Question

Given a distributive lattice $A$ we can look at $Spec(A)$, whose points are prime ideals and its open sets are given similarly to the Zariski topology on Spec of a ring. That is, the basis of open sets is composed of sets of the form $D(I)=\{p~\mathrm{prime~in~A}:I\nsubseteq p\}$.

So, given a prime ideal, it is not hard to show that its complement is a prime filter. Hence there is a set bijection between the set of prime ideals and the set of prime filters. Does anyone know, if we force this bijection to be a homeomorphism based on the topology on $Spec(A)$, is there a nice description of the open basis elements on the set of prime filters of $A$?

Note: Perhaps this question is purely lattice theory. I guess it depends on your point of view. Please add or remove tags as necessary.

Thanks!

Jon

Answer 1

If you just take the basis of sets $D(I)$ that you gave for the space of prime ideals and transport it via the bijection you gave, you obviously get a basis for the space of prime filters. It consists of the sets $M(I)=\{p \text{ prime filter}:p\cap I\neq\varnothing\}$. Clearly, this $M(I)$ is the union, over all $a\in I$, of the sets that Ben called $D(a)$ in his answer. So his base and mine (which is really yours) generate the same topology. Actually, it seems that your base of $D(I)$'s is not just a base but the whole topology (and therefore the same goes for the $M(I)$'s).

Answer 2

I believe that if $a\in A$, one defines $D(a)$ to be the set of all prime filters containing $a$ and these give the open sets of the topology. More details can be found in Johnstone's Stone Spaces book when he does Stone duality between distributive lattices and coherent spaces.