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?
 A: They are isomorphic as abelian groups under addition, but not as rings. In $A:=n \Bbb Z$, we have that
$$\tag1 \forall x,\,\exists y,\,x\cdot x=\underbrace{y+y+\ldots+y}_{k\text{ summands}}$$
is a theorem if and only if $k$ is a divisor of $n$ (in particular for $k=n$ but not for any $k>n$). Hence $n\Bbb Z$ and $m\Bbb Z$ can be distinguished by the set of $k$ for which $(1)$ is true. If they were isomorphic, these sets would be equal.
A: 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!
A: 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 $m \Bbb Z \to m\Bbb Z$.
