# Tensor, exterior, symmetric powers over fields of nonzero characteristic

I was reading Fulton and Harris' discussion of exterior and symmetric products as quotient spaces of tensor products in their rep theory book when I noticed that they made this claim (the emphasis is mine):

The exterior powers $\bigwedge^n V$ and symmetric powers $\operatorname{Sym}^n V$ can also be realized as subspaces of $V^{\otimes n}$, assuming, as we have throughout, that the ground field has characteristic 0.

Why must the ground field be characteristic zero for this to be true? In particular, I don't see any problem with the quotient construction of the exterior and symmetric products for a vector space with nonzero characteristic; if there is no problem with this construction, then the inclusion of these products into the tensor product ought to be well-defined as well. Am I missing something obvious here?

The usual construction of embedding the symmetric power into the tensor power involves dividing by $n!$. This can't always be done in positive characteristic (in fact there are only finitely many $n$ where it is valid).
• Thank you. It's pretty apparent there were some silly assumptions I made - in particular, composing the natural inclusion and the natural projection does not necessarily create a projection map into the image of the inclusion. Also, the projection map into the exterior and symmetric subspaces must be idempotent by definition, which necessitates division by $n!$. May 31 '15 at 23:52