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I want to solve an exercise from Liu's book Algebraic Geometry and Arithmetic Curves, namely exercise 4.9 in chapter 2: Let $X$ be a Noetherian scheme. Show that the set of points $x\in X$ such that $\mathcal{O}_{X,x}$ is reduced is open.

I already did the following: Because $X$ is Noetherian, we can write $X=\bigcup_{i=1}^n \operatorname{Spec}(A_i)$ where $A_i$ is a Noetherian ring. If we denote by $Z$ the set of points where $\mathcal{O}_{X,x}$ is reduced, then $Z$ is open if and only if $Z\cap \operatorname{Spec}(A_i)$ is open. So we can suppose $X=\operatorname{Spec}(A)$ with $A$ a Noetherian ring.

Then we also know that $\mathcal{O}_{X,x}=A_P$ with $P$ the prime ideal corresponding to the point $x$. So $\mathcal{O}_{X,x}$ is Noetherian too.

Can someone help me solving this problem?

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    This usually boils down to showing the following. Let $A$ be a local noetherian ring. Then, $A$ is reduced if and only if $A_p$ is reduced for all prime ideals $p$.2012-12-02

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Let $N$ be the nilradical of $A$. First observe that $O_{X,x}$ is reduced if and only if $NO_{X,x}=0$. Then show that $NO_{X,x}=0$ implies the same is true in some open neighborhood of $x$.

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    Yes, locally noetherian is enough.2012-12-04