What is a projective resolution of $\mathbb{Z}/m\mathbb{Z}$ as a $\mathbb{Z}/n\mathbb{Z}$-module? Let $m,n$ be positive integers with $m\mid n$. I want to compute Ext and Tor groups involving $\mathbb{Z}/m\mathbb{Z}$, and I think the easiest way is to find a projective resolution of it as a $\mathbb{Z}/n\mathbb{Z}$-module. I'm not sure how to do this. I think the way to start is to try to build a free resolution.
I have the following exact sequence of $\mathbb{Z}/n\mathbb{Z}$-modules.
$$
 0 \to \frac{n}{m}(\mathbb{Z}/n \mathbb{Z}) \hookrightarrow \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z} \to 0
$$
The first map is $x \mapsto x$, the second map is $x \mapsto mx$ and the third map is $x \mapsto x \bmod m$. Please correct me if this is not an exact sequence, or one of these maps is not $\mathbb{Z}/n\mathbb{Z}$-linear. 
My question is, is this projective? I don't think it is, because the first nonzero term isn't free. If this resolution isn't projective, how can I come up with one?
 A: If $n=p^r$ and $m=p^i\enspace(1\le i<r)$, you have a short non-split exact sequence:
$$0\longrightarrow \mathbf Z/p^i \mathbf Z\xrightarrow{\;\times\,p^{r-i}} \mathbf Z/p^r\mathbf Z\longrightarrow \mathbf Z/p^i \mathbf Z\longrightarrow 0$$
You can replicate this exact sequence on the left ad libitum, which shows $\mathbf Z/p^i \mathbf Z$ has infinite projective dimension over  $\mathbf Z/p^r\mathbf Z$, and a projective (actually free) resolution is:
$$\dots \xrightarrow{\;\times\,p^{r-i}} \mathbf Z/p^r \mathbf Z\xrightarrow{\;\times\,p^{r-i}} \mathbf Z/p^r \mathbf Z\xrightarrow{\;\times\,p^{r-i}} \mathbf Z/p^r\mathbf Z\longrightarrow \mathbf Z/p^i \mathbf Z\longrightarrow 0$$
I conjecture the same is true if some prime factor  of $m$ is a multiple factor of $n$, since it is known that $\mathbf Z/n\mathbf Z$ has global dimension $0$ if $n$ is square-free, $\infty$ otherwise.
A: Thanks to Lord Shark the Unknown for this answer to my other question How to compute Ext and Tor of $(\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/m\mathbb{Z})$ over the ring $\mathbb{Z}/n\mathbb{Z}$?.
The following is a projective resolution of $\mathbb{Z}/m\mathbb{Z}$ as a $\mathbb{Z}/n\mathbb{Z}$-module.
$$
\ldots \to \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z} \to 0
$$
The map $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$ is $x \mapsto x \bmod m$. The next map (going from the left) $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ is multiplication by $m$. The next map is multiplication by $n/m$. From then on, the maps alternate between multiplication by $m$ and multiplication by $n/m$, so the resolution is periodic.
A: Correct me if I am wrong, but I don't see a general answer here. E.g. 


*

*if $\mathbb{Z}_2$ is a $\mathbb{Z}_4$-module, then the projective dimension of $\mathbb{Z}_2$ is infinite; as described by @Bernard above.

*if $\mathbb{Z}_2$ is a $\mathbb{Z}_6$-module, then it is projective as a direct summand of the free module $\mathbb{Z}_6$ and therefore has a finite projective dimension.


Maybe the question could be more specified.
