Though I suspect that this is a folklore result, I've been unable to find a proof by digital book searching. I also mention (my apologies) that I am new to noncommutative algebras theory and I am not familiar with the literature.

Let $0 \to M \to N \to P \to 0$ be a short exact sequence of left $A$-modules ($A$ some arbitrary ring). If $M$ and $P$ are semisimple, prove that $N$ is semisimple.

Closest I could find is the (well-known) fact that submodules/quotients of semisimple modules are semisimple.

I would be grateful for any indication or relevant reference.

  • 1
    $\begingroup$ Where did you get that "result"? I think that's not true. For example, let $R$ be the subalgebra (with unit) of $\mathbb C^{2\times 2}$ generated by $A = \begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}$. $R$ acts from the left on $N := \mathbb C^2$. $M := \begin{pmatrix}\mathbb C \\ 0 \end{pmatrix}$ is an $R$-submodule of $N$. The quotient $P:=N/M$ is isomorphic to $M$. We have the exact sequence $0 \rightarrow M \rightarrow N \rightarrow P \rightarrow 0$. But from the Jordan Normal Form Theorem we see that $N$ can't be semisimple. $\endgroup$ – jflipp Nov 16 '14 at 15:22
  • $\begingroup$ Umm... it seems you are right. I was trying to prove that $R/I$, $R/J$ semisimple imply $R/I \cap J$ semisimple with the standard "short exact sequence" argument. $\endgroup$ – Chindea Filip Nov 16 '14 at 18:11

The sequence $$0\to R/(I \cap J) \to R/I \oplus R/J \to R/(I+J) \to 0 $$ is exact. Since $R/I$ and $R/J$ are semisimple, $R/I \oplus R/J $ is semisimple, and then $R/(I+J)$ is semisimple.


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