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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?

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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.

Dietrich Burde
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  • Thank you. I think i mixed it up a little, finite here means finite number of elements and not finitely generated? –  Sep 17 '17 at 16:25
  • 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? –  Sep 17 '17 at 16:29
  • For this see this duplicate, or this one. – Dietrich Burde Sep 17 '17 at 16:51
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    Not that it changes the answer in this specific case, but some extra care is needed. The stated result is when working in the category of all the modules. Restricting to a subcategory can in some cases introduce new projectives. – Tobias Kildetoft Sep 17 '17 at 16:55
  • @TobiasKildetoft, that was what I was thinking, I am a little confused: when considering the homomorphisms in my question, are we assuming that $M, N$ are also finite $\mathbb{Z}$-modules or can they be any $\mathbb{Z}$-modules? –  Sep 17 '17 at 17:07
  • @Sodan They must be finite. We never wish to consider anything outside the category. – Tobias Kildetoft Sep 17 '17 at 17:18