# Twisted differential operators in several variables

Pierre Houédry

###### Abstract.

We introduce new concepts in order to develop a general formalism for twisted
differential operators in several variables. We investigate the notion of
twisted coordinates on Huber rings that allows us to build various rings of
twisted differential operators and compare them. We show that there exists an
equivalence between modules endowed with a twisted connection and modules
endowed with an action of the twisted derivatives. This work is in line with
the recent developments in $p$-adic Hodge cohomology and prismatic cohomology.

###### Contents

  1. 1 Huber rings
  2. 2 Topologically étale morphisms
  3. 3 Reminders on filtered modules
  4. 4 Twisted principal parts in several variables
  5. 5 Coordinates
  6. 6 Twisted differential forms
  7. 7 Derivations
  8. 8 Twisted differential operators of infinite level

## Introduction

A $q$-difference equation is a relation of the form

$E(x,f,\partial_{q}(f),\ldots,\partial_{q}^{n}(f))=0$

between an unknown function $f$ and its iterations by the $q$-difference
operator

$\partial_{q}(f)(x)=\frac{f(qx)-f(x)}{(q-1)x}.$

The primary motivation for studying this equation is a practical one. The aim
is to discretize a differential equation to make it easier to manipulate. This
process is called “deformation”. The process of recovering a differential
equation from a $q$-difference equation, by making $q$ tend towards 1, is
called “confluence”. The interesting problem, from a practical point of view,
is therefore to measure the difference between the solutions of the equation
discretized by deformation and that of the original differential equation.

This problem can be tackled from an arithmetic perspective, i.e. in the world
of $p$-adic numbers. In 2000, Lucia Di Vizio proved a $q$-analogue of
Grothendieck’s $p$-curvature conjecture ([DV02]). Shortly afterwards, in 2004,
Yves André and Lucia Di Vizio studied the phenomenon of confluence. (cf.
[ADV04]). They compute the Tannakian group of $q$-difference equations over
the Robba ring endowed with a Frobenius action and show that it is equal to
the Tannakian group of differential equations over the Robba ring. At the same
time, during his doctoral thesis, Andrea Pulita was also interested in these
problems and obtained an equivalence between these categories by removing the
assumptions on $q$ from [ADV04] for equations on the affine line. He did so by
stating a principle called propagation principle. This principle, when
verified, is then sufficient to demonstrate the equivalence between confluence
and deformation on rings of $p$-adic functions. Andrea Pulita generalizes his
results to curves and to actions of operators more general than multiplication
by $q$ (cf. [Pul08]).

These different works show that in the $p$-adic framework in one variable, the
applied mathematician’s dream is realized: the solutions of the discretized
equation are the same as the solutions of the original differential equation
after confluence.

Interest in this theory increased when Bhargav Bhatt, Matthew Morrow and Peter
Scholze published their work (cf. [BMS18]) on the $p$-adic Hodge cohomology.
Their construction allows specialization to de Rham cohomology and étale
cohomology. They showed that this cohomology can be computed locally using
$q$-difference theory. Bhatt and Scholze also introduced the notions of prism
and prismatic cohomology ([BS22]) to unify the cohomologies they had defined
before. Once again, it appears that $q$-difference equations can be used to
compute prismatic cohomology.

In the past years, Bernard Le Stum, Adolfo Quirós and Michel Gros have created
a more general theory (cf. [LSQ15], [LSQ18b], [LSQ18a], [LSQ20], [GLS14] and
[GLQ19]) that embraces the previous results on $p$-adic $q$-difference
equations ([ADV04], [Pul08]). They generalize the notion of differential
operators and work on affinoid algebras. They build their theory in analogy
with the formalism developed in EGA-IV (cf [Gro67]). It is an approach
oriented by the classical yoga of crystalline cohomology. For $R$ a
commutative ring, a deformed $R$-algebra is a commutative $R$-algebra $A$ with
a $R$-algebra endomorphism $\sigma$. They then define the notion of
$\sigma$-derivation: on an $A$-module $M$ it is the given of a $R$-linear
application that satisfies the twisted Leibniz rule:
$D(xs)=D(x)s+\sigma(x)D(s).$ If an element $x$ of $A$ is a $\sigma$-coordinate
in the sense that there exists a $\partial_{\sigma}$-basis of $A$-derivations
such that $\partial_{\sigma}(x)=1$, then it is possible to define the algebra
of $\text{D}_{\sigma}$-derivations. This gives rise to the concept of
$\sigma$-differential modules, i.e. modules endowed with an action by the
$\partial_{\sigma}$ derivation. The aim is then to compare $\text{D}_{\sigma}$
with $\text{D}_{\tau}$ where $\tau$ is another $R$-linear endomorphism of $A$
such that $x$ is also a $\tau$-coordinate. Differential operators are usually
the dual of a ring of formal functions. Thus, to understand,
$\text{D}_{\sigma}$ we need to understand the notion of $\sigma$-divided
power. Another strategy is to replace $\text{D}_{\sigma}$ by a Grothendieck
ring of differential operators $\text{D}^{(\infty)}_{\sigma}$. The classical
approach then works particularly well for this generalization. The idea is to
compare these rings for different endomorphisms and show, under certain
conditions, that the categories of modules with an action by operators are
equivalent and that this equivalence preserves the solutions.

However, their techniques do not extend to several variables. As in classical
calculus going from differential equations to partial differential equations
is highly non-trivial. Matthew Morrow and Takeshi Tsuji have developed a
several variables theory ([MT21]) in the $q$-difference case. The present
article is a foundational work that makes it possible to generalize the
previous theories considering the case of several variables and with more
general twists than the multiplication by $q$. It is also worth mentioning the
work of Claude Sabbah ([Sab93]) which, even if it is done over $\mathbb{C}$,
is closely related to the work carried out here.

The first two sections are dedicated to Huber rings and étale morphisms. We
start by recalling the definition of a Huber ring defined in [Hub93]. We then
devote a few pages to diverse constructions associated with these rings:
completion, tensor product and rational localization. All these constructions
are useful for including in our multivariate theory the results so far
demonstrated in the one-variable case. In the second section, we propose the
concept of topologically étale morphism. This definition is an affine variant
of [Abb10, Definition 2.3.13]. We also introduce the notion of good étale
morphism. When working on an adic ring, a topologically étale morphism is
automatically good étale. The idea behind this is that we are going to give
ourselves coordinates that satisfy certain properties related to our
differentials, and we want to be able to extend these properties. Under
certain conditions, rational localization and the passage through a good étale
morphism permit us to extend the characteristics of the coordinates.

We then follow a similar path as in the classical situation to develop the
twisted differential calculus. However, if we provide $A$ with many
automorphisms $\underline{\sigma}=(\sigma_{1},\ldots,\sigma_{d})$, there is no
natural notion of twisted power of $I$ that allows us to implement the same
formalism in several variables. Our work, consist in finding the right
definitions to be able to define the usual rings of differential operators so
that we can extend as much results of the one variable case to the several
variables case as we can. The approach we develop in this work is to proceed
in the opposite direction to the one of [LSQ20]: given a Huber ring $R$ and an
$R$-aldic algebra $A$ endowed with continuous linear $R$-endomorphisms
$\underline{\sigma}=(\sigma_{1},\ldots,\sigma_{d})$ we first specify the
module of twisted Taylor series $P_{A,(n)_{\underline{\sigma}}}$ before
defining the module of twisted differential forms
$\Omega^{1}_{A,\underline{\sigma}}$. Those definitions rely on a choice of
coordinates who will need to satisfy some good properties for extending the
one variable case to the several variables case. The core definition that
allows this generalization is the notion of $\underline{\sigma}$-coordinates.

######  Definition.

Elements $x_{1},\ldots,x_{d}\in A$ are called $\underline{\sigma}$-coordinates
over $A$ if, for every $n\in\mathbb{N}$, there exists an unique adic morphism
of $R$-algebras

$\displaystyle\Theta_{A,(n)_{\underline{\sigma}}}\colon A$ $\displaystyle\to
P_{A,(n)_{\underline{\sigma}}}$ $\displaystyle x_{i}$ $\displaystyle\mapsto
x_{i}+\xi_{i}$

such that the composition of the projection on $A$ with
$\Theta_{A,(n)_{\underline{\sigma}}}$ is the identity. The morphism
$\Theta_{A,(n)_{\underline{\sigma},\underline{x}}}$ is said to be the $n$-th
Taylor morphism of $A$ with respect to $\underline{\sigma}$ and
$\underline{x}$.

Under this condition we can define the module of twisted differential
operators $\Omega^{1}_{A,\underline{\sigma}}$ and recover some operators
$\partial_{\underline{\sigma},1}(f),\ldots,\partial_{\underline{\sigma},d}(f)$.
There is no apparent reasons for $\partial_{\underline{\sigma},i}(f)$ to be a
$\sigma_{i}$-derivation. We need to add an hypothesis for this. It is then
natural to add the requirement

$\forall i=1,\ldots,d,\forall f\in
A,~{}\sigma_{i}(f)-f=(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f).$

Coordinates satisfying those conditions are called classical. It allows us to
prove the following theorem

######  Theorem.

The $\underline{\sigma}$-coordinates $x_{1},\ldots,x_{d}$ are classical
coordinates if and only if $\forall i\in\\{1,\ldots,d\\},$
$\partial_{\underline{\sigma},i}$ is a $\sigma_{i}$-derivation.

Given $\underline{\sigma}$-coordinates we can define the notion of twisted
connection and we have that:

######  Theorem.

When the coordinates $x_{1},\ldots,x_{d}$ are classical, there exists an
equivalence of categories

$\nabla_{\underline{\sigma}}\emph{-Mod}(A)\simeq\emph{T}_{A,\underline{\sigma}}\emph{-Mod}(A).$

between $A$-modules with a twisted connection and $A$-modules with an action
of the derivations.

Adding a mild hypothesis over the coordinates we can define the ring of
twisted differentials operators of infinite level
$\text{D}_{\underline{\sigma}}^{(\infty)}$. In the case where our coordinates
are $\underline{q}$-coordinates, that is there exists $q_{1},\ldots,q_{q}\in
R$ such that $\sigma_{i}(x_{j})=q_{i}^{\delta_{i,j}}x_{j}$, and $\forall
n\in\mathbb{N},\frac{q_{i}^{n}-1}{q_{i}-1}\in R^{\times}$, we have the
following:

######  Theorem.

There exists an equivalence of categories

$\nabla_{\underline{\sigma}}^{\emph{Int}}\text{-}\emph{Mod}(A)\simeq\emph{D}_{A,\underline{\sigma}}^{(\infty)}\text{-}\emph{Mod}.$

where $\nabla_{\underline{\sigma}}^{\emph{Int}}\text{-}\emph{Mod}(A)$
designates the $A$-modules endowed with an integrable twisted connection.

Thus, the definition of $\underline{\sigma}$-coordinates seems to be the
appropriate approach to construct the theory of twisted differential operators
in several variables as we are able to recover, under some hypotheses that
appears naturally, the usual properties of twisted differential calculus.

In what follows, we will assume that every ring is commutative and unitary.

##  1\. Huber rings

This section recalls some constructions introduced in [Hub93].

######  Definition 1.1.

Let $A$ be a topological ring.

  1. (1)

We say that $A$ is a _Huber ring_ if there exists an open subring $A_{0}$ of
$A$ and a finitely generated ideal $I_{0}$ of $A_{0}$ such that
$\\{I_{0}^{n}\\}_{n\in\mathbb{N}}$ is a fundamental system of neighbourhoods
of 0 in $A_{0}$. In that case, we say that $A_{0}$ is a ring of definition,
$I_{0}$ is an ideal of definition and that $(A_{0},I_{0})$ is a couple of
definition .

  2. (2)

We say that $A$ is _adic_ in the case where it is a Huber ring and we can
choose $A_{0}=A$.

  3. (3)

We say that $A$ is _discrete_ in the case where it is a Huber ring and we can
choose $A_{0}=A$ and $I_{0}=0$.

######  Examples.

  1. (1)

The ring of integers $\mathbb{Z}$ endowed with the discrete topology is a
discrete ring.

  2. (2)

For a prime number $p$, the ring of $p$-adic integers

$\mathbb{Z}_{p}=\left\\{\sum_{n\geq 0}a_{n}p^{n},~{}0\leq a_{n}<p\right\\}$

endowed with the $p$-adic topology is an adic ring: the ideal $(p)$ is an
ideal of definition.

  3. (3)

Let $A$ be a Huber ring. For every $n\in\mathbb{N}$, the ring
$A[X_{1},\ldots,X_{n}]$ can be endowed with a structure of Huber ring that
induces the topology on $A$: we can choose
$(A_{0}[X_{1},\ldots,X_{n}],I_{0}A_{0}[X_{1},\ldots,X_{n}])$ as a couple of
definition.

  4. (4)

Let $K$ be a _non-archimedean field_ : by that we mean a topological field for
a non-trivial non-archimedean absolute value $|\cdot|$. There exists an
element $\pi$ of $K$ such that $0<|\pi|<1$. The field $K$ is a Huber ring: the
set $\left\\{|x|\leq 1\right\\}$ is a ring of definition and $(\pi)$ is an
ideal of definition.

  5. (5)

Let $d\in\mathbb{N}$. In the previous situation, we consider the following
ring

$A=\left\\{\sum_{\underline{k}\in\mathbb{Z}^{d}}a_{\underline{k}}\underline{X}^{\underline{k}},a_{\underline{k}}\in
K,|a_{\underline{k}}|\prod_{i=1}^{d}\max(1,|\pi|^{k_{i}})\rightarrow 0\text{
when }\underline{k}\rightarrow+\infty\right\\},$

endowed with the topology given by the Gauss norm

$\left\lVert\sum_{\underline{k}\in\mathbb{Z}^{d}}a_{\underline{k}}\underline{X}^{\underline{k}}\right\rVert=\max\left\\{\sup_{|\underline{k}|\geq
0}|a_{\underline{k}}|,\sup_{|\underline{k}|<0}|\pi|^{|\underline{k}|}|a_{\underline{k}}|\right\\}.$

With this topology, this ring is a Huber ring. A couple of definition is given
by

$\displaystyle
A_{0}=\left\\{\sum_{\underline{k}\in\mathbb{Z}^{d}}a_{\underline{k}}\underline{X}^{\underline{k}}\in
A\text{ such that }|a_{\underline{k}}|\leq 1\text{ if }|\underline{k}|\geq
0,~{}|\pi^{|\underline{k}|}a_{\underline{k}}|\leq 1\text{ if
}|\underline{k}|<0\right\\},$ $\displaystyle
I_{0}=\left\\{\sum_{\underline{k}\in\mathbb{Z}^{d}}a_{\underline{k}}\underline{X}^{\underline{k}}\in
A\text{ such }|a_{\underline{k}}|<1\text{ if }|\underline{k}|\geq
0,~{}|\pi^{|\underline{k}|}a_{\underline{k}}|<1\text{ if
}|\underline{k}|<0\right\\}.$

######  Definition 1.2.

Let $A$ and $B$ be two Huber rings. A morphism of Huber rings $u:A\rightarrow
B$ is a morphism of rings which is continuous. This is equivalent to require
that there exist a ring of definition $A_{0}$ of $A$ (resp. $B_{0}$ of $B$)
and an ideal of definition $I_{0}\subset A_{0}$ (resp. $J_{0}\subset B_{0}$)
such that $u(A_{0})\subset B_{0}$ and $u(I_{0})\subset J_{0}$. In the case
where $u(I_{0})B_{0}$ is an ideal of definition of $B$, the morphism is said
to be adic.

######  Examples.

  1. (1)

Every morphism of discrete rings is adic.

  2. (2)

The identity $\mathbb{Z}\rightarrow\mathbb{Z}$ is not continuous if
$\mathbb{Z}$ is endowed with the $p$-adic topology on the left side and the
discrete topology on the right side.

  3. (3)

If $A$ is a discrete ring and $A[T]$ is endowed with the $T$-adic topology,
then the canonical morphism $A\rightarrow A[T]$ is continuous but not adic.

######  Proposition 1.3.

Let $u:A\rightarrow B$ be an adic morphism of Huber rings. If $A_{0}$ and
$B_{0}$ are rings of definition of $A$ and $B$ such that $u(A_{0})\subset
B_{0}$, then, for every ideal of definition $I_{0}$ of $A_{0}$, the ideal
$u(I_{0})B_{0}$ is an ideal of definition of $B$.

###### Proof.

We refer the reader to [Mor19, Proposition II.1.3.3]. ∎

######  Definition 1.4.

If $R\rightarrow A$ is a morphism of Huber rings, $A$ is said to be a Huber
$R$-algebra. If the morphism is adic $A$ is said to be a $R$-adic alegbra.

######  Remarks.

  1. (1)

An $R$-adic algebra is not necessarily an adic ring.

  2. (2)

Let $R$ be a Huber ring with $(A_{0},I_{0})$ as a couple of definition. If $M$
is a finite $A$-module, we can endow it with the universal topology for the
continuous $A$-linear morphisms $M\rightarrow N$ into topological $A$-modules.
In that case, every $A_{0}$-module $M_{0}$ of $M$ that generates $M$ is
necessarily open in $M$ and the topology of $M_{0}$ is the $I_{0}$-adic
topology.

Later, it will be useful to have a semi-norm on our rings to investigate
matters of convergence. For this reason we will need Tate rings. Those rings
are Huber rings whose topology can be defined by an ultrametric
submutiplicative norm.

######  Definition 1.5.

Let $A$ be a Huber ring. We say that $A$ is a _Tate ring_ if it has a
topologically nilpotent unit.

######  Examples.

  1. (1)

The ring $\mathbb{Q}$ endowed with the $p$-adic topology is a Tate ring.

  2. (2)

It is possible to endow every ring $A$ with the $p$-adic topology and in that
case $A\left[\frac{1}{p}\right]$ is a Tate ring.

  3. (3)

Every non-archimedean field is a Tate ring.

######  Proposition 1.6.

The topology of a Huber ring can always be defined by a submultiplicative
semi-norm. In the case of a Tate ring, the topology can be defined by a
submultiplicative norm.

###### Proof.

Let $A$ be a Huber ring and $(A_{0},I_{0})$ a couple of definition. For an
element $f$ of $A$ we set

$\lVert f\rVert=\begin{cases}e^{n}&\text{if }fI_{0}^{n}\subset A_{0}\text{ but
}fI_{0}^{n-1}\not\subset A_{0},\\\ e^{-n}&\text{if }f\in I_{0}^{n}\text{ but
}f\not\in I^{n+1},\\\ 0&\text{if }f\in\bigcap I_{0}^{n}.\end{cases}$

We then have, $A_{0}=\left\\{f\in A,\lVert f\rVert\leq 1\right\\}$ and
$I_{0}=\left\\{f\in A,\lVert f\rVert<1\right\\}$. ∎

######  Definition 1.7.

A submultiplicative norm $\lVert~{}\rVert$ on a ring $A$ is said to be
_contractive_ with respect to an endomorphism $\varphi$ of $A$ if $\forall
x\in A,~{}\lVert\varphi(x)\rVert\leq\lVert x\rVert$.

######  Definition 1.8.

Let $A$ be a topological ring. A subset $E$ of $A$ is

  1. (1)

bounded if for every neighbourhood $U$ of $0$ in $A$ there exists an open
neighbourhood $V$ of $0$ such that $ax\in U$ for every $a\in E$ and $x\in V$,

  2. (2)

power-bounded if $\bigcup_{n\geq 1}E(n)$ is bounded, where
$E(n)=\left\\{\prod\limits_{i=1}^{n}f_{i},~{}f_{i}\in E\right\\}$,

  3. (3)

topologically nilpotent if for every neighbourhood $W$ of $0$ in $A$, there
exists a positive integer $N$ such that $E(n)\subset W$ for $n\geq N$.

If $E$ is a singleton $\\{f\\}$ we say that $f$ is power-bounded (resp.
topologically nilpotent) if $E$ is.

######  Proposition 1.9.

Let $A$ be a topological ring and $E=\\{f_{1},\ldots,f_{r}\\}$ be a finite set
of $A$ then,

$E\text{ is power-bounded}\Leftrightarrow\forall
i\in\\{1,\ldots,r\\},f_{i}\text{ is power-bounded}.$

###### Proof.

Only the converse requires details. By definition, for every neighbourhood $U$
of $0$ in $A$ there exists a neighbourhood $W_{r}$ of $0$ in $A$ such that for
every $n_{r}\in\mathbb{N}$

$f_{r}^{n_{r}}W_{r}\subset U.$

By induction, there exists a sequence $(W_{1},\ldots,W_{r-1})$ of
neighbourhoods of $0$ in $A$ such that for every
$(n_{1},\ldots,n_{r-1})\in\mathbb{N}^{r-1}$

$\forall i=1,\ldots,r-1,~{}f_{i}^{n_{i}}W_{i}\subset W_{i+1}.$

Therefore,

$f_{1}^{n_{1}}\ldots f_{r}^{n_{r}}W_{1}\subset U.$

This completes the proof. ∎

######  Proposition 1.10.

Let $A$ be a Tate ring and $A_{0}$ be a ring of definition of $A$. Then
$A_{0}$ contain a topologically nilpotent unit $\pi$ of $A$ and
$A=A_{0}\left[\frac{1}{\pi}\right]$.

###### Proof.

For a topologically nilpotent unit $\pi$ of $A$ there exists $m\in\mathbb{N}$
such that $\pi^{m}\in A_{0}$. Every element $a$ of $A$ satisfies

$\lim_{n\to\infty}\pi^{n}a=0.$

Therefore, for some $n$ we have $\pi^{n}a\in A_{0}$. This gives $a\in
A_{0}\left[\frac{1}{\pi}\right]$. The converse is clear. ∎

The tensor product of two Huber rings doesn’t make sense in general. For the
need of extending the scalars on rings of series and polynomials, we introduce
here some conditions for the tensor product to exist in the category of Huber
rings.

Let $R$ be a Huber ring, $A$ and $B$ be two $R$-adic algebra. Choose a couple
of definition $(R_{0},I)$ of $R$ and rings of definition $A_{0}$ and $B_{0}$
of $A$ and $B$ such that the image of $R_{0}$ in $A$ is contained in $A_{0}$
and the image of $R_{0}$ in $B$ is contained in $B_{0}$. Let $J$ denote the
ideal of $\text{Im}(A_{0}\otimes_{R_{0}}B_{0}\rightarrow A\otimes_{R}B)$
generated by the image of $I$.

######  Proposition 1.11.

With the previous notations, if we endow
$\emph{Im}(A_{0}\otimes_{R_{0}}B_{0}\rightarrow A\otimes_{R}B)$ with the
$J$-adic topology and if we equip $A\otimes_{R}B$ with the unique structure of
topological group for which $\emph{Im}(A_{0}\otimes_{R_{0}}B_{0}\rightarrow
A\otimes_{R}B)$ is an open subgroup then $A\otimes_{R}B$ is a Huber ring with
$(\emph{Im}(A_{0}\otimes_{R_{0}}B_{0}\rightarrow A\otimes_{R}B),J)$ as a
couple of definition.

###### Proof.

We refer the reader to [Mor19, Proposition II.3.2.1]. ∎

Later, for matters of convergence we will need to consider complete
topological rings.

######  Proposition 1.12.

Let $A$ be a Huber ring and $(A_{0},I)$ be a couple of definition of $A$. Set
$\hat{A}=\varprojlim_{n\geq 0}A/I^{n}A_{0}$ (as an abelian group) Then:

  1. (1)

If we put the unique topology on $\hat{A}$ for which $\hat{A_{0}}$ is an open
subgroup, then $\hat{A}$ is a complete topological group.

  2. (2)

There is a unique ring structure on $\hat{A}$ that makes the canonical map
$A\rightarrow\hat{A}$ continuous, and $\hat{A}$ is a topological ring.

  3. (3)

The ring $\hat{A}$ is a Huber ring and $(\hat{A_{0}},I_{0}\hat{A_{0}})$ is a
couple of definition of $\hat{A}$. Moreover, the canonical map
$A\rightarrow\hat{A}$ is adic.

  4. (4)

The canonical map $\hat{A_{0}}\otimes_{A_{0}}A\rightarrow\hat{A}$ is an
isomorphism.

  5. (5)

For every natural integer $n$, the induced map
$\hat{A_{0}}/I^{n}\hat{A_{0}}\rightarrow A_{0}/I^{n}A_{0}$ is an isomorphism.

###### Proof.

We refer the reader to [Mor19, Theorem II.3.1.8, Corollary II.3.1.9]. ∎

######  Definition 1.13.

If $A$ is a Huber ring, the Huber ring $\hat{A}$ defined in proposition 1.12
is called the _completion_ of $A$.

######  Remarks.

  1. (1)

If $A$ is a complete Huber ring then every ring of definition of $A$ is also
complete: a ring of definition is closed in $A$ because it is an open subgroup
of $A$.

  2. (2)

A ring with the discrete topology is a complete adic ring. There exists a
fully faithful functor from the category of rings into the category of
complete adic rings.

One technique that we will use to extends the results obtained requires to
define the rational localization of a Huber ring.

######  Proposition 1.14 ([Hub93, Section 3, page 467]).

Let $A$ be a Huber ring. Let $f_{0},\ldots,f_{r}$ be elements of $A$ that
generate an open ideal of $A$. There exists a unique topological ring
$A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$ and continuous morphism
$A\rightarrow A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$ universal for
the following property: if $B$ is a Huber ring and $u:A\rightarrow B$ is a
continuous morphism such that

$u(f_{0})\in B^{\times}\text{ and }\forall
i=1,\ldots,r,~{}u(f_{i})u(f_{0})^{-1}\text{ is power-bounded in $B$},$

then, there exists a unique continuous morphism extending $u$ to
$A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$.

###### Proof.

We refer the reader to [Wed19, Proposition and Definition 5.51]. ∎

######  Remarks.

  1. (1)

The construction gives
$A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)=A\left[\frac{1}{f_{0}}\right]$
as rings. If $(A_{0},I_{0})$ is a couple of definition of $A$ then
$A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$ is a Huber ring admitting

$\left(A_{0}\left[\frac{f_{1}}{f_{0}},\ldots,\frac{f_{r}}{f_{0}}\right],I_{0}\left[\frac{f_{1}}{f_{0}},\ldots,\frac{f_{r}}{f_{0}}\right]\right)$

as a couple of definition. For more details we refer the reader to [Wed19,
Proposition 6.21]. By construction, the set
$\left\\{\frac{f_{i}}{f_{0}},~{}1\leq i\leq r\right\\}$ is power-bounded in
$A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$. The completion of
$A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$ will be denoted by
$A\left\\{\frac{f_{1},\ldots,f_{r}}{f_{0}}\right\\}$ or
$A\left\\{\underline{f}/f_{0}\right\\}$ when no confusion arises.

######  Definition 1.15.

Let $A$ be a Huber ring and $f_{0},\ldots,f_{r}$ be elements of $A$ that
generate an open ideal of $A$. We say that
$A\left\\{\underline{f}/f_{0}\right\\}$ is a _regular rational localization_
of $A$ if the morphism $A\rightarrow A\left\\{\underline{f}/f_{0}\right\\}$
preserves regular elements.

######  Example.

Let $K$ be a non-archimedean field. Denote by $\lVert\cdot\rVert$ its absolute
value. Consider two elements $a$ and $b$ of $K$ such that $0<\lVert
a\rVert\leq\lVert b\rVert$. We can consider the ring
$K[X]\left\\{\frac{X^{2},ab}{bX}\right\\}$. As a topological ring, we have

$K[X]\left\\{\frac{X^{2},ab}{bX}\right\\}=\left\\{\sum\limits_{n\in\mathbb{Z}}a_{n}X^{n},~{}a_{n}\in
K,~{}\begin{cases}\lVert a_{n}\rVert\lVert b\rVert^{n}\rightarrow 0\text{ when
}n\rightarrow+\infty\\\ \lVert a_{n}\rVert\lVert a\rVert^{n}\rightarrow
0\text{ when }n\rightarrow-\infty\\\ x\end{cases}\right\\}.$

This ring comes with the usual norm

$\left\lVert\sum a_{n}X^{n}\right\rVert:=\max\left\\{\lVert a_{n}\rVert\lVert
b\rVert^{n},n\geq 0;\lVert a_{n}\rVert\lVert a\rVert^{n},n\leq 0\right\\}.$

Choosing an endormophism $\sigma$ of
$K[X]\left\\{\frac{X^{2},ab}{bX}\right\\}$ is equivalent to choosing an
element $\sigma(X)$ such that $\lVert
a\rVert\leq\lVert\sigma(X)\rVert\leq\lVert b\rVert$. We recover the Robba ring
$\mathcal{R}$ defined in [ADV04] by setting

$\mathcal{R}_{K}=\varinjlim_{\lVert
a\rVert<1}\varprojlim_{\|b\rVert<1}K[X]\left\\{\frac{X^{2},ab}{bX}\right\\}$

##  2\. Topologically étale morphisms

We shall define topologically étale morphisms. In this class of morphisms we
will take a deeper look at good étale morphisms and rational localizations.
Those will be the two types of étale morphisms along which we will be able to
extend the coordinates we will work with.

######  Definition 2.1.

A morphism of Huber rings $\varphi:A\rightarrow B$ is topologically of finite
type (resp. of finite presentation) if it is adic and if there exists rings of
definition $A_{0}$ and $B_{0}$ of $A$ and $B$ such that $\varphi(A_{0})\subset
B_{0}$ and

  1. (1)

for every ideal of definition $I\subset A_{0}$, the induced map

$A_{0}/I\rightarrow B_{0}/\varphi(I)B_{0}$

is of finite type (resp. finite presentation),

  2. (2)

$A\otimes_{A_{0}}\widehat{B_{0}}\rightarrow\widehat{B}$ is of finite type
(resp. finite presentation).

######  Remark.

This definition is a variant of the definition 2.3.13 of [Abb10] that
specifies only to formal schemes, which corresponds to adic rings in the
affine case. The conditions here are weaker than in the algebraic version of
the definition 1.8.18 of [Abb10]. In the case where $A$ is a noetherian ring
those two definitions are equivalent.

######  Examples.

  1. (1)

For discrete rings, this definition is simply the classical algebraic notion
of finite type (resp. finite presentation).

  2. (2)

Let $\varphi:A\rightarrow B$ be a morphism of finite type (resp. finite
presentation) between two rings. For an ideal of finite type $I$ of $A$, it is
possible to put the $I$-adic topology on $A$ and $B$. Then, the morphism
$\varphi$ is topologically of finite type (resp. finite presentation).

  3. (3)

Let $A$ be a Huber ring and $f_{0},\ldots,f_{r}$ be elements of $A$ that
generate an open ideal, the morphism

$A\rightarrow A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$

is topologically of finite presentation. Indeed, by construction it is adic
and for every couple of definition $(A_{0},I)$ of $A$, the morphism

$A_{0}/I\rightarrow\left(A_{0}/I\right)\left[\frac{1}{f_{0}}\right]$

is of finite presentation. The second condition is automatically satisfied.

######  Proposition 2.2.

Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be two morphisms between adic
rings.

  1. (1)

If $f$ and $g$ are topologically of finite type (resp. finite presentation),
so is $g\circ f$.

  2. (2)

If $g\circ f$ is topologically of finite type and $g$ is adic then, $f$ is
topologically of finite type.

  3. (3)

If $g\circ f$ is topologically of finite presentation and $g$ is topologically
of finite type then, $f$ is topologically of finite presentation.

###### Proof.

This is the affine version of the proposition 2.3.18 of [Abb10]. ∎

######  Proposition 2.3.

A morphism between adic rings $A\rightarrow B$ is topologically of finite
presentation if and only if the completed morphism
$\widehat{A}\rightarrow\widehat{B}$ is topologically of finite presentation.

###### Proof.

Let us first consider the converse. If $\widehat{A}\rightarrow\widehat{B}$ is
topologically of finite presentation, then, for every ideal of definition $I$
of $A$, the ideal $I\widehat{A}$ is an ideal of definition of $\widehat{A}$ by
the proposition 1.12. The same proposition implies the following isomorphisms:

$\widehat{A}/I\widehat{A}\simeq A/I\text{ and }\widehat{B}/I\widehat{B}\simeq
B/IB.$

That ensures that for every ideal of definition $I$ of $A$, the morphism
$A/I\rightarrow B/I$ is of finite type. Let us now show the implication. If
$I^{\prime}$ is an ideal of definition of $\widehat{A}$ then for every ideal
of definition $I$ of $A$ there exists $n\in\mathbb{N}$ such that
$I^{n}\widehat{A}\subset I^{\prime}$. It is then enough to look at the
following commutative diagram

$\textstyle{\widehat{A}/I^{n}\widehat{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\widehat{B}/I^{n}\widehat{B}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\widehat{A}/I^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\widehat{B}/I^{\prime}}$

Using proposition 1.12 we can deduce that the above morphism is topologically
of finite presentation, implying than the one at the bottom is also of finite
presentation. The second condition can be easily checked. ∎

######  Proposition 2.4.

Let $\varphi:A\rightarrow B$ be a morphism between adic rings, the following
statements are equivalent:

  1. (1)

The morphism $\varphi$ is topologically of finite type (resp. finite
presentation).

  2. (2)

For every ideal of definition $I$ of $A$, the induced map

$A/I\rightarrow B/\varphi(I)B$

is of finite type (resp. finite presentation).

###### Proof.

The converse $(2)\Rightarrow(1)$ is clear. It remains to prove
$(1)\Rightarrow(2)$. Without loss of generality, we will assume that the rings
$A$ and $B$ are complete. By hypothesis, the morphism is adic. Let $A_{0}$ and
$B_{0}$ be rings of definition of $A$ and $B$ that satisfy the conditions of
the definition. For every ideal of definition $I\subset A_{0}$ of $A$, the
induced map

$A_{0}/I\rightarrow B_{0}/\varphi(I)B_{0}$

is of finite type. By extension of scalars and using the second condition of
the definition, we get that the map

$A/I\simeq A\otimes_{A_{0}}A_{0}/I\rightarrow
A\otimes_{A_{0}}B_{0}/\varphi(I)B_{0}\rightarrow B/\varphi(I)B$

is of finite type. ∎

######  Lemma 2.5.

Let $A$, $B$ and $A^{\prime}$ be adic rings, $A\rightarrow B$ be a topological
morphism of finite presentation and $A\rightarrow A^{\prime}$ and adic
morphism then, $A^{\prime}\rightarrow A^{\prime}\otimes_{A}B$ is topologically
of finite presentation.

###### Proof.

The morphism $A\rightarrow A^{\prime}$ being adic, there exists an ideal of
definition $J$ of $A$ such that the image of $J$ generates an ideal of
definition of $A^{\prime}$. Moreover, the induced map

$A/J\rightarrow B/JB$

is of finite presentation. It implies that the map

$A^{\prime}/JA^{\prime}\simeq
A^{\prime}/JA^{\prime}\otimes_{A/J}A/J\rightarrow
A^{\prime}/JA^{\prime}\otimes_{A/J}B/J\simeq(A^{\prime}\otimes_{A}B)/J(A^{\prime}\otimes_{A}B)$

is of finite presentation. ∎

######  Lemma 2.6.

Let $R$ be an adic ring, $A$, $A^{\prime}$, $B$ and $B^{\prime}$ be $R$-adic
algebras. If $A\rightarrow B$ and $A^{\prime}\rightarrow B^{\prime}$ are
topological morphisms of finite presentation, then
$A\otimes_{R}A^{\prime}\rightarrow B\otimes_{R}B^{\prime}$ is topologically of
finite presentation.

###### Proof.

By extension of scalars, using lemma 2.5, the morphisms

$A\otimes_{R}A^{\prime}\rightarrow B\otimes_{R}A^{\prime}\text{ and
}B\otimes_{R}A^{\prime}\rightarrow B\otimes_{R}B^{\prime}$

are topologically of finite presentation. Using proposition 2.2, with the
morphism $B\otimes_{R}A^{\prime}\rightarrow B\otimes_{R}B^{\prime}$, we obtain
that the morphism $A\otimes_{R}A^{\prime}\rightarrow B\otimes_{R}B^{\prime}$
is topologically of finite presentation. ∎

######  Definition 2.7.

A morphism of adic rings $A\rightarrow B$ is formally étale if for every
commutative diagram of adic rings

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C/\mathfrak{c}}$

where $C$ is a complete adic ring and $\mathfrak{c}$ is nilpotent, there
exists a unique morphism $B\rightarrow C$ making the diagram commutes. The
morphism $A\rightarrow B$ is topologically étale if it is formally étale and
topologically of finite presentation.

######  Examples.

  1. (1)

In the case of discrete rings, this is simply the classical algebraic notion
of étales morphisms.

  2. (2)

Let $\varphi:A\rightarrow B$ be an étale morphism between rings. For an ideal
of finite type $I$ of $A$, it is possible to put the $I$-adic topology on $A$
and $B$. The morphism $\varphi$ is then adic, topologically of finite type and
formally étale. It is topologically étale.

  3. (3)

Let $A$ be an adic ring and $f_{0},\ldots,f_{r}$ be elements of $A$ that
generate an open ideal, the morphism

$A\rightarrow A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$

is topologically étale. We have shown in the remark below the proposition 2.1
that this morphism is topologically of finite presentation. It is then enough
to prove that this morphism is formally étale. If we choose a commutative
diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C/\mathfrak{c}}$

where $C$ is a complete adic ring and $\mathfrak{c}$ is nilpotent, we can
easily check that $A\rightarrow C$ satisfies the necessary conditions to apply
the universal property of the rational localization, which allows us to obtain
a unique lifting from $A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$ to $C$
that makes the diagram commute.

######  Proposition 2.8.

Let $A$ and $B$ be two adic rings and $\varphi:A\rightarrow B$ be an adic
morphism. The following statements are equivalent:

  1. (1)

The morphism $\varphi$ is formally étale.

  2. (2)

For every diagram of adic rings

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C/\mathfrak{c}}$

where $C$ is a complete adic ring and $\mathfrak{c}$ is a topologically
nilpotent ideal, there exists a unique morphism $B\rightarrow C$ that makes
the diagram commute.

###### Proof.

The converse $(2)\Rightarrow(1)$ is clear as a nilpotent ideal is
topologically nilpotent. It remains to prove $(1)\Rightarrow(2)$. Let us
consider a diagram of adic rings

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C/\mathfrak{c}}$

where $C$ is a complete adic ring and $\mathfrak{c}$ is a topologically
nilpotent ideal. For every ideal of definition $I$ of $A$ and every natural
integer $k$ this diagram induces the following commutative diagram

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C/I^{k}C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C/I^{k}C/\mathfrak{c}.}$

The ideal $\mathfrak{c}$ is nilpotent in $C/I^{k}C$ because $I^{k}C$ is an
ideal of definition of $C$ by proposition 1.3. Moreover, $C/I^{k}C$ is
complete for the discrete topology. Therefore, we obtain a lifting
$B\rightarrow C/I^{k}C.$ Going to the completion, we obtain a morphism
$B\rightarrow C$ and we can easily check that it is adic. ∎

######  Proposition 2.9.

Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be two morphisms of adic rings.
If $f$ and $g$ are topologically étale then $g\circ f$ is topologically étale.

###### Proof.

We easily check that the composition of formally étale morphisms is formally
étale. The proposition can then be deduced from the proposition 2.2. ∎

######  Proposition 2.10.

Let $R$ be an adic ring, $\varphi:A\rightarrow C$ and $\psi:B\rightarrow D$ be
two morphisms between adic rings. If $A$, $B$, $C$ and $D$ are $R$-adic
algebras and the morphisms $\varphi$ and $\psi$ are topologically étale then
$\varphi\otimes_{R}\psi:A\otimes_{R}B\rightarrow C\otimes_{R}D$ is
topologically étale.

###### Proof.

Choose a diagram of Huber $R$-algebras

$\textstyle{A\otimes_{R}B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C\otimes_{R}D\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E/\mathfrak{e}}$

where $E$ is complete and $\mathfrak{e}$ is nilpotent. This diagram induces
the following diagrams

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E/\mathfrak{e}}$
$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{D\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E/\mathfrak{e}}$

By hypothesis there exist unique liftings $C\rightarrow E$ and $D\rightarrow
E$ that make those diagrams commute. By universal property there exists a
unique map $C\otimes_{R}D\rightarrow E$ that makes the first diagram commute.
The fact that the morphism is topologically of finite presentation follows
from proposition 2.6. ∎

######  Definition 2.11.

A morphism $\varphi:A\rightarrow B$ of Huber rings is said to be good étale if
there exist rings of definition $A_{0}$ and $B_{0}$ of $A$ and $B$ satisfying
$\varphi(A_{0})\subset B_{0}$ such that the induced morphism
$\varphi_{0}:A_{0}\rightarrow B_{0}$ is topologically étale and
$B=A\otimes_{A_{0}}B_{0}$. The triplet $(\varphi_{0},A_{0},B_{0})$ is said to
be a good model of $\varphi$.

######  Examples.

  1. (1)

In the case of discrete rings this is just the notion of étale morphism.

  2. (2)

For a prime number $p\neq 2$, the morphism

$\mathbb{Q}_{p}[T]\rightarrow\mathbb{Q}_{p}[X,Y]/(Y^{2}-X^{3}+X),~{}T\mapsto
X$

is not good étale, where the topology on both sides is the $p$-adic topology.

  3. (3)

Let $A$ be a Huber ring and $f_{0},\ldots,f_{r}$ be elements of $A$ that
generate an open ideal, the morphism

$A\rightarrow A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$

is good étale.

######  Proposition 2.12.

Let $A$ and $B$ be two adic rings and $\varphi:A\rightarrow B$ a morphism. If
$\varphi$ is good étale then it is formally étale.

###### Proof.

Let us consider a commutative diagram of adic rings

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C/\mathfrak{c}}$

with $C$ complete and $\mathfrak{c}$ nilpotent. For rings of definition
$A_{0}$ and $B_{0}$ of $A$ and $B$ satisfying the requirements of the
definition of a good étale morphism, we deduce the following diagram

$\textstyle{A_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{C/\mathfrak{c}.}$

By hypothesis, the exists a unique lifting $B_{0}\rightarrow C$ that can be
extended in a unique way using the fact that $B=A\otimes_{A_{0}}B_{0}$. ∎

######  Definition 2.13.

Let $A$ be a Huber ring. An endomorphism $\sigma$ of $A$ is said to be
_infinitesimal_ over a subset $V$ of $A$ if there exists a topologically
nilpotent subset $U$ of $A$ such that $(\sigma-\text{Id})(V)\subset U$ and
$\sigma(V)\subset V$.

######  Example.

Let $A$ be a Huber ring. Denote by $\lVert\cdot\rVert$ its submultiplicative
semi-norm and let $A_{0}$ be a ring of definition. For an element $q$ of $A$
such that $\lVert q-1\rVert<1$

$A[X]\rightarrow A[X],~{}X\mapsto qX$

is infinitesimal over $A_{0}[X]$.

######  Proposition 2.14.

Let $B$ be a complete Huber ring and $u:A\rightarrow B$ be a good étale
morphism of Huber rings with $(u_{0},A_{0},B_{0})$ as a good model. If
$\sigma_{A}$ is an endomorphism of $A$ infinitesimal over $A_{0}$, then,
$\sigma_{A}$ can be extended in an unique way to an endomorphism $\sigma_{B}$
of $B$ infinitesimal over $B_{0}$.

###### Proof.

Denote by $J$ the ideal of $A$ generated by $(\sigma_{A}-\text{Id})(A_{0})$.
The morphism $\sigma_{A}$ is infinitesimal over $A_{0}$, hence, $J\subset
A_{0}$. The ideal $JB_{0}$ is topologically nilpotent over $B_{0}$. To obtain
the lift it is enough to consider the diagram below

$\textstyle{A_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{u\circ\sigma_{A}}$$\scriptstyle{u}$$\textstyle{B_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{B_{0}/JB_{0}.}$

The morphism $A\rightarrow B$ is good étale, so, there exists a unique lift
$B_{0}\rightarrow B_{0}$. By extension of the scalars this morphism can be
extended in an endomorphism of $B$ that satisfies the desired properties. ∎

######  Definition 2.15.

Let $R$ be a Huber ring and $A$ be a Huber $R$-algebra. Elements
$x_{1},\ldots,x_{d}\in A$ are said to be _good étales coordinates_ if there
exists a good étale morphism

$\varphi:R[X_{1},\ldots,X_{d}]\rightarrow A,~{}X_{i}\mapsto x_{i}.$

######  Proposition 2.16.

Let $R$ be an adic ring, $A$ and $B$ be adic $R$-algebras that are adic rings.
If $A$ and $B$ are endowed with good étales coordinates étales
$x_{1},\ldots,x_{d}$ and $y_{1},\ldots,y_{e}$, then, $x_{1}\otimes
1,\ldots,x_{d}\otimes 1,1\otimes y_{1},\ldots,1\otimes y_{e}$ are good étale
coordinates over $A\otimes_{R}B$.

###### Proof.

This is a consequence of the proposition 2.10. ∎

##  3\. Reminders on filtered modules

In this section we recall some general results about filtered modules that
will be useful for our work.

######  Definition 3.1.

Let $A$ be a ring. A $A$-module $M$ is said to be _filtered_ if we can put an
exhaustive decreasing filtration indexed by natural integers
$\text{Fil}_{\bullet}M$ on it. For every $i\in N$, we denote by
$\text{Gr}_{i}M=\text{Fil}_{i}M/\text{Fil}_{i+1}M$ the associated graduation.

######  Proposition 3.2.

Let $A$ be a ring, $M$ and $N$ be two filtered $A$-modules. There exists a
natural filtration defined for $k\in\mathbb{N}$ by

$\emph{Fil}_{k}(M\otimes_{R}N)=\sum_{i+j=k}\emph{Im}\left(\emph{Fil}_{i}M\otimes\emph{Fil}_{j}N\rightarrow
M\otimes N\right).$

that gives rise to a natural morphism

$\bigoplus_{i+j=k}\emph{Gr}_{i}M\otimes\emph{Gr}_{j}N\rightarrow\emph{Gr}_{k}\left(M\otimes
N\right).$

Moreover, if for every $i,j\in\mathbb{N}$, $M/\emph{Fil}_{i}M$ and
$N/\emph{Fil}_{j}N$ are flat modules over $A$, this morphism is an
isomorphism.

###### Proof.

By definition, for $i+j=k$ fixed there exists a map

$\text{Fil}_{i}M\otimes\text{Fil}_{j}N\rightarrow M\otimes
N\rightarrow\text{Gr}_{k}(M\otimes N).$

The image of $\text{Fil}_{i+1}M\otimes\text{Fil}_{j}N$ is zero by this map,
and going to the quotient this gives a morphism

$\text{Gr}_{i}M\otimes\text{Fil}_{j}N\rightarrow\text{Gr}_{k}(M\otimes N).$

Noticing that the image of $\text{Gr}_{i}M\otimes\text{Fil}_{j+1}N$ is zero by
this morphism we obtain, by going to the quotient, a morphism

$\psi_{i,j}:\text{Gr}_{i}M\otimes\text{Gr}_{j}N\rightarrow\text{Gr}_{k}(M\otimes
N).$

Taking the direct sum, we have a morphism

$\bigoplus_{i+j=k}\text{Gr}_{i}M\otimes\text{Gr}_{j}N\rightarrow\text{Gr}_{k}(M\otimes
N)$

which is the desired morphism. Assume now that for every $i,j\in\mathbb{N}$,
$M/\text{Fil}_{i}M$ and $N/\text{Fil}_{j}N$ are flat modules over $A$. We want
to construct an inverse. For $i+j=k$ fixed, we can consider the composed
morphism

$\phi_{i,j}:\text{Fil}_{k}(M\otimes N)\hookrightarrow M\otimes N\rightarrow
M/\text{Fil}_{i+1}M\otimes N/\text{Fil}_{j+1}N.$

We can then compose with the surjection

$\rho_{1}:M/\text{Fil}_{i+1}M\otimes N/\text{Fil}_{j+1}N\rightarrow
M/\text{Fil}_{i}M\otimes N/\text{Fil}_{j+1}N$

This gives the zero map. Indeed, given an element in the image of $\phi_{i,j}$
it can be written under the form
$\sum_{v+u=k}\overline{m_{u}}\otimes\overline{n_{v}}$ with
$m_{u}\in\text{Fil}_{u}M$ and $n_{v}\in\text{Fil}_{v}N$. It is a simple matter
to check that either $u\geq i$ or $v\geq j+1$. Such an element is sent to 0 by
$\rho_{1}$. The image of $\phi_{i,j}$ is in the kernel of $\rho_{1}$. We can
then use the flatness of $N/\text{Fil}_{j+1}N$ to compute this kernel. By
definition the following sequence is exact

$0\rightarrow\text{Gr}_{i}M\rightarrow M/\text{Fil}_{i+1}M\rightarrow
M/\text{Fil}_{i}M\rightarrow 0.$

By flatness of $N/\text{Fil}_{j+1}N$, the sequence obtained by applying the
tensor product is also exact

$0\rightarrow\text{Gr}_{i}M\otimes N/\text{Fil}_{j+1}N\rightarrow
M/\text{Fil}_{i+1}M\otimes N/\text{Fil}_{j+1}N\rightarrow
M/\text{Fil}_{i}M\otimes N/\text{Fil}_{j+1}N\rightarrow 0.$

Then, $\text{ker}(\rho_{1})=\text{Gr}_{i}M\otimes N/\text{Fil}_{j+1}N$ and
there exists a morphism

$\phi_{i,j}^{\prime}:\text{Fil}_{k}(M\otimes
N)\rightarrow\text{Gr}_{i}M\otimes N/\text{Fil}_{j+1}N.$

Let us proceed in the same way and consider the surjection

$\rho_{2}:\text{Gr}_{i}M\otimes
N/\text{Fil}_{j+1}N\rightarrow\text{Gr}_{i}M\otimes N/\text{Fil}_{j}N.$

A similar argument as before gives a morphism

$\phi_{i,j}^{\prime\prime}:\text{Fil}_{k}(M\otimes
N)\rightarrow\text{Gr}_{i}M\otimes\text{Gr}_{j}N.$

The same reasoning on the indexes allows us to go to the quotient

$\text{Gr}_{k}\left(M\otimes
N\right)\rightarrow\text{Gr}_{i}M\otimes\text{Gr}_{j}N.$

Taking again the direct sum, we obtain a morphism

$\text{Gr}_{k}(M\otimes
N)\rightarrow\bigoplus_{i+j=k}\text{Gr}_{i}M\otimes\text{Gr}_{j}N.$

It is an inverse. ∎

######  Lemma 3.3.

Let $M$ and $N$ be two $A$-modules endowed, respectively, with exhaustive
decreasing filtrations $\emph{Fil}_{\bullet}M$ and $\emph{Fil}_{\bullet}N$. If
$M\rightarrow N$ is an isomorphism of filtered modules that induces for every
$i$ an isomorphism $\emph{Gr}_{i}M\rightarrow\emph{Gr}_{i}N$. Then, for every
$i\in\mathbb{N}$

$\emph{Fil}_{i}M\simeq\emph{Fil}_{i}N.$

###### Proof.

We proceed by induction. For $k=0$, the result is clearly true. Assume then
that the statement is true for some $k\in\mathbb{N}^{*}$. We have the
following commutative diagram

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\text{Fil}_{k+1}M\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\text{Fil}_{k}M\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\text{Gr}_{k}M\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\text{Fil}_{k+1}N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\text{Fil}_{k}N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\text{Gr}_{k}N\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

The center and right vertical morphisms are isomorphisms, so, by the Five
lemma $\text{Fil}_{k+1}M\rightarrow\text{Fil}_{k+1}N$ is an isomorphism. ∎

##  4\. Twisted principal parts in several variables

######  Definition 4.1.

Let $R$ be a Huber ring, $A$ an $R$-adic algebra and $d\in\mathbb{N}$. The
$R$-algebra $A$ is said to be _twisted of order $d$_ if it is endowed with a
sequence of $R$-linear continuous endomorphisms
$\underline{\sigma}=\left(\sigma_{A,1},\ldots,\sigma_{A,d}\right)$ that
commute.

Set $P_{A}=A\otimes_{R}A$. We extend $\sigma_{i}$ to $P_{A}$ by setting
$\sigma_{i}(a\otimes b)=\sigma_{i}(a)\otimes b$. If no confusion can arise, we
will simply write $\sigma_{i}$ instead of $\sigma_{A,i}$. The data of the ring
and the set of endomorphisms will be denoted under the form of a couple
$(A,\underline{\sigma})$.

######  Examples.

Let $R$ be a Huber ring and $A$ be an $R$-adic algebra. The following examples
are twisted algebras

  1. (1)

Let $h_{1},\ldots,h_{d}$ be elements of $A$, the ring $A[\underline{X}]$ with
$\underline{\sigma}=\left(\sigma_{1},\ldots,\sigma_{d}\right)$, defined by
$\sigma_{i}(X_{j})=X_{j}+\delta_{i,j}h_{i}$, is a twisted $R$-algebra of order
$d$. This is the situation of finite differences.

  2. (2)

Let $l\in\mathbb{N}$, the ring $A[\underline{X}]$ endowed with
$\underline{\sigma}=\left(\sigma_{1},\ldots,\sigma_{d}\right)$, defined by

$\forall a\in A,\forall
i,j=1,\ldots,d,~{}\sigma_{i}(aX_{j})=aX_{i}^{l^{\delta_{i,j}}},$

is a twisted $R$-algebra of order $d$. This is the case of Malher’s
differences.

  3. (3)

In the case of two variables $A=R[X_{1},X_{2}]$, we can look at the case
$\sigma_{1}=\sigma_{2}$, $\sigma_{1}\left(\sum
a_{i,j}X_{1}^{i}X_{2}^{j}\right)=\sum a_{i,j}X_{2}^{i+j}$ and
$\sigma_{2}\left(\sum a_{i,j}X_{1}^{i}X_{2}^{j}\right)=\sum
a_{i,j}X_{1}^{i+j}$.

######  Definition 4.2.

Let $R$ be a Huber ring, $(A,\underline{\sigma_{A}})$ and
$(B,\underline{\sigma_{B}})$ be twisted Huber $R$-algebras of order $d$. A
morphism of twisted Huber $R$-algebras of order $d$, $u:A\rightarrow B$ is a
morphism of Huber $R$-algebras such that

$\forall i=1,\ldots,d,~{}\sigma_{B,i}\circ u=u\circ\sigma_{A,i}.$

In what follows we consider a Huber ring $R$ and a twisted $R$-adic algebra
$(A,\underline{\sigma})$ of order $d$. We also consider some elements
$\underline{x}=(x_{1},\ldots,x_{d})$ of $A$ on which we will later put various
conditions. We will use the same notation $\sigma_{i}$ for the endomorphism of
the ring of polynomials $A\left[\underline{\xi}\right]$ defined by

$\forall
i,j\in\\{1,\ldots,d\\},~{}\sigma_{i}(\xi_{j})=\begin{cases}\xi_{i}+x_{i}-\sigma_{i}(x_{i}),&\text{if
}j=i\\\ \xi_{j},&\text{otherwise.}\end{cases}$

######  Definition 4.3.

For $n\in\mathbb{N}$, the $A$-module
$P_{A,(n)_{\underline{\sigma},\underline{x}}}$ of twisted principal parts of
order $n$ is the $A$-module defined by

$P_{A,(n)_{\underline{\sigma},\underline{x}}}:=A[\underline{\xi}]\Big{/}\left(\underline{\xi}^{(\underline{k})_{\underline{\sigma},\underline{x}}}\textrm{
such that }|\underline{k}|=n+1\right)$

where for $\underline{k}\in\mathbb{N}^{d}$

$\underline{\xi}^{(\underline{k})_{\underline{\sigma},\underline{x}}}:=\xi_{1}^{(k_{1})_{\sigma_{1}}}\ldots\xi_{d}^{(k_{d})_{\sigma_{d}}}\textrm{
and
}\xi_{i}^{(k_{i})_{\sigma_{i}}}:=\xi_{i}\sigma_{i}(\xi_{i})\ldots\sigma_{i}^{k_{i}-1}(\xi_{i}).$

This gives

(1)
$\underline{\xi}^{(\underline{k})_{\underline{\sigma},\underline{x}}}=\prod_{i=1}^{d}\prod_{j=0}^{k_{i}-1}\left(\xi_{i}+x_{i}-\sigma_{i}^{j}(x_{i})\right).$

In the case where $d=1$, this is the module specified by Le Stum and Quirós in
[LSQ18a, Definition 1.5].

######  Remark.

As remarked in the examples following the definition 1.1, the ring of
polynomials $A[\underline{\xi}]$ can be endowed with a structure of Huber
rings inducing the topology on $A$. The ring
$P_{A,(n)_{\underline{\sigma},\underline{x}}}$ endowed with the quotient
topology is also a Huber ring. Here and subsequently we will put this topology
on it.

######  Example.

When $A=R\left[X_{1},X_{2}\right]$, and
$i,j=1,2,~{}\sigma_{i}(X_{j})=q_{i}^{\delta_{i,j}}X_{j}$ where $q_{1}$ and
$q_{2}$ are elements of $R$, we have the following description

$P_{A,(n)_{\underline{\sigma},\\{X_{1},X_{2}\\}}}=A[\xi_{1},\xi_{2}]\Big{/}\left(\prod_{i=1}^{2}\prod_{j=0}^{k_{i}-1}(\xi_{i}+(1-q_{i}^{j})X_{i})\textrm{
such that }k_{1}+k_{2}=n+1\right).$

######  Proposition 4.4.

For every natural integer $n$, the composed morphism

$A[\underline{\xi}]_{\leq n}\hookrightarrow
A[\underline{\xi}]\twoheadrightarrow
P_{A,(n)_{\underline{\sigma},\underline{x}}}$

is an isomorphism.

###### Proof.

Consider the morphism of $A$-modules

(2) $\displaystyle A[\underline{\xi}]$ $\displaystyle\rightarrow
A[\underline{\xi}]$ $\displaystyle\underline{\xi}^{\underline{k}}$
$\displaystyle\mapsto\underline{\xi}^{(\underline{k})_{\underline{\sigma},\underline{x}}}.$

Formula 1 ensures that for all $\underline{k}\in\mathbb{N}^{d}$,
$\underline{\xi}^{(\underline{k})_{\underline{\sigma},\underline{x}}}$ is of
degree $|\underline{k}|$. Hence, the morphism 2 preserves the degree and
induces for every $n\in\mathbb{N}$ a morphism

$A[\underline{\xi}]_{\leq n}\rightarrow A[\underline{\xi}]_{\leq n},$

where $A[\underline{\xi}]_{\leq n}$ designates the set of polynomials of
degree less or equal to $n$. For $n=0$, this morphism is an isomorphism.
Proceeding by induction, we assume that it is an isomorphism for some natural
integer $n$. It induces a morphism

$A[\underline{\xi}]_{\leq n+1}/A[\underline{\xi}]_{\leq n}\rightarrow
A[\underline{\xi}]_{\leq n+1}/A[\underline{\xi}]_{\leq n}.$

This morphism is an isomorphism. Indeed, given
$\underline{k}=(k_{1},\ldots,k_{d})\in\mathbb{N}^{d}$ such that
$|\underline{k}|=n+1$ by formula 1 we have
$\underline{\xi}^{(\underline{k})_{\underline{\sigma},\underline{x}}}\equiv\xi_{1}^{k_{1}}\ldots\xi_{d}^{k_{d}}\text{
mod }A[\underline{\xi}]_{\leq n}$. The following diagram

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A[\underline{\xi}]_{\leq
n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A[\underline{\xi}]_{\leq
n+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A[\underline{\xi}]_{\leq
n+1}/A[\underline{\xi}]_{\leq
n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A[\underline{\xi}]_{\leq
n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A[\underline{\xi}]_{\leq
n+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A[\underline{\xi}]_{\leq
n+1}/A[\underline{\xi}]_{\leq
n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

allows us to conclude, using the Five lemma, that the morphism

$A[\underline{\xi}]_{\leq n+1}\rightarrow A[\underline{\xi}]_{\leq n+1}$

is an isomorphism. This establishes that the morphism 2 is an isomorphism. ∎

######  Remark.

The module $P_{A,(n)_{\underline{\sigma},\underline{x}}}$ is a free $A$-module
of basis
$\left\\{\underline{\xi}^{(\underline{k})_{\underline{\sigma},\underline{x}}}\right\\}_{|\underline{k}|\leq
n}.$

######  Proposition 4.5.

Let $u:A\rightarrow B$ be a morphism of twisted $R$-algebras of order $d$
then,

$B\otimes_{A}P_{A,(n)_{\underline{\sigma_{A}},\underline{x}}}\simeq
P_{B,(n)_{\underline{\sigma_{B}},\underline{u(x)}}}.$

###### Proof.

By definition,

$\displaystyle B\otimes_{A}P_{A,(n)_{\underline{\sigma_{A}},\underline{x}}}$
$\displaystyle=B\otimes_{A}A[\underline{\xi}]\Big{/}\left(\underline{\xi}^{(\underline{k})_{\underline{\sigma_{A}},\underline{x}}}\textrm{
such that }|\underline{k}|=n+1\right)$ $\displaystyle\simeq
B[\underline{\xi}]\Big{/}\left(\underline{\xi}^{(\underline{k})_{\underline{\sigma_{B}},\underline{u(x)}}}\textrm{
such that
}|\underline{k}|=n+1\right)=P_{B,(n)_{\underline{\sigma_{B}},\underline{u(x)}}}.\qed$

######  Definition 4.6.

The elements $x_{1},\ldots,x_{d}$ are called $\underline{\sigma}$-coordinates
over $A$ if, for every $n\in\mathbb{N}$, there exists a unique adic morphism
of $R$-algebras

$\displaystyle\Theta_{A,(n)_{\underline{\sigma},\underline{x}}}\colon A$
$\displaystyle\to P_{A,(n)_{\underline{\sigma},\underline{x}}}$ $\displaystyle
x_{i}$ $\displaystyle\mapsto x_{i}+\xi_{i}$

such that the composition of the projection on $A$ with
$\Theta_{A,(n)_{\underline{\sigma},\underline{x}}}$ is the identity. The
morphism $\Theta_{A,(n)_{\underline{\sigma},\underline{x}}}$ is called the
$n$-th Taylor morphism of $A$ with respect to $\underline{\sigma}$ and
$\underline{x}$.

######  Remark.

Let $\underline{q}=(q_{1},\ldots,q_{d})$ be elements of $R$. In the situation
where

$\forall
i,j\in\\{1,\ldots,d\\},\sigma_{i}(x_{j})=\begin{cases}q_{i}x_{i}&\text{ if
}i=j\\\ x_{j}&\text{ otherwise,}\end{cases}$

and $x_{1},\ldots,x_{d}$ are $\underline{\sigma}$-coordinates, we will say
that they are $\underline{q}$-coordinates to emphasize the link with the
elements $q_{1},\ldots,q_{d}$. In the case $d=1$, this is the situation
described in [ADV04].

######  Example.

When $\sigma_{1}=\ldots=\sigma_{d}=\text{Id}_{A}$ and there exists an étale
morphism

$R[\underline{X}]\rightarrow A,~{}X_{i}\mapsto x_{i},$

then, $x_{1},\ldots,x_{d}$ are $\underline{\sigma}$-coordinates over $A$.
Indeed, for every natural integer $n$ the following diagram is commutative

$\textstyle{R[\underline{X}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{X_{i}\mapsto
x_{i}+\xi_{i}}$$\scriptstyle{X_{i}\mapsto x_{i}}$
$\textstyle{P_{A,(n)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$
$\textstyle{P_{A,(n)_{\underline{\sigma},\underline{x}}}/(\xi_{1},\ldots,\xi_{d})}$

As we are in the case where $\sigma_{1}=\ldots=\sigma_{d}=\text{Id}_{A}$, the
elements $\xi_{i}$ are nilpotent elements of
$P_{A,(n)_{\underline{\sigma},\underline{x}}}$. The first vertical arrow being
formally étale, there exists a unique lift $A\rightarrow
P_{A,(n)_{\underline{\sigma},\underline{x}}}$ that satisfies the desired
property.

Let $x_{1},\ldots,x_{d}$ be $\underline{\sigma}$-coordinates over $A$.
Consider for $n\in\mathbb{N}$ the morphism

$\displaystyle\tilde{\Theta}_{A,(n)_{\underline{\sigma},\underline{x}}}:$
$\displaystyle P_{A}\rightarrow P_{A,(n)_{\underline{\sigma},\underline{x}}}$
$\displaystyle a\otimes b\mapsto
a\Theta_{A,(n)_{\underline{\sigma},\underline{x}}}(b)$

that extends $\Theta_{A,(n)_{\underline{\sigma},\underline{x}}}$ to $P_{A}$ by
$A$-linearity. For $i=1,\ldots,d,$

$\tilde{\Theta}_{A,(n)_{\underline{\sigma},\underline{x}}}(1\otimes
x_{i}-x_{i}\otimes 1)=\xi_{i}.$

The proposition 4.4 ensures that this morphism is surjective. We define

$I_{A}^{(n+1)_{{\underline{\sigma},\underline{x}}}}:=\textrm{ker}\left(\tilde{\Theta}_{A,(n)_{\underline{\sigma},\underline{x}}}\right).$

By the first isomorphism theorem

$P_{A}/I_{A}^{{(n+1)_{\underline{\sigma},\underline{x}}}}\simeq
P_{A,(n)_{\underline{\sigma},\underline{x}}}.$

######  Remarks.

  1. (1)

If we denote

$p_{2}^{*}:A\rightarrow P_{A},~{}f\mapsto 1\otimes f$

and by $\pi$ the projection $P_{A}\twoheadrightarrow
P_{A,(n)_{\underline{\sigma},\underline{x}}}$, then, by the uniqueness of
$\Theta_{A,(n)_{\underline{\sigma},\underline{x}}}$ the following diagram is
commutative

$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{2}^{*}}$$\scriptstyle{\Theta_{A,(n)_{\underline{\sigma},\underline{x}}}}$$\textstyle{P_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A,(n)_{\underline{\sigma},\underline{x}}}}$

  2. (2)

For every $n\in\mathbb{N}$, there exists, by factorization, a morphism

$\phi_{n}:P_{A,(n)_{\underline{\sigma},\underline{x}}}\rightarrow
P_{A,(n-1)_{\underline{\sigma},\underline{x}}}$

that makes the diagram commute

$\textstyle{A[\underline{\xi}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A,(n-1)_{\underline{\sigma},\underline{x}}}}$$\textstyle{P_{A,(n)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{n}}$

Using the uniqueness of
$\tilde{\Theta}_{A,(n-1)_{\underline{\sigma},\underline{x}}}$ we can easily
check that

$\phi_{n}\circ\tilde{\Theta}_{A,(n)_{\underline{\sigma},\underline{x}}}=\tilde{\Theta}_{A,(n-1)_{\underline{\sigma},\underline{x}}}.$

  3. (3)

The diagram below is commutative

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{A}^{(n)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A,(n-1)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{A}^{(n+1)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A,(n)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

######  Lemma 4.7.

Let $x_{1},\ldots,x_{d}$ be $\underline{\sigma}$-coordinates such that for
$i=1,\ldots,d,~{}\sigma_{i}(x_{i})\in R[x_{i}]$. For every $n,m\in\mathbb{N}$,
there exists a unique morphism of algebras

$\delta_{n,m}\colon P_{A,(n+m)_{\underline{\sigma},\underline{x}}}\rightarrow
P_{A,(n)_{\underline{\sigma},\underline{x}}}\otimes_{A}^{\prime}P_{A,(m)_{\underline{\sigma},\underline{x}}},~{}\xi_{i}\mapsto\xi_{i}\otimes
1+1\otimes\xi_{i}$

where $\otimes_{A}^{\prime}$ means that the action on the right is given by
$\Theta_{A,(n)_{\underline{\sigma},\underline{x}}}$.

###### Proof.

There exists a unique morphism of $A$-algebras

$\delta:A[\underline{\xi}]\rightarrow
P_{A,(n)_{\underline{\sigma},\underline{x}}}\otimes_{A}^{\prime}P_{A,(m)_{\underline{\sigma},\underline{x}}},~{}\xi_{i}\mapsto
1\otimes\xi_{i}+\xi_{i}\otimes 1.$

For $i=1,\ldots,d$ denote by $\mathfrak{a}_{i,n}$ the free sub-module of basis
$\left\\{\xi_{i}^{(k)_{\underline{\sigma},\underline{x}}},k\geq n\right\\}$.
Let us show by induction the following inclusion:

$\delta(\mathfrak{a}_{i,n})\subset\sum\limits_{j=0}^{n}\mathfrak{a}_{i,j}\otimes_{A}\mathfrak{a}_{i,n-j}.$

For $n=1$, it is clear. Assume that it holds for a certain $n\in\mathbb{N}$.
For $k\geq n+1$, we have:

$\delta\left(\xi_{i}^{(k)_{\underline{\sigma},\underline{x}}}\right)=\delta\left(\xi_{i}^{(n)_{\underline{\sigma},\underline{x}}}\right)\prod_{j=n}^{k-1}\delta\left(\sigma_{i}^{j}(\xi_{i})\right).$

Moreover, as ideals we have,
$\prod_{j=n}^{k-1}\delta\left(\sigma_{i}^{j}(\xi_{i})\right)\subset\delta\left(\sigma_{i}^{n}(\xi_{i})\right).$
Having disposed of this preliminary step in the induction, we can now
demonstrate that

$\delta\left(\sigma_{i}^{n}(\xi_{i})\right)\subset
P_{A,(n)_{\underline{\sigma},\underline{x}}}\otimes_{A}^{\prime}\sigma_{i}^{n-j}(\xi_{i})+\sigma_{i}^{j}(\xi_{i})\otimes_{A}^{\prime}P_{A,(m)_{\underline{\sigma},\underline{x}}}.$

As $\delta$ is $A$-linear and $\sigma_{i}(x_{i})\in R[x_{i}]$ it is enough to
verify this inclusion over $R[x_{i}]$, we will here identify $\xi_{i}$ to
$1\otimes x_{i}-x_{i}\otimes 1$ by the isomorphism
$P_{A}/I_{A}^{{(n+1)_{\underline{\sigma},\underline{x}}}}\simeq
P_{A,(n)_{\underline{\sigma},\underline{x}}}$. Proposition 3.3 of [LSQ18a]
allows us to obtain the inclusion. Hence,

$\displaystyle\delta\left(\xi_{i}^{(k)_{\underline{\sigma},\underline{x}}}\right)$
$\displaystyle\subset\sum\limits_{j=0}^{n}\left(\mathfrak{a}_{i,j}\otimes_{A}\mathfrak{a}_{i,n-j}\right)\left(P_{A,(n)_{\underline{\sigma},\underline{x}}}\otimes_{A}^{\prime}\sigma_{i}^{n-j}(\xi_{i})+\sigma_{i}^{j}(\xi_{i})\otimes_{A}^{\prime}P_{A,(m)_{\underline{\sigma},\underline{x}}}\right)$
$\displaystyle\subset\sum\limits_{j=0}^{n}\left(\mathfrak{a}_{i,j}\otimes_{A}\mathfrak{a}_{i,n-j}\sigma_{i}^{n-j}(\xi_{i})+\mathfrak{a}_{i,j}\sigma_{i}^{j}(\xi_{i})\otimes_{A}\mathfrak{a}_{i,n-j}\right)$
$\displaystyle\subset\sum\limits_{j=0}^{n}\left(\mathfrak{a}_{i,j}\otimes_{A}\mathfrak{a}_{i,n+1-j}+\mathfrak{a}_{i,j+1}\otimes_{A}\mathfrak{a}_{i,n-j}\right)$
$\displaystyle\subset\sum\limits_{j=0}^{n+1}\mathfrak{a}_{i,j}\otimes_{A}\mathfrak{a}_{i,n+1-j}.$

Denote by $\mathfrak{a}_{n}$ the free sub-module of base
$\left\\{\underline{\xi}^{(\underline{k})_{\underline{\sigma},\underline{x}}},|\underline{k}|\geq
n\right\\}$. The task is now to prove that

$\delta(\mathfrak{a}_{n})\subset\sum\limits_{j=0}^{n}\mathfrak{a}_{j}\otimes_{A}\mathfrak{a}_{n-j}.$

The equality
$\mathfrak{a}_{n}=\sum_{|\underline{k}|=n}\prod_{i=1}^{d}\mathfrak{a}_{i,k_{i}}$
gives

$\displaystyle\delta(\mathfrak{a}_{n})$
$\displaystyle=\delta\left(\sum_{|\underline{k}|=n}\prod_{i=1}^{d}\mathfrak{a}_{i,k_{i}}\right)$
$\displaystyle\subset\sum_{|\underline{k}|=n}\prod_{i=1}^{d}\delta(\mathfrak{a}_{i,k_{i}})$
$\displaystyle\subset\sum_{|\underline{k}|=n}\prod_{i=1}^{d}\sum\limits_{j=0}^{k_{i}}\mathfrak{a}_{k_{i},j}\otimes_{A}\mathfrak{a}_{i,k_{i}-j}$
$\displaystyle\subset\sum_{|\underline{k}|=n}\sum_{|\underline{j}|\leq|\underline{k}|}\prod_{i=1}^{d}(\mathfrak{a}_{i,j_{i}}\otimes_{A}\mathfrak{a}_{i,k_{i}-j_{i}})$
$\displaystyle\subset\sum_{|\underline{k}|=n}\sum_{|\underline{j}|\leq|\underline{k}|}\left(\prod_{i=1}^{d}\mathfrak{a}_{i,j_{i}}\otimes_{A}\prod_{i=1}^{d}\mathfrak{a}_{i,k_{i}-j_{i}}\right)$
$\displaystyle\subset\sum_{|\underline{k}|=n}\sum_{|\underline{j}|\leq|\underline{k}|}\left(\mathfrak{a}_{|\underline{j}|}\otimes_{A}\mathfrak{a}_{|\underline{k-j}|}\right)$
$\displaystyle\subset\sum_{j=0}^{n}\mathfrak{a}_{j}\otimes_{A}\mathfrak{a}_{n-j}$

For $0\leq i\leq n+m+1$, either $i>n$ and in that case
$\mathfrak{a}_{i}\subset\mathfrak{a}_{n+1}$ or $i\leq n$ and in that case
$m+n+1-i>m$ and so $\mathfrak{a}_{n+m+1-i}\subset\mathfrak{a}_{m+1}$. Hence,

$\delta(\mathfrak{a}_{n+m+1})\subset
A[\underline{\xi}]\otimes_{A}\mathfrak{a}_{n+1}+\mathfrak{a}_{m+1}\otimes_{A}A[\underline{\xi}].$

Going to the quotient, we obtain the map we wanted.∎

######  Remark.

The map $\delta_{n,m}$ is automatically continuous.

If $A$ and $B$ are two $R$-adic algebras such that $B$ is endowed with
continuous $R$-linear endomorphisms $\tau_{1},\ldots,\tau_{e}$ and
$\underline{\tau}$-coordinates $y_{1},\ldots,y_{e}$ then, it is possible to
define an endomorphism $(\sigma\cup\tau)_{i}$ on $A\otimes_{R}B$ by setting

$(\sigma\cup\tau)_{i}(a\otimes b)=\begin{cases}\sigma_{i}(a)\otimes b&\text{if
}1\leq i\leq d\\\ a\otimes\tau_{i-d}(b)&\text{otherwise.}\end{cases}$

As before, we will use the same notation for the endomorphism
$(\sigma\cup\tau)_{i}$ of $(A\otimes_{R}B)[\xi_{1},\ldots,\xi_{d+e}]$ defined
by

$(\sigma\cup\tau)_{i}(\xi_{j})=\begin{cases}\xi_{i}+x_{i}\otimes
1-\sigma_{i}(x_{i})\otimes 1&\text{if }i=j\text{ and }1\leq i\leq d\\\
\xi_{i}+1\otimes y_{i}-1\otimes\tau_{i-d}(y_{i-d})&\text{if }i=j\text{ and
}d+1\leq i\leq d+e\\\ \xi_{j}&\text{otherwise.}\end{cases}$

For the next two propositions, we assume that $(A,\underline{\sigma})$ is a
twisted $R$-adic algebra of order $d$ endowed with
$\underline{\sigma}$-coordinates $x_{1},\ldots,x_{d}$

######  Proposition 4.8.

If $(B,\underline{\tau})$ is a twisted $R$-adic algebra of order $e$ endowed
with $\underline{\tau}$-coordinates $y_{1},\ldots,y_{e}$ then, $x_{1}\otimes
1,\ldots,x_{d}\otimes 1,1\otimes y_{1},\ldots 1\otimes y_{e}$ are
$\underline{\sigma\cup\tau}$-coordinates over $A\otimes_{R}B$.

###### Proof.

It is here enough to apply the universal property of the tensor product. We
have the following morphisms:

$A\xrightarrow{\Theta_{A,(n)_{\underline{\sigma},\underline{x}}}}P_{A,(n)_{\underline{\sigma},\underline{x}}}\rightarrow
P_{A\otimes_{R}B,(n)_{\underline{\sigma\cup\tau},\underline{x\otimes
1},\underline{1\otimes y}}}$

and

$B\xrightarrow{\Theta_{B,(n)_{\underline{\sigma},\underline{y}}}}P_{B,(n)_{\underline{\sigma},\underline{y}}}\rightarrow
P_{A\otimes_{R}B,(n)_{\underline{\sigma\cup\tau},\underline{x\otimes
1},\underline{1\otimes y}}}.$

We can construct a bilinear map

$A\times B\rightarrow
P_{A\otimes_{R}B,(n)_{\underline{\sigma\cup\tau},\underline{x\otimes
1},\underline{1\otimes y}}}.$

Hence, there exists a unique map

$A\otimes_{R}B\rightarrow
P_{A\otimes_{R}B,(n)_{\underline{\sigma\cup\tau},\underline{x\otimes
1},\underline{1\otimes y}}}.$

satisfying the desired properties. ∎

######  Proposition 4.9.

Let $(B,\underline{\tau})$ be a twisted $R$-adic algebra of order $e$ endowed
with $\underline{\tau}$-coordinates $y_{1},\ldots,y_{e}$. If $A$ and $B$ are
flat over $R$ then, for every $n\in\mathbb{N}$

$I_{A\otimes_{R}B}^{(n)_{\underline{\sigma\cup\tau},\underline{x\otimes
1,1\otimes
y}}}\simeq\sum\limits_{i+j=n}I_{A}^{(i)_{\underline{\sigma},\underline{x}}}\otimes
I_{B}^{(j)_{\underline{\tau},\underline{y}}}.$

###### Proof.

We can put an exhaustive decreasing filtration over $P_{A}$: for
$n\in\mathbb{N}$, we set
$\text{Fil}_{n}P_{A}=I_{A}^{(n)_{\underline{\sigma},\underline{x}}}$.
Moreover, the next diagram is commutative and his rows and columns are exact

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\oplus_{|\underline{k}|=n}A\underline{\xi}^{(\underline{k})_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{A}^{(n+1)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A,(n)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I_{A}^{(n)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A,(n-1)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{I_{A}^{(n)_{\underline{\sigma},\underline{x}}}/I_{A}^{(n+1)_{\underline{\sigma},\underline{x}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

By the Snake lemma,
$\oplus_{|\underline{k}|=n}A\underline{\xi}^{(\underline{k})_{\underline{\sigma},\underline{x}}}\simeq
I_{A}^{(n)_{\underline{\sigma},\underline{x}}}/I_{A}^{(n+1)_{\underline{\sigma},\underline{x}}}=\text{Gr}_{n}P_{A}$.
In the same way consider the exhaustive decreasing filtration
$\text{Fil}_{n}P_{B}=I_{B}^{(n)_{\underline{\tau},\underline{y}}}$ and its
associated graduation
$\text{Gr}_{n}P_{B}\simeq\oplus_{|\underline{k}|=n}B\underline{\xi}^{(\underline{k})_{\underline{\tau},\underline{y}}}$.
The isomorphism

$P_{A}\otimes_{R}P_{B}\rightarrow P_{A\otimes_{R}B}$

induces an isomorphism

$\text{Gr}_{k}\left(P_{A}\otimes_{R}P_{B}\right)\simeq\text{Gr}_{k}P_{A\otimes_{R}B}.$

Indeed, $P_{A}/I_{A}^{(i)_{\underline{\sigma},\underline{x}}}$ is a free
$A$-module and $P_{B}/I_{B}^{(j)_{\underline{\tau},\underline{y}}}$ is a free
$B$-module. As $A$ and $B$ are flat over $R$, we get that
$P_{A}/I_{A}^{(i)_{\underline{\sigma},\underline{x}}}$ and
$P_{B}/I_{B}^{(j)_{\underline{\tau},\underline{y}}}$ are flat over $R$.
Propositions 3.2 and 3.3 allow to conclude. ∎

##  5\. Coordinates

In this section, we study the requirements for preserving the property of our
coordinates when we apply a good étale morphism or a rational localization to
them.

We write here the hypothesis that will be used along this section: $R$ is a
Huber ring and $(A,\underline{\sigma})$ is a twisted $R$-adic algebra of order
$d$ (definition 4.1). From now on, to lighten the notations, coordinates
$\underline{x}$ will be forgotten in the indexes.

######  Proposition 5.1.

Assume that $A$ is complete. Let $x_{1},\ldots,x_{d}$ be good étale
coordinates over $A$ and $(\varphi_{0},R_{0},A_{0})$ be the associated good
model. If $\forall i=1,\ldots,d$, $\sigma_{i}$ is infinitesimal over $A_{0}$
then, $x_{1},\ldots,x_{d}$ are $\underline{\sigma}$-coordinates over $A$.

###### Proof.

By definition, there exists a good étale morphism $R[\underline{X}]\rightarrow
A$. We apply here the same strategy as in the proof of proposition 2.14. The
induced morphism $\sigma_{0}:R_{0}[\underline{X}]\rightarrow A_{0}$ is
topologically étale. For every $1\leq i\leq d$,
$(\sigma_{i}-\text{Id})(A_{0})$ is topologically nilpotent. Consider the
diagram below

$\textstyle{R_{0}[\underline{X}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{X_{i}\mapsto\xi_{i}+x_{i}}$$\textstyle{P_{A_{0},(n)_{\underline{\sigma}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{A_{0},(n)_{\underline{\sigma}}}/\left(\left(\xi_{1},\ldots,\xi_{d}\right)+J\right)}$

where $J=\left(\sum_{i=1}^{d}(\sigma_{i}-\text{Id})(A_{0})\right)$. As the
morphism $R[\underline{X}]\rightarrow A$ is good étale it gives us a unique
lifting $A_{0}\rightarrow P_{A,(n)_{\underline{\sigma}}}$ that can be extended
using that $A=A_{0}\otimes_{R_{0}}R$. ∎

######  Definition 5.2.

The elements $x_{1},\ldots,x_{d}$ are said to be _good $\sigma$-coordinates_
if they are good étale coordinates and $\underline{\sigma}$-coordinates.

######  Proposition 5.3.

Let $u:A\rightarrow B$ be a good étale morphism of complete Huber $R$-algebras
and $x_{1},\ldots,x_{d}$ be good $\underline{\sigma}$-coordinates, with
$(\varphi_{0},R_{0},A_{0})$ for associated good model, such that there exists
a ring of definition $B_{0}$ of $B$ such that $(u_{|A_{0}},A_{0},B_{0})$ is a
good model for $u$. If for $i=1,\ldots,d,$ the endomorphism $\sigma_{i}$ is
infinitesimal over $A_{0}$, then, it extends to $B$ in a unique way in an
endomorphism $\sigma_{i,B}$. Moreover, $u(x_{1}),\ldots,u(x_{d})$ are good
$\underline{\sigma_{B}}$-coordinates over $B$.

###### Proof.

The result can be easily deduced from the propositions 5.1 and 2.14. ∎

######  Definition 5.4.

Elements $x_{1},\ldots,x_{d}$ of $A$ are said to be _kind_ (resp. _strong_)
coordinates with respect to $\underline{\sigma}$ if for every natural integer
$1\leq i\leq d$ and $0\leq k<n$: $\sigma_{i}^{n}(x_{i})-\sigma^{k}_{i}(x_{i})$
is not a zero divisor (resp. is invertible) in $A$.

######  Example.

In the case where $\sigma_{i}(x_{j})=q_{i}^{\delta_{i,j}}x_{i}$ with $q_{i}\in
R$, the coordinates are kind if and only if $q_{i}^{k}-1$ is regular in $R$.
This is equivalent to requiring that $q_{i}$ is not a root of unity. The
coordinates will be strong when $q_{i}^{k},~{}q_{i}^{n-k}-1$ are invertible
elements of $R$ and $x_{1},\ldots,x_{d}$ are invertible elements of $A$. This
relates to the situation described in [ADV04], where the considered ring $R$
is a field $K$ of characteristic $0$, complete with respect to a non-
archimedean absolute value $|\cdot|$e, with residue field of characteristic
$p>0$ and for an interval $I\subset\mathbb{R}_{>0}$ the $R$-algebra twisted of
order $1$ is

$\mathcal{A}_{K}(I)=\left\\{\sum_{n\in\mathbb{Z}}a_{n}x^{n},a_{n}\in
K,\lim_{\pm\infty}|a_{n}|\rho^{n}=0~{}\forall\rho\in I\right\\}$

endowed with the endomorphism

$\displaystyle\sigma_{q}\colon\mathcal{A}_{K}(I)$
$\displaystyle\to\mathcal{A}_{K}(I)$ $\displaystyle f(x)$
$\displaystyle\mapsto f(qx),$

where $q$ is an element of $K$ such that $|q-1|<1$ and $q$ is not a root of
unity.

For $n\in\mathbb{N}$, we write
$\alpha^{n}_{d}:=\\#\left\\{\underline{k}\in\mathbb{N}^{d},~{}|\underline{k}|\leq
n\right\\}=\binom{n+d}{d}$

######  Proposition 5.5.

If $A$ is endowed with kind (resp. strong) coordinates $x_{1},\ldots,x_{d}$
then, for every natural integer $n$, there exists an injective (resp.
bijective) morphism of $A$-algebras

$\pi_{n}:P_{A,(n)_{\underline{\sigma}}}\rightarrow A^{\alpha^{n}_{d}},$

which compounds are given by

$\pi_{\underline{k}}:P_{A,(n)_{\underline{\sigma}}}\rightarrow
A,~{}\xi_{i}\rightarrow\sigma_{i}^{k_{i}}(x_{i})-x_{i},$

for $\underline{k}\in\mathbb{N}^{d}$ such that $|\underline{k}|\leq n$.

###### Proof.

A direct computation gives

$\pi_{\underline{k}}\left(\sigma_{i}^{j}\left(\xi_{i}\right)\right)=\pi_{\underline{k}}\left(\xi_{i}-\sigma_{i}^{j}(x_{i})+x_{i}\right)=\sigma_{i}^{k_{i}}(x_{i})-\sigma_{i}^{j}(x_{i}),$

which is zero if $k_{i}=j$ and regular (resp. invertible) otherwise. We can
deduce that

$\pi_{\underline{k}}\left(\underline{\xi}^{(\underline{l})_{\underline{\sigma}}}\right)=\prod_{i=1}^{d}\prod_{0\leq
j_{i}<l_{i}}(\sigma_{i}^{k_{i}}(x_{i})-\sigma_{i}^{j_{i}}(x_{i}))$

is zero if there exists $i$ such that $k_{i}<l_{i}$, and regular (resp.
invertible) otherwise. Hence,

$\pi_{\underline{k}}\left(\underline{\xi}^{(\underline{l})_{\underline{\sigma}}}\right)=0\text{
if }\underline{k}<_{\text{lex}}\underline{l}$

and

$\pi_{\underline{k}}\left(\underline{\xi}^{(\underline{l})_{\underline{\sigma}}}\right)\in
A^{\text{reg}}~{}(\text{resp.}A^{\times})\text{ si
}\underline{k}=\underline{l}.$

Then, if we consider $P_{A,(n)_{\underline{\sigma}}}$ with basis
$\left\\{\underline{\xi}^{(\underline{l})_{\underline{\sigma}}}\right\\}_{|\underline{l}|\leq
n}$ and $A^{\alpha^{n}_{d}}$ with the canonical basis, as $A$-algebras, the
matrix of $\pi_{n}$ express in those bases, indexed with the lexicographic
order, is lower triangular with regular (resp. invertible) elements on the
diagonal. Thus , we have $\text{det}(\pi_{n})\in A^{\text{reg}}$ (resp.
$A^{\times})$. ∎

######  Remark.

For every $\underline{k}\in\mathbb{N}^{d}$, the morphisms
$\pi_{\underline{k}}$ are finite morphisms as morphisms of finite
$A$-algebras.

######  Definition 5.6.

Elements $x_{1},\ldots,x_{d}$ of $A$ are said to be _transverse coordinates_
with respect to $\underline{\sigma}$ if for every $i\neq j,$
$\sigma_{i}(x_{j})=x_{j}$.

######  Example.

In [ADV04] the variable $x$ is clearly a transverse coordinate.

######  Lemma 5.7.

Let $x_{1},\ldots,x_{d}$ be transverse coordinates over $A$. For every natural
integer $n$ and every $\underline{k}\in\mathbb{N}^{d}$ such that
$|\underline{k}|\leq n$, the diagram below commutes

$\textstyle{R[\underline{X}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{X_{i}\mapsto
x_{i}}$$\scriptstyle{X_{i}\mapsto\xi_{i}+x_{i}}$$\textstyle{P_{A,(n)_{\underline{\sigma}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{\underline{k}}}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\underline{\sigma}^{\underline{k}}}$$\textstyle{A}$

###### Proof.

For $\underline{k}\in\mathbb{N}^{d}$, as the coordinates are transverse we
have

$\pi_{\underline{k}}(x_{i}+\xi_{i})=x_{i}+(\sigma^{k_{i}}_{i}(x_{i})-x_{i})=\sigma^{k_{i}}_{i}(x_{i})=\underline{\sigma}^{\underline{k}}(x_{i}).\qed$

######  Proposition 5.8.

Assume that $A$ is complete. Let $x_{1},\ldots,x_{d}$ be good étale
coordinates associated to a good model $(\varphi_{0},R_{0},A_{0})$. Assume
that for $i=1,\ldots,d$, the endomorphism $\sigma_{i}$ is infinitesimal over
$A_{0}$, then, for $i=1,\ldots,d,$ then endomorphism $\sigma_{i}$ is uniquely
determined by its restriction to $R[\underline{X}]$: if $\tau_{i}$ is a
continuous $R$-linear endomorphism of $A$ such that
$\tau_{i_{|R[\underline{X}]}}=\sigma_{i_{|R[\underline{X}]}}$ then
$\tau_{i}=\sigma_{i}$.

###### Proof.

Given the following diagram

$\textstyle{R_{0}[\underline{X}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{X_{i}\mapsto\sigma_{i}(x_{i})}$$\scriptstyle{X_{i}\mapsto
x_{i}}$$\textstyle{A_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A_{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A_{0}/(\sigma_{i}-\text{Id})(A_{0})}$

where the left vertical arrow sends $X_{j}$ on $\sigma_{i}(x_{j})$. The first
horizontal morphism being supposed formally étale, there exists a unique
lifting from $A_{0}$ to $A_{0}$ which is equal to $\sigma_{i}$ over
$R_{0}[\underline{X}]$. This lifting can be extended to $A$ using the fact
that $A=A_{0}\otimes_{R_{0}}R$ and we can now check that it is $\sigma_{i}$. ∎

The following lemma is a general result over commutative rings.

######  Lemma 5.9.

Let $A\rightarrow B$ be an injective morphism between commutative rings then
$A^{\times}=A\cap B^{\times}$.

###### Proof.

This comes from the fact that $\text{Spec}(B)\rightarrow\text{Spec}(A)$
([Aut23, Chapter 10, Lemma 10.36.17]) and the fact that a function is
invertible if and only if it doesn’t have any zeros. ∎

######  Lemma 5.10.

Let $A\rightarrow B$ be an injective morphism of Huber algebras. If the
induced morphism $A^{0}\rightarrow B^{0}$ is integral then, $A^{0}=A\cap
B^{0}$.

###### Proof.

If $f\in A\cap B^{0}$ then, in particular, $f\in B^{0}$. and as
$A^{0}\rightarrow B^{0}$ is integral, it implies that $f$ is integral over
$A^{0}$. However, as $f\in A$ and $A^{0}$ is integrally closed in $A$,
necessarily $f\in A^{0}$. ∎

######  Lemma 5.11.

Let $f_{0},\ldots,f_{r}$ be elements of $A$ that generate an open ideal of
$A$. If $A$ is complete, $\forall i=1,\ldots,d,$ the endomorphism $\sigma_{i}$
extends to $A\left\\{\underline{f}/f_{0}\right\\}$ and for every natural
integer $n$

$\Theta_{A,(n)_{\underline{\sigma}}}(f_{0})\in
P_{A,(n)_{\underline{\sigma}}}\left(\underline{f}/f_{0}\right)^{\times}\text{
and }$ $\forall
i\in\\{1,\ldots,r\\},~{}\Theta_{A,(n)_{\underline{\sigma}}}(f_{i})/\Theta_{A,(n)_{\underline{\sigma}}}(f_{0})\in
P_{A,(n)_{\underline{\sigma}}}\left(\underline{f}/f_{0}\right)^{0},$

then, $x_{1},\ldots,x_{d}$ are $\underline{\sigma}$-coordinates over
$A\left\\{\underline{f}/f_{0}\right\\}$.

###### Proof.

It follows directly from the fact that we want to check that the endomorphism

$A\rightarrow P_{A,(n)_{\underline{\sigma}}}\rightarrow
P_{A,(n)_{\underline{\sigma}}}\left(\underline{f}/f_{0}\right)\rightarrow
P_{A,(n)_{\underline{\sigma}}}\left\\{\underline{f}/f_{0}\right\\}$

can be extended to $A\left\\{\underline{f}/f_{0}\right\\}$. ∎

######  Proposition 5.12.

Suppose that $A$ is complete. Let $x_{1},\ldots,x_{d}$ be kind and transverse
good $\underline{\sigma}$-coordinates associated to a good model
$(R_{0},A_{0})$. Assume that for $i=1,\ldots,d$, the endomorphism $\sigma_{i}$
is infinitesimal over $A_{0}$. Let $f_{0},\ldots,f_{r}$ be elements of $A$
that generate an open ideal of $A$. If $A\left\\{\underline{f}/f_{0}\right\\}$
is a regular rational localization of $A$ and $\forall i=1,\ldots,d$

$\sigma_{i}(f_{0})\in A\left(\underline{f}/f_{0}\right)^{\times}\emph{ and
}\forall j\in\\{1,\ldots,r\\},~{}\sigma_{i}(f_{j})/\sigma_{i}(f_{0})\in
A\left(\underline{f}/f_{0}\right)^{0},$

then, $\underline{\sigma}$ extends to $A\left\\{\underline{f}/f_{0}\right\\}$
and $x_{1},\ldots,x_{d}$ are $\underline{\sigma}$-coordinates over
$A\left\\{\underline{f}/f_{0}\right\\}$.

###### Proof.

By universal property and continuity, for every $i=1,\ldots,d$, the
endomorphism $\sigma_{i}$ extends in an endomorphism of
$A\left\\{\underline{f}/f_{0}\right\\}$. The coordinates $x_{1},\ldots,x_{d}$
are kind over $A\left\\{\underline{f}/f_{0}\right\\}$ because the morphism
$A\rightarrow A\left\\{\underline{f}/f_{0}\right\\}$ preserves the regular
elements. By universal property of $A\left\\{\underline{f}/f_{0}\right\\}$
they are also transverse. By lemma 5.11, the coordinates $x_{1},\ldots,x_{d}$
are $\underline{\sigma}$-coordinates if and only if for every natural integer
$n$

$\displaystyle\Theta_{A,(n)_{\underline{\sigma}}}(f_{0})\in
P_{A,(n)_{\underline{\sigma}}}\left(\underline{f}/f_{0}\right)^{\times}\text{
and }$ $\displaystyle\forall i=1,\ldots,r\text{ on a
}\Theta_{A,(n)_{\underline{\sigma}}}(f_{i})/\Theta_{A,(n)_{\underline{\sigma}}}(f_{0})\in
P_{A,(n)_{\underline{\sigma}}}\left(\underline{f}/f_{0}\right)^{0}.$

Thanks to lemmas 5.9 and 5.10 this is equivalent to ask that

$\displaystyle\left(\pi_{n}\circ\Theta_{A,(n)_{\underline{\sigma}}}\right)(f_{0})\in\left(A\left(\underline{f}/f_{0}\right)^{{d+n}\choose{n}}\right)^{\times}$
$\displaystyle\text{ and }\forall
i\in\\{1,\ldots,r\\},\left(\pi_{n}\circ\Theta_{A,(n)_{\underline{\sigma}}}\right)(f_{i})/\left(\pi_{n}\circ\Theta_{A,(n)_{\underline{\sigma}}}\right)(f_{0})\in\left(A\left(\underline{f}/f_{0}\right)^{{d+n}\choose{n}}\right)^{0}.$

We now apply lemmas 5.7 and 5.8 to obtain that for $|\underline{k}|\leq n$ the
diagram below commutes

|  |
---|---|---
$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Theta_{A,(n)_{\underline{\sigma}}}}$$\textstyle{P_{A,(n)_{\underline{\sigma}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{\underline{k}}}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\left(\underline{f}/f_{0}\right)}$$\textstyle{R[\underline{X}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma_{\underline{k}}}$

and check that our hypotheses are sufficient to obtain the result. ∎

######  Corollary 5.13.

Moreover,

$A\left\\{\underline{f}/f_{0}\right\\}\otimes_{A}P_{A,(n)_{\underline{\sigma}}}\simeq
P_{A\left\\{\underline{f}/f_{0}\right\\},(n)_{\underline{\sigma}}}.$

###### Proof.

By proposition 5.12 the endomorphisms $\underline{\sigma}$ extend to
$A\left\\{\underline{f}/f_{0}\right\\}$ and $x_{1},\ldots,x_{d}$ are
$\underline{\sigma}$-coordinates over $A\left\\{\underline{f}/f_{0}\right\\}$.
It suffices to apply proposition 4.5 to conclude. ∎

##  6\. Twisted differential forms

We keep the hypothesis of the previous section: $R$ is a Huber ring and
$(A,\underline{\sigma})$ is a twisted $R$-adic algebra of order $d$
(definition 4.1) Here and subsequently we assume that there exists some
elements $\underline{x}=(x_{1},\ldots,x_{d})$ of $A$ that are
$\underline{\sigma}$-coordinates (definition 4.6).

######  Definition 6.1.

The $A$-module of _twisted differential forms_ of $A$ over $R$ is

$\Omega^{1}_{A,\underline{\sigma}}:=I_{A}^{(1)_{\underline{\sigma}}}/I_{A}^{(2)_{\underline{\sigma}}}.$

######  Remarks.

  1. (1)

The isomorphism obtained by the Snake lemma in proposition 4.9 gives that
$\Omega^{1}_{A,\underline{\sigma}}$ is a free module of rank $d$ admitting
$(1\otimes x_{i}-x_{i}\otimes 1)_{i=1,\ldots,d}$ as a basis. The proof of this
proposition also gives that for $i=1,\ldots,r,$ the element $\xi_{i}$ of
$P_{A,(1)_{\underline{\sigma}}}$ can be identified with the class of $1\otimes
x_{i}-x_{i}\otimes 1$ in $P_{A}/I_{A}^{(2)_{\underline{\sigma}}}$.

  2. (2)

With the notations of the proof of the proposition 4.7, we have

$\Omega^{1}_{A,\underline{\sigma}}=\mathfrak{a}_{1}/\mathfrak{a}_{2}.$

  3. (3)

The isomorphism $P_{A,(1)_{\underline{\sigma}}}\simeq
P_{A}/I_{A}^{(2)_{\underline{\sigma}}}$ allows us to identify
$\Omega^{1}_{A,\underline{\sigma}}$ with an ideal of
$P_{A,(1)_{\underline{\sigma}}}$.

######  Proposition 6.2.

The following sequence of $A$-modules,

$0\rightarrow\Omega^{1}_{A,\underline{\sigma}}\rightarrow
P_{A,(1)_{\underline{\sigma}}}\rightarrow A\rightarrow 0$

where the first morphism $\Omega^{1}_{A,\underline{\sigma}}\rightarrow
P_{A,(1)_{\underline{\sigma}}}$ is the inclusion and the morphism
$P_{A,(1)_{\underline{\sigma}}}\rightarrow A$ is the multiplication, is a
split exact sequence.

###### Proof.

The fact that the sequence is exact comes from the construction. To show that
it is split exact, it is enough to consider $A\rightarrow
P_{A}/I_{A}^{(2)_{\underline{\sigma}}}\simeq P_{A,(1)_{\underline{\sigma}}}$
that sends an element $a$ to $a\otimes 1$. ∎

######  Proposition 6.3.

Let $A$ and $B$ be two twisted $R$-algebras of order $d$ and $u:A\rightarrow
B$ a morphism of twisted $R$-algebras of order $d$. Assume that
$u(x_{1}),\ldots,u(x_{d})$ are $\underline{\sigma_{B}}$-coordinates, then,
there exists an isomorphism

$B\otimes_{A}\Omega^{1}_{A,\underline{\sigma_{A}}}\simeq\Omega^{1}_{B,\underline{\sigma_{B}}}.$

###### Proof.

Applying the tensor product, using propositions 4.5 and 6.2 we get

$0\rightarrow B\otimes_{A}\Omega^{1}_{A,\underline{\sigma_{A}}}\rightarrow
P_{B,(1)_{\underline{\sigma_{B}}}}\rightarrow B\rightarrow 0.\qed$

Let us recall the following general result.

######  Lemma 6.4.

Let $f:A\rightarrow B$ and $g:C\rightarrow D$ be surjective morphisms of
$R$-modules. The following equality is satisfied

$\emph{ker}(f\otimes
g)=\emph{Im}\left(A\otimes\emph{ker}(g)+\emph{ker}(f)\otimes C\rightarrow
A\otimes_{R}C\right).$

######  Proposition 6.5.

Let $A$ and $B$ be two twisted $R$-algebras of order $d$ . Assume $A$ and $B$
are flat over $R$ and there exists a morphism of twisted $R$-algebras of order
$d$, $u:A\rightarrow B$. If $u(x_{1}),\ldots,u(x_{d})$ are
$\underline{\sigma_{B}}$-coordinates then, there exists a natural isomorphism

$\Omega^{1}_{A,{\underline{\sigma}}}\otimes_{R}B\oplus
A\otimes_{R}\Omega^{1}_{B,{\underline{\tau}}}\simeq\Omega^{1}_{A\otimes_{R}B,{\underline{\sigma\cup\tau}}}.$

###### Proof.

Let
$\pi_{A}:I_{A}^{(1)_{\underline{\sigma}}}\rightarrow\Omega^{1}_{A,{\underline{\sigma}}}$
and
$\pi_{B}:I_{B}^{(1)_{\underline{\tau}}}\rightarrow\Omega^{1}_{B,{\underline{\tau}}}$
be the canonical surjections. Consider the map

$\textstyle{\text{Gr}_{1}P_{A\otimes_{R}B}\simeq
I_{A}^{(1)_{\underline{\sigma}}}\otimes_{R}P_{B}+P_{A}\otimes_{R}I_{A}^{(1)_{\underline{\tau}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{A}\otimes\Delta_{B}+\Delta_{A}\otimes\pi_{B}}$$\textstyle{\Omega^{1}_{A,{\underline{\sigma}}}\otimes_{R}B\oplus
A\otimes_{R}\Omega^{1}_{B,{\underline{\tau}}}}$

which, by proposition 6.4, admits as kernel

$I_{A}^{(2)_{\underline{\sigma}}}\otimes_{R}P_{B}+P_{A}\otimes_{R}I_{B}^{(2)_{\underline{\tau}}}+I_{A}^{(1)_{\underline{\sigma}}}\otimes_{R}I_{B}^{(1)_{\underline{\tau}}}$

By proposition 4.9 we get

$\displaystyle\Omega^{1}_{A\otimes_{R}B,{\underline{\sigma\cup\tau}}}$
$\displaystyle=I_{A\otimes_{R}B}^{(1)_{\underline{\sigma\cup\tau}}}/I_{A\otimes_{R}B}^{(2)_{\underline{\sigma\cup\tau}}}$
$\displaystyle\simeq\left(I_{A}^{(1)_{\underline{\sigma}}}\otimes_{R}P_{B}+P_{A}\otimes_{R}I_{A}^{(1)_{\underline{\tau}}}\right)/\left(I_{A}^{(2)_{\underline{\sigma}}}\otimes_{R}P_{B}+P_{A}\otimes_{R}I_{B}^{(2)_{\underline{\tau}}}+I_{A}^{(1)_{\underline{\sigma}}}\otimes_{R}I_{B}^{(1)_{\underline{\tau}}}\right)$
$\displaystyle\simeq\Omega^{1}_{A,{\underline{\sigma}}}\otimes_{R}B\oplus
A\otimes_{R}\Omega^{1}_{B,{\underline{\tau}}}.\qed$

##  7\. Derivations

We keep the hypothesis of the previous section: $R$ is a Huber ring,
$(A,\underline{\sigma})$ is a twisted $R$-adic algebra of order $d$
(definition 4.1) and there exists some elements
$\underline{x}=(x_{1},\ldots,x_{d})$ of $A$ that are
$\underline{\sigma}$-coordinates (definition 4.6).

######  Definition 7.1.

Let $1\leq i\leq d$. We recall from [LSQ18b] that a _$\sigma_{i}$ -derivation_
$D$ of $A$ is an $R$-linear morphism from $A$ into an $A$-module $M$ that
verifies the twisted Leibniz rule:

$\forall x,y\in A,~{}D(xy)=xD(y)+\sigma_{i}(y)D(x).$

######  Remark.

Denote by $(e_{i})_{i=1,\ldots,d}:=((1\otimes x_{i}-x_{i}\otimes
1)^{*})_{i=1,\ldots,d}$ the basis of
$\textrm{Hom}_{A}\left(\Omega^{1}_{A,\underline{\sigma}},A\right)$ dual to the
basis $(1\otimes x_{i}-x_{i}\otimes 1)_{i=1,\ldots,d}$ of
$\Omega^{1}_{A,\underline{\sigma}}$. Set

$\text{d}:A\rightarrow\Omega^{1}_{A,\underline{\sigma}},~{}f\mapsto\Theta_{A,(1)_{\underline{\sigma}}}(f)-f.$

This map permits to define

$\forall
i\in\\{1\ldots,d\\},~{}\partial_{\underline{\sigma},i}:=e_{i}\circ\text{d}$

We will denote by $\text{Der}_{\underline{\sigma}}(A,A)$ the $A$-module
generated by
$\partial_{\underline{\sigma},1}\ldots,\partial_{\underline{\sigma},d}$. This
module is free over $A$: given $a_{1},\ldots,a_{d}$ some elements of $A$ such
that

$\sum_{i=1}^{d}a_{i}\partial_{\underline{\sigma},i}=0$

We can compute the value of this sum in $x_{1},\ldots,x_{d}$, which gives
$a_{1}=\ldots=a_{d}=0.$ Write
$\text{T}_{A,\underline{\sigma}}:=\text{Der}_{\underline{\sigma}}(A,A)$ and
$\text{T}_{A,\underline{\sigma}}\text{-Mod}(A)$ the category of $A$-modules
endowed with an action by the derivations.

######  Lemma 7.2.

The following formula is satisfied

$\forall f,g\in
A,~{}\partial_{\underline{\sigma},i}(fg)=f\partial_{\underline{\sigma},i}(g)+(g+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(g)))\partial_{\underline{\sigma},i}(f).$

###### Proof.

By definition of the maps
$\partial_{\underline{\sigma},1},\ldots,\partial_{\underline{\sigma},d}$ we
have a ring morphism

$\Theta_{A,(1)_{\underline{\sigma}}}:A\rightarrow
P_{A,(1)_{\underline{\sigma}}},~{}f\mapsto
f+\sum\limits^{d}_{i=1}\partial_{\underline{\sigma},i}(f)\xi_{i}.$

In $P_{A,(1)_{\underline{\sigma}}}$ for $i\neq j$, $\xi_{i}\xi_{j}=0$ and
$\sigma_{i}\left(\xi_{i}\right)\xi_{i}=0$ which implies

$\xi_{i}(\xi_{i}+x_{i}-\sigma_{i}(x_{i}))=0$

Hence,

$\xi_{i}^{2}=(\sigma_{i}(x_{i})-x_{i})\xi_{i}.$

Now for $f,g\in A$, on one side we have
$\Theta_{A,(1)_{\underline{\sigma}}}(fg)=fg+\sum\limits^{d}_{i=1}\partial_{\underline{\sigma},i}(fg)\xi_{i}$
and on the other side

$\displaystyle\Theta_{A,(1)_{\underline{\sigma}}}(fg)$
$\displaystyle=\Theta_{A,(1)_{\underline{\sigma}}}(f)\Theta_{A,(1)_{\underline{\sigma}}}(g)$
$\displaystyle=\left(f+\sum\limits^{d}_{i=1}\partial_{\underline{\sigma},i}(f)\xi_{i}\right)\left(g+\sum\limits^{d}_{i=1}\partial_{\underline{\sigma},i}(g)\xi_{i}\right)$
$\displaystyle=fg+g\sum\limits^{d}_{i=1}\partial_{\underline{\sigma},i}(f)\xi_{i}+f\sum\limits^{d}_{i=1}\partial_{\underline{\sigma},i}(g)\xi_{i}+\sum\limits_{i,j=1}^{d}\partial_{\underline{\sigma},i}(f)\partial_{\underline{\sigma},j}(g)\xi_{i}\xi_{j}$
$\displaystyle=fg+\sum\limits^{d}_{i=1}(f\partial_{\underline{\sigma},i}(g)+g\partial_{\underline{\sigma},i}(f))\xi_{i}+\sum\limits_{i=1}^{d}\partial_{\underline{\sigma},i}(g)\partial_{\underline{\sigma},i}(f)(\sigma_{i}(x_{i})-x_{i})\xi_{i}$
$\displaystyle=fg+\sum\limits_{i=1}^{d}(f\partial_{\underline{\sigma},i}(g)+(g+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(g)))\partial_{\underline{\sigma},i}(f))\xi_{i}$

We get the formula by identification. ∎

######  Remark.

For $1\leq i\leq d$, the map $\partial_{\underline{\sigma},i}$ is
automatically continuous as a compound of
$\Theta_{A,(1)_{\underline{\sigma}}}$ which is continuous by hypothesis.

Unlike the situation in [LSQ20], it is not direct that the canonical maps that
we defined are $\underline{\sigma}$-derivations. We have to add a hypothesis
on our coordinates for it to be the case.

######  Definition 7.3.

The $\underline{\sigma}$-coordinates $x_{1},\ldots,x_{d}$ are _classical_ if

$\forall i=1\ldots,d,~{}\forall f\in
A,~{}\sigma_{i}(f)=f+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f).$

######  Remark.

When $\sigma_{i}(x_{i})-x_{i}$ is invertible, we recover the formula

$\partial_{\underline{\sigma},i}(f)=\frac{\sigma_{i}(f)-f}{\sigma_{i}(x_{i})-x_{i}}.$

######  Theorem 7.4.

The $\underline{\sigma}$-coordinates $x_{1},\ldots,x_{d}$ are classical
coordinates if and only if $i\in\\{1,\ldots,d\\},$
$\partial_{\underline{\sigma},i}$ is a $\sigma_{i}$-derivation.

###### Proof.

This follows from lemma 7.2. ∎

######  Proposition 7.5.

Assume that A is complete. If $x_{1},\ldots,x_{d}$ are good étale coordinates
over $A$ associated to a good model $(u_{0},R_{0},A_{0})$ such that $\forall
i=1,\ldots,d,~{}\sigma_{i}$ is infinitesimal over $A_{0}$ then,
$x_{1},\ldots,x_{d}$ are classical coordinates.

###### Proof.

Set

$\forall i=1,\ldots,d,\forall f\in
A,~{}\sigma_{i}^{\prime}(f)=f+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f).$

We start by showing that $\sigma_{i}^{\prime}$ is a morphism of rings: for
$f,g\in A$

$\displaystyle\sigma_{i}^{\prime}(fg)$
$\displaystyle=fg+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(fg)$
$\displaystyle=fg+(\sigma_{i}(x_{i})-x_{i})(f\partial_{\underline{\sigma},i}(g)+\sigma_{i}^{\prime}(g)\partial_{\underline{\sigma},i}(f))$
$\displaystyle=f(g+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(g))+(\sigma_{i}(x_{i})-x_{i})\sigma_{i}^{\prime}(g)\partial_{\underline{\sigma},i}(f)$
$\displaystyle=(f+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f))\sigma_{i}^{\prime}(g)$
$\displaystyle=\sigma_{i}^{\prime}(f)\sigma_{i}^{\prime}(g).$

We here use the formula of lemma 7.2. We easily check that

$\sigma_{i|_{R_{0}[\underline{X}]}}=\sigma_{i|_{R_{0}[\underline{X}]}}^{\prime}$

and that $\sigma_{i}^{\prime}=\sigma_{i}\text{ mod
}\left(\left(\sigma_{i}-\text{Id})(A_{0}\right)\right)$. We can then apply
proposition 5.8 to obtain that $\sigma_{i}=\sigma_{i}^{\prime}$. ∎

The upcoming definition comes from classical calculus.

######  Definition 7.6.

The twisted $R$-algebra of order $d,$ $A$ endowed with
$\underline{\partial}=\left(\partial_{\underline{\sigma},1},\ldots,\partial_{\underline{\sigma},d}\right)$
is said to be _twisted of order $d$ of type Schwarz_ if the following
equalities are satisfied

$\forall
i,j=1,\ldots,d,~{}\partial_{\underline{\sigma},i}\circ\partial_{\underline{\sigma},j}=\partial_{\underline{\sigma},j}\circ\partial_{\underline{\sigma},i}\textrm{
and }\sigma_{i}\circ\sigma_{j}=\sigma_{j}\circ\sigma_{i},$ $\forall i\neq
j,~{}\sigma_{i}\circ\partial_{\underline{\sigma},j}=\partial_{\underline{\sigma},j}\circ\sigma_{i}.$

######  Proposition 7.7.

If the $\underline{\sigma}$-coordinates $x_{1},\ldots,x_{d}$ are classical and
$\forall i=1,\ldots,d,$ $\sigma_{i}(x_{i})\in R[x_{i}]$ then, for every $i\neq
j,$ $\partial_{\underline{\sigma},j}(\sigma_{i}(x_{i}))=0$.

###### Proof.

Set $\sigma_{i}(x_{i})=\sum_{k=1}^{m}a_{k}x_{i}^{k}$ with $a_{k}\in R$. We do
a direct computation using the $R$-linearity and the twisted Leibniz rule

$\displaystyle\partial_{\underline{\sigma},j}(\sigma_{i}(x_{i}))$
$\displaystyle=\partial_{\underline{\sigma},j}\left(\sum_{k=1}^{m}a_{k}x_{i}^{k}\right)$
$\displaystyle=\sum_{k=1}^{m}a_{k}\partial_{\underline{\sigma},j}(x_{i}^{k})$

We then proceed by induction to show that for every $k$ and every $i\neq j$,
$\partial_{\underline{\sigma},j}(x_{i}^{k})=0$. For $k=0$, it is the classical
trick

$\partial_{\underline{\sigma},i}(1)=\partial_{\underline{\sigma},i}(1\cdot
1)=2\partial_{\underline{\sigma},i}(1),$

where $\partial_{\underline{\sigma},i}(1)=0$. Assume the property to be true
for some integer $k$. As the coordinates are assumed to be classical, we have

$\partial_{\underline{\sigma},j}(x_{i}^{k+1})=\partial_{\underline{\sigma},j}(x_{i}^{k}\cdot
x_{i})=x_{i}^{k}\partial_{\underline{\sigma},j}(x_{i})+\sigma_{j}(x_{i})\partial_{\underline{\sigma},j}(x_{i}^{k})=0.\qed$

######  Proposition 7.8.

If the $\underline{\sigma}$-coordinates $x_{1},\ldots,x_{d}$ are classical and
$\forall i=1,\ldots,d,$ $\sigma_{i}(x_{i})\in R[x_{i}]$ then $A$ is a twisted
$R$-algebra of order $d$ of type Schwarz.

###### Proof.

The endomorphisms $\sigma_{i}$ commute by hypothesis. Let us show that the
derivatives commute. The diagram below is commutative

|  |  |
---|---|---|---
$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$d$\textstyle{P_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta}$$\scriptstyle{\tilde{\Theta}_{A,(2)_{\underline{\sigma}}}}$$\scriptstyle{p_{1}}$$\scriptstyle{p_{2}}$$\textstyle{P_{A}\otimes_{A}^{\prime}P_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tilde{\Theta}_{A,(1)_{\underline{\sigma}}}\otimes\tilde{\Theta}_{A,(1)_{\underline{\sigma}}}}$$\textstyle{A\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}}$$\textstyle{P_{A,(2)_{\underline{\sigma}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p_{1}^{\prime}}$$\scriptstyle{p_{2}^{\prime}}$$\scriptstyle{\delta_{1,1}}$$\textstyle{P_{A,(1)_{\underline{\sigma}}}\otimes_{A}^{\prime}P_{A,(1)_{\underline{\sigma}}}}$

where we have set

$\text{d}:A\rightarrow P_{A},~{}a\mapsto 1\otimes a-a\otimes 1,$
$p_{1}:P_{A}\rightarrow P_{A}\otimes_{A}^{\prime}P_{A},~{}a\otimes b\mapsto
a\otimes b\otimes 1\otimes 1$ $\text{and }p_{2}:P_{A}\rightarrow
P_{A}\otimes_{A}^{\prime}P_{A},~{}a\otimes b\mapsto 1\otimes 1\otimes a\otimes
b.$

And by analogy $p_{2}^{\prime}$ and $p_{1}^{\prime}$ are the morphisms
associated to the morphism $P_{A,(2)_{\underline{\sigma}}}\rightarrow
P_{A,(1)_{\underline{\sigma}}}$. Above, $\otimes_{A}^{\prime}$ denotes that
the action and the right is given by $p_{1}$ and the action on the left is
given by $p_{2}$. Below, $\otimes_{A}^{\prime}$ indicates that the action on
the right is given by $\Theta_{A,(2)_{\underline{\sigma}}}$. Let us now set

$d_{1}\colon=p_{2}-\delta+p_{1}\text{ and
}d_{2}\colon=p_{2}^{\prime}-\delta_{1,1}+p_{1}^{\prime}.$

We can easily check that $d_{1}\circ d=0$ and hence that
$d_{2}\circ\left(\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}\right)=0$.
Therefore, for every $f\in A$

$\delta_{1,1}\left(\left(\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}\right)(f)\right)=p_{1}^{\prime}\left(\left(\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}\right)(f)\right)+p_{2}^{\prime}\left(\left(\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}\right)(f)\right).$

Using the morphism $P_{A,(2)_{\underline{\sigma}}}\rightarrow
P_{A,(1)_{\underline{\sigma}}}$ we deduce that the element
$\left(\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}\right)(f)$ can be written
under the form

$\sum\limits_{i}\partial_{\underline{\sigma},i}(f)\xi_{i}+\sum\limits_{i<j}g_{i,j}\xi_{i}\xi_{j}+\sum\limits_{i}g_{i}\xi_{i}^{(2)_{\underline{\sigma}}}$

where $g_{i}$ and $g_{i,j}$ are elements of $A$. Thus,

$\delta_{1,1}\left(\left(\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}\right)(f)\right)=\delta_{1,1}\left(\sum\limits_{i}\partial_{\underline{\sigma},i}(f)\xi_{i}+\sum\limits_{i<j}g_{i,j}\xi_{i}\xi_{j}+\sum\limits_{i}g_{i}\xi_{i}^{(2)_{\underline{\sigma}}}\right).$

Using that
$\xi_{i}^{(2)_{\underline{\sigma}}}=\xi_{i}(\xi_{i}+x_{i}-\sigma_{i}(x_{i}))=0$
in $P_{A,(1)_{\underline{\sigma}}}$ we get:

$\displaystyle\delta_{1,1}\left(\xi_{i}^{(2)_{\underline{\sigma}}}\right)$
$\displaystyle=\delta_{1,1}(\xi_{i}(\xi_{i}+x_{i}-\sigma_{i}(x_{i})))=\delta_{1,1}(\xi_{i}^{2}+(x_{i}-\sigma_{i})\xi_{i})$
$\displaystyle=(1\otimes^{\prime}\xi_{i}+\xi_{i}\otimes^{\prime}1)^{2}+(x_{i}-\sigma_{i}(x_{i}))\xi_{i}\otimes^{\prime}1+(x_{i}-\sigma_{i}(x_{i}))\otimes^{\prime}\xi_{i}$
$\displaystyle=\xi_{i}^{2}\otimes^{\prime}1+2(\xi_{i}\otimes^{\prime}\xi_{i})+1\otimes^{\prime}\xi_{i}^{2}+(x_{i}-\sigma_{i}(x_{i}))\xi_{i}\otimes
1+(x_{i}-\sigma_{i}(x_{i}))\otimes^{\prime}\xi_{i}$
$\displaystyle=-(x_{i}-\sigma_{i}(x_{i}))\xi_{i}\otimes^{\prime}1+(2\xi_{i})\otimes^{\prime}\xi_{i}+1\otimes^{\prime}\xi_{i}^{2}+(x_{i}-\sigma_{i}(x_{i}))\xi_{i}\otimes
1$ $\displaystyle\hskip
14.22636pt+(x_{i}-\sigma_{i}(x_{i}))\otimes^{\prime}\xi_{i}$
$\displaystyle=(2\xi_{i})\otimes^{\prime}\xi_{i}-1\otimes^{\prime}(x_{i}-\sigma_{i}(x_{i}))\xi_{i}+(x_{i}-\sigma_{i}(x_{i}))\otimes^{\prime}\xi_{i}$
$\displaystyle=(2\xi_{i})\otimes^{\prime}\xi_{i}-\text{d}(x_{i}-\sigma_{i}(x_{i}))\otimes^{\prime}\xi_{i}$
$\displaystyle=(2\xi_{i})\otimes^{\prime}\xi_{i}-\xi_{i}\otimes^{\prime}\xi_{i}+\text{d}(\sigma_{i}(x_{i}))\otimes^{\prime}\xi_{i}$
$\displaystyle=(\xi_{i}+\text{d}(\sigma_{i}(x_{i}))\otimes^{\prime}\xi_{i}$
$\displaystyle=\left(\xi_{i}+\partial_{\underline{\sigma},i}(\sigma_{i}(x_{i}))\xi_{i}\right)\otimes^{\prime}\xi_{i}$
$\displaystyle=(1+\partial_{i,\underline{\sigma}}(\sigma_{i}(x_{i})))\xi_{i}\otimes^{\prime}\xi_{i}.$

Hence,

$\displaystyle\delta_{1,1}\left(\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id})(f)\right)$
$\displaystyle=\delta_{1,1}\left(\sum\limits_{i}\partial_{\underline{\sigma},i}(f)\xi_{i}+\sum\limits_{i<j}g_{i,j}\xi_{i}\xi_{j}+\sum\limits_{i}g_{i}\xi_{i}^{(2)_{\underline{\sigma}}}\right)$
$\displaystyle=\sum\limits_{i}\partial_{\underline{\sigma},i}(f)\left(\xi_{i}\otimes^{\prime}1+1\otimes^{\prime}\xi_{i}\right)+\sum_{i<j}g_{i,j}\xi_{i}\otimes^{\prime}\xi_{j}+\sum_{i>j}g_{j,i}\xi_{i}\otimes^{\prime}\xi_{j}$
$\displaystyle+\sum_{j=1}^{d}g_{j}\left(1+\sum_{i=1}^{d}\partial_{\underline{\sigma},i}(\sigma_{j}(x_{j}))\right)\xi_{i}\otimes^{\prime}\xi_{j}$
$\displaystyle=\sum\limits_{i}\partial_{i,\underline{\sigma}}(f)\xi_{i}\otimes^{\prime}1+\sum\limits_{j}\partial_{\underline{\sigma},j}(f)\otimes^{\prime}\xi_{j}+\sum\limits_{i<j}(g_{i,j}+g_{j}\partial_{\underline{\sigma},i}(\sigma_{j}(x_{j})))\xi_{i}\otimes^{\prime}\xi_{j}$
$\displaystyle+\sum\limits_{i>j}(g_{j,i}+g_{j}\partial_{\underline{\sigma},i}(\sigma_{j}(x_{j})))\xi_{i}\otimes^{\prime}\xi_{j}+\sum\limits_{i=1}^{d}g_{i}(1+\partial_{\underline{\sigma},i}(\sigma_{i}(x_{i})))\xi_{i}\otimes^{\prime}\xi_{i}.$

We have used again that $\xi_{i}\xi_{j}=\xi_{i}^{(2)_{\underline{\sigma}}}=0$
in $P_{A,(1)_{\underline{\sigma}}}$. On the other side, we can simply compute
$p_{1}^{\prime}\left((\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}\right)(f)$
and
$p_{2}^{\prime}\left((\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}\right)(f))$.
Which gives us

$p_{1}^{\prime}\left((\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}\right)(f)=\sum\limits_{i}\partial_{\underline{\sigma},i}(f)\xi_{i}\otimes^{\prime}1$

and

$\displaystyle
p_{2}^{\prime}\left((\Theta_{A,(2)_{\underline{\sigma}}}-\text{Id}\right)(f))$
$\displaystyle=\sum\limits_{j}1\otimes^{\prime}\partial_{\underline{\sigma},j}(f)\xi_{j}$
$\displaystyle=\sum\limits_{j}\Theta_{A,(1)_{\underline{\sigma}}}\left(\partial_{\underline{\sigma},j}(f)\right)\otimes^{\prime}\xi_{j}$
$\displaystyle=\sum\limits_{j}\left(\partial_{\underline{\sigma},j}(f)+\sum\limits_{i}\partial_{\underline{\sigma},i}(\partial_{\underline{\sigma},j}(f))\xi_{i}\right)\otimes^{\prime}\xi_{j}$
$=\sum\limits_{j}\partial_{\underline{\sigma},j}(f)\otimes^{\prime}\xi_{j}+\sum\limits_{i<j}\partial_{\underline{\sigma},i}(\partial_{\underline{\sigma},j}(f))\xi_{i}\otimes^{\prime}\xi_{j}+\sum\limits_{i>j}\partial_{\underline{\sigma},j}(\partial_{\underline{\sigma},i}(f))\xi_{i}\otimes^{\prime}\xi_{j}+\sum\limits_{i}\partial^{2}_{j,\underline{\sigma}}(f)\xi_{i}\otimes^{\prime}\xi_{i}.$

By identification on the coefficients, we obtain the following relations

$\forall
i<j,~{}g_{ij}=\partial_{\underline{\sigma},i}(\partial_{\underline{\sigma},j}(f))\partial_{\underline{\sigma},i}(\sigma_{j}(x_{j}))g_{j}=\partial_{\underline{\sigma},j}(\partial_{\underline{\sigma},j}(f))-\partial_{\underline{\sigma},j}(\sigma_{i}(x_{i}))g_{i}$

and

$\forall
i,~{}(1+\partial_{\underline{\sigma},i}(\sigma_{i}(x_{i})))g_{i}=\partial^{2}_{i,\underline{\sigma}}(f).$

As for every $i,$ $\sigma_{i}(x_{i})\in R[x_{i}]$ we have

$\forall i\neq j,~{}\partial_{\underline{\sigma},j}(f)(\sigma_{i}(x_{i}))=0.$

To prove that the derivatives commute with the endomorphisms $\sigma_{i}$ we
can use the formula of definition 7.3. On one side we have

$\sigma_{i}(\partial_{\underline{\sigma},j}(f))=\partial_{\underline{\sigma},j}(f)+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(\partial_{\underline{\sigma},j}(f))=\partial_{\underline{\sigma},j}(f)+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},j}(\partial_{\underline{\sigma},i}(f)).$

and on the other side,

$\displaystyle\partial_{\underline{\sigma},j}(\sigma_{i}(f))$
$\displaystyle=\partial_{\underline{\sigma},j}(f+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f))$
$\displaystyle=\partial_{\underline{\sigma},j}(f)+\partial_{\underline{\sigma},j}((\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f))$
$\displaystyle=\partial_{\underline{\sigma},j}(f)+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},j}(\partial_{\underline{\sigma},i}(f))+\sigma_{j}(\partial_{\underline{\sigma},i}(f))\partial_{\underline{\sigma},j}(\sigma_{i}(x_{i})-x_{i})$
$\displaystyle=\partial_{\underline{\sigma},j}(f)+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},j}(\partial_{\underline{\sigma},i}(f)).\qed$

The next proposition ensures that given classical coordinates on the
polynomial ring, those coordinates remain classical on a rational localization
of the polynomial ring.

######  Proposition 7.9.

Suppose that $A$ is complete. Let $x_{1},\ldots,x_{d}$ be classical, kind,
transverse and good $\underline{\sigma}$-coordinates associated to a good
model $(R_{0},A_{0})$. Assume that for $i=1,\ldots,d$, the endomorphism
$\sigma_{i}$ is infinitesimal over $A_{0}$. Let $f_{0},\ldots,f_{r}$ be
elements of $A$ that generate an open ideal of $A$. If
$A\left\\{\underline{f}/f_{0}\right\\}$ is a regular rational localization of
$A$ and $\forall i=1,\ldots,d$

$\sigma_{i}(f_{0})\in A\left(\underline{f}/f_{0}\right)^{\times}\emph{ and
}\forall j\in\\{1,\ldots,r\\},~{}\sigma_{i}(f_{j})/\sigma_{i}(f_{0})\in
A\left(\underline{f}/f_{0}\right)^{0},$

then, $x_{1},\ldots,x_{d}$ are classical coordinates over
$A\left\\{\underline{f}/f_{0}\right\\}$.

###### Proof.

From proposition 5.12 we have that for $i=1,\ldots,d,$ the endomorphism
$\sigma_{i}$ extends to an endomorphism $\tilde{\sigma}_{i}$ of
$A\left\\{\underline{f}/f_{0}\right\\}$ such that $x_{1},\ldots,x_{d}$ are
$\underline{\tilde{\sigma}}$-coordinates over
$A\left\\{\underline{f}/f_{0}\right\\}$. Recall that as rings we have
$A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)=A[1/f_{0}]$. For
$i\in\\{1,\ldots,d\\}$, to extend the $\sigma_{i}$-derivation
$\partial_{\underline{\sigma},i}$ in a $\tilde{\sigma}_{i}$-derivation
$\tilde{\partial}_{\underline{\tilde{\sigma}},i}$ it is necessary to satisfy
the Leibniz rule, meaning we should have:

$\displaystyle\tilde{\partial}_{\underline{\tilde{\sigma}},i}\left(\frac{1}{f_{0}}f_{0}\right)=$
$\displaystyle\tilde{\partial}_{\underline{\tilde{\sigma}},i}\left(\frac{1}{f_{0}}\right)f_{0}+\tilde{\sigma}_{i}\left(\frac{1}{f_{0}}\right)\partial_{\underline{\sigma},i}(f_{0})$
$\displaystyle=\tilde{\partial}_{\underline{\tilde{\sigma}},i}\left(\frac{1}{f_{0}}\right)f_{0}+\frac{1}{\sigma_{i}(f_{0})}\partial_{\underline{\sigma},i}(f_{0}).$

However, this quantity must be equal to zero which implies

$\tilde{\partial}_{\underline{\tilde{\sigma}},i}\left(\frac{1}{f_{0}}\right)=\frac{-\partial_{\underline{\sigma},i}(f_{0})}{\sigma_{i}(f_{0})f_{0}}.$

The operator defined in this fashion is an extension of
$\partial_{\underline{\sigma},i}$ to
$A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$ and is unique by
construction. We now show that this operator is continuous to extend it to
$A\left\\{\underline{f}/f_{0}\right\\}$. By hypothesis, the endomorphism
$\sigma_{i}$ is continuous. If $I_{0}$ is an ideal of definition of $A$ then,
there exists a natural integer $m$ such that
$\sigma_{i}\left(I_{0}^{m}\right)\subset I_{0}$. Moreover, a basis of
neighbourhoods of 0 in $A\left(\frac{f_{1},\ldots,f_{r}}{f_{0}}\right)$ is
given by

$(J_{n})_{n\in\mathbb{N}}:=\left(I_{0}^{n}A_{0}\left[f_{1}/f_{0},\ldots,f_{r}/f_{0}\right]\right)_{n\in\mathbb{N}}.$

We proceed by induction to show that

$\tilde{\partial}_{\underline{\tilde{\sigma}},i}(J_{m+1})\subset J_{1}.$

Let $\frac{1}{f_{0}^{n}}ab_{1}b_{2}$ with $n\in\mathbb{N},~{}a\in
A_{0},~{}b_{1}\in I_{0}^{m}$ and $b_{2}\in I_{0}$. Thus,

$\displaystyle\tilde{\partial}_{\underline{\tilde{\sigma}},i}\left(\frac{1}{f_{0}^{n}}ab_{1}b_{2}\right)=\tilde{\partial}_{\underline{\tilde{\sigma}},i}\left(\frac{1}{f_{0}^{n}}\right)ab_{1}b_{2}+\frac{1}{\sigma_{i}(f_{0}^{n})}\left(\partial_{\underline{\tilde{\sigma}},i}(ab_{1})b_{2}+\sigma_{i}(a)\sigma_{i}(b_{1})\partial_{\underline{\tilde{\sigma}},i}(a_{2})\right)$

This is our claim. Again by induction,
$\tilde{\partial}_{\underline{\tilde{\sigma}},i}(J_{N(m+1)})\subset J_{N}$ for
every natural integer $N$. We conclude that the operator
$\tilde{\partial}_{\underline{\tilde{\sigma}},i}$ is continuous. By
completion, it extends to $A\left\\{\underline{f}/f_{0}\right\\}$ and this is
automatically a $\tilde{\sigma}_{i}$-derivation. We can easily check that the
operator $\tilde{\partial}_{\underline{\tilde{\sigma}},i}$ is equal to the
operator $\partial_{\underline{\tilde{\sigma}},i}$. The fact that the
coordinates $\underline{x}$ remain classical over
$A\left\\{\underline{f}/f_{0}\right\\}$ follows from the construction of our
maps. ∎

######  Definition 7.10.

The $\underline{\sigma}$-coordinates $x_{1},\ldots,x_{d}$ are said to be
_symmetrical_ if

$\forall i\in\\{1,\ldots,d\\},\sigma_{i}(x_{i})\in R[x_{i}]\text{ and }$
$\forall n\in\mathbb{N},\forall f\in
A,~{}\delta_{n,m}\left(\Theta_{A,(n+m)_{\underline{\sigma}}}(f)\right)=1\otimes^{\prime}\Theta_{A,(m)_{\underline{\sigma}}}(f).$

Where $\otimes^{\prime}$ means that the action on the left is given by
$\Theta_{A,(n)_{\underline{\sigma}}}$.

######  Proposition 7.11.

Assume that the coordinates are symmetrical. Set

$\displaystyle\delta\colon P_{A}$ $\displaystyle\rightarrow
P_{A}\otimes_{A}P_{A}$ $\displaystyle a\otimes b$ $\displaystyle\mapsto
a\otimes 1\otimes 1\otimes b$

then,

$\delta\left(I_{A}^{(n+m+1)_{\underline{\sigma}}}\right)\subset
I_{A}^{(n+1)_{\underline{\sigma}}}\otimes P_{A}+P_{A}\otimes
I_{A}^{(m+1)_{\underline{\sigma}}}.$

###### Proof.

To obtain the assertion it is enough to show that the diagram below is
commutative

$\textstyle{P_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta}$$\scriptstyle{\tilde{\Theta}_{A,(n+m)_{\underline{\sigma}}}}$$\textstyle{P_{A}\otimes_{A}P_{A}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tilde{\Theta}_{A,(n)_{\underline{\sigma}}}\otimes~{}\tilde{\Theta}_{A,(m)_{\underline{\sigma}}}}$$\textstyle{P_{A,(n+m)_{\underline{\sigma}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\delta_{n,m}}$$\textstyle{P_{A,(n)_{\underline{\sigma}}}\otimes_{A}P_{A,(m)_{\underline{\sigma}}}}$

which is direct by assumption on the coordinates. This implies that

$\delta\left(I_{A}^{(n+m+1)_{\underline{\sigma}}}\right)\subset\text{ker}\left(\tilde{\Theta}_{A,(n)_{\underline{\sigma}}}\otimes\tilde{\Theta}_{A,(m)_{\underline{\sigma}}}\right).$

We now conclude by applying lemma 6.4 ∎

######  Definition 7.12.

When the coordinates $\underline{x}$ are classical, a _twisted connection_ on
an $A$-module $M$ is the data of a map

$\nabla_{\underline{\sigma}}:M\rightarrow
M\otimes_{A}\Omega^{1}_{\underline{\sigma}}$

verifying that the induced map

$\Theta_{M}:M\rightarrow
M\otimes_{A}P_{A,(1)_{\underline{\sigma}}},~{}s\rightarrow s\otimes
1+\nabla_{\underline{\sigma}}(s)$

is $\Theta_{A,(1)_{\underline{\sigma}}}$-linear:

$\forall f\in A,~{}\forall s\in
M,\Theta_{M}(fs)=\Theta_{A,(1)_{\underline{\sigma}}}(f)\Theta_{M}(s).$

######  Remark.

For $n\in\mathbb{N}^{*}$, set
$\Omega^{n}_{\underline{\sigma}}=\bigwedge^{n}\Omega^{1}_{\underline{\sigma}}.$
A twisted connection $\nabla_{\underline{\sigma}}$ on an $A$-module $M$ give a
sequence of $A$-linear maps

$\nabla_{n,\underline{\sigma}}:M\otimes_{A}\Omega^{n}_{\underline{\sigma}}\rightarrow
M\otimes_{A}\Omega^{n+1}_{\underline{\sigma}}$

by setting $\nabla_{n,\underline{\sigma}}(m\otimes
w)=m\otimes\text{d}(w)+\nabla_{\underline{\sigma}}(m)\wedge(-1)^{n}w$.

######  Definition 7.13.

A twisted connection $\nabla_{\underline{\sigma}}$ on an $A$-module $M$ is
said to be integrable if
$\nabla_{2,\underline{\sigma}}\circ\nabla_{\underline{\sigma}}=0.$

######  Remark.

On an $A$-module $M$, it is possible to construct the de Rham complex

$\text{DR}(A)=\left[M\rightarrow
M\otimes_{A}\Omega^{1}_{\underline{\sigma}}\rightarrow
M\otimes_{A}\Omega^{2}_{\underline{\sigma}}\rightarrow\dots\right].$

The de Rham cohomology is the cohomology of this complex.

We denote by $\nabla_{\underline{\sigma}}$-$\text{Mod}(A)$ the set of
$A$-modules endowed with a twisted connection and by
$\nabla_{\underline{\sigma}}^{\text{Int}}$-$\text{Mod}(A)$ the set of
$A$-modules endowed with an integrable twisted connection.

######  Theorem 7.14.

When the coordinates $x_{1},\ldots,x_{d}$ are classical, there exists an
equivalence of categories

$\nabla_{\underline{\sigma}}\emph{-Mod}(A)\simeq\emph{T}_{A,\underline{\sigma}}\emph{-Mod}(A).$

###### Proof.

Let us first construct a functor

$\nabla_{\underline{\sigma}}\text{-}\text{Mod}(A)\rightarrow\text{T}_{A,\underline{\sigma}}\text{-}\text{Mod}(A).$

Let $M$ be an $A$-module endowed with a twisted connection
$\nabla_{\underline{\sigma}}$. For an element $s$ of $M$ we write

$\nabla_{\underline{\sigma}}(s)=\sum_{i=1}^{d}\nabla_{i}(s)\otimes\xi_{i}.$

The condition on $\Theta_{M}$ can be translated by: for every $f\in A,~{}s\in
M$

$fs\otimes 1+\sum_{i=1}^{d}\nabla_{i}(fs)\otimes\xi_{i}=\left(f\otimes
1+\sum_{i=1}^{d}\partial_{\underline{\sigma},i}(f)(1\otimes\xi_{i})\right)\left(s\otimes
1+\sum_{i=1}^{d}\nabla_{i}(s)\otimes\xi_{i}\right)$

Computing the right term we get

$\displaystyle fs\otimes
1+\sum_{i=1}^{d}f\nabla_{i}(s)\otimes\xi_{i}+\sum_{i=1}^{d}\partial_{\underline{\sigma},i}(f)s\otimes\xi_{i}+\left(\sum_{i=1}^{d}\partial_{\underline{\sigma},i}(f)(1\otimes\xi_{i})\right)\left(\sum_{i=1}^{d}\nabla_{i}(s)\otimes\xi_{i}\right)$
$\displaystyle=fs\otimes
1+\sum_{i=1}^{d}\left(f\nabla_{i}(s)+\partial_{\underline{\sigma},i}(f)s\right)\otimes\xi_{i}+\sum_{i,j=1}^{d}\partial_{\underline{\sigma},i}(f)\nabla_{j}(s)\otimes\xi_{i}\xi_{j}.$

We recall the following relations in $\Omega^{1}_{A,\underline{\sigma}}$

$\xi_{i}\xi_{j}=0\text{ for }i\neq j\text{ and
}\xi_{i}^{2}=(\sigma_{i}(x_{i})-x_{i})\xi_{i}.$

Thus,

$\sum_{i,j=1}^{d}\partial_{\underline{\sigma},i}(f)\nabla_{j}(s)\otimes\xi_{i}\xi_{j}=\sum_{i=1}^{d}(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f)\nabla_{i}(s)\otimes\xi_{i}.$

By identification, for $i=1,\ldots,d,$

$\displaystyle\nabla_{i}(fs)$
$\displaystyle=f\nabla_{i}(s)+\partial_{\underline{\sigma},i}(f)s+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f)\nabla_{i}(s)$
$\displaystyle=\partial_{\underline{\sigma},i}(f)s+(f+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f))\nabla_{i}(s)$
$\displaystyle=\partial_{\underline{\sigma},i}(f)s+\sigma_{i}(f)\nabla_{i}(s).$

Hence, given a twisted connection $\nabla_{\underline{\sigma}}$ we recover
$\sigma_{i}$-derivations. and so, an action by
$\text{T}_{A,\underline{\sigma}}$. We now wish to construct an inverse functor

$\text{T}_{A,\underline{\sigma}}\text{-Mod}\rightarrow\nabla_{\underline{\sigma}}\text{-Mod}(A).$

Assume $M$ is endowed with an action by the derivations
$\partial_{\underline{\sigma},1},\ldots,\partial_{\underline{\sigma},d}$. Set
$\nabla_{\underline{\sigma}}(s)=\sum\limits_{i=1}^{d}\partial_{\underline{\sigma},i}(s)\otimes\xi_{i}$.
We want to check that the induced map $\Theta_{M}$ is
$\Theta_{A,(1)_{\underline{\sigma}}}$-linear.

$\displaystyle\Theta_{A,(1)_{\underline{\sigma}}}(f)\Theta_{M}(s)$
$\displaystyle=\left(f+\sum_{i=1}^{d}\partial_{\underline{\sigma},i}(f)(1\otimes\xi_{i})\right)\left(s\otimes
1+\sum_{i=1}^{d}\partial_{\underline{\sigma},i}(s)\otimes\xi_{i}\right)$
$\displaystyle=fs\otimes+\sum_{i=1}^{d}(f\partial_{\underline{\sigma},i}(s)+s\partial_{\underline{\sigma},i}(f))\otimes\xi_{i}+\sum_{i,j=1}^{d}\partial_{\underline{\sigma},i}(f)\partial_{\underline{\sigma},i}(s)\otimes\xi_{i}\xi_{j}$
$\displaystyle=fs\otimes+\sum_{i=1}^{d}(f\partial_{\underline{\sigma},i}(s)+s\partial_{\underline{\sigma},i}(f))\otimes\xi_{i}+\sum_{i=1}^{d}(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f)\partial_{\underline{\sigma},i}(s)\otimes\xi_{i}$
$\displaystyle=fs\otimes
1+\sum_{i=1}^{d}((f+(\sigma_{i}(x_{i})-x_{i})\partial_{\underline{\sigma},i}(f))\partial_{\underline{\sigma},i}(s)+s\partial_{\underline{\sigma},i}(f))\otimes\xi_{i}$
$\displaystyle=fs\otimes
1+\sum_{i=1}^{d}(\sigma_{i}(f)\partial_{\underline{\sigma},i}(s)+s\partial_{\underline{\sigma},i}(f))\otimes\xi_{i}$
$\displaystyle=fs\otimes
1+\sum_{i=1}^{d}\partial_{\underline{\sigma},i}(fs)\otimes\xi_{i}=\Theta_{M}(fs).$

This establishes our claim. ∎

##  8\. Twisted differential operators of infinite level

We keep the hypothesis of the previous section: $R$ is a Huber ring,
$(A,\underline{\sigma})$ is a twisted $R$-adic algebra of order $d$
(definition 4.1) and there exists some elements
$\underline{x}=(x_{1},\ldots,x_{d})$ of $A$ that are
$\underline{\sigma}$-coordinates (definition 4.6). In this section we will
also assume that the coordinates $x_{1},\ldots,x_{d}$ are symmetrical
(definition 7.10). This will be useful to definie the composition of twisted
differential operators.

######  Definition 8.1.

Let $M$ and $N$ be $A$-modules. A _twisted differential operator of order at
most $n$_ is an $R$-linear morphism $\phi:M\rightarrow N$ which
$A$-linearization $\tilde{\phi}$ factorizes throught
$P_{A,(n)_{\underline{\sigma}}}$. This gives the diagram

$\textstyle{P_{A}\otimes_{A}M\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sim}$$\scriptstyle{\tilde{\phi}}$$\textstyle{A\otimes_{R}M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N}$$\textstyle{P_{A,(n)_{\underline{\sigma}}}\otimes_{A}^{\prime}M\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

where $\otimes_{A}^{\prime}$ designates that the action on the left is given
by $\Theta_{A,(n)_{\underline{\sigma}}}$. This condition can be translated by
the fact that $\tilde{\phi}$ is zero over
$I_{A}^{(n+1)_{\underline{\sigma}}}$. We denote by
$\text{Diff}_{n,\underline{\sigma}}(M,N)$ the set of twisted differential
operators of order at most $n$.

By definition
$\text{Diff}_{n,\underline{\sigma}}(M,N)\simeq\text{Hom}_{A}(P_{A,(n)_{\underline{\sigma}}}\otimes_{A}^{\prime}M,N)$.
In the case where $M=N$ we will simply write
$\text{Diff}_{n,\underline{\sigma}}(M)$. We also set

$\text{Diff}_{\underline{\sigma}}(M,N)=\varinjlim_{n}\text{Hom}_{A}\left(P_{A,(n)_{\underline{\sigma}}}\otimes_{A}^{\prime}M,N\right)\text{
and
}\text{D}^{(\infty)}_{A,\underline{\sigma}}=\text{Diff}_{\underline{\sigma}}(A).$

Recall that for an element $q$ of $R$ and every $n\in\mathbb{N}$, we define
the $q$-analogue of $n$ as

$(n)_{q}:=1+q+\ldots+q^{n-1}.$

It verifies $\lim\limits_{q\mapsto 1}(n)_{q}=n.$ In a similar manner, we can
define the $q$-analogue of the factorial of $n$ by setting:

$(n)_{q}!:=(2)_{q}\ldots(n)_{q}.$

######  Remarks.

  1. (1)

A standard basis of $\text{Diff}_{n,\underline{\sigma}}(A)$ is given by
$\partial^{[\underline{k}]}_{\underline{\sigma}}$ dual to the image of
$\underline{\xi}^{(\underline{k})_{\underline{\sigma}}}$ in
$P_{A,(n)_{\underline{\sigma}}}$ for $\underline{k}\in\mathbb{N}^{d}$ such
that $|\underline{k}|\leq n$.

  2. (2)

Let $\underline{q}=\\{q_{1},\ldots,q_{d}\\}$ be elements of $R$. In the case
where $x_{1},\ldots,x_{d}$ are classical $\underline{q}$-coordinates, by
corollary 6.2 of [LSQ18a], we have

$\forall i\in\\{1,\ldots,d\\},~{}\forall k\in\mathbb{N},~{}\forall z\in
A,\partial^{k}_{q_{i}}(z)=(k)^{!}_{q_{i}}\partial_{q_{i}}^{[k]}(z).$

For $\underline{k}=(k_{1},\ldots,k_{d})$, set

$\partial_{\underline{\sigma}}^{\underline{k}}:=\partial_{\underline{\sigma},1}^{k_{1}}\circ\ldots\circ\partial_{\underline{\sigma},d}^{k_{d}}.$

Thus, by the previous relation,

(3)
$\partial_{\underline{\sigma}}^{\underline{k}}=(k_{1})_{q_{1}}^{!}\ldots(k_{d})_{q_{d}}^{!}\partial_{\underline{\sigma}}^{[\underline{k}]}.$

######  Proposition 8.2.

Let $u:A\rightarrow B$ be a flat morphism of twisted $R$-algebras of order
$d$. If $B$ is $A$-module of finite presentation and
$u(x_{1}),\ldots,u(x_{d})$ are $\underline{\sigma_{B}}$-coordinates the, there
exists an isomorphism

$\emph{Diff}_{\underline{\sigma_{B}}}(B\otimes_{A}M,B\otimes_{A}N)\simeq
B\otimes_{A}\emph{Diff}_{\underline{\sigma_{A}}}(M,N).$

###### Proof.

By commutativity of the tensor product with inductive limits, we get

$B\otimes_{A}\text{Diff}_{\underline{\sigma_{A}}}(M,N)=B\otimes_{A}\varinjlim\text{Hom}_{A}(P_{A,(n)_{\underline{\sigma}}}\otimes_{A}^{\prime}M,N)=\varinjlim
B\otimes_{A}\text{Hom}_{A}(P_{A,(n)_{\underline{\sigma}}}\otimes_{A}^{\prime}M,N).$

We then apply [Bou98, Chapitre 1, Section 2, Sous-section 10, Proposition 11]
to obtain

$B\otimes_{A}\text{Diff}_{\underline{\sigma_{A}}}(M,N)\simeq\varinjlim\text{Hom}_{B}(B\otimes_{A}P_{A,(n)_{\underline{\sigma}}}\otimes_{A}^{\prime}M,B\otimes_{A}N).$

Proposition 4.5 allows us to conclude. ∎

######  Proposition 8.3.

Let $M,~{}N$ and $L$ be $A$-modules. For natural integers $m$ and $n$, given
$\varphi\in\emph{Diff}_{n,\underline{\sigma}}(M,N)$ and
$\psi\in\emph{Diff}_{m,\underline{\sigma}}(L,M)$ then
$\varphi\circ\psi\in\emph{Diff}_{m+n,\underline{\sigma}}(L,N)$. In particular,
$\emph{Diff}_{\underline{\sigma}}(M)$ is a subring of $\emph{Hom}_{R}(M)$.

###### Proof.