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Let $X$ be a quasi-compact, separated scheme with a single closed point.

Is $X$ necessarily affine, and thus isomorphic to the spectrum of a local ring?

I could not think of a counter-example; is there one?

Bruno Joyal
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    I think you should require $X$ to be connected, otherwise the disjoint union of a scheme without closed points and the spectrum of a local ring is a counterexample. – Yuchen Liu Oct 23 '13 at 22:48
  • Dear @YuchenLiu, you are right. I have edited the question. Thank you! – Bruno Joyal Oct 23 '13 at 23:37
  • You should also require X to be separated. – Cantlog Oct 24 '13 at 11:48
  • Dear @Cantlog I suspect so as well. Do you have a non-separated counterexample? – Bruno Joyal Oct 24 '13 at 14:42
  • Karl Schwede wrote up an example of an affine scheme without closed points (easy to find on google), by constructing an affine scheme with a single closed point and a sequence of infinitely many non-closed points, and removing the closed point. I'm not sure if it works, but a candidate may be to glue two copies of this affine scheme with the closed point along their closed points. – Andrew Oct 24 '13 at 20:24
  • Dear @Andrew, I do not know how to glue schemes along closed subsets... – Bruno Joyal Oct 25 '13 at 12:21
  • Take a local scheme $X$ such that the complement $U$ of the closed point has no closed point in $U$. Let $V$ be a strict non-empty open subset of $U$. Glue $X$ and $U$ along $V$. – Cantlog Oct 25 '13 at 16:25
  • @Cantlog You are right. I have changed the hypotheses to "quasi-compact and separated", as they seem reasonable. – Bruno Joyal Oct 30 '13 at 22:17
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    You forgot the connectedness. I think MathOverflow is more appropriate for this question. – Cantlog Oct 31 '13 at 11:10
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    Dear @Cantlog: 'quasi-compact' also rules out Yuchen Liu's example, and I feel that it is a more natural condition. Perhaps you are right about MathOverflow. – Bruno Joyal Oct 31 '13 at 15:01
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    Sorry I didn't pay attention to this hypothesis. Then any non-empty closed subset has a closed point and the answer is easy (I let you find out a solution, and maybe replace with less restrictive condition). – Cantlog Oct 31 '13 at 16:58
  • @Cantlog You are right, there is an easy argument. So, if we want the question to be more interesting, do you think connectedness and separatedness are the right hypotheses? – Bruno Joyal Oct 31 '13 at 19:51
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    Dear Bruno, quasi-compact is too strong, and connectdess and separatedness are necessary without quasi-compactness. But I don't know if this is sufficient to conclude $X$ is affine. – Cantlog Oct 31 '13 at 20:40
  • Dear @Cantlog, I am still thinking about this question once in a while. By the way, thanks for the bounty on the "rigid field" question! You definitely didn't have to do that. :) – Bruno Joyal Nov 21 '13 at 04:13

1 Answers1

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Well, I am answering my own question. As Cantlog points out, under the present assumptions, the question has a simple answer.

Let $X$ be a quasi-compact scheme with a single closed point $x_0$. Let $U$ be an open affine containing $x_0$. In a quasi-compact scheme $X$, every nonempty closed set contains a closed point of $X$. This implies that $X-U$ must be empty, since it does not contain $x_0$. So $X$ is affine. (Separatedness was not even necessary.)

I still do not know whether the question has a positive answer if $X$ is only assumed to be connected and separated. It seems like a more serious question.

Bruno Joyal
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