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?

  • $\begingroup$ Hint: for each $x \in Z$, try to find an open neighborhood $U$ such that $\mathcal{O}_{X,y}$ is also reduced for all $y \in U$ (hence $U \subset Z$ and you can cover $Z$ with open sets this way). $\endgroup$ – user314 Dec 2 '12 at 13:24
  • $\begingroup$ 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 $p$. $\endgroup$ – Harry Dec 2 '12 at 15:26

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 $x$.

  • $\begingroup$ Actually, that was exactly my problem: I can't construct an open neighborhood of $x$ where the ring of germs is reduced. $\endgroup$ – Alies Dec 3 '12 at 13:14
  • 1
    $\begingroup$ @Alies: let $f_1,\dots, f_n$ be a system of generators of $N$. Then $NO_{X,x}=0$ if and only if $(f_i)_x=0$ for all $i$. Now, if $f\in A$ and $f_x=0$, can you find an open neighborhood of $x$ in which $f=0$ ? $\endgroup$ – user18119 Dec 3 '12 at 21:03
  • $\begingroup$ thank you very much! I didn't know how to use Noetherianity, but know I do! Thanks again! $\endgroup$ – Alies Dec 4 '12 at 16:37
  • $\begingroup$ Is it true that we didn't use quasi-compactness? So this statement would be true for locally Noetherian schemes as well? $\endgroup$ – Alies Dec 4 '12 at 17:00
  • $\begingroup$ Yes, locally noetherian is enough. $\endgroup$ – user18119 Dec 4 '12 at 19:19

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