11
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ I am completely stuck on it.I have no idea how to proceed at all $\endgroup$
    – gumti
    Jun 30 '13 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 '13 at 16:44
12
$\begingroup$

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?

$\endgroup$
0
14
$\begingroup$

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).

$\endgroup$
2
  • 3
    $\begingroup$ When lookin for an isomorphism, we may assume wlog. $n>m$. $\endgroup$ Jun 30 '13 at 17:26
  • 1
    $\begingroup$ Indeed so, @HagenvonEitzen. Thanks. $\endgroup$
    – DonAntonio
    Jun 30 '13 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.