Basis of a subset of finitely generated torsion-free module

Question

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:

Thanks in advance

Answer 1

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 $x_i$).