6
$\begingroup$

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 $A\otimes_k A$?

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.