Can someone give a reasonably explicit example of an irreducible one-dimensional scheme with no closed points?
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$\begingroup$ @StevenLandsburg no, I am OK with the axiom of choice (and I am not sure that examples of this spirit depend on it). What I mean is that one could probably give an answer like "take this well-known example of a scheme with no closed point and take the closure of this point", but this closure is not necessarily easy to compute. What I want is fairly concrete presentation of the example. $\endgroup$– user137767Commented Apr 17, 2019 at 16:16
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1$\begingroup$ Every noetherian topological space has a closed point - just keep taking smaller and smaller closures of points, eventually you will find a closed point. For a one-dimensional scheme, you find that every non-generic point is closed. $\endgroup$– WojowuCommented Apr 17, 2019 at 16:19
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$\begingroup$ Apologies, I kind of forgot what the definition of noetherian. You only need to assume your space has no infinite descending chains of irreducible closed subsets for the argument to work (which follows trivially from finite-dimensionality). Also, as Eric notes in his answer, you need to assume the space is $T_0$ (which schemes always are) $\endgroup$– WojowuCommented Apr 17, 2019 at 16:28
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1$\begingroup$ If this question was interesting enough that an answer received 5 upvotes, then what reasonable grounds do we have for downvoting the question? $\endgroup$– Mark FischlerCommented Apr 17, 2019 at 18:25
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$\begingroup$ @MarkFischler the question is kind of way too trivial and I am ashamed of having asked it. The proper policy, I believe, is not to give any answers to an off-topic question and to close it asap (and if there any answers, even good ones, to down-vote them to discourage answering off-topic questions). So in this case, down-votes for the question are justified and up-votes for the answer are not, I think. $\endgroup$– user137767Commented Apr 17, 2019 at 18:28
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1 Answer
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If $X$ is a 1-dimensional scheme with no closed points, let $x\in X$ be a point. Since $x$ is not closed, then there is some point $y\neq x$ in its closure. Since $y$ is not closed, there is a point $z\neq y$ in its closure. Now we have a chain $$\overline{\{z\}}\subset\overline{\{y\}}\subset\overline{\{x\}}$$ of irreducible closed subsets. This contradicts the assumption that $X$ is 1-dimensional.
More generally, the same argument shows any nonempty finite dimensional $T_0$ space has a closed point (the $T_0$ assumption is used to guarantee that the inclusions as above are strict).
answered Apr 17, 2019 at 16:24
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$\begingroup$ How do you know that $\overline{\{y\}}\subsetneq\overline{\{x\}}$? $\endgroup$– rmdmc89Commented Apr 3, 2020 at 19:46
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$\begingroup$ That follows from the fact that $x\neq y$ since any scheme is $T_0$. $\endgroup$ Commented Apr 3, 2020 at 20:01