A projective covering of an $R$-module $M$ is an epimorphism $\pi:P\rightarrow M$ s.t. $P$ is a projective $R$-module and $\textrm{Ker}(\pi)$ is co-essential in $P.$ The existence theorem for projective coverings says: every module over a finite-dimensional algebra has got a projective covering, i.e. any finite-dimensional algebra is (left and right) perfect.
I wonder if there are rings $R$ s.t.
- Only projective (free) $R$-modules has got projective coverings?
- Every finitely generated left $R$-module has got a projective covering but there is a left module which hasn't got?
- $R$ is left perfect but not right perfect?
(I know that, for instance, every module over a field is projective, so in this case only projective modules has got projective coverings; but this is not the example I expect :))
Yes, there's an elementary theorem that over any semiprimitive ring, the only modules with projective covers are already projective. See T.Y. Lam's First course in noncommutative rings example 5 p 361.
Yeah, that is the difference between a semiperfect ring and a left perfect ring. Here's the DaRT query. For posterity, such an example is $k[[x]]$ where $k$ is a field.
Yes, it is known to be an asymmetric condition. DaRT query. Specifically if you take the subset of $M_\mathbb N(F)$ for a field $F$ of infinite matrices which only have finitely many nonzero entries above the diagonal, and generate the subring generated by those elements and the identity matrix, you get a left-not-right perfect ring.