By Wai Kiu Chan

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This implies that Q takes on both positive and negative rational numbers. The follow theorem is a consequence of the Strong Approximation Theorem. 2 Let L be a lattice on a nondegenerate quadratic space V over Q. If V is indefinite 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 ) = Lp }, which is a finite set. Let S be the set of primes p for which p ∈ T and Lp is not the Zp -lattice spanned by the basis for V that defines p .

51 Now, we can find a lattice J on V such that Jq = M (q). By the Invariant Factor Theorem of finitely generated modules over PID, there exists a basis e1 , . . , en for L and rational numbers r1 , . . , rn such that r1 e1 , . . , rn en is a basis for J. For any i, let ti be the q-part of ri . ✷ A rational number a is said to be represented by the genus of L if V∞ represents a and Lp represents a for every p. This is a necessary condition for a to be represented by L. 9 If a is represented by the genus of L, then a is represented by some lattice in gen(L).

14 it follows that Zp v is an orthogonal summand 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 Z2 u + Z2 v is s(L)-modular and therefore is an orthogonal summand of L. ✷ 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 is called a Jordan decomposition of L, and each Li is a Jordan component.

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