It is known that if $R$ is a PID, then a finitely generated $R$-module being torsion-free implies that it is free, which results from the Structural Theorem; and the converse also holds if $R$ is a Noetherian domain.
My question is, if $R$ is given as just a domain (without the Noetherian condition), does the converse still hold, i.e. if a finitely generated $R$-module being torsion-free implies that it is free, $R$ is a PID?
I conjecture that this is wrong, where for example taking $R = x\mathbb{Q}[x] + \mathbb{Z}$ which is Bezout could partially fix the problem in $\mathbb{Z}[x]$; but the proof is not clear to me.
Best Answer
As mentioned here, among integral domains, the condition that "every f.g. torsion-free module is free" is equivalent to being a Bezout domain.
So any non-Noetherian Bezout domain (like the one you suggested) will furnish a counterexample.