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!
Best Answer
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