Can anyone help with the following problem:
Prove that the rings $\displaystyle 2\mathbb{Z}$ and $\displaystyle 3\mathbb{Z}$ are not isomorphic?
Be grateful for help
Peter
Let $\displaystyle f:2\mathbb{Z} \to 3\mathbb{Z}$ be a ring isomorphism. Then $\displaystyle f(2) \in 3\mathbb{Z}$. Thus $\displaystyle f(2) = 3 x$, $\displaystyle x \in \mathbb{Z}$.
Therefore $\displaystyle f(4) = f(2+2) = f(2)+f(2) = 6x$ and $\displaystyle f(4)=f(2\cdot2)=f(2) f(2)=9x^2$.
So, $\displaystyle 6x=9x^2 \Rightarrow x=0$ or $\displaystyle x=2/3$.
$\displaystyle x=0$ means $\displaystyle f(2)=0$, so $\displaystyle f(2\lambda ) = f(2)f(\lambda )=0$, $\displaystyle \forall \lambda \in \mathbb{Z}$ which means that $\displaystyle f $ is not an isomorphism,
and $\displaystyle x=2/3$ with $\displaystyle x\in \mathbb{Z}$ is Absurd.
So, the rings $\displaystyle 2\mathbb{Z}$ and $\displaystyle 3\mathbb{Z}$ are not isomorphic.