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 $A$.

  • $\begingroup$ 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. $\endgroup$ – Oria Gruber Apr 5 '14 at 0:29
  • $\begingroup$ I think $\;A\;$ must be commutative Noetherian...? $\endgroup$ – DonAntonio Apr 5 '14 at 0:37
  • $\begingroup$ humm I don't know but we have not yet seen the Noetherian ring.... $\endgroup$ – jenny Apr 5 '14 at 1:15
  • $\begingroup$ Please try to use the search function before you ask questions, especially for ones like these that are exercise in a lot of books $\endgroup$ – rschwieb Apr 5 '14 at 1:35

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


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