Isotropic quadratic forms

Question

Let $q$ be a quadratic form of dimension $n$ over a field $F$, and let $\lambda\in F$ be a scalar. Suppose $q^\prime=q\perp -\lambda q$ is isotropic. Does there exist a quadratic form $q_1$ of dimension $n-1$ such that $q^\prime\cong q_1\perp -\lambda q_1 \perp \mathbb{H}$, where $\mathbb{H}$ is a hyperbolic plane?

Answer 1

Even if you assume that $q$ is non-degenerate and $\lambda\in F^\times$, it does not work. Say that $q=\langle 1,\lambda\rangle$ where $\lambda\in F^\times$ is not a square. Then $q'=q\perp \langle -\lambda\rangle q$ is hyperbolic, but cannot be written $q'\simeq \langle a,-\lambda a\rangle \perp \mathbb{H}$ for any $a\in F^\times$ (as $\langle a,-\lambda a\rangle$ is not hyperbolic).

Answer 2

There is a theorem that can be stated. We suppose that $\langle 1, -\lambda \rangle q$ is isotropic. Then there is a subform $q_1$ of $q$ such that $\langle 1, -\lambda \rangle q \cong \langle 1, -\lambda \rangle q_1 \perp r \mathbb{H}$, where $\langle 1, -\lambda \rangle q_1$ is anisotropic and $r = \dim(q) - \dim(q_1)$.

Here is an outline of the proof with a reference for further details. Since $\langle 1, -\lambda \rangle q$ is isotropic, there exists a subform $\beta_1$ of $q$ with $\dim(\beta_1) = 2$ such that $\langle 1, -\lambda \rangle \beta_1$ is hyperbolic. That is, $\langle 1, -\lambda \rangle \beta_1 \cong 2 \mathbb{H}$. This follows from Proposition 2.2 in R. Elman, T. Y. Lam, Quadratic forms and the $u$-Invariant I, Math. Zeit., Volume 131 (1973), 283-304. This result is also stated in Lam's textbook "Introduction to Quadratic Forms over Fields" in Chapter XI, Exercise 11, page 421.

Write $q \cong \beta_1 \perp \gamma_1$.
If $\langle 1, -\lambda \rangle \gamma_1$ is isotropic, then repeat the argument. Eventually we can write $q \cong (\beta_1 \perp \cdots \perp \beta_r) \perp \gamma$ where each $\langle 1, -\lambda \rangle \beta_i$ is hyperbolic and $\langle 1, -\lambda \rangle \gamma$ is anisotropic. Let $q_1 = \gamma$ and let $\beta = \beta_1 \perp \cdots \perp \beta_r$. Then $\langle 1, - \lambda \rangle q \cong \langle 1, -\lambda \rangle q_1 \perp \langle 1, -\lambda \rangle \beta \cong \langle 1, -\lambda \rangle q_1 \perp r \mathbb{H}$. We have $r = \dim(\beta)$, which is even because each $\dim(\beta_i) = 2$.

The counterexample given above shows that we cannot always take $r = 1$. We now see that $r$ is always even.