# Minimal prime ideals consist of zerodivisors [duplicate]

I don't find the proof for this little demonstration ...

Let $P$ be a minimal prime ideal of $A$. Show that $P$ is contained in the set of zero divisors of $A$.

• Hint: if $\xi$ is the set of all ideals in which every element is a zero-divisor, then $\xi$ has maximal elements, and those maximal elements are prime ideals. It follows that the set of all zero-divisors is a union of prime ideals. – Oria Gruber Apr 5 '14 at 0:29
• I think $\;A\;$ must be commutative Noetherian...? – DonAntonio Apr 5 '14 at 0:37
• humm I don't know but we have not yet seen the Noetherian ring.... – jenny Apr 5 '14 at 1:15
• Please try to use the search function before you ask questions, especially for ones like these that are exercise in a lot of books – rschwieb Apr 5 '14 at 1:35

1) The only prime ideal of the localization $\;A_p\;$ is $\;pA_p\;$
2) We have that $\;x\in p\implies \frac x1\in pA_p\;$ is nilpotent