Let $k$ be a algebraically closed field and suppose that $A$ and $B$ are finite dimensional $k$-algebras. If we assume that $A$ is a symmetric $k$-algebra and $A\otimes_k I$ is a projective $A\otimes_k B$-module for some $B$-module $I$, is it true that $I$ must be a projective $B$-module? Or could someone provide me with any counterexample?
1 Answer
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We have in that case (algebraically closed is important) that $\operatorname{pdim} M \otimes_K N= \operatorname{pdim} M + \operatorname{pdim} N$ and thus if $I$ is not projective then $A \otimes_K I$ is not projective.
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$\begingroup$ Thanks for your useful answer. $\endgroup$ Commented Apr 8, 2022 at 8:15