10

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

Zev Chonoles
  • 129,973
gumti
  • 319

2 Answers2

13

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).

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
12

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?

Espen Nielsen
  • 3,863
  • 17
  • 31
  • 1
    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 $m=n$ – Micheal Brain Hurts Jun 16 '22 at 21:11
  • What about the case $\phi(m)=-n,$ like in Arthur's answer, since both $n$ and $-n$ generate $n\Bbb Z$? – PinkyWay Dec 04 '23 at 06:30