# Adic spaces

Katharina Hübner<EMAIL_ADDRESS>Robert Mayer Straße 6-8, 60325
Frankfurt

###### Abstract.

Building up on the theory of Huber pairs presented in John Bergdall’s lecture
([Ber24]) we explain the construction of adic spaces. We study some important
classes of adic spaces such as rigid analytic spaces and formal schemes and
show the connections between them. In the course of the lecture we will
illustrate the respective concepts with the fundamental examples of the open
and closed disc and the affine line.

###### Contents

  1. 1 Introduction
  2. 2 The adic spectrum of a Huber pair
  3. 3 The structure (pre-)sheaf
  4. 4 Adic spaces
  5. 5 Local Huber pairs
  6. 6 Analytic adic spaces
  7. 7 Morphisms of finite type
  8. 8 Formal schemes and generic fibers
  9. 9 The fiber product of adic spaces
  10. 10 Analytification

##  1\. Introduction

These lecture notes are based on a series of lectures at the Spring school
“Non-archimedean geometry and Eigenvarieties” in March 2023 in Heidelberg. The
objective of the first three courses was to give an introduction to the theory
of adic spaces. Like schemes they are locally ringed spaces but with
additional structure. Their structure sheaf carries a topology and every point
comes with a valuation on its residue field. One could say that the foundation
of scheme theory is commutative algebra, i.e., the theory of rings and
modules. In this spirit the foundation of the theory of adic spaces would be
“topological” commutative algebra, i.e. the theory of topological rings and
modules. In order to construct adic spaces we only need to consider
topological rings of a certain kind, so called Huber rings. Moreover, in order
to encode the valuations on residue fields we need to study subrings of
integral elements of a Huber ring. A Huber ring $A$ together with a subring
$A^{+}$ of integral elements is called a Huber pair and denoted $(A,A^{+})$.
An example of a Huber ring would be $\mathds{Q}_{p}$ and an example of a ring
of integral elements therein would be $\mathds{Z}_{p}$. So
$(\mathds{Q}_{p},\mathds{Z}_{p})$ is a Huber pair.

This foundation on topological rings has been laid in the first lecture by
John Bergdall. In this course we build on top of that and start with the
construction of the adic spectrum of a Huber pair together with its structure
sheaf. This defines affinoid adic spaces. We can then glue affinoid adic
spaces together to obtain adic spaces, our main objects of study. The category
of adic spaces encompasses many different types of spaces. In fact, rigid
analytic varieties, formal schemes, and algebraic varieties can all be found
as full subcategories in the category of adic spaces. We will throw a
spotlight at each of these incarnations of adic spaces and show connections
between them. In the process we will get to know many of the fundamental
examples of adic spaces such as the closed and open unit disc and the affine
line.

##  2\. The adic spectrum of a Huber pair

For a Huber pair $(A,A^{+})$ we define its adic spectrum. As a set it is
defined as

$\operatorname{\mathrm{Spa}}(A,A^{+})=\\{v:A\to\Gamma\cup\\{0\\}~{}\text{continuous
valuation}\mid v(A^{+})\leq 1\\}/\sim,$

where "$\sim$" refers to equivalence of valuations. In other words

$\operatorname{\mathrm{Spa}}(A,A^{+})=\\{v\in\operatorname{\mathrm{Cont}}(A)\mid
v(A^{+})\leq 1\\}.$

If $A^{+}=A^{\circ}$, we often just write $\operatorname{\mathrm{Spa}}(A)$
instead of $\operatorname{\mathrm{Spa}}(A,A^{\circ})$.

For a point $x\in\operatorname{\mathrm{Spa}}(A,A^{+})$ we use the following
notation for the corresponding valuation:

$\displaystyle A\longrightarrow$ $\displaystyle~{}\Gamma_{x}\cup\\{0\\}$
$\displaystyle a\longmapsto$ $\displaystyle~{}|a(x)|.$

This notation suggests that we should think of taking the valuation of $a$
corresponding to a point $x$ as evualuating $a$ at $x$ in the same way a
function is evaluated at a point.

The _support_ of a valuation $v$ on a ring $A$ is the prime ideal

$\operatorname{\mathrm{supp}}v:=\\{a\in A\mid v(a)=0\\}$

In fact a valuation $v$ on $A$ can equivalently be viewed as a valuation of
the field $k(\operatorname{\mathrm{supp}}v)$. We just need to convince
ourselves that $v$ naturally defines a valuation $\bar{v}$ of
$A/\operatorname{\mathrm{supp}}v$. Since $\bar{v}$ takes the value zero only
if the argument is zero, we can extend $\bar{v}$ to the residue field
$k(\operatorname{\mathrm{supp}}v)=\mathrm{Frac}(A/\operatorname{\mathrm{supp}}v)$
by setting

$\bar{v}(\frac{a}{b})=\frac{\bar{v}(a)}{\bar{v}(b)}.$

Conversely, any valuation $\bar{v}$ of a residue field $k(\mathfrak{p})$ for
$\mathfrak{p}\in\operatorname{\mathrm{Spec}}A$ gives rise to a valuation on
$A$ by composing with the natural map:

$v:A\longrightarrow k(\mathfrak{p})\overset{\bar{v}}{\longrightarrow}\Gamma.$

This valuation has support $\mathfrak{p}$ and the two constructions are
inverse to each other.

######  Example 2.1.

  * •

The space
$\operatorname{\mathrm{Spa}}(\mathds{Q}_{p})=\operatorname{\mathrm{Spa}}(\mathds{Q}_{p},\mathds{Z}_{p})$,
has only one point, the $p$-adic valuation of $\mathds{Q}_{p}$. The trivial
valuation is not continuous.

  * •

In
$\operatorname{\mathrm{Spa}}(\mathds{Z}_{p})=\operatorname{\mathrm{Spa}}(\mathds{Z}_{p},\mathds{Z}_{p})$
there are two points, the $p$-adic valuation of $\mathds{Q}_{p}$ and the
trivial valuation of $\mathds{F}_{p}=\mathds{Z}_{p}/p\mathds{Z}_{p}$.

  * •

$\operatorname{\mathrm{Spa}}(\mathds{Q},\mathds{Z})$ has one point for each
prime $p$ representing the $p$-adic valuation and an additional point for the
trivial valuation.

  * •

Let $(k,k^{\circ})$ be an algebraically closed non-archimedean field, e.g.
$k=\mathds{C}_{p}$. The structure of the open unit disc over $k$

$\mathds{D}_{k}=\operatorname{\mathrm{Spa}}(k\langle
T\rangle)=\operatorname{\mathrm{Spa}}(k\langle T\rangle,k^{\circ}\langle
T\rangle)$

is already quite complicated. The easiest type of points is of the following
form: Take $\alpha\in k$, with norm $|\alpha|\leq 1$ and $r\in(0,1]$. This
defines a point $x_{\alpha,r}\in\mathds{D}_{k}$ given by

$|f(x_{\alpha,r})|=\max_{i}|a_{i}|r^{i}$

for $f=\sum a_{i}(T-\alpha)^{i}$. The full classification of points of
$\mathds{D}_{k}$ can be found in [Ber24], §1.4.4.

Let us study in some more detail the adic spectrum of an _affinoid field_. By
this we mean a Huber pair $(k,k^{+})$, where $k$ is a field with valuation
ring $k^{+}$ and such that $k$ is endowed either with the valuation topology
or the discrete topology. In case $k$ carries the valuation topology and
$k^{+}\neq k$, we need the extra condition that the valuation be _microbial_ ,
i.e., that $k^{+}$ contains a nonzero topologically nilpotent element (see
[Hub96], Definition 1.1.4). But we are on the safe side if we assume that the
rank of $k^{+}$ is finite (which is fine for all our examples). If the
topology of $k$ is discrete, we call the affinoid field $(k,k^{+})$ discrete
and otherwise analytic. The standard example of an analytic affinoid field is
$(k,k^{\circ})$ for a nonarchimedean field $k$.

Remember that the valuation topology associated with the valuation ring
$k^{+}$ is the topology defined by the metric111This is not a metric in the
classical sense as it might not take values in $\mathds{R}$ (in case the
valuation has higher rank) but otherwise satisfies all properties of a metric

$d(x,y):=|x-y|,$

where $|\cdot|$ denotes the valuation associated with $k^{+}$. So the sets

$\\{x\in k\mid|x|<\gamma\\}$

for varying $\gamma\in|k^{\times}|$ form a fundamental system of open
neighborhoods of $0\in k$.

######  Exercise 2.2.

The sets

$\\{x\in k\mid|x|\leq\gamma\\}$

for $\gamma\in|k^{\times}|$ are open and also form a fundamental system of
open neighborhoods of $0$.

At this point we can explain why we need the valuation ring $k^{+}$ to be
microbial in the analytic case. Suppose this is the case and let $\varpi$ be a
topologically nilpotent element.

######  Exercise 2.3.

The valuation topology on $k$ coincides with the topology defined by the pair
of definition $(k^{\circ},(\varpi))$. Moreover $k^{+}$ is open in $k$ and
contained in $k^{\circ}$ (which implies that we can alternatively take the
pair of definition $(k^{+},(\varpi))$). In particular, $(k,k^{+})$ is indeed a
Huber pair.

######  Lemma 2.4.

The subring $k^{\circ}\subseteq k$ of powerbounded elements is a valuation
ring of rank $1$ with maximal ideal $k^{\circ\circ}$.

###### Proof.

In general, every subring of $k$ containing a valuation ring of $k$ (in our
case $k^{+}$) is again a valuation ring. Here is the argument: Suppose $x\in
k$ is not contained in $k^{\circ}$. Then it is not contained in $k^{+}$
either. Since $k^{+}$ is a valuation ring, $1/x$ is contained in $k^{+}$,
hence in $k^{\circ}$.

In order to see that $k^{\circ\circ}$ is the maximal ideal of $k^{\circ}$, we
have to check that every element in $k^{\circ}\setminus k^{\circ\circ}$ is a
unit of $k^{\circ}$. If $x\in k^{\circ}$ is not topologically nilpotent, there
is an open neighborhood $U\subseteq k$ of zero not containing any power of
$x$. Unravelling the definition of the valuation topology we can equivalently
formulate this in terms of the valuation $|\cdot|$ of $k^{+}$: There is
$\gamma\in|k^{\times}|$ such that all powers of $x$ are bounded below by
$\gamma$:

$|x^{n}|\geq\gamma\quad\forall n\in\mathds{N}.$

This implies

$\left|\left(\frac{1}{x}\right)^{n}\right|\leq\frac{1}{\gamma}\quad\forall
n\in\mathds{N},$

so the powers of $1/x$ are bounded by $1/\gamma$. This translates to $1/x$
being powerbounded in the valuation topology. We conclude that $1/x$ is
contained in $k^{\circ}$, hence $x$ is invertible.

It remains to convince ourselves that $k^{+}$ has rank $1$, or equivalently
Krull dimension $1$. We claim that $(0)\neq k^{\circ\circ}$ and these are the
only prime ideals of $k^{\circ}$. The inequality $(0)\neq k^{\circ\circ}$
follows from the existence of the topologically nilpotent unit $\varpi$. Being
a unit just means $\varpi\neq 0$, so $\varpi\in
k^{\circ\circ}\setminus\\{0\\}$. We already know that $k^{\circ}$ is a
valuation ring (hence in particular local) with maximal ideal
$k^{\circ\circ}$. Any other prime ideal needs to be contained in
$k^{\circ\circ}$. Therefore it suffices to show that the radical ideal
generated by any nonzero element $x$ of $k^{\circ\circ}$ equals
$k^{\circ\circ}$. To see this we take any other element $y\in k^{\circ\circ}$
and use its topological nilpotence to find $n\in\mathds{N}$ such that

$|y^{n}|\leq|x|$

This implies that $a:=y^{n}/x$ is contained in $k^{+}$, hence in $k^{\circ}$.
In other words, $y$ is contained in the radical ideal generated by $x$. ∎

######  Remark 2.5.

The valuation $|\cdot|^{\circ}$ corresponding to $k^{\circ}$ can be obtained
from $|\cdot|$ in the following way. Remember that an element $\gamma$ of a
totally ordered group $\Gamma$ is _cofinal_ if for any other element
$\delta\in\Gamma$ there is $n\in\mathds{N}$ such that

$\gamma^{n}\leq\delta.$

Note that $x\in k^{\times}$ is topologically nilpotent if and only if $|x|$ is
cofinal in $|k^{\times}|$. Now we convince ourselves that the subset

$\Delta:=\\{\gamma\in|k^{\times}|\mid\text{neither $\gamma$ nor $1/\gamma$ is
cofinal in $|k^{\times}|$}\\}$

of $|k^{\times}|$ is a convex subgroup. Then $\Gamma:=|k^{\times}|/\Delta$ is
a totally ordered group and we claim that $|\cdot|^{\circ}$ is given by the
composite

$k\overset{|\cdot|}{\longrightarrow}|k^{\times}|\cup\\{0\\}\twoheadrightarrow\Gamma\cup\\{0\\}.$

Notice that $x\in k^{\times}$ is powerbounded if and only if $1/x$ is _not_
topologically nilpotent. In this way we identify $k^{\circ}$ with the set of
topologically nilpotent elements $k^{\circ\circ}$ together with those elements
$x\in k^{\times}$ such that neither $x$ nor $1/x$ is topologically nilpotent.
This is precisely the valuation ring corresponding to the above defined
valuation.

Affinoid fields are important as their adic spectrum is the prototype of
something like a point in an adic space. This is not a hundred percent precise
since $\operatorname{\mathrm{Spa}}(k,k^{+})$ may contain several points as the
following example shows.

######  Example 2.6 (The adic spectrum of an affinoid field).

We want to understand the adic spectrum of an affinoid field $(k,k^{+})$. We
first take a look at all continuous valuations. In case the topology of $k$ is
discrete, there are no restrictions and any valuation ring of $k$ defines a
continuous valuation. For example the continuos valuations of $\mathds{Q}$
(with the discrete topology) are precisely all $p$-adic valuations for
different prime numbers $p$ together with the trivial valuation (this follows
from Ostrowski’s theorem).

Let us now assume that the valuation ring $k^{+}$ is non-trivial, i.e. not
equal to $k$ and $k$ carries the valuation topology. In other words
$(k,k^{+})$ is analytic. It is not hard to see that the valuation
corresponding to a valuation ring $R$ is continuous if and only if $R$ is
contained in the subring of powerbounded elements $k^{\circ}\subseteq k$. In
our setting, where the topology on $k$ is the valuation topology defined by
the valuation ring $k^{+}$, $k^{\circ}$ is a valuation ring of rank $1$ (see
Lemma 2.4). Having an inclusion of valuation rings

$R\subseteq k^{\circ}\subseteq k$

automatically implies that $k^{\circ}$ is the localization of $R$ at a prime
ideal (in our case a prime ideal of height $1$ since $k^{\circ}$ has rank
$1$). On the one hand, the image of $R$ under

$\pi:k^{\circ}\twoheadrightarrow k^{\circ}/k^{\circ\circ}$

is a valuation ring $\bar{R}$ and $R=\pi^{-1}(\bar{R})$. On the other hand the
preimage of any valuation ring of $k^{\circ}/k^{\circ\circ}$ under $\pi$ is a
valuation ring corresponding to a continuous valuation. In this way we see
that the continuous valuations of $k$ correspond to the valuations of
$k^{\circ}/k^{\circ\circ}$. Via this correspondence $k^{\circ}$ corresponds to
the trivial valuation on $k^{\circ}/k^{\circ\circ}$.

Remember that there are no non-trivial valuations on a finite field.
Therefore, if $k^{\circ}/k^{\circ\circ}$ is an algebraic extension of a finite
field, there is only one continuous valuation of $k$, namely the one
corresponding to $k^{\circ}$. This is the case for $k=\mathds{Q}_{p}$ or
$k=\mathds{C}_{p}$.

Finally let us describe the adic spectrum
$\operatorname{\mathrm{Spa}}(k,k^{+})$. By definition it consists of all
(equivalence classes of) continuous valuations whose valuation ring contains
$k^{+}$. These correspond to valuation rings $R$ in between $k^{+}$ and
$k^{\circ}$:

$k^{+}\subseteq R\subseteq k^{\circ}.$

All valuation rings of this kind are totally ordered by inclusion and can be
obtained by localizing $k^{+}$ at a prime ideal. In this case the rank of $R$
equals the height of the prime ideal. If the topology of $k$ is discrete,
$k^{\circ}=k$ and any prime ideal of $k^{+}$ is allowed. If $(k,k^{+})$ is
analytic, $k^{\circ}$ is of rank $1$ and thus only the prime ideals of height
greater or equal to $1$ occur. We conclude that
$\operatorname{\mathrm{Spa}}(k,k^{+})$ has $\operatorname{\mathrm{rk}}k^{+}$
many points if $(k,k^{+})$ is analytic and $\operatorname{\mathrm{rk}}k^{+}+1$
many points if $(k,k^{+})$ is discrete.

For abelian topological groups and commutative topological rings there is a
general theory of completion (see [Bou71], III §3). In our situation the
completion $\hat{A}$ of a Huber ring $A$ is obtained by $I$-adically
completing the ring of definition $A_{0}$ (where $I\subseteq A_{0}$ is an
ideal of definition) and then taking the tensor product:

$\hat{A}=\hat{A}_{0}\otimes_{A_{0}}A.$

We obtain a ring of integral elements $\hat{A}^{+}\subseteq\hat{A}$ by taking
the topological closure of $A^{+}$ in $\hat{A}$. Then $\hat{A}^{+}$ is
automatically integrally closed in $\hat{A}$ as can be seen by approximation.

######  Lemma 2.7.

Let $(A,A^{+})$ be a Huber pair with completion $(\hat{A},\hat{A}^{+})$. Then
we have a natural identification

$\operatorname{\mathrm{Spa}}(\hat{A},\hat{A}^{+})\cong\operatorname{\mathrm{Spa}}(A,A^{+}).$

###### Proof.

The crucial point is that the adic spectrum only contains _continuous_
valuations. A continuous valuation

$v:A\to\Gamma$

naturally extends to the completion of $A$. For a sequence
$(a_{n})_{n\in\mathds{N}}$ in $A$ converging to $a\in\hat{A}$ the continuity
of the valuation $v$ implies that for any $\gamma\in\Gamma$ the valuation of
the difference $v(a_{n}-a_{m})$ is less than $\gamma$ for $m$ and $n$ big
enough. If the sequence $(v(a_{n}))_{n\in\mathds{N}}$ becomes arbitrarily
small for $n\to\infty$, we set $v(a)=0$.

Otherwise we claim that $(v(a_{n}))_{n\in\mathds{N}}$ stabilizes. By
assumption there is $C\in\Gamma$ with $C>0$ such that

$v(a_{n})\geq C$

for $n$ big enough. Let $\gamma\in\Gamma$ with $0<\gamma<C$. Then for $n$ and
$m$ big enough, $v(a_{n}-a_{m})<\gamma$ and therefore

$v(a_{m})=v(a_{n}+(a_{m}-a_{n}))=v(a_{n}).$

Here we used that

$v(a_{m}-a_{n})<\gamma<C\leq a_{n}$

such that we even get an equality in the strong triangle inequality. Hence we
can set $v(a)$ equal to the constant value $v(a_{n})$ for $n$ big enough.

Of course one needs to check that the construction does not depend on the
chosen sequence $(a_{n})_{n\in\mathds{N}}$ and indeed defines a continuous
valuation on $\hat{A}$. ∎

We equip $\operatorname{\mathrm{Spa}}(A,A^{+})$ with the subspace topology
from $\operatorname{\mathrm{Cont}}(A)$. This is the topology generated by the
subsets of $\operatorname{\mathrm{Spa}}(A,A^{+})$ of the form

$R\left(\frac{f_{1},\ldots,f_{n}}{g}\right)=\\{x\in\operatorname{\mathrm{Spa}}(A,A^{+})\mid|f_{i}(x)|\leq|g(x)|\neq
0~{}\text{for}~{}i=1,\ldots,n\\},$

where $f_{1},\ldots,f_{n},g$ are elements of $A$ generating an _open_ ideal of
$A$. These subsets are called _rational subsets_. Note that if $A$ is a Tate
ring, i.e. contains a topologically nilpotent unit $\varpi$, the only open
ideal is $A$ itself because some power of $\varpi$ would have to be contained
in it.

If the elements $f_{1},\ldots,f_{n},g$ do _not_ generate an open ideal, the
corresponding subset of $\operatorname{\mathrm{Spa}}(A,A^{+})$ is not rational
(by definition) but still open. This follows from the following exercise.

######  Exercise 2.8.

Let $f_{1},\ldots,f_{n},g$ be arbitrary elements of $A$. For each
$r\in\mathds{N}$ let $E_{r}$ be a finite set of generators of $I^{r}$ (where
$I$ is an ideal of definition). Then

$R\left(\frac{f_{1},\ldots,f_{n}}{g}\right)=\bigcup_{r\in\mathds{N}}R\left(\frac{{\\{f_{1},\ldots,f_{n}\\}}\cup
E_{r}}{g}\right)$

and the set

$\\{f_{1},\ldots,f_{n},g\\}\cup E_{r}$

generates an open ideal of $A$.

In Lemma 2.7 we have seen that $\operatorname{\mathrm{Spa}}(A,A^{+})$ is
naturally isomorphic to $\operatorname{\mathrm{Spa}}(\hat{A},\hat{A}^{+})$. A
priori it is not clear, however, whether
$\operatorname{\mathrm{Spa}}(A,A^{+})$ and
$\operatorname{\mathrm{Spa}}(\hat{A},\hat{A}^{+})$ have the same rational
subsets. There could be more of them in
$\operatorname{\mathrm{Spa}}(\hat{A},\hat{A}^{+})$ because we can consider
rational subsets defined by $f_{1},\ldots,f_{n},g\in\hat{A}$ that are not all
contained in $A$. Luckily it turns out that we can approximate
$f_{1},\ldots,f_{n},g$ by elements of $A$ without changing the corresponding
rational subset. This is ensured by the following proposition.

######  Proposition 2.9.

Suppose $f_{1},\ldots,f_{n},g\in\hat{A}$ generate an open ideal. Then there is
an open neighborhood $U$ of $0$ in $\hat{A}$ such that for all
$f^{\prime}_{i}\in f_{i}+U$ and $g^{\prime}\in g+U$ the ideal of $\hat{A}$
generated by $f^{\prime}_{1},\ldots,f^{\prime}_{n},g^{\prime}$ is open and

$R\left(\frac{f_{1},\ldots,f_{n}}{g}\right)=R\left(\frac{f^{\prime}_{1},\ldots,f^{\prime}_{n}}{g^{\prime}}\right).$

###### Proof.

[Hub93], Lemma 3.10 ∎

As $A$ is dense in $\hat{A}$ we can choose
$f^{\prime}_{1},\ldots,f^{\prime}_{n},g^{\prime}$ as in the lemma to be
contained in $A$ and then the corresponding rational subset $U$ is defined by
elements in $A$. However, this does not yet ensure that
$f^{\prime}_{1},\ldots,f^{\prime}_{n},g^{\prime}$ generate an open ideal _in
$A$_. But we can extend the set $\\{f^{\prime}_{1},\ldots,f^{\prime}_{n}\\}$
by elements $f^{\prime}_{n+1},\ldots,f^{\prime}_{k}$ that are less than or
equal to $g^{\prime}$ on $U$ and such that the ideal of $A$ generated by
$f^{\prime}_{1},\ldots,f^{\prime}_{k},g$ is open (see [Hub93], Lemma 3.11).

######  Example 2.10.

Let $k$ be a nonarchimedean field. For simplicity we assume it is
algebraically closed. Then

$\mathds{D}_{k}=\operatorname{\mathrm{Spa}}(k\langle T\rangle,k^{\circ}\langle
T\rangle)$

is called the _adic unit disc_ over $k$. It can also be described as the
subset

$\\{x\in\operatorname{\mathrm{Cont}}(k\langle T\rangle)\mid|T(x)|\leq
1,~{}|k^{\circ}(x)|\leq 1\\}$

The second condition ensures that the restriction of $x$ to $k$ coincides with
the given valuation of $k$. Without this condition $x|_{k}$ could be any
continuous valuation of $k$. As we have seen in Example 2.6 the corresponding
valuation ring could be a proper subring of $k^{\circ}$ (this happens if the
rank of $x|_{k}$ is greater than $1$). The condition $|k^{\circ}(x)|\leq 1$
forces $k^{\circ}$ to be contained in the valuation ring implying that the
valuation ring of $x|_{k}$ equals $k^{\circ}$. If $k=\mathds{C}_{p}$, this is
automatic as its residue field is $\bar{\mathds{F}}_{p}$, an algebraic
extension of a finite field (see Example 2.6).

For $\alpha\in k^{\circ}$ and $r\in|k^{\times}|$ the rational subset

$\mathds{D}_{k}(\alpha,r):=\\{x\in\mathds{D}_{k}\mid|(T-\alpha)(x)|\leq r\\}$

is a disc of radius $r$ around $\alpha$. Here $\alpha$ is considered as a
“classical point”, i.e. the valuation

$k\langle
T\rangle\overset{T\mapsto\alpha}{\longrightarrow}k\overset{|\cdot|}{\longrightarrow}\mathds{R}_{>0}\cup\\{0\\}.$

The “boundary ring”

$\mathds{S}_{k}(0,1):=\\{x\in\mathds{D}_{k}\mid|T(x)|=1\\}=\\{x\in\mathds{D}_{k}\mid|T(x)|\leq
1,1\leq|T(x)|\neq 0\\}$

is also a rational subset. One might expect that the union of
$\mathds{S}_{k}(0,1)$ with all discs $\mathds{D}_{k}(0,r)$ of radius $r<1$ is
all of $\mathds{D}_{k}$. However the union misses one point: Consider the
abelian group $\mathds{R}_{>0}\times\mathds{R}_{>0}$. We endow it with the
lexicographical ordering, i.e.

$(a_{1},b_{1})<(a_{2},b_{2})\Leftrightarrow
a_{1}<a_{2}~{}\text{or}~{}(a_{1}=a_{2}~{}\text{and}~{}b_{1}<b_{2}).$

Now we pick an element $\epsilon\in\mathds{R}_{>0}\times\mathds{R}_{>0}$ of
the form $(1,\epsilon_{2})$ with $\epsilon_{2}<1$. So $\epsilon$ is an element
that is less than $1=(1,1)$ but infinitesimally close to $1$ in the sense that
$(r_{1},r_{2})<\epsilon$ whenever $r_{1}<1$. Via the embedding

$\mathds{R}_{>0}\hookrightarrow\mathds{R}_{>0}\times\mathds{R}_{>0},\qquad
r\mapsto(r,1)$

we can consider the valuation on $k$ as a valuation
$k\to(\mathds{R}_{>0}\times\mathds{R}_{>0})\cup\\{0\\}$. We now define a
valuation

$x_{0,1-}:k\langle
T\rangle\longrightarrow(\mathds{R}_{>0}\times\mathds{R}_{>0})\cup\\{0\\}$

that maps $f=\sum_{i}a_{i}T^{i}\in k\langle T\rangle$ to

$|f(x_{0,1-})|=\max_{i}|a_{i}|\epsilon^{i}.$

This is indeed an element of $\mathds{D}_{k}$: It is continuous as for any
pseudouniformizer $\varpi\in k$ we have that $|\varpi(x_{0,1-})|=|\varpi|$ is
cofinal in $\mathds{R}_{>0}\times\mathds{R}_{>0}$ and $|\varpi f(x_{0,1-})|<1$
for any $f\in k^{\circ}\langle T\rangle$ (see the criterion for continuity in
[Ber24], Proposition 1.2.7.1). Moreover $|f(x_{0,1-})|\leq 1$ for all $f\in
k^{\circ}\langle T\rangle$, so $x_{0,1-}\in\mathds{D}_{k}$. However,

$|T(x_{0,1-})|=\epsilon$

and

$r<\epsilon<1~{}\forall~{}r\in[0,1).$

So neither is $x_{0,1-}$ contained in the ring $\mathds{S}_{k}(0,1)$ nor in
any $\mathds{D}_{k}(0,r)$, and this is indeed the only point of
$\mathds{D}_{k}$ not contained in this union, as follows from the
classification of the points of $\mathds{D}_{k}$, which is explained in
[Ber24], §1.4.4.

We want to remark here that the point $x_{0,1-}$ defined in this way does not
depend on the choice of $\epsilon=(1,\epsilon_{2})$ as long as
$\epsilon_{2}<1$. The valuations for different choices of $\epsilon$ are
equivalent. This is why $\epsilon$ does not show up in the index of
$x_{0,1-}$. By the minus sign we indicate that $\epsilon<1$, or equivalently
that we take a radius that is infinitesimally smaller than $1$. If we choose
$\epsilon>1$ we obtain a different valuation that will play a role in Example
7.12.

##  3\. The structure (pre-)sheaf

We want to define on $X=\operatorname{\mathrm{Spa}}(A,A^{+})$ a sheaf of
complete topological rings ${\mathscr{O}}_{X}$ together with a subsheaf
${\mathscr{O}}_{X}^{+}$ providing ${\mathscr{O}}_{X}$ with an integral
structure satisfying

${\mathscr{O}}_{X}(X)=\hat{A},\qquad{\mathscr{O}}_{X}^{+}(X)=\hat{A}^{+}$

The construction starts by assigning to each rational subset $U$ a pair of
rings $(A_{U},A_{U}^{+})$. This defines two presheavs, ${\mathscr{O}}_{X}$ and
${\mathscr{O}}_{X}^{+}$, on the category of rational subsets. It is in the
same spirit as the construction of the structure sheaf on an affine scheme.
Unfortunately there are Huber pairs $(A,A^{+})$ such that ${\mathscr{O}}_{X}$
does not satisfy the sheaf axioms. The problem lies in the completion of rings
that is involved in the construction. So a priori we only construct a
structure presheaf. We will then discuss the main instances when
${\mathscr{O}}_{X}$ is actually a sheaf.

For a rational subset

$U=R\left(\frac{f_{1},\ldots,f_{n}}{g}\right)=\\{x\in
X\mid|f_{i}(x)|\leq|g(x)|\neq 0~{}\forall i=1,\ldots,n\\}$

as above we consider the pair of rings

(1)
$\left(A\left[\frac{1}{g}\right],A^{+}\left[\frac{f_{1}}{g},\ldots,\frac{f_{n}}{g}\right]^{N}\right)$

where $(\cdot)^{N}$ stands for taking the integral closure in
$A_{g}=A[\frac{1}{g}]$ ($N$ for normalization). We equip it with a topology in
the following way. Let $A_{0}$ be a ring of definition and $I$ an ideal of
definition for the Huber ring $A$. Then we turn $A_{g}$ into a Huber ring by
setting its ring of definition equal to

$A_{0}\left[\frac{f_{1}}{g},\ldots,\frac{f_{n}}{g}\right]$

and its ideal of definition to

$IA_{0}\left[\frac{f_{1}}{g},\ldots,\frac{f_{n}}{g}\right].$

One checks that with these definitions the pair of rings defined in (1) is
indeed a Huber pair. We then take completions to obtain a complete Huber pair
$(A_{U},A_{U}^{+})$.

######  Lemma 3.1.

The homomorphism of Huber pairs $(A,A^{+})\to(A_{U},A_{U}^{+})$ induces an
injection

$\operatorname{\mathrm{Spa}}(A_{U},A_{U}^{+})\hookrightarrow\operatorname{\mathrm{Spa}}(A,A^{+})$

whose image is the rational subset $U$.

Note that Lemma 3.1 allows us to view the rational subset $U$ as the adic
spectrum of the Huber pair $(A_{U},A_{U}^{+})$.

###### Proof.

It is clear that the morphism is injective since $A_{U}$ is the completion of
$A_{g}$ and the valuations of $A_{g}$ form a subset of the valuations of $A$.
Also a valuation $x$ of $A$ extends to $A_{g}$ if and only if $|g(x)|\neq 0$.
Moreover $x$ is less than or equal to $1$ on

$A^{+}\left[\frac{f_{1},\ldots,f_{n}}{g}\right]^{N}$

if and only if $|f_{i}(x)|\leq|g(x)|$ for $i=1,\ldots,n$. This already shows
that the image of $\operatorname{\mathrm{Spa}}(A_{U},A_{U}^{+})$ is contained
in $U$. In order to show the converse inclusion, we need to check that for any
$x\in U$ its extension to $A_{g}$ is continuous. Remember that we defined

$\left(A_{0}\left[\frac{f_{1},\ldots,f_{n}}{g}\right],IA_{0}\left[\frac{f_{1},\ldots,f_{n}}{g}\right]\right)$

to be a pair of definition for $A_{g}$ for a given pair of definition
$(A_{0},I)$ of $A$. By the criterion for continuity explained in [Ber24],
Proposition 1.2.7.1, it is enough to convince ourselves that $|x(b)|$ is
cofinal in the value group for $b\in I$ and $|x(a)|<1$ for all $a\in
IA_{0}[\frac{f_{1},\ldots,f_{n}}{g}]$. This follows directly from the
condition $x\in U$. ∎

We now define the presheaf ${\mathscr{O}}_{X}$ and its subpresheaf
${\mathscr{O}}_{X}^{+}$ on the category of rational subsets by

${\mathscr{O}}_{X}(U):=A_{U},\qquad{\mathscr{O}}_{X}^{+}(U):=A_{U}^{+}.$

As mentioned before, there are examples where the sheaf condition is not
satified (see Example 3.5 below). We call a Huber pair $(A,A^{+})$ _sheafy_ if
the structure presheaf ${\mathscr{O}}_{X}$ on
$X=\operatorname{\mathrm{Spa}}(A,A^{+})$ is a sheaf.

######  Lemma 3.2.

Let $(A,A^{+})$ be a sheafy Huber pair and set
$X=\operatorname{\mathrm{Spa}}(A,A^{+})$. Then ${\mathscr{O}}_{X}^{+}$ is a
sheaf.

###### Proof.

For a rational open covering $U=\bigcup_{i}U_{i}$ of a rational open
$U\subseteq X$ we consider the diagram

${{\mathscr{O}}_{X}(U)}$${\prod_{i}{\mathscr{O}}_{X}(U_{i})}$${\prod_{ij}{\mathscr{O}}_{X}(U_{i}\cap
U_{j})}$${{\mathscr{O}}_{X}^{+}(U)}$${\prod_{i}{\mathscr{O}}_{X}^{+}(U_{i})}$${\prod_{ij}{\mathscr{O}}_{X}^{+}(U_{i}\cap
U_{j}).}$

The upper row is exact because $(A,A^{+})$ is sheafy. It follows immediately
that the left lower arrow is injective. Exactness in the middle is also clear
once we know the following formula, which will be discussed in John’s lecture
(Theorem 3.3):

${\mathscr{O}}_{X}^{+}(U)=\\{a\in{\mathscr{O}}_{X}(U)\mid|a(x)|\leq
1~{}\forall~{}x\in U\\}.$

∎

It turns out that in basically all known cases where we can show sheafiness of
the structure presheaf the proof can be enhanced just a bit to show in
addition that the structure sheaf is acyclic. By this we mean explicitly that
the sheaf cohomology groups

$H^{i}(X,{\mathscr{O}}_{X})$

vanish for $i>0$. The following theorem summarizes the known results. We will
discuss them below and also explain all the adjectives showing up in the
hypothesis.

######  Theorem 3.3.

Let $(A,A^{+})$ be a Huber pair satisfying one of the following hypothesis.

  1. (i)

The topology on $A$ is discrete,

  2. (ii)

$A$ has a noetherian ring of definition,

  3. (iii)

$A$ is a strongly noetherian Tate ring,

  4. (iv)

$(A,A^{+})$ is stably uniform.

Then $(A,A^{+})$ is sheafy and the structure sheaf is acyclic.

The easiest class of sheafy Huber pairs is the class of Huber pairs with
discrete topology. In this case there is no completion involved and the sheaf
condition translates to the sheaf condition for fundamental opens in affine
schemes. So the proof of the theorem is straight forward (see [Wed19], Theorem
8.28 (c)).

The second case, where $A$ has a noetherian ring of definition, is often
referred to as the “formal scheme setting”. The reason behind this
nomenclature is that there is a fully faithful functor from locally noetherian
formal schemes to adic spaces (see [Hub94], Proposition 4.2). At the level of
affine formal schemes it sends the formal spectrum
$\operatorname{\mathrm{Spf}}(A^{+})$ to
$\operatorname{\mathrm{Spa}}(A^{+},A^{+})$. The affinoids in the essential
image of this functor satisfy condition (ii) of Theorem 3.3. Note that not all
affinoids satisfying condition (ii) come from formal schemes. This is only the
case for Huber pairs $(A,A^{+})$ with $A=A^{+}$. However, they are closely
enough related to formal schemes to make the sheafiness proof for locally
noetherian formal schemes work in this case (see [Hub94], Theorem 2.5).

We will now treat the case of Tate Huber pairs, which is most relevant to us.
Remember that a Huber ring is Tate if it contains a topologically nilpotent
unit. This is the case for the classes of Huber pairs described in (iii) and
(iv) in Theorem 3.3.

Let us explain the terminology. A Tate ring $A$ is called _strongly
noetherian_ if $A\langle T_{1},\ldots,T_{n}\rangle$ is noetherian for any
$n\geq 0$. Looking at the condition for $n=0$ this tells us in particular that
$\hat{A}$ is noetherian. An equivalent condition for strong noetherianness is
that all $A$-algebras that are topologically of finite type are noetherian. We
call a Tate Huber pair $(A,A^{+})$ strongly noetherian if $A$ is strongly
noetherian. Examples include all Tate rings of the form

$k\langle T_{1},\ldots,T_{n}\rangle/I$

for a nonarchimedean field $k$ and any ideal $I\subseteq k\langle
T_{1},\ldots,T_{n}\rangle$. This boils down to the classical result from rigid
geometry that $k\langle T_{1},\ldots,T_{n}\rangle$ is noetherian ([BGR84], §
5.2.6, Theorem 1). The proof is by no means trivial and involves the
Weierstraß division and preparation theorems.

A Tate ring $A$ is _uniform_ if the subset of power bounded elements
$A^{\circ}$ is bounded. A Huber pair $(A,A^{+})$ is called _stably uniform_ if
for all rational subsets $U\subseteq X=\operatorname{\mathrm{Spa}}(A,A^{+})$
the Huber ring ${\mathscr{O}}_{X}(U)$ is uniform.

There are strongly noetherian Tate Huber pairs that are not stably uniform and
vice versa. For instance for any nonarchimedean field $k$ with
pseudouniformizer $\varpi$ the Huber ring

$A:=k\langle T\rangle/(T^{2})$

is strongly noetherian but not uniform. The reason is that all elements
$\varpi^{-n}T$ for $n\in\mathds{N}$ are nilpotent, hence power bounded.
However, none of the sets

$\left(\varpi^{m}k^{\circ}\langle
T\rangle/(T^{2})\right)\cdot\\{\varpi^{-n}T\mid n\in\mathds{N}\\}$

for $m\in\mathds{N}$ is contained in the open $k^{\circ}\langle
T\rangle/(T^{2})$ as would be required if $\\{\varpi^{-n}T\mid
n\in\mathds{N}\\}$ were bounded.

In fact the same argument shows that more generally a Tate ring $A$ with pair
of definition $(A_{0},(\varpi))$ is _not_ uniform as soon as it contains a
nilpotent element $a$ such that

$\\{\varpi^{n}a\mid n\in\mathds{Z}\\}\nsubseteq A_{0}.$

On the other hand, a large class of non-noetherian stably uniform Tate Huber
pairs is given by perfectoid Huber pairs $(A,A^{+})$:

######  Proposition 3.4.

Any perfectoid Huber pair $(A,A^{+})$ is stably uniform.

###### Proof.

A perfectoid ring is uniform by definition (see Ben’s notes, Definition
2.1.1). Moreover, by Theorem 3.1.4 in Ben’s notes, for any rational open
$U\subseteq\operatorname{\mathrm{Spa}}(A,A^{+})$ the ring $A_{U}$ is again
perfectoid, hence uniform. We conclude that perfectoid Huber pairs are stably
uniform. ∎

Let us get back to the question of sheafiness for Tate Huber pairs. The
following example shows that there are indeed non-sheafy Tate Huber pairs.

######  Example 3.5 ([BV18], §4.1).

Let $k$ be a complete non-Archimedean field and

$A=k[T,T^{-1},Z]/(Z^{2}).$

We turn $A$ into a Huber $k$-algebra by considering the ring of definition
$A_{0}$ generated as an $k^{\circ}$-module by

$\varpi^{|n|}T^{n},\qquad\varpi^{-|n|}T^{n}Z,\quad\forall~{}n\in\mathds{Z}.$

Then $A_{0}$ is not noetherian as

$ZA\cap A_{0}=(Z,\frac{Z}{\varpi T},\frac{Z}{\varpi^{2}T^{2}},\ldots)$

is not finitely generated.

We consider the cover of $X=\operatorname{\mathrm{Spa}}(A,A^{\circ})$ by the
two opens

$U=\\{x\in X\mid|T(x)|\leq 1\\},\qquad V=\\{x\in X\mid|T(x)|\geq 1\\}.$

We claim that

${\mathscr{O}}_{X}(X)\to{\mathscr{O}}_{X}(U)\oplus{\mathscr{O}}_{X}(V)$

is not injective. We have

$\displaystyle{\mathscr{O}}_{X}(X)$
$\displaystyle=\hat{A}=\lim_{n}A/(\varpi^{n}A_{0})$
$\displaystyle{\mathscr{O}}_{X}(U)$
$\displaystyle=\lim_{n}A/(\varpi^{n}A_{0}[T])$
$\displaystyle{\mathscr{O}}_{X}(V)$
$\displaystyle=\lim_{n}A/(\varpi^{n}A_{0}[T^{-1}]).$

As $\varpi^{-n}Z\notin A_{0}$ for any $n\in\mathds{N}$, we know that $Z\neq 0$
in ${\mathscr{O}}_{X}(X)$. But

$\displaystyle\varpi^{-n}Z$ $\displaystyle=(\varpi^{-n}T^{-n}Z)T^{n}\in
A_{0}[T],$ $\displaystyle\varpi^{-n}Z$
$\displaystyle=(\varpi^{-n}T^{n}Z)T^{-n}\in A_{0}[T^{-1}].$

Hence, $Z=0$ on $U$ and on $V$.

In Example 3.5 even though $A$ is noetherian, it is not strongly noetherian.
Neither is $A$ uniform because $Z$ is nilpotent and $\varpi^{-m}Z$ for $m\geq
1$ is not contained in $A_{0}$.

Let us now outline the general strategy to study the question of sheafiness
for a Tate Huber pair. As mentioned before we show sheafiness and acyclicity
at the same time. Let $X=\operatorname{\mathrm{Spa}}(A,A^{+})$ for a Tate
Huber pair $(A,A^{+})$. For an open covering $U=\bigcup_{i}U_{i}$ for rational
subsets $U$ and $U_{i}$ we consider the sequence

(2)
$0\to{\mathscr{O}}_{X}(U)\to\prod_{i}{\mathscr{O}}_{X}(U_{i})\to\prod_{ij}{\mathscr{O}}_{X}(U_{i}\cap
U_{j})\to\prod_{ijk}{\mathscr{O}}_{X}(U_{i}\cap U_{j}\cap U_{k})\to\ldots$

Exactness at the first two terms gives the sheaf condition for the covering
$U=\bigcup_{i}U_{i}$ and exactness at all other terms is equivalent to the
vanishing of the C̆ech cohomology groups

$\check{H}^{i}(U,{\mathscr{O}}_{X})$

for $i>0$. It is a general fact (see [SP, Tag 01EV]) that in order to show
that the sheaf cohomology groups $H^{i}(X,{\mathscr{O}}_{X})$ vanish, it
suffices to show that all C̆ech cohomology groups as above for all coverings
$U=\bigcup_{i}U_{i}$ vanish.

In the proof for sheafiness the first step is to reduce to the case of very
simple coverings.

######  Definition 3.6.

Let $(A,A^{+})$ be a Tate Huber pair and set
$X=\operatorname{\mathrm{Spa}}(A,A^{+})$.

  * •

Let $t_{1},\ldots,t_{n}$ be elements of $A$ generating the unit ideal. The
_standard rational covering_ generated by $t_{1},\ldots,t_{n}$ is the covering
of $X$ by

$U_{i}=R\left(\frac{t_{1},\ldots,t_{n}}{t_{i}}\right),\quad\text{where
}i\in\\{1,\ldots,n\\}.$

  * •

Let $t_{1},\ldots,t_{n}$ be in $A$. The _standard Laurent covering_ generated
by $t_{1},\ldots,t_{n}$ is the covering of $X$ by

$U_{I}=\\{x\in X\mid|t_{i}(x)|\leq 1~{}\forall i\in I,~{}|t_{i}|\geq
1~{}\forall~{}i\notin I\\},\quad\text{where }I\subseteq\\{1,\ldots,n\\}.$

  * •

A _simple_ Laurent covering is a standard Laurent covering generated by one
element.

It is clear that Laurent coverings are indeed coverings. In case of standard
rational coverings this is ensured by the condition that $t_{1},\ldots,t_{n}$
generate the unit ideal: For $x\in X$ we take $t_{i}$ such that $|t_{i}(x)|$
is maximal among all $|t_{j}(x)|$. Because $t_{1},\ldots,t_{n}$ generate the
unit ideal, we must have $|t_{i}(x)|\neq 0$. By definition this implies $x\in
U_{i}$.

######  Lemma 3.7.

Let $(A,A^{+})$ be a complete Tate Huber pair and
$X=\operatorname{\mathrm{Spa}}(A,A^{+})$.

  1. (i)

Every covering of $X$ has a refinement by a standard rational covering.

  2. (ii)

For every standard rational covering ${\mathscr{U}}$ of $X$ there exists a
Laurent covering ${\mathscr{V}}$ of $X$ such that for every
$V\in{\mathscr{V}}$ the covering ${\mathscr{U}}_{V}:=\\{V\cap U\mid
U\in{\mathscr{U}}\\}$ is a standard rational covering generated by units.

  3. (iii)

For every standard rational covering ${\mathscr{U}}$ of $X$ generated by
units, there is a Laurent covering ${\mathscr{V}}$ of $X$ refining
${\mathscr{U}}$.

###### Proof.

See [Mor19], Proposition IV.2.3.3. ∎

Lemma 3.7 tells us that it is enough to show exactnes of sequence (2) for
Laurent coverings. Indeed if we start with some covering of $X$, we can refine
it by a standard rational covering by part (i) of the lemma. It is formal to
check that exactness of sequence (2) for a refinement of the covering implies
exactness for the original covering. We are thus reduced to deal with a
standard rational covering ${\mathscr{U}}$. By part (ii) of Lemma 3.7 we can
find a Laurent covering ${\mathscr{V}}$ of $X$ such that for every
$V\in{\mathscr{V}}$ the restriction ${\mathscr{U}}_{V}$ is a standard rational
covering of $V$ generated by units. Now we can convince ourselves that it is
enough to prove exactness of sequence (2) for the Laurent covering
${\mathscr{V}}$ and for each of the restricted coverings ${\mathscr{U}}_{V}$.
But standard rational coverings that are generated by units can be refined by
Laurent coverings by part (iii) of the lemma. We are thus reduced to Laurent
coverings. For a Laurent covering ${\mathscr{U}}$ generated by
$t_{1},\ldots,t_{n}$ we consider the simple Laurent covering $X=W_{-}\cup
W_{+}$ with

$W_{-}=\\{x\in X\mid|t_{n}(x)|\leq 1\\},\qquad W_{+}=\\{x\in
X\mid|t_{n}(x)|\geq 1\\}.$

The restrictions ${\mathscr{U}}_{W_{-}}$ and ${\mathscr{U}}_{W_{+}}$ are
Laurent coverings generated by $t_{1},\ldots,t_{n-1}$. We can thus use
induction to conclude that it is enough to prove exactness of sequence (2) for
simple Laurent coverings.

We summarize our arguments in the following lemma.

######  Lemma 3.8.

Let $(A,A^{+})$ be a Tate Huber pair and set
$X=\operatorname{\mathrm{Spa}}(A,A^{+})$. Suppose that for any rational open
$W\subseteq\operatorname{\mathrm{Spa}}(A,A^{+})$ and any simple Laurent
covering $W=W_{-}\cup W_{+}$, the sequence

(3)
$0\to{\mathscr{O}}_{X}(W)\to{\mathscr{O}}_{X}(W_{-})\oplus{\mathscr{O}}_{X}(W_{+})\to{\mathscr{O}}_{X}(W_{-}\cap
W_{+})\to 0$

is exact. Then $(A,A^{+})$ is sheafy and ${\mathscr{O}}_{X}$ is acyclic.

We can now discuss case (iii) of Theorem 3.3, where $A$ is a strongly
noetherian Tate ring. The result goes back to Tate’s acyclicity theorem first
formulated in the language of Tate’s rigid analytic spaces. Huber gave a proof
in the more general setting of strongly noetherian Tate Huber pairs (see
[Hub94], Theorem 2.2) using the same approach.

###### Idea of the proof.

First note that being strongly noetherian is stable under passing to a
rational open subset

$W=R\left(\frac{f_{1},\ldots,f_{n}}{g}\right)\subseteq\operatorname{\mathrm{Spa}}(A,A^{+}).$

with $(f_{1},\ldots,f_{n})=A$. The reason is that we can write
${\mathscr{O}}_{X}(W)$ as

${\mathscr{O}}_{X}(W)=A\langle T_{1},\ldots,T_{n}\rangle/(gT_{i}-f_{i}).$

Therefore, we need to check the hypothesis of Lemma 3.8 only for
$W=X=\operatorname{\mathrm{Spa}}(A,A^{+})$. We consider a simple Laurent
covering defined by an element $t\in A$, i.e., the covering of $X$ by

$W_{-}=\\{x\in X\mid|t(x)|\leq 1\\},\qquad W_{+}=\\{x\in X\mid|t(x)|\geq
1\\}.$

We have

$\displaystyle{\mathscr{O}}_{X}(X)$ $\displaystyle=\hat{A}$
$\displaystyle{\mathscr{O}}_{X}(W_{-})$ $\displaystyle=A\langle
T\rangle/(T-t)$ $\displaystyle{\mathscr{O}}_{X}(W_{+})$
$\displaystyle=A\langle S\rangle/(tS-1)$
$\displaystyle{\mathscr{O}}_{X}(W_{-}\cap W_{+})$ $\displaystyle=A\langle
S,T\rangle/(T-t,ST-1)$

The strongly noetherian assumption enters when showing that

$\hat{A}\to A\langle T\rangle/(T-t)\times A\langle S\rangle/(tS-1)$

is faithfully flat, hence injective. Showing the remaining exactness
properties comes down to an explicit computation. ∎

Finally we want to explain the main argument for sheafiness in case (iv) of
Theorem 3.3, where $(A,A^{+})$ is stably uniform. The proof is due to Kevin
Buzzard and Alain Verberkmoes and we refer to their article ([BV18],
Proposition 12) for the details.

###### Idea of the proof.

By definition any rational subset of $X=\operatorname{\mathrm{Spa}}(A,A^{+})$
is again stably uniform. This reduces us to checking the hypothesis of Lemma
3.8 for a simple Laurent covering of $W=X$. Explicitly this is a covering of
the form

$X=\\{x\in X\mid|t(x)|\leq 1\\}\cup\\{x\in X\mid|t(x)|\geq 1\\}=:W_{-}\cup
W_{+}.$

We consider the following topological rings with topology given by the
respective ring of definition and ideal of definition generated by $\varpi$:

$\displaystyle A_{-}=A,$ $\displaystyle\qquad(A_{-})_{0}=A_{0}[t]$
$\displaystyle A_{+}=A[1/t],$ $\displaystyle\qquad(A_{+})_{0}=A_{0}[1/t]$
$\displaystyle A_{\pm}=A[1/t],$
$\displaystyle\qquad(A_{\pm})_{0}=A_{0}[t,1/t].$

(Here we are a bit sloppy because the homomorphims $\phi:A\to A[1/t]$ might
not be injective, so we should write $(A_{+})_{0}=\phi(A_{0})[1/t]$ etc.) Then

$\displaystyle{\mathscr{O}}_{X}(X)$ $\displaystyle=\hat{A}$
$\displaystyle{\mathscr{O}}_{X}(W_{-})$ $\displaystyle=\widehat{A_{-}}$
$\displaystyle{\mathscr{O}}_{X}(W_{+})$ $\displaystyle=\widehat{A_{+}}$
$\displaystyle{\mathscr{O}}_{X}(W_{-}\cap W_{+})$
$\displaystyle=\widehat{A_{\pm}}$

The sequence

$0\to A\overset{\epsilon}{\longrightarrow}A_{-}\oplus
A_{+}\overset{\delta}{\longrightarrow}A_{\pm}\to 0$

is obviously exact and completion gives the sequence (3). However, completion
might destroy exactness if $\epsilon$ and $\delta$ are not strict (quotient
topology on image coming from the source is the same as subspace topology
coming from the target). The map $\delta$ is always strict because it is
surjective and the image of $(A_{-})_{0}\oplus(A_{+})_{0}$ is $(A_{\pm})_{0}$
and thus open; therefore the sequence (3) is exact if and only if $\epsilon$
is strict. The subspace topology on $A$ induced from $A_{-}\oplus A_{+}$ is
given by the ring of definition $S_{0}=(A_{-})_{0}\cap(A_{+})_{0}$ and
strictness means that the sets $\varpi^{n}A_{0}$ are open with respect to the
topology defined by $S_{0}$. This boils down to checking that there is
$m\in\mathds{N}$ such that

$\varpi^{m}S_{0}\subseteq A_{0}$

One can show that $S_{0}\subseteq A^{\circ}$. By hypothesis $A$ is uniform, so
$A^{\circ}$, hence $S_{0}$, is bounded. This implies the claim. ∎

######  Corollary 3.9.

Every perfectoid Huber pair $(A,A^{+})$ is sheafy and the structure sheaf is
acyclic.

###### Proof.

This follows from case (iv) of Theorem 3.3 since perfectoid Huber pairs are
stably uniform by Proposition 3.4. ∎

##  4\. Adic spaces

In this section we explain how to glue adic spectra of sheafy Huber pairs to
obtain adic spaces. The process is similar to what is done for defining
schemes. Remember that a scheme is a locally ringed space that is locally
isomorphic to the spectrum of a ring. An adic space is also a locally ringed
space but it has additional structure. So instead of the category of locally
ringed spaces we work in the category of locally $v$-ringed spaces, which is
defined as follows.

  * •

Objects are triples $(X,{\mathscr{O}}_{X},\\{v_{x}\\}_{x\in X})$, where
$(X,{\mathscr{O}}_{X})$ is a locally topologically ringed space and for each
$x\in X$ we have an equivalence class of continuous valuations $v_{x}$ on the
local ring ${\mathscr{O}}_{X,x}$ whose support is the maximal ideal.

  * •

Morphisms

$(X,{\mathscr{O}}_{X},\\{v_{x}\\}_{x\in
X})\longrightarrow(Y,{\mathscr{O}}_{Y},\\{v_{y}\\}_{y\in Y})$

are morphisms of locally topologically ringed spaces
$(X,{\mathscr{O}}_{X})\to(Y,{\mathscr{O}}_{Y})$ such that for a point $x\in X$
mapping to $y\in Y$ we have $v_{x}|_{{\mathscr{O}}_{Y,y}}=v_{y}$.

The collection of valuations $\\{v_{x}\\}_{x\in X}$ determines a subsheaf
${\mathscr{O}}_{X}^{+}$ of ${\mathscr{O}}_{X}$ by setting for an open
$U\subseteq X$

${\mathscr{O}}_{X}^{+}(U)=\\{a\in{\mathscr{O}}_{X}(U)\mid v_{x}(a_{x})\leq
1~{}\forall~{}x\in U\\}.$

Every sheafy Huber pair $(A,A^{+})$ determines a locally $v$-ringed space by
setting $X=\operatorname{\mathrm{Spa}}(A,A^{+})$ and considering the triple
$(X,{\mathscr{O}}_{X},\\{v_{x}\\}_{x\in X})$, where the valuations $v_{x}$ are
determined by $x$ in the following way. For every rational open neighborhood
$U\subseteq X$ of $x$, $x$ comes from a unique valuation on
${\mathscr{O}}_{X}(U)$ (see Lemma 3.1). In the colimit over all neighborhoods
we obtain the valuation $v_{x}$ on ${\mathscr{O}}_{X,x}$. We call a locally
$v$-ringed space arising in this way an _affinoid adic space_.

A morphism of $v$-ringed spaces $(X,{\mathscr{O}}_{X},\\{v_{x}\\}_{x\in
X})\to(Y,{\mathscr{O}}_{Y},\\{v_{y}\\}_{y\in Y})$ is an _open immersion_ if it
induces a homeomorphism of $X$ onto an open subspace $U\subseteq Y$ and an
isomorphism

$(X,{\mathscr{O}}_{X},\\{v_{x}\\}_{x\in
X})\overset{\sim}{\longrightarrow}(U,{\mathscr{O}}_{Y}|_{U},\\{v_{u}\\}_{u\in
U}).$

Inclusions of rational opens in an affinoid naturally give rise to open
immersions of locally $v$-ringed spaces.

######  Definition 4.1.

An adic space is a locally $v$-ringed space
$(X,{\mathscr{O}}_{X},\\{v_{x}\\}_{x\in X})$ that admits an open covering
$X=\bigcup_{i}U_{i}$ by affinoid adic spaces.

Adic spaces can be obtained from affinoid adic spaces by glueing just in the
same way as for schemes.

######  Lemma 4.2.

Suppose we are given

  * •

an index set $I$,

  * •

for every $i\in I$ an adic space $U_{i}$,

  * •

for every $i,j\in I$ open subspaces $U_{ij}\subseteq U_{i}$ and
$U_{ji}\subseteq U_{j}$ and an isomorphism $\varphi_{ij}:U_{ij}\to U_{ji}$

satisfying $\varphi_{ij}^{-1}(U_{ji}\cap U_{jk})=U_{ij}\cap U_{ik}$ and the
cocyle condition: For $i,j,k\in I$ the diagram

${U_{ij}\cap U_{ik}}$${U_{ki}\cap U_{kj}}$${U_{ji}\cap
U_{jk}}$$\scriptstyle{\varphi_{ik}}$$\scriptstyle{\varphi_{ij}}$$\scriptstyle{\varphi_{jk}}$

commutes. Then there is an adic space $X$ and open immersions
$\varphi_{i}:U_{i}\to X$ satisfying

  * •

$X=\bigcup_{i}\varphi(U_{i})$,

  * •

$\varphi_{i}$ induce isomorphisms
$U_{ij}\overset{\sim}{\to}\varphi_{i}(U_{i})\cap\varphi_{j}(U_{j})$,

  * •

the diagram

${U_{ij}}$${U_{ji}}$${X}$$\scriptstyle{\varphi_{ij}}$$\scriptstyle{\varphi_{i}}$$\scriptstyle{\varphi_{j}}$

commutes.

###### Proof.

That the underlying topological space of $X$ can be glued from the $U_{i}$ is
clear. As all morphisms are morphisms of locally $v$-ringed spaces, they
respect the structure sheaf and the valuations. We thus obtain a structure
sheaf on $X$ and also a set of valuations $\\{v_{x}\\}_{x\in X}$. Since each
$U_{i}$ is locally isomorphic to an affinoid adic space, the same holds for
$X$. ∎

######  Example 4.3 (The affine line).

Let $k$ be a nonarchimedean field with pseudouniformizer $\varpi\in k$. For
$n\in\mathds{N}$ the closed (adic) disc of radius $|\varpi^{-n}|$ is defined
as

$\mathds{D}_{k}(0,|\varpi^{-n}|)=\operatorname{\mathrm{Spa}}(k\langle\varpi^{n}T\rangle,k^{\circ}\langle\varpi^{n}T\rangle),$

where $\varpi^{n}T$ is to be treated as a variable. So actually the discs
$\mathds{D}_{k}(0,|\varpi^{-n}|)$ are all isomorphic but we write them like
this because of the way they are contained in each other. The notation is
justified by looking at classical points: For $\alpha\in k$ the valuation

$x_{\alpha}:k[T]\overset{T\mapsto\alpha}{\longrightarrow}k\overset{|\cdot|}{\longrightarrow}\mathds{R}_{\geq
0}$

extends to a continuous valuation on $k\langle\varpi^{n}T\rangle$ precisely if

$|\varpi^{n}T(x_{\alpha})|\leq 1\iff|\alpha|\leq\varpi^{-n}.$

For $m\geq n$ we have natural homomorphisms

(4)
$(k\langle\varpi^{m}T\rangle,k^{\circ}\langle\varpi^{m}T\rangle)\longrightarrow(k\langle\varpi^{n}T\rangle,k^{\circ}\langle\varpi^{n}T\rangle)$

mapping $\varpi^{m}T$ to $\varpi^{m-n}(\varpi^{n}T)$. The corresponding
morphism of discs

$\mathds{D}_{k}(0,|\varpi^{-n}|)\to\mathds{D}_{k}(0,|\varpi^{-m}|)$

is precisely the inclusion of the disc of radius $|\varpi^{-n}|$ into the disc
of radius $|\varpi^{-m}|$. In fact, $\mathds{D}_{k}(0,|\varpi^{-n}|)$ can be
identified with the rational subset

$\\{x\in\mathds{D}_{k}(0,|\varpi^{-m}|)\mid|T(x)|\leq\varpi^{-n}\\}.$

So we have an ascending sequence of discs whose radii become arbitrarily
large. Since the inclusions are open immersions, we can glue the discs to
obtain the _affine line_ :

${\mathds{A}}_{k}^{1}=\bigcup_{n}\mathds{D}_{k}(0,|\varpi^{-n}|).$

As an adic space the affine line is not affinoid. This can be seen by noting
that the discs $\mathds{D}_{k}(0,|\varpi^{-n}|)$ cover ${\mathds{A}}_{k}^{1}$
but this covering does not have a finite subcovering. However, all affinoid
adic spaces are quasi-compact.

Let us compute the global sections of the structure sheaf of
${\mathds{A}}_{k}^{1}$. The sheaf property gives us an exact sequence

${\mathscr{O}}_{{\mathds{A}}_{k}^{1}}({\mathds{A}}_{k}^{1})\longrightarrow\prod_{m\in\mathds{N}}{\mathscr{O}}_{{\mathds{A}}_{k}^{1}}(\mathds{D}_{k}(0,|\varpi^{-m}|))\longrightarrow\prod_{m,n}{\mathscr{O}}_{{\mathds{A}}_{k}^{1}}(\mathds{D}_{k}(0,|\varpi^{-n}|)\cap\mathds{D}_{k}(0,|\varpi^{-m}|)).$

For $m\geq n$ the disc $\mathds{D}_{k}(0,|\varpi^{-n}|)$ is contained in
$\mathds{D}_{k}(0,|\varpi^{-m}|)$. This simplifies the right hand term.
Throwing out redundant factors we obtain an exact sequence

${\mathscr{O}}_{{\mathds{A}}_{k}^{1}}({\mathds{A}}_{k}^{1})\longrightarrow\prod_{m\in\mathds{N}}{\mathscr{O}}_{{\mathds{A}}_{k}^{1}}(\mathds{D}_{k}(0,|\varpi^{-m}|))\overset{\varphi}{\longrightarrow}\prod_{m>n}{\mathscr{O}}_{{\mathds{A}}_{k}^{1}}(\mathds{D}_{k}(0,|\varpi^{-n}|)),$

where $\varphi$ maps $(\alpha_{m})_{m\in\mathds{N}}$ to

$(\alpha_{m}|_{\mathds{D}_{k}(0,|\varpi^{-n}|)}-\alpha_{n})_{m\geq n}$

The kernel of $\varphi$ identifies with the projective limit

$\lim_{m\in\mathds{N}}{\mathscr{O}}_{A_{k}^{1}}(\mathds{D}_{k}(0,|\varpi^{-m}|))$

with transition maps given by (4). The rings $k\langle\varpi^{m}T\rangle$ are
all subrings of the ring of formal power series $k\llbracket T\rrbracket$.
Hence, the transition maps are injective and the above limit can be viewed
just as an intersection taking place in $k\llbracket T\rrbracket$.

Remember that

$\displaystyle k\langle\varpi^{m}T\rangle$
$\displaystyle=\left\\{\sum_{i=0}^{\infty}a_{i}\varpi^{im}T^{i}\in k\llbracket
T\rrbracket\mid a_{i}\to 0\right\\}$
$\displaystyle=\left\\{\sum_{i=0}^{\infty}b_{i}T^{i}\in k\llbracket
T\rrbracket\mid b_{i}\varpi^{-im}\to 0\right\\}$

are the power series converging when inserting $x\in k$ with
$|x|\leq|\varpi^{-m}|$. The intersection of these for all $m\in\mathds{N}$ are
the power series that converge everywhere. They are also called _entire power
series_ and denoted by $k\\{\\{T\\}\\}$. Now what do they look like? Of course
the polynomial ring is contained in $k\\{\\{T\\}\\}$. To construct an infinite
series in $k\\{\\{T\\}\\}$, we just need to come up with coefficients that
decrease very rapidly, faster than any polynomial. An example would be

$\sum_{i=0}^{\infty}\varpi^{i^{2}}T^{i}.$

Indeed, for any $m\in\mathds{N}$

$\varpi^{i^{2}}\varpi^{-im}=\varpi^{i^{2}-im}$

converges to zero as $i$ goes to infinity since $\varpi$ is topologically
nilpotent and $i^{2}-im\to\infty$.

######  Example 4.4 (The open unit disc).

The _open unit disc_ $\overset{\circ}{\mathds{D}}_{k}(0,1)$ over a
nonarchimedean field $k$ is defined as the union of all closed discs with
smaller radius taken inside $\mathds{D}_{k}(0,1)$ (or ${\mathds{A}}_{k}^{1}$).
If $\varpi\in k$ is a pseudouniformizer, we can write it as the union

$\overset{\circ}{\mathds{D}}_{k}(0,1)=\bigcup_{m\in\mathds{N}}\mathds{D}_{k}(0,|\varpi^{1/m}|)$

A priori $\mathds{D}_{k}(0,|\varpi^{1/m}|)$ is only well defined if $\varpi$
has an $m$-th root but we can make sense of it by defining it as the rational
subset

$\mathds{D}_{k}(0,|\varpi^{1/m}|):=\\{x\in\mathds{D}_{k}(0,1)\mid|T^{m}(x)|\leq|\varpi|\\}$

of $\mathds{D}_{k}(0,1)=\operatorname{\mathrm{Spa}}(k\langle T\rangle)$. We
have seen in Example 2.10 that $\overset{\circ}{\mathds{D}}_{k}(0,1)$ is _not_
equal to the subset

$\\{x\in\mathds{D}_{k}(0,1)\mid|T(x)|<1\\}$

of $\mathds{D}_{k}(0,1)$. The above subset cointains one point more than
$\overset{\circ}{\mathds{D}}_{k}(0,1)$, namely the point $x_{0,1-}$ described
in Example 2.10.

Let us compute the global sections of the structure sheaf of
$\overset{\circ}{\mathds{D}}:=\overset{\circ}{\mathds{D}}_{k}(0,1)$. We start
by determining the global sections of $\mathds{D}_{k}(0,|\varpi^{1/m}|)$. By
definition
${\mathscr{O}}_{\overset{\circ}{\mathds{D}}}(\mathds{D}_{k}(0,|\varpi^{1/m}|))$
is the completion of $k[T]$ (or $k\langle T\rangle$) with respect to the
topology defined by the pair of definition

$(k^{\circ}[T,T^{m}/\varpi],(\varpi)).$

The completion can be taken inside the power series ring $k\llbracket
T\rrbracket$. To justify this first note that

${\mathscr{O}}_{\overset{\circ}{\mathds{D}}}(\mathds{D}_{k}(0,|\varpi^{1/m}|))=k\langle
T,S\rangle/(\varpi S-T^{m}).$

We have a natural inclusion

$k\langle T,S\rangle\longrightarrow k\llbracket T,S\rrbracket$

In order to deduce an inclusion

$k\langle T,S\rangle/(\varpi S-T^{m})\longrightarrow k\llbracket
T,S\rrbracket/(\varpi S-T^{m})=k\llbracket T\rrbracket$

we need to argue that

$\mathfrak{p}:=(\varpi S-T^{m})k\llbracket T\rrbracket\cap k\langle
T,S\rangle$

equals the ideal of $k\langle T,S\rangle$ generated by $\varpi S-T^{m}$.
First note that the polynomial $\varpi S-T^{m}$ is irreducible in $k\langle
T,S\rangle$, which can easily be checked explicitly. Therefore it generates a
prime ideal since $k\langle T,S\rangle$ is a unique factorization domain (see
[BGR84], § 5.2.6, Theorem 1), and it is clearly contained in $\mathfrak{p}$.
The equality follows if we show that $\mathfrak{p}$ is not a maximal ideal,
because $k\langle T,S\rangle$ has dimension $2$. But $k\langle
T,S\rangle/\mathfrak{p}$ is not a field since it embeds into $k\llbracket
T,S\rrbracket/(\varpi S-T^{m})=k\llbracket T\rrbracket$ and the image contains
$T$, which is not a unit.

Having identified the two ideals we get the desired inclusion

${\mathscr{O}}_{\overset{\circ}{\mathds{D}}}(\mathds{D}_{k}(0,|\varpi^{1/m}|))=k\langle
T,S\rangle/(\varpi S-T^{m})\longrightarrow k\llbracket T,S\rrbracket/(\varpi
S-T^{m})=k\llbracket T\rrbracket.$

It is now not hard anymore to identify the image of this embedding:

${\mathscr{O}}_{\overset{\circ}{\mathds{D}}}(\mathds{D}_{k}(0,|\varpi^{1/m}|))=\left\\{\sum_{i=0}^{\infty}a_{i}T^{i}\mid
a_{i}^{m}\varpi^{i}\to
0\right\\}=\left\\{\sum_{i=0}^{\infty}a_{i}T^{i}\mid\lvert
a_{i}\rvert\cdot\lvert\varpi\rvert^{i/m}\to 0\right\\}.$

Taking the intersection of them for varying $m\in\mathds{N}$ yields the global
sections of the open unit disc. They can be identified as the subring of the
formal power series ring $k\llbracket T\rrbracket$ consisting of all power
series that converge when inserting an element of $\bar{k}$ (the algebraic
closure of $k$) of norm less than $1$.

######  Example 4.5 (The projective line).

We want to explain how to glue two closed unit discs to obtain the projective
line. Let us start with a nonarchimedean field $k$ with pseudouniformizer
$\varpi$ and two discs

$\mathds{D}_{1}:=\operatorname{\mathrm{Spa}}(k\langle
T\rangle,k^{\circ}\langle
T\rangle),\qquad\mathds{D}_{2}:=\operatorname{\mathrm{Spa}}(k\langle
S\rangle,k^{\circ}\langle S\rangle).$

We want to glue these two spaces along $\\{|T|=1\\}$ and $\\{|S|=1\\}$ such
that $T$ identifies with $S^{-1}$. Following Lemma 4.2 this is done by
specifying a glueing isomorphism

$\mathds{S}_{1}:=\\{x\in\mathds{D}_{1}\mid|T(x)|=1\\}\overset{\sim}{\longrightarrow}\mathds{S}_{2}:=\\{x\in\mathds{D}_{2}\mid|S(x)|=1\\}.$

This can be done on the level of coordinate rings by writing down an
isomorphism of the corresponding Huber pairs:

$\displaystyle(k\langle S,S^{-1}\rangle,k^{\circ}\langle S,S^{-1}\rangle)$
$\displaystyle\longrightarrow(k\langle T,T^{-1}\rangle,k^{\circ}\langle
T,T^{-1}\rangle)$ $\displaystyle S$ $\displaystyle\longmapsto T^{-1}.$

Since there are only two spaces to be glued, there is no cocycle condition to
check. We call the resulting space the _projective line_ and denote it by
${\mathds{P}}_{k}^{1}$.

Let us compute the global sections of the structure sheaf. They are defined by
the exact sequence

$\displaystyle{\mathscr{O}}_{{\mathds{P}}_{k}^{1}}({\mathds{P}}_{k}^{1})\longrightarrow
k\langle T\rangle\oplus k\langle S\rangle$ $\displaystyle\longrightarrow
k\langle T,T^{-1}\rangle,$ $\displaystyle\left(\sum a_{i}T^{i},\sum
b_{i}S^{i}\right)$ $\displaystyle\longmapsto\sum a_{i}T^{i}-\sum b_{i}T^{-i}$

coming from the sheaf condition. If

$\sum_{i=0}^{\infty}a_{i}T^{i}-\sum_{i=0}^{\infty}b_{i}T^{-i}=0,$

all coefficients $a_{i}$ and $b_{i}$ for $i>0$ have to vanish and we must have
$a_{0}=b_{0}$. We conclude that

${\mathscr{O}}_{{\mathds{P}}_{k}^{1}}({\mathds{P}}_{k}^{1})=k,$

just as for the scheme theoretic projective line. Of course we still need to
justify the name _projective line_ for the space we have constructed. This
will become more transparent in Section 10, where we talk about
analytification of algebraic varieties.

Finally we want to explain how to embed the affine line into the projective
line. For this we need to construct compatible empbeddings

$\mathds{D}_{k}(0,|\varpi^{-m}|)\longrightarrow{\mathds{P}}_{k}^{1}.$

For $m=0$ we can just take the natural isomorphism

$\mathds{D}_{k}(0,1)\overset{\sim}{\longrightarrow}\mathds{D}_{1}.$

For $m>0$ we cover $\mathds{D}_{k}(0,|\varpi^{-m}|)$ by two rational subsets

$\mathds{D}_{k}(0,|\varpi^{-m}|)=\mathds{D}_{k}(0,1)\cup\\{x\in\mathds{D}_{k}(0,|\varpi^{-m}|)\mid|T(x)|\geq
1\\}.$

We map $\mathds{D}_{k}(0,1)$ isomorphically to $\mathds{D}_{1}$ via the
natural map and define

$\\{x\in\mathds{D}_{k}(0,|\varpi^{-m}|)\mid|T(x)|\geq
1\\}\longrightarrow\mathds{D}_{2}$

by the homomorphism of Huber pairs

$\displaystyle(k\langle S\rangle,k^{\circ}\langle S\rangle)$
$\displaystyle\longrightarrow(k\langle\varpi^{m}T,T^{-1}\rangle,k^{\circ}\langle\varpi^{m}T,T^{-1}\rangle)$
$\displaystyle S$ $\displaystyle\longmapsto T^{-1}.$

We leave it to the reader to check that these maps are compatible and glue to
an open immersion

$\iota:{\mathds{A}}_{k}^{1}\hookrightarrow{\mathds{P}}_{k}^{1}.$

It is clear that $\mathds{D}_{1}$ lies in the image of $\iota$ being the image
of $\mathds{D}_{k}(0,1)\subseteq{\mathds{A}}_{k}^{1}$. Moreover the image of
$\mathds{D}_{k}(0,|\varpi^{-m}|)$ intersected with $\mathds{D}_{2}$ is the
subset

$\\{x\in\mathds{D}_{2}\mid|S(x)|\geq|\varpi^{m}|\\}$

Hence, the image of $\iota$ intersected with $\mathds{D}_{2}$ consists of all
points $x\in\mathds{D}_{2}$ such that $|S(x)|>0$ (remember that $\varpi$ is
topologically nilpotent). If $|S(x)|=0$, then $S$ lies in the support of the
valuation $x$, i.e. $x$ comes from a valuation of

$k\langle S\rangle/(S)=k$

Since $x$ needs to take values less than or equal to $1$ on $k^{\circ}$, there
is only one such continuous valuation, namely the valuation of $k$ we started
with. This single point $x$ that does not lie in the image of $\iota$ is a
classical point of $\mathds{D}_{2}$. It is the point $0\in\mathds{D}_{2}$ to
be precise. Since $\mathds{D}_{1}$ is glued to $\mathds{D}_{2}$ by mapping
$T\to S^{-1}$, this corresponds to $T$ being equal to $\infty$ (loosely
speaking). This is the reason why we call this point $\infty$.

##  5\. Local Huber pairs

In the theory of schemes taking the intersection of all open neighborhoods
corresponds to forming the localization at the corresponding prime ideal at
the level of rings. More precisely, for a point $x$ of a scheme $X$, we can
describe the localization $X_{x}$ of $X$ at $x$ (which is defined as the
intersection of all open neighborhoods of $x$) in the following way. Choose
any open affine neighborhood $\operatorname{\mathrm{Spec}}A$ of $x$. Then $x$
corresponds to a prime ideal $\mathfrak{p}\subset A$ and
$X_{x}=\operatorname{\mathrm{Spec}}A_{\mathfrak{p}}$. So in particular, local
schemes are spectra of local rings.

In this section we want to develop a similar correspondence for adic spaces.
Since the material is not thoroughly treated in the literature, we spell out
all the proofs and technical details. Things are a bit more subtle for adic
spaces because there are two rings involved and we need to take into account
the topology of the rings. It is clear what we mean by the localization
$X_{x}$ of an adic space $X$ at a point $x$ just as a topological space: the
intersection of all open neighborhoods of $x$. We will see that we can endow
$X_{x}$ with the structure of an affinoid adic space such that it is the limit
of all open neighborhoods of $x$ in an appropriate sense. The corresponding
Huber pair will be a _local Huber pair_ , so let us start by specifying what
we mean by this.

If $A$ is a ring with a valuation $v$, we can endow $A$ with the valuation
topology, which is the coarsest topology such that the subsets

$\\{a\in A\mid v(a)<\gamma\\}$

are open for varying $\gamma\in\Gamma_{v}$.

######  Definition 5.1.

A Huber pair $(A,A^{+})$ is called local222In [Hüb21], Definition 6.1, appears
a different definition of local Huber pair. The boundedness of
$\mathfrak{m}^{+}$ in [Hüb21] leads to $A^{+}$ being a ring of definition,
which might be too restrictive. Moreover, the topology in [Hüb21] might be
coarser than the valuation topology leading to $|\cdot|$ not being continuous.
But then $\operatorname{\mathrm{Spa}}(A,A^{+})$ is empty, a case we want to
exclude. if the following conditions are satisfied:

  * •

$A$ and $A^{+}$ are local,

  * •

$A^{+}$ is the preimage of a valuation ring $k_{A}^{+}$ of the residue field
$k_{A}$ of $A$,

  * •

the valuation $v_{A}$ of $A$ defined by $k_{A}^{+}$ is continuous.

A homomorphism $(A,A^{+})\to(B,B^{+})$ of local Huber pairs is local if $A\to
B$ and $A^{+}\to B^{+}$ are local ring homomorphisms.

Actually the condition on $A^{+}$ to be local follows from the second
condition, that $A^{+}$ be the preimage of $k_{A}^{+}$. We just included it to
be able to speak of local homomorphisms $A^{+}\to B^{+}$.

Let us take a look at the topological conditions in the definition. The
requirement that $v_{A}$ be continuous ensures that
$v_{A}\in\operatorname{\mathrm{Spa}}(A,A^{+})$, otherwise the adic spectrum of
$(A,A^{+})$ would in fact be empty. We can rephrase this condition by saying
that the valuation topology defined by $v_{A}$ needs to be coarser than the
given topology on $A$.

For what follows we fix a local Huber pair $(A,A^{+})$. It is a rather
peculiar but very important property of local Huber pairs that the maximal
ideal $\mathfrak{m}$ of $A$ is contained in $A^{+}$ and is indeed a prime
ideal of $A^{+}$. This just follows from the fact that $A^{+}$ is the preimage
in $A$ of a subring of $k_{A}=A/\mathfrak{m}$. We also want to point out that

$k_{A}^{+}=A^{+}/\mathfrak{m}.$

The valuation ring $k_{A}^{+}$ defines a valuation $|\cdot|$ on $A$ such that

$A^{+}=\\{a\in A\mid|a|\leq 1\\}.$

######  Exercise 5.2.

If $(A,A^{+})$ is local, then $A$ is the localization of $A^{+}$ at
$\mathfrak{m}$.

The prime spectrum of $A^{+}$ is of a very special form. Since $A$ is a
localization of $A^{+}$, we already know that

$\operatorname{\mathrm{Spec}}A\hookrightarrow\operatorname{\mathrm{Spec}}A^{+}.$

In order to understand the prime ideals of $A^{+}$ that do not come from $A$,
the following lemma is crucial.

######  Lemma 5.3.

Let $(A,A^{+})$ be a local Huber pair and $\mathfrak{a}\subseteq A^{+}$ an
ideal. Then either $\mathfrak{a}\subseteq\mathfrak{m}$ or
$\mathfrak{m}\subseteq\mathfrak{a}$. If $\mathfrak{m}\subsetneq\mathfrak{a}$
and in addition $\mathfrak{a}$ is finitely generated, then $\mathfrak{a}$ is
principal.

###### Proof.

Suppose that $\mathfrak{a}\nsubseteq\mathfrak{m}$. We want to show
$\mathfrak{m}\subseteq\mathfrak{a}$, so let us take $m\in\mathfrak{m}$ and try
to show that $m\in\mathfrak{a}$. By assumption there is $a\in\mathfrak{a}$
which is not contained in $\mathfrak{m}$. Considered as an element of $A$, $a$
is invertible and inside $A$ we can form the product $a^{-1}m$. This is
contained in the maximal ideal $\mathfrak{m}$, which in turn is contained in
$A^{+}$. We obtain

$m=a\cdot(a^{-1}m)\in\mathfrak{a}.$

Thus we have $\mathfrak{m}\subsetneq\mathfrak{a}$.

Now assume in addition that $\mathfrak{a}$ is finitely generated, by
$a_{1},\ldots,a_{n}$, say. We consider the valuation $|\cdot|$ on $A$ induced
by $k_{A}^{+}$. Choose $i$ such that $|a_{i}|$ is maximal. Since
$\mathfrak{m}\subsetneq\mathfrak{a}$, the valuation $|a_{i}|$ must be greater
than zero. In particular, $a_{i}$ is not contained in $\mathfrak{m}$ and thus
invertible in $A$. The elements $a_{j}/a_{i}\in A$, $j=1,\ldots,n$ all have
valuation $\leq 1$ by construction, i.e. they lie in $A^{+}$, so that
$a_{j}\in a_{i}A^{+}$ for all $j$ and therefore $\mathfrak{a}$ is principal. ∎

This lemma tells us that the spectrum of $A^{+}$ consists of the spectrum of
$A$ plus additional prime ideals strictly containing $\mathfrak{m}$. These
prime ideals are totally ordered by inclusion and identify with the spectrum
of $k_{A}^{+}$ (minus the zero ideal).

The topology of a Huber pair $(A,A^{+})$ is particularly nice if $(A,A^{+})$
is uniform. Then $A^{+}$ is bounded and thus a ring of definition. Hence the
results of 5.3 apply to any ideal of definition $I\subseteq A^{+}$.

In this setting we use the following notation. If $I$ is contained in
$\mathfrak{m}$, we speak of the _formal case_ (formal for formal scheme). If
in addition, $I=(0)$, we are in the _discrete case_. Otherwise, if
$I\nsubseteq\mathfrak{m}$ (i.e. $\mathfrak{m}\subsetneq I$) we refer to this
situation as the _analytic case_.

We want to convince ourselves that in the analytic case the resulting topology
of $A$ coincides with the valuation topology. Let us consider a uniform
analytic local Huber pair $(A,A^{+})$. In particular, the corresponding
valuation $v_{A}$ is non-trivial or in other words $A^{+}\neq A$.

######  Exercise 5.4.

Let $I\subseteq A^{+}$ be an ideal of definition. Then $I$ is principal and
generated by a topologically nilpotent element $\varpi$.

Let $A_{\mathrm{val}}$ be the abstract ring $A$ endowed with the valuation
topology. Since $v_{A}$ is continuous (in the topology of $A$), the identity
on $A$ induces a continuous ring homomorphism

$\psi:A\longrightarrow A_{\mathrm{val}}.$

The topologically nilpotent element $\varpi$ from 5.4 maps to a topologically
nilpotent element in $A_{\mathrm{val}}$. Put differently, $\varpi$ is also
topologically nilpotent with respect to the valuation topology. But this
precisely means that the valuation $v_{A}$ is microbial.

Now we can show that $\psi$ is in fact a homeomorphism. To achieve this we
need to convince ourselves that the ideals $(\varpi^{n})$ of $A^{+}$ are open
in the valuation topology of $A$. But this follows from the identification

$(\varpi^{n})=\\{a\in A\mid v_{A}(a)\leq v_{A}(\varpi^{n})\\}$

and Exercise 2.2. We summarize our results in the following proposition.

######  Proposition 5.5.

Let $(A,A^{+})$ be a uniform analytic local Huber pair. Then $A$ carries the
valuation topology.

######  Corollary 5.6.

Let $(A,A^{+})$ be a uniform local Huber pair. Then $(k_{A},k_{A}^{+})$ (with
the induced topology) is an affinoid field.

######  Lemma 5.7.

The completion $(\hat{A},\hat{A}^{+})$ of a uniform local Huber pair
$(A,A^{+})$ is local. In case $(A,A^{+})$ is analytic, the completion
$(\hat{A},\hat{A}^{+})$ coincides with the completion of $(k_{A},k_{A}^{+})$
with respect to the valuation topology.

###### Proof.

If $(A,A^{+})$ is formal, we first convince ourselves that also $(A,IA)$ is a
pair of definition. The crucial point is that $I\subseteq\mathfrak{m}$, whence

$IA\subseteq\mathfrak{m}\subseteq A^{+}.$

This implies

$I^{2}\subseteq I^{2}A\subseteq I,$

or in other words, the topology defined by $I^{n}$ coincides with the topology
defined by $(IA)^{n}$.

Now the assertion of the lemma (in the formal case) essentially follows from
the fact that the $IA$-adic completion $\hat{A}$ of the local ring $A$ is
local with the same residue field. The closure of the maximal ideal
$\mathfrak{m}$ in $\hat{A}$ is the maximal ideal $\mathfrak{m}_{\hat{A}}$ of
$\hat{A}$. Since $\mathfrak{m}_{A}\subseteq A^{+}$, this implies that
$\hat{A}^{+}$ contains $\mathfrak{m}_{\hat{A}}$. Moreover,

$\hat{A}^{+}/\mathfrak{m}_{\hat{A}}\supseteq
A^{+}/\mathfrak{m}_{A}=k_{A}^{+},$

hence $\hat{A}^{+}$ contains the preimage of $k_{A}^{+}$ in $\hat{A}$. But
this preimage is closed and contains $A^{+}$, from which we conclude that it
coincides with $\hat{A}^{+}$.

Let us now consider the analytic case. By Exercise 5.4 we can take a pair of
definition of the form $(A^{+},(\varpi))$ for a topologically nilpotent unit
$\varpi$ of $A$. Taking into account that $\varpi$ is not contained in
$\mathfrak{m}$, Lemma 5.3 tells us that all powers $(\varpi^{n})$ for
$n\in\mathds{N}$ contain $\mathfrak{m}$. Therefore, the completion of
$(A,A^{+})$ factors through

$(A/\mathfrak{m},A^{+}/\mathfrak{m})=(k_{A},k_{A}^{+}).$

The completion of $(k_{A},k_{A}^{+})$ we need is the one with respect to the
topology defined by the pair of definition $(k_{A}^{+},(\bar{\varpi}))$, where
$\bar{\varpi}$ is the image of $\varpi$ in $k_{A}^{+}$. But we have seen in
Exercise 2.3 that this topology coincides with the valuation topology. For the
valuation topology we know that the completion $\hat{k}_{A}$ of $k_{A}$ is
again a valued field with valuation ring $\hat{k}_{A}^{+}$, which implies all
claims. ∎

The denomination _local_ is justified for a local Huber pair by the following
result.

######  Proposition 5.8.

A complete Huber pair $(A,A^{+})$ with ring of definition $A^{+}$ is local if
and only if $\operatorname{\mathrm{Spa}}(A,A^{+})$ is local as a topological
space (it has a unique closed point and all other points specialize to it).
Furthermore, an adic space is local (as a topological space) if and only if it
is the spectrum of a local Huber pair.

###### Proof.

Let $(A,A^{+})$ be a local Huber pair. We claim that
$\operatorname{\mathrm{Spa}}(A,A^{+})$ is local with closed point $x$ given by
the valuation ring $k_{A}^{+}\subseteq k_{A}$. Since
$\operatorname{\mathrm{Spa}}(A,A^{+})$ is spectral, it suffices to show that
any open neighborhood of $x$ equals $\operatorname{\mathrm{Spa}}(A,A^{+})$. In
fact it is enough to check this for a rational open neighborhood

$U:=R\left(\frac{f_{1},\ldots,f_{n}}{g}\right)=\\{y\in\operatorname{\mathrm{Spa}}(A,A^{+})\mid|f_{i}(y)|\leq|g(y)|\neq
0~{}\forall i=1,\ldots,n\\}$

of $x$ defined by elements $f_{1},\ldots,f_{n},g\in A$ generating the unit
ideal. Since $x$ is supposed to be contained in $U$, it satisfies

$|f_{i}(x)|\leq|g(x)|\neq 0$

for $i=1,\ldots,n$. The condition $|g(x)|\neq 0$ translates to
$g\notin\mathfrak{m}_{A}$, so $g$ is invertible in $A$. This implies that
$v(g)\neq 0$ for _any_ valuation of $A$. The inequalities can now be
rearranged to

$\left|\frac{f_{i}}{g}(x)\right|\leq 1,$

or in other words $f_{i}/g\in A^{+}$. But any point
$y\in\operatorname{\mathrm{Spa}}(A,A^{+})$ satisfies $|a(y)|\leq 1$ for all
$a\in A^{+}$, in particular for $a=f_{i}/g$. We conclude that all points of
$\operatorname{\mathrm{Spa}}(A,A^{+})$ are contained in $U$.

Conversely suppose that $(A,A^{+})$ is a complete Huber pair such that
$\operatorname{\mathrm{Spa}}(A,A^{+})$ is local with closed point $x$. We
first show that $A$ is local with maximal ideal
$\mathfrak{m}:=\operatorname{\mathrm{supp}}x$. Suppose that
$\mathfrak{m}^{\prime}\subseteq A$ is a maximal ideal different from
$\mathfrak{m}$. Since $\mathfrak{m}^{\prime}$ is a maximal ideal of a complete
Huber ring it is possible to construct a continuous valuation $x^{\prime}$
with support $\mathfrak{m}^{\prime}$ lying in
$\operatorname{\mathrm{Spa}}(A,A^{+})$ (see [Hub94], Lemma 1.4). But
$x^{\prime}$ does not specialize to $x$. This can be seen by choosing an
element $g$ of $\mathfrak{m}^{\prime}$ that is not contained in
$\mathfrak{m}$. Then the open subset

$\\{x\in\operatorname{\mathrm{Spa}}(A,A^{+})\mid|g(x)|\neq 0\\}$

of $\operatorname{\mathrm{Spa}}(A,A^{+})$ contains $x$ but not $x^{\prime}$.
To conclude, $x^{\prime}$ does not specialize to $x$, which gives a
contradiction to our assumption that $A$ is not local. Thus we now know that
$A$ is local with maximal ideal $\mathfrak{m}$. We denote its residue field by
$k_{A}$.

In order to prove that $A^{+}$ is the preimage in $A$ of a valuation ring of
$k_{A}$, it suffices to check the following: Every $a\in A$ with $a\notin
A^{+}$ is a unit in $A$ and $1/a\in A^{+}$ (the reader can convince themselves
that this criterion holds or consult [KZ96], Theorem 2.5). For such an $a$ we
consider the rational subset

$R\left(\frac{a}{1}\right)=\\{y\in\operatorname{\mathrm{Spa}}(A,A^{+})\mid|a(y)|\leq
1\\}.$

Its complement is nonempty as

$a\notin A^{+}=\\{b\in A\mid|b(y)|\leq 1~{}\forall
y\in\operatorname{\mathrm{Spa}}(A,A^{+})\\}.$

Since $\operatorname{\mathrm{Spa}}(A,A^{+})$ is local with closed point $x$,
this implies that $x$ is contained in the complement, or in other words

$|a(x)|>1.$

First this implies $|a(x)|\neq 0$, whence $a\notin\mathfrak{m}$, so $a\in
A^{\times}$. Then we get

$\left|\frac{1}{a}(x)\right|<1,$

so

$x\in
R\left(\frac{1}{a}\right)=\\{y\in\operatorname{\mathrm{Spa}}(A,A^{+})\mid
1\leq|a(y)|\\}=\left\\{y\in\operatorname{\mathrm{Spa}}(A,A^{+})\mid\left|\frac{1}{a}(y)\right|\leq
1\right\\}.$

We conclude that this rational subset is all of
$\operatorname{\mathrm{Spa}}(A,A^{+})$ and thus $1/a\in A^{+}$.

The condition on the topology of $A$ follows automatically as $x$ is a
continuous valuation, which means that the valuation topology on $A$ defined
by $x$ is coarser than the given topology.

Finally any local adic space needs to be affinoid as adic spaces are locally
affinoid. Therefore, the second assertion of the proposition follows from the
first one. ∎

We now have a basic understanding of local Huber pairs. Our goal is to
interpret $X_{x}$ for a point $x$ of an adic space $X$ as the adic spectrum of
a local Huber pair $(A,A^{+})$. This local Huber pair $(A,A^{+})$ should
morally be given by $({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+})$.
Unfortunately, $({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+})$ is not uniform,
in general. While it is true that
$({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+})$ is local, we run into trouble
when it comes to the topology. The colimit topology on ${\mathscr{O}}_{X,x}$
might be finer than the valuation topology of the valuation corresponding to
$x$. But we don’t want to use the colimit topology because it is quite messy
and does not reflect the geometry of $X_{x}$.

We can solve this issue by resorting to the concept of limit of adic spaces
defined in [Hub96], Definition 2.4.2. Following Huber an adic space $X$
together with compatible maps

$\pi_{i}:X\longrightarrow X_{i}$

is called a limit of a projective system of adic spaces $(X_{i})_{i\in I}$ if
the following two assertions hold.

  1. (i)

The morphisms $\pi$ induce an identification $X=\lim_{i}X_{i}$ as topological
spaces.

  2. (ii)

For every $x\in X$ there is an open neighborhood $U\subseteq X$ such that the
image of

$\operatorname*{colim}_{(i,V)}{\mathscr{O}}_{X_{i}}(V)\longrightarrow{\mathscr{O}}_{X}(U)$

is dense, where $(i,V)$ runs over all pairs with $i\in I$ and $V$ an open
subset of $X_{i}$ containing the image of $U$.

We then write

$X\sim\lim_{i\in I}X_{i}.$

Using “$\sim$” instead of “$=$” reflects the fact that this is not a
projective limit in the categorical sense. For that we would need the union in
part (ii) to _equal_ ${\mathscr{O}}_{X}(U)$. But it turns out that the
categorical limit is too restrictive in many situations and the above defined
concept of limit serves our purposes.

Our strategy now is to write down a nice local Huber pair $(A,A^{+})$, define

$X_{x}:=\operatorname{\mathrm{Spa}}(A,A^{+}),$

and show that

$X_{x}\sim\lim_{U}U,$

where the limit runs over all open neighborhoods $U$ of $x$.

We start with an affinoid open neighborhood
$\operatorname{\mathrm{Spa}}(B,B^{+})$ of $x$. Let $\mathfrak{p}\subseteq B$
denote the support of $x$. We define $A:=B_{\mathfrak{p}}$ and consider the
natural projection

$\pi:A=B_{\mathfrak{p}}\twoheadrightarrow k(\mathfrak{p})=k(x)$

The point $x$ comes with a valuation of $k(x)$ whose valuation ring we denote
by $k(x)^{+}$. Then we set

$A^{+}:=\\{a\in A\mid\pi(a)\in k(x)^{+}\\}.$

We still need to specify a topology on $A$. Let $(B_{0},J)$ be a pair of
definition for $B$ and assume $J\subseteq B^{+}$ (otherwise we replace $J$ by
a power). We then take

$I:=JA^{+}\subseteq A^{+}$

as an ideal of definition for $A$ (with ring of definition $A^{+}$). This is
the point where we possibly make the topology coarser than the colimit
topololgy, in which $A^{+}$ might possibly not be bounded.

With this topology $(A,A^{+})$ is a local Huber pair and

$(B,B^{+})\longrightarrow(A,A^{+})$

is continuous.

######  Exercise 5.9.

If $(B,B^{+})\to(B^{\prime},B^{\prime+})$ is a rational localization with
$x\in\operatorname{\mathrm{Spa}}(B^{\prime},B^{\prime+})$, we obtain a local
homomorphism $(A,A^{+})\to(A^{\prime},A^{\prime+})$ of the corresponding
localizations at $x$. This homomorphism induces an isomorphism on completions.

$(\hat{A},\hat{A}^{+})\to(\hat{A}^{\prime},\hat{A}^{\prime+}).$

The above exercise shows that the completion of $(A,A^{+})$ (and thus
$\operatorname{\mathrm{Spa}}(A,A^{+})$) is independent of the choice of the
affinoid neighborhood $\operatorname{\mathrm{Spa}}(B,B^{+})$ of $x$.

######  Proposition 5.10.

Let $x$ be a point of an adic space $X$ and $(A,A^{+})$ the local Huber pair
constructed above. Setting $X_{x}:=\operatorname{\mathrm{Spa}}(A,A^{+})$ we
have

$X_{x}\sim\lim_{U}U,$

where the limit runs over all open neighborhoods $U$ of $x$.

###### Proof.

Let us first verify that $X_{x}$ is a limit at the level of topological
spaces. We consider the natural morphisms

$\operatorname{\mathrm{Spa}}(A,A^{+})\overset{\iota}{\longrightarrow}\operatorname{\mathrm{Spa}}(B,B^{+})\longrightarrow
X,$

where $\operatorname{\mathrm{Spa}}(B,B^{+})$ is the affinoid open used in the
construction of $(A,A^{+})$. It is clear that $\iota$ is injective as a
valuation of $A=B_{\mathfrak{p}}$ is uniquely determined by its restriction to
$B$. Moreover, its image is contained in the set of all generalizations of $x$
as $\operatorname{\mathrm{Spa}}(A,A^{+})$ is local and its closed point maps
to $x$.

Now let us pick a point $y$ specializing to $x$ and show that it lies in the
image of $\operatorname{\mathrm{Spa}}(A,A^{+})$. We start by checking that
$\operatorname{\mathrm{supp}}y$ is contained in
$\mathfrak{p}=\operatorname{\mathrm{supp}}x$. For
$b\in\operatorname{\mathrm{supp}}y$ we consider the rational subset

$R\left(\frac{b}{b}\right)=\\{z\in\operatorname{\mathrm{Spa}}(B,B^{+})\mid|b(z)|\neq
0\\}.$

If $x$ were contained in it, then also $y$ would be contained but $|b(y)|=0$.
Therefore, $x$ is not contained, whence $|b(x)|=0$, i.e. $b\in\mathfrak{p}$.

By what we have just seen the valuation $y$ extends from $B$ to $A$. Moreover,
$y$ takes values less or equal to $1$ on $A^{+}$ as $x$, hence $y$, is
contained in the rational subsets

$R\left(\frac{a}{1}\right)=\\{z\in\operatorname{\mathrm{Spa}}(B,B^{+})\mid|a(z)|\leq
1\\}$

for all $a\in A^{+}$.

The critical point now is to check that $y$ is a continuous valuation with
respect to the topology of $A$. Let $(B_{0},I)$ be a pair of definition for
$B$. Then $(A^{+},IA^{+})$ is a pair of definition for $A$. By [Ber24],
Proposition 1.2.7.1, we need to check that $|b(y)|$ for $b\in I$ are cofinal
in $\Gamma_{y}$ and that $|ba(y)|<1$ for all $b\in I$ and $a\in A^{+}$. The
first condition holds because $y$ is a continuous valuation of $B$ and the
second condition because $|a(y)|\leq 1$ for all $a\in A^{+}$.

Finally we need to convince ourselves that $\iota$ induces a homeomorphism to
the set of all generalizations of $x$ (and not just a bijection). We consider
the image via $\iota$ of a rational open

$R\left(\frac{f_{1},\ldots,f_{n}}{g}\right)$

of $\operatorname{\mathrm{Spa}}(A,A^{+})$. Multiplying $f_{1},\ldots,f_{n}$
and $g$ by a unit of $A$ does not change the rational subset. Hence we can
clear denominators by multiplying these elements with some $b\in
B\setminus\mathfrak{p}$ to achieve that they are all contained in the image of
$B$. We pick preimages $f^{\prime}_{1},\ldots,f^{\prime}_{n}$ and $g^{\prime}$
in $B$. Then

$\iota(R\left(\frac{f_{1},\ldots,f_{n}}{g}\right))=R\left(\frac{f^{\prime}_{1},\ldots,f^{\prime}_{n}}{g^{\prime}}\right)\cap\\{y\in\operatorname{\mathrm{Spa}}(B,B^{+})\mid
y\rightsquigarrow x\\}.$

The set

$R\left(\frac{f^{\prime}_{1},\ldots,f^{\prime}_{n}}{g}\right)=\\{y\in\operatorname{\mathrm{Spa}}(B,B^{+})\mid|f^{\prime}_{i}(y)|\leq|g^{\prime}(y)|\neq
0\\}$

might not be rational because we cannot guarantee that
$f^{\prime}_{1},\ldots,f^{\prime}_{n}$ and $g$ generate the unit ideal in $B$.
Nonetheless it is still open (see Exercise 2.8). This finishes the proof that
$\iota$ is a homeomorphism onto the set of generalizations of $x$.

It remains to show that for any rational open
$V=\operatorname{\mathrm{Spa}}(C,C^{+})\subseteq\operatorname{\mathrm{Spa}}(A,A^{+})$
the image of

$\operatorname*{colim}_{\iota(V)\subseteq
U}{\mathscr{O}}_{X}(U)\longrightarrow(\hat{C},\hat{C}^{+})$

is dense. We may assume that $C$ is a localization $A_{f}$ for some $f\in A$.
We claim that in this case $C$ lies in the image of the colimit. Indeed, an
element $c$ of $C$ can be written as $a/f^{n}$ for $a\in A$ and
$n\in\mathds{N}$. In turn, $a$ can be written as $b/s$ for $b\in B$ and $s\in
B\setminus\mathfrak{p}$. This combines to

$c=\frac{b}{f^{n}s}.$

For all points $y\in V$ we have $|f^{n}s(y)|\neq 0$. Therefore,

$\iota(V)\subseteq\bigcup_{m\in\mathds{N}}R\left(\frac{E_{m}}{f^{n}s}\right),$

where $E_{m}$ is a finite set of generators of $I^{m}$. Since $\iota(V)$ is
quasicompact, it is contained already in a finite union. If we assume
$B_{0}\subseteq B^{+}$ (which we can always achieve by replacing $B_{0}$ by
$B_{0}\cap B^{+}$), the union is an ascending union and thus

$\iota(V)\subseteq R\left(\frac{E_{m}}{f^{n}s}\right)$

for some $m$. The corresponding map on global sections of the structure sheaf
is the completion of

$B\left[\frac{1}{f^{n}s}\right]\longrightarrow A_{f}=C,$

whose image obviously contains $c=b/(f^{n}s)$. ∎

######  Remark 5.11.

The Huber pair of local rings $({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+})$
is local but not necessarily uniform if it is equipped with the colimit
topology. However, $X_{x}$ is uniform by construction. Forcing
${\mathscr{O}}_{X,x}^{+}$ to be a ring of definition,
$({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+})$ becomes a uniform local Huber
pair and then

$X_{x}=\operatorname{\mathrm{Spa}}({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+}).$

In particular, the completion of
$({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+})$ coincides with the completion
of the Huber pair $(A,A^{+})$ constructed above.

In the following when we work with
$({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+})$ we will always assume that
${\mathscr{O}}_{X,x}$ carries the topology with pair of definition
$({\mathscr{O}}_{X,x}^{+},I{\mathscr{O}}_{X,x}^{+})$, where $I$ is an ideal of
definition coming from some affinoid open neighborhood of $x$. In particular,

$X_{x}=\operatorname{\mathrm{Spa}}({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+})$

and the global sections of the structure sheaf of $X_{x}$ are given by the
completion

$(\hat{{\mathscr{O}}}_{X,x},\hat{{\mathscr{O}}}_{X,x}^{+})$

##  6\. Analytic adic spaces

Analytic adic spaces are those one would say are closest to rigid analytic
varieties but without finiteness assumptions.

######  Definition 6.1.

  * •

Let $(A,A^{+})$ be a Huber pair. A point
$x\in\operatorname{\mathrm{Spa}}(A,A^{+})$ is _analytic_ if the prime ideal
$\operatorname{\mathrm{supp}}x$ of $A$ is _not_ open.

  * •

A point $x$ of an adic space $X$ is _analytic_ if it has an affinoid
neighborhood $\operatorname{\mathrm{Spa}}(A,A^{+})$ such that $x$ is analytic
as a point of $\operatorname{\mathrm{Spa}}(A,A^{+})$.

  * •

An adic space is _analytic_ if all its points are analytic.

Let us explain the intuition behind this definition. We take a point $x$ of an
adic space $X$ and consider the (uniformized) local Huber pair
$({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+})$. Then we have the following:

######  Exercise 6.2.

With the above notation $x$ is analytic if and only if
$({\mathscr{O}}_{X,x},{\mathscr{O}}_{X,x}^{+})$ is analytic or, in other
words, the residue field $k(x)$ carries the valuation topology. Moreover, the
points with support $\operatorname{\mathrm{supp}}x$ correspond precisely to
the continuous valuations of $k(x)$.

We learn from the exercise that only in the analytic case there is a
continuity condition involved for valuations of $k(x)$. Or in other words, for
a Huber pair $(A,A^{+})$ and an open prime ideal $\mathfrak{p}$, _any_
valuation of the residue field $k(\mathfrak{p})$ defines a continuous
valuation of $A$. In order to be an element of
$\operatorname{\mathrm{Spa}}(A,A^{+})$ it just needs to satisfy the condition
that it is less or equal to one on $A^{+}$.

Let us get back to the study of analytic points and look at some examples.

######  Example 6.3.

$\operatorname{\mathrm{Spa}}(\mathds{Z}_{p},\mathds{Z}_{p})$ has two points:
the $p$-adic valuation on $\mathds{Q}_{p}$ and the trivial valuation on
$\mathds{F}_{p}=\mathds{Z}_{p}/p\mathds{Z}_{p}$. The former point is analytic
and the latter point is not.

The analyticity of a point $x\in X$ does not depend on the choice of an
affinoid open neighborhood $\operatorname{\mathrm{Spa}}(A,A^{+})$ of $x$. This
is because for a rational subset
$\operatorname{\mathrm{Spa}}(B,B^{+})\subseteq\operatorname{\mathrm{Spa}}(A,A^{+})$,
the homomorphism $A\to B$ is adic, which is defined as follows.

######  Definition 6.4.

  * •

A homomorphism of Huber rings $\varphi:R\to S$ is _adic_ if $R$ and $S$ have
rings of definition $R_{0}$ and $S_{0}$ with $\varphi(R_{0})\subseteq S_{0}$
such that for an ideal of definition $I_{R}\subseteq R_{0}$ the ideal
$\varphi(I_{R})S_{0}$ is an ideal of definition for $S$.

  * •

A morphism of adic spaces is _adic_ if it is locally defined by an adic
morphism of Huber pairs.

In other words one could say that for an adic homomorphism $R\to S$, the
topology on $S$ is defined by the topology on $R$. In particular, an ideal
$J\subset S$ is open if and only if its preimage in $R$ is open. This settles
the issue brought up above that analyticity is independent of the choice of
affinoid open neighborhood.

We want to remark further that the critical point in the above definition is
not that we can find rings of definition with $\varphi(R_{0})\subseteq S_{0}$.
This is possible for any homomorphism of Huber rings. The crucial point is
that $\varphi(I_{R})S_{0}$ is an ideal of definition.

######  Proposition 6.5.

Let $x$ be a point of an adic space $X$. The following are equivalent.

  1. (1)

$x$ is analytic,

  2. (2)

$x$ has an affinoid open neighborhood $\operatorname{\mathrm{Spa}}(A,A^{+})$
such that $A$ is a Tate ring (i.e. has a topologically nilpotent unit).

###### Proof.

Suppose $x$ has an open affinoid neighborhood
$\operatorname{\mathrm{Spa}}(A,A^{+})$ such that
$\operatorname{\mathrm{supp}}x\subseteq A$ is not open. Let
$I=(a_{1},\ldots,a_{r})$ be an ideal of definition in a ring of definition
$A_{0}$ of $A$. We can order the generators $a_{i}$ in such a way that the
value $|a_{1}(x)|$ is the maximum of all $|a_{i}(x)|$. This maximum cannot be
zero as otherwise all $a_{i}$ would be contained in
$\operatorname{\mathrm{supp}}x$ and thus $\operatorname{\mathrm{supp}}x$ would
be open. Hence $x$ is contained in the rational subset

$R\left(\frac{a_{1},\ldots,a_{r}}{a_{1}}\right)=\\{y\in\operatorname{\mathrm{Spa}}(A,A^{+})\mid|a_{i}(y)|\leq|a_{1}(y)|\neq
0~{}\forall~{}i=1,\ldots,f\\}.$

The corresponding Huber pair is the completion of

$(A_{a_{1}},A^{+}[\frac{a_{2}}{a_{1}},\ldots,\frac{a_{r}}{a_{1}}]^{N}),$

where $(\cdot)^{N}$ refers to taking the integral closure in $A_{a_{1}}$.
Since $a_{1}\in I$, it is topologically nilpotent. Its image in $A_{a_{1}}$ is
also topologically nilpotent and moreover, it is a unit. Therefore $A_{a_{1}}$
is a Tate ring.

For the converse direction we just need to note that a Tate ring with
topolologically nilpotent unit $\varpi$ does not have open prime ideals as
every open ideal has to contain the unit $\varpi$. ∎

From this characterization we see that rigid analytic varieties (defined in
the next section) and perfectoid spaces (discussed in Ben Heuer’s lecture
([Heu23]) are analytic.

######  Lemma 6.6.

Let $f:X\to Y$ be a morphism of Huber pairs. Then $f$ takes non-analytic
points to non-analytic points.

###### Proof.

We may assume that $f$ is given by a homomorphism of Huber pairs
$\varphi:(A,A^{+})\to(B,B^{+})$. A non-analytic point $x$ has open support
$\operatorname{\mathrm{supp}}x$. Its image in $Y$ has support
$\varphi^{-1}(\operatorname{\mathrm{supp}}x)$, which is again open. ∎

It can happen, however, that a morphism takes analytic points to non-analytic
points. For instance, take a Tate ring $R$ with ring of integral elements
$R^{+}$. Let $R_{\mathit{disc}}$ be the same ring $R$ but endowed with the
discrete topology and set $R_{\mathit{disc}}^{+}=R^{+}$. Then the identity
induces a homomorphism of Huber pairs
$(R_{\mathit{disc}},R_{\mathit{disc}}^{+})\to(R,R^{+})$. Consider the
corresponding morphism of adic spaces

$\operatorname{\mathrm{Spa}}(R,R)\to\operatorname{\mathrm{Spa}}(R_{\mathit{disc}},R_{\mathit{disc}}^{+}).$

All points of $\operatorname{\mathrm{Spa}}(R,R^{+})$ are analytic and all
points of
$\operatorname{\mathrm{Spa}}(R_{\mathit{disc}},R_{\mathit{disc}}^{+})$ are
non-analytic. Something like this cannot happen for adic morphisms:

######  Lemma 6.7.

Every adic morphism $f:X\to Y$ takes analytic points to analytic points.

###### Proof.

This follows from the observation that for an adic homomorphism of Huber pairs
$\varphi:(R,R^{+})\to(S,S^{+})$ an ideal $J\subseteq S$ is open if and only if
$\varphi^{-1}(J)$ is open. ∎

######  Lemma 6.8.

Every morphism of analytic adic spaces is adic.

###### Proof.

It suffices to consider a morphism induced by a homomorphism of Tate rings
$\varphi:(R,R^{+})\to(S,S^{+})$ (see Proposition 6.5). Let $R_{0}$ and $S_{0}$
be rings of definition with $\varphi(R_{0})\subseteq S_{0}$. The ideal
generated by any topologically nilpotent unit $\varpi$ of $R$ which is
contained in $R_{0}$ is an ideal of definition of $R_{0}$. But
$\varphi(\varpi)$ is a topologically nilpotent unit of $S$ that is contained
in $S_{0}$, so it generates an ideal of definition. ∎

We learn from Lemma 6.6 and Lemma 6.7 that adic morphisms take analytic points
to analytic points and non-analytic points to non-analytic points. In
particular, for an adic space $X$ over
$\operatorname{\mathrm{Spa}}(\mathds{Z}_{p},\mathds{Z}_{p})$ whose structure
morphism $X\to\operatorname{\mathrm{Spa}}(\mathds{Z}_{p},\mathds{Z}_{p})$ is
adic, the analytic points are those mapping to the $p$-adic valuation of
$\mathds{Q}_{p}$ and the non-analytic points those mapping to the trivial
valuation of $\mathds{F}_{p}$. However, there are relevant examples of adic
spaces over $\operatorname{\mathrm{Spa}}(\mathds{Z}_{p},\mathds{Z}_{p})$ with
non-adic structure morphism.

######  Example 6.9.

Consider the Huber ring $\mathds{Z}_{p}\llbracket T\rrbracket$ with pair of
definition $(\mathds{Z}_{p}\llbracket T\rrbracket,(p,T))$. We want to study
the adic space

${\mathscr{X}}:=\operatorname{\mathrm{Spa}}(\mathds{Z}_{p}\llbracket
T\rrbracket)=\operatorname{\mathrm{Spa}}(\mathds{Z}_{p}\llbracket
T\rrbracket,\mathds{Z}_{p}\llbracket T\rrbracket)$

over $\operatorname{\mathrm{Spa}}(\mathds{Z}_{p})$. Its structure morphism is
not adic as the ideal of definition for $\mathds{Z}_{p}\llbracket T\rrbracket$
has in addition to $p$ the generator $T$. Let us denote by $\eta$ the point of
$\operatorname{\mathrm{Spa}}(\mathds{Z}_{p})$ corresponding to the $p$-adic
valuation and by $s$ the point corresponding to the trivial valuation of
$\mathds{F}_{p}$. As the notation suggests, $\eta$ is the generic point and
$s$ the closed point of $\operatorname{\mathrm{Spa}}(\mathds{Z}_{p})$.
Moreover, as noted before, $\eta$ is analytic and $s$ is non-analytic.

From Lemma 6.6 we know that all points mapping to $\eta$ are supposed to be
analytic. Indeed, the supports of these points are contained in the generic
fiber of

$\operatorname{\mathrm{Spec}}\mathds{Z}_{p}\llbracket
T\rrbracket\longrightarrow\operatorname{\mathrm{Spec}}\mathds{Z}_{p},$

which is the set of all prime ideals _not_ containing $p$. But such a prime
ideal cannot be open as open prime ideals contain $(p,T)$.

Among the points mapping to $s$ there are many analytic points and only one
non-analytic point. Namely, the support of a non-analytic point has to contain
$(p,T)$. But $(p,T)$ is a maximal ideal with residue field $\mathds{F}_{p}$,
which has only the trivial valuation.

For an adic space $X$ we denote by $X_{a}$ the subset of analytic points and
by $X_{na}$ the set of non-analytic points. Then Proposition 6.5 immediately
implies the following:

######  Lemma 6.10.

The set of analytic points $X_{a}$ of an adic space $X$ is open.

######  Exercise 6.11.

Let $X=\operatorname{\mathrm{Spa}}(A,A^{+})$ be an affinoid adic space, and
fix a pair of definition $(A_{0},I)$ as well as generators
$s_{1},\ldots,s_{n}$ for $I$. Prove that

$X_{a}=\bigcup_{i=1}^{n}R\left(\frac{s_{1},\ldots,s_{n}}{s_{i}}\right).$

In particular, for any quasicompact adic space $X$, the subset $X_{a}$ is even
quasicompact open.

######  Example 6.12.

Let us illustrate this by picking up Example 6.9. The analytic locus of
${\mathscr{X}}=\operatorname{\mathrm{Spa}}(\mathds{Z}_{p}\llbracket
T\rrbracket)$ is given by the union of two affinoids:

${\mathscr{X}}_{a}=R\left(\frac{p}{T}\right)\cup R\left(\frac{T}{p}\right).$

The rational subset $R(T/p)$ is contained in the generic fiber (the preimage
of $\eta$) as all of its points $x\in R(T/p)$ satisfy $|p(x)|\neq 0$. However,
it does not cover all of the generic fiber as some points of the latter might
satisfy $|p(x)|\neq 0$ but $|p(x)|<|T(x)|$. The generic fiber is open, having
the following description

${\mathscr{X}}_{\eta}=\\{x\in X\mid|p(x)|\neq 0\\}.$

But it is not rational as the ideal generated by $p$ is not open. We can cover
it by rational open subsets in the following way:

${\mathscr{X}}_{\eta}=\bigcup_{n\geq 1}R\left(\frac{T^{n}}{p}\right).$

These sets are indeed rational as the ideals $(p,T^{n})$ are open. Moreover,
they cover ${\mathscr{X}}_{\eta}$ for the following reason. Since $T$ is
topologically nilpotent, every open neighborhood of $0$ in
$\mathds{Z}_{p}\llbracket T\rrbracket$ contains a power of $T$. This implies
that for a continuous valuation $v$ of $\mathds{Z}_{p}\llbracket T\rrbracket$
with $v(p)\neq 0$ (i.e., any point $v$ of ${\mathscr{X}}_{\eta}$), there is
$n\in\mathds{N}$ with

$v(T^{n})<v(p).$

Thus the above union of rationals equals ${\mathscr{X}}_{\eta}$.

At last we want to point out that we have come across the space
${\mathscr{X}}_{\eta}$ before. This might be more obvious when writing the
rational subsets in the covering of ${\mathscr{X}}_{\eta}$ as spectra of Huber
pairs:

$R\left(\frac{T^{n}}{p}\right)=\operatorname{\mathrm{Spa}}\left(\mathds{Z}_{p}\llbracket
T\rrbracket\left[\frac{1}{p}\right],\mathds{Z}_{p}\llbracket
T\rrbracket\left[\frac{T^{n}}{p}\right]\right).$

This might not yet look very familiar but it will once we have computed the
completion of $(\mathds{Z}_{p}\llbracket
T\rrbracket[1/p],\mathds{Z}_{p}\llbracket T\rrbracket[T^{n}/p])$. Note that a
pair of definition is given by

$\left(\mathds{Z}_{p}\llbracket
T\rrbracket\left[\frac{T^{n}}{p}\right],(p,T)\right).$

Observing that

$T^{n}=\frac{T^{n}}{p}\cdot p\in(p),$

we realize that actually also $(p)$ is an ideal of definition. The $p$-adic
completion of $\mathds{Z}_{p}\llbracket T\rrbracket[T^{n}/p]$ can be computed
as follows. Modulo $p^{i}$ every element of $\mathds{Z}_{p}\llbracket
T\rrbracket[T^{n}/p]$ has a representative which is a polynomial (of degree
less than $in$). Therefore, the completion of $\mathds{Z}_{p}\llbracket
T\rrbracket[T^{n}/p]$ is the same as the $p$-adic completion of

$\mathds{Z}_{p}[T]\left[\frac{T^{n}}{p}\right]=\mathds{Z}_{p}[T,S]/(pS-T^{n}),$

which equals

$\mathds{Z}_{p}\langle T,S\rangle/(pS-T^{n}).$

Inverting $p$ we get the completion of $\mathds{Z}_{p}\llbracket
T\rrbracket[1/p]$:

$\mathds{Q}_{p}\langle T,S\rangle/(pS-T^{n}).$

Now we can recognize that

$R\left(\frac{T^{n}}{p}\right)=\operatorname{\mathrm{Spa}}(\mathds{Q}_{p}\langle
T,S\rangle/(pS-T^{n}),\mathds{Z}_{p}\langle T,S\rangle/(pS-T^{n}))$

is the disc of radius $p^{1/n}$ and the union over all $n\in\mathds{N}$ gives
the open unit disc:

${\mathscr{X}}_{\eta}=\bigcup_{n=1}^{\infty}\mathds{D}_{\mathds{Q}_{p}}(0,p^{1/n})=\overset{\circ}{\mathds{D}}_{\mathds{Q}_{p}}(0,1).$

See Example 4.4 for the discussion of the open unit disc.

##  7\. Morphisms of finite type

In this section we explain what it means for a morphism of adic spaces to be
of finite type. It is too restrictive to require the corresponding ring maps
to be of finite type. Already the Tate algebra

$k\langle T\rangle=\left\\{\sum_{n\geq 0}a_{n}T^{n}\in k\llbracket
T\rrbracket\mid a_{n}\to 0\right\\}$

over a nonarchimedean field $k$ is not finitely generated but $k\langle
T\rangle$ is supposed to be a topological analogue of the ring of polynomials
$k[T]$. Note that we can state the property of a ring homomorphism $R\to S$ to
be of finite type by requiring $S$ to be a quotient of a polynomial ring over
$R$ in finitely many variables. This notion can be transported to the realm of
Huber rings by considering topological quotients:

######  Definition 7.1.

A homomorphism $R\to S$ of complete Huber rings is a _quotient map_ if it is
surjective, continuous, and open. A homomorphism $(R,R^{+})\to(S,S^{+})$ of
Huber pairs is a _quotient map_ if $R\to S$ is a quotient map and $S^{+}$ is
the relative integral closure of the image of $R^{+}$ in $S$.

Any closed ideal $I\subseteq R$ defines a quotient map from $R$ to $S:=R/I$
equipped with the quotient topology. We can enhance this to a quotient map of
Huber pairs by defining $S^{+}$ to be the integral closure of $R^{+}/(I\cap
R^{+})$ in $S$. We often denote this quotient Huber pair by $(R,R^{+})/I$.

Now a straightforward replacement for ring homomorphisms of finite type in the
topological setting is the following.

######  Definition 7.2.

A homomorphism $R\to S$ of Huber rings is _strictly of topologically finite
type_ if there is a quotient map

$R\langle T_{1},\ldots,T_{n}\rangle\twoheadrightarrow\hat{S}$

of $\hat{R}$-algebras.

However, this class of $R$-algebras results to be too small to cover all
important constructions. Even an (algebraically) finitely generated
$R$-algebra might not be a quotient of any Tate algebra $R\langle
T_{1},\ldots,T_{n}\rangle$. In order to see this we take a look at rings of
definition.

######  Exercise 7.3.

Let $R$ and $S$ be Huber rings and

$\pi:R\langle T_{1},\ldots,T_{n}\rangle\twoheadrightarrow S$

a quotient map. If $R_{0}\subseteq R$ is a ring of definition, then
$\pi(R_{0}\langle T_{1},\ldots,T_{n}\rangle)$ is a ring of definition of $S$.

We can now give a simple example of a finite type $R$-algebra that is not
strictly of topologically finite type.

######  Example 7.4.

We consider the Huber ring $\mathds{Z}_{p}$ and the $\mathds{Z}_{p}$-algebra
$\mathds{Q}_{p}$. Since $\mathds{Q}_{p}=\mathds{Z}_{p}[1/p]$, it is finitely
generated over $\mathds{Z}_{p}$. However, the following argument shows that it
is not strictly of topologically finite type. Suppose there is a quotient map

$\mathds{Z}_{p}\langle
T_{1},\ldots,T_{n}\rangle\twoheadrightarrow\mathds{Q}_{p}$

The ring $\mathds{Z}_{p}$ serves as its own ring of definition. It thus
follows from Exercise 7.3 that $\mathds{Q}_{p}$ is its own ring of definition,
which is not true. Therefore, $\mathds{Q}_{p}$ cannot be strictly of
topologically finite type over $\mathds{Z}_{p}$.

One can also see this more explicitly by checking that for any
$a\in\mathds{Q}_{p}\setminus\mathds{Z}_{p}$ the homomorphism

$\displaystyle\mathds{Z}_{p}[T]$ $\displaystyle\longrightarrow\mathds{Q}_{p}$
$\displaystyle T$ $\displaystyle\longmapsto a$

is _not_ continuous, where $\mathds{Z}_{p}[T]$ is endowed with the $p$-adic
topology. Otherwise we would get a homomorphism

$\mathds{Z}_{p}\langle T\rangle\longrightarrow\mathds{Q}_{p}$

mapping $T$ to $a$. But $a^{-1}\in\mathds{Z}_{p}$ and $a^{-1}T-1$ is contained
in the kernel (it would in fact generate the kernel). However, in
$\mathds{Z}_{p}\llbracket T\rrbracket$ we have the identity

$\frac{1}{a^{-1}T-1}=\sum_{i=0}^{\infty}a^{-i}T^{i}$

and the series is in fact contained in $\mathds{Z}_{p}\langle T\rangle$ as
$|a|>1$. Therefore, $a^{-1}T-1$ is invertible and generates the unit ideal. We
would get a factorization

$\mathds{Z}_{p}\langle T\rangle\longrightarrow\mathds{Z}_{p}\langle
T\rangle/(a^{-1}T-1)=0\longrightarrow\mathds{Q}_{p},$

a contradiction.

The example suggests that in addition to quotients of Tate algebras one should
include quotients of more general $R$-algebras. These generalized Tate
algebras are constructed as follows. We take finite subsets
$M_{1},\ldots,M_{n}$ of $R$ such that for any $i$ and any $r\in\mathds{N}$ the
ideal

$M_{i}^{r}R:=\langle m_{1}\cdots m_{r}a\mid m_{1},\ldots,m_{r}\in
M_{i},~{}a\in R\rangle$

is open (the angle brackets stand for the subgroup generated by the listed
elements). Often we write $M$ for the tuple $(M_{1},\ldots,M_{n})$ of finite
sets. We associate with each such tuple $M$ of finite sets an $R$-algebra of
_weighted convergent power series_

$\displaystyle R\langle T\rangle_{M}$ $\displaystyle:=R\langle
T_{1},\ldots,T_{n}\rangle_{M}$
$\displaystyle:=\left\\{\sum_{i\in\mathds{N}^{n}}a_{i}T^{i}\in\hat{R}\llbracket
T\rrbracket\mid\forall~{}0\in U\subseteq\hat{R}~{}\text{open}:~{}a_{i}\in
M^{i}U~{}\text{for almost all~{}$i$}\right\\}$

Here we use the short notation $T$ for the collection of the variables
$T_{1},\ldots,T_{n}$ and set

$T^{i}:=T^{i_{1}}\cdots T^{i_{n}}$

for $i\in\mathds{N}^{n}$. Likewise, for an open neighborhood $U\subseteq R$ of
$0$, we write

$M^{i}U:=M_{1}^{i_{1}}\cdots M_{n}^{i_{n}}U.$

With this notation we often replace the condition that $M_{i}^{r}R$ be open
for all $i=1,\ldots,n$ and for all $r\in\mathds{N}$ by the equivalent
condition that $M^{i}R$ be open for all $i\in\mathds{N}^{n}$. Let us give this
property a name: We call a tuple $M=(M_{1},\ldots,M_{n})$ _voluminous_ if
$M^{i}R$ is open for all $i\in\mathds{N}^{n}$.

######  Example 7.5.

We can rewrite the Huber rings $k\langle\varpi^{k}T\rangle$ from Example 4.3
as

$k\langle\varpi^{k}T\rangle=k\langle T\rangle_{\\{\varpi^{k}\\}}.$

We need to give $R\langle T\rangle_{M}$ the structure of a Huber ring.
Starting with a pair of definition $(R_{0},I)$ for $R$ we specify the ring of
definition

$R_{0}\langle T\rangle_{M}:=\left\\{\sum_{i\in\mathds{N}^{n}}a_{i}T^{i}\in
R\langle T\rangle\mid a_{i}\in M^{i}\hat{R}_{0}~{}\forall
i\in\mathds{N}\right\\}$

and take the ideal of definition to be the ideal of $R_{0}\langle
T\rangle_{M}$ generated by $I$. Then for every open subgroup $U$ of $R$ the
subgroup

$U\langle T\rangle_{M}:=\left\\{\sum_{i\in\mathds{N}^{n}}a_{i}T^{i}\in
R\langle T\rangle\mid a_{i}\in M^{i}\hat{U}~{}\forall
i\in\mathds{N}^{n}\right\\}$

is open in $R\langle T\rangle_{M}$. In fact the collection of all $U\langle
T\rangle_{M}$ as $U$ varies over the open subgroups of $R$ is a fundamental
system of open neighborhoods of zero. Finally we define the ring of integral
elements

$R\langle T\rangle_{M}^{+}$

to be the integral closure of $R^{+}\langle T\rangle_{M}$ in $R\langle
T\rangle_{M}$.

######  Definition 7.6.

Let $R\to S$ be a homomorphism of Huber rings. Then $S$ is of _topologically
finite type_ over $R$ if there is a voluminous tuple $M=(M_{1},\ldots,M_{n})$
of finite subsets of $R$ and a quotient map

$R\langle T\rangle_{M}\twoheadrightarrow S$

of $R$-algebras.

######  Example 7.7.

Now luckily $\mathds{Q}_{p}$ is of topologically finite type over
$\mathds{Z}_{p}$. Indeed

$\mathds{Q}_{p}\cong\mathds{Z}_{p}\langle T\rangle_{\\{p\\}}/(pT-1).$

The crucial point is that

$\frac{1}{pT-1}=\sum_{i=0}^{\infty}p^{i}T^{i}\in\mathds{Z}_{p}\llbracket
T\rrbracket$

is not contained in

$\mathds{Z}_{p}\langle
T\rangle_{\\{p\\}}=\left\\{\sum_{i=0}^{\infty}a_{i}T^{i}\mid\forall
r\in\mathds{N}:~{}a_{i}\in p^{i+r}\mathds{Z}_{p}~{}\text{for almost
all}~{}i\in\mathds{N}\right\\}.$

Therefore $(pT-1)$ is not the unit ideal in $\mathds{Z}_{p}\langle T\rangle$
and dividing out $(pT-1)$ results in inverting $p$ in $\mathds{Z}_{p}$.

We get back the familiar ring of converging power series by choosing
$M_{i}=\\{1\\}$ for all $i$:

$R\langle T_{1},\ldots,T_{n}\rangle=R\langle
T\rangle_{(\\{1\\},\ldots,\\{1\\})}.$

In particular, all $R$-algebras that are of strictly topologically finite type
are also of topologically finite type.

It turns out that we need to consider the more complicated $R$-algebras
$R\langle T\rangle_{M}$ only if we leave the realm of Tate rings. This is not
obvious by just looking at the definitions. We need to take a detour via an
alternative description of $R$-algebras (strictly) of topologically finite
type.

######  Proposition 7.8.

For a homomorphism $\varphi:R\to S$ of Huber rings such that $S$ is complete
the following are equivalent:

  1. (i)

$S$ is (strictly) of topologically finite type over $R$,

  2. (ii)

There are rings of definition $R_{0}\subseteq R$ and $S_{0}\subseteq S$ with
$\varphi(R_{0})\subseteq S_{0}$ such that $R_{0}\to S_{0}$ is strictly of
topologically finite type and $S$ is finitely generated over (equal to)
$\varphi(R)\cdot S_{0}$.

  3. (iii)

For every open subring $R_{0}\subseteq R$ there is an open subring
$S_{0}\subseteq S$ with $\varphi(R_{0})\subseteq S_{0}$ such that $R_{0}\to
S_{0}$ is strictly of topologically finite type and $S$ is finitely generated
over (equal to) $\varphi(R)\cdot S_{0}$.

Moreover, in the non-strict case we have the following equivalent
characterization.

  1. (iv)

$\varphi$ is adic, there are rings of definition $R_{0}\subseteq R$ and
$S_{0}\subseteq S$ with $\varphi(R_{0})\subseteq S_{0}$ and finite subsets
$M\subseteq S$ and $N\subseteq S_{0}$ such that $\varphi(R)[M]$ is dense in
$S$ and $\varphi(R_{0})[N]$ is dense in $S_{0}$.

###### Proof.

We refer to [Lud20], Proposition 2.2 and 2.3. ∎

######  Corollary 7.9.

Any homomorphism $\varphi:R\to S$ of Tate rings of topologically finite type
is strictly of topologically finite type.

###### Proof.

We may assume that $S$ is complete. Then by Proposition 7.8 there are rings of
definition $R_{0}\subseteq R$ and $S_{0}\subseteq S$ with
$\varphi(R_{0})\subseteq S_{0}$ such that $R_{0}\to S_{0}$ is strictly of
topologically finite type and $S$ is finitely generated over $\varphi(R)\cdot
S_{0}$. Let $\varpi\in R_{0}$ be a topologically nilpotent unit. Its image
$\varpi_{S}$ in $S$ is a topologically nilpotent unit as well and moreover
contained in $S_{0}$. By [Ber24], §1.1.3, we have

$S=S_{0}\left[\frac{1}{\varpi_{S}}\right]\subseteq\varphi\left(R_{0}\left[\frac{1}{\varpi}\right]\right)\cdot
S_{0}=\varphi(R)\cdot S_{0},$

so $S$ is in fact strictly of topologically finite type over $R$. ∎

Now we want to include the rings of integral elements into the picture. A
homomorphism of Huber rings $(R,R^{+})\to(S,S^{+})$ should be topologically of
finite type if $S$ is topologically of finite type over $R$ plus some
additional finiteness condition on $R^{+}\to S^{+}$. There are several
variants that play a role, which lead to different finiteness notions. The
following two are the most common:

######  Definition 7.10.

Let $\varphi:(R,R^{+})\to(S,S^{+})$ be a homomorphism of complete Huber pairs.
Then $\varphi$ is

  * •

_weakly of topologically finite type_ if $R\to S$ is of topologically finite
type.

  * •

of _topologically finite type_ if there is a quotient map

$(R\langle T\rangle_{M},R\langle T\rangle_{M}^{+})\twoheadrightarrow(S,S^{+})$

of Huber pairs over $(R,R^{+})$ for some tuple $M$ of finite subsets of $R$
such that $M^{i}U$ is open for every open neighborhood $U$ of zero in $R$ and
every $i\in\mathds{N}^{n}$.

######  Example 7.11.

Let $(R,R^{+})$ be a complete Huber pair.

  * •

For a rational open $U\subseteq\operatorname{\mathrm{Spa}}(R,R^{+})$, the
Huber pair $(R_{U},R_{U}^{+})$ is of topologically finite type over
$(R,R^{+})$.

  * •

$(R\langle T\rangle,R^{+}\langle T\rangle)$ is of topologically finite type
over $(R,R^{+})$.

  * •

Let $(R_{\mathit{disc}},R_{\mathit{disc}}^{+})=(R,R^{+})$ but with the
discrete topology. Then the identity induces a (continuous) homorphism of
Huber pairs $(R_{\mathit{disc}},R_{\mathit{disc}}^{+})\to(R,R^{+})$, which in
general is _not_ of topologically finite type.

After these simple examples let us take a look at a more sophisticated one
that shows the difference between topologically finite type and weak
topologically finite type.

######  Example 7.12.

Let $(k,k^{\circ})$ be an analytic field with pseudouniformizer $\varpi$. We
consider the adic unit disc

$\mathds{D}_{k}=\operatorname{\mathrm{Spa}}(k\langle T\rangle,k^{\circ}\langle
T\rangle).$

As a subspace of ${\mathds{A}}_{k}^{1}$ (or a disc of bigger radius, for
instance $\mathds{D}_{k}(0,\varpi^{-1})$) it is not closed even though it
looks like a closed unit disc and it is even called by that name. It turns out
there is one point in its closure that is not contained in $\mathds{D}_{k}$
itself: Here we return to the constructions we discussed in Example 2.10. We
consider the ordered group $\mathds{R}_{>0}\times\mathds{R}_{>0}$ but this
time choose $\epsilon=(1,\epsilon_{2})\in\mathds{R}_{>0}\times\mathds{R}_{>0}$
with $\epsilon_{2}>1$. The corresponding valuation

$\displaystyle x_{0,1+}:k\langle T\rangle$
$\displaystyle\longrightarrow\mathds{R}_{>0}\times\mathds{R}_{>0}$
$\displaystyle\sum_{i}a_{i}T^{i}$
$\displaystyle\longmapsto\max_{i}|a_{i}|\epsilon^{i}$

is not contained in $\mathds{D}_{k}$ as

$|T(x_{0,1+})|=\epsilon>1.$

But it is still a continuous valuation of $k\langle T\rangle$ by Proposition
1.2.7.1 in [Ber24]: We have that $|\varpi(x_{0,1+})|=|\varpi|$ is cofinal in
the value group and for every element $\sum_{i}a_{i}T^{i}$ in the ring of
definition $k^{\circ}\langle T\rangle$ we look at

$|\varpi\left(\sum_{i}a_{i}T^{i}\right)(x_{0,1+})|=\max_{i}|\varpi
a_{i}|\epsilon^{i}.$

The terms $|\varpi a_{i}|$ are of the form
$(r_{1},1)\in\mathds{R}_{>0}\times\mathds{R}_{>0}$ with $r_{1}<1$. So the
product looks like $(r_{1},\epsilon_{2}^{i})<1$.

The closure $\overline{\mathds{D}}_{k}$ of $\mathds{D}_{k}$ turns out to be an
affinoid adic space that is weakly of topologically finite type but not of
topologically finite type. We define

$A^{+}=\left\\{\sum_{i}a_{i}T^{i}\in k\langle T\rangle\mid|a_{0}|\leq
1,~{}|a_{i}|<1~{}\forall~{}i\geq 1\right\\}.$

This is open as $\varpi k^{\circ}\langle T\rangle\subseteq A^{+}$, and let us
check that it is integrally closed: If $x$ is integral over $A^{+}$, it is
contained in $k^{\circ}\langle T\rangle$. Its reduction $\bar{x}$ in

$\tilde{k}[T]=k^{\circ}\langle T\rangle/\mathfrak{m}_{k}k^{\circ}\langle
T\rangle$

($\mathfrak{m}_{k}$ is the maximal ideal of $k^{\circ}$ and $\tilde{k}$ its
residue field) is integral over

$A^{+}/(\mathfrak{m}_{k}k^{\circ}\langle T\rangle)=\tilde{k}.$

Hence $\bar{x}\in\tilde{k}$. By definition this implies $x\in A^{+}$.

Now let us compute the valuation of $\sum_{i}a_{i}T^{i}\in A^{+}$:

$|\left(\sum_{i}a_{i}T^{i}\right)(x_{0,1+})|=\max_{i}|a_{i}|\epsilon^{i}.$

For $i=0$ we obtain $|a_{0}|\leq 1$ and for $i\geq 1$ we get
$(|a_{i}|,\epsilon^{i})<1$, so
$x_{0,1+}\in\operatorname{\mathrm{Spa}}(k\langle T\rangle,A^{+})$. In fact it
is true that

$\overline{\mathds{D}}_{k}=\operatorname{\mathrm{Spa}}(k\langle
T\rangle,A^{+}).$

But $(k\langle T\rangle,A^{+})$ is not of topologically finite type over
$(k,k^{\circ})$.

######  Definition 7.13.

A morphism of adic spaces $Y\to X$ is _locally (weakly) of finite type_ if
locally on $X$ and $Y$ it is given by a morphism of affinoids induced by a
homomorphism of Huber pairs that is (weakly) of topologically finite type. It
is _of (weakly) finite type_ if in addition it is quasi-compact.

######  Proposition 7.14.

Every morphism of adic spaces that is locally weakly of finite type is adic.

###### Proof.

This follows from characterization (iv) in Proposition 7.8 of homomorphisms of
Huber pairs of topologically finite type. ∎

######  Definition 7.15.

Let $(k,k^{\circ})$ be an analytic field. A rigid analytic variety over
$(k,k^{\circ})$ is an adic space $X$ of finite type over $(k,k^{\circ})$.

For those familiar with the more classical language of Tate’s rigid analytic
varieties we want to remark the following. There is a fully faithful functor
from Tate’s rigid analytic varieties to adic spaces sending $\mathrm{Sp}(R)$
to $\operatorname{\mathrm{Spa}}(R,R^{\circ})$. Its essential image is (modulo
some quasi-separatedness issues) given by the rigid analytic varieties defined
above (see [Hub94], Proposition 4.4).

##  8\. Formal schemes and generic fibers

Let $A$ be a noetherian ring endowed with the $I$-adic topology for a
(finitely generated) ideal $I$. In particular, this is a Huber ring which is
its own ring of definition. Its _formal spectrum_
${\mathscr{X}}:=\operatorname{\mathrm{Spf}}A$ is the set of all open prime
ideals endowed with the Zariski topology. So as a topological space
${\mathscr{X}}$ is isomorphic to $\operatorname{\mathrm{Spec}}A/I$. What makes
the difference is the structure sheaf ${\mathscr{O}}_{{\mathscr{X}}}$: For
$f\notin\sqrt{I}$ its sections over the corresponding fundamental open
$\operatorname{\mathrm{Spf}}A_{f}\subseteq{\mathscr{X}}$ are

${\mathscr{O}}_{{\mathscr{X}}}(\operatorname{\mathrm{Spf}}A_{f})=\widehat{A_{f}},$

where the completion is the $I$-adic completion. In particular,
${\mathscr{O}}_{{\mathscr{X}}}({\mathscr{X}})=\widehat{A}$. Formal spectra
$({\mathscr{X}},{\mathscr{O}}_{{\mathscr{X}}})$ as above can be glued to
obtain _locally noetherian formal schemes_. For details see [SGA7.1].

######  Theorem 8.1 ([Hub94], Proposition 4.2).

There is a fully faithful functor from the category of locally noetherian
formal schemes to the category of adic spaces. It maps
$\operatorname{\mathrm{Spf}}A$ to
$\operatorname{\mathrm{Spa}}(A)=\operatorname{\mathrm{Spa}}(A,A)$.

By the above theorem we can consider formal schemes as adic spaces. More
precisely we can identify them as the locally noetherian adic spaces $X$ that
can be covered by affinoid opens of the form
$\operatorname{\mathrm{Spa}}(A,A)$.

For a formal scheme ${\mathscr{X}}$ we write
${\mathscr{X}}^{\operatorname{\mathrm{ad}}}$ for the adic space associated to
it according to the theorem. Let us try to understand the relation between
${\mathscr{X}}$ and ${\mathscr{X}}^{\operatorname{\mathrm{ad}}}$. We will
construct functorial morphisms of topologically locally ringed spaces

$({\mathscr{X}},{\mathscr{O}}_{{\mathscr{X}}})\overset{\iota_{\mathscr{X}}}{\longrightarrow}({\mathscr{X}}^{\operatorname{\mathrm{ad}}},{\mathscr{O}}_{{\mathscr{X}}^{\operatorname{\mathrm{ad}}}}^{+})\overset{r_{\mathscr{X}}}{\longrightarrow}({\mathscr{X}},{\mathscr{O}}_{\mathscr{X}})$

such that $r_{\mathscr{X}}\circ\iota_{\mathscr{X}}$ is the identity. It is
enough to give the construction in the affinoid case, i.e., where
${\mathscr{X}}=\operatorname{\mathrm{Spf}}(A)$. Then $\iota_{\mathscr{X}}$
maps an open prime ideal $\mathfrak{p}\subseteq A$ to the trivial valuation of
$k(\mathfrak{p})$.

######  Exercise 8.2.

For a rational open subset
$\operatorname{\mathrm{Spa}}(B,B^{+})\subseteq{\mathscr{X}}^{\operatorname{\mathrm{ad}}}$
we have

$\iota_{\mathscr{X}}^{-1}(\operatorname{\mathrm{Spa}}(B,B^{+}))=\operatorname{\mathrm{Spf}}B,$

where $B$ is equipped with the $IB$-adic topology for an ideal of definition
$I\subseteq B^{+}$. In particular, the preimage is empty if $IB=B$ is
invertible in $B$.

The exercise shows that $\iota_{\mathscr{X}}$ is continuous. We now want to
define the homomorphism of sheaves
${\mathscr{O}}_{{\mathscr{X}}^{\operatorname{\mathrm{ad}}}}^{+}\to(\iota_{{\mathscr{X}}})_{*}{\mathscr{O}}_{\mathscr{X}}$.
Using Exercise 8.2 this comes down to specifying for every open affinoid
$\operatorname{\mathrm{Spa}}(B,B^{+})$ of
${\mathscr{X}}^{\operatorname{\mathrm{ad}}}$ a homomorphism

$B^{+}={\mathscr{O}}_{{\mathscr{X}}^{\operatorname{\mathrm{ad}}}}^{+}(\operatorname{\mathrm{Spa}}(B,B^{+}))\longrightarrow{\mathscr{O}}_{\mathscr{X}}(\iota_{\mathscr{X}}^{-1}(\operatorname{\mathrm{Spa}}(B,B^{+}))={\mathscr{O}}_{\mathscr{X}}(\operatorname{\mathrm{Spf}}B)=B.$

It is easy to check that by taking this homomorphism to be the natural
inclusion, $\iota_{\mathscr{X}}$ becomes a morphism of topologically locally
ringed spaces.

In order to define the map $r_{\mathscr{X}}$ we consider a point
$x\in{\mathscr{X}}^{\operatorname{\mathrm{ad}}}$, which defines a valuation
ring $k(x)^{+}$ of $k(x)$. Since $|A(x)|\leq 1$, we have a natural map $A\to
k(x)^{+}$ leading to

$\operatorname{\mathrm{Spec}}k(x)^{+}\longrightarrow\operatorname{\mathrm{Spec}}A.$

Then $r_{\mathscr{X}}(x)$ is the image of the closed point of
$\operatorname{\mathrm{Spec}}k(x)^{+}$, also called the _center_ of $x$. Note
that the continuity of $x$ ensures that $r_{\mathscr{X}}(x)$ is indeed an open
prime ideal of $A$. Now it is not hard to check that for an affine open
$\operatorname{\mathrm{Spf}}B\subseteq\operatorname{\mathrm{Spf}}A$ we have

$r_{\mathscr{X}}^{-1}(\operatorname{\mathrm{Spf}}B)=\operatorname{\mathrm{Spa}}(B,B).$

We define the sheaf homomorphism
${\mathscr{O}}_{\mathscr{X}}\to(r_{\mathscr{X}})_{*}{\mathscr{O}}_{{\mathscr{X}}^{\operatorname{\mathrm{ad}}}}$
by taking over an open affinoid $\operatorname{\mathrm{Spf}}B$ the identity of
$B$. This defines $r_{\mathscr{X}}$ and obviously we have
$r_{\mathscr{X}}\circ\iota_{\mathscr{X}}=\mathrm{id}$.

The existence of $r_{\mathscr{X}}$ and $\iota_{\mathscr{X}}$ suggests that
${\mathscr{X}}$ should be thought of as a deformation retract of
${\mathscr{X}}^{\operatorname{\mathrm{ad}}}$. Although
${\mathscr{X}}^{\operatorname{\mathrm{ad}}}$ has many more points, it carries
more or less the same information as ${\mathscr{X}}$.

Let $(k,k^{\circ})$ be an analytic field. We consider the special case of
formal schemes over $k^{\circ}$. More precisely, we study the category of
formal schemes that are locally of the form $\operatorname{\mathrm{Spa}}(A,A)$
for a $k^{\circ}$-algebra $A$ of formally finite type. For a noetherian ring
$A$ with the $I$-adic topology for some ideal $I\subseteq A$ the standard
example of an $A$-algebra of formally finite type is

$\hat{A}\llbracket T_{1},\ldots,T_{n}\rrbracket\langle
X_{1},\ldots,X_{m}\rangle$

for $m,n\in\mathds{N}\cup\\{0\\}$. We endow this ring with the $J$-adic
topology, where $J$ is the ideal generated by $I$ and $T_{1},\ldots,T_{n}$.

######  Exercise 8.3.

Let $A$ be a noetherian ring with an ideal $I\subseteq A$ and
$m,n\in\mathds{N}\cup\\{0\\}$. We consider the polynomial ring

$B:=A[T_{1},\ldots,T_{n},X_{1},\ldots,X_{m}]$

endowed with the $J$-adic topology, where $J$ is the ideal generated by $I$
and $T_{1},\ldots,T_{n}$. Then

$\hat{B}=\hat{A}\llbracket T_{1},\ldots,T_{n}\rrbracket\langle
X_{1},\ldots,X_{m}\rangle.$

Note that $A\to B$ is adic if and only if $n=0$. This in turn is the case if
and only if $B$ is of topologically finite type over $A$. Using the standard
building blocks $\hat{A}\llbracket T\rrbracket\langle X\rangle$ we can now
define $A$-algebras of formally finite type.

######  Definition 8.4.

A morphism $f:{\mathscr{X}}\to{\mathscr{Y}}$ of locally noetherian formal
schemes is of _formally finite type_ if it is locally on ${\mathscr{X}}$ and
${\mathscr{Y}}$ given by a morphism of affinoids

$\operatorname{\mathrm{Spa}}(B,B)\longrightarrow\operatorname{\mathrm{Spa}}(A,A)$

induced by a ring homomorphism $A\to B$, such that there is a quotient map

$A\llbracket T_{1},\ldots,T_{n}\rrbracket\langle
X_{1},\ldots,X_{m}\rangle\twoheadrightarrow\hat{B}$

of $\hat{A}$-algebras.

In this setting an important construction is the (rigid analytic) generic
fiber. Classically, one would do this explicitly on affine opens and then
check that the construction glues. In the language of adic spaces the
definition is straight forward:

######  Definition 8.5.

Let $(k,k^{\circ})$ be an analytic field with pseudouniformizer $\varpi\in
k^{\circ}$. Let ${\mathscr{X}}$ be a formal scheme of formally finite type
over $k^{\circ}$ considered as an adic space over
$\operatorname{\mathrm{Spa}}(k^{\circ},k^{\circ})$. The generic fiber of
${\mathscr{X}}$ is defined as the open subspace

${\mathscr{X}}_{\eta}=\\{x\in{\mathscr{X}}||\varpi(x)|\neq 0\\}.$

If ${\mathscr{X}}$ is a (classical) formal scheme we write
${\mathscr{X}}_{\eta}$ for
$({\mathscr{X}}^{\operatorname{\mathrm{ad}}})_{\eta}$.

Note that $\operatorname{\mathrm{Spa}}(k^{\circ},k^{\circ})$ has two points.
The generic point is open and cut out precisely by the condition $|\varpi|\neq
0$. Hence we see that ${\mathscr{X}}_{\eta}$ is nothing but the preimage in
${\mathscr{X}}$ of the generic point of
$\operatorname{\mathrm{Spa}}(k^{\circ},k^{\circ})$.

If ${\mathscr{X}}$ is even of topologically finite type over $k^{\circ}$, the
inclusion of the generic fiber is quasi-compact. In order to see this we may
assume ${\mathscr{X}}$ is affinoid, i.e.,
${\mathscr{X}}=\operatorname{\mathrm{Spa}}(A,A)$ for a $\varpi$-adic ring $A$.
Then the ideal of $A$ generated by $\varpi$ is open, and thus

${\mathscr{X}}_{\eta}=R\left(\frac{\varpi}{\varpi}\right)$

is a rational subset.

######  Example 8.6.

We consider the polynomial ring $k^{\circ}[T]$ over the valuation ring
$k^{\circ}$ of a nonarchimedean field $k$ with the induced topology from $k$.
This is the $\varpi$-adic topology for some pseudouniformizer $\varpi\in
k^{\circ}$. The completion of $k^{\circ}[T]$ is $k^{\circ}\langle T\rangle$.
We want to determine the generic fiber of

${\mathscr{X}}:=\operatorname{\mathrm{Spa}}(k^{\circ}\langle T\rangle).$

By definition it is the rational subset

$R\left(\frac{\varpi}{\varpi}\right)=\\{x\in{\mathscr{X}}\mid|\varpi(x)|\neq
0\\}.$

We obtain the corresponding Huber pair by inverting $\varpi$ in the first
component and adjoining $\varpi/\varpi=1$ in the second component (so nothing
happens there):

$(k^{\circ}\langle T\rangle[\frac{1}{\varpi}],k^{\circ}\langle
T\rangle[\frac{\varpi}{\varpi}])=(k\langle T\rangle,k^{\circ}\langle
T\rangle).$

The adic spectrum of this Huber pair is the closed unit disc
$\mathds{D}_{k}(0,1)$.

######  Example 8.7.

In Example 6.12 we have already computed the generic fiber of
${\mathscr{X}}=\operatorname{\mathrm{Spa}}(\mathds{Z}_{p}\llbracket
T\rrbracket)$:

${\mathscr{X}}_{\eta}=\bigcup_{n=1}^{\infty}R\left(\frac{T^{n}}{p}\right)=\overset{\circ}{\mathds{D}}_{\mathds{Q}_{p}}(0,1).$

Here one can see that ${\mathscr{X}}_{\eta}$ is not quasi-compact due to the
fact that $\mathds{Z}_{p}\llbracket T\rrbracket$ is not of topologically
finite type over $\mathds{Z}_{p}$.

Starting with a rigid analytic variety $X$ over $(k,k^{\circ})$ (i.e., an adic
space locally of finite type over $k$) we can always construct a formal scheme
${\mathscr{X}}$ over $\operatorname{\mathrm{Spa}}(k^{\circ},k^{\circ})$ with
${\mathscr{X}}_{\eta}=X$. We call such a space ${\mathscr{X}}$ a _formal
model_ of $X$. Formal models are almost never unique (only if $X$ is quasi-
finite over $k$). We can construct new formal models from an existing one -
let us call it ${\mathscr{X}}$ \- by performing an _admissible blowup_. This
is a blowup of ${\mathscr{X}}$ with respect to a sheaf of ideals
${\mathscr{I}}$ such that $\varpi\in{\mathscr{I}}$. It turns out that
admissible blowups are the only thing that stand in the way of uniqueness of
formal models.

######  Theorem 8.8 (Raynaud).

There is an equivalence of categories

$\left\\{\begin{subarray}{c}\text{qc $\varpi$-torsion free formal
$k^{\circ}$-schemes of tft}\\\ \text{localized by admissible formal
blowups}\end{subarray}\right\\}\overset{\sim}{\longrightarrow}\big{\\{}\begin{subarray}{c}\text{qcqs
rigid}\\\ \text{$k$-spaces}\end{subarray}\big{\\}}.$

induced by mapping a formal scheme ${\mathscr{X}}$ to its generic fiber
${\mathscr{X}}_{\eta}$.

The theorem lies the foundation for the formal model approach to
nonarchimedean geometry. Instead of rigid analytic spaces one can study formal
schemes (up to admissible blowup). This strategy has been adopted by many
researchers, see [BL93], [FK06].

Theorem 8.8 was originally formulated in the language of formal schemes and
Tate’s rigid analytic varieties. In this setting the construction of the rigid