Hochster's theorem on spectral spaces

Remarks

One direction is a direct check. $\operatorname{Spec} R$ is quasi-compact, the basic open sets $D(f) \cong \operatorname{Spec} R_f$ are quasi-compact, and a quasi-compact open set is a finite union of them, so finite intersections stay quasi-compact by $D(f) \cap D(g) = D(fg)$. An irreducible closed set is $V(\mathfrak p)$ for a unique prime $\mathfrak p$, which is its generic point. The other direction, building a ring from a spectral space, is the hard part of the theorem.

Example: a finite $T_0$ space is spectral, so it is the spectrum of some ring. Non-example: the real line with its usual topology is not spectral; it is not quasi-compact. A Hausdorff spectral space is a Stone space, and every Stone space is the spectrum of a Boolean ring. An infinite Stone space is never the spectrum of a Noetherian ring.

Going deeper

For a student

The conditions make the class of spectra purely topological, with no reference to rings. To show that a space is homeomorphic to a spectrum, one may check the three conditions instead of finding a ring. Spectral spaces also appear in logic and lattice theory, as the spaces dual to bounded distributive lattices.