# If the factor of a finitely generated module is free then submodule is also finitely generated

All rings are commutative, associative and with 1.

Consider short exact sequence $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ of $R$-modules. How to show that if $M$ is finitely generated and $M''$ is free then $M'$ is also finitely generated?

• Are you familiar with the notion of a split short exact sequence? Sep 15 '14 at 14:01
• Oh, @KeenanKidwell thanks a lot guys! I know about splitting, this solves the question immediately. I was just trying to show it explicitly, but it is not necessary. Sep 15 '14 at 14:09

If $M''$ is free then the sequence is split, so $M'$ is a direct summand of $M$. In particular, $M'$ is isomorphic to a factor module of $M$ and thus it is finitely generated.