I would appreciate any help with the following problem. let $R$ be a commutative ring (with $1$). I need to show that the following are equivalent
i) for every prime ideal $P$, the localization $R_P$ is an integral domain
ii) for every maximal ideal $M$, $R_M$ is an integral domain
iii) for $x,y \in R$ such that $xy = 0$ we have $Ann(x) + Ann(y) = R$ where $Ann$ denotes the annihilator ideal.
Clearly, i) implies ii), and I showed that ii) implies iii). If iii) were false, then we can find $x,y \in R$ with $xy = 0$ and $Ann(x) + Ann(y) \subset M$ for some maximal ideal $M$. Then, in $R_M$ we have $\frac{xy}{1} = \frac{0}{1}$. Now, we are assuming that $R_M$ is an integral domain, hence either $\frac{x}{1} = \frac{0}{1}$ or $\frac{y}{1} = \frac{0}{1}$. Therefore, there exist $s,t \in R-M$ such that either $sx = 0$ or $ty = 0$. But $sx = 0$ means that $s \in Ann(x)$, hence $s \in M$, which contradicts the fact that $s \in R-M$. Similarly, $ty = 0$ leads to a contradiction. So ii) implies iii).
I am having problems proving that iii) implies i). Let $P$ be a prime ideal of $R$ and suppose that $\frac{x}{s} \cdot \frac{y}{t} =\frac{0}{1}$. This means that there exists $z \in R-P$ with $zxy = 0$. Now, I want to show that either $\frac{x}{s}$ or $\frac{y}{t}$ is equal to $0$, so I attempted to show that either $zx = 0$ or $zy = 0$. From the given assumption we have $Ann(zx) + Ann(y) = R$, but I do not see how to conclude that $Ann(zx) = R$ (or maybe this approach is not even correct).
I am studying this on my own, so this is not a HW problem. I would appreciate any suggestions. Thank you in advance.