Looking for help, Aluffi Exercise 5.13, Chapter 6: characterization of PIDs

Question

I quote:

"Let $M$ be a finitely generated module over an integral domain $R$.

Prove that if $R$ is a PID, then $M$ is torsion-free if and only if it is free. Prove that this property characterizes PIDs. (Cf. Exercise 4.3.)"

I am having trouble on the second part. Btw, Exercise 4.3 says that an integral domain $R$ is a PID if and only if its ideals are free as $R$-modules, which is easy to see.

I am not sure if I am interpreting Aluffi correctly, but I think I am. Here is how I have written the problem:

Let $R$ be an integral domain. IF

($\forall M$ finitely generated $R$-modules): ($M$ is torsion-free $\Longleftrightarrow$ every ideal of $R$ is free as an $R$-module),

THEN

$R$ is a PID.

---

I figured I was supposed to prove that every ideal $I$ of $R$ is finitely generated, i.e. $R$ is Noetherian, so that $I$ being torsion-free (it is a submodule of the free module $R^1$) implies $I$ is a free $R$-module. But I have made pretty much zero progress...

Thanks for your help.

Answer 1

There are commutative domains (for example, any non-Noetherian valuation domain) which are not PIDs but for which every finitely generated torsion-free module is free.

Answer 2

It should be clear that free always implies torsion-free. So it suffices you assume $R$ is a domain where every torsion-free module is free. As noted in the comments, one should assume $R$ is noetherian. Now every ideal $\mathfrak a$ of $R$ is certainly torsion-free, so it must be free. Assume this is generated by more than one element. Let $a_1,a_2$ be distinct generators. Then $a_2a_1-a_1a_2=0$ is a dependence relation $b_1a_1+b_2a_2=0$, which implies $b_1=b_2=0$, contradicting $a_1,a_2$ are part of a basis. Note I didn't assume the ideal was finitely generated.

ADD One can show that in a domain where every finitely generated ideal is principal, torsion-free is equivalent to flatness. So you have a nice result that for PIDs, free equals projective equals flat.