# Is any connected dimension $0$ scheme affine?

Let $$X$$ be a connected Scheme of dimension $$0$$. Is $$X$$ necessarily affine ?

I know this is true if $$X$$ is Noetherian (even without assuming $$X$$ is connected). But what happens if $$X$$ is not Noetherian ?

Here is a counterexample. Let $$C$$ be the Cantor set (or, any Stone space with no isolated points). For each pair of distinct points $$x,y\in C$$, let $$D_{xy}$$ be the quotient of $$C$$ by the equivalence relation that identifies $$x$$ and $$y$$. Let $$U_{xy}=C\setminus\{x,y\}$$ and let $$V_{xy}$$ be the image of $$U_{xy}$$ in $$D_{xy}$$. Note that the quotient map $$C\to D_{xy}$$ restricts to a homeomorphism $$U_{xy}\to V_{xy}$$.

Now fix a field $$k$$, let $$A$$ be the ring of locally constant functions $$C\to k$$, and let $$B_{xy}$$ be the ring of locally constant functions $$D_{xy}\to k$$. There is a natural homeomorphism $$\operatorname{Spec} A\cong C$$ sending a point of $$A$$ to the ideal of functions which vanish at it, and similarly $$\operatorname{Spec} B_{xy}\cong D_{xy}$$. The quotient map $$C\to D_{xy}$$ induces a morphism $$\operatorname{Spec} A\to \operatorname{Spec} B_{xy}$$ which restricts to an isomorphism of schemes between $$U_{xy}$$ and $$V_{xy}$$, considered as open subschemes of $$\operatorname{Spec} A$$ and $$\operatorname{Spec} B_{xy}$$. Now let $$X$$ be the scheme obtained by gluing together $$\operatorname{Spec} A$$ and $$\operatorname{Spec} B_{xy}$$ for all pairs $$x,y$$ along these isomorphisms between $$U_{xy}$$ and $$V_{xy}$$. We will identify the copy of $$\operatorname{Spec} A$$ in $$X$$ with $$C$$ and the copy of $$\operatorname{Spec} B_{xy}$$ in $$X$$ with $$D_{xy}$$.

(If you have trouble visualizing this, it is analogous to the "line with doubled origin", except instead of "doubling" a single point of $$C$$, we have taken every pair of points in $$C$$ and "doubled" them but glued together their doubled versions.)

It is clear that $$X$$ is $$0$$-dimensional, since it is obtained by gluing together $$0$$-dimensional affine schemes. Clearly $$X$$ is not affine (for instance, it is not quasicompact since it is obtained by gluing together infinitely many affine open sets, all of which are irredundant). But I claim $$X$$ is connected. Indeed, suppose $$X=G\cup H$$ is a nontrivial partition of $$X$$ into open sets. Note that $$C$$ is dense in $$X$$ (since $$V_{xy}$$ is dense in $$D_{xy}$$ and has been identified with $$U_{xy}\subset C$$), so $$G\cap C$$ and $$H\cap C$$ are both nonempty. Let $$x\in G\cap C$$ and $$y\in H\cap C$$. Since $$C$$ has no isolated points and $$G$$ and $$H$$ are open, $$x$$ is not isolated in $$G\cap C$$ and $$y$$ is not isolated in $$H\cap C$$. But this means $$x$$ is in the closure of $$G\cap U_{xy}$$ and $$y$$ is in the closure of $$H\cap U_{xy}$$. It follows that the common image of $$x$$ and $$y$$ is in the closure of both $$G\cap D_{xy}$$ and $$H\cap D_{xy}$$, and thus is in both $$G$$ and $$H$$ since they are closed. This is a contradiction since $$G$$ and $$H$$ were disjoint.

If this construction seems kind of ridiculous, note that any affine $$0$$-dimensional scheme is totally disconnected (see If $R$ is zero-dimensional, then $\mathrm{Spec}(R)$ is Hausdorff and totally disconnected). So, to get a connected $$0$$-dimensional scheme with more than $$1$$ point, you have to somehow glue together a bunch of totally disconnected spaces along open sets to get a total space which is connected. The idea of the construction above is to start with the Cantor set and then glue on pieces which kill all the open partitions $$C=G\cup H$$ by making the closures of $$G$$ and $$H$$ intersect in the pieces that were glued on. Note that the non-Hausdorffness of the construction is crucial, since if our scheme were Hausdorff then all the clopen sets in any affine open would remain closed in the whole space by compactness.

• I can't quickly tell whether $D$ has something to do with all the $D_{xy}$ or whether $D$ is an arbitrary, particular $D_{xy}$. May 6, 2019 at 17:45
• Oops, that was just a mistake and I meant to write $D_{xy}$. Fixed now. May 6, 2019 at 18:18
• Thanks ... say, do you think the answer would be yes if there existed a monomorphism $X \to Y$ for some Affine, Noetherian scheme $Y$ of Krull dimension $1$ ? May 6, 2019 at 18:46
• I think so: I think having a monomorphism to any affine scheme forces $X$ to be Hausdorff. I haven't checked the details though. May 6, 2019 at 20:27
• Is your $X$ quasi-compact ? May 14, 2019 at 21:52