# One-dimensional scheme with no closed points

Can someone give a reasonably explicit example of an irreducible one-dimensional scheme with no closed points?

• @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. – user137767 Apr 17 '19 at 16:16
• 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. – Wojowu Apr 17 '19 at 16:19
• 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) – Wojowu Apr 17 '19 at 16:28
• If this question was interesting enough that an answer received 5 upvotes, then what reasonable grounds do we have for downvoting the question? – Mark Fischler Apr 17 '19 at 18:25
• @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. – user137767 Apr 17 '19 at 18:28

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).

• How do you know that $\overline{\{y\}}\subsetneq\overline{\{x\}}$? – rmdmc89 Apr 3 at 19:46
• That follows from the fact that $x\neq y$ since any scheme is $T_0$. – Eric Wofsey Apr 3 at 20:01

There cannot be such an example: first of all, a scheme is covered by affine schemes, and secondly, every nonempty affine scheme has at least one closed point.

The second point follows from one of the basic facts in commutative algebra that, every nonzero commutative ring with 1 always has a maximal ideal.

• I think you are assuming that the schemes are quasi-compact. – Francesco Polizzi Apr 17 '19 at 16:09
• I am sorry for being dense, but where exactly did you use the hypothesis of dimension 1 (which is necessary)? How do I see that a point that is closed in an affine open is closed in the entire space (e.g. this is not true for the generic point of the spectrum of a DVR, you probably have additional assumptions)? – user137767 Apr 17 '19 at 16:10
• @StepanBanach You don't need to use the dimension 1 assumption. – Jinhyun Park Apr 17 '19 at 16:11
• There exist non-quasicompact schemes without closed points, see math.stackexchange.com/questions/692049/… – Francesco Polizzi Apr 17 '19 at 16:11
• A point which is closed in an open subset needn't be closed in the whole space. – Wojowu Apr 17 '19 at 16:21