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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!

user26857
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2 Answers2

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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.

Community
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user26857
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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.

egreg
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  • After reading my comments the OP has asked this as a new question: http://math.stackexchange.com/questions/1335828/quotient-module-and-submodule – user26857 Jun 23 '15 at 10:06