Question about answer about quadratic forms on MO I have a question regarding this MO answer: 
The answer says that in characteristic $2$, we cannot obtain a quadratic form from a bilinear form. I thought it was the other way around and now I am confused. 
I thought that if $2 \neq 0$ then we have a bijection between quadratic $q$ and bilinear forms $b$, by getting $q(x) = b(x,x)$ from $b$ and $b(x,y) = \frac{1}{2}(q(x+y) - q(x) - q(y))$.
But if $2 = 0$, we cannot divide by $2$ hence we cannot go from quadratic forms to bilinear ones, but we can still go the other way.
Would you help me resolve my confusion? Thanks a lot.
 A: As you know, given a bilinear form $b(x,y)$, you can make a quadratic form by $q(x) = b(x,x)$. You can also make a bilinear form from a quadratic form $q(x)$ by setting $$b(x,y) = q(x+y) - q(x) - q(y).$$ Both of these operations work in arbitrary characteristic and are invariant under coordinate changes. However, they are not inverses of each other. You can easily check that a round trip of either bilinear to quadratic to bilinear or quadratic to bilinear to quadratic will introduce a factor of 2. Thus, if 2 is invertible, you can introduce a factor of one half into the definition of the bilinear form and get a bijection between quadratic and bilinear forms. On the other hand, if the characteristic is 2 then you can still convert in either direction, but both directions are neither surjective nor injective.
In the linked post, I understand the author's statement that "you can't go the other way around" not to mean that there's no way to convert a bilinear form to a quadratic form, but that there's no way to invert the map from quadratic forms to bilinear forms.
A: Given a bilinear form $b(x,y)$, the form $x \mapsto b(x,x)$ is a quadratic form. For example, the inner product on $\mathbb{F}_2^n$ given by $( \{x_i\}, \{y_i\}) = \sum x_i y_i$ leads to the quadratic form $\sum x_i^2$, just as in characteristic $\neq 2$. But to get the bilinear form back from the quadratic form (the "norm"), you need the polarization identity, which involves factors of two.
A: In some areas of mathematics (group theory in particular) a quadratic form for characteristic 2 will be defined in the following way $Q:V \to k$ with
1) $Q(av)=a^2Q(v)$ for all $a \in k$ and $v \in V$ and
2) the map defined by $b(x,y)=Q(x+y)-Q(x)-Q(y)$ is bilinear
admittedly the minus signs in 2 are a little suprifulous but are meant to make it "work" in odd characteristic as well
