Why isn't it true that every finitely generated module is also projective? I know this can't be true, but I'm not sure why.
Let $M$ be a finitely generated $A$-module, say with $n$ generators. Then there is a surjection $f: R^n \to M$ in the obvious way, so that $M \oplus \ker f = R^n$. The definition of a projective module is one such that there exists a module $N$ so that $M \oplus N$ is free. In this case, $N = \ker f$ is chosen.
This 'proof' shows that the finitely generated module is always free, but I know this is wrong - why?
 A: Welcome to MSE!
Consider the homomorphism of $\mathbb{Z}$-modules $\pi : \mathbb{Z} \to \mathbb{Z}/2$. This is a surjection, but there is no module (abelian group) $M$ so that $\mathbb{Z} \cong M \oplus \mathbb{Z}/2$. Indeed, $\mathbb{Z}$ has no element of order $2$.
This is a "standard" reason for the lack of splittings - if $f : R^m \to M$ is a surjection, then we identify $R^m / K \cong M$ (where $K$ is the kernel of $f$). Of course, when we mod out by $K$ we are introducing relations among the elements of $M$ - that is what quotienting does, after all. If these relations are not present in $R^m$, then no splitting can exist.
Moreover, most of the time the relations in $M$ will not be present in $R^m$. After all, $R^m$ is free, and therefore satisfies only the relations which it must satisfy. So, as a point of intuition, if there is a splitting $g : M \to R^m$, then $K$ can't have added any relations that weren't already there. So the relations in $M$ look like the relations in $R^m$, we just killed some other part of $R^m$ in restricting our attention to $M$.
This is some (informal) justification for this property being equivalent to being a projective module, that is, $M \oplus K \cong R^m$.

I hope this helps ^_^
