Let $A$ be a commutative ring with unity different from $0$. Let $D(A)$ be the set of those prime ideals $\mathfrak{p}$ of $A$ which satisfy $$(*) \mbox{ there is } a\in A \mbox{ s.t }. \mathfrak{p} \mbox{ is a minimal among prime ideals containing }(0:a).$$
Q. Show that $x$ is a zero divisor if and only if $x\in\mathfrak{p}$ for some prime ideal in $D(A)$.
I was able to prove one way of the proposition:
(1) $x$ is zero divisor $\Rightarrow$ $x\neq 0$, $xy=0$ for some $y\neq 0$. So the ideal $(0:y)$ is a proper non-zero ideal of $A$.
(2) Then $(0:y)$ will be contained in some prime ideal of $A$ (Zorn's lemma)
(3) Hence (Zorn's) $(0:y)$ will be contained in a minimal among prime ideals containing $(0:y)$.
Clearly $x\in (0:y)\subset \mathfrak{p}$.
I didn't get any direction for the reverse way of the proposition. Any hint?