Let $A$ be a commutative ring with unity that is not an integral domain and $\mathcal{P}$
be any prime ideal of $A$. Then I know that $A_{\mathcal{P}}$ is not
an integral domain using the correspondence between prime ideals of $A$
that does not meet $A \backslash \mathcal{P}$ and prime ideals of $A_{\mathcal{P}}$.
Just from curiosity, I wanted to prove it by a different approach, namely starting from
$f, g \in A$ both non-zero and $fg=0$, I wanted to construct two non zero
elements of $A_{\mathcal{P}}$ that multiply to $0$. Could someone possibly give me
a hand on how one could do this? Thanks!
Best Answer
Your claim is not true. Take $A := \prod_{i \geq 1} A_i,$ where each $A_i = \mathbb Z/2\mathbb Z.$ Then localization of $A$ at every prime ideal is $\mathbb Z/2\mathbb Z,$ which is a field. (Being integral domain is not a local property.)