Proof about finitely generated torsion-free R-module M is free, where R is a PID

Question

Proof about finitely generated torsion-free $R$-module $M$ is free, where $R$ is a PID.

Can I prove by the following way?

Proof by induction on $n$, where $M=\langle v_1,...,v_n\rangle$.

If $n=1$, then $M$ is cyclic, and it's easy to see $M\cong R$, thus $M$ is free.

For the inductive step, let $M=\langle v_1,...,v_{n+1}\rangle$ and define $S=\langle v_{n+1}\rangle$. And as M is torsion free thus its submodule M/S is torsion free. And clearly M/S is generated by n elements. Thus by inductive hypothesis, M/S is free. Thus M/S is a projective module. Thus by the sequence: 0$\to S \to M \to M/S \to 0$, we can get $M\cong S\oplus M/S$. And we know S is free, thus M is free.

Is there a mistake in my proof? Thank you!

Answer 1

The usual proof uses the Structure theorem for finitely generated modules over a principal ideal domain.

But there is an alternative proof based on the fact that the finitely generated torsion-free modules over an integral domain are isomorphic to a submodule of a free module of finite rank. (For a proof see here.) Since over a PID the submodules of free modules are also free, you can conclude.

Answer 2

Your proof is incorrect, sorry. The module $M/S$ is not a submodule of $M$ and indeed need not be torsion-free.

Example. Consider $R=\mathbb{Z}$; then $\mathbb{Z}=\langle 2,3\rangle$, but $\mathbb{Z}/\langle 3\rangle$ is not torsion-free.