When is $\mathbb{Z}/m\mathbb{Z}$ a module over $\mathbb{Z}/n\mathbb{Z}?$

Question

I'm trying to find a torsion free finite module: let $M$ be such $R$-module. If for every $0\ne a\in M$ and every $0\ne r\in R$ we have $ra\ne 0$ then $$ra=sa\implies (r-s)a=0\implies r=s$$ so the cyclic module $R\langle a\rangle$ generated by any $a\ne0$ has as many elements as the ring itself, so it must have a finite number of elements. The go to for finite rings (and modules) are $\mathbb{Z}/n\mathbb{Z}$ but, for example $\mathbb{Z}/3\mathbb{Z}$ is not a $\mathbb{Z}/2\mathbb{Z}$-module. When does this happen? I'm tempted to say that $\mathbb{Z}/m\mathbb{Z}$ is a $\mathbb{Z}/n\mathbb{Z}$ if and only if $n$ divides $m$. The "if" part is obvious.

If the product is defined directly $(\overline{a},\overline{b})\mapsto\overline{ab}$ for $\overline{a}\in\mathbb{Z}/n\mathbb{Z}$ and $\overline{b}\in\mathbb{Z}/m\mathbb{Z}$ then $R\langle\overline{1}\rangle=\{\overline{0},\overline{1},\dots,\overline{n}\}$ is not a submodule of $\mathbb{Z}/m\mathbb{Z}$. If the operation is defined by the monomorphism $\mathbb{Z}/n\mathbb{Z}\hookrightarrow\mathbb{Z}/m\mathbb{Z}$ we are already assuming that $n$ divides $m$ (and the module wouldn't be torsion free) so this is no help.

Is there a non-trivial answer? Also, is there a finite and torsion free module? Thanks

Answer 1

$\newcommand{\intmod}[1]{\Bbb Z/#1\Bbb Z}\newcommand\End{\operatorname {End}}$First recall that $ \End (\intmod m)\cong\intmod m$ as rings. Moreover $\intmod m $ is a $\intmod n $-module if and only if there exists a ring homomorphism $\intmod n\to\End (\intmod m )$, hence this is true if and only if $m|n $.

Finally note that every integral domain is a torsion-free module over itself.

Answer 2

Another approach:

$\mathbf Z/m\mathbf Z$ is a $\mathbf Z$-module, and it becomes a $\mathbf Z/n\mathbf Z$-module if and only if $n$ annihilates its generator $[1]_m=1+m\mathbf Z$, i.e. $n\cdot 1=n\in m\mathbf Z\:$ — in plain language, if and only if $m$ divides $n$.