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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.

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  • $\begingroup$ (After fixing a typo) I find your proof to (i) pretty clear. $\endgroup$
    – user26857
    Commented Dec 8, 2014 at 8:50

1 Answer 1

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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.

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