In the ring $\mathbb{Z}_m$ the ideal $(n) \cong n\mathbb{Z}/m\mathbb{Z}$?

Question

How does one show in the ring $\mathbb{Z}_m$ that the ideal $(n) \cong n\mathbb{Z}/m\mathbb{Z}$?

I only know this for $n=1$ but it’s meant to be true for all $n \in \mathbb{Z}_m$

Answer 1

$(n)/(m)$ doesn't even make sense unless $(m) \subseteq (n)$, which is equivalent to $n | m$, so I'll assume this. But your statement is false unless $n = 0$ or $m = 0$. If neither is 0, then $(n)/(m) \cong \mathbb Z/(m/n)\mathbb Z$. This is a finite group, as $m/n \neq 0$. However, $(n)$ is infinite as $n \neq 0$ so these are not isomorphic.

Answer 2

Actually this is true only if $n\mid m$. Indeed $$n\cdot\mathbf Z/m\mathbf Z=(n\mathbf Z+m\mathbf Z)/m\mathbf Z=\gcd(m,n)\mathbf Z/m\mathbf Z.$$