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.


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.


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.


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