If A and B are Morita equivalent matrices and A is a symmetric algebra (in the sense that A is isomorphic with A^* as bimodules), is it true that B is also symmetric?
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4$\begingroup$ I'm assuming $A,B$ are finite dimensional, then I believe the answer is yes. The endofunctor $- \otimes_A A^*$ on the category of f.g. $A$-modules has an intrinsic description so it is preserved by Morita equivalence, and you're asking whether it's isomorphic to the identity functor. A useful keyword here is "Nakayama functor", see e.e. Schweigert-Woike arxiv.org/abs/2103.15772 . $\endgroup$Adrien– Adrien2026-04-13 15:02:48 +00:00Commented Apr 13 at 15:02
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2 Answers
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Yes, here is a textbook reference: Corollary 4.3 in Chapter IV. in the book "Frobenius algebras I" by Skowronski and Yamagata.
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My favorite proof uses the fact that being symmetric is equivalent to the fact that, for finitely generated modules $M$ and finitely generated projective modules $P$, there is a natural isomorphism between $\operatorname{Hom}(P,M)$ and the vector space dual of $\operatorname{Hom}(M,P)$, a description of being symmetric that clearly only depends on the module category.
A variation of this proof also shows that being symmetric is derived invariant.
