TITLE: What's special about characteristic 2?
QUESTION [8 upvotes]: I'm trying to get the big picture of how bilinear forms and quadratic forms relate over fields $F$ with $char(F) = 2$ and fields with $char(F) \neq 2$. 
What I gather so far is that if $char(F) \neq 2$ then there is a one-to-one correspondence between quadratic forms and bilinear forms so that the corresponding theories are equivalent. However, if $char(F) = 2$, while it's still possible to define a bilinear form in terms of a quadratic form it's no longer possible to recover the quadratic form from it.  
Is this irreversibility what distinguishes the case $char(F) = 2$ from $char(F) \neq 2$ or is there a more fundamental difference that I'm not aware of?

REPLY [6 votes]: The correct statement should be: when characteristic $\neq2$, there is a one-to-one correspondence between quadratic forms and symmetric bilinear forms. We can recover the symmetric bilinear form from a quadratic form as follows:
$$
b(\mathbf{x},\mathbf{y}) = \frac12\left(
q(\mathbf{x}+\mathbf{y}) - q(\mathbf{x}) - q(\mathbf{y})
\right).
$$
With characteristic $2$, the factor $\frac12$ does not exist. In fact, if we write $b(\mathbf{x},\mathbf{y})=\mathbf{y}^TB\mathbf{x}$, then any cross-term in $b(\mathbf{x},\mathbf{x})$ must vanish because $b_{ij}x_ix_j+b_{ji}x_jx_i=2b_{ij}x_ix_j=0$. Hence $b(\mathbf{x},\mathbf{x})$ is necessarily equal to $\sum_i b_{ii}x_i^2$. In other words, with characteristic 2, no symmetric bilinear form can induce a quadratic form with nonvanishing cross terms.