# No projective modules in category.

I have an exercise that reads:

Let $C$ be a the category of all finite $\mathbb{Z}$-modules, prove that there are no projective modules in $C$.

So, in order for $P$ to not be projective $\mathbb{Z}$-module I must prove that for every surjection $g: P \to M$ and every $f: N \to M$ it can't exist a homomorphism $h: P \to N$ such that $f \circ h = g$. My question is, as we are working in $C$ are both the modules $N$ and $M$ also assumed to be finite $\mathbb{Z}$-modules? Also, is a proof my contradiction a good idea?

Since $\mathbb{Z}$ is a PID, projective modules $P$ are free $\mathbb{Z}$-modules. However, since $P$ is finite of order $n$, we have $nP=0$, so that $P$ is not free - see this MSE-question.
• Yes, finite $\mathbb{Z}$-module, i.e., having finitely many elements. – Dietrich Burde Sep 17 '17 at 16:27
• Can u provide any proof or idea for a proof for the fact that $P$ is free because $\mathbb{Z}$ is a PID? – user117449 Sep 17 '17 at 16:29