# Reduced scheme and closed points

In The Geometry of Schemes by Eisenbud and Harris, Exercise I-32 asks one to show that a scheme $X$ is reduced if and only if every local ring $\mathcal{O}_{X,p}$ is reduced for closed points $p \in X$. However, this does not seem to work in general, since $X$ may not have enough closed points. What additional hypotheses on $X$ do I need for such an assertion to hold?

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Not all of us have the book at hand. Can you give more information on the question? – Anweshi Jan 17 '10 at 16:17
Exercise I-32. A scheme X is reduced if and only if every affine open subscheme of X is reduced, if and only if every local ring O_{X,p} is reduced for closed points p ∈ X. What is it that does not work? – Alberto García-Raboso Jan 17 '10 at 16:22
Brunoh, since you seem to be a beginner in scheme theory, I find it praiseworthy that you noticed this subtle mistake in a book written by such eminent algebraic geometers. Moreover this discussion is a service to our community, for which we should be very grateful to you. +1 – Georges Elencwajg Jan 17 '10 at 20:06
Hi,David and Kevin: I have just checked Eisenbud-Harris carefully from page 21 where they define schemes to page 26 where this treacherous exercise lurks: there is no standing assumption. They are just plain wrong. Could someone somehow somewhat downvote them, please :) – Georges Elencwajg Jan 17 '10 at 20:21
+1. I'm a bit late to this one, pity I missed it first time. But, fwiw, I also remember getting confused at that exercise, and the following paper helped me construct a counterexample via a scheme without closed points: math.utah.edu/~schwede/Papers/SchemeWithoutPoints.pdf. – George Lowther Nov 9 '10 at 22:58

There do exist schemes without a closed point, yes. (Liu, exercises 3.3.26/27)

But under some very reasonable additional conditions - I think quasi-compactness will be sufficient, if you are happy with using Zorn's lemma - the result holds. Use/prove the existence of a closed point, and the fact that localizing a reduced ring still gives you a reduced ring.

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Yes, quasi-compactness is enough. A quasi-compact scheme always has a closed point. If you consider a non-closed point, you can pick a closed point in its closure and use the fact that the stalk at the original point is a localization of the stalk at the closed point. This is a fairly standard trick. – Adam Topaz Jan 17 '10 at 18:50
That's exactly what I meant. – Wanderer Jan 17 '10 at 19:45
@AdamTopaz Could you please precise what this "standard trick" is exactly? Thanks! – ACL Apr 22 at 8:58

It seems to me that looking at closed points only is not sufficient since they are not always a dense set of X ...

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They do not always exist :) – Wanderer Jan 17 '10 at 16:48
My point was only, after reading this very clear and pedagogical book, to emphasize that an errata to this exercise was obviously necessary. The condition to look at closed points is not sufficient. Very reasonable conditions should be added (but quasi-compactness does not seem sufficient - must be enough closed points, like a dense set). I did not find a counter example nevertheless (like a non reduced scheme without closed points ...) Thank you all for your explanations, but what did i do to deserve a -1 in my question ? – brunoh Jan 17 '10 at 17:20
Don't get hurt that you got a neg vote. It happens sometimes. I was not the negvoter; but I suppose you would not have got it if you had included the exercise in your question. Make questions clear. At least now if you can edit and include the exercise, it would be nice. – Anweshi Jan 17 '10 at 19:12
Thank you very much for your explanation : I was not hurt, just trying to understand my mistake. I think you are right. – brunoh Jan 17 '10 at 19:52

I think that any quasi-compact, non-empty sober topological space has a closed point (a topological space $X$ is sober if every irreducible closed subset of $X$ has a unique generic point; any scheme is sober). So, let $X$ be such a space. Let $\mathscr F$ be the set of irreducible closed subsets of $X$, ordered by reverse inclusion. The set $\mathscr F$ is inductive; indeed, let $(F_i)_{i\in I}$ be a totally ordered family of irreducible closed subsets of $X$. Let's first prove that its intersection is non-empty. If it were empty then by quasi-compactness (in its dual version, involving intersection of closed subsets), they would exist a finite subset $J$ of $I$ such that $\bigcap _{i\in J}F_i=\varnothing$. But then we get a contradiction: if $J=\varnothing$, then $\bigcap _{i\in J}F_i=X$ and the latter is non-empty by assumption; and if $J$ is non-empty, $\bigcap _{i\in J}F_i=F_i$ for some $i\in J$, hence is non-empty. Now if $x$ is a point in $\bigcap F_i$, then $\overline{\{x\}}$ is an irreducible closed subset of $X$ contained in all the $F_i$'s. Therefore $\mathscr F$ is inductive. It thus has a maximal element $G$. Since $X$ is sober, its closed irreducible subset $G$ has a unique generic point $\eta$. Let $x\in G$. By maximality of $G$ one has $\overline{\{x\}}=G$, and $x=\eta$. As a consequence, $G=\{\eta\}$ and $\eta$ is closed.

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Yes you are right. Look at the comment of Adam Topaz. – brunoh Apr 21 at 18:00
@brunoh Note that Antoine Ducros has been careful enough to fill both the missing hypothesis in the statemetn and the details in the proof. :-) – ACL Apr 21 at 22:10
@ACL yep. That is why I gave him +1 for his effort. – brunoh Apr 21 at 23:35
Thank you ACL! I have just slightly rewritten the proof. – Antoine Ducros Apr 22 at 6:56
In fact, every $T_0$ quasi compact topological space has a closed point. By Zorn lemma there is a minimal closed subspace, which is a singleton by $T_0$. – Uri Bader Apr 22 at 12:00

Brunoh:

1) If $X$ is a quasi-compact scheme such that $\mathscr O_{X,x}$ is reduced for every closed point $x$, then $X$ is reduced. Indeed, let $y\in X$. The scheme $\overline{\{y\}}$ is a closed subscheme of $X$, hence is quasi-compact, and non-empty because it contains $y$. It thus has a closed point $x$, which is closed in $X$ as well. Now $\mathscr O_{X,y}$ is a localization of $\mathscr O_{X,x}$, hence is reduced because so is $\mathscr O_{X,x}$ by assumption.

2) Let $k$ be a field and let $v$ be the valuation on $k(X_i)_{i\in \mathbb Z_{> 0}}$ defined by the composition of the successive discrete valuations provided by the $X_i$'s. Let $X$ be the spectrum of the corresponding valuation ring. Then topologically, $X=\{x_0,\ldots, x_n,\ldots\}\bigcup \{x_\infty\}$ where every $x_i$ specializes to $x_{i+1}$, and where $x_\infty$ is the unique closed point (the point $x_0$ is the generic one, and $x_i$ corresponds to the prime ideal generated by $X_i$). Now if you remove $x_\infty$ you get an open subscheme $U$ of $X$, without any closed point. Of course, $U$ is reduced, but $U\times_k \mathrm{Spec}\; k[\epsilon]$ (with $\epsilon\neq 0$ and $\epsilon^2=0$) is not reduced, and homeomorphic to $U$.

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I don't think quasi-compactness is enough,for Noether scheme it is true. in a noether scheme, every point P has a closed point in its closure, so ..... but i don't find a necessary and sufficient condition

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If you look closely at the hint given by Adam Topaz, quasi-compactness appears for me clearly sufficient because you have enough closed points "next" to each point (in the closure), then you use the standard little trick ... – brunoh Apr 13 '11 at 18:38