A finitely generated torsion-free module A over a PID R is free

A finitely generated torsion-free module A over a PID R is free.
Proof
(Hungerford 221)

If A=0, then A is free of rank 0. Now assume A\neq 0. Let X be a finite set of nonzero generators of A. If x\in X, then rx=0 (r\in R) if and only if r=0 since A is torsion-free.

Consequently, there is a nonempty subset S=\{x_1,\dots,x_k\} of X that is maximal with respect to the property: \displaystyle r_1x_1+\dots+r_kx_k=0\ (r_i\in R) \implies r_i=0\ \text{for all}\ i.

The submodule F generated by S is clearly a free R-module with basis S. If y\in X-S, then by maximality there exist r_y,r_1,\dots,r_k\in R, not all zero, such that r_yy+r_1x_1+\dots+r_kx_k=0. Then r_yy=-\sum_{i=1}^kr_ix_i\in F. Furthermore r_y\neq 0 since otherwise r_i=0 for every i.

Since X is finite, there exists a nonzero r\in R (namely r=\prod_{y\in X-S}r_y) such that rX=\{rx\mid x\in X\} is contained in F:

If y_i\in X-S, then ry=r_{y_1}\dots r_{y_n}y_i\in F since r_{y_i}y_i\in F. If x\in S, then clearly rx\in F since F is generated by S.

Therefore, rA=\{ra\mid a\in A\}\subset F. The map f:A\to A given by a\mapsto ra is an R-module homomorphism with image rA. Since A is torsion-free \ker f=0, hence A\cong rA\subset F. Since a submodule of a free module over a PID is free, this proves A is free.

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