Relationship between tensor product and wedge product

Over a real vector space $V$ one can regard the second exterior power $\Lambda^2 V$ as either a subspace of $V\otimes V$ (the space of antisymmetric tensors) or as a quotient space of $V\otimes V$ (by factoring out the subspace generated by the $v\otimes v$). Over $\Bbb R$ these definitions are equivalent; the equivalence class $[v\otimes w]$ corresponding to the tensor $\frac12(v\otimes w-w\otimes v)$.

But over a vector space of characteristic 2, or over a ring where $2$ is not invertible, these definitions can vary. I would say the right definition is as a quotient of $V\otimes V$. But let $V$ be a vector space of dimension $n$ over the field $\Bbb F_2$. Here antisymmetry is the same as symmetry, so the antisymmetric tensors in $V\otimes V$ have dimension $\frac12n(n+1)$. However the "right" definition as a quotient gives the dimension of $\Lambda^2(V)$ as $\frac12n(n-1)$.

Of course all these considerations extend to $\Lambda^k(V)$. Over a field of characteristic zero, it matters not which approach we take.