Why is the definition of the adic spectrum $\operatorname{Spa}\,(A,A^+)$ the "right" definition?

Question

I'm currently going through a number of expository accounts of Huber's adic spaces in order to start understanding perfectoid spaces and I'd like to understand the motivation behind the definition of adic spaces.

Take $A$ a topological ring containing an open subring $A_0$ with subspace topology the $I$-adic topology for a finitely generated ideal $I\subseteq A_0$. We call such a ring Huber. Let $A^+$ be an open, integrally closed subring of $A$ contained in $A^\circ=\{a\in A:\{a^n\}_{n\geq0}\text{ is bounded} \}$. We define the adic spectrum of a ``Huber pair" $(A,A^+)$ to be

$$\operatorname{Spa}\,(A,A^+):=\{|\cdot|_x\in\operatorname{Cont}(A):|a|_x\leq1,\:\forall a\in A^+\}$$ where $\operatorname{Cont}\,(A)$ is the space of continuous valuations on $A$. A valuation $|\cdot|:A\longrightarrow\Gamma_{|\cdot|}\cup\{0\}$ is continuous if the valuation topology is coarser than the given topology i.e., $\{a\in A:|a|<\gamma\}$ is open in $A$ for all $\gamma\in\Gamma_{|\cdot|}$.

I am wondering why this is the right definition to be using. First, what is a compelling reason to restrict ourselves to continuous valuations? Second, what added flexibility does the $A^+$ give us that justifies using $\operatorname{Spa}\,(A,A^+)$ instead of $\operatorname{Spa}\,(A)=\operatorname{Spa}\,(A,A^\circ)$ or just $\operatorname{Cont}\,(A)$?

Answer 1

You aren't the first to ask these questions. You won't be the last.

First, to set the stage let me point out that definitions are neither right nor wrong. They are only right for a purpose or wrong for a purpose. Huber's papers "A generalization..." proposes to define adic spaces $\operatorname{Spa}(A,A^+)$ as a new model for rigid analytic geometry. So, that is the context in which we discuss right and wrong.

The rigid space associated with the affine ring $\mathbb{Q}_p$ is a point. But, the bare naked valuation spectrum $\operatorname{Spv}(\mathbb{Q}_p)$ contains too much. For one, it contains at least two points: the $p$-adic norm and the trivial norm. Maybe you can just ignore the trivial norm? But no, there is a lot more. In fact, if $L \supseteq K$ is a field extension then a valuation on $K$ extends to a valuation on $L$. End of story. Concretely, the restriction map $\operatorname{Spv}(\mathbb{Q}_p) \rightarrow \operatorname{Spv}(\mathbb{Q})$ is surjective. The $3$-adic norm on $\mathbb{Q}$ extends in some crazy way to the $2$-adic numbers $\mathbb{Q}_2$.

The answer to your first question is: the valuation spectrum is not suited to the "simple" job of modeling a point in rigid geometry. (However! The abstract valuation theory has a huge role to play in proving theorems.)

The $A^+$ question lies deeper in the theory, so it is good to remember why some bound is imposed at all. Beyond points, the next adic space to understand is the the $p$-adic unit disc. In rigid geometry, the ring corresponding to the the unit disc (over $\mathbb{C}_p$) is the Tate algebra $\mathbb{C}_p\langle X \rangle$ of series

$$ f(X) = a_0 + a_1X + a_2X^2 + \dotsb \;\;\; a_i \in \mathbb{C}_p \text{ and } \lim_{i\rightarrow\infty} a_i = 0. $$

For $x \in \mathbb{C}_p$ with $|x|_p\leq 1$ you get a point in $\operatorname{Cont}(\mathbb{C}_p\langle X \rangle)$ by $|f|_x := |f(x)|_p$. These points satisfy $|X|_x = |x|_p \leq 1$. Remembering what we are doing, it occurs to us that modeling the unit disc whould probably involve a bound like $|X|\leq 1$ on the valuations. Therefore, we define

$$ D = \{\text{continuous valuations } |\cdot| \text{ on $\mathbb{C}_p\langle X \rangle$ } :|X| \leq 1\}. $$

Now we have two comments.

• Is the bound $|X|\leq 1$ automatic? No. The inclusion $D \subseteq \mathrm{Cont}(\mathbb{C}_p\langle X \rangle)$ is strict. You can see why in Section 1.5 of Weinstein's adic space notes cited below. (And many other places.) So, the bound we impose on $X$ really cuts out a subset of the continuous spectrum.
• Imposing a bound on a function is a mathematical mess. Geometrically, the coordinate $X$ is a choice. Another valid coordinate could be $Y=X+p$. A theory where theorems are proven would rather impose a bound on an honest algebraic object. In this particular example, the space $D$ I've defined is the same as what is usually called the adic unit disc $D = \operatorname{Spa}(\mathbb{C}_p\langle X \rangle, \mathcal{O}_{\mathbb{C}_p}\langle X \rangle)$, whereupon the bound $|f|\leq 1$ is valid for all $f \in \mathcal{O}_{\mathbb{C}_p}\langle X \rangle$.

Thinking about passing from functions to algebra can partially explain where the $A^+$ comes from. In practice, we impose bounds on functions. To prove theorems, we replace these bounds with bounds on the smallest possible rings containing those functions largest ring containing the functions and for which the bound imposed automatically extends. That ring turns out to be one of these open and integrally closed subrings $A^+$. (See Section 10.3 of Conrad's notes in his perfectoid seminar.) But, it has no reason to be $A^{\circ}$ itself. For instance, in $\mathrm{Cont}(\mathbb{C}_p\langle X\rangle)$ we do have the automatic bound $|n|\leq 1$ for all $n \in \mathbb{Z}$ and the corresponding smallest ring $A^+$ will be something smaller than $A^{\circ} = \mathcal{O}_{\mathbb{C}_p}\langle X \rangle$. (See Example 11.3.14 in the 11th lecture of Conrad's perfectoid seminar.)

A second explanation for the role of $A^+$ is getting closer to making technical arguments in theorem proofs. In Section 3.3 of the Berkeley notes of Scholze--Weinstein you see phrasing saying $A^{\circ}$ has issues with being "...stable under rational subsets". Here is what that is referring to. Within Huber's theory there are various constructions to make on the rings $A$. One of them is a localization process $A \mapsto B = A(T/s)$. You can read about this in Lecture 7 of Conrad's perfectoid seminar. Given a pair $(A,A^+)$ there is a natural choice of pair $(B,B^+)$ that is the analog of algebraic localization in scheme theory. However, it is not always true that if $A^+ = A^{\circ}$ then $B^+ = B^{\circ}$. This kind of issue happens all the time in the constructions with Huber's theory. Tensor products and residue field suffer from similar challenges. The residue field issue is somehow most convincing to me: a topological field $k$ can have its topology defined by many different valuations, but $k^{\circ}$ can be the valuation ring for only one of them. See the start of Lecture 11 from Conrad's notes of his perfectoid seminar. (And Example 11.4.3 of the same notes deal with the localization issue just mentioned.)

Huber, Roland, A generalization of formal schemes and rigid analytic varieties, Math. Z. 217, No. 4, 513-551 (1994). ZBL0814.14024.

Scholze, Peter; Weinstein, Jared, Berkeley lectures on (p)-adic geometry, Annals of Mathematics Studies 207. Princeton, NJ: Princeton University Press (ISBN 978-0-691-20209-9/hbk; 978-0-691-20208-2/pbk; 978-0-691-20215-0/ebook). x, 250 p. (2020). ZBL1475.14002.

Weinstein, Jared, Adic spaces, Cais, Bryden (ed.), Perfectoid spaces. Lectures from the 20th Arizona winter school, University of Arizona, Tuscon, AZ, USA, March 11–17, 2017. With an introduction by Peter Scholze. Providence, RI: American Mathematical Society (AMS). Math. Surv. Monogr. 242, 1-43 (2019). ZBL1451.14083.