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Let $A$ be a commutative ring with $1$. $D(A)$ is the set of prime ideals $p$ which satisfy the following condition: there exists $a\in A$ such that $p$ is minimal in the set of prime ideals containing $(0:a)$, where $(0:a)=\{r\in A:ra=0\}$. I want to prove that

$x\in A$ is a zero divisor iff $x\in p$ for some $p\in D(A)$

One direction is easy: suppose $x$ is a zero divisor. Then $\exists y\not =0$ such that $xy=0$, so $x\in (0:y)$. Since $y\not =0$, $(0:y)\not=(1)$. Then the set of prime ideals containing $(0:y)$ is not empty and hence has a minimal one which contains $x$.

But how to prove another direction?

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I'll give you a series of statements, by combining them you can arrive at a proof of what you want (although there might be other ways too). Let $A$ be a commutative ring with unity.

  • The radical of an ideal $I$, i.e. $\sqrt{I}$ is the intersection of all prime ideals containing $I$.

  • Given a prime ideal $\mathfrak{p}\subset A$, the localization $A_\mathfrak{p}$ is a local ring.

  • There is a one-to-one correspondence between (prime, radical) ideals of $A_\mathfrak{p}$ and (prime, radical) ideals of $A$ contained in $\mathfrak{p}$.

Prove that: If $a$ is a zero-divisor and $\mathfrak{p}$ the minimal prime ideal containing $(0:a)$, then $\sqrt{(0:a)A_\mathfrak{p}}=\mathfrak{p}A_\mathfrak{p}$. Use this to prove what you want.

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  • $\begingroup$ I think it is always ture that $\sqrt {(0:a)A_{p}}=pA_{p}$ $\endgroup$ – Mike Apr 27 '18 at 14:14
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I think this is a minor variant of Hamed's answer.

The statement is an immediate consequence of the following fact:

If $\mathfrak a$ is an ideal of a commutative ring $A$ with $1$, and if $\mathfrak p$ is a minimal prime over $\mathfrak a$, then $$ \mathfrak p\subset\bigcup_{a\notin\mathfrak a}\ (\mathfrak a:a). $$ Proof. Let $x$ be in $\mathfrak p$. As $\mathfrak p_{\mathfrak p}$ is the radical of $\mathfrak a_{\mathfrak p}$, there is a positive integer $n$ and an $s$ in $A\setminus\mathfrak p$ such that $x^ns\in\mathfrak a$. We can assume that $n$ is minimum for this condition. This gives $x\,(x^{n-1}s)\in\mathfrak a$ and $x^{n-1}s\notin\mathfrak a$.

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