Every module is isomorphic to a free quotient I am trying to prove that every module is isomorphic to a free quotient. Is the following a proof for that statement?
Let $M$ be an $R$-module. Choose $S \subset M$ such that $S$ generates $M$ with the generating set $\{s_i \}$. If $s_n$ is a generator that can be obtained from the other, i.e if $s_n = \sum r_is_i$, then remove $s_n$ from $\{s_i \}$. Now consider $\phi: (S) \to M$ given by $s \mapsto s$, then $\phi$ is surjective and $(S)/ \ker \phi \cong M$. Furthermore $(S)/ \ker \phi$ is free since we have killed every linear combination that equals $0$. Is this correct?
 A: As an example why this does not work, consider the $\mathbb{Z}$-module $M = \mathbb{Z}/4\mathbb{Z}$. It is generated by $S = \{1 + 4 \mathbb{Z}\}$ and as there is only one element in $S$ we cannot do any elimination. But then $(S) = M$ which is definitely not free.
From a more general point, note that you are trying to take a submodule $N = (S)$ of $M$, restrict the identity map $M \to M$ to a map $N \to M$ and want this to be surjective. Well, this will only work if $N = M$. Also, it will always have a trivial kernel, so using this to get $M$ as the quotient of some free module is bound to fail.
But there is a way to improve your idea to get the result. Instead of taking the submodule generated by $S$ in $M$, you can take $R^S$, the free $R$-module with basis $S$. If $S$ is a generating set (you do not have to do any elimination process here), then the natural map $R^S \to M$ induced by the restriction of the identity $M \to M$ to a map $S \to M$ will give you the result. (Note that you can even take $S = M$ and then you do not have to think about generating sets at all).
