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Let A be a Noetherian ring, Is the set of prime ideals $\{p\in \operatorname{Spec} A| A_p$ is a reduced ring $\}$ an open subset of $\operatorname{Spec} A$ in Zariski topology?

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A similar problem: if $A$ is a noetherian ring, then the set of prime ideals $P$ of $A$ such that $A_P$ is a domain is an open subset of $\mathrm{Spec} A$. –  Andrea Sep 28 '11 at 20:26
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up vote 6 down vote accepted

Yes.

The key to your question is the easy to prove equality (in case of emergency: Atiyah-Macdonald, Corollary (3.12)) $$Nil(A_{\mathfrak p})= (Nil (A))_{\mathfrak p}$$

Since $A_{\mathfrak p}$ reduced means $Nil (A_{\mathfrak p})=0$, you are asking in effect about the set of points ${\mathfrak p}\in Spec(A)$ such that $(Nil (A))_{\mathfrak p}=0$.
This set is the complement of the closed set $supp (Nil(A))=V(Ann(Nil (A)))$ and is thus open.

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Thanks, the proof is elegant! –  Yubin Sep 27 '11 at 11:28
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