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I am working on some homology and I just want to check if my thoughts are correct, I am working on projective modules over the ring $\mathbb{Z}_8$ and I want to show that $\mathbb{Z}_4$ is projective as a module over our ring, I know I just need to find a module such that their direct sum is free.

My thought is $\mathbb{Z}_2\oplus \mathbb{Z}_4$ being the answer with the homomorphism $$\varphi(a\oplus b)=4a+b +8\mathbb{Z}$$ It is evidently an monomorphism and equally so an epimorphism to me (I leave out the formal proof here) so it would be an isomorphism.

Or did I perhaps miss something that makes what I deem "evident" that is false?

Been answered, forgot the order of elements

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  • $\begingroup$ The bad news is that any element of $\mathbb Z_2\oplus \mathbb Z_4$ has order at most $4$, so your function is not a $\mathbb Z_8$ homomorphism. $\endgroup$ – Quang Hoang Oct 9 '15 at 7:03
  • $\begingroup$ Drats =< I felt I missed something, do you know any module there that would be projective? $\endgroup$ – Zelos Malum Oct 9 '15 at 7:07
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    $\begingroup$ That's the problem. Every element of $\mathbb Z_4$ has order at most $4$, so $\mathbb Z_4$ cannot be a direct summand of a free $\mathbb Z_8$. $\endgroup$ – Quang Hoang Oct 9 '15 at 7:23
  • $\begingroup$ Well I meant more are there ANY projective modules of $\mathbb{Z}_8$ that is not trivial? not neccisery for $\mathbb{Z}_4$ $\endgroup$ – Zelos Malum Oct 9 '15 at 7:41
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    $\begingroup$ @QuangHoang, well, $\mathbb Z_2$ is projective over $\mathbb Z_6$, so there is more to that, no? $\endgroup$ – Mariano Suárez-Álvarez Oct 9 '15 at 8:17
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$\newcommand{\Z}{\mathbb{Z}}\Z_4$ is not projective over $\Z_8$. Indeed, consider the following exact sequence: $$0 \to \Z_2 \to \Z_8 \to \Z_4 \to 0.$$ The map $i : \Z_2 \to \Z_8$ maps $1$ to $4$, and the map $p : \Z_8 \to \Z_4$ is the quotient map. Then this exact sequence is not split, i.e. there's no $s : \Z_4 \to \Z_8$ such that $p \circ s = \operatorname{id}_{\Z_4}$. To see this, note that a map $s : \Z_4 \to \Z_8$ is uniquely determined by $s(1)$; since $1 \in \Z_4$ has order $4$, $s(1)$ must have an order dividing $4$. It follows that $s(1) = 2k$ for some $k \in \Z_8$, and so $p(s(1))$ cannot equal $1$.

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  • $\begingroup$ Instead of torsion-free you probably want to say faithful or with vanishing annihilator. $\endgroup$ – Mariano Suárez-Álvarez Oct 9 '15 at 8:11
  • $\begingroup$ @MarianoSuárez-Alvarez Sorry, I got my definitions mixed-up. I'm correcting it. $\endgroup$ – Najib Idrissi Oct 9 '15 at 8:12
  • $\begingroup$ If $R=k\times k$ is a direct product of two fields, and $M=Re_1$ is the ideal generated by $e_1=(1,0)$, then $M$ is projective (the ring is semisimple, in fact) yet not faithful. The claim in your last paragraph need $R$ to be a domain, or something along those lines (and $\mathbb Z_8$ is not one :-) ) $\endgroup$ – Mariano Suárez-Álvarez Oct 9 '15 at 8:13
  • $\begingroup$ @MarianoSuárez-Alvarez You're completely right, I definitely made a mistake here... In fact I think $\Z_4$ is torsion-free over $\Z_8$, but not faithful, but as you point out a projective module isn't necessarily faithful. $\endgroup$ – Najib Idrissi Oct 9 '15 at 8:15
  • $\begingroup$ I find the lack of faith disturbing, jokes aside thank you. Made things clearer $\endgroup$ – Zelos Malum Oct 9 '15 at 8:22
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One way to proceed is to notice that $\mathbb Z_8$ is a local ring, so that its finitely generated projective modules are in fact free. Of course, this implies that finitely generated modules have at least $8$ elements.

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  • $\begingroup$ So the only one possible is $\mathbb{Z}_8$ which is free and ergo projective, I had hoped for something a wee bit more exotic. $\endgroup$ – Zelos Malum Oct 9 '15 at 8:20

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