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?

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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.