# Finitely generated torsion free module is free

I have to show that if $R$ is an Euclidean domain and $M$ is a finitely generated torsion-free $R$-module, then $M$ is free using the Structure Theorem. I can prove it using basic facts but can't see how to apply the Structure Theorem. Does it suffice to show that Tor(M) is isomorphic to the direct sum of $\frac{R}{<d_i>}$ where the $d_i$'s are the invariant factors from the Structure Theorem and if yes, why?

Cheers

• The structure theorem for finitely generated modules over PIDs says that $M$ is isomorphic to a direct sum $T \oplus F$, where $T$ is a finite direct sum of modules of the form $R/aR$ for $a \in R$, and $F$ is a finite direct sum of copies of $R$. Then $M$ is torsion free if and only if $T = 0$. – D_S Jun 7 '17 at 17:49