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?

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.
