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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 $n$?

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    $\begingroup$ Both rings have one of two possible distinguished elements which generate their underlying abelian group. The behavior of this element differs between the two rings if $m \neq n$; specifically, its square behaves differently. $\endgroup$ – Qiaochu Yuan Jun 22 '16 at 4:29
  • $\begingroup$ @QiaochuYuan I am a beginner in ring theory. Could you please explain better? $\endgroup$ – Qwerty Jun 22 '16 at 4:31
  • $\begingroup$ @Qwerty: then do the group theory problem first. $\endgroup$ – Hurkyl Jun 22 '16 at 4:34
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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 $m \neq n$.

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  • $\begingroup$ +1 for your answer! But is My answer correct? please check.. $\endgroup$ – Qwerty Jun 22 '16 at 7:04
  • $\begingroup$ @Qwerty You mean if it's correct that to show it's isomorphism is to show it's onto and has zero kernel? Surely it's true as long as you are considering ring morphisms (with or without identity). $\endgroup$ – Censi LI Jun 22 '16 at 7:08
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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?

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  • $\begingroup$ @QiaochuYuan Please consider correcting me if I am wrong. $\endgroup$ – Qwerty Jun 22 '16 at 6:29
  • $\begingroup$ @JonasMeyer Have I done right? I am a beginner and please help.. $\endgroup$ – Qwerty Jun 22 '16 at 6:30

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