There is this claim in Scharlau's "Quadratic and Hermitian forms",
Every 4-dimensional isotropic quadratic space of determinant 1 is hyperbolic.
How can we prove it? I know that any 2-dimensional isotropic quadratic space is hyperbolic, I understand the proof but I don't see how I could extend the result to four dimensions.
I got it, after some help from my teacher. It's not hard, I was just missing on a couple of elementary results that are useful.
Let $(V,q)$ be such an $F$-quadratic space. As it is isotropic, then $V$ contains a hyperbolic plane, and as hyperbolic planes are non-degenerate, it admits an orthogonal complement. So, $$q\simeq \langle 1,-1,e,f\rangle$$ for some $e,f\in F^*$.
Comparing (square classes of) determinants, we get that $-ef=\alpha^2$ for some $\alpha\in F^*$. Dividing the last component of the diagonalization of $q$ as above by $\alpha^2$ and then multiplying it by $e^2$, we get $$q\simeq \langle 1,-1,e,-e \rangle \simeq 2\mathbb H$$ as $\langle \omega,-\omega\rangle \simeq \mathbb H$ for any $\omega\in F^*$. Thus the assertion is proved.