When do the symmetric tensors of a commutative algebra form a Frobenius algebra?

Abstract:

For a commutative $k$-algebra $B$, consider the subalgebra ${({B^{ \otimes n}})^{{S_n}}}$ of the $n$th tensor power of $B$, formed by the tensors invariant under arbitrary permutations of the indices. Necessary and sufficient conditions are found for ${({B^{ \otimes n}})^{{S_n}}}$ to be Frobenius. When ${\dim _k}B \ne 2$, these say that $B$ is Frobenius and $n!$! is invertible in $k$, unless $B$ is separable. Some additional cases occur for two-dimensional algebras in positive characteristic, depending on the divisibility of $n + 1$.