Is there a commutative ring $R$ such that $\mathrm{Spec}(R)$ is homeomorphic to $\Bbb C$, both endowed with the Zariski topologies?

In other words, is $\Bbb C$ a spectral space, when it is endowed with the Zariski (i.e. cofinite) topology?

I know that $\mathrm{Spec}(\Bbb C[x])$ can be seen as $\Bbb C \cup \{(0)\}$, and that $\mathrm{SpecMax}(\Bbb C[x])$ is homeomorphic to $\Bbb C$. This question is closely related. We need to check the following conditions:

  • $X$ is sober. Since we have the cofinite topology, I think this is verified.

  • $X$ is compact. The compactness necessary condition seems to be verified.

  • If $U,V\subseteq X$ are compact open sets, then $U\cap V$ is also compact. I had more trouble to check this property, and also the next one.

  • The compact open subsets of $X$ form a basis for the topology of $X$.

I tried also to read this, but it didn't answer my question. Anyway, if $\Bbb C$ happens to be a spectral space, what would be a corresponding ring $R$? Its Krull dimension has to be infinite. $\color{white}{\text{In some sense, we could try to change $\Bbb C[X]$ into $R$ by removing the fact that it is an integral domain...}}$

Any comment will be appreciated. Thank you!

  • 5
    $\begingroup$ Isn't $\mathbb{C}$ still irreducible? It certainly can't be written as the union of two clsoed subsets. It has no generic point now though. $\endgroup$ – Alex Youcis Nov 5 '16 at 10:09

As Alex Youcis points out in his comment, $\mathbb{C}$ with the cofinite topology is not sober, since the whole space is irreducible but has no generic point.

But since you asked about the other conditions in the definition of spectral space: Note that our space is Noetherian. Equivalently, (exercise) every subset is compact. So the other conditions are trivially satisfied: the open sets are closed under intersection and form a basis for the space.

Finally, even if $\mathbb{C}$ were sober, I don't see why a corresponding ring would have infinite Krull dimension. The Krull dimension is one less than the length of a maximal chain of irreducible closed sets. Here the only irreducible closed sets are the whole space and the points, so we should expect Krull dimension $1$.

  • 2
    $\begingroup$ Or, since every point is closed, we might expect Krull dimension $0$ instead (since every prime should be maximal). Of course, this contradiction (the ring needs to have both dimension $0$ and dimension $1$) is just the non-sobriety of the cofinite topology causing trouble again. $\endgroup$ – Eric Wofsey Nov 6 '16 at 5:07
  • 2
    $\begingroup$ It occurs to me that I should perhaps say "drunkenness" instead of "non-sobriety". $\endgroup$ – Eric Wofsey Nov 6 '16 at 22:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.