# 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.

If $M=\langle e_1,\ldots,e_n\rangle$ is 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 a 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 it has a maximal element $\{x_1,\ldots, x_s\}$, $1\le s\le n$
• 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