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

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!

• Your proof is correct. – Crostul Jun 22 '15 at 12:41
• @Crostul Please explain me why $M/S$ is a submodule of $M$ as the OP claims. – user26857 Jun 22 '15 at 15:32
• $M/S$ is free by inductive hypothesis, hence the sequence splits. This implies that $M \cong S \oplus M/S$. – Crostul Jun 22 '15 at 15:40
• @Crostul $M$ is $(n+1)$-generated, $S$ is generated by one of those generators, so $M/S$ is torsion-free (???) and by the induction hypothesis it's free. Not so clear, right? – user26857 Jun 22 '15 at 20:30
• @user26857 you are right, I made a mistake. But why not just talking with me directly? – 6666 Jun 23 '15 at 6:06

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.