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An $R$-module $M$ is called quasi-projective if $\text{Hom}_R(M,M)\to\text{Hom}_R(M,N)$ is surjective for every surjective homomorphism $M\twoheadrightarrow N$.

What are the rings $R$ for which every quasi-projective $R$-module is projective? Does there exist such a ring which is not semisimple?

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    $\begingroup$ E.g. $R$ being a field? $\endgroup$
    – Włodzimierz Holsztyński
    Commented Jun 24, 2016 at 3:22
  • $\begingroup$ I do not know. I do not have any prove for it. I f you have any idea, may I ask you please let me know. Thank you $\endgroup$
    – Najmeh Dehghani
    Commented Jun 24, 2016 at 14:35
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    $\begingroup$ Every module (i.e. linear space) in a category of linear spaces over a fixed field is free hence projective. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Jun 25, 2016 at 0:07
  • $\begingroup$ Yes. In fact over a semsimple ring R, every module is projective. But my question what is necessarily condition on R provided every quasi-projective module is projective. $\endgroup$
    – Najmeh Dehghani
    Commented Jun 26, 2016 at 7:19

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Every simple module is trivially quasi-projective, and if every simple $R$-module is projective then $R$ is semisimple. So semisimple rings are the only rings for which quasi-projective implies projective.

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  • $\begingroup$ @Jeremy Rickard : Do you have a reference (or a proof) for the fact that projectivity of simple modules implies semi-simplicity? $\endgroup$
    – abx
    Commented Jun 26, 2016 at 13:26
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    $\begingroup$ @abx If $M$ is a non-semisimple right $R$-module, pick $m\in M$ not in the socle, and let $I<R$ be a maximal right ideal containing the annihilator of $m$. Then $R\to R/I$ is a non-split epimorphism to a simple module, and so $R/I$ is a non-projective simple module. $\endgroup$
    – Jeremy Rickard
    Commented Jun 27, 2016 at 3:37

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