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Proceedings of the American Mathematical Society

ISSN 1088-6826(online) ISSN 0002-9939(print)

 
 

 

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
DOI: https://doi.org/10.1090/S0002-9939-1982-0656088-5
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 $n + 1$.


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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