Facts about alternating bilinear maps on vector spaces

Let $F$ be a field, let $V$ be an $n$-dimensional $F$-vector space, and let $f : V \times V \rightarrow W$ be a bilinear map, where $W$ is an $F$-vector space.

1. Prove that if $f$ is an alternating $F$-bilinear map on $V$ then $f(x,y) = -f(y,x)$ for all $x,y \in V$.
2. Suppose $\mathsf{char}\ F \neq 2$. Prove that $f(x,y)$ is an alternating bilinear map on $V$ if and only if $f(x,y) = -f(y,x)$ for all $x,y \in V$.
3. Suppose $\mathsf{char}\ F = 2$. Prove that every alternating bilinear form $f(x,y)$ on $V$ is symmetric. (I.e. $f(x,y) = f(y,x)$ for all $x,y \in V$.) Prove that there exist symmetric bilinear maps which are not alternating.

Let $x,y \in V$, and suppose $f$ is an alternating bilinear map. Now $0 = f(x-y,x-y)$ $= f(x,x) - f(x,y) - f(y,x) + f(y,y)$, so that $f(x,y) = -f(x,y)$.

Suppose $\mathsf{char}\ F \neq 2$; in particular, 2 is a unit in $F$. If $f$ is bilinear such that $f(x,y) = -f(y,x)$ for all $x,y \in V$, then in particular we have $f(x,x) = -f(x,x)$, so that $2f(x,x) = 0$. Thus $f(x,x) = 0$ for all $x \in V$, and so $f$ is alternating. Conversely, if $f$ is alternating then by part (1) above we have $f(x,y) = -f(y,x)$ for all $x, y \in V$.

Now suppose $\mathsf{char}\ F = 2$. Note that $(x+y) \otimes (x+y) = x \otimes x - x \otimes y + y \otimes x - y \otimes y$. Mod $\mathcal{A}^2(V)$, we have $x \otimes y - y \otimes x = 0$. In particular, the submodule $\mathcal{C}^2(V)$ generated by all tensors of the form $x \otimes y - y \otimes x$ is contained in $\mathcal{A}^2(V)$.

We have already seen that $\mathsf{dim}_F\ \mathcal{S}^2(V) = {{n+1} \choose {n-1}} = \frac{n(n+1)}{2}$ and $\mathsf{dim}_F\ \bigwedge^2(V) = {n \choose 2} = \frac{n(n-1)}{2}$. Thus the containment $\mathcal{C}^2(V) \subsetneq \mathcal{A}^2(V)$ is proper.