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

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?

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It is perfectly clear in this context, but is a good practice to explicitely say “non-unital ring” (or “rng” or something) at least once when you talk about rings without 1. –  PseudoNeo Jul 26 '12 at 8:51
It's OK, just don't forget to say why $\varphi(2)\neq 0$ (since for example, it is $1-1$ and $\varphi(0)=0$ –  Belgi Jul 26 '12 at 8:53
We can remark that the same proof shows that there is as few isomorphisms between the $n \mathbb Z$ as possible: $n\mathbb Z$ is isomorphic to $m\mathbb Z$ if and only if $|n|=|m|$ and the only automorphism of $n\mathbb Z$ is the identity. –  PseudoNeo Jul 26 '12 at 9:06

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 $3\mathbb{Z}$.
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