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?


1 Answer 1


Here is an example of two spectral topologies with the same specialization order.

Every boolean space is homeomorphic to the prime spectrum of some commutative ring (boolean spaces are spectral, and Hochster showed that spectral spaces are precisely the spaces arising as prime spectra of commutative rings).

The specialization order for boolean spaces is trivial: every point is closed. So two boolean spaces will have the same specialization order if they have the same cardinality.

So take two countable boolean spaces that are not homeomorphic, for example $$X = \{0\} \cup \{ 1/n : n = 1,2,3,\dots \} \subseteq \mathbb{R}$$ and $Y = X \sqcup X$.

The associated commutative rings in this example will be non-noetherian of Krull dimension zero.

  • $\begingroup$ Oh, thank you. I should have seen that. $\endgroup$ Commented Dec 22, 2019 at 14:31

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