Is Socle of evey locally Noetherian module a direct summand Definition: An $R$-module $M$ is called locally Noetherian if, any finitely generated submodule of $M$ is Noetherian.
Question: Let $R$ be a ring with identity and $M$ be a  locally Noetherian $R$-module. Is $soc(M)$   a direct summand of $M$.
My attempt: Let $S=soc(M)$=sum of  simple submodules of $M$. Since each simple submodule is finitely generated $R$-module, by hypothesis, $S$ is a sum of Noetherian simple $R$-modules. Now, how should I proceed?

You can't prove it, because it's false. An obvious counterexample is $\mathbb{Z}/4\mathbb{Z}$ which is Noetherian, hence locally Noetherian, but indecomposable.
If you want a counterexample with a locally Noetherian module which is not Noetherian, take the Prüfer group $\mathbb{Z}(p^\infty)$ for any prime $p$. This is indecomposable as well.
Both examples are Artinian modules and the socle of an Artinian module is nonzero and never a direct summand except in the trivial case when the module is semisimple and finitely generated.
