# $\mathbb Z/n\mathbb Z$ is not a projective module

I want to show that $\mathbb Z/n\mathbb Z$ is not projective for $n\geq 2$. I choose the exact sequence $\mathbb Z\stackrel{\pi}\rightarrow\mathbb Z/n\mathbb Z\rightarrow 0,$ and from $\mathbb Z/n\mathbb Z$ to $\mathbb Z/n\mathbb Z$ choose the identity map, and let $\phi$ :$\mathbb Z/n\mathbb Z \rightarrow \mathbb Z$, if I can show there is no way the diagram is commuted, then it's done, namely $\text{id}=\phi\circ \pi$ can not be true for any $\phi$, but I am confused here, why it's not. Hope someone can help me with that, thanks in advance.

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Projectives are direct summands of free. Is this (torsion)free? – YACP Jan 4 at 22:29
Z is $\mathbb Z$, right? (Try to use the right mathematical symbols!) – YACP Jan 5 at 9:49

If $\phi:\mathbb Z/n\mathbb Z\to \mathbb Z$ is a homomorphism, then $\phi(i)$ must have finite order for all $i\in\mathbb Z/n\mathbb Z$, since $i$ has finite order. But the only element of $\mathbb Z$ with finite order is $0$. So $\phi$ must be trivial, hence $\phi\circ \pi\ne \mathrm{id}$.
More directly, an abelian group (i.e., a $\,\Bbb Z-$module) is projective iff it is a free abelian group, and clearly a torsion abelian group cannot be free.
 You don't need projective <=> free (over PID) in order to see that projective => torsionfree. – Martin Brandenburg Jan 4 at 22:34 Your answer has more meaning to me, for I don't know how to prove Z and the countable sum of Z is not free. Can you give me some help about this? – user53800 Jan 4 at 22:35 @MartinBrandenburg, I don't understand your comment. Anyway, I think that's straighter. – DonAntonio Jan 4 at 22:46 @YACP Thanks a lot, I see that you have already give me a link there :) – user53800 Jan 4 at 22:47 @user53800, the result you're asking about is Baer's proof of the Baer-Specker group $\,\Bbb Z^{\Bbb N}\,$. The proof can be found in Kaplansky's book "Infinite Abelian Groups", but also in the following PDF: tinyurl.com/bkjfrxn . This group always intrigued me... – DonAntonio Jan 4 at 23:03