Quadratic forms over finite fields I'm reading some very old papers (by Birch et al) on quadratic forms and I don't get the following point: 

If $f$ is a quadratic form in $X_1,X_2,\cdots,X_n$ over a
  finite field, then one can change variables such that $f$ can be written as $\sum_{i = 1}^s Y_{2i - 1}Y_{2i} + g$, where $g$ is a quadratic form which involves
  variables other than
  $Y_1,Y_2,\cdots,Y_{2s}$ and has order at
  most 2 (i.e. can be written using at
  most two linear forms).

So either this is a well-known result - but I don't find a reference - or either this is easy to see, but in that case I'm just missing the point. By the way, is this really true in characteristic 2?  
(And in fact I'm not sure what role is played by the fact that the field is finite...)
 A: 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://alpha.math.uga.edu/~pete/4400ChevalleyWarning.pdf
For Witt Decomposition:
http://alpha.math.uga.edu/~pete/quadraticforms.pdf
A: 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.
