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Given a partially ordered set $P$, I'm interested in what is known about when $P$ is the prime spectrum of some (not necessarily commutative, not necessarily unital) ring: i.e., when does there exist a ring $R$ having $\mathrm{Spec}(R) \cong P$ (as an order isomorphism).

Obviously some conditions will be needed on $P$, for example, that every descending chain has a greatest lower bound (because the intersection of a descending chain of prime ideals is prime). From Bergman (personal correspondence, and http://math.berkeley.edu/~gbergman/papers/pm_arrays.pdf), every finite partially ordered set can occur as a subset of the prime ideals of some commutative ring. From other as yet unpublished work, a partially ordered set can occur as precisely the prime spectrum of a (non-commutative, not necessarily unital) ring in case: (a) it has the D.C.C., (b) it is chain-finite, and (c) the set of elements covered by any given element is countable. However, none of those conditions are necessary.

Is anyone aware of any more results on this subject?

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I immediately thought of Hochster's thesis (ams.org/journals/tran/1969-142-00/S0002-9947-1969-0251026-X/…), which characterizes the topological spaces in the image of $\operatorname{Spec}$, but I'm sure you've seen it. Is looking for compatible topologies of this sort on $P$ just as hard? –  Dylan Moreland Jul 22 '11 at 20:14
    
I had not seen it, actually. Thank you! The topological view on Spec(R) is new to me, so I haven't necessarily seen anything in particular. –  Chris Smith Jul 22 '11 at 21:22

1 Answer 1

In H.A. Priestley, ''Spectral Sets'' (1994), a partially ordered set P is called spectral if it occurs as the specialization order of a spectral topology. The cited paper is a survey of the known results: in particular note Theorem 1.1: a poset is spectral iff it is profinite, iff it is the spectrum of a distributive lattice.

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