In a reduced ring the set of zero divisors equals the union of minimal prime ideals. 
If $R$ is a reduced commutative ring with identity, why is the set $Z$ of zero divisors the union of minimal prime ideals?

I know that $Z$ is a union of associated primes, and that the intersection of all the prime ideals  is the zero ideal  in case $R$ is reduced, but can not reach any relationship between this information and the statement of my question.
Any help would be thanked!
 A: Let $x\in R$ be a zero-divisor. We want to show that $x$ belongs to a minimal prime ideal. Let $y\in R$, $y\ne 0$ such that $xy=0$. If $x$ does not belong to any minimal prime, then from $xy=0\in\mathfrak p$, for $\mathfrak p$ a minimal prime, we get $y\in\mathfrak p$, so $y$ belongs to any minimal prime hence it is nilpotent and thus $y=0$, a contradiction.
The other containment follows from the well known fact that every minimal prime consists of zero-divisors.
A: This is because when $R$ is reduced, all associated prime ideals are minimal and hence $\operatorname{Ass} R= \operatorname{Min} R$.
Some details: The above assertion is true for noetherian rings. However, it remains true that the set of zero divisors in a reduced ring (not necessarily noetherian) ring is the union of all minimal prime ideals. To prove is, I'll use the weakly associated prime ideals, which is the relevant notion in the non-noetherian case (for noetherian rings,the two notions are the same).
Consider the natural mapping
$$R\longrightarrow \bigoplus_{\mathfrak p\in\operatorname{Min} R}R/\mathfrak p.$$
Its kernel is the nilradical of $R$ so that, since $R$ is reduced, it is injective, hence
$$\operatorname{Ass} R\subset \operatorname{Ass}\Bigl(\bigoplus_{\mathfrak p\in\operatorname{Min} R}R/\mathfrak p\Bigr)=\bigcup_{\mathfrak p\in\operatorname{Min} R}\operatorname{Ass}R/\mathfrak p =\operatorname{Min} R $$
so that $$Z(R)=\bigcup\limits_{\mathfrak p\in\operatorname{Ass}R}\mathfrak p\subset \bigcup\limits_{\mathfrak p\in\operatorname{Min} R}\mathfrak p$$
Conversely, let $\mathfrak p\in\operatorname{Min} R$ and $x\in\mathfrak p$. Consider the multiplicative set $$S=\bigl\{sx^k\mid s\in R\smallsetminus\mathfrak p,k\in\mathbf{N}\bigr\}.$$
Claim: $0\in S$. Note first $S$ contains $R\smallsetminus\mathfrak p$ and $x$. Now, if it were not true, there would be a prime ideal $\mathfrak p'$ disjoint from $S$, so $\mathfrak p'\subset \mathfrak p$, and the inclusion is strict since $x\notin \mathfrak p'$. But this contradicts the minimality of $\mathfrak p$.
Thus there exists an element $s\notin \mathfrak p$ and a natural number $k>0$ such that $sx^k=0$, but $sx^{k-1}\ne 0$. This proves every element of a minimal prime $\mathfrak p$ is a zero-divisor, and concludes the proof that
$$Z(R)=\bigcup\limits_{\mathfrak p\in\operatorname{Min} R}\mathfrak p.$$
A: Let $Z(R)$ denote the set of zero divisors of $R$, and let $\{P_i\}$ denote the set of minimal primes. Suppose that $x\in R$ is a zero divisor. Choose some nonzero $y\in R$ such that $xy = 0$. Since $R$ is reduced, the intersection of all primes is just $\{ 0\}$. However, every prime contains a minimal prime so $\cap_i P_i = 0$ as well. Hence $y\not\in P_i$ for some minimal prime $P_i$. But $xy = 0\in P_i$ so $x\in P_i$. Thus, $Z(R)\subseteq \cup_i P_i$.
We have to prove now that minimal primes consist of zero divisors to show the reverse inclusion.
Let $P$ be a minimal prime of $R$. Let $S = \{ xy : x\not\in P, y\not\in Z(R) \}$. Then $S$ is multiplicatively closed and does not contain zero. Let $Q$ be a prime ideal maximal with respect to being disjoint from $S$; then $Q$ is prime. By definition of $S$, no element of $Q$ can be written as $1\cdot y$ where $y$ is not a zero divisor, so $Q$ consists of zero divisors.
By definition of $S$, no element of $Q$ can be written as $x\cdot 1$ with $x\not\in P$. Hence $Q\subseteq P$ and so $P = Q$ by minimality of $P$. QED
