I came over a question in ring theory which I am not being able to proceed upon:
When is the ring $m\mathbb{Z}$ isomorphic to the ring $n\mathbb{Z}$, where $m, n \in \mathbb{N}$?
I know that to show isomorphism, I need to show that it is onto and the kernel consists of only $\{0\}_{m\mathbb{Z}}$ but I am not being able to write things down properly. Will someone help? What is the relation between $m$ and $n$?
On
So I came up with something following @QiaochuYuan 's approach. Suppose there exists an isomporphism $\psi :m\Bbb{Z}\to n\Bbb{Z}$
$$\psi(km)=pn\ \ (k,p\in\Bbb{Z})$$ $$\implies\psi(\ \underbrace{km+km+\cdots+km}_{km \text{ times}} )=kmpn=\psi((km)^2)=p^2n^2$$ $$\implies km=pn$$
As there must exist a $p$ for every $k$, therefore
Firstly if $(m,n)=1$ then this ismorphism can't exist.
Secondly if $m=\alpha n\ \ \ \alpha \ne1$ $$k\alpha=p\implies p\text{ can take values only in the set\{$\pm \alpha,\pm2\alpha,\cdots$\} and not in entire $\Bbb{Z}$} $$Hence not onto so no isomorphism.
Similarly if $n=\alpha m\ \ \ \alpha \ne1$ $$k=\alpha p\implies k\text{ can take values only in the set\{$\pm \alpha,\pm2\alpha,\cdots$\} and not in entire $\Bbb{Z}$}\implies Ker(\psi)\ne \{0\}$$Hence not an isomorphism.
Is this right or have I missed or over assumed something?
A ring isomorphism between $m\mathbb Z$ and $n\mathbb Z$, if exists, must be an isomorphism of underlying addive groups. Both of $(m\mathbb Z, +)$, $(n\mathbb Z, +)$ being infinite cyclic groups, there are only two such morphism, namely $km\mapsto kn, \forall k\in\mathbb Z$ and $km\mapsto -kn, \forall k\in\mathbb Z$. It's straightforward to verify that both morphisms don't preserve multiplicative structure if $m \neq n$.