It is clear that if $R$ is a ring with identity and $e\in R$ is an idempotent then $Re$ is a direct summand of $R$ while $R$ is a generator in the category of left $R$-modules. I have my question when $R$ is semiperfect: is $Re$ (isomorphic to) a direct summand of "every" generator of the left $R$-modules? Thanks in advance!
2026-04-01 15:11:39.1775056299
Generators over semiperfect rings
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If $R=M_2(k)$ is the ring of $2\times 2$ matrices over a field, then the $2$-dimensional left module of column vectors is a generator, but doesn't have $R$ (i.e., $Re$ for $e=1$) as a direct summand.
More generally, if $R$ contains any indecomposable projective module $P$ as a direct summand with multiplicity greater than one, then you can choose a generator $Q$ and an idempotent $e$ so that $Re$ contains $P$ with greater multiplicity than $Q$ does, so that $Re$ can't be a direct summand of $Q$.