Assume $P$ is a right projective $R$-module. Is $P$, viewed as a left $\mathrm{End}_R(P)$-module, projective as well? If not, under what conditions does it hold?


I am trying to establish Morita equivalence between a finite-dimensional $k$-algebra $A$, $k$ a field, and its associated basic algebra $A^b \simeq eAe$ for certain idempotent $e \in A$ (more specifically, such that if $e=e_1+ \dots +e_n$ is a decomposition of $e$ into sum of pairwise orthogonal primitive idempotents, then $e_1A, \dots, e_nA$ is a representative set of indecomposable direct summands of A, up to isomorphism).

According to my notes, this should be done via the module $eA$ (that is, use $eA$ as the balanced bimodule which is progenerator in both relevant categories): since $eAe \simeq \mathrm{End}_A(eA)$, $eA$ is a left-right $(eAe,A)$-bimodule. It is not difficult to see that it is balanced, $eA_A$ is a progenerator of $\text{Mod-}A$ and $_{eAe}eA$ is a generator of $eAe\text{-Mod}$. What I fail to see is that it is a projective $eAe$-module.

  • $\begingroup$ If $P$ is a generator, then it is projective as a module over its endomorphism ring. $\endgroup$ – egreg Jun 3 '14 at 21:49

The key fact is that a generator $P_R$ is finitely generated projective over its endomorphism ring.

Let $S=\operatorname{End}(P_R)$ and consider an epimorphism $P_R^n\to R_R$, which exists because $P_R$ is a generator and $R_R$ is finitely generated; this epi splits because $R_R$ is projective. Then, if we apply to it the functor $\operatorname{Hom}_R(-,{}_SP)$, we get a split mono $$ \operatorname{Hom}_R(R,{}_SP)\to\operatorname{Hom}_R(P^n,{}_SP) $$ which is a split mono ${}_SP\to {}_SS^n$, so ${}_SP$ is projective (and finitely generated).

Also the converse is true: if $P_R$ is a finitely generated projective module, then it is a generator over its endomorphism ring. In this case, consider a split epi $R_R^n\to P_R$ and apply again $\operatorname{Hom}_R(-,{}_SP)$ to get a split mono ${}_SS\to{}_SP^n$. So ${}_SS$ is an epimorphic image of a direct sum of copies of ${}_SP$, which is then a generator.

  • $\begingroup$ Thanks. By the way, is the "finitely generated" assumption needed for something other than the converse statement? $\endgroup$ – Pavel Čoupek Jun 4 '14 at 7:01
  • $\begingroup$ @PavelC There is no “finitely generated” assumption in the direct statement; it's necessary for the converse, because $_SS$ a summand of a product of copies of $_SP$ doesn't imply $_SP$ is a generator. $\endgroup$ – egreg Jun 4 '14 at 7:06

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