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!
Best Answer
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.