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

Question

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

Answer 1

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).

Answer 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?

Answer 3

Here is an alternate proof that I think is more straightforward to think about:

Let us say for contradiction that there exists an isomorphism $\phi$ from $<m>$ to $<n>$. As previous answers have noted, it must be true that $\phi(m) = n$ else it would not be surjective and so not an isomorphism. So, our general isomorphism here is $\phi(mx) = nx$. Now, note that $\phi(mn) = nn$. But also notice that $\phi(mm) = \phi(m)\phi(m) = nn$. So, we have that $\phi(mm) = \phi(mn)$. Since $\phi$ is an isomorphism, $\phi^{-1}$ is also an isomorphism in the reverse direction. Using this we have $\phi^{-1}(\phi(mm)) = \phi^{-1}(\phi(mn))$. Simplifying yields $mm = mn$. Algebra yields $m = n$. But this is a contradiction because $m$ and $n$ were known to be distinct.

Answer 4

Let the group index $|RR : R|$ denote the $spread$ of a rng $R$. Since $m\mathbb Z$ and $n\mathbb Z$ have different spreads $m$ and $n$ respectively, they cannot be isomorphic.