Are $m\mathbb{Z} \cong n\mathbb{Z}$ as rings for arbitrary $m,n \in \mathbb{N}$? Are $m\mathbb{Z} \cong n\mathbb{Z}$ as rings for arbitrary $m,n \in \mathbb{N}$?
$\alpha: m\mathbb{Z} \rightarrow n\mathbb{Z}$ 
$\alpha(m)=n$.
Then $\alpha(ma) = \alpha(mb) \rightarrow a=b$ so $\alpha$ is injective.
It can also be shown that this map is surjective, as well as a homomorphism.
So, they are isomorphic then?

More generally, consider a ring homomorphism $f:m \Bbb Z \to n\Bbb Z$.
Then, following the previous answers, we have
$$
f(m^2) = f(m+\cdots+m) = m f(m)
$$
and
$$
f(m^2) = f(m\cdot m) = f(m) f(m)
$$
Therefore, $m f(m) = f(m) f(m)$ and so


*

*If $f(m)=0$, then $f=0$.

*If $f(m)\ne0$, then $m=f(m) \in n\Bbb Z$, and so $n$ divides $m$.
Thus, $f$ is surjective iff $m=\pm n$. We can't have $m=-n$ because $mf(m)=f(m^2) = f(m)^2 \ge 0$. Therefore, $m=n$.
Bottom line: the only surjective ring homomorphism $m \Bbb Z \to n\Bbb Z$ is the identity $m \Bbb Z \to m\Bbb Z$.
