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...)



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.


