Free $R$-module question 
Let $R$ be a ring with identity element. Let $M$ be a finitely generated $R$-module. Show that there is a free $R$-module $F$ and a submodule $K\subseteq F$ such that $M\cong F/K$ as $R$-modules. 

My first idea is that I know if $F$ is a $R$-module and $K$ a submodule of $F$ then $F/K$ is an $R$-module. 
I also feel like I should be playing around with $M/M_{tor}$ and I know for whatever free module $F$ I need a surjective mapping $\phi$ from $F$ to $M$ and then look at $ker(\phi)$. But beyond that I haven't an idea where to begin.

Let $I = \{[m]\,|\, m\in M\}$ be an index set indexed by the elements of $M$. Consider $F := \bigoplus_{[m]\in I}R.[m]$, the free $R$-module spanned by the symbols $[m]$ for $m\in M$. Then we have  a natural $R$-linear map
$$
f\colon F\longrightarrow M,\quad \sum_{[m]\in I} \lambda_{[m]}.[m] \longmapsto \sum_{[m]\in I}\lambda_{[m]}\cdot m,
$$
where the sums are finite. Obviously, $f$ is surjective. Setting $K:= \ker(f)$, we obtain by the homomorphism theorem for $R$-modules an isomorphism $F/K\cong \operatorname{im}(f) = M$.
