# 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?

• The so that... sentence needs justification (and is in fact incorrect). Consider $f:\Bbb Z \rightarrow \Bbb Z /2\Bbb Z$. Mar 29 '20 at 1:20
• You seem to have thought that, whenever $M$ admits a surjection from some other module, then $M$ is a direct summand of that other module. That property of a module $M$ is in fact equivalent to projectivity. Mar 29 '20 at 2:14

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, $$M \oplus K \cong R^m$$.