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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?

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  • $\begingroup$ My gut says maybe $\endgroup$ – Zelos Malum Oct 1 '16 at 4:05
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Yes.

If $M$ is not free, then the result is trivial.

Assume that $M$ is free. Then $M$ contains a copy of $R$, so it is sufficient to prove the claim for $R$.

Since $R$ is not a PID, then either it contains a non-principal ideal, or it is not an integral domain. In the latter case, let $a$ be a zero-divisor; then the submodule of $R$ generated by $a$ is not free.

Finally, assume that $R$ is an integral domain, and let $I$ be a non-principal ideal of $R$. If $I$ were a free $R$-module, then it would admit a basis $(x_j)_{j\in J}$, with $J$ a set containing at least two elements. But then, for any $i\neq j$ in $J$, we would have $$ x_ix_j - x_jx_i = 0, $$ contradicting the linear independence of $x_i$ and $x_j$. Thus $I$ is not a free module.

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