Show that if $m$ and $n$ are distinct positive integers, then $m\mathbb{Z}$ is not ring-isomorphic to $n\mathbb{Z}$.

Can I get some help to solve this problem

  • $\begingroup$ I am completely stuck on it.I have no idea how to proceed at all $\endgroup$
    – gumti
    Jun 30, 2013 at 16:39
  • 2
    $\begingroup$ OK let's start by trying to show that $\mathbb{Z}$ is not ring isomorphic to $2\mathbb{Z}$ $\endgroup$
    – Amr
    Jun 30, 2013 at 16:44

2 Answers 2


Assume you have an isomorphism $\phi: m\mathbb{Z} \rightarrow n\mathbb{Z}$, $m\neq n$.

Since $m$ is a generator of $m\mathbb{Z}$, $\phi$ is determined by its value on $m$, which must be $n$ if $\phi$ is to be a bijection. How can you from this derive a contradiction?

  • $\begingroup$ To complete this, $ \phi$ is a group isomorphism then it should send the generator to the generator as if $ \phi(m)=nk$ then $k$ must be $1$ otherwise since it is surjective there is $\tilde m$ such that $ \phi(\tilde m)=n$ then $nk= \phi(\tilde m k)= \phi(m)$ by injectivity $m=\tilde m k$ but $m$ is such least positive integer then $k=1$ is must. Above shows $ \phi(m)=n$ then $ \phi(mk)=nk$ is required morphism. It contradicts since $n(mkk')= \phi(mk mk')= \phi(mk) \phi(mk')=n^2kk'$ then $nkk'(m-n)=0$ since $nkk'\neq 0$ and $n\mathbb Z$ is a domain implies $m=n$ $\endgroup$ Jun 16, 2022 at 21:11

Hints: suppose we have a ring homomorphism

$$\phi:m\Bbb Z\to n\Bbb Z\;,\;\;\text{with}\;\;\phi(m)=nz$$

But then

$$n^2z^2=\phi(m)^2=\phi(m^2)=\phi(\underbrace{m+m+\ldots+m}_{m\;\text{ times}})=m\phi(m)=mnz\implies m=nz$$

and this already is a contradiction if $\,n\nmid m\,$ , but even if $\,n\mid m\,$ then

$$\forall\,x\in\Bbb Z\;,\;\;\phi(mx)=xnz=xm\in m\Bbb Z\lneqq n\Bbb Z $$

and thus we have problems with $\,\phi\,$ being surjective (fill in details).

  • 3
    $\begingroup$ When lookin for an isomorphism, we may assume wlog. $n>m$. $\endgroup$ Jun 30, 2013 at 17:26
  • 1
    $\begingroup$ Indeed so, @HagenvonEitzen. Thanks. $\endgroup$
    – DonAntonio
    Jun 30, 2013 at 17:28
  • $\begingroup$ how can you assume there is only this kind of homomorphisms? $\endgroup$ Jun 16, 2022 at 19:59
  • $\begingroup$ Is there any other "kind" of homomorphism between these rings? Any function in fact must take an element of $\;m\Bbb Z\;$ to an element of $\;n\Bbb Z\;$ and thus the first line. The second line uses the fact that this function must in fact be a homomorphism of rings... $\endgroup$
    – DonAntonio
    Jun 17, 2022 at 9:23

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