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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 $\phi\circ \pi\neq\text{id}$ for any $\phi$, but I am confused here, why it's not. Hope someone can help me with that, thanks in advance.

janindunethmal
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user53800
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    Projectives are direct summands of free. Is this (torsion)free? –  Jan 04 '13 at 22:29
  • Z is $\mathbb Z$, right? (Try to use the right mathematical symbols!) –  Jan 05 '13 at 09:49
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    $\mathbb{Z}$ is indecomposable, because any two nonzero subgroups have non trivial intersection. So the only projective quotients of $\mathbb{Z}$ are $\mathbb{Z}$ and ${0}$. – egreg Nov 26 '15 at 21:59

2 Answers2

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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 $\phi\circ \pi\ne \mathrm{id}$.

Alex Becker
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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.

DonAntonio
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    You don't need projective <=> free (over PID) in order to see that projective => torsionfree. – Martin Brandenburg Jan 04 '13 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 04 '13 at 22:35
  • @MartinBrandenburg, I don't understand your comment. Anyway, I think that's straighter. – DonAntonio Jan 04 '13 at 22:46
  • @YACP Thanks a lot, I see that you have already give me a link there :) – user53800 Jan 04 '13 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: http://tinyurl.com/bkjfrxn . This group always intrigued me... – DonAntonio Jan 04 '13 at 23:03
  • @DonAntonio Thanks very much, I have been confused of this for a long time. By the way, why the torsion abelian group can not be free, can you show me – user53800 Jan 04 '13 at 23:13
  • @user53800, because it fulfills a non-trivial relation, which (abelian) free groups cannot. – DonAntonio Jan 04 '13 at 23:17
  • @DonAntonio while can you show me by the definition, namely that we can not find a basis for the torsion group? – user53800 Jan 04 '13 at 23:24
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    Of course: if for all $,x\in A,$ there exists $,n\in\Bbb N,,,s.t.,,,nx=0,$ ,then we can always find a non-trivial linear integral combination $,a_1x_1+...+a_mx_m=0,,,,x_i\in A,,,,m_i\in \Bbb Z,$...In fact, it is not hard to see the group isn't abelian free even if there is one single non-trivial torsion element in the group – DonAntonio Jan 04 '13 at 23:41