New PDF release: Arithmetic of Quadratic Forms

By Wai Kiu Chan

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Extra resources for Arithmetic of Quadratic Forms

Example text

Proof. 9. 11. Suppose that L is anisotropic. Let x be a basis vector of L. Since L is unimodular, there exists y ∈ L = L# such that B(x, y) = 1. Then {x, y} must be a basis for L and the corresponding symmetric matrix for L is 2a 1 . 1 2b 46 If a or b is divisible by 2, then d(L) = 4ab − 1 ≡ −1 mod 8. So L is isotropic which is a contradiction. 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 spaces over Q2 , we see that 2 1 V ∼ , = 1 2 2 1 2 1 .

Since p > 2, d(L1 ) = d(K1 ) ×2 in Z× p /Zp . ✷ The above theorem does not hold when p = 2. For example, consider the Z2 -lattice L with Jordan decomposition 2 1 L∼ ⊥ −2 . = 1 2 Suppose that {x, y, z} is a basis which gives the symmetry matrix on the right. Then the vectors {x + z, y + z} spans a sublattices which is isometric to H. Therefore, L has another Jordan decomposition 0 1 L∼ ⊥ 6. = 1 0 Obviously, the leading components of the two Jordan decompositions are not isometric. 5 Let L = L1 ⊥ · · · ⊥ Lt = K1 ⊥ · · · ⊥ Ks be two Jordan decompositions of a Z2 -lattice L.

Then there exists a nondegenerate lattice K such that L = K ⊥ rad(L). Proof. It is clear that if K exists then it must be nondegenerate. It remains to demonstrate the 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 that rv ∈ rad(L). 8 In the decomposition L = K ⊥ rad(L), the lattice K is not unique, but its isometry 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 )).

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Arithmetic of Quadratic Forms by Wai Kiu Chan

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