Question

For each positive integer $k$, let $k\mathrm{\mathbb{Z}}$denote the ring of all integer multiples of$k$ (see Exercise 6 of Section 3.1). Prove that if $m\ne n$, then $m\mathrm{\mathbb{Z}}$is not isomorphic to$n\mathrm{\mathbb{Z}}$ .

Short answer

It is proved that, $m\mathrm{\mathbb{Z}}$and$n\mathrm{\mathbb{Z}}$ are isomorphic. But, by contraposition, if $m\ne n$, then$m\mathrm{\mathbb{Z}}$ is not isomorphic to$n\mathrm{\mathbb{Z}}$ .

Step-by-step solution

Determine k∈-1,1 .

Let’s assume that$m\mathrm{\mathbb{Z}}$ and$n\mathrm{\mathbb{Z}}$ are isomorphic$\left(k\mathrm{\mathbb{Z}}=\left\{kI\overline{)I\in \mathrm{\mathbb{Z}}}\right\}\right)$ that means there exists an isomorphism $f:m\mathrm{\mathbb{Z}}\to n\mathrm{\mathbb{Z}}$.

By using Theorem 3.10, $f\left(0\right)=0$.

Assume that$f\left(m\right)=kn\left(m\in m\mathrm{\mathbb{Z}}\right)$

Then, for arbitrary$I\in \mathrm{\mathbb{Z}}$

$f\left(lm\right)=If\left(m\right)=I\left(kn\right)$

By observation,

$...,f\left(-2m\right)=-2\left(kn\right),f\left(-m\right)=\left(-1\right)kn,f\left(0\right)=0,f\left(m\right)=\left(1\right)kn,f\left(2m\right)=\left(2\right)kn$

Therefore, $\left(f\right)=\left(kn\right)\mathrm{\mathbb{Z}}$.

On other side, as $f$is surjective, Image$\left(f\right)=n\mathrm{\mathbb{Z}}$

Thus,$n\mathrm{\mathbb{Z}}-\left(kn\right)\mathrm{\mathbb{Z}}$

Further simplify, assume that$k\notin \left\{-1,1\right\}$

But then,

$kn\overline{)n\Rightarrow n\in \left(kn\right)\mathrm{\mathbb{Z}}}$

As$n\in n\mathrm{\mathbb{Z}}$ it has$\left(kn\right)\mathrm{\mathbb{Z}}\not\subset n\mathrm{\mathbb{Z}}$ that is $\left(kn\right)\mathrm{\mathbb{Z}}\ne n\mathrm{\mathbb{Z}}$. This is a contradiction.

Thus,$k\in \left\{-1,1\right\}$ .

If m≠n, then mℤ is not an isomorphic to nℤ 

Case 1: $k-1$that is$f\left(m\right)-n$

Then,$f\left(lm\right)=\ln ,\forall l\in \mathrm{\mathbb{Z}}$

Now, let’s assume that,$f\left(2m\right)$ and$f\left(3m\right)$

$\begin{array}{rcl}f\left(2m\right)f\left(3m\right)& =& 2n·3n=6n^2\\ f\left(2m·3m\right)& =& f\left(\left(6m\right)m\right)=\left(6m\right)n=6mn\end{array}$

Thus,$f$ is homomorphism, it can be:

$\begin{array}{rcl}6n^2& =& 6mn/:6n\ne 0\\ n& =& m\end{array}$

Therefore, $m=n$.

Case 2:$k-1$ that is$f\left(m\right)-n$

Then, for every integer If$\left(lm\right)=-\ln$

From case 1,$m=n$ .

Thus, $m\mathrm{\mathbb{Z}}$and$n\mathrm{\mathbb{Z}}$ are isomorphic.

But by contraposition if$m\ne n$ , then $m\mathrm{\mathbb{Z}}$is not isomorphic to$n\mathrm{\mathbb{Z}}$ .