Why in this theorem the field of characteristic equal to $2$ excluded?

Question

Theorem:

Let $V$ be a vector space over $\mathbb{F}$ (where it is not of characteristic two). Then, there is a one-to-one correspondence between symmetric bilinear forms and quadratic forms.

I already proved that. Why here do we exclude the field of characteristic equal to two? What happens in that case? Thanks.

Answer 1

The standard proof is this: if $B$ is a symmetric bilinear form and you define $q(v)=B(v,v)$, then you get a quadratic form. And how can you get $B$ from $q$ (in order to prove that this correspondence is one-to-one)? It turns out that$$B(v,w)=\frac12\bigl(B(v+w,v+w)-B(v,v)-B(w,w)\bigr).$$Problem: you can't divide by $2$ in a field whose characteristic is $2$, because in such a field $2=1+1=0$.