# Is a (non-commutative) quasi-Frobenius regular ring semisimple?

Let $$R$$ be a non-commutative ring which is quasi-Frobenius and regular. Is $$R$$ a semisimple ring?

Recall:

• quasi-Frobenius means a right $$R$$-module is projective if and only if it's injective (among many possible characterizations),
• regular means every $$R$$-module has finite projective dimension.

I believe this is true in the commutative case, since then $$R$$ is a product of local Artinian rings (Thm. 15.27 in Lam's Lectures on modules and rings), and regular local rings are integral domains (Corollary 10.14 in Eisenbud's Commutative algebra), so all the maximal ideals of those Artinian rings will be $$0$$ and $$R$$ will in fact be a product of fintely many fields. But none of these results seem to apply for the non-commutative case.

• That is a very unusual usage of "regular" by the way. Thanks for including it. I would not have supposed that meaning. Oct 22 '19 at 15:48

As shown here, since projectives are injectives, you can always say that a module over a QF ring having finite projective dimension is already projective. Therefore the projective dimension of a module over a QF ring is either $$0$$ or $$\infty$$.

So adding your second condition amounts to every module being projective, and thus you have a semisimple ring.