# 8. Show 2Z not ring-isomorphic to 3Z

Show that ${2\mathbb{Z} \ncong 3\mathbb{Z}}$.

Assume ${f: 2\mathbb{Z} \xrightarrow{\cong} 3\mathbb{Z}}$. In ${2\mathbb{Z}}$, ${4=2+2=2*2}$. Thus ${f(4)=f(2)+f(2)=2f(2)}$ and also ${f(4)=f(2)*f(2)=f(2)^2}$. But ${2f(2)=f(2)^2\Longrightarrow f(2)*(f(2)-2)=0}$. Since ${f}$﻿ is by hypothesis an isomorphism, ${f(2)\neq 0}$, so ${f(2)=2}$. But ${2 \notin 3\mathbb{Z}}$, a contradiction. ${\hfill \mbox{\raggedright \rule{.07in}{.1in}}}$

This is a follow-on to Problem 7. I found it on the Internet. If at first glance the answer does not seem intuitive, your mind is probably confusing set isomorphism with ring isomorphism. The proof is short and sweet and so has appeal to me.

Note to students surfing the web to find answers to assigned problem sets: best not to copy my stuff; I am terrible at math.