In The Geometry of Schemes by Eisenbud and Harris, Exercise I32 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?

2$\begingroup$ Not all of us have the book at hand. Can you give more information on the question? $\endgroup$– AnweshiJan 17, 2010 at 16:17

2$\begingroup$ Exercise I32. 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? $\endgroup$– Alicia GarciaRabosoJan 17, 2010 at 16:22

2$\begingroup$ 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 quasicompactness 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 ? $\endgroup$– brunohJan 17, 2010 at 17:21

12$\begingroup$ 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 $\endgroup$– Georges ElencwajgJan 17, 2010 at 20:06

8$\begingroup$ Hi,David and Kevin: I have just checked EisenbudHarris 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 :) $\endgroup$– Georges ElencwajgJan 17, 2010 at 20:21
5 Answers
There do exist schemes without a closed point, yes. (Liu, exercises 3.3.26/27)
But under some very reasonable additional conditions  I think quasicompactness 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.

5$\begingroup$ Yes, quasicompactness is enough. A quasicompact scheme always has a closed point. If you consider a nonclosed 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. $\endgroup$– user1884Jan 17, 2010 at 18:50


1$\begingroup$ @AdamTopaz Could you please precise what this "standard trick" is exactly? Thanks! $\endgroup$– ACLApr 22, 2016 at 8:58
Brunoh:
1) If $X$ is a quasicompact 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 quasicompact, and nonempty 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 $U$.
It seems to me that looking at closed points only is not sufficient since they are not always a dense set of X ...


$\begingroup$ 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 quasicompactness 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 ? $\endgroup$– brunohJan 17, 2010 at 17:20

$\begingroup$ 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. $\endgroup$– AnweshiJan 17, 2010 at 19:12

1$\begingroup$ Thank you very much for your explanation : I was not hurt, just trying to understand my mistake. I think you are right. $\endgroup$– brunohJan 17, 2010 at 19:52
I think that any quasicompact, nonempty 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 nonempty. If it were empty then by quasicompactness (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 nonempty by assumption; and if $J$ is nonempty, $\bigcap _{i\in J}F_i=F_i$ for some $i\in J$, hence is nonempty. 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.

1$\begingroup$ Yes you are right. Look at the comment of Adam Topaz. $\endgroup$– brunohApr 21, 2016 at 18:00

$\begingroup$ @brunoh Note that Antoine Ducros has been careful enough to fill both the missing hypothesis in the statemetn and the details in the proof. :) $\endgroup$– ACLApr 21, 2016 at 22:10

$\begingroup$ @ACL yep. That is why I gave him +1 for his effort. $\endgroup$– brunohApr 21, 2016 at 23:35

$\begingroup$ Thank you ACL! I have just slightly rewritten the proof. $\endgroup$ Apr 22, 2016 at 6:56

$\begingroup$ 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 $T_0$. $\endgroup$ Apr 22, 2016 at 12:00
I don't think quasicompactness 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

1$\begingroup$ If you look closely at the hint given by Adam Topaz, quasicompactness 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 ... $\endgroup$– brunohApr 13, 2011 at 18:38