7
$\begingroup$

The Gorenstein symmetric conjecture states that for Artin algebras $A$ one has the the regular module has finite injective dimension as a right module if and only if it has finite injective dimension as a left module. Let $id(M)$ denote the injective dimension of a module $M$.

Question: Is there a general (noetherian if possible) ring $R$ with $id(R_R) \neq id(_{R}R)$?

I would expect that there is an easy counterexample or that this question has been considered before somewhere.

$\endgroup$
6
$\begingroup$

By Lemma A of

Zaks, Abraham, Injective dimension of semi-primary rings, J. Algebra 13, 73-86 (1969). ZBL0216.07001,

if $R$ is Noetherian and $\text{id}(R_R)$ and $\text{id}(_RR)$ are both finite, then they are equal.

In

Kirkman, E.; Kuzmanovich, J.; Small, L., Finitistic dimensions of Noetherian rings, J. Algebra 147, No. 2, 350-364 (1992). ZBL0765.16004.

the authors comment that they know of no example of a Noetherian ring with finite injective dimension on one side and infinite injective dimension on the other side, so there's probably no easy counterexample.

For non-Noetherian rings, there are well known examples of rings that are self-injective on one side but not on the other: for example, the ring of endomorphisms of a countable-dimensional vector space.

$\endgroup$

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.