Integral domain and localization

Question

Let $ A $ be a commutative ring with unit. Show that the following are equivalent:

1. $ \forall \mathfrak{p} \in \operatorname{Spec}(A) $, $ A_\mathfrak{p} $ is an integral domain.

2. $ \forall \mathfrak{m} \in \operatorname{SpecMax}(A) $, $ A_\mathfrak{m} $ is an integral domain.

3. $ \forall x, y \in A $ such that $ xy = 0 $, then $ \operatorname{Ann}(x) + \operatorname{Ann}(y) = A $.

I already showed that $ 1 \leftrightarrow 2 \rightarrow 3 $

Any hint to show that $ 3 \rightarrow (1 \text{ or } 2) $?

Answer 1

Let show $3\Rightarrow 1$. Let $\mathfrak p$ be a prime ideal in $A$. If $\frac{a}{b}\cdot \frac{c}{d}=0$ in $A_{\mathfrak p}$, then $s ac =0$ for some $s\notin \mathfrak p$.

By hypothesis, there is $\alpha, \beta\in A$ such that $1=\alpha+\beta$ with $\alpha sa= \beta c=0$. On the other hand either $\alpha\notin \mathfrak p$ or $\beta\notin \mathfrak p$. In the first case you get $\frac{a}{1}=0$ in $A_{\mathfrak p}$, while in the second case you get $\frac c 1=0$. This proves $A_{\mathfrak p}$ is integral as we want.