# Arithmetic of Quadratic Forms, Edition: version 26 Feb 2016 by Wai Kiu Chan

By Wai Kiu Chan

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**Sample text**

An ) is not 1, then d−1 e 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 exists x ∈ L, but not in Re, and n = 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 (a1 x1 + · · · + an xn ) n which implies that n divides gcd(a1 , . . , an ). In particular, gcd(a1 , . . 4 A fractional ideal of F is a subset of F of the form (a) = {ar : r ∈ R} for some a ∈ F × .

Xn ) be the quadratic form associated with this basis. Then a is represented by L if and only if the diophantine equation f (x1 , . . , xn ) = a has a solution (x1 , . . , xn ) ∈ Rn , not only in F n . The fundamental question is still the representation problem which asks for an effective determination of Q(L), the set of elements of F represented by L. Two lattices M and L are isometric, written M ∼ = L, if there exists an isometry σ : F M −→ F L such that σ(M ) = L. It is clear that isometric lattices represent the same set of elements of F .

In terms of symmetric matrix −9 1 0 L∼ = 1 7 0 . 0 0 2 Since d(L) = −27 , Lp is unimodular for all primes p ≥ 3. In particular, Lp represents 1 for 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 following elementary proof is due to D. Zagier. e. there exist integers x, y, z such that −9x2 + 2xy + 7y 2 + 2z 2 = 1. We may rewrite this equation as 2z 2 − 1 = (x − y)2 + 8(x − y)(x + y).