**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.1**

Prove that the rings $2\mathbb{Z}$ and $3\mathbb{Z}$ are not isomorphic.

Solution: Suppose $\varphi : 2\mathbb{Z} \rightarrow 3\mathbb{Z}$ is an isomorphism. Then $\varphi(2) = 3a$ for some integer $a$. Now $$\varphi(4) = \varphi(2 \cdot 2) = \varphi(2) \cdot \varphi(2) = 9a^2,$$ and also $$\varphi(4) = \varphi(2 + 2) = \varphi(2) + \varphi(2) = 6a,$$ so that $6a = 9a^2$. Since $\mathbb{Z}$ is an integral domain, we have $2 = 3a$. However, this equation has no solutions in the integers, and we have a contradiction. Thus no such isomorphism exists.