Rings $2\Z$ and $3\Z$ are Not Isomorphic

Problems and solutions of ring theory in abstract algebra

Problem 170

Prove that the rings $2\Z$ and $3\Z$ are not isomorphic.
 
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Definition of a ring homomorphism.

Let $R$ and $S$ be rings.

  • A homomorphism is a map $f:R\to S$ satisfying
    1. $f(a+b)=f(a)+f(b)$ for all $a, b \in R$, and
    2. $f(ab)=f(a)f(b)$ for all $a, b \in R$.
  • A bijective ring homomorphism is called an isomorphism.
  • If there is an isomorphism from $R$ to $S$, then we say that rings $R$ and $S$ are isomorphic (as rings).

Proof.

Suppose that the rings are isomorphic. Then we have a ring isomorphism
\[f:2\Z \to 3\Z.\] Let us put $f(2)=3a$ for some integer $a$. Then we compute $f(4)$ in two ways.
First we have
\[f(4)=f(2+2)=f(2)+f(2)=3a+3a=6a.\]

Next we have
\[f(4)=f(2\cdot 2)=f(2)\cdot f(2)=3a\cdot 3a=9a^2.\] These are equal and hence we have
\[6a=9a^2.\] The only integer solution is $a=0$.

But then we have $f(0)=0=f(2)$, which contradicts that $f$ is an isomorphism (hence in particular injective).
Therefore, there is no such isomorphism $f$, thus the rings $2\Z$ and $3\Z$ are not isomorphic.


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