Show that any finitely generated torsion free module is a free module..

Any help is appreciated ..

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- Jan 18th 2011, 03:36 AMmath.djModules over P.I.D
Show that any finitely generated torsion free module is a free module..

Any help is appreciated .. - Jan 18th 2011, 04:28 AMTheArtofSymmetry
See Hungerford's algebra book p221.

The sketch of the proof is

Let A be a finitely generated torsion-free module over PID R. Let X be a finite set of nonzero generators of A. Let S be a maximal subset of X such that the submodule F generated by the inclusion $\displaystyle i:S \rightarrow X$ is free. For each $\displaystyle y_i \in X \setminus S$, there exists $\displaystyle 0 \neq r_i \in R$ such that $\displaystyle r_iy_i \in F$. Let $\displaystyle r=\prod_i r_i$. It follows that $\displaystyle rA \subset F$. Since A is torsion-free, $\displaystyle f:A \rightarrow A$ given by $\displaystyle a \mapsto ra$ is a R-module homomorphism whose kernel is trivial. So, $\displaystyle A \cong rA$. Verify that rA is a submodule of A and a submodule of a free R-module for PID R is free. Thus, A is free.