About the module structure of $\mathbb{Z}/m\mathbb{Z}$. Let $m, n$ be arbitrary natural numbers. Let $R=\mathbb{Z}/n\mathbb{Z}$, and let $S=\mathbb{Z}/m\mathbb{Z}$.
For which values we can equip $S$ with an $R$-module structure?

I know that if $m \mid n$, then the multiplication by $\dfrac{n}{m}$ is a group homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to itself, and the image is isomorphic to $\mathbb{Z}/m\mathbb{Z}$ as a group.
This map does not preserve multiplication, and so this map is not a ring homomorphism. How should I prove that there is not any ring homomorphism between $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}/m\mathbb{Z}$, for $m\neq n$?
I don't have any idea about the module structure.
 A: $\def\Z{\mathbb{Z}}$
Suppose that $\mathbb{Z}/n\mathbb{Z}$ is a $\mathbb{Z}/m\mathbb{Z}$-module. Then
$$
(0+n\Z)(1+m\Z)=0+m\Z,
$$
but also
$$
(0+n\Z)(1+m\Z)=(n+n\Z)(1+m\Z)=n[(1+n\Z)(1+m\Z)]=n(1+m\Z)=n+m\Z.
$$
So $0+m\Z=n+m\Z$, that tells you that $m$ divides $n$, say $n=mk$. So this is necessary.
Now, we show that $m$ divides $n$ is sufficient. In order to prove that $\Z/m\Z$ is a $\Z/n\Z$-module we need to verify

*

*$(a+b)x=ax+bx$


*$a(x+y)=ax+ay$


*$a(bx)=(ab)x$


*$1x=x$
for all $a,b\in\Z/n\Z$ and all $x,y\in\Z/m\Z$.
But if $a=a'$ (mod $n$), then $a-a'=nt=mkt$, and so $a=a'$ (mod $m$). From this fact 1), 2) and 3) are true because $\Z/m\Z$ is a ring. 4) is also trivially true.
In conclusion $\Z/m\Z$ is a $\Z/n\Z$-module iff $m$ divides $n$.
A: Following the answer by @AntonioFicarra we know that "if $\mathbb{Z}/m\mathbb{Z}$ is a $\mathbb{Z}/n\mathbb{Z}$-module, then $m \mid n$.
Note that if $A\otimes_{\mathbb{Z}} B$ is an $A$-module. suppose that $m \mid n$, then $\mathbb{Z}/n\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}$ is a $\mathbb{Z}/n\mathbb{Z}$-module. Now note that $\mathbb{Z}/n\mathbb{Z}\otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}=\mathbb{Z}/m\mathbb{Z}$, so we can conclude that $\mathbb{Z}/m\mathbb{Z}$ is a $\mathbb{Z}/n\mathbb{Z}$-module.
