Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be a commutative ring. The specialization preorder on $\mathrm{Spec}(R)$ is given by $\mathfrak{p} \prec \mathfrak{q} \Leftrightarrow \mathfrak{p} \in \overline{\{\mathfrak{q}\}} \Leftrightarrow \mathfrak{q} \subseteq \mathfrak{p}$. Is it possible to recover the topology on $\mathrm{Spec}(A)$ from this preorder?

If $A$ is noetherian (or just $A_{red}$ noetherian, which has the same spectral space), then the closed subsets are the finite unions of irreducible closed subsets, and the irreducible closed subsets are precicely those of the form $\{\mathfrak{q} : \mathfrak{q} \prec \mathfrak{p}\}$ for some $\mathfrak{p}$. Thus, in this case, we may recover the topology.

Is it possible to do so in general? More formally, assume that $X$ is a set, endowed with two spectral topologies. Assume that their specialization preorders are the same. Does this imply that the topologies coincide? Probably not. But what are interesting additional assumptions on these topologies (not on the rings) which make it true?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.