# Is there a ring whose spectrum is homeomorphic to $\Bbb C$ with the Zariski topology?

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!

• 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. – 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.
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 $1$.
• 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. – Eric Wofsey Nov 6 '16 at 5:07