# Basis of a subset of finitely generated torsion-free module

Based on the comments of rschwieb's answer in this question asked recently: Can we contruct a basis in a finitely generated module.

Let $$R$$ be an integral domain. Let $$M=\langle e_1,\ldots,e_n\rangle$$ be a finitely generated torsion-free $$R$$-module. I'm trying to construct a free submodule $$F$$, i.e, isomorphic to $$R^s$$ for some $$s$$, finding a subset $$S=\{e_1,\ldots,e_s\}$$ such that $$S$$ is a maximal independent subset of $$M$$, then $$S$$ generates this free submodule $$F$$ of $$M$$ with basis $$S$$.

I'm asking that because I didn't understand why Peter Clark in his commutative algebra pdf wrote this: Let's call $$S$$ the L.I. subsets of $$\{x_1,\ldots,x_n\}$$. Since $$M$$ is torsion-free, for $$x_1\in M,r\in R$$, we have $$rx_1=0$$ iff $$r=0$$. Then $$S$$ is non-empty.
Since $$\{x_1,\ldots,x_n\}$$ is finite, then $$S$$ has a maximal element $$\{x_1,\ldots, x_s\}$$, $$1\le s\le n$$ (after possibly relabeling the $$x_i$$).
• I think this isn't quite enough for Prof. Clark's claim that the given $s$ is unique, but this can be obtained by tensoring with $R$'s fraction field and invoking the well-definedness of dimension for a vector space over a field. On the other hand, I don't see that Prof. Clark actually needs $s$'s uniqueness in the present proof. – Ben Blum-Smith Dec 17 '15 at 20:31