7.3.1

Prove that $2\mathbb{Z} \not\cong 3\mathbb{Z}$.

$\textit{Proof}$ : Suppose to the contrary that $2\mathbb{Z} \cong 3\mathbb{Z}$. Then $\exists \varphi: 2\mathbb{Z} \rightarrow 3\mathbb{Z}$ s.t. $\varphi$ is a ring isomorphism. $$\varphi(0)=\varphi(0+0)=\varphi(0)+\varphi(0)$$ We can subtract $\varphi(0)$ from both sides since $(3 \mathbb{Z},+)$ is an Abelian group. So, $$\varphi(0)=0$$ Now, $$\varphi(0)= \varphi(1+1)=\varphi(1)+\varphi(1)=0$$ What is $\varphi(1)$? In $3 \mathbb{Z}$, $0+0=0$, $1+1=1$ and $2+2 \equiv 1$. From above, $\varphi(1)+\varphi(1)=0$ so the only possibility is $\varphi(1)=0$.

$\varphi(0)=0$ and $\varphi(1)=0 \Longrightarrow \varphi$ is not an injective map, which contradicts the assumption that $\varphi$ was an isomorphism.

$\therefore$ $2 \mathbb{Z} \not \cong 3 \mathbb{Z}$ $\blacksquare$