most recent 30 from http://mathoverflow.net 2013-05-21T19:59:04Z http://mathoverflow.net/feeds/question/71016 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71016/constructing-rings-with-a-desired-prime-spectrum Chris Smith 2011-07-22T19:57:17Z 2012-07-03T09:22:01Z

<p>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).</p> <p>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 <a href="http://math.berkeley.edu/~gbergman/papers/pm_arrays.pdf" rel="nofollow">http://math.berkeley.edu/~gbergman/papers/pm_arrays.pdf</a>), every finite partially ordered set can occur as a <em>subset</em> 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 From other as yet unpublished work, a partially ordered set can occur as precisely the prime spectrum of a (non-commutative, not necessarily unital) ring in case: (a) it has the D.C.C., (b)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.</p> <p>Is anyone aware of any more results on this subject?</p>

http://mathoverflow.net/questions/71016/constructing-rings-with-a-desired-prime-spectrum/97642#97642 Sam 2012-05-22T07:42:55Z 2012-05-22T07:42:55Z

<p>In <a href="http://www.sciencedirect.com/science/article/pii/0022404994900086" rel="nofollow">H.A. Priestley, ''Spectral Sets'' (1994)</a>, a partially ordered set P is called <em>spectral</em> 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.</p>