# Module such that every finitely generated submodule is semisimple [closed]

Is there an example of a module $$M$$ (over a commutative ring) that is not free, and such that each of its finitely generated submodule is semisimple (i.e. such that any submodule of any finitely generated submodule $$N$$ of $$M$$ is a direct factor of $$N$$) ?

## closed as off-topic by YCor, Jan-Christoph Schlage-Puchta, Sean Lawton, user44191, David HandelmanJun 13 at 14:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – YCor, Sean Lawton, user44191, David Handelman
If this question can be reworded to fit the rules in the help center, please edit the question.

• Take $M=(A/\mathfrak{m})^\oplus n$ for any maximal ideal $\mathfrak{m}$, – Mohan Jun 13 at 0:14
• Sounds like a mixture of 2 questions, none being research-level (1) find [a ring and] a non-free semisimple module (this is very easy) (2) if every f.g. submodule of $M$ is semisimple then is $M$ semisimple (yes, classical and easy). – YCor Jun 13 at 6:23

@Mohan has already given an example in the comments. If you ask that the ring $$A$$ injects into $$End_A(M)$$, then here is an example. Let $$M=\oplus {\mathbb Z}/p{\mathbb Z}$$ be the $$\mathbb Z$$-module (= an abelian group). Here $$p$$ runs over all primes. Then $$M$$ is not free and every finitely generated submodule is of the form $$\oplus _{p \in S} {\mathbb Z}/p{\mathbb Z}$$ where $$S$$ is a finite set of primes $$p$$. Hence $$S$$ is semi-simple.
• Thank you. Unless I am mistaken, the module $M$ is also semisimple. Is it also possible to find a example where $M$ is non-semisimple ? – Jon-S Jun 13 at 3:29
• If $N$ is the sum of all simple submodules of $M$, then it is an easy exercise that $N$ is semi-simple. But if $v\in M\setminus N$, then $Av$ is a sum of simple modules by your assumptions, and hence $Av$ lies in $N$, contradiction. Hence $M=N$. – Venkataramana Jun 13 at 4:32