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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
The second equivalence should require some very mild finite type hypothesis, because there are schemes without closed points. But there may well be standing hypotheses in Eisenbud's book that make this work. – David Speyer Jan 17 '10 at 17:12
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
up vote 10 down vote accepted

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

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 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

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