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Lets take a ring $R$ and an ideal $\mathfrak p \subset R$, and call them an L-pair (just for brevity) if $\mathfrak p$-adic completion of any projective module is again projective (as R-module); and L-ring is a ring which is an L-pair with any of its finitely generated ideals. (I'm not sure, but have a feeling that looking at non-f.g. ideals is not right)

What L-rings do look like? I'm more interested in nontrivial necessary conditions than sufficient ones. (do not have much hope for full characterisation)

P. S. Ring being right perfect + left coherent + gd $\leq 2$ is equivalent to "projectives closed under arbitrary limits". I tried to cook up an example of an L-ring which does not look like that (in particular, non-artinian) to no avail, and would be happy to see one.

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    $\begingroup$ In response to the P.S., $k[x]/(x^2)$, where $k$ is a field, is an example that is right perfect and left coherent, but has infinite global dimension. $\endgroup$
    – Jeremy Rickard
    Commented Nov 4, 2020 at 9:35
  • $\begingroup$ Well, yeah, of course. My "not look like that" included being not artinian, so we have some actual completion going on, and preserving projectives is not super obvious. $\endgroup$
    – Denis T
    Commented Nov 4, 2020 at 10:58
  • $\begingroup$ Regarding the PS: From the context it seems that you suggest that "projectives closed under arbitrary limits" implies that $R$ is an L-ring. Why is that? The standard way to get completion is to take an inverse limit over certain quotients, which will typically not be projective themselves... $\endgroup$
    – Pavel Čoupek
    Commented Nov 4, 2020 at 14:58
  • $\begingroup$ @PavelČoupek I double-checked my "proof" that this implies L-ness and found an error. Nevertheless, every commutative example I know so far is perfect+coherent. $\endgroup$
    – Denis T
    Commented Dec 4, 2020 at 19:22

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