By Wai Kiu Chan

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**Example text**

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 . Note that S is also a finite set. By the Strong Approximation Theorem, there exists σ ∈ O (V ) such that (i) for all p ∈ T ∪ S, σ p = 1; (ii) for all p ∈ T , σ and σp are sufficiently close so that σ(Lp ) = σp (Lp ); 58 (iii) for all p ∈ S, σ and 1Vp are sufficient close so that σ(Lp ) = Lp .

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 case τv+σ(v) τv (v) = σ(v). So in either case there exists ρ ∈ O(V ) which is a product of one or two symmetries such that ρ(v) = σ(v). Let W be the orthogonal complement of v. Then ρ−1 σ is in O(W ) and induction hypothesis implies that ρ−1 σ is a product of symmetries τv1 · · · τvk in O(W ). 16 The proof of the above proposition can be modified to show that if p > 2 and L is a modular Zp -lattice, then O(L) is also generated by symmetries.

Then Zp z + Zp y ⊆ L and so b/k ∈ (a), that is, (b/a) ⊆ rz ry . But rx is chosen to be maximal. Therefore, rx ry = (b/a) which implies that the scale of Zp kx + Zp k y is precisely (b/2). Hence Zp kx + Zp k y splits K. 13 Let L be an (a)-maximal lattice on a nondegenerate quadratic space V over Qp . Then m 0 a/2 ∼ L= ⊥ L0 a/2 0 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.