I'm reading a paper that translates between formal geometry and rigid geometry.

In particular, this paper begins with two rigid analytic spaces $A$ and $C$ (each coming from a scheme over $\mathbb{Z}_p$), both defined over $B$, and finds the formal completion of each, calling them $\mathfrak{A}$, $\mathfrak{C}$, and $\mathfrak{B}$ respectively. It then forms the fiber product $X = \mathfrak{A} \times_{\mathfrak{B}} \mathfrak{C}$, and proceeds to treat it as a rigid analytic space, including forming the fiber product of $X$ with another rigid analytic space over $C$.

My question is, are the viewpoints of formal geometry and of rigid geometry essentially the same? The formal geometry doesn't seem to do anything else in the entire paper, except be the source for the map $C \to B$. But it's also used in the main reference for this section! Is there a subtle advantage that formal geometry affords us, or some difference in how the fiber product is formed, or is it a quirk of the original paper that this one carried over?

Is there some way to add in the algebraic geometry over $\mathbb{Z}_p$ that the spaces we're talking about come from?


1 Answer 1


No, these are not the same thing. Formal schemes are to rigid-analytic spaces as $\mathbf{Z}_p$-schemes are to $\mathbf{Q}_p$-schemes.

The book Lectures in Formal and Rigid Geometry by Bosch is an excellent and friendly reference on this subject - take a look especially at sections 7.4 and 8.3.

In particular, let $K$ be a non-archimedean field (i.e. a field complete with respect to some $\mathbf{R}_{>0}$-valued multiplicative norm) and let $\mathscr{O}_K$ be its valuation ring. Then to any "reasonable" $\mathscr{O}_K$-formal scheme $\mathfrak{X}$, we can associate a rigid-analytic "generic fiber" $X = \mathfrak{X}_K$. (This is literally the generic fiber in the broader context of adic spaces, which subsume both formal schemes and rigid-analytic varieties).

We say that a formal scheme $\mathfrak{X}$ with $X = \mathfrak{X}_K$ is a formal model of $X$. It is a deep theorem of Raynaud that formal models of (reasonable) rigid-analytic spaces always exist, and are unique up to the operation of "admissible formal blowing up" (more precisely, the category of reasonable rigid-analytic spaces over $K$ is equivalent to the localization of the category of reasonable formal schemes over $\mathscr{O}_K$ with respect to this operation).

One warning: this "generic fiber" operation is not compatible with the usual one for schemes under analytification and formal completion.

For example, consider the affine $\mathbf{Z}_p$-line $\mathrm{Spec}(\mathbf{Z}_p[T])$. Its generic fiber is the affine $\mathbf{Q}_p$-line $\mathrm{Spec}(\mathbf{Q}_p[T])$. The analytification is the rigid-analytic affine line, which includes all elements of $\mathbf{Q}_p$ as $\mathbf{Q}_p$-points.

On the other hand, the formal completion of $\mathrm{Spec}(\mathbf{Z}_p[T])$ at $p$ is the formal unit ball $\mathrm{Spf}(\mathbf{Z}_p\{T\} := \varprojlim \mathbf{Z}/p^n[T])$. The generic fiber of this is the rigid-analytic unit ball, given by the max-spectrum of the ring $\mathbf{Q}_p\{T\} = \mathbf{Z}_p\{T\}[1/p]$. The $\mathbf{Q}_p$-points of this space are the elements of $\mathbf{Z}_p$.


For completeness (i.e. in case my advisor is reading this), let me add a few details:

Let $\varpi \in \mathscr{O}_K$ be a pseudo-uniformizer, i.e. a non-zero element with $|\varpi| < 1$. We define rings of restricted power series over $\mathscr{O}_K$ to be $$\mathscr{O}_K\{T_1, \ldots, T_n\} := \varprojlim_n (\mathscr{O}_K/\varpi^n)[T_1, \ldots, T_n] = \left\{\sum_\alpha a_\alpha T^\alpha \in \mathscr{O}_K[[T_1, \ldots, T_n]] \mid a_\alpha \rightarrow 0\right\} $$ Then we define $$ K\{T_1, \ldots, T_n\} := \mathscr{O}_K\{T_1, \ldots, T_n\}[1/\varpi] = \left\{ \sum_\alpha a_\alpha T^\alpha \in K[[T_1, \ldots, T_n]] \mid a_\alpha \rightarrow 0\right\} $$

The "formal closed unit ball" in $n$ variables over $\mathscr{O}_K$ is $\mathrm{Spf} \mathscr{O}_K\{T_1, \ldots, T_n\}$. Note that it is the formal completion of affine $n$-space over $\mathscr{O}_K$ at the special fiber $\{\varpi = 0\}$.

A formal scheme $\mathfrak{X}$ over $\mathscr{O}_K$ is admissible if it is locally of the form $\mathrm{Spf} A$ where $A = \mathscr{O}_K\{T_1, \ldots, T_n\}/I$ with $I$ a finitely generated ideal, and $A$ has no $\varpi$-torsion. Essentially, this means that $\mathfrak{X}$ is locally the $\varpi$-adic completion of a flat finite-type $\mathscr{O}_K$-scheme.

The rigid-analytic closed unit ball in $n$ variables over $K$ is $\mathrm{Sp} K\{T_1, \ldots, T_n\}$ (as a set, it consists of maximal ideals in this ring). Rigid-analytic spaces are constructed by gluing together spaces of the form $\mathrm{Sp} (K\{T_1, \ldots, T_n\}/I)$, where $I$ is any ideal (automatically finitely generated and closed).

The generic fiber functor sends $\mathrm{Spf} (\mathscr{O}_K\{T_1, \ldots, T_n\}/I)$ to $\mathrm{Sp} (K\{T_1, \ldots, T_n\}/I)$. In particular, it sends the formal closed unit ball to the rigid-analytic closed unit ball. It extends to a functor from admissible formal schemes to rigid-analytic spaces.

Raynaud's theorem applies once we add additional (very mild) compactness assumptions on both sides: the rigid spaces must be "quasiseparated and quasi-paracompact", and the formal schemes must be "quasi-paracompact".

An admissible formal blowing up of an admissible formal scheme is a certain sort of blowup along a closed subset supported on the special fiber.

  • 2
    $\begingroup$ I should add that the generic fiber functor commutes with fiber product, so if $\mathfrak{A}, \mathfrak{B}, \mathfrak{C}$ are formal models for tigid-analytic spaces $A,B,C$, we have $(\mathfrak{A} \times_{\mathfrak{B}} \mathfrak{C})_K =A \times_B C$ $\endgroup$
    – dorebell
    Nov 21, 2019 at 8:43
  • $\begingroup$ What do you mean when you write $\mathbb{Q}_p\{T\}$ with curly braces? Is this power series? Power series whose terms go to 0? $\endgroup$
    – Jon Aycock
    Nov 21, 2019 at 16:07
  • 1
    $\begingroup$ The definition is given above, but yes, this is power series whose terms go to $0$. You can also think of it as the completion of $\mathbf{Q}_p[T]$ in the $p$-adic topology. $\endgroup$
    – dorebell
    Nov 21, 2019 at 18:04

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