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?
Does this proof that $2\mathbb{Z}\not\cong 3\mathbb{Z}$ work?
3
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abstract-algebra
ring-theory
rngs
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0It 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. – 2012-07-26
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2It's OK, just don't forget to say why $\varphi(2)\neq 0$ (since for example, it is $1-1$ and $\varphi(0)=0$ – 2012-07-26
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0We 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. – 2012-07-26