up : p ∈ S there exists u ∈ J with∏

p∈S(Q(qsMp) + au2

p) =∏p∈S

(Q(qsMp) + au2).

Suppose that t ∈ Q(gen(L′)) and that t ≥ au2 for all u ∈ J . Then for each p ∈ S, thereexists u ∈ J such that

t− au2 ∈ Q(qsMp).

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For each ` 6∈ S, t− au2 ∈ Z` since L′` is unimodular and u ∈ Z`. But qsM` is a unimodularlattice of rank ≥ 3 and ` is odd. Therefore, Q(qsM`) = Z` and hence t − au2 ∈ Q(qsM`)for all ` 6∈ S. Consequently, it follows from part (a) that

t− au2 ∈ Q(gen(qsM)) ⊆ Q(M).

As a result, t ∈ Q(M ⊥ 〈a〉).2

Proof of Theorem 11.6. Let L be a Z-lattice on a positive definite quadratic space V ofdimension at least 4 over Q. Choose a prime q - 2d(L) such that the Witt index of Vq isat least 2. When dim(V ) ≥ 5, q can be chosen to be any prime not dividing 2d(L). Whendim(V ) = 4, Dirichlet’s theorem implies that there are infinitely many primes ` which arecongruent to 1 mod p3 for all p | 2d(L). By the Quadratic Reciprocity Law, d(L) is a squaremod `. We then choose q to be one of these `.

Let p ∈ S. For any α ∈ Q(Lp), α ∈ Q(Zpu) for some u ∈ Lp with |Q(u)|p maximal.Since Q(Lp) is compact and each Q(Zpu) is open, there exists a finite subset Ip of Lp suchthat

Q(Lp) =⋃up∈Ip

Q(Zpup).

Let up ∈ Ip : p ∈ S be given. Choose t ∈ Q with the following properties:

(1) t ∈ Z×2p Q(up) for all p ∈ S;

(2) the prime factors of the denominator of t are in S;

(3) if p 6∈ S, then t is either a prime or a unit in Zp.

We can construct such a t as follows. Let Q(up) = papεp, where εp ∈ Z×p . By the ChineseRemainder Theorem, there exists an integer k such that

k∏`∈S\p

`a` ≡ εp mod p3, for all p ∈ S.

By Dirichlet’s theorem there exists a prime p 6∈ S such that

p ≡ k mod∏`∈S

`3

Thenp∏`∈S

`a` ∈ papεpZ×2p = Q(up)Z×2

p .

We then sett = p

∏`∈S

`a`

and we are done.

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From (1), it is clear that t ∈ Q(Lp) for all p ∈ S. For any p 6∈ S, p is odd and Lpis unimodular; hence t ∈ Q(Lp). Consequently, t ∈ Q(gen(L)). It follows form Theorem7.21 that t ∈ Q(spn+(L)). In other words, there exists K ∈ spn+(L) such that t ∈ Q(K).Let ` be a prime not in S such that ` - t and the Witt index of V` is at least 2. ThenZ[1/`]K ∼= Z[1/`]L by Theorem 8.4, and hence there exists v ∈ V such that Q(v) = t andv ∈ Lp for all p 6= `.

Choose an integer e ≥ 0 such that `ev is a primitive vector in L. We shall construct asublattice M of L of rank rank(L)− 1. For the sake of convenience, let J be the rank onelattice spanned by `ev. We construct M by specifying its localization Mp for each prime p.Let W be the orthogonal complement of QJ in V .

If p 6∈ S, p 6= ` and p - t: since Jp is unimodular, it splits Lp. We let M(p) be the orthogonalcomplement of Jp in Lp. Then θ(O+(M(p))) ⊇ Z×p and Q(qsM(p) ⊥ Lp) = Zp becauseqsM(p) is unimodular of rank ≥ 3.

If p ∈ S: let L′(p) be a maximal lattice which contains Lp∩Wp. Then paL′(p) ⊆ Lp∩Wp forsome a ≥ 0. LetM(p) be the Zp-lattice paL′(p). SinceM(p) is maximal, θ(O+(M(p))) ⊇ Z×pby Proposition 11.2. Also, Q(qsM(p) ⊥ Jp) ⊇ Q(Jp) = Q(Zpup).

If p = `: since the Witt index of V` is at least 2 and L` is unimodular, L` = H1 ⊥ H2 ⊥ K,where H1 and H2 are hyperbolic planes and K is a unimodular Z`-lattice. Also, since `evis primitive in L`, there exists z ∈ L` such that B(z, `ev) = 1. Let G be the binary Z`lattice spanned by z and `ev. If Q(`ev) is not a unit, then G is a hyperbolic plane andhence L` = G ⊥ H with H ∼= H2 ⊥ K. In this case, we let M(`) = I ⊥ H where I is theorthogonal complement of J` in G.

If Q(`ev) is a unit, then J` splits L`. Therefore, L` = J` ⊥ E = H1 ⊥ H2 ⊥ K, whereE is a unimodular Z`-lattice. Let w be a primitive vector of H1 such that Q(w) = Q(`ev).By Lemma 11.3, there exists an isometry φ ∈ O(L`) such that φ(w) = `ev. Hence

E ∼= H2 ⊥ K ⊥ ( orthogonal complement of w in H1 ).

In this case, let M(`) be E.In any case, θ(O+(M(`))) ⊇ Zp since M(`) has a rank 2 unimodular component. Also,

Q(qsM(`)) ⊇ Q(qsM(`)) = Z`.

If p 6∈ S and p | t: by the construction of t, we see that p2 - t. Therefore, v is primitivein Lp. Let z ∈ Lp such that B(z, v) = 1. Then the binary sublattice spanned by v and z,denoted H, is a hyperbolic plane. So, Lp = H ⊥ G for some unimodular Zp-lattice G. Also,H has a vector w satisfying B(w, v) = 0 and Q(w) = −Q(v) = −t. Let M(p) be Zpw ⊥ G.Then θ(O+(M(p))) ⊇ Z×p and Q(qsM(p) ⊥ Jp) = Zp.

Note that or almost all primes p, M(p) is the orthogonal complement of Jp in Lp. ByTheorem 7.8, there exists a Z-lattice M on W such that Mp = M(p) for all p. Sinceθ(O+(M(p))) ⊇ Z×p for all p, the genus of M contains only one spinor genus. Note that alltogether we have constructed only finitely many t and M .

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Let α ∈ Qr(gen(L)). Then at each p ∈ S, α ∈ Q(Zpup) for some up ∈ Jp. For p 6∈ S,Lp is unimodular; hence α ∈ Zp. Let t, J and M be as constructed as above. Thenα ∈ Qr(gen(qsM ⊥ J)). By Lemma 11.6, all but finitely many α are in Q(L).2

12 Classes and Genera of Representations

Throughout this section, R is either Z or Zp for some prime p, and F is the field of fractionsof R. An R-lattice L is said to be integral if s(L) ⊆ R.

12.1 Representations of Lattices

Definition 12.1 Let L and M be R-lattices. A representation of L by M is an injectiveR-homomorphism σ : L→M such that Q(φ(v)) = Q(v) for all v ∈ L.

This generalizes the representation of an element in R by a lattice. For, let a ∈ R, L bethe rank 1 R-lattice 〈a〉, and M be an R-lattice. Then having a representation of L by Mis equivalent to the existence of a vector x ∈M with Q(x) = a.

Definition 12.2 Two representations φi : L → Mi (i = 1, 2) of R-lattices are in the sameclass if there exists an isometry σ : M1 →M2 such that φ2 = σφ1.

Proposition 12.3 Suppose that L and M are nondegenerate R-lattices. If φ : L → M isa representation, then there exists an R-lattice M ′ such that L ⊆ M ′ and φ is in the sameclass of the inclusion L →M ′.

Proof. Let U be a quadratic space which is isometric to φ(FL)⊥, and let τ : U → φ(FL)⊥

be an isometry. Then φ ⊥ τ is an isometry from FL ⊥ U to FM . Let M ′ be (φ ⊥ τ)−1(M).Then L ⊆M ′ and φ is in the same class of the inclusion L →M ′. 2

Lemma 12.4 Suppose that L and M are unimodular Zp-lattices, p > 2. Then there is onlyat most one class of representations of L by M .

Proof. Let φi : L→M , i = 1, 2, be two representations of L by M . Then φ1(L) and φ2(L)are two primitive sublattices of M . Let τ : φ1(L) → φ2(L) be an isometry. By Theorem11.4, there exists an isometry σ ∈ O(M) such that σ|φ1(L) = τ . Then σφ1 = φ2, hence φ1

and φ2 are in the same class.2

Lemma 12.5 Suppose that L and M are nondegenerate R-lattices with L ⊆ M . Let Nbe the orthogonal complement of L in M . Then d(N) | d(M)d(L). In fact, N#/N is ahomomorphic image of a subgroup of M#/M ⊕ L#/L.

Proof. This is just Lemma 7.2. Since M/N is torsion free, N is a primitive sublattice ofM . So, the restriction map induces a surjective homomorphism

M#/(L ⊥ N)→ N#/N.

On the other hand, the map M#/(L ⊥ N) −→ L#/L⊕M#/M which sends x to (P (x), x)where P is the orthogonal projection from FM onto FL, is injective.2

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Theorem 12.6 Suppose that L is a nondegenerate R-lattice. Given any positive integer nand element d ∈ R \ 0, there are only finitely many classes of representations φ : L→Mwith rank(M) = n and d(M) = d.

Proof. Let φ : L → M be one of such representations. If N is the orthogonal complementof φ(L) in M , then

φ(L) ⊥ N ⊆M ⊆M# ⊆ (φ(L) ⊥ N)#.

By Lemma 12.5, d(N) divides d(M)d(L). Therefore, there are only finitely many isometryclasses of N . Let Ni be a complete set of representatives of these classes. For each i,there are only finitely many integral R-lattices Mij with rank(Mij) = n, d(Mij) = d, and

L ⊥ Ni ⊆Mij ⊆ (L ⊥ Ni)#.

Since there are only finitely many Ni, the set Mij is finite. Let φij : L → Mij be therepresentation induced by the inclusion L ⊥ Ni →Mij . We claim that every representationof L by some lattice M with rank(M) = n and d(M) = d is in the class of some φij .

Let ρ : L→M be such a representation. Let N be the orthogonal complement of ρ(L)in M . Then N must be isometric to Ni for some i. Let τi : N → Ni be an isometry, and σbe the isometry φ−1 ⊥ τi from φ(L) ⊥ N to L ⊥ Ni. Then σ((φ(L) ⊥ N)#) = (L ⊥ Ni)

#,and hence σ(M) = Mij for some j. It is immediate that σρ = φij .2

The above theorem is false if d(L) = 0. Consider a Z-lattice L = Zx ∼= 〈0〉, and ahyperbolic plane M with a basis e, f whose associated symmetric matrix is ( 0 1

1 0 ). Thenx 7→ nf , n ∈ Z \ 0, are all different representations. The orthogonal group of M is clearlyfinite because ±e,±f are the only primitive isotropic vectors in M and any isometry ofL must send e (or f) to a primitive isotropic vector.

Lemma 12.7 Let M1 and M2 be two R-lattices on a quadratic space over F . Then [O(Mi) :O(M1) ∩O(M2)] is finite for i = 1, 2.

Proof. There exist two nonzero elements a and b in F such that bM1 ⊆ M2 ⊆ aM1. Wehave a natural homomorphism

O(M1)→ AutZ(aM1/bM1).

Let G be the kernel of this homomorphism. Then [O(M1) : G] is finite since aM1/bM1 is afinite group. The elements of G acts trivially on M2/bM1, hence G ⊆ O(M2). as a result,G ⊆ O(M1) ⊆ O(M2) and thus [O(M1) : O(M1)∩O(M2)], which is small than [O(M1) : G],is finite. Similarly, [O(M2) : O(M1) ∩O(M2)] is also finite.2

Theorem 12.8 Let L be a nondegenerate Z-lattice. If M is a positive definite Z-lattice orQM is the hyperbolic plane, then there are only finitely many representations of L by M .

Proof. Since there are only finitely many classes of representations of L by M , it suffices toshow that O(M) is finite when M is positive definite or QM is the hyperbolic plane.

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Suppose that M is positive definite. Let e1, . . . , en be an orthogonal basis for QMwith Q(ei) ≤ Q(ei+1) for all i, and let M ′ = Ze1 ⊥ · · · ⊥ Zen. Then |O(M ′)| is finite. ByLemma 12.7, |O(M)| is finite.

If QM is the hyperbolic plane, let M ′ be a lattice on QM with symmetric matrix ( 0 11 0 ).

Then |O(M ′)|, and hence |O(M)| as well, is finite.2

Remark 12.9 It can be shown that if M is an indefinite Z-lattice and QM is not thehyperbolic plane, then O(M) is an infinite group.

12.2 Genus of Representations

Definition 12.10 Two representations φ : L → M and φ′ : L → M ′ of Z-lattices are inthe same genus if there are isometries σp : Mp →M ′p for all primes p such that φ′ = σpφ forall p.

For a prime p and a pair of Z-lattices L and M , let c(Lp,Mp) be the number of classesof representations of Lp by Mp. By Theorem 11.4, c(Lp,Mp) is equal to 1 for almost all p.

Theorem 12.11 Let L and M be nondegenerate Z-lattices. The number of genera of rep-resentations of L by lattices in gen(M) is equal to

∏p c(Lp,Mp).

Proof. Let G be the set of all genera of representations of L by lattices in gen(M). For each p,let Cp be the set of classes of representations of Lp byMp. Let φ : L→M ′ be a representationwith M ′ ∈ gen(M). For each prime p, φ induces representations φ : Lp → M ′p. Letσp : M ′p → Mp be an isometry. Then σpφ : Lp → Mp is a representation. Different choicesof the isometry σp result in different representations of Lp by Mp that are in the same class.So we have a well-defined map

Φ : G −→∏p

Cp.

Let φ : L → K and τ : L → N be two representations of L by lattices in gen(M).Suppose that for each p, the classes of representations of Lp by Mp induced by φ and τ arethe same. Then for each p, there are isometries σp, ψp and ρp which make the followingdiagram commute

Lpφ−−−−→ Kp

σp−−−−→ MpyψpLp

τ−−−−→ Npρp−−−−→ Mp

Let ηp = ρ−1p ψpσp. Then τ = ηpφ. Therefore, φ and τ are in the same genus. This shows

that Φ is injective.Suppose that a representation φp : Lp →Mp is given for each p. Let T be a finite set of

primes satisfying

(a) c(Lp,Mp) = 1 for all p 6∈ T ;

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(b) Lp ⊆Mp for all p 6∈ T .

Let W and V be the Q-spaces spanned by L and M , respectively. By the Hasse Principle,there is an isometry sending W into V . So we may assume that L is a Z-lattice in V . Foreach p, it follows from Witt’s extension theorem that there exists an isometry σp ∈ O(Vp)such that σp|Lp = φp. Define a Z-lattice M ′ on V by specifying its localization as follows:

M ′p =

Mp for all p 6∈ T ;σ−1p (Mp) for all p ∈ T .

Then at each p ∈ T ,Lp = σ−1

p (φp(Lp)) ⊆M ′p.

Note that for each p 6∈ T , Lp ⊆ Mp = M ′p. Therefore, L ⊆ M ′. Let j : L → M ′ be theinclusion. Then for each p 6∈ T , φp and j are in the same class because c(Lp,Mp) = 1. Foreach p ∈ T , σp(φp(x)) = x for all x ∈ Lp; thus j = σ−1

p φp for each p ∈ T . Consequently, jand φp are in the same class for all p. This proves that Φ is surjective.2

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