Partial Orders realized by Prime Ideals on commutative rings

Question

Is there a general criterion for which partial orders can be realized by the prime ideals of commutative rings (like we have for topological spaces - https://en.wikipedia.org/wiki/Spectral_space)?

And in general, what constraints on the partial order are created by additional classical properties of rings - such as being a UFD, principal domain, noetherian, etc...?

Examples for some obvious constraints -

1. Fields always have just one prime ideal - 0.
2. Local rings have just one maximal ideal.
3. Noetherian rings have only a finite number of minimal primes, and no infinitely ascending or descending sequence of prime ideals.
4. Every ring has a minimal prime and a maximal prime.

(Question originally posted in m.se https://math.stackexchange.com/questions/1628013/partial-orders-that-can-be-realized-by-prime-ideals-of-commutative-rings)

Answer 1

The following characterization follows easily from the general theory of spectral spaces, though it isn't exactly the most explicit criterion to apply in practice.

Theorem (Hochster, Proposition 12 of this paper): Let $X$ be a poset. Then $X$ is isomorphic to the poset of prime ideals of a commutative ring iff it is order-isomorphic to a closed subset of $\{0,1\}^V$ for some set $V$ (putting the discrete topology on $\{0,1\}$ and the product topology on $\{0,1\}^V$, and the usual order on $\{0,1\}$ and the product order on $\{0,1\}^V$).

(Given an order-isomorphism $X\cong \operatorname{Spec}(R)$, the set $V$ can be taken to be the set of compact open subsets in the Zariski topology on $X$, and the Zariski topology can be recovered as the product topology on $\{0,1\}^V$, where you topologize $\{0,1\}$ such that $\{0\}$ is open and $\{1\}$ is not. Note that every poset embeds in some $\{0,1\}^V$, so the key point of this result is the requirement that it be a closed subset of $\{0,1\}^V$, which is a sort of compactness condition.)