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I am trying to show that $R$ is semisimple if and only if every R module is projective. What I know is that if $R$ is semisimple then every $R$ module $M$ is semisimple which implies $M$ can be expressed as a direct sum of simple modules i.e $R=\sum_{i\in I} Ra_i$ for some $a_i\in M$ then I am not sure how to use this to deduce $M$ is projective. I also know that $R$ is artinian but I don't know if this is relevant. There is already an answer to this question but it involved advanced techinques I am not familiar with.

Can anyone show me an easy way to see this

Thanks in advance

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  • $\begingroup$ Advanced techniques like Zorn's Lemma, for instance? $\endgroup$ – Lord Shark the Unknown Apr 27 '18 at 5:49
  • $\begingroup$ Zorns lemma is fine I was talking about global dimension or projective dimension $\endgroup$ – TheGeometer Apr 27 '18 at 5:52
  • $\begingroup$ To show all modules are projective, you need that each submodule of a module has a complement. Given each module is a sum of simples, that's a fairly straightforward Zorn's lemma argument. $\endgroup$ – Lord Shark the Unknown Apr 27 '18 at 5:54
  • $\begingroup$ Related but arguably not something to use as a duplicate. $\endgroup$ – rschwieb Apr 30 '18 at 13:36
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I guess you should know these charaterisations of a simple $R$-module $M$:

  • $M$ is a sum of simple $R$-modules.

  • $M$ is a direct sum of simple $R$-modules.

  • For each submodule $N$ of $M$, $N$ admits a complement $M'$, i.e., there is a submodule $M'$ such that $M=N\oplus M'$.

Now back to your problem. $R$ is semisimple if and only if any (left or right) $R$-module $M$ is semisimple. This is equivalent to each $M$ being peojective, as there is always a free $R$-module $F$ such that $F\twoheadrightarrow M$ and $M$ having a complement means that the epimorphism admits a section $s\colon M\to F$.

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  • $\begingroup$ Thank you very much for your response I guess we can also use the characteraisation that M is projective if and only is a direct summand of a free module and so since it has a complement in F we are done ? $\endgroup$ – TheGeometer Apr 27 '18 at 6:37
  • $\begingroup$ @TheGeometer Yeah, surely this way also works~ $\endgroup$ – josephz Apr 27 '18 at 6:48

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