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Tengo un ejercicio 2.4.9 del Qing Liu Algebraic Geometry and Arithmetic Curves. La primera parte dice lo siguiente:

Si $\DeclareMathOperator{\Spec}{Spec} X$ esquema noetheriano, mostrar que $B=\left\{{x\in{X} \mid O_{X,x}  \text{ reducido}}\right\}$ es abierto.

Prueba:

Como $X$ es noetheriano entonces $X=\bigcup_i{X_i}$ unión finita de abiertos afines $X_i$ tales que $O_X(X_i)$ son anilos noetherianos. Podemos suponer que $X=\Spec(A)$ y $A=O_X(X)$ anillo noetheriano. Luego tenemos que probar que el conjunto $C=\left\{{p\in{\Spec(A) \mid A_p  \text{ reducido}}}\right\}$ es abierto. Tengo que probar que para algún $f\in{A}$, $D(f)=C$. Sabemos que $D(f)=\Spec(A)-V(\langle f \rangle)$ donde $V(\langle f \rangle)=\left\{{p\in{\Spec(A) \mid \langle f \rangle\subset{p} }}\right\}$.


I have an exercise 2.4.9 of Qing Liu's Algebraic Geometry and Arithmetic Curves. The first part says the following:

If $X$ is a noetherian scheme, show that $B = \{x \in X \mid O_{X,x} \text{ is reduced}\}$ is open.

Proof:

Since $X$ is noetherian, then $X = \bigcup_i X_i$ is a finite union of open affines $X_i$ such that $O_X(X_i)$ are noetherian rings. We can assume that $X = \Spec(A)$ and $A = O_X(X)$ is a noetherian ring. Then we have to prove that the set $C = \{p \in \Spec(A) \mid A_p \text{ is reduced}\}$ is open. I have to prove that for any $f \in A$, $D(f) = C$. We know that $D(f) = \Spec(A) - V(\langle f \rangle)$, where $V(\langle f \rangle) = \{p \in \Spec(A) \mid \langle f \rangle \subseteq p\}$.

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This is an expanded version of my original comment. As you have already noted we may take $X$ to be affine, so let $X=Spec A$ with $A$ noetherian. Let $I$ be the nilradical of $A$, that is $I=\{x\in A:\exists n, x^n=0\}$. As $A$ is noetherian, $I$ has finitely many associated primes $\mathfrak{p}_1,\ldots,\mathfrak{p}_n$. Associated primes have the property that that for any prime $\mathfrak{q}\subseteq A$, we have $IA_\mathfrak{q}\neq 0$ if and only if for some $\mathfrak{p}_i$ among the associated primes $\mathfrak{p}_i\subseteq \mathfrak{q}$.

Now you can check that for any prime $\mathfrak{q}$,$IA_{\mathfrak{q}}$ is the nilradical of $A_\mathfrak{q}$. Thus the nilradical of $A_\mathfrak{q}$ is zero (which is to say that $A_\mathfrak{q}$ is reduced) precisely when $\mathfrak{q}$ is not in $\bigcup_i V(\mathfrak{p}_i)$, which is precisely when $\mathfrak{q}\in X-\bigcup_i V(\mathfrak{p}_i)$. Now $\bigcup_i V(\mathfrak{p}_i)$ is a finite union of closed subsets so is closed. Thus we have shown that the reduced locus is open.

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