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An $R$-module $M$ is said to be a semisimple module if it is a direct sum of its simple submodules.

I've proved the following equivalent characterization for semisimple modules:

$M$ is semisimple iff every submodules of $M$ is a direct summand.

From the above proposition we can easily show that every submodule and factor module of a semisimple module $M$ are alosa semisimple.But the converse is not always true.

I want to find a module $M$ whose proper submodules and nonzero factor modules are semisimple but $M$ is not semisimple.

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  • $\begingroup$ Apparently you mean proper submodules and factor modules by nonzero submodules? Or else the entire module is included in things you have assumed are semisimple. $\endgroup$ – rschwieb Apr 3 '18 at 18:31
  • $\begingroup$ Not "nonzero factor modules", a "zero factor module" will be interpreted as "not $M/M$", and not "not $M/\{0\}$" as you desire. $\endgroup$ – rschwieb Apr 3 '18 at 18:38
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For the ring $R=\mathbb R[x]/(x^2)$, $M=R$ as a right $R$ module has this property.

I worked out that $M$ has to be a local module whose maximal ideal is the socle for the submodule condition to hold. But then I noticed this example before considering what the other half of the conditions entailed.

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Let $k$ be a field and let $R \subseteq \operatorname{M}_2(k)$ be the $k$-subalgebra of upper triangular matrices. Then $M = k^2$ is an $R$-module via matrix multipliation. The only submodules of $M$ apart from $0, M$ is the one-dimensional $k$-linear subspace $N :=\langle e_1 \rangle$. It then follows that both $N$ and $M/N$ are one-dimensional over $k$, and therefore simple as $R$-modules. But $M$ itself is not semisimple because $N$ has no direct complement.

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