The localization of the ring $\mathbb{Z} \times \mathbb{Z}$ at every prime ideal is an integral domain

Question

I want to show that the localization of the ring $\mathbb{Z} \times \mathbb{Z}$ at every prime ideal is an integral domain.

$\mathbb{Z} \times \mathbb{Z}$ is not an integral domain since $(0,1)\cdot(1,0)=(0,0)$. I think $(\mathbb{Z} \times \mathbb{Z})_P$ wouldn't be an integral domain for similar reason. Can anyone explain what difference the localization can make here? Thanks in advance for your help.

Answer 1

Every localization of $\Bbb{Z}\times\Bbb{Z}$ is an integral domain because every localization of $\Bbb{Z}$ is an integral domain, because $\Bbb{Z}$ is an integral domain. This follows immediately from the following simple fact:

If $R$ and $S$ are commutative unital rings and $P\subset R\times S$ is a prime ideal, then, after switching $R$ and $S$ if necessary, we have $P=Q\times S$, where $Q\subset R$ is a prime ideal, from which it follows that $$(R\times S)_P\cong R_Q.$$