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.

I think the following may help. 
Say $\;M=\langle m_1,...,m_n\rangle_R\;$ , with $\;\mathcal X:=\{m_1,...,m_n\}\;$ a minimal generator set for $\;M\;$ as 
$\;R $ - module, and let $\;F:=F(m_1,...,m_n)\;$ be the free $\;R $ - module generated by $\;\mathcal X\;$ .
Define the function $\;f:\mathcal X\to M\;,\;\;f(m_i):=m_i\;$ , so by the universal property of free 
$\;R $ - modules, there exists a unique $\;R $ - homomorphism $\;\phi: F\to M\;$ extending $\;f\;$.
Clearly $\;\phi\;$ is surjective and by the isomorphisms theorems we get $\;F/\ker\phi\cong M\;$
