Is every submodule of a free Noetherian module free?

Question

Let $R$ be a ring and $M$ be a free $R$-module and $N$ be an $R$-submodule of $M$.

If $M$ is Noetherian, is $N$ free?

If $R$ is a principal ideal domain, then $M$ is Noetherian and $N$ is free. This fact makes me curious of that whether the Noetherian property of $M$ actually implies that $N$ is free.

Answer 1

Take $R$ a noetherian ring (e.g. $R=\mathbb Z[X]$) and $I$ a non-principal ideal (e.g. $I=(2,X)$).