Is $m\mathbb{Z}$ not isomorphic to $n\mathbb{Z}$ when $m\neq n$?

Question

Exercise from "Abstarct Algebra: An Introduction" by T.W.Hungerford.

For each positive integer $k,$ let $k\mathbb{Z}$ denote the ring of all integer multiples of $k$. Prove that if $m\neq n$, then $m\mathbb{Z}$ is not isomorphic to $n\mathbb{Z}$.

I do understand that I should find some property $P$ that should be preserved by an isomorphism, however, just one of $m\mathbb{Z}$ and $n\mathbb{Z}$ had this property. Simply I can not find one. To me all properties that hold in $m\mathbb{Z}$, also hold in $n\mathbb{Z}$. Any suggestions what $P$ I should check?

Answer 1

Suppose $f:m\mathbb{Z} \to n\mathbb{Z}$ is our isomorphism map. Then $f(m) = nk$ for some $k \neq 0$, and $f(m^2) = n^2k^2$. That squaring should be suspicious.

On the other hand, $f(m^2) = f(m + m + \ldots + m) = mf(m) = mnk$.

Thus $m = nk$. (So we've already shown it if $n \nmid m$).

Since $m \neq n$, $k \neq 1$ - ring isomorphisms will take generators to generators. $m$ generates $m\mathbb{Z}$, but $nk$ for $k > 1$ does not generate $n\mathbb{Z}$.

Answer 2

An isomorphism of rings must especially be an isomorphism of the additive groups underlying the rings. Since the groups $m\Bbb Z$ and $n\Bbb Z$ are both cyclic, given a group isomorphism $f:n\Bbb Z \to m\Bbb Z$ it must map a generator to a generator. So either $f(n) = m$ or $f(n) = -m$. Show that in both of these cases $f(n^2) \neq (f(n))^2$, proving that it is not a ring homomorphism at all (assuming that $m \neq n$).