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Let $(R,\frak{m})$ be a commutative Cohen-Macaulay local ring and $M$ and $N$ two maximal Cohen-Macaulay modules of Krull dimension $d$. Is $M\otimes_RN$ Cohen-Macaulay?

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  • $\begingroup$ I doubt this would be true. For the tensor product to be CM, usually one needs at least one of M, N to have something to say about the homological dimension. $\endgroup$
    – user782932
    May 26, 2022 at 19:21

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No. Tensor products resist being Cohen-Macaulay outside of exceedingly strong hypotheses. For a class of examples, let $R$ be a local domain of dimension $1$ that is not a DVR and let $I$ be a nonprincipal ideal. Then $I$ is maximal Cohen-Macaulay as an $R$-module since it embeds into $R$ and we are in dimension $1$. But $I \otimes_R I$ cannot be Cohen-Macaulay. Indeed, if $x,y$ is part of a minimal generating set of $I$, one may observe that $x \otimes y-y \otimes x$ is a nonzero torsion element of $I \otimes_R I$.

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