Hasse Invariant for Integral Quadratic Forms

Question

On Page 50 of Serre's "A Couse in Arithmetic", it is mentioned that a non-degenerate quadratic form over $Z_p$ can be splitted into an orthogonal direct sum of $Z_p$-modules of rank $1$ for $p>2$ and of rank no greater than $2$ for $p=2$. I had some difficulties in finding a proof for this. Could anyone please provide a concise proof? Furthermore, how do we prove the form of the Hasse invariant for $p=2$ after the decomposition process? Serre mentioned a paper by Cassels but I could not find an English version. Thanks in advance!

Answer 1

Given a quadratic form $q(u)=b(u,u)$ with $b$ a symmetric bilinear form $U\to \Bbb{Q}_p$ where $U$ is a free $\Bbb{Z}_p$-module (the $p$-adic integers) of rank $r$ and $p\ne 2$.

Take $w\in U$ such that $q(w)$ is of minimal $p$-adic valuation. Let $$M=\{ u\in U, b(w,u)=0\}$$

For $z\in U$, $b(w,z) = \frac12 (q(w+z)-q(w)-q(z))$ so $v_p(b(w,z))\ge v_p(q(w))$.

Whence $$z = \frac{b(w,z)}{q(w)}w+(z-\frac{b(w,z)}{q(w)}w)\in \Bbb{Z}_p w \oplus M$$ where $\oplus$ is a direct sum of $\Bbb{Z}_p$-modules and quadratic spaces, as $$\forall \alpha \in \Bbb{Z}_p, \forall m\in M,\qquad q(\alpha w+m)=q(\alpha w)+q(m)$$

$M$ is a finitely generated sub-module of $U$ so it is free, of rank $r-1$.

Repeating with $M$ instead of $U$ until $q$ is identically zero on $M$ gives a $\Bbb{Z}_p$-basis $w_1,\ldots,w_r$ of $U$ such that $$q(\sum_{j=1}^r \alpha_j w_j)=\sum_{j=1}^r \alpha_j^2 q( w_j)$$