I'm studying quadratic forms and I have to prove that every quadratic module $(V,Q)$ has an orthogonal basis, where $V$ is a vectorial space on $\mathbb{K}$ and $Q$ a quadratic form on $V$. I'm following Serre's proof in "A course in Arithmetic".
Assuming that $dim(V) \ge 1$, let $x$ be a non isotropic element of $V$ and let $H$ be the orthogonal complement of the subspace $\mathbb{K}x$. Then $Q$ is non degenerate on $\mathbb{K}x$ so $x \notin H$ and $\mathbb{K}x \cap H = \{0\}$. I need to say that $V=\mathbb{K}x \oplus H$, and I know this is true when $Q$ is nondegenerate on $V$ and on $\mathbb{K}x$, but in this case I don't assume that $V$ is non degenerate. So how can I say that $dim(V)=1+dim(H)$?
Thanks for help!
you can first suppose that $Q$ is degenerate. Then you can write $V=V^0 \hat{\oplus} U$ where $U$ is a supplementary subspace of $V^0$. By induction you know that $U$ have an orthogonal basis and $V^0$ also (by definition). The union of both is an orthogonal basis of $V$.
Then you can prove the case with $Q$ nondegenerate :) .