Berthelot functor, rigid analytic space

Question

If $X=\operatorname{Spec} A$, where $A$ is a noetherien, complete local ring, with a finite residual field $\mathbb{F}_p$. We can associate to $A$ a rigid analytic space with two different ways, we can use Berthelot functor or the method of Raynaud. These methods rise to same rigid space when the scheme is proper over $W(\mathbb{F}_p)$.

If $\mathfrak{p} \in \operatorname{Spec}A(\mathbb{C}_p)$, can we compare the completed local ring of $\operatorname{Spec}A$ and the completed local ring of the rigid analytification $(\operatorname{Spec}A)^{rig}$ of the affine scheme $\operatorname{Spec}A$?

Can we compare the global sections $H^{0}((\operatorname{Spec}A)^{rig},\mathcal{O}_{(\operatorname{Spec}A)^{rig}})$ and the ring $A$?

Answer 1

Let me say something related to your first question (the local one) in a slightly different setting. Let $\mathscr{A}$ be a $\mathbb{Q}_p$-affinoid algebra and consider its reduction $\tilde{\mathscr{A}} = \mathscr{A}^\circ/\mathscr{A}^{\circ\circ}$. (The ring $\mathscr{A}^\circ$ is close to your $A$.) We have a reduction map $r \colon X \to \tilde{X}$, where $X$ denotes the rigid spectrum of $\mathscr{A}$ and $\tilde{X}$ the spectrum of $\tilde{\mathscr{A}}$ (a scheme).

Let $x$ be a rigid point of $X$, i.e. a point in a finite extension of $\mathbb{Q}_p$. In the case where $\mathscr{A}$ is distinguished (in your case, it is equivalent to requiring that $\mathscr{A}$ is reduced and $\|\mathscr{A}\|_{sup} = |k|$) and $X$ is equidimensional, S. Bosch proved that the completed local ring at the point $r(x)$ is isomorphic to the ring of functions that are bounded by 1 on the tube of $r(x)$, i.e. $r^{-1}(r(x))$ (which is an open set containing $x$).

In case you are interested, S. Bosch's result holds over an arbitrary non-archimedean complete valued with non-trivial valuation. In the discretely valued case, F. Martin reproved it in arxiv.org/abs/1510.01178 (using de Jong's results as in nfdc23's answer) and removed the equidimensionality assumption.

Answer 2

The setup of the question is not general enough:

(i) you mean to work with Spf rather than Spec,

(ii) Raynaud's construction doesn't apply to the formal scheme Spf($A$) for such $A$,

(iii) the compatibility you allude to between the Berthelot and Raynaud constructions in the proper case is not what you meant to say; you meant to refer to a compatibility between analytification of algebraic schemes over the non-archimedean field and the Berthelot construction of a certain formal scheme over the valuation ring (the agreement of the Berthelot and Raynaud constructions when both make sense is quick from the definitions and doesn't involve properness hypotheses),

(iv) I think you meant to consider a "classical point" over an arbitrary finite extension of $\mathbf{Q}_p$ rather than a $\mathbf{C}_p$-point (the latter is vastly more general than the former).

In general, since you don't mention specific references or intended applications, what follows is a guess as to the questions you intended to ask. Parts of what I say may already be known to you. All of the essential content in what follows is developed very elegantly from scratch in section 7 of deJong's IHES 82 paper http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1995__82_/PMIHES_1995__82__5_0/PMIHES_1995__82__5_0.pdf

Let $K$ be a field complete for a nontrivial discrete valuation (e.g., $\mathbf{Q}_p$), with valuation ring $O$ having residue field $k$ and uniformizer $\pi$. Let $\mathfrak{X}$ be a locally noetherian formal scheme over ${\rm{Spf}}(O)$ such that the associated ordinary scheme $\mathfrak{X}_{\rm{red}}$ obtained by killing the coherent ideal sheaf of locally topologically nilpotent sections of $\mathscr{O}_{\mathfrak{X}}$ is locally of finite type over $k$. For example, if $\mathfrak{X} = {\rm{Spf}}(A)$ is affine (so $A$ is complete for the topology defined by some ideal) then it is an elementary exercise in commutative algebra to check that this is equivalent to the condition that $A$ is a topological quotient of $B_{n,m} := O\{t_1,\dots,t_n\}[\![y_1,\dots,y_m]\!]$ equipped with the $(\pi, y_1, \dots, y_m)$-adic topology.

Informally, we want to associate to ${\rm{Spf}}(B_{n,m})$ the "analytification" given by $$\{|t| \le 1\}^n \times \{|y| < 1\}^m$$ and then passing to analytic subspaces to handle/define ${\rm{Spf}}(A)^{\rm{rig}}$ for a general $A$ and then globalize via gluing. The bare-hands approach is a nightmare to globalize rigorously due to the intervention of coordinates, so deJong/Berthelot use a slick coordinate-free way to define ${\rm{Spf}}(A)^{\rm{rig}}$ for any $A$ as above that naturally gives the desired result for $A = B_{n,m}$ and globalizes well (and is compatible with fiber products) and recovers Raynaud's construction when $\mathfrak{X}$ is locally of finite type over ${\rm{Spf}}(O)$. It seems you probably already know that if $X$ is a separated flat scheme of finite type over $O$ and $\widehat{X}$ is its formal completion along the special fiber then there is a natural quasi-compact open immersion $\widehat{X}^{\rm{rig}} \hookrightarrow (X_K)^{\rm{an}}$ which is an isomorphism if $X$ is $O$-proper.

The end of Lemma 7.1.9 in deJong's paper gives that if $\mathfrak{X} = {\rm{Spf}}(A)$ is affine then the underlying set of $\mathfrak{X}^{\rm{rig}}$ is in natural bijection with the set of maximal ideals in $A \otimes_O K$ (for the Raynaud setup this assertion is a tautology, but not for the more general Berthelot setup), and the associated completed local rings are naturally isomorphic. In other words, if $x \in \mathfrak{X}^{\rm{rig}}$ then via the natural $O$-algebra map $$A = \Gamma(\mathfrak{X}, \mathscr{O}_{\mathfrak{X}}) \rightarrow \Gamma(\mathfrak{X}^{\rm{rig}}, \mathscr{O}_{\mathfrak{X}^{\rm{rig}}})$$ that comes out of the construction we get an induced map $A \otimes_O K \rightarrow \mathscr{O}_{\mathfrak{X}^{\rm{rig}},x}$ for which the contraction of the maximal ideal of the target is the associated maximal ideal $\mathfrak{m}_x \subset A \otimes_O K$, and the induced local map $$(A \otimes_O K)_{\mathfrak{m}_x} \rightarrow \mathscr{O}_{\mathfrak{X}^{\rm{rig}},x}$$ induces an isomorphism between completions. That is the only affirmative answer I can imagine for your first question (concerning comparison of local rings).

This comparison of completed local rings implies immediately that if $A$ as above is $O$-flat then the natural map $$A = \Gamma(\mathfrak{X}, \mathscr{O}_{\mathfrak{X}}) \rightarrow \Gamma(\mathfrak{X}^{\rm{rig}}, \mathscr{O}_{\mathfrak{X}^{\rm{rig}}})$$ is injective, and your last question seems to be asking if we can describe this subring in terms of the analytic space $\mathfrak{X}^{\rm{rig}}$. There are some subtleties since one can imagine situations where $A$ is not normal but $A \otimes_O K$ is normal (so in particular $A$ is reduced), and hence (by excellence considerations, via excellence theorems of Kiehl and Valabrega) $\mathfrak{X}^{\rm{rig}}$ is normal: in such cases the module-finite (!) normalization of $A$ gives rise to the same analytic space (since Berthelot's functor preserves finiteness of morphisms: see Proposition 7.2.1(d) of deJong's paper).

Consequently, the most "reasonable" case in which to try to recover $A$ from $\mathfrak{X}^{\rm{rig}}$ is when $A$ is normal. And in deJong's paper this is answered definitively: by Proposition 7.3.6 if $A$ is normal then it is (what else?) exactly the ring of power-bounded global functions on $\mathfrak{X}^{\rm{rig}}$, Theoprem 7.4.1 of that paper gives a generalization beyond the affine case, and in remark 7.4.2 one finds a conjecture beyond the normal case for when equality should hold.

Remark: The proof of Theorem 7.4.1 (and Theorem 7.5.2) rests on 7.1.13 that is the topic of the erratum http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1998__87_/PMIHES_1998__87__175_0/PMIHES_1998__87__175_0.pdf,