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?

It is not ring isomorphic. I will give a particular case and you will easily develop for general case. Take $n=2$ and $m=3$ and consider $f:3 \Bbb Z \to 2\Bbb Z$. If it were a ring isomorphism, then it maps generator to generator. So assume for example, $f(3)=2$ Now  $$f(9)=f(3+3+3)=2+2+2=6$$ and $$f(9)=f(3 \times 3)=f(3)f(3)=4$$ so we get $4=6$. This absurdity proves the result!
