The following theorems appear many places in books and on this site also.
Sub-modules of a free module are free, provided the ring of scalars is P.I.D.
If $R$ is not P.I.D. then sub-module of a free $R$-module may not be free.
I have gone through the proof of theorem as well as counterexample. But my next question comes from these two facts:
Let $M$ is an $R$-module (and assume that $M$ has proper sub-modules). If $R$ is not a P.I.D., then is it necessary that there exists a sub-module of $M$ which is not free?