Let $R$ be a PID and $M$ a module. Show the following:

(i) If $M$ is finitely generated and $S$ is a free submodule with $M/S$ torsion-free, then $M$ is free.

(ii) If $M$ is torsion-free and $M/S$ is finitely generated of torsion for all submodule $S \neq 0$ of $M$, then $M \cong R$. Deduce that if an infinite abelian group satisfies every non zero subgroup has finite index then $G \cong \mathbb Z$.

I think I could solve (i) but I am not so sure if it is correct so I'll write my solution:

I know that if $R$ is a PID and $M$ is a f.g. torsion-free module, then $M$ is free. So, using this, I've tried to show that $M$ is torsion-free. So suppose there is $r \neq 0, m \in M$ such that $rm=0$. Consider the projection $$\pi: M \to M/S$$ $$ m \to \overline{m}$$

Then $0=\pi(rm)=r\pi(m)$. Since $r \neq 0$ and $M/S$ is torsion-free, it must be $\pi(m)=0 \implies m \in S$. But $S$ is a free submodule which implies $S$ is torsion-free, as $rm=0$ and $r \neq 0$, it follows $m=0$. It follows directly that $M$ is torsion-free.

I have no idea what to do in (ii). As for the last part of (ii), any abelian group is a $\mathbb Z$-module. I don't see why $G/S$ is of torsion (if we have this, then we are under the hypothesis of first part of (ii) and we can apply the proposition).

Any suggestions would be appreciated.

  • $\begingroup$ (After fixing a typo) I find your proof to (i) pretty clear. $\endgroup$ – user26857 Dec 8 '14 at 8:50

If consider $S$ a non-zero cyclic submodule, from $M/S$ finitely generated you get $M$ finitely generated, so $M$ is free. Now you can use this answer.

For an abelian group $G$ and $S$ a non-zero subgroup, having finite index means that $G/S$ is finite, or a finite abelian group is finitely generated and torsion.

| cite | improve this answer | |

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.