Let $S$ be a Noetherian semiperfect ring (https://en.m.wikipedia.org/wiki/Perfect_ring). Let $R$ be a module finite associative $S$-algebra. Then, is $R$ also a semiperfect ring? (Clearly, $R$ is Noetherian). If this is not true in general, what if I also assumed $S$ is commutative?
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$\begingroup$ Previously posted on MSE. $\endgroup$– Jeremy RickardCommented Dec 30, 2023 at 10:08
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1 Answer
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No, even if $S$ is commutative. There may be easier counterexamples, but ...
There are commutative Noetherian local (and therefore semiperfect) rings $S$ with a finitely generated indecomposable module $M$ whose endomorphism ring is not local. For example, in
Evans, E. Graham jun., Krull-Schmidt and cancellation over local rings, Pac. J. Math. 46, 115-121 (1973). ZBL0272.13006.
an example due to Swan is given.
Let $R=\operatorname{End}_S(M)$, and let $\operatorname{add}(M)$ be the category of direct summands of finite direct sums of copies of $M$.
The functor $\operatorname{Hom}_S(M, -)$ is an equivalence of categories from $\operatorname{add}(M)$ to the category of finitely generated projective right $R$-modules.
Since $\operatorname{add}(M)$ does not satisfy the Krull-Schmidt theorem, neither does the category of finitely generated projective right $R$-modules. But the category of finitely generated projectives for a semiperfect ring does satisfy the Krull-Schmidt theorem, so $R$ is not semiperfect.
answered Dec 30, 2023 at 11:11
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$\begingroup$ Thanks for the answer ... could you please explain why $add(M)$ does not satisfy Krull-Schmidt? $\endgroup$– unoCommented Jan 5 at 23:39
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$\begingroup$ @uno To satisfy Krull-Schmidt means that every object is a finite direct sum of indecomposable objects with local endomorphism rings. But $M$ is indecomposable and doesn’t have a local endomorphism ring. See Krull-Schmidt category. $\endgroup$ Commented Jan 6 at 8:22