I was reading one of the possible definitions of semisimple modules, which is "for every submodule $N$ of $M$, there exists a submodule $P$ such that $M=N \bigoplus P$. After reading this I immediately thought that a particular case of semisimple modules are simple modules: if $M$ is simple then the only submodules are $0$ and $M$, and $M=0 \bigoplus M$, so the simple modules satisfy the conditions of the definition of semisimple modules.

However, I've also read that not all simple rings are semisimple rings, but a simple ring in particular is a simple module, so how can it be that it satisfied the definition of semisimple module but is not semisimple?

I am very confused by this, I would appreciate if someone could clear up my confusion, thanks in advance.

  • $\begingroup$ That definition of semisimple leaves out a rather important part: that the direct factors must be simple modules in itself. $\endgroup$ – Timbuc Oct 27 '14 at 5:24
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    $\begingroup$ @Timbuc This definition is equivalent to yours. $\endgroup$ – user98602 Oct 27 '14 at 5:34

Be careful. A simple ring $R$ is "A ring with no proper nontrivial two-sided ideals", but a simple module is "A module with no proper nontrivial submodules". A semisimple ring $R$ is "A ring that is semisimple as an $R$-module". Every ring that's not a division ring has left ideals (and thus is not simple as a module over itself). Simple rings do not need to be simple as modules over themselves. (For instance, consider $R = M_2(\Bbb Q)$.)

You are correct that every simple module is a semisimple module. But the notion of simple rings and simple modules don't coincide.

  • $\begingroup$ One last doubt: suppose the ring is simple but also commutative, then is it also semisimple seen as an R-module with the "natural" action on itself? $\endgroup$ – user16924 Oct 27 '14 at 5:38
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    $\begingroup$ @user16924 Yes; commutative simple rings are fields, and thus simple as modules over themselves. (Proof: let $a \in R$; then $Ra$ is a two-sided ideal, because $R$ is commutative; because $R$ is simple, $Ra = R$, and thus there's some $b$ with $ba=ab = 1$. So all nonzero elements have an inverse.) $\endgroup$ – user98602 Oct 27 '14 at 5:40

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