Let $V$ be a finite dimensional vector space and let $S^d(V)$ be the degree $d$ part of the symmetric algebra on $V$.
Is it true that $S^d(V^*)^*$ is canonically isomorphic to $S^d(V)$?
What (I think) I know: $S^d(V^*)$ are homogeneous polynomials of degree $d$ on $n=\dim{V}$ variables with coefficients in the base field $k$, because a basis of $V^*$ is given by the $n$ coordinate functions $x_i$ which are the dual basis of any fixed basis for $V$. So the symmetric algebra $S(V^*)$ really is the polynomial ring $k[x_1,...,x_n]$, right? (Remark: in this description of the symmetric algebra I had to make a choice, namely fixing a basis of $V$.)
What I don't know: I think I don't really understand $S(V)$. Should I just think of it as $k[e_1,...,e_n]$, regarding the $e_i$'s just as formal symbols?
In any case, for both descriptions I had to choose a basis. So I was hoping that there may be a canonical isomorphism as in the case of $V\cong (V^*)^*$. Is this true?