Canonical way to construct $R$ with $\mathrm{Spa}(A, A^+) \simeq \mathrm{Spec}(R)$?

Question

In this Scholze-Weinstein's note on $p$-adic geometry, there's a Huber's theorem (Theorem 2.3.3): Adic spectrum is spectral, i.e. $\mathrm{Spa}(A, A^+)$ is homeomorphic to $\mathrm{Spec}( R)$ for some $R$ (which says that $\mathrm{Spa}$ is not that weird space). Is there a canonical/functorial way to attach $R$ from $A$? I can't find a proof in the lecture note (maybe I missed), and I also checked Huber's original paper (Continuous Valuations), but I'm not sure the proof gives canonical construction or not.

Answer 1

The question is too vague to discount some modification for which the answer 'yes'. That said, as someone who very regularly uses adic spaces, I can tell you the answer is morally no and, even if some bizarre modification exists which makes it yes, that this never gets used in practice.

Definitions

As Aphelli alluded to in the comments, the correct definition of a spectral space is as follows.

Definition: A topological space $X$ is called spectral if it is

• quasi-compact,
• quasi-separated (i.e., the intersection of two quasi-compact open subsets is quasi-compact),
• it has a basis of quasi-compact open subsets,
• and it is sober (i.e., every closed irreducible subset has a unique generic point).

A topological space $X$ is called locally spectral if it is sober and has an open cover by spectral spaces.

One then has the following well-known theorem of Mel Hochster.

Theorem ([Hochster]): A topological space $X$ is spectral if and only if there exists some ring $R$ such that $X$ is homeomorphic to $\mathrm{Spec}(R)$.

This ring is not functorial in an way. In my opinion, one should think of Hochster's theorem as a comfort result -- it tells you that the topology of spectral spaces is precisely as hard as the Zariski topology of schemes. But, it doesn't really help you in practice.

PSA

In fact, I actually think that it is harmful psychologically to think this way. In particular, very rarely in algebraic geometry to we consider the type of rings $R$ such that $\mathrm{Spec}(R)$ is (homeomorphic to) the typical space that shows up in rigid geometry. Namely, these spaces are essentially never locally topologically Noetherian.

Even when we consider non-Noetherian rings in algebraic geometry, they tend to be topologically Noetherian. For example even in rigid geometry we often consider rings of finite presentation over something like $\mathcal{O}_{\mathbb{C}_p}/p^n$, which are not Noetherian, but are topologically Noetherian.

To drive the point in, I want to point point out several (related) degeneracies of the most basic of adic spaces: the closed disk $\mathbb{B}^1_{\mathbb{C}_p}=\mathrm{Spa}(\mathbb{C}_p\langle t\rangle,\mathcal{O}_{\mathbb{C}_p}\langle t\rangle)$.

Non-clopen partially proper open subsets

The closed disk contains an open subset which is closed under specializations but not closed. Namely, we can consider the open disk $$\mathbb{D}^1_{\mathbb{C}_p}=\bigcup_{0<r<1}\{|x|\leqslant r\}\subseteq\mathbb{B}^1_{\mathbb{C}_p}.$$ That this is closed under specializations is not very hard, and I leave it as an exercise.

That said, this definitely not something that normally happens in algebraic geometry. To qualify this, let us call an open subset $U$ of a topological space $X$ retro-compact if the inclusion $U\to X$ is quasi-compact.

Proposition ([Fujiwara--Kato, Chapter 0, Corollary 2.2.27]): Let $X$ be a locally spectral space, and $U$ a retro-compact open subset. Then, $$\overline{U}=\bigcup_{x\in U}\overline{\{x\}}.$$

In other words, this says that for locally spectral spaces, the closure of a retro-compact open subset is precisely adding in all specializations of points of $U$. So, if $U$ is closed under specializations, then $U$ is closed.

Corollary: Suppose that $X$ is a locally topologically Noetherian (locally spectral) topological space and $U\subseteq X$ is an open subset closed under specializations. Then, $U$ is closed. In particular, if $X$ is also connected, then $U=X$.

Proof: By the previous proposition, it suffices to show that $U$ is retro-compact. But, as $U\to X$ being quasi-compact is local on the target, we may assume that $X$ is topologically Noetherian. Then, this is Tag 04ZA. $\blacksquare$

So, this sort of phenomenon never happens in the typical spaces we see in algebraic geometry.

Valuativeness

The space $\mathbb{B}^1_{\mathbb{C}_p}$ is a valuative topological space, in the sense of [Fujiwara--Kato]. This means that in addition to being locally spectral, for any point $x$ the set $G_x$ of generalizations of $x$ is totally ordered.

On the face of things, that may not immediately ring alarm bells, but it should. Think about how you might get a chain of generalizations in something like $\mathbb{A}^2_{\mathbb{C}}$. We have the following generalization relationship

$$(x,y)\prec (x) \prec (0),$$

corresponding to the origin being contained in the irreducible closed subvariety given by the $y$-axis, then being contained in the full space $\mathbb{A}^2_{\mathbb{C}}$. But, we have another obvious chain of generalizations

$$(x,y)\prec (y) \prec (0),$$

corresponding to the origin being contained in the irreducible closed subvariety given by the $x$-axis, then being contained in the full space $\mathbb{A}^2_{\mathbb{C}}$. The points $(x)$ and $(y)$ are not comparable. The set $G_{(x,y)}$ of generalizations is not totally ordered.

This should put into perspective out how degenerate this property is from the viewpoint of typical algebraic geometry. For varieties a closed point is essentially always going to have many irreducible closed subvarieties containing it (think about all the curves!) which are not contained in each other. This property is very bizarre from that perspective.

Maximal $T_1$-quotients

Recall that a topological space $X$ is called $T_1$ if for any two points $x$ and $y$ of $X$, there exists a open subsets $U$ and $V$ of $X$ containing $x$ but not $y$, and $y$ but not $x$, respectively. For a topological space $X$, we can form a maximal $T_1$-quotient, denoted $[X]$, given essentially by modding out $X$ by identifying points which are topologically indistinguishable.

I leave it to you to show that $[X]$ is essentially always a point for 'reasonable' topological spaces in algebraic geometry. For instance, it is a good exercise you to show that $[\mathbb{A}^2_\mathbb{C}]$ is a point.

On the other hand, $[\mathbb{B}^1_{\mathbb{C}}]$ is very large, and is, in fact, Hausdorff. This is typical in rigid geometry. Here $[\mathbb{B}^1_{\mathbb{C}_p}]$ is what is called the 'Berkovich space' associated to $\mathbb{B}^1_{\mathbb{C}_p}$, and is an infinitely branching metric tree.

Some positive results

Let me just send by pointing out two results that may help.

The first is another result of Hochster from the same paper, that is occasionally useful for getting one's hand on (arbitrary) spectral spaces.

Theorem ([Hochster]): Let $X$ be a topological space. Then, $X$ is spectral if and only if $X$ is homeomorphic to $\varprojlim T_i$, where $T_i$ are finite $T_0$-spaces.

Here recall $X$ is called $T_0$ if for any two points $x$ and $y$ of $X$, there exists an open subset $U$ of $X$ containing one of $x$ or $y$, but not the other.

On the other hand, there is the following theorem, called Stone's representation theorem, which says almost precisely what you want if you change rings to distributive lattices (see [Fujiwara--Kato, Chapter 0, §2.2.(b)] for definitions and notation).

Theorem ([Fujiwara--Kato, Chapter 0, Theorem 2.2.8]): The functor $$\mathrm{Spec}(-)\colon \left\{\begin{matrix}\text{Distributive lattices}\\ \text{and lattice homomorphisms}\end{matrix}\right\}\to \left\{\begin{matrix}\text{Spectral spaces}\\ \text{and quasi-compact maps}\end{matrix}\right\},$$ is an equivalence, with quasi-inverse given by sending $X$ to the distributive lattice $\mathbf{QCOuv}(X)$ of quasi-compact open subsets.

References

[Fujiwara--Kato] Fujiwara, K. and Kato, F., 2018. Foundations of rigid geometry I.

[Hochster] Hochster, M., 1969. Prime ideal structure in commutative rings. Transactions of the American Mathematical Society, 142, pp.43-60.