$\textbf{Proposition 1.15}$ If $A$ is a ring(commutative with unity) and if $D$ is the collection of zero-divisors of $A$, then $D = \cup_{x \neq 0} \sqrt{Ann(x)}$.
$\textbf{Proof:}$ $D = \sqrt{D} = \sqrt{\cup_{x \neq 0}Ann(x)} = \cup_{x \neq 0} \sqrt{Ann(x)}$
Everything is fine except the first equality: $D \subset \sqrt{D}$ is trivial. We only need to show that $\sqrt{D} \subset D$. If $x \in \sqrt{D}$ then for some $n \in \Bbb N $, $x^n \in D$(choose the least such $n$) and therefore there exists $y \neq 0$ such that $x^ny = 0$. So we get that $x(x^{n-1}y) = 0$. But unless $A$ is a domain, we can't guarantee that $x^{n-1}y \neq 0$ in which case $x \in D$ completing proof. Kindly shed some light on this matter.
So you have $x^ny=0$. If $x^{n-1}y\neq 0$, then you're done. Otherwise, $x^{n-1}y=0$. Now, if $x^{n-2}y\neq 0$, then you're done. Otherwise,....