Let $A$ be a commutative ring. Suppose $P \subset A$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors.
This can be proved using localization, when $A$ is noetherian: $A_P$ is local artinian, so every element of $PA_P$ is nilpotent. Hence every element of $P$ is a zero-divisor. (As Matt E has observed, when $A$ is nonnoetherian, one can still use a similar argument: $PA_P$ is the only prime in $A_P$, hence is the radical of $A_P$ by elementary commutative algebra.)
Can this be proved without using localization?
Best Answer
Denote set complements in $\rm A $ by $\rm\,\bar T = A - T.\, $ Consider the monoid $\rm\,S\,$ generated by $\rm\,\bar P\,$ and $\rm\,\bar Z,\ $ for $\rm\,Z = $ all zero-divisors in $\rm A $ (including $0).\,$ $\rm\,0\not\in S\ $ (else $\rm\, 0 = ab,$ $\rm\ a\in \bar P,$ $\rm\ b\in \bar Z\ $ $\rm \Rightarrow b\in Z),\,$ so we can enlarge $\,0\,$ to an ideal $\rm\,Q\,$ maximally disjoint from $\rm\,S.\, $ Since $\rm\,S\,$ is a monoid, $\rm\,Q\,$ is prime. $\rm\, S\,\supset\, \bar P \cup \bar Z\ \Rightarrow\ Q \subset \bar S \subset P\cap Z,\, $ so by minimality of $\rm\,P\,$ we infer $\rm\, P = Q \subset Z.\quad$ QED