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# Is $Sym^n (V^*) \cong Sym^n (V)^\ast$ naturally in positive characteristic?

Background/motivation

It is a classical fact that we have a natural isomorphism $Sym^n (V^*) \cong Sym^n (V) ^\ast$ for vector spaces $V$ over a field $k$ of characteristic 0. One way to see this is the following.

On the one hand elements of $Sym^n (V^*)$ are symmetric powers of degree n of linear forms on $V$, so they can be identified with homogeneous polynomials of degree n on $V$. On the other hand elements of $Sym^n (V) ^\ast$ are linear functionals on $Sym^n V$; by the universal property of $Sym^n V$ these correspond to n-multilinear symmetric forms on $V$. The isomorphism is then as follows.

An n-multilinear symmetric form $\phi$ corresponds to the homogeneous polynomial $p(v) = \phi(v, \dots, v)$. In the other direction to a polynomial $p(v)$ we attach the multinear form obtained by polarization $\phi(v_1, \dots, v_n) = \frac{1}{n!}\sum_{I \subset [n]} (-1)^{n - \sharp I} p(\sum_{i \in I} v_i)$. Here $[n]$ is the set $\lbrace 1, \dots, n \rbrace$.

Problem

Of course this will not work for $n$ greater than the characteristic of $k$ if the latter is positive.

One can expect that an isomorphism $Sym^n (V^*) \cong Sym^n (V) ^\ast$ holds also in positive characteristic, and that this should be trivially true by using the universal properties of the symmetric powers. The problem is that if I try to define a natural map between the two spaces using the universal properties I have at some point to divide by $n!$ anyway.

Still there may be some natural isomorphism that I cannot see. Or maybe there is not a natural isomorphism, but I don't know how to prove this.

Is there a natural isomoprhism $Sym^n (V^*) \cong Sym^n (V)^\ast$ in positive characteristic?