# Berthelot functor, rigid analytic space

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 $A$?

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.

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,

• For the properness I mean that I we take a proper, flat scheme $X$ over $\mathbb{Z}_p$ and if $\mathfrak{X}$ is the formal completion along the special fibre, the generic fibre $X^{rig}$ in the sense of Raynaud of $\mathfrak{X}$ is a rigid space and it is the same given by the functor Berthelot $X^{ana}$. If the scheme is not proper but separated, we have only an open immersion $X^{rig}\rightarrow X^{ana}$, and we need the valuation criteria of proper schemes to get the surjection. – Adel BETINA Jan 28 '16 at 7:52
• Your 2nd comment repeats the mix-up I allude to in my point (iii). For formal schemes locally of finite type over the dvr, which are the only ones relevant in your 2nd comment, the Raynaud and Berthelot constructions always coincide, and properness is not relevant. There is a third construction being considered here, separate from both of those, namely the analytification of the algebraic generic fiber $X_{\mathbf{Q}_p}$, and the quasi-compact open immersion goes $\mathfrak{X}^{\rm{rig}} \rightarrow (X_{\mathbf{Q}_p})^{\rm{an}}$, an equality for proper $\mathfrak{X}$. Please look at (iii). – nfdc23 Jan 28 '16 at 9:53
• The result of Remark 7.4.2 is worked out by Christian Kappen and Florent Martin in Theorem 2.1 of arxiv.org/abs/1510.01178. – Jérôme Poineau Jan 28 '16 at 10:05