If $m$ divides $n$, find a free resolution of $\mathbb{Z}/m$ as a $\mathbb{Z}/n$-module. 
If $m$ divides $n$, find a free resolution of $\mathbb{Z}/m$ as a $\mathbb{Z}/n$-module.

I have tried this one and got $0 \leftarrow \mathbb{Z}/m \leftarrow \mathbb{Z}/n \leftarrow \mathbb{Z}/n$.
The map $\mathbb{Z}/n \to \mathbb{Z}/n$ is multiplication by $m$, but I couldn't go further. Please help me.
 A: If $m = n$, then $0 \to \mathbb{Z}/n\mathbb{Z} \xrightarrow{\operatorname{id}} \mathbb{Z}/n\mathbb{Z} \to 0$ is a free resolution. If $m < n$, it will be useful to write $n = km$, where $k > 1$.
The first step of the resolution looks like $F_0 \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0$ for some free $\mathbb{Z}/n\mathbb{Z}$-module $F_0$. Exactness at $\mathbb{Z}/m\mathbb{Z}$ tells us that $\epsilon : F_0 \to \mathbb{Z}/m\mathbb{Z}$ is surjective, as such, $F_0 \neq \{0\}$, the free $\mathbb{Z}/n\mathbb{Z}$-module of rank zero. Let's see if we can take $F_0$ to be the free $\mathbb{Z}/n\mathbb{Z}$-module of rank one, i.e. $F_0 = \mathbb{Z}/n\mathbb{Z}$. As $\mathbb{Z}/n\mathbb{Z}$ is cyclic, any ring homomorphism $\epsilon : \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$ is completely determined by $\epsilon(1) \in \mathbb{Z}/m\mathbb{Z}$. Note however, as $n1 = 0$ in $\mathbb{Z}/n\mathbb{Z}$, we need to choose $\epsilon(1) \in \mathbb{Z}/m\mathbb{Z}$ such that $n\epsilon(1) = 0$, but that's true for any choice of $\epsilon(1)$ as $n\epsilon(1) = km\epsilon(1) = 0$ in $\mathbb{Z}/m\mathbb{Z}$. Therefore, we can complete the first step of the resolution by choosing $\epsilon(1) \in \mathbb{Z}/m\mathbb{Z}$; let's choose $\epsilon(1) = 1$.
The second step of the resolution looks like $F_1 \xrightarrow{d_1} \mathbb{Z}/n\mathbb{Z} \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0$ for some free $\mathbb{Z}/n\mathbb{Z}$-module $F_1$. Exactness at $\mathbb{Z}/n\mathbb{Z}$ tells us that the image of $d_1$ is equal to the kernel of $\epsilon$. Note that
$$\ker\epsilon = \{a \in \mathbb{Z}/n\mathbb{Z} \mid \epsilon(a) = 0\} = \{a \in \mathbb{Z}/n\mathbb{Z} \mid a\epsilon(1) = 0\} = \{a \in \mathbb{Z}/n\mathbb{Z} \mid a1 = 0\}.$$
As $a1 = 0$ in $\mathbb{Z}/m\mathbb{Z}$ if and only if $m \mid a$, $\ker\epsilon = \langle m\rangle = \{0, m, 2m, \dots, (k-1)m\} \cong \mathbb{Z}/k\mathbb{Z}$. Therefore, the image of $d_1$ is isomorphic to $\mathbb{Z}/k\mathbb{Z}$. In particular, $F_1$ can't be $\{0\}$; remember, $k > 1$. As before, let's try $F_1 = \mathbb{Z}/n\mathbb{Z}$. We would like to define a map $d_1 : \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ such that $\operatorname{im}d_1 = \langle m\rangle$. Again, as $\mathbb{Z}/n\mathbb{Z}$ is cyclic, it is enough to choose $d_1(1) \in \mathbb{Z}/n\mathbb{Z}$, and then $\operatorname{im}d_1 = \langle d_1(1)\rangle$. To ensure $\langle d_1(1)\rangle = \langle m\rangle$, choose $d_1(1) = m$. The map $d_1$ is simply multiplication by $m$.
The third step of the resolution looks like $F_2 \xrightarrow{d_2} \mathbb{Z}/n\mathbb{Z} \xrightarrow{\times\ m} \mathbb{Z}/n\mathbb{Z} \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0$ for some free $\mathbb{Z}/n\mathbb{Z}$-module $F_2$. Exactness at the first $\mathbb{Z}/n\mathbb{Z}$ tells us that the image of $d_2$ is equal to the kernel of multiplication by $m$. Note that $\ker(\times m) = \{a \in \mathbb{Z}/n\mathbb{Z}\ \mid am = 0\}$. As $am = 0$ in $\mathbb{Z}/n\mathbb{Z}$ if and only if $k \mid a$, $\ker(\times m) = \langle k\rangle = \{0, k, 2k, \dots, (m-1)k\} \cong \mathbb{Z}/m\mathbb{Z}$. Therefore the image of $d_2$ is isomorphic to $\mathbb{Z}/m\mathbb{Z}$. In particular, $F_2$ can't be $\{0\}$. As before, let's try $F_2 = \mathbb{Z}/n\mathbb{Z}$. We would like to define a map $d_2 : \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ such that $\operatorname{im}d_2 = \langle k\rangle$. Again, as $\mathbb{Z}/n\mathbb{Z}$ is cyclic, it is enough to choose $d_2(1) \in \mathbb{Z}/n\mathbb{Z}$, and then $\operatorname{im}d_2 = \langle d_2(1)\rangle$. To ensure $\langle d_2(1)\rangle = \langle k\rangle$, choose $d_2(1) = k$. The map $d_2$ is simply multiplication by $k$.
The fourth step of the resolution looks like $F_3 \xrightarrow{d_3} \mathbb{Z}/n\mathbb{Z} \xrightarrow{\times\ k} \mathbb{Z}/n\mathbb{Z} \xrightarrow{\times\ m} \mathbb{Z}/n\mathbb{Z} \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0$ for some free $\mathbb{Z}/n\mathbb{Z}$-module $F_3$. Exactness at the first $\mathbb{Z}/n\mathbb{Z}$ tells us that the image of $d_3$ is equal to the kernel of multiplication by $k$. Note that $\ker(\times k) = \{a \in \mathbb{Z}/n\mathbb{Z} \mid ak = 0\}$. As $ak = 0$ in $\mathbb{Z}/n\mathbb{Z}$ if and only if $m \mid a$, $\ker(\times k) = \langle m\rangle = \{0, m, 2m, \dots, (k-1)m\} \cong \mathbb{Z}/k\mathbb{Z}$. Therefore the image of $d_3$ is isomorphic to $\mathbb{Z}/k\mathbb{Z}$. In particular, $F_3$ can't be $\{0\}$. As before, let's try $F_3 = \mathbb{Z}/n\mathbb{Z}$. We would like to define a map $d_3 : \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ such that $\operatorname{im}d_3 = \langle m\rangle$. This is exactly what we had to do when determining $d_1$ above. There we found that we could take $d_1$ to be multiplication by $m$; for the exact same reasons, we can take $d_3$ to be multiplication by $m$.
As we saw in determining $d_3$ above, the construction starts to repeat itself. We can complete the resolution by taking all the free modules to be rank one, and the differentials (the maps $d_i$) alternating between multiplication by $m$ and multiplication by $k$. That is, a free resolution of $\mathbb{Z}/m\mathbb{Z}$ as a $\mathbb{Z}/n\mathbb{Z}$-module is
$$\dots \xrightarrow{\times\ k} \mathbb{Z}/n\mathbb{Z} \xrightarrow{\times m} \mathbb{Z}/n\mathbb{Z} \xrightarrow{\times\ k} \mathbb{Z}/n\mathbb{Z} \xrightarrow{\times m} \mathbb{Z}/n\mathbb{Z} \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0.$$
