# Is every poset the poset of prime ideals of a ring?

The answer to this question, as it is, is trivially false, for one necessary condition is the existence of maximal element(s), i.e., maximal ideals exist and are prime.

My question was inspired from Exercise 1.8 of Atiyah-Macdonald's book in Commutative Algebra:

The set of prime ideals of a nonzero ring has minimal elements with respect to inclusion.

Thus the above is also a necessary condition. My question is if there are more necessary conditions which are also sufficient, i.e.,

Are there necessary and sufficient conditions that a poset must possess so that it is the poset of a ring's spectrum.

Answers (possibly partial answers) are welcome for both necessary and sufficient conditions as well as imposing restrictions on the ring. For example, the poset corresponding to Noetherian rings must obey the ACC. If the poset is a lattice, then the ring must be local. Do nice conditions occur assuming our ring is Artinian or a Dedekind domain?

• József Pelikán told me that every finite poset can be realized as the spectrum of a ring (although I don't have a reference). A crucial point is that you can obtain closed and open subspaces by ring quotients and localizations, respectively. So the real challenge is building a ring with a big, complicated spectrum with the posets you want as subspaces. I enjoyed constructing examples by hand for some small posets; it's a nice exercise. Apr 29, 2011 at 7:51
• @Kopp, The papers below due to Hochster and Lewis do have results for finite posets. Although it will take time for me to read the papers, I had a glance. Apr 29, 2011 at 10:22