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Suppose the ring $R$ considered as a left $R$-module is a direct sum $R=L_1\oplus L_2\oplus ...\oplus L_n$ such that $L_i$ are simple modules (no nonzero proper submodules) and such that $L_i=Re_i$ for $e_i\in R$ with

(a) $e_ie_j=0$ if $i\neq j$

(b) $e_i^2 = e_i$ for all $i$ (idempotent)

(c) $\Sigma _{i=1}^n e_i = 1$

We want to show that every $R$-module is completely reducible; that is there are simple submodules $N_i\subset M$ with $M=\oplus N_i$.

My game plan for the proof is the following:

I have a theorem that says $M$ is completely reducible is equivalent to saying $M$ is semisimple or that $M$ is generated by its simple submodules.

I have been racking my brain for about an hour and I haven't been able to come up with anything. I've tried looking up solutions online and haven't had any luck. Help would be greatly appreciated.

I've seen plenty of proofs that seem to be proving the opposite way - that is starting with completely reducible and proving $R$ is a direct sum, but can't find anything in the direction I'm trying to prove.

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  • $\begingroup$ Do you already know that quotients of semisimple modules are semisimple? $\endgroup$
    – Hanno
    Apr 15, 2016 at 5:08
  • $\begingroup$ Yes! We proved that in class. $\endgroup$
    – Mark
    Apr 15, 2016 at 5:11

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Let $R$ be semisimple as an $R$-module und $M$ an $R$-module. Then every free $R$-module is also semisimple. Since $M$ is a quotient of a free $R$-module, $M$ is also semisimple.

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  • $\begingroup$ So since were given that $R$ is a direct sum of simple modules, then we can say $R$ is semisimple, correct? $\endgroup$
    – Mark
    Apr 15, 2016 at 5:49
  • $\begingroup$ Yes, isn't that your definition of being “semisimple”? $\endgroup$
    – Claudius
    Apr 15, 2016 at 5:54
  • $\begingroup$ Yes, just clarifying. thank you this makes a lot of sense and is a lot easier than I thought it was going to be. $\endgroup$
    – Mark
    Apr 15, 2016 at 5:56

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