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Let $A$ be a ring and consider modules on the left. Let $M$ and $S$ be $A$-modules and suppose $S$ is a finitely generated semisimple module. Is it true that $\operatorname{Hom}_A(M,S)$ is a semisimple $(\operatorname{End}_A(M))^{op}$-module? If not, what properties should $M$ have in order for the statement above to hold?

I have been thinking about this for a while, and have come up with an apparent proof that $\operatorname{Hom}_A(M,S)$ is a semisimple $(\operatorname{End}_A(M))^{op}$-module. I was not able to formalise this very well, so I am not sure if my proof is right.

We can reduce this to the case of $S$ being simple. In this case all non-zero elements in $\operatorname{Hom}_A(M,S)$ are surjections. I tried to pick a non-zero homomorphism $f$ in $\operatorname{Hom}_A(M,S)$ and to factor an arbitrary element in $\operatorname{Hom}_A(M,S)$ by $f$.

I would appreciate very much if someone could help me with this. Thanks in advance!

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  • $\begingroup$ Hi: I'm rusty on all of this, so I wanted to ask. Are we sure we can reduce to the case $S$ is simple? If $Hom(M,-)$ necessarily preserved colimits, I would agree (since that would say $Hom(M,\oplus S_i)\cong \oplus Hom(M,S_i)$. But I thought I remember that Hom functors do not necessarily preserve colimits... maybe there is an exceptional case, here? $\endgroup$ – rschwieb Oct 3 '13 at 13:33
  • $\begingroup$ Thanks for noticing this. I forgot to add that I'm only interested in the case where $S$ is finitely generated semisimple. Suppose $S$ is a finitely generated semisimple. It's probably false that $N=\operatorname{Hom}_A(M,S)$ is semisimple in general. However I am thinking what properties should $M$ have in order for $N$ to be semisimple. $\endgroup$ – user98284 Oct 3 '13 at 14:10
  • $\begingroup$ OK perfect then :) The reduction certainly works for f.g. semisimple modules. $\endgroup$ – rschwieb Oct 3 '13 at 16:19

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