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?
Best Answer
The answer is no (and well-known to people working in the representation theory of algebraic groups in positive characteristic). In fact for $V$ finite dimensional and of dimension $>1$ the two vector spaces are not isomorphic as $GL(V)$-modules ($GL(V)$ is either considered naively as an abstract group when the field $k$ is infinite or as an algebraic group in the general case) for $n=p$ equal to the characteristic.
Under the assumption that $V$ is finite dimensional we may instead formulate the problem as the impossibility of having a $GL(V)$-isomorphism $Sym^n(V) \cong Sym^n(V^\ast)^\ast$. Now, we have an injective $GL(V)$-map $V^{(p)} \to Sym^p(V)$ given by $v \mapsto v^p$, where $V^{(p)}=k\bigotimes_kV$, where $k$ acts on the left hand side through the $p$'th power (concretely if we choose a basis for $V$ then the action on $V=k^m$ is given by the group homomorphism $GL_m(k) \to GL_m(k)$ which takes $(a_{ij})$ to $(a^p_{ij})$). As $\dim V > 1$ we have $\dim V^{(p)}=\dim V < \dim Sym^p(V)$ so that the inclusion is proper.
It is easily verified that $V^{(p)}$ is irreducible and it is in fact the unique irreducible submodule of $S^p(V)$. This can be seen by starting with an arbitrary non-zero element $f$ of $S^p(V)$ and then acting on it by suitable linear combinations of the action of elementary matrices of $GL(V)$ until one arrives at a non-zero element of (the image of) $V^{(p)}$. This is more easily understood if one uses the fact that we have an action of an algebraic and consider the induced action by its Lie algebra. Choosing again a basis of $V$ we have elements $x_i\partial/\partial x_j$ whose action on a monomial are very visible. In this way it is clear that starting with any monomial of degree $p$ one may apply a sequence of such operators to obtain a non-zero multiple of a monomial of the form $x_k^p$. This plus some thought shows the statement.
Assume now that we have a $GL(V)$-isomorphism $Sym^p(V) \cong Sym^p(V^\ast)^\ast$. Dualising the inclusion $V^{\ast(p)} \hookrightarrow Sym^p(V^\ast)$ and composing with the isomorphism we got a quotient map $Sym^p(V) \to V^{(p)}$. It is easy to see that in Jordan-Hölder sequence of $Sym^p(V)$ $V^{(p)}$ so that the composite $V^{(p)} \to Sym^p(V) \to V^{(p)}$ must be an isomorphism and hence the inclusion $V^{(p)} \hookrightarrow Sym^p(V)$ is split, contradicting that $V^{(p)}$ is the unique simple submodule.