Part of this question (asked by someone else) for semiperfect rings has circulated a few weeks [on math.se](https://math.stackexchange.com/q/4164710/29335) but without much attention. I think it might be above the threshold of difficulty to be on mathoverflow, but you can let me know if I should move it.


Let $R$ be a ring and $e$ be an idempotent element of $R$ (that is, $e^2=e$).


1. If $R$ is semiperfect, is $eRe$ necessarily semiperfect?
2. If $R$ is right perfect, is $eRe$ necessarily right perfect?


[This wiki page has the definitions, if you are not familiar.](https://en.wikipedia.org/wiki/Perfect_ring). "$eRe$" is the so-called **corner ring**.


I was a bit surprised I could not turn up the answer for this quickly in the literature or through my own computations.


I'm familiar with the characterization of both conditions via projective covers, and also for the characterization of semiperfect rings using the existence of a finite decomposition of $1$ into orthogonal local idempotents, and the characterization of right perfect rings as satisfying the DCC on left principal ideals.


But they don't seem to avail me since I don't understand the relationships of $eRe$ modules to $R$ modules. I am not even very clear on what can be said about the relationship of right ideals of $R$ and $eRe$.



Are corner rings of (semi)perfect rings (semi)perfect?
Rowen shows in Lemma 2.7.34 of his book Ring Theory that $R$ is right perfect iff for each idempotent $e$ one has both $eRe$ and $(1-e)R(1-e)$ are right perfect. Hence $R$ right perfect implies $eRe$ is right perfect.


Here is a proof for the semiperfect case. Note that $eR$ is a finitely generated projective module and hence by Corollary 24.14 of Lam's First Course in Noncommutative Rings, $eR$ is isomorphic to a finite direct sum of $e\_iR$, say $i=1,\ldots, n$, with the $e\_i$ primitive idempotents. Since $eRe=\mathrm{End}\_R(eR)$, we can find orthogonal $f\_1,\ldots, f\_n$ idempotents of $eRe$ with $e=f\_1+\cdots+f\_n$ and $f\_iR\cong e\_iR$. Moreover, $f\_ieRef\_i=f\_iRf\_i\cong \mathrm{End}\_R(e\_iR)\cong e\_iRe\_i$ is local. Thus $eRe$ is semiperfect.


**Added:** To make this more self-contained, let me sketch Lam's proof. Note that $eR\to eR/eJ(R)$ is a projective cover (with $J(R)$ the radical). On the other hand, since $R/J(R)$ is semisimple, $eR/eJ(R)$ is isomorphic to a finite direct sum of simples $\bigoplus\_{i=1}^n e\_iR/e\_iJ(R)$. Hence the finite direct sum of the $e\_iR$ also are a projective cover and hence since any two projective covers are isomorphic, $eR\cong e\_1R\oplus\cdots\oplus e\_nR$.


You can also prove this by showing $eRe$ is semilocal (this is proved in Rowen) and showing that idempotents lift under $eRe\to eRe/eJ(R)e$ using some trick I forget to get your lift in $R$ to be conjugate to one in $eRe$.


