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 $\lbrace 1, \dots, n \rbrace$.


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?

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    $\begingroup$ No: I think that if $V$ is $2$-dimensional in characteristic $p$, then $Sym^p(V)$ (considered as a $GL(V)$-module) has a two-dimensional submodule, but no two-dimensional quotient. $\endgroup$
    – moonface
    Feb 15, 2010 at 16:38

3 Answers 3


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.


The dual to symmetric powers of a projective module of finite rank over any commutative ring is most elegantly expressed in terms of divided powers (thereby also "explaining" why over a field of nonzero characteristic there are subtleties once the dimension reaches the characteristic, and why the version in characteristic zero has factorials all over the place in the denominators). It is very nicely explained in an appendix to the book by Berthelot-Ogus on crystalline cohomology.


As several MO'ers have noted, the answer is in general no, contrary to what I falsely claimed in my previous post. To make amends for writing nonsense before, let me point you to Eisenbud, Commutative Algebra with a View, where this issue is nicely discussed in A.2.4. What turns out to be true, as Marty had hinted at, is that the graded dual of the symmetric algebra is naturally isomorphic to the divided power algebra of the dual. When you are working over $\mathbf{Q}$, this divided power algebra is isomorphic to the symmetric algebra, but in general it need not even be noetherian.

  • $\begingroup$ I don't think this works... it starts going awry with the phrase "canonical inclusions $V \rightarrow V \times \cdots \times V$". We're talking about multilinear maps, and really tensor products, so there is no such inclusion. I think the answer should include something about divided power algebras. $\endgroup$
    – Marty
    Feb 15, 2010 at 15:41
  • $\begingroup$ I don't understand the construction of f. As far as I can see, the only way to get a canonical linear inclusion of V into $V^{\otimes n}$ is to fix a nonzero element u in V and send v to $u \otimes u \otimes \cdots \otimes v \otimes \cdots \otimes u$. You should also be bothered that you always send \phi to a product of linear forms; most elements in Sym^n are not such products (for dim V > 2.) $\endgroup$ Feb 15, 2010 at 15:45
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    $\begingroup$ Moreover the second map is not symmetric. You then have to symmetrize, and this is where the n! appears again. $\endgroup$ Feb 15, 2010 at 16:05
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    $\begingroup$ I edited my previous post (which was nonsense) to be more useful. My apologies if this makes the resulting comments seem a little out of context. $\endgroup$
    – B. Cais
    Feb 15, 2010 at 17:16

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