Quadratic forms over finite fields

I will sketch below a standard argument to show what you need, because I find it very neat!

Let $V$ be a finite dimensional vector space over a field $k$ and let $q \colon V \to k$ be a quadratic form on $V$. Denote by $b$ the symmetric bilinear form associated to $q$: thus for vectors $v,w \in V$ define $b(v,w) := q(v+w)-q(v)-q(w)$. Suppose that $v$ is a non-singular zero of $q$. Since $v$ is non-singular, it follows that there is a $w' \in W$ such that $\alpha := b (v , w') \neq 0$. Let $w := \frac{1}{\alpha^2} (\alpha w' - q(w') v)$; it is immediate to check that $q(w)=0$ and $b (v , w) = 1$. Observe that the ``orthogonal complement'' of $v,w$ with respect to the form $q$ has codimension two and does not contain the span of $v,w$. Thus, we conclude that we can find a basis of $V$ such that $q(x_1,\ldots,x_n) = x_1 x_2 + q'$, where $q'$ is a quadratic form over a space of dimension two less than the dimension of $V$.

The statement about finite fields follows at once, since over a finite field, any quadratic form in three or more variables admits a non-trivial zero. This is a consequence of the Chevalley-Warning Theorem. More generally, any field such that quadratic forms in three variables always admit a zero has the property you need, e.g. any $C_1$-field would work.

Let me try a similar explanation with different words. (Note that my explanation does not cover characteristic $2$.)

A quadratic form is nondegenerate if any of its associated symmetric matrices has nonzero determinant. (Alternately, if the associated bilinear form $B(x,y) = q(x+y) - q(x) - q(y)$ is nondegenerate in the usual sense: $B(x,y) = 0 \ \forall y \in K \implies x = 0$.)

Let $K$ be a field of characteristic different from $2$. The hyperbolic plane is the special quadratic form

H(x,y) = xy.

(As with any quadratic form over $K$, it can be diagonalized: $\frac{1}{2} x^2 - \frac{1}{2} y^2$.)

A nondegenerate quadratic form $q(x_1,\ldots,x_n)$ is isotropic if there exist $a_1,\ldots,a_n \in K$, not all $0$, such that $q(a_1,\ldots,a_n) = 0$ and otherwise anisotropic.

Witt Decomposition Theorem: Any quadratic form $q$ can be written as an orthogonal direct sum of an identically zero quadratic form, an anistropic quadratic form, and some number of hyperbolic planes. In particular, any isotropic quadratic form $q(x_1,...,x_n)$ can be written, after a linear change of variables, as $x_1 x_2 + q(x_3,...,x_n)$.

For your purposes, you might as well assume your quadratic form is nondegenerate -- otherwise, it simply involves more variables than actually appear!

Now over a finite field, the Chevalley-Warning theorem implies that any nondegenerate quadratic form in at least three variables is isotropic, so that by Witt Decomposition, you can split off a hyperbolic plane. If you still have at least three variables, you can do this again. Repeated application gives your result.

References:

For Chevalley-Warning:

http://math.uga.edu/~pete/4400ChevalleyWarning.pdf

For Witt Decomposition:

http://math.uga.edu/~pete/quadraticforms.pdf