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

Authors: Annetta G. Aramova and Luchezar L. Avramov
Journal: Proc. Amer. Math. Soc. 85 (1982), 299-304
MSC: Primary 13E10
MathSciNet review: 656088
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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$.

References [Enhancements On Off] (What's this?)

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Keywords: Frobenius algebra, zero-dimensional Gorenstein ring, symmetric tensors, invariants of a finite group of automorphisms
Article copyright: © Copyright 1982 American Mathematical Society

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