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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 $A$.

user26857
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jenny
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  • 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 05 '14 at 00:29
  • I think $;A;$ must be commutative Noetherian...? – DonAntonio Apr 05 '14 at 00:37
  • humm I don't know but we have not yet seen the Noetherian ring.... – jenny Apr 05 '14 at 01:15
  • 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 05 '14 at 01:35

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Hints (assuming commutative Noetherian):

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

DonAntonio
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