Equivalence of exterior algebra definitions

Question

For a finite-dimensional vector space $V$, let $A$ be the subspace of $V\otimes V$ generated by all elements of the form $v\otimes v$ for some $v\in V$ and let $B$ be the subspace generated by all elements of the form $v\otimes w+w\otimes v$. I then learned that I can define $$V\wedge V:=V\otimes V/A$$ or $$V\wedge V:=V\otimes V/B$$ I was told that this equivalence only held when the field that $V$ is over is not of characteristic two. I believe the problem arises when one tries to show that $A\subset B$. Could someone shed light on this?

Answer 1

You're right that the problem arises when you try to show $A\subseteq B$.

To show $B\subseteq A$ we note that for arbitrary $v$ and $w$, the tensors $v\otimes v$,$w\otimes w$ and $(v+w)\otimes (v+w)$ are all in $A$. Since $A$ is a subspace, it also contains

$$(v+w)\otimes (v+w) -v\otimes v - w\otimes w=v\otimes w+w\otimes v,$$

and so $B\subseteq A$.

To show $A\subseteq B$ when the characteristic isn't $2$, note that for arbitrary $v$, $2v\otimes v$ is in $B$ (by setting $v=w$ in $v\otimes w+w\otimes v$). So when $2$ is invertible we have that $v\otimes v$ is in $B$.

In characteristic $2$ we in fact have that $v\otimes v$ is never in $B$ when $v$ is nonzero. To see this, pick some covector $\theta$ such that $\theta(v)=1$. Then consider the map $\theta\otimes\theta$. It sends any vector of the form $u\otimes w+w\otimes u$ to $\theta(u)\theta(w)+\theta(w)\theta(u)=2\theta(u)\theta(w)=0$, and so it sends all of $B$ to $0$. But $(\theta\otimes\theta)(v\otimes v)=\theta(v)\theta(v)=1$.