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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?

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