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