Let $k$ be a field, and let $A$ be a commutative noetherian $k$-algebra.

If a finitely generated $A$-module $M$ has finite injective dimension over $A$, does this imply that $M\otimes_k M$ has finite injective dimension over $A\otimes_k A$?


1 Answer 1


If you are willing to assume that:

1) $A$ has a dualizing complex $R$.

2) $A\otimes_k A$ is noetherian of finite Krull dimension.

Then the answer is yes:

One can show that under these conditions, $R\otimes_k R$ is a dualizing complex over $A\otimes_k A$.

Now, a finitely generated module $M\otimes_k M$ has finite injective dimension if and only if its dual $RHom_{A\otimes_k A}(M\otimes_k M, R\otimes_k R) \cong RHom_A(M,R)\otimes_k RHom_A(M,R)$ has finite projective dimension, and this is easy to verify.


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