Prove that mZ, nZ is a ring isomorphic iff m = n

검색으로 많이 찾아봤지만 제대로된 증명이 없어서 직접 작성했다.
isomorphic은 symmetric하다는 점을 이용하면 가장 간단하게 증명할 수 있다.

$\text{Let }R, S \text{ be rings. then,}\\ R \text{ is isomorphic to }S \iff S \text{ is isomorphic to }R$

증명 : isomorphism은 bijective이므로 inverse function existence가 보장된다. inverse function 역시 isomorphism인 것을 보이는 것은 쉽다.

이제 다음을 증명하자.

$\text{Let } m,n \in \Z_{>0}. \\m\Z, n\Z \text{ are isomorphic} \iff m=n$

$\text{Proof.}\\$
$(\Leftarrow)$
$\text{Trivial by identity function } I:m\Z\to m\Z \\$
$(\Rightarrow) \\\text{Suppose that there is a ring isomorphism }f:m\Z\to n\Z.$
$f(m^2)=f(m)f(m)$
$f(m^2)=f(m) + \cdots + f(m) = mf(m) \quad (\text{Consider }m \text{ as scalar})$

$\text{Since }{n\Z}$$\text{ is an integral domain, we have }m=f(m).$
$\text{This implies that }n|m. \qquad (1)$

$\text{Now, Consider a ring isomorphism }g:n\Z \to m\Z.$
$g(n^2)=g(n)g(n)$
$g(n^2)=g(n) + \cdots + g(n) = ng(n) \quad (\text{Consider }n \text{ as scalar})$

$\text{Since }m\Z \text{ is an integral domain, we have }n=g(n).$
$\text{This implies that }m|n. \qquad (2)$

$\therefore m=n$$\quad \Box$