Ring with spectrum homeomorphic to a given topological space

I would like to ask whether given a topological space $X$, we can find a commutative ring with unity $R$ such that $\operatorname{Spec} R$ (together with the Zariski topology) is homeomorphic to $X$.

Since the spectrum is a compact space, this is obviously only possible if $X$ is compact. Furthermore, from this answer we obtain that for spectra, $T_1$ already implies Hausdorff.

How many more restrictions must we impose? Can we give a characterisation of when a topological space is a spectrum of a ring?

• en.wikipedia.org/wiki/Spectral_space Apr 14 '16 at 8:23
• Not exactly: $X$ is quasi-*compact and *Kolmogorov (T0). Melvin Hochster's dissertation for his Ph. D. at Princeton was about this problem (1967). Apr 14 '16 at 8:29
• @Bernard There are different terminologies. I call a space compact iff you call it quasi-compact. Apr 14 '16 at 8:30
• Sure. I use Bourbaki's terminology. Apr 14 '16 at 8:31
• Bourbaki books are great! And of course, in their terminology, you are quite right. Apr 14 '16 at 8:38

A topological space which is homeomorphic to the spectrum of a ring is called a spectral space. Spectral spaces were characterized intrinsically by Melvin Hochster in his thesis:

Theorem (Hochster): Let $X$ be a topological space. Then $X$ is spectral iff it satisfies the following conditions:

• $X$ is sober.

• $X$ is compact.

• If $U,V\subseteq X$ are compact open sets, then $U\cap V$ is also compact.

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

It is not hard to show that every spectral space satisfies these conditions (note that the compact open subsets of $\operatorname{Spec} A$ are just the finite unions of distinguished open sets). The reverse direction is much more difficult; see Theorem 6 of this paper of Hochster's for details.

• Now both answers are correct, but I can only accept one... Apr 14 '16 at 8:39
• In any case, I upvoted both answers. Apr 14 '16 at 8:43