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 Handelman Jun 13 at 14:35

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    $\begingroup$ Take $M=(A/\mathfrak{m})^\oplus n$ for any maximal ideal $\mathfrak{m}$, $\endgroup$ – Mohan Jun 13 at 0:14
  • $\begingroup$ 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). $\endgroup$ – 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.

  • $\begingroup$ 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 ? $\endgroup$ – Jon-S Jun 13 at 3:29
  • $\begingroup$ 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 $M=N$. $\endgroup$ – Venkataramana Jun 13 at 4:32

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