TITLE: the reduced locus of a Noetherian ring
QUESTION [5 upvotes]: 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?

REPLY [8 votes]: 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.