Does this proof that $2\mathbb{Z}\not\cong 3\mathbb{Z}$ work?

Question

I'm trying to show $2\mathbb{Z}\not\cong 3\mathbb{Z}$ as rings. Suppose $\phi$ is an isomorphism. Then $\phi(2+2)=\phi(2\cdot 2)$, which implies $2\phi(2)=\phi(2)\phi(2)$. Then $2=\phi(2)$, which is impossible since $2\not\in 3\mathbb{Z}$. Is it this simple, or am I over stepping something?

Answer 1

your proof is correct, here I am writing in step by step,,Suppose their is a ring isomorphism $f:2\mathbb{Z}\rightarrow 3\mathbb{Z}$. Then $f(2)=3n$ for some in integer $n$. Now use the ring isomomorphism property that $f(a+b)=f(a)+f(b)$ and $f(a.b)=f(a).f(b)$ so $f(4)=f(2+2)=f(2)+f(2)=3n+3n$. Also $f(4)=f(2)f(2)=3n\times3n=9n^2$ But then comparing the two expressions we have for $f(4)$, we obtain $3n+3n=9n^2$ but this implies $n=0$ and we have $f(2)=0$ (since $f(2)=3n$ for some $n$ as we said above), however, we have $f(0)=0$ for ring ismorphism,so no such $f$ can exist and we must conclude that $2\mathbb{Z}$ is NOT isomorphic to $3\mathbb{Z}$.