This question already has an answer here:
Let $R$ be a commutative ring with identity. Then $R^n$ ($n$ is some positive integer) is a module over $R$ in a natural way. It is free and finitely generated. Now let $M$ be a finitely generated submodule of $R^n$. Does it follow that $M$ is free? My intuition leads me to believe that this is true, but I am not too sure.