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Given a commutative ring $R$, let $\operatorname{Spec} R$ denote its set of prime ideals, equipped with the Zariski topology. Let $\operatorname{Spm} R$ denote the maximal spectrum of $R$, which is the topological subspace of $\operatorname{Spec} R$ made from the maximal ideals. It is fairly easy to prove that $\operatorname{Spec} R$ is a sober topological space. Now is $\operatorname{Spec} R$ the soberification of $\operatorname{Spm} R$?

For the soberification we probably need that every prime ideal of $R$ is an intersection of maximal ideals, in other words that $R$ is a Jacobson ring.

KReiser
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V. Semeria
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    Well, suppose $R$ is a local ring. Then it has a unique maximal ideal... – Zhen Lin May 27 '22 at 14:10
  • @ZhenLin Thanks, that confirms we really need $R$ to be a Jacobson ring. With that extra hypothesis, do you know if Spec $R$ is the soberification of Spm $R$? – V. Semeria May 27 '22 at 14:22

1 Answers1

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The points of the soberification $\hat{X}$ of a topological space $X$ can be identified with the irreducible closed subsets of $X$. So we are off to a good start: the underlying set of the soberification of $\operatorname{Spm} R$ can be identified with the underlying set of $\operatorname{Spec} R$, if $R$ is Jacobson.

What about the topology? A subset $\hat{F} \subseteq \hat{X}$ is closed if and only if there is a closed subset $F \subseteq X$ such that $\hat{F}$ is the set of irreducible closed subsets (of $X$) contained in $F$. But that means the topology of the soberification of $\operatorname{Spm} R$ agrees with the Zariski topology under the identification of the points previously described.

In general, I believe the soberification of $\operatorname{Spm} R$ can be identified with $\operatorname{Spec} R / J$, where $J$ is the Jacobson radical of $R$.

Zhen Lin
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    The last sentence is probably false, because the Jacobson radical is a global property of the ring (the intersection of all maximal ideals). However in the proof of soberification we need each radical ideal to be the intersection of the maximal ideals that contain it. – V. Semeria May 28 '22 at 07:49
  • More generally, let $X$ be a Jacobson sober topological space. If $X_0\subset X$ is the subset of closed points, then the inclusion $X_0\to X$ is the soberification of $X_0$.

    This is because (i) a soberification of a topological space $A$ is the same as a quasi-homeomorphism $A\to S$, where $S$ is sober (a quasi-homeomorphism is a continuous map $f : A \to B$ such that $V\subset B\mapsto f^{-1}(V)\subset A$ is a bijection between the open sets of $B$ and $A$), and (ii) a space $X$ is Jacobson if and only if $X_0\to X$ is a quasi-homeomorphism, where $X_0\subset X$ are the closed points.

     – Elías Guisado Villalgordo Mar 09 '23 at 11:04
  • Conversely, if $X$ is a T₁ space and $X\to S$ is a soberification, then $X$ is a homeomorphism onto $S_0$, the set of closed points of $S$. In particular, the soberification of a T₁ space is a Jacobson space. – Elías Guisado Villalgordo Mar 15 '23 at 13:47
  • Combining these two topological results with 00G3, it follows that for a ring $R$ we have that $R$ is Jacobson if and only if $\operatorname{Spm} R\to\operatorname{Spec} R$ is a soberification. – Elías Guisado Villalgordo Mar 15 '23 at 13:49