I'm trying to find conditions for $m$ so $\mathbb{Z}_n$ could be seen as a $\mathbb{Z}_m$-module via $(a,b) \mapsto ab \in \mathbb{Z}_n$ where $a \in \mathbb{Z}_m$ and $b \in \mathbb{Z}_n$.
I proved that $n$ divides $m$ but I can't find any other condition, is there anything I'm missing?
Thank you!
Edit: some typos
First of all, you want the operation $\Bbb Z_m \times \Bbb Z_n \ni (\hat a, \bar b) \mapsto \overline{ab} \in \Bbb Z_n$ to be well-defined. If $\hat c = \hat a$ and $\bar d = \bar b$ then there exist $k,l \in \Bbb Z$ such that $c = a + km$ and $d = b + ln$, which implies that
$$\overline{cd} = \overline{ab + aln + bkm + klmn} = \overline{ab} + \overline{bkm} .$$
If $n \mid m$ then $\overline{cd} = \overline{ab}$ and the multiplication with scalars is well-defined. It is then straightforward to check that this operation has all the required properties, without any further assumption. So far, the condition $n \mid m$ is sufficient.
Let us show that it is also necessary: if $\Bbb Z_n$ is a $\Bbb Z_m$-module then, in particular, it is true that
$$\bar 0 = \hat 0 \cdot \bar 1 = \hat m \cdot \bar 1 = \bar m ,$$
whence it follows that $n \mid m$.
It follows that the only necessary and sufficient condition for $\Bbb Z_n$ to be a $\Bbb Z_m$-module is $n \mid m$.