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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

$\newcommand{\Z}{\mathbb{Z}}\Z_4$ 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 $1$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.