Can we contruct a basis in a finitely generated module

Question

Let $M=\langle e_1,\ldots,e_n\rangle$ be a finitely generated $R$-module.

My question is can we 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 know this is true in vector spaces, but in general free modules?

Thanks in advance

Answer 1

$\Bbb Z/3\Bbb Z$ is a finitely generated $\Bbb Z$ module, but it does not contain any free $\Bbb Z$ modules $\Bbb Z^s$ for any integer $s>0$.

Answer 2

In general, the construction you're looking for is not possible, unless you count a trivial solution which I'll explain in a moment. Consider the $\mathbb Z$-module $\mathbb Z/(4)\times\mathbb Z/(4)$ with generating set $S=\{(1,0),(0,1)\}$. No submodule of $S$ is isomorphic to $\mathbb Z$ or $\mathbb Z^2$.

On the other hand, if you allow the "rank $0$ module" $\{0\}$ and define the span of $\varnothing$ to be $\{0\}$ (this is a standard definition) then you can pick the zero submodule and the empty subset of $S$ to make your construction. But that's not very interesting.