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To help myself prepare for an upcoming exam, I've been working on various problems that focus on the topics we have been dealing with in class. I came across this one:

Let $M$ be a right $R$-module such that $M / Soc(M)$ is finitely generated. Show that $M = A \oplus B$, where $A_R$ is finitely generated, and $B_R$ is semisimple.

Any help would be greatly appreciated. Thanks!

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1 Answer 1

up vote 6 down vote accepted

For the canonical surjection $f \colon M \to M/Soc(M)$, choose a preimage for each of the finitely many generators of $M/Soc(M)$. These generate a finitely generated submodule $A \subseteq M$. Notice that $M/Soc(M) = f(A) = (A + Soc(M))/Soc(M)$, which implies that $A + Soc(M) = M$.

Now $A \cap Soc(M)$ is a submodule of the semisimple module $Soc(M)$, so there is a submodule $B$ of $Soc(M)$ such that $Soc(M) = (A \cap Soc(M)) \oplus B$. I'll leave it to you to finish the proof by showing that $A + B = M$ and $A \cap B = 0$. (Or, if you're stumped, leave a comment and I'll complete this answer.)

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