endomorphism $G\colon k\\{T\\}\to k\\{T\\}$ such that $\|G(f)\|\leq\|f\|$.
Since we have a closed embedding
$\mathcal{M}(A)\hookrightarrow\bar{\mathbb{D}}^{d}(0,1):=\mathcal{M}(k\\{T\\})$,
we are reduced to proving $\operatorname{h_{top}}(g)=0$ where
$g\colon\bar{\mathbb{D}}^{d}(0,1)\to\bar{\mathbb{D}}^{d}(0,1)$ is the map
induced by $G$.

For any $\varepsilon\in(0,1]$, consider the ideal
$\mathfrak{m}_{\varepsilon}:=\\{|z|<\varepsilon\\}\subset k^{\circ}$, and the
ring $k_{\varepsilon}:=k^{\circ}/\mathfrak{m}_{\varepsilon}$. Observe that
(see Lemma 5.1)
$A_{\varepsilon}:=k^{\circ}\\{T\\}/\mathfrak{m}_{\varepsilon}\\{T\\}=k_{\varepsilon}[T]$
is a finitely generated $k_{\varepsilon}$-algebra. We have a canonical
reduction map
$\operatorname{red}_{\varepsilon}\colon\bar{\mathbb{D}}^{d}(0,1)\to\operatorname{Id}(A_{\varepsilon})$
which sends a multiplicative semi-norm $x\in\mathcal{M}(k\\{T\\})$ to the
image of the ideal $\\{f\in k^{\circ}\\{T\\},\,|f(x)|<\varepsilon\\}$ in
$A_{\varepsilon}$ (note that $\\{f\in
k^{\circ}\\{T\\},\,|f(x)|<\varepsilon\\}$ contains
$\mathfrak{m}_{\varepsilon}\\{T\\}$).

Note that given any $f\in k^{\circ}\\{T\\}$, we have (for the case
$\varepsilon=1$ see [Ber90, Lemma 2.4.1])

$\displaystyle\operatorname{red}_{\varepsilon}^{-1}\\{\mathfrak{A}\in\operatorname{Id}(A_{\varepsilon}),\,\tilde{f}\in\mathfrak{A}\\}$
$\displaystyle=\\{|f|<\varepsilon\\}$
$\displaystyle\operatorname{red}_{\varepsilon}^{-1}\\{\mathfrak{A}\in\operatorname{Id}(A_{\varepsilon}),\,\tilde{f}\notin\mathfrak{A}\\}$
$\displaystyle=\\{|f|\geq\varepsilon\\}$

where $\tilde{f}$ denotes the image of $f$ in $A_{\varepsilon}$. It follows
that $\operatorname{red}_{\varepsilon}$ is not continuous. However, as the
usual reduction map $\operatorname{red}_{1}$,
$\operatorname{red}_{\varepsilon}$ is anti-continuous with respect to the
Zariski topology on $\operatorname{Id}(A_{r})$ defined in Section 3.4.4.

The map $G$ is a bounded $k$-algebra homomorphism which maps
$k^{\circ}\\{T\\}$ to itself so that
$G(\mathfrak{m}_{r}\\{T\\})\subset\mathfrak{m}_{r}\\{T\\}$, hence $G$ descends
to a continuous map
$g_{\varepsilon}\colon\operatorname{Id}(A_{\varepsilon})\to\operatorname{Id}(A_{\varepsilon})$
satisfying $\operatorname{red}_{\varepsilon}\circ
g=g_{\varepsilon}\circ\operatorname{red}_{\varepsilon}$.

######  Lemma 4.5.

Let $\mathfrak{U}$ be a finite open cover of $\bar{\mathbb{D}}^{d}(0,1)$. Then
for every sufficiently small $\varepsilon\in|k^{*}|$, there is a finite
constructible (hence clopen) cover $\mathfrak{V}$ of
$\operatorname{Id}(A_{\varepsilon})$ such that
$\\{\operatorname{red}_{\varepsilon}^{-1}(U),\,U\in\mathfrak{V}\\}$ refines
$\mathfrak{U}$.

Assume Lemma 4.5. Pick any finite open cover $\mathfrak{U}$ and choose an open
cover $\mathfrak{V}$ as in the previous lemma. Then we have
$h(g,\mathfrak{U})\leq h(g_{\varepsilon},\mathfrak{V})$. By Corollary 4.2 we
have $\operatorname{h_{top}}(g_{\varepsilon})=0$, which concludes the proof of
Theorem D.

###### Proof of Lemma 4.5.

To any choice of elements $f_{i},g_{j}\in k^{\circ}\\{T\\}$, and
$a_{i},b_{j}\in k^{*}$ we attach the set:

(14) $U(f<a,\,g\geq
b):=\bigcap_{i=1}^{m}\left\\{|f_{i}|<|a_{i}|\right\\}\bigcap_{j=1}^{n}\left\\{|g_{j}|\geq|b_{j}|\right\\}~{}.$

Choose any $\varepsilon=|c|\in|k^{*}|$ smaller than
$\min\\{|a_{i}|,|b_{j}|\\}$, and observe that
$f^{\prime}_{i}:=\frac{c}{a_{i}}f_{i}$, $g^{\prime}_{j}:=\frac{c}{b_{j}}g_{j}$
all belong to $k^{\circ}\\{T\\}$. It follows that we have $U(f<a,\,g\geq
b)=\operatorname{red}_{\varepsilon}^{-1}(V)$ where $V$ is the open set of
$\operatorname{Id}(A_{\varepsilon})$ given by

$V:=\bigcap_{i=1}^{m}\left\\{\mathfrak{A}\in\operatorname{Id}(A_{\varepsilon}),\,\widetilde{f^{\prime}_{i}}\in\mathfrak{A}\right\\}\bigcap_{j=1}^{n}\left\\{\mathfrak{A}\in\operatorname{Id}(A_{\varepsilon}),\,\widetilde{g^{\prime}_{j}}\notin\mathfrak{A}\right\\}~{}.$

Now pick any finite open cover $\mathfrak{U}$ of $\bar{\mathbb{D}}^{d}(0,1)$.
For each $x$, choose a neighborhood $U_{x}$ of the form (14) which is
contained in an element of $\mathfrak{U}$ (the existence of such a $U_{x}$
follows from [Ber90, Remarks 2.2.2]). By compactness of
$\bar{\mathbb{D}}^{d}(0,1)$ we may extract a finite subcover
$U_{x_{1}},\ldots,U_{x_{n}}$ which refines $\mathfrak{U}$.

By our previous argument, we may find $\varepsilon>0$ so small so that for
each $i$ we have $U_{x_{i}}=\operatorname{red}_{\varepsilon}^{-1}(V_{i})$ for
some open subset $V_{i}$ of $\operatorname{Id}(A_{\varepsilon})$. This
completes the proof. ∎

##  5\. Endomorphisms having good reduction

Our aim is to globalize the arguments of the previous section when a self-map
of an affinoid domain is replaced by an endomorphism of a $k^{\circ}$-model of
a projective variety. This will prove Theorem B.

To that end, we develop in §5.1 the theory of $\varepsilon$-reduction of any
strictly affinoid domain defined over an algebraically closed complete
metrized field $(k,|\cdot|)$. We then extend this $\varepsilon$-reduction
functor to projective models $\mathfrak{X}$ over $k^{\circ}$ in §5.3–5.4. The
proof of Theorem B is given in §5.5. It is based on an argument of absolute
Noetherian approximation, that we develop with some details.

###  5.1. $\varepsilon$-reduction of strictly affinoid domains

Let $A$ be any strictly affinoid $k$-algebra. Recall the definition of the
spectral norm $\rho(a):=\lim_{n}\|a^{n}\|^{1/n}$, and that we set
$A^{\circ}:=\\{\rho\leq 1\\}$.

######  Lemma 5.1.

For any strictly affinoid $k$-algebra $A$ and for any $1\geq\varepsilon>0$,
then

$A_{\varepsilon}:=A^{\circ}/\\{\rho<\varepsilon\\}$

is a finitely generated $k_{\varepsilon}$-algebra with unit.

###### Proof.

When $A=k\\{T\\}$ is the Tate algebra, then
$A_{\varepsilon}=k^{\circ}\\{T\\}/\mathfrak{m}_{\varepsilon}\\{T\\}=k_{\varepsilon}[T]$
so that the result is clear. When $A$ is reduced, since $k$ is algebraically
closed, then it follows from [BGR84, §6.4.3] that there exists an epimorphism
$\alpha\colon k\\{T\\}\to A$ such that $\rho$ is equal to the residual norm.
We get a surjective morphism $k_{\varepsilon}[T]\to A_{\varepsilon}$ and
$A_{\varepsilon}$ is finitely generated over $k_{\varepsilon}$. Finally, when
$A$ is not reduced, we conclude by observing that
$A_{\varepsilon}=B_{\varepsilon}$ where $B$ is the quotient of $A$ by its
nilradical. ∎

######  Remark 5.1.

The previous proof works for any stable fields in the sense of [BGR84, §3.6].
When $\varepsilon=1$, then it is known that $\tilde{A}:=A_{1}$ is finitely
generated over $\tilde{k}$ without any assumption on $k$. It remains open
whether $A_{\varepsilon}$ is finitely generated over $k_{\varepsilon}$ when
$k$ is not necessarily stable.

Pick any multiplicative semi-norm $x\in\mathcal{M}(A)$, and set
$\mathfrak{A}_{\varepsilon}(x):=\\{a\in A^{\circ},\,|a(x)|<\varepsilon\\}$ so
that $\mathfrak{A}_{\varepsilon}(x)\cap k^{\circ}=\mathfrak{m}_{\varepsilon}$.
Since $\mathfrak{A}_{\varepsilon}(x)\supset\\{\rho<\varepsilon\\}$, the ideal
$\mathfrak{A}_{\varepsilon}(x)$ projects onto an ideal
$\tilde{\mathfrak{A}}_{\varepsilon}(x)$ of $A_{\varepsilon}$ which satisfies
$\tilde{\mathfrak{A}}_{\varepsilon}(x)\cap k_{\varepsilon}=(0)$. We get a
canonical reduction map
$\operatorname{red}_{\varepsilon}\colon\mathcal{M}(A)\to\operatorname{Id}(A_{\varepsilon})$
sending $x$ to $\tilde{\mathfrak{A}}_{\varepsilon}(x)$.

Recall from §3.4.4 that the Zariski topology on
$\operatorname{Id}(A_{\varepsilon})$ is generated by open sets of the form:

$U:=\\{\mathfrak{A}\in\operatorname{Id}(A_{\varepsilon}),\,a\not\in\mathfrak{A}\\}~{},$

so that
$\operatorname{red}_{\varepsilon}^{-1}(U)=\\{x\in\mathcal{M}(A),\,|a(x)|\geq\varepsilon\\}$
is closed. It follows that $\operatorname{red}_{\varepsilon}$ is anti-
continuous from the affinoid space $\mathcal{M}(A)$ to
$(\operatorname{Id}(A_{\varepsilon}),\tau^{\operatorname{zar}})$.

######  Remark 5.2.

When $\varepsilon=1$, the image of the reduction map
$\operatorname{red}_{1}(A)$ is precisely the spectrum of $\tilde{A}$, see
[BGR84, 7.1.5/4], [Ber90, §2.4]. The description of
$\operatorname{red}_{\varepsilon}(A)$ is unclear when $\varepsilon<1$. In
general $\operatorname{red}_{\varepsilon}(A)$ is not closed in
$\operatorname{Id}(A_{\varepsilon})$ endowed either with the Zariski or the
constructible topology.

######  Remark 5.3.

It is instructive to compute the $\varepsilon$-reduction of
$\bar{\mathbb{D}}^{1}(0,1)=\mathcal{M}(k\\{T\\})$. Given any $a\in k^{\circ}$
and any $r\in[0,1]$, denote by $\zeta(a,r)$ the point in
$\bar{\mathbb{D}}^{1}(0,1)$ defined by
$|P(\zeta(a,r))|=\sup_{\bar{B}(a,r)}|P|$.

Direct computations show that $\operatorname{red}_{\varepsilon}(\zeta(a,r))$
is the prime ideal in $k_{\varepsilon}[T]$ generated by $(T-a)$ if
$r<\varepsilon$; whereas for any $r\geq\varepsilon$,
$\operatorname{red}_{\varepsilon}(\zeta(a,r))$ is not a prime ideal. Indeed,
one has
$\operatorname{red}_{\varepsilon}(\zeta(a,r))=\\{\sum_{i}a_{i}(T-a)^{i},\,a_{0}=0,|a_{1}|r<\varepsilon\\}$.
Also, we have
$\operatorname{red}_{\varepsilon}(\zeta(a,r))=\operatorname{red}_{\varepsilon}(\zeta(\alpha,\rho))$
iff either $\zeta(a,r)=\zeta(\alpha,\rho)$, or $r,\rho<\varepsilon$ and
$|a-\alpha|<\varepsilon$.

Observe that these computations imply that
$\operatorname{red}_{\varepsilon}(\mathcal{M}(k\\{T\\}))$ is _not_ a closed
subset of $\operatorname{Id}(k_{\varepsilon}[T])$ (at least when $k$ is
spherically complete).

The $\varepsilon$-reduction map is functorial. For any bounded homomorphism of
$k$-algebras $f^{*}\colon A\to B$ inducing an analytic map
$f\colon\mathcal{M}(B)\to\mathcal{M}(A)$, we have $\rho(f(a))\leq\rho(a)$
hence we get a canonical map
$f_{\varepsilon}\colon\operatorname{Id}(B_{\varepsilon})\to\operatorname{Id}(A_{\varepsilon})$
making the following diagram commutative:

(15)
$\textstyle{\mathcal{M}(B)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\operatorname{red}_{\varepsilon}}$$\textstyle{\mathcal{M}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{red}_{\varepsilon}}$$\textstyle{\operatorname{Id}(B_{\varepsilon})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{\varepsilon}}$$\textstyle{\operatorname{Id}(A_{\varepsilon})}$

The map $f_{\varepsilon}$ is continuous for both the Zariski and constructible
topologies.

Let us finish this section by the following generalization of Lemma 4.5.

######  Lemma 5.2.

Let $\mathfrak{U}$ be a finite open cover of $\mathcal{M}(A)$. Then for every
sufficiently small $\varepsilon\in|k^{*}|$, there is a finite constructible
cover $\mathfrak{V}$ of
$(\operatorname{Id}(A_{\varepsilon}),\tau^{\operatorname{con}})$ such that
$\\{\operatorname{red}_{\varepsilon}^{-1}(U),\,U\in\mathfrak{V}\\}$ refines
$\mathfrak{U}$.

###### Proof.

As in the proof of Lemma 5.1, we may suppose that $A$ is reduced and choose an
epimorphism $\alpha^{*}\colon k\\{T\\}\to A$ such that the spectral norm on
$A$ is equal to the residual norm. This induces a closed embedding
$\alpha\colon\mathcal{M}(A)\to\mathcal{M}(k\\{T\\})$, and an epimorphism
$k_{\varepsilon}[T]\to A_{\varepsilon}\to 0$ of $k_{\varepsilon}$-algebras.
For each $U\in\mathfrak{U}$ choose an open set $U^{\prime}$ of
$\mathcal{M}(k\\{T\\})$ such that $U^{\prime}\cap\mathcal{M}(A)=U$, and
consider the open cover
$\mathfrak{U}^{\prime}:=\\{U^{\prime}\\}\cup\\{\mathcal{M}(k\\{T\\})\setminus\mathcal{M}(A)\\}$
of $\mathcal{M}(k\\{T\\})$. By Lemma 4.5, for sufficiently small
$\varepsilon\in|k^{*}|$, there exists a finite constructible cover
$\mathfrak{V}^{\prime}$ of
$(\operatorname{Id}(k_{\varepsilon}[T]),\tau^{\operatorname{con}})$ such that
$\\{\operatorname{red}_{\varepsilon}^{-1}(U),\,U\in\mathfrak{V}^{\prime}\\}$
refines $\mathfrak{U}^{\prime}$. Since
$\alpha^{-1}(\mathfrak{U}^{\prime})=\mathfrak{U}$ by construction, it follows
from (15) that the cover
$\\{\operatorname{red}_{\varepsilon}^{-1}(\alpha_{\varepsilon}^{-1}(U)),\,U\in\mathfrak{V}^{\prime}\\}$
refines $\mathfrak{U}$. We conclude the proof by setting
$\mathfrak{V}:=\alpha_{\varepsilon}^{-1}(\mathfrak{V}^{\prime})$. ∎

###  5.2. Primary ideals

There is a technical difficulty in globalizing the $\varepsilon$-reduction
using the sets $\operatorname{Id}(A_{\varepsilon})$ as above. We refer to
Remark 5.8 below for a discussion on this problem. In order to properly
globalize the $\varepsilon$-reduction and to develop smoothly our descent
argument, we will work instead with the set of primary ideals in
$A_{\varepsilon}$. In this section, we discuss briefly some general properties
of this set.

Recall that an ideal $I$ is primary if $fg\in I$ and $f\notin I$ implies
$g^{n}\in I$ for some $n\geq 1$. Any prime ideal is primary, and the radical
of a primary ideal is prime.

######  Definition 5.1.

Let $R$ be any commutative ring with unit. Then we let ${\rm P}(R)$ be the set
of all primary ideals in $\operatorname{Id}(R)$ and
$\operatorname{\bar{P}}(R)$ be its closure (for the constructible topology) in
$\operatorname{Id}(R)$.

The set of primary ideals is a countable intersection of clopen sets. As the
examples below show, it is neither closed, nor open, nor dense in general.

######  Example 5.1.

Let $k$ be an algebraically closed field, and set $R=k[x,y]$.

Then $I=(xy)$ is not a primary ideal, but the sequence of primary ideals
$(x^{n},y^{n},xy)$ converges to $I$ in $\operatorname{Id}(R)$. This shows that
the set of primary ideals is not closed in general.

Take $I^{\prime}=(x)$. Then a basis for open neighbourhoods of $I^{\prime}$
are those of the form $U=\\{J\in\operatorname{Id}(R),f_{i}\in J,g_{j}\notin
J\\}$ (where $f_{i}$ divides $x$). Each such $U$ contains two maximal ideals
$J_{1}$ and $J_{2}$ for which the product $J_{1}\cdot J_{2}$ is not primary
and belongs to $U$. This proves the set of primary ideals is not open in
general.

Consider now $I^{\prime\prime}=(xy(x+y-1))$. Then the set
$\\{J\in\operatorname{Id}(R),xy\notin J,y(x+y-1)\notin J,x(x+y-1)\notin
J,xy(x+y-1)\in J\\}$ is an open neighborhood of $I^{\prime\prime}$ but
contains no primary ideals. Indeed if $J$ would be a primary ideal lying in
this set then we could find an integer $n>0$ such that $x^{n},y^{n}$ and
$(x+y-1)^{n}$ belong to $J$, a contradiction. This shows that the set of
primary ideals is not dense in general.

Let $f$ be any non-nilpotent element in $R$. Recall that
$R[f^{-1}]=R[T]/(1-Tf)$ is an overring of $R$ so that we have two natural maps
$\operatorname{res}\colon\operatorname{Id}(R[f^{-1}])\to\operatorname{Id}(R)$
and
$\operatorname{ext}\colon\operatorname{Id}(R)\to\operatorname{Id}^{+}(R[f^{-1}])$
defined by $\operatorname{res}(I)=I\cap R$ and $\operatorname{ext}(I)=I\cdot
R[f^{-1}]$ respectively. Beware that if $f^{n}\in I$ for some
$I\in\operatorname{Id}(R)$ then $\operatorname{ext}(I)=R$.

######  Lemma 5.3.

The map
$\operatorname{res}\colon\operatorname{Id}(R[f^{-1}])\to\operatorname{Id}(R)$
is injective and its image consists of those ideals $I\in\operatorname{Id}(R)$
such that if $gf^{n}\in I$ for some $g\in R$ and some $n\geq 0$, then $g\in
I$.

###### Proof.

We first prove the second statement. Suppose $I=J\cap R$ for some ideal $J$ of
$R[f^{-1}]$. If $gf^{n}\in I$ for some $g\in R$ and some $n\geq 0$, then $g\in
J$ since $f$ is invertible in $R[f^{-1}]$. It follows that $g\in I$.

Conversely suppose $I$ is an ideal in $R$ such that $gf^{n}\in I$ for some
$g\in R$ and some $n\geq 0$ implies $g\in R$. We shall prove that
$\operatorname{ext}(I)\cap R=I$. It is clear that $\operatorname{ext}(I)\cap
R\supset I$. So pick $h\in\operatorname{ext}(I)\cap R$ and write $h=g/f^{n}$
with $g\in I$. Then $hf^{n}\in I$ hence $h\in I$.

The former arguments imply $\operatorname{ext}\circ\operatorname{res}(J)=J$
for any $J\in R[f^{-1}]$. This implies $\operatorname{res}$ to be injective. ∎

Observe that $\operatorname{res}(I)$ is a prime (resp. primary) ideal whenever
$I$ is prime (resp. primary).

######  Lemma 5.4.

A primary ideal $I\subset R$ belongs to
$\operatorname{res}(\operatorname{Id}(R[f^{-1}]))$ if and only if $f^{n}\notin
I$ for all $n\geq 0$.

###### Proof.

Suppose $I=J\cap R$ for some $J\in\operatorname{Id}(R[f^{-1}])$, and $f^{n}$
belongs to $I$ for some $n>0$. Then the previous lemma implies $1\in I$, a
contradiction.

Conversely, suppose $I$ is primary and $f^{n}\notin I$ for all $n\geq 0$.
Suppose that $gf^{n}\in I$ for some $g\in R$ and some $n\geq 0$. If $g\notin
I$, then $f^{nm}\in I$ for some $m$ since $I$ is primary, a contradiction.
Hence $g\in I$, and the previous lemma implies $I=J\cap R$ for some
$J\in\operatorname{Id}(R[f^{-1}])$. ∎

######  Proposition 5.5.

For any commutative ring $R$ with unit, and any non-nilpotent element $f\in
R$,
$\operatorname{res}\colon\operatorname{\bar{P}}(R[f^{-1}])\to\operatorname{\bar{P}}(R)$
induces a homeomorphism onto its image which is compact in
$\operatorname{\bar{P}}(R)$.

Moreover, for any finite family of non-nilpotent elements
$f_{1},\cdots,f_{m}\in R$ generating the unit ideal, the set
$\\{\operatorname{res}(\operatorname{\bar{P}}(R[f_{i}^{-1}])\\}$ forms a
finite compact cover of $\operatorname{\bar{P}}(R)$.

###### Proof.

By Lemma 5.3, the map $\operatorname{res}$ is injective. Since it is also
continuous (for the constructible topology), the restriction of
$\operatorname{res}$ to any compact subset of $\operatorname{Id}(R[f^{-1}])$
induces a homeomorphism onto its image. By construction,
$\operatorname{\bar{P}}(R[f^{-1}])$ is compact in
$\operatorname{Id}(R[f^{-1}])$, and the first statement follows.

We now prove the second statement. Suppose $f_{1},\cdots,f_{m}$ generate the
unit ideal, and pick any primary ideal $I$. Suppose that $I$ does not belong
to the union of
$\\{\operatorname{res}(\operatorname{\bar{P}}(R[f_{i}^{-1}])\\}$. Then, by the
previous lemma, for each index $i$, we have $f_{i}^{n_{i}}\in I$ for some
integer $n_{i}$. Set $n=\max n_{i}$. Then $f^{n}_{1},\cdots,f^{n}_{m}\in I$,
but it is also easy to show that they generate the unit ideal. This is a
contradiction. ∎

######  Remark 5.4.

The previous result is not true when $\operatorname{\bar{P}}(R)$ is replaced
by $\operatorname{Id}(R)$. Indeed, in general
$\\{\operatorname{Id}(R[f_{i}^{-1}])\\}$ does not form a cover of
$\operatorname{Id}(R)$ if $(f_{1},\cdots,f_{n})$ generates the unit ideal
(take $R={\mathbb{Z}}$, $f_{1}=2$ and $f_{2}=3$ for instance).

######  Remark 5.5.

To simplify notation, in the sequel we shall always identify
$\operatorname{\bar{P}}(R[f^{-1}])$ with its image inside
$\operatorname{\bar{P}}(R)$.

We conclude this section with the following observations.

######  Lemma 5.6.

For any strictly affinoid algebra $A$ and any $\varepsilon>0$, we have
$\operatorname{red}_{\varepsilon}(\mathcal{M}(A))\subset{\rm
P}(A_{\varepsilon})$.

###### Proof.

Pick any $x\in\mathcal{M}(A)$. Then
$\operatorname{red}_{\varepsilon}(x)=\\{f\in
A_{\varepsilon},|f(x)|<\varepsilon\\}$ is a primary ideal. Indeed suppose
$fg\in\operatorname{red}_{\varepsilon}(x)$ and
$f\notin\operatorname{red}_{\varepsilon}(x)$. Then $|fg(x)|<\varepsilon$ and
$|f(x)|\geq\varepsilon$, hence $|g(x)|<1$. Hence we can find $n>0$ such that
$|g^{n}(x)|<\varepsilon$, and $g^{n}\in\operatorname{red}_{\varepsilon}(x)$. ∎

To simplify notation, write $\bar{R}_{\varepsilon}(A)$ for the closure (in
$\operatorname{\bar{P}}(A_{\varepsilon})$) of the set
$\operatorname{red}_{\varepsilon}(\mathcal{M}(A))$.

######  Proposition 5.7.

Let $A$ be any strictly affinoid algebra $A$. For any $\varepsilon\in|k^{*}|$,
and any $f\in A^{\circ}$, we have $\bar{R}_{\varepsilon}(A\langle
f^{-1}\rangle)=\bar{R}_{\varepsilon}(A)\cap\operatorname{\bar{P}}(A_{\varepsilon}[f^{-1}])$,
and this set is a clopen subset of $\bar{R}_{\varepsilon}(A)$.

###### Proof.

Write $\varepsilon=|a|$ for some $a\in k^{*}$. Observe first that

$\operatorname{red}_{\varepsilon}(\mathcal{M}(A\langle
f^{-1}\rangle))=\operatorname{red}_{\varepsilon}(\mathcal{M}(A))\cap\operatorname{\bar{P}}(A_{\varepsilon}[f^{-1}])=\operatorname{red}_{\varepsilon}(\mathcal{M}(A))\cap\operatorname{Id}(A_{\varepsilon}[f^{-1}])~{}.$

We now prove

$\operatorname{red}_{\varepsilon}(\mathcal{M}(A))\cap\operatorname{Id}(A_{\varepsilon}[f^{-1}])=\\{I\in\operatorname{red}_{\varepsilon}(\mathcal{M}(A)),af\notin
I\\}~{}.$

Suppose $I=\operatorname{red}_{\varepsilon}(x)$ for some $x\in\mathcal{M}(A)$.
Observe that $f^{n}\notin I$ for all $n$, if and only if $|f(x)|=1$, and the
latter is equivalent to $|(af)(x)|=\varepsilon$. By Lemma 5.4, this proves
$I\in\operatorname{Id}(\mathcal{M}(A\langle f^{-1}\rangle))$ if and only if
$af\notin I$ as required.

Since the set of ideals $I\in\operatorname{Id}(\mathcal{M}(A)))$ such that
$af\notin I$ is clopen, the closure of
$\\{I\in\operatorname{red}_{\varepsilon}(\mathcal{M}(A)),af\notin I\\}$ equals
$\\{I\in\bar{R}_{\varepsilon}(A),af\notin I\\}$ which implies

$\bar{R}_{\varepsilon}(A\langle
f^{-1}\rangle)=\\{I\in\bar{R}_{\varepsilon}(A),af\notin
I\\}\subset\bar{R}_{\varepsilon}(A)\cap\operatorname{\bar{P}}(A_{\varepsilon}[f^{-1}]).$

Pick
$I\in\bar{R}_{\varepsilon}(A)\cap\operatorname{\bar{P}}(A_{\varepsilon}[f^{-1}])$.
Observe that $a\notin I$. (Indeed, for any $x\in\mathcal{M}(A)$ we have
$\operatorname{red}_{\varepsilon}(x)\cap k_{\varepsilon}=(0)$. Hence, for any
ideal $J\in\bar{R}_{\varepsilon}(A)$ we have $J\cap k_{\varepsilon}=(0)$, so
that $a\notin J$ since $a\in k_{\varepsilon}$.)

Now the set $\\{J\in\operatorname{Id}(A),a\notin J\\}$ is both open and closed
hence the closure of the set of primary ideals of $A_{\varepsilon}[f^{-1}]$
not containing $a$ is equal to
$\operatorname{\bar{P}}(A_{\varepsilon}[f^{-1}])\cap\\{J\in\operatorname{Id}(A),a\notin
J\\}$. It follows that

$\displaystyle\operatorname{\bar{P}}(A_{\varepsilon}[f^{-1}])\cap\\{J\in\operatorname{Id}(A),a\notin
J\\}$ $\displaystyle=\overline{\\{J\in{\rm P}(A_{\varepsilon}),f^{n}\notin
J~{}\text{for all}~{}n,a\notin J\\}}$
$\displaystyle\subseteq\overline{\\{J\in{\rm P}(A_{\varepsilon}),af\notin
J\\}}$

and we conclude that $af\notin I$. This proves
$\bar{R}_{\varepsilon}(A)\cap\operatorname{\bar{P}}(A_{\varepsilon}[f^{-1}])=\bar{R}_{\varepsilon}(A\langle
f^{-1}\rangle)$ and concludes the proof. ∎

###  5.3. Formal affinoid subdomains

Let $A$ be any strictly affinoid $k$-algebra. An affinoid subdomain $V$ of
$\mathcal{M}(A)$ is a compact subset $V\subset\mathcal{M}(A)$ such that there
exists a bounded morphism of affinoid $k$-algebras $\alpha\colon A\to A_{V}$
satisfying the following universal property. Given any bounded morphism of
affinoid $k$-algebras $\beta\colon A\to B$ such that
$\beta^{*}(\mathcal{M}(B))\subset V$, then we may write
$\beta=\theta\circ\alpha$ for some unique bounded morphism $\theta\colon
A_{V}\to B$.

For any elements $a,b\in A$ generating the unit ideal, and for any positive
$r,s\in|k^{*}|$, the Weierstrass domain $\\{x\in\mathcal{M}(A),\,|a(x)|\leq
r\\}$, the Laurent domain $\\{x\in\mathcal{M}(A),\,|a(x)|\leq r,|b(x)|\geq
s\\}$, and the rational domain $\\{x\in\mathcal{M}(A),\,|a(x)|\leq
r|b(x)|\\}$, are all affinoid subdomains of $\mathcal{M}(A)$. In each case, it
is possible to describe the affinoid algebra of the affinoid subdomain
explicitely in terms of $A$ and the data $a,b,r,s$. For instance, we have

$\\{x\in\mathcal{M}(A),\,|b(x)|\geq 1\\}=\mathcal{M}\left(A\langle
b^{-1}\rangle\right)$

where $A\langle b^{-1}\rangle=A\\{T\\}/(Tb-1)$ is endowed with the residual
norm, and the ring $A\\{T\\}:=\\{\sum_{n}a_{n}T^{n},\,a_{n}\in A,|a_{n}|\to
0\\}$ with the Gauss norm.

Following [Ber90, §4.3], we say that an affinoid subdomain $V$ of
$\mathcal{M}(A)$ is formal when the morphism induced at the level of their
(standard) reductions
$\operatorname{Spec}\tilde{A}_{V}\to\operatorname{Spec}\tilde{A}$ is an open
immersion. It follows from [BGR84, 7.2.6] that for any $f\in A^{\circ}$ such
that $\|f\|_{\sup}=1$, then the Laurent domain $V:=\\{|f|\geq
1\\}=\\{|f|=1\\}$ is a formal affinoid subdomain so that
$\tilde{A}_{V}=\tilde{A}[\tilde{f}^{-1}]$. Conversely, by [Bo77, Satz 1.2],
any formal affinoid subdomain of $\mathcal{M}(A)$ is a finite union of domains
of the form $\\{|f_{i}|\geq 1\\}$ with $f_{i}\in A^{\circ}$.

######  Lemma 5.8.

For any formal affinoid subdomain $V\subset\mathcal{M}(A)$ and for any
$\varepsilon>0$, the canonical map
$\operatorname{Id}((A_{V})_{\varepsilon})\to\operatorname{Id}(A_{\varepsilon})$
induces a homeomorphism from
$(\operatorname{\bar{P}}((A_{V})_{\varepsilon}),\tau^{\operatorname{con}})$
onto a closed subset of
$(\operatorname{\bar{P}}(A_{\varepsilon}),\tau^{\operatorname{con}})$.

###### Proof.

Let us first consider the case $V=\\{|f|\geq 1\\}$ with $f\in A^{\circ}$. By
[BGR84, 7.2.6/2] the morphism $\sigma\colon A\to A_{V}$ satisfies
$|\sigma(a)|_{\sup}=\lim_{n\to\infty}|f^{n}a|_{\sup}\leq|a|_{\sup}$ so that we
have a canonical morphism $\sigma_{\varepsilon}\colon
A_{\varepsilon}\to(A_{V})_{\varepsilon}$. Since $f$ is a unit in
$A_{V}^{\circ}$, we get the following commutative diagram

$\textstyle{A_{\varepsilon}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sigma_{\varepsilon}}$$\scriptstyle{\phi}$$\textstyle{(A_{V})_{\varepsilon}}$$\textstyle{A_{\varepsilon}[f_{\varepsilon}^{-1}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\imath}$

where $f_{\varepsilon}$ denotes the class of $f$ in $A_{\varepsilon}$. Note
that the lemma follows from Proposition 5.5 as soon as we can prove that
$\imath$ is an isomorphism.

To see that $\imath$ is injective, suppose that
$\imath(af_{\varepsilon}^{-n})=0$ with $a\in A_{\varepsilon}$ and $n\geq 0$.
Then $\imath(a)=0$ so that $\sigma_{\varepsilon}(a)=0$. Choose $b\in
A^{\circ}$ representing $a$. Since $|\sigma(b)|_{\sup}<\varepsilon$, we have
$|f^{m}b|_{\sup}<\varepsilon$ for some integer $m$, hence
$f_{\varepsilon}^{m}a=0$ in $A_{\varepsilon}$, which implies $a=0$ in
$A_{\varepsilon}[f_{\varepsilon}^{-1}]$, and $\imath$ is injective.

Pick any $b\in A^{\circ}_{V}$. Observe that $A[f^{-1}]$ is dense (for the sup-
norm) in $A_{V}=A\langle f^{-1}\rangle$, hence we can find $a\in A$ and
$n\in{\mathbb{N}}^{*}$ such that $|f^{n}b-\sigma(a)|_{\sup}<\varepsilon$. But
this implies $\sigma(a)\in A_{V}^{\circ}$, and $b=f^{-n}\sigma(a)$ modulo
$\\{\rho<\varepsilon\\}$. It follows that $\imath$ is surjective, and the
claim is proven.

Suppose now that $V$ is a general formal affinoid subdomain of
$\mathcal{M}(A)$. Then $V$ is a finite union of affinoid subdomains
$V_{i}=\\{|f_{i}|\geq 1\\}$ with $f_{i}\in A^{\circ}$. By what precedes we
have
$\operatorname{\bar{P}}((A_{V_{i}})_{\varepsilon})=\operatorname{\bar{P}}((A_{V})_{\varepsilon}[f_{i,\varepsilon}^{-1}])$.

We claim that $\operatorname{\bar{P}}((A_{V_{i}})_{\varepsilon})$ covers
$\operatorname{\bar{P}}((A_{V})_{\varepsilon})$. To see this, consider the
usual reduction map
$\operatorname{red}\colon\mathcal{M}(A_{V})\to\operatorname{Spec}(\tilde{A}_{V})$,
and denote by $\tilde{f_{i}}$ the image of $f_{i}$ in $\tilde{A}_{V}$. Since
$V=\cup\\{|f_{i}|\geq 1\\}$, we obtain

$\operatorname{red}^{-1}(\cap\\{\tilde{f_{i}}=0\\})=\cap\\{|f_{i}|<1\\}=\emptyset$

so that $\tilde{f_{i}}$ generates the unit ideal of $\tilde{A}_{V}$, and
therefore $f_{i}$ generates the unit ideal of $A_{V}^{\circ}$, and we conclude
using Proposition 5.5.

Now the preceding argument shows that the restriction of the natural map

$\sigma_{\varepsilon}^{*}\colon\operatorname{\bar{P}}((A_{V})_{\varepsilon})\to\operatorname{\bar{P}}(A_{\varepsilon})$

to $\operatorname{\bar{P}}((A_{V_{i}})_{\varepsilon})$ is continuous and
injective. To conclude the proof it remains to show that $\phi$ is injective
on $\operatorname{\bar{P}}((A_{V})_{\varepsilon})$. Pick
$x,y\in\operatorname{\bar{P}}((A_{V})_{\varepsilon})$ such that
$\sigma_{\varepsilon}^{*}(x)=\sigma_{\varepsilon}^{*}(y)\in\operatorname{Id}(A_{\varepsilon})$.
Then $x$ and $y$ are two ideals of $(A_{V})_{\varepsilon}$ and we may find an
index $i$ such that $f_{i,\varepsilon}\notin x$. If $f_{i,\varepsilon}\notin
x$ and $f_{i,\varepsilon}\in y$, then we have
$f_{i,\varepsilon}\notin\sigma_{\varepsilon}^{*}(x)$ and
$f_{i,\varepsilon}\in\sigma_{\varepsilon}^{*}(y)$ hence
$\sigma_{\varepsilon}^{*}(x)\neq\sigma_{\varepsilon}^{*}(y)$. Hence
$f_{i,\varepsilon}\notin y$ and $x=y$ since $\phi$ is injective on
$\operatorname{Id}((A_{V_{i}})_{\varepsilon})$. ∎

###  5.4. The functor of $\varepsilon$-reduction on $k^{\circ}$-models

Let $X$ be any projective variety over $k$, and $\mathfrak{X}$ a model of $X$
over $k^{\circ}$ (in the sense of §2.2). Recall that there is a reduction map
$\operatorname{red}_{\mathfrak{X}}\colon
X^{\operatorname{an}}\to\mathfrak{X}_{s}$. In this section, we globalize the
construction of the previous section, and build an $\varepsilon$-reduction map
associated to $\mathfrak{X}$ for any $0<\varepsilon\leq 1$. Let us first
explain how we construct the space $\mathfrak{X}[\varepsilon]$.

#### 5.4.1.

The first step of the construction is to endow $X^{\operatorname{an}}$ with a
canonical formal structure in the sense of [Bo77] (see also the discussion in
[Gu98, §1] or [Ber90, §6.4]), that is, with a covering by affinoid domains
$V_{i}$ such that for each $i,j$, $V_{ij}:=V_{i}\cap V_{j}$ is a formal
affinoid subdomain of $V_{i}$.

To do so we proceed as follows. Since $\mathfrak{X}$ is projective, we may fix
once and for all an embedding of $k^{\circ}$-scheme
$\imath\colon\mathfrak{X}\to\mathbb{P}^{N}_{k^{\circ}}$. In particular, we
have an embedding of the special fiber
$\imath_{s}\colon\mathfrak{X}_{s}\to\mathbb{P}^{N}_{\tilde{k}}$, and an
embedding of the analytifications $\imath_{\operatorname{an}}\colon
X^{\operatorname{an}}\to\mathbb{P}^{N,\operatorname{an}}_{k}$. All these
morphisms fit into the following commutative diagram:

$\textstyle{X^{\operatorname{an}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\imath_{\operatorname{an}}}$$\scriptstyle{\operatorname{red}_{\mathfrak{X}}}$$\textstyle{\mathbb{P}^{N,\operatorname{an}}_{k}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{red}_{\mathbb{P}^{N}_{k^{\circ}}}}$$\textstyle{\mathfrak{X}_{s}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\imath_{s}}$$\textstyle{\mathbb{P}^{N}_{\tilde{k}}}$

Take homogeneous coordinates $[z_{0}:\cdots:z_{N}]$ on $\mathbb{P}^{N}$, and
set $U_{i}:=\\{z_{i}\neq 0\\}$. This defines a finite cover
$\mathfrak{U}=\\{U_{i}\\}$ of $\mathbb{P}^{N}_{k^{\circ}}$ by affine subscheme
over $k^{\circ}$, and we get an induced finite cover
$\mathfrak{V}_{s}=\\{V_{i,s}\\}$ of $\mathfrak{X}_{s}$ by affine subschemes
given by $V_{i,s}:=\imath_{s}^{-1}(U_{i}\cap\mathbb{P}^{N}_{\tilde{k}})$.

######  Lemma 5.9.

For any $i$, $V_{i}:=\operatorname{red}_{\mathfrak{X}}^{-1}(V_{i,s})$ is an
affinoid domain in $X^{\operatorname{an}}$.

Moreover, for any principal open affine subscheme $W_{s}$ of $V_{i,s}$, the
set $W:=\operatorname{red}_{\mathfrak{X}}^{-1}(W_{s})$ is a formal affinoid
subdomain of $V_{i}$.

###### Proof.

To simplify notation, let $i=0$. In the chart $U_{0}=\\{z_{0}\neq 0\\}$, we
may write $\mathfrak{X}=\operatorname{Spec}\mathcal{A}$ where
$\mathcal{A}:=k^{\circ}[z_{1},\cdots,z_{N}]/\mathfrak{I}$ where $\mathfrak{I}$
is a finitely generated ideal. Then
$\operatorname{red}_{\mathfrak{X}}^{-1}(V_{0,s})$ is the set of multiplicative
semi-norms $x$ on $\mathcal{A}\otimes_{k^{\circ}}k$ such that $|f(x)|\leq 1$
for all $f\in\mathcal{A}$.

Let $\hat{\mathcal{A}}:=\projlim_{\pi\in k^{\circ\circ}}\mathcal{A}/(\pi)$ be
the formal completion of $\mathcal{A}$, and set
$\hat{\mathcal{A}}_{k}:=\hat{\mathcal{A}}\otimes_{k^{\circ}}k$. One checks
that $\hat{\mathcal{A}}=k^{\circ}\langle
z_{1},\cdots,z_{N}\rangle/\mathfrak{I}$, hence $\hat{\mathcal{A}}_{k}=k\langle
z_{1},\cdots,z_{N}\rangle/\mathfrak{I}$ is a $k$-algebra which becomes an
affinoid algebra when endowed with the residue norm $|\cdot|_{\mathrm{res}}$
coming from the morphism $k\langle
z_{1},\cdots,z_{N}\rangle\to\hat{\mathcal{A}}_{k}$.

Observe that $\mathcal{A}\otimes_{k^{\circ}}k$ is a dense subring of
$\hat{\mathcal{A}}_{k}$. Any norm on $\mathcal{A}\otimes_{k^{\circ}}k$ bounded
by $1$ on $\mathcal{A}$ is bounded by $|\cdot|_{\mathrm{res}}$ hence defines a
point in the Berkovich spectrum of the affinoid algebra
$\hat{\mathcal{A}}_{k}$. Conversely any norm in this spectrum restricts to a
norm on $\mathcal{A}\otimes_{k^{\circ}}k$ bounded by $1$ on $\mathcal{A}$.
This proves $\operatorname{red}_{\mathfrak{X}}^{-1}(V_{0,s})$ is equal to the
Berkovich spectrum of $\hat{\mathcal{A}}_{k}$, and is an affinoid domain.

Suppose now that $\tilde{f}\in\tilde{k}[V_{i,s}]$ and $W_{s}=\\{\tilde{f}\neq
0\\}\subset V_{0,s}$. Let $f$ be a lift of $\tilde{f}$ to $\mathcal{A}$. Then
$\operatorname{red}_{\mathfrak{X}}^{-1}(W_{s})=\\{|f|\geq 1\\}$ is a formal
affinoid subdomain of $V_{0}$. ∎

The previous lemma implies that $\mathfrak{V}:=\\{V_{i}\\}$ is a finite cover
of $X^{\operatorname{an}}$ by affinoid domains such that $V_{ij}$ is a formal
affinoid subdomain of $V_{i}$ for all $i$ and $j$, and therefore defines a
formal structure on $X^{\operatorname{an}}$. Declare that a formal structure
$\mathfrak{V}^{\prime}=\\{V^{\prime}_{j}\\}$ on $X^{\operatorname{an}}$ is
equivalent to $\mathfrak{V}$ if $V^{\prime}_{j}\cap V_{i}$ is a formal
affinoid subdomain of $V_{i}$ for all $i$ and $j$.

We claim that the formal structure defined by $\mathfrak{U}$ on
$X^{\operatorname{an}}$ is independent on the choice of an embedding into
$\mathbb{P}^{N}_{k^{\circ}}$. Indeed, choose another embedding
$X^{\operatorname{an}}\subset\mathbb{P}^{N^{\prime},\operatorname{an}}_{k}$,
and consider the corresponding finite covers $\\{V^{\prime}_{j,s}\\}$ of
$\mathfrak{X}_{s}$, and $\\{V^{\prime}_{j}\\}$ of $X^{\operatorname{an}}$
respectively. Note that $V^{\prime}_{j,s}\cap V_{i,s}$ is a affine open
subscheme of $V_{i,s}$ for all $i$ and $j$ so that again by the previous
lemma, $V^{\prime}_{j}\cap V_{i}$ is a formal affinoid subdomain for all
$i,j$.

#### 5.4.2.

Let us first recall the construction of the reduction associated to the formal
structure defined by $\mathfrak{X}$ following [Bo77, Gu98]. Recall we are
given a finite cover $\mathfrak{V}=\\{V_{i}\\}$ by affinoid domains such that
for all $i$ and $j$, we have $V_{ij}$ is a formal affinoid subdomain of
$V_{i}$.

For each $i$, consider the quotient algebra
$A_{i}:=\mathcal{O}(V_{i})^{\circ}/\mathcal{O}(V_{i})^{\circ\circ}$. By
[BGR84, 6.3.4/3], this is a reduced $\tilde{k}$-algebra of finite type, hence
$\tilde{V}_{i}:=\operatorname{Spec}(A_{i})$ is an affine variety over
$\tilde{k}$. Since $V_{ij}$ is a formal affinoid subdomain, its reduction
$\tilde{V}_{ij}$ is an open affine subscheme of $\tilde{V}_{i}$. We may thus
patch the $\tilde{V}_{i}$ along their intersections $\tilde{V}_{ij}$ to define
a $\tilde{k}$-scheme of finite type $\mathfrak{X}\langle 1\rangle$. The
canonical reduction maps on affinoid algebras also patch together, and defines
an anti-continuous surjective map $\operatorname{red}_{1}\colon
X^{\operatorname{an}}\to\mathfrak{X}\langle 1\rangle$.

The construction of $\mathfrak{X}\langle 1\rangle$ and
$\operatorname{red}_{1}$ do not depend on the choice of covering defining the
formal structure induced by $\mathfrak{X}$.

######  Remark 5.6.

There is a canonical finite surjective morphism $\mathfrak{X}\langle
1\rangle\to\mathfrak{X}_{s}$, which is an isomorphism when $\mathfrak{X}_{s}$
is reduced, see [GRW17, §2.11] and [Me19, §4.1] for a more detailed discussion
of these facts.

#### 5.4.3.

We generalize the previous discussion to the case $\varepsilon$ is any
positive real number less than $1$. For each $i$, endow
$\operatorname{\bar{P}}(\mathcal{O}(V_{i})_{\varepsilon})$ with its
constructible topology. By Lemma 5.8, the restriction map
$\mathcal{O}(V_{i})\to\mathcal{O}(V_{ij})$ induces a continuous and injective
map

$\operatorname{\bar{P}}(\mathcal{O}(V_{ij})_{\varepsilon})\to\operatorname{\bar{P}}(\mathcal{O}(V_{i})_{\varepsilon}).$

We set $\mathfrak{X}[\varepsilon]$ to be the disjoint union of
$\operatorname{\bar{P}}(\mathcal{O}(V_{i})_{\varepsilon})$ modulo the relation
that two points
$x_{i}\in\operatorname{\bar{P}}(\mathcal{O}(V_{i})_{\varepsilon})$ and
$x_{j}\in\operatorname{\bar{P}}(\mathcal{O}(V_{j})_{\varepsilon})$ are
identified whenever there exists
$x_{ij}\in\operatorname{\bar{P}}(\mathcal{O}(V_{ij})_{\varepsilon})$ whose
images in $\operatorname{\bar{P}}(\mathcal{O}(V_{i})_{\varepsilon})$ and
$\operatorname{\bar{P}}(\mathcal{O}(V_{j})_{\varepsilon})$ are $x_{i}$ and
$x_{j}$ respectively.

We endow $\mathfrak{X}[\varepsilon]$ with the quotient topology which is the
finest one making all maps
$\operatorname{\bar{P}}(\mathcal{O}(V_{j})_{\varepsilon})\to\mathfrak{X}[\varepsilon]$
continuous. Note that patching finitely many compact sets always yields a
compact space (the Hausdorff property can be easily seen using the Tietze-
Urysohn’s theorem). It follows that $\mathfrak{X}[\varepsilon]$ is a compact
topological space.

By Lemma 5.6 the $\varepsilon$-reduction maps
$\operatorname{red}_{\varepsilon}\colon\mathcal{M}(\mathcal{O}(V_{i}))\to\operatorname{Id}(\mathcal{O}(V_{j})_{\varepsilon})$
constructed in §5.1 take its values in
$\operatorname{\bar{P}}(\mathcal{O}(V_{j})_{\varepsilon})$ and patch together,
and yield an anti-continuous map

$\operatorname{red}_{\varepsilon}\colon
X^{\operatorname{an}}\to\mathfrak{X}[\varepsilon].$

#### 5.4.4.

The $\varepsilon$-reduction is functorial in the following sense. If
$\mathfrak{X}$ and $\mathfrak{Y}$ are two models of $X$ and $Y$ respectively,
and $\mathfrak{f}\colon\mathfrak{X}\to\mathfrak{Y}$ is a morphism of schemes
over $k^{\circ}$, then there is a unique map
$f_{\varepsilon}\colon\mathfrak{X}[\varepsilon]\to\mathfrak{Y}[\varepsilon]$
so that for any open affine subschemes $\tilde{V}\subset\mathfrak{X}_{s}$ and
$\tilde{W}\subset\mathfrak{Y}_{s}$ such that
$\tilde{V}\subset\mathfrak{f}^{-1}(\tilde{W})$, the map
$f_{\varepsilon}|_{\operatorname{\bar{P}}(\mathcal{O}(V)_{\varepsilon})}$ is
the canonical map
$\operatorname{\bar{P}}(\mathcal{O}(V)_{\varepsilon})\to\operatorname{\bar{P}}(\mathcal{O}(W)_{\varepsilon})$.
Note that the following diagram is commutative:

(16)
$\textstyle{X^{\operatorname{an}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{\operatorname{red}_{\varepsilon}}$$\textstyle{Y^{\operatorname{an}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\operatorname{red}_{\varepsilon}}$$\textstyle{\mathfrak{X}[\varepsilon]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{\varepsilon}}$$\textstyle{\mathfrak{Y}[\varepsilon]}$

The map $f_{\varepsilon}$ is continuous for both the Zariski and the
constructible topology.

It follows from the functoriality that $\mathfrak{X}[\varepsilon]$ does not
depend on the choice of affine cover of $\mathfrak{X}_{s}$. Note that for each
affine subscheme $\tilde{U}$ of $\mathfrak{X}_{s}$, the natural map
$\operatorname{\bar{P}}(\mathcal{O}(U)_{\varepsilon})\to\mathfrak{X}[\varepsilon]$
is a continuous injective map.

######  Remark 5.7.

Pick any ring $R$. Then we may introduce the following natural topology
$\tau_{z}$ on $\operatorname{\bar{P}}(R)$, generated by sets of the form

$\operatorname{\bar{P}}(R[f^{-1}])=\\{I\in\operatorname{\bar{P}}(R),\text{ if
}gf^{n}\in I\text{ for some }n\geq 0,\text{ then }g\in I\\},$

where $f$ ranges over all elements in $R$. Note that the restriction of
$\tau_{z}$ on $\operatorname{Spec}R\subseteq\operatorname{\bar{P}}(R)$ is the
Zariski topology. Endowing each set
$\operatorname{\bar{P}}(\mathcal{O}(V_{j})_{\varepsilon})$ with this topology
induces a topology $\tau_{z}$ on $\mathfrak{X}[\varepsilon]$. The topological
space $(\mathfrak{X}[\varepsilon],\tau_{z})$ carries a canonical structure
sheaf and naturally fits into the category of ringed spaces that are locally
isomorphic to $(\operatorname{\bar{P}}(R),\tau_{z},R)$ for a ring $R$. We
propose to call these objects $\imath$-schemes. There is a natural equivalence
of categories between schemes and $\imath$-schemes. It would be interesting to
investigate in further details the properties of this new category.

######  Remark 5.8.

We choose to patch the closure of the set of primary ideals
$\operatorname{\bar{P}}(A_{\varepsilon})$ in $A_{\varepsilon}$ instead of
patching directly $\operatorname{Id}(A_{\varepsilon})$ because for the latter
set Proposition 5.5 (hence Lemma 5.8) is not valid.

#### 5.4.5.

For any $\varepsilon>0$, denote by $\bar{R}_{\varepsilon}(\mathfrak{X})$ the
closure of the image by $\operatorname{red}_{\varepsilon}$ of
$X^{\operatorname{an}}$. This is a compact subspace of
$\mathfrak{X}[\varepsilon]$.

Suppose that $\varepsilon\in|k^{*}|$, and let $V_{i}$ be a finite cover by
affinoid domains of $X^{\operatorname{an}}$ as before. Then

$\bar{R}_{\varepsilon}(\mathfrak{X})=\cup_{i}\bar{R}_{\varepsilon}(\mathcal{O}(V_{i}))\subset\cup_{i}\operatorname{\bar{P}}(\mathcal{O}(V_{i})_{\varepsilon})=\mathfrak{X}[\varepsilon].$

It follows from Proposition 5.7 that each set
$\bar{R}_{\varepsilon}(\mathcal{O}(V_{i}))\cap\bar{R}_{\varepsilon}(\mathcal{O}(V_{j}))$
is a clopen set in $\bar{R}_{\varepsilon}(\mathcal{O}(V_{i}))$ so that
$\bar{R}_{\varepsilon}(\mathcal{O}(V_{i}))$ forms a clopen cover of
$\bar{R}_{\varepsilon}(\mathfrak{X})$.

######  Remark 5.9.

The space $\bar{R}_{\varepsilon}(\mathcal{O}(V_{i}))$ does not depend only on
$A_{\varepsilon}$ which makes the descent argument in the next section more
involved. This explains why we prefered to work with
$\mathfrak{X}[\varepsilon]$ instead of $\bar{R}_{\varepsilon}(\mathfrak{X})$.

###  5.5. Proof of Theorem B

Recall the setting. Let $X$ be any projective variety over $k$, and
$\mathfrak{X}$ a model of $X$ over $k^{\circ}$. Let $f\colon X\to X$ be a
(regular) algebraic self-map whose extension
$\mathfrak{f}\colon\mathfrak{X}\to\mathfrak{X}$ remains regular. We want to
prove that the induced map $f\colon X^{\operatorname{an}}\to
X^{\operatorname{an}}$ has topological entropy $0$.

######  Proposition 5.10.

For any finite open cover $\mathfrak{U}$ of $X^{\operatorname{an}}$, there
exists $\varepsilon\in|k^{*}|$ and a finite clopen cover $\mathfrak{V}$ of
$\bar{R}_{\varepsilon}(\mathfrak{X})$ such that
$\operatorname{red}_{\varepsilon}^{-1}(\mathfrak{V})$ refines $\mathfrak{U}$.

###### Proof.

Pick any finite affine cover $V_{j}$ of $\mathfrak{X}_{s}$. For each $j$,
$\\{U_{i}\cap\operatorname{red}_{\mathfrak{X}}^{-1}(V_{j})\\}$ is a finite
open cover of the affinoid domain
$\operatorname{red}_{\mathfrak{X}}^{-1}(V_{j})$. By Lemma 5.2, there exist
$\varepsilon\in|k^{*}|$ and a finite constructible cover $\mathfrak{V}_{i}$ of
$(\operatorname{Id}(\mathcal{O}(V_{j})_{\varepsilon}),\tau^{\operatorname{con}})$
such that $\operatorname{red}_{\varepsilon}^{-1}(\mathfrak{V}_{j})$ is a
refinement of $\\{U_{i}\cap\operatorname{red}_{\mathfrak{X}}^{-1}(V_{j})\\}$.

Since $\bar{R}_{\varepsilon}(\mathcal{O}(V_{i}))$ is a clopen subset of
$\bar{R}_{\varepsilon}(\mathfrak{X})$ (see §5.4.5),
$\mathfrak{V}:=\bigcup_{i}(\mathfrak{V}_{i}\cap\bar{R}_{\varepsilon}(\mathcal{O}(V_{i})))$
forms a finite clopen cover of $\bar{R}_{\varepsilon}(\mathfrak{X})$ such that
$\operatorname{red}_{\mathfrak{X}}^{-1}(\mathfrak{V})$ refines $\mathfrak{U}$
as required. This concludes the proof. ∎

Let $\mathfrak{U}$ be a finite open cover of $X^{\operatorname{an}}$. By
Proposition 5.10, there is $\varepsilon\in|k^{*}|$ and a finite clopen cover
$\mathfrak{V}$ of $\bar{R}_{\varepsilon}(\mathfrak{X})$ such that
$\operatorname{red}_{\varepsilon}^{-1}(\mathfrak{V})$ refines $\mathfrak{U}$.
As in Section 1, for each $n\in{\mathbb{N}}$, we write
$\mathfrak{U}_{n}=\mathfrak{U}\vee\cdots\vee f^{-(n-1)}(\mathfrak{U})$ and
$\mathfrak{V}_{n}=\mathfrak{V}\vee\cdots\vee
f_{\varepsilon}^{-(n-1)}(\mathfrak{V})$. Since every element in
$\mathfrak{V}_{n}$ is open and closed, a subset of $\mathfrak{V}_{n}$ covers
$\operatorname{red}_{\varepsilon}(X^{\operatorname{an}})$ if and only if it
covers $\bar{R}_{\varepsilon}(\mathfrak{X})$. Hence

$h(f,\mathfrak{U})\leq
h(f_{\varepsilon}|_{\bar{R}_{\varepsilon}(\mathfrak{X})},\mathfrak{V})\leq\operatorname{h_{top}}(f_{\varepsilon}|_{\bar{R}_{\varepsilon}(\mathfrak{X})}).$

We conclude the proof by the following proposition.

######  Proposition 5.11.

For any $1\geq\varepsilon>0$, the topological entropy of the map
$f_{\varepsilon}\colon\mathfrak{X}[\varepsilon]\to\mathfrak{X}[\varepsilon]$
is $0$. In particular, the topological entropy of the induced map
$f_{\varepsilon}\colon\bar{R}_{\varepsilon}(\mathfrak{X})\to\bar{R}_{\varepsilon}(\mathfrak{X})$
is also $0$.

The idea is to reduce the problem to a Noetherian situation where Theorem 4.2
can be applied. To do so, we develop an argument of absolute Noetherian
approximation analog to the one used in the proof of Theorem 4.2.

####  5.5.1. Gluing data

Let $R$ be a commutative ring with unit. A Gluing Data over $R$ (or (GD) for
short) is the following collection of data (1) – (7) subject to the conditions
(a) – (f) below:

  * (1)

a family of $R$-algebras $A_{i}$ indexed by a finite set $I$;

  * (2)

for every pair $i,j\in I$, $h_{i,j}\in A_{i}$ with $h_{i,i}=1$ for $i\in I;$

  * (3)

for every pair $i,j\in I$, isomorphisms $\alpha_{i,j}^{*}\colon A_{j,i}\to
A_{i,j}$, with $A_{i,j}:=A_{i}[h_{i,j}^{-1}]$;

  * (4)

for every $i,j,k\in I$, isomorphisms $\alpha_{i,j,k}^{*}\colon A_{j,i,k}\to
A_{i,j,k}$, with $A_{i,j,k}:=A_{i}[(h_{i,j}h_{i,k})^{-1}]$;

  * (5)

for every $i\in I$, a finite subset $S_{i}$ of $A_{i}$, and a map $S_{i}\to
A_{i}$, $s\mapsto h_{s}$;

  * (6)

for every $i\in I$, a map $\sigma_{i}\colon S_{i}\to I$, and for every $s\in
S_{i}$, a morphism $\theta_{i,s}^{*}\colon A_{\sigma_{i}(s)}\to
A_{i}[s^{-1}],$ $i\in I,s\in S_{i}$;

  * (7)

for every $i\in I$, and $s_{1},s_{2}\in S_{i},$ a morphism
$\theta^{*}_{i,s_{1},s_{2}}\colon A_{\sigma_{i}(s_{1}),\sigma_{i}(s_{2})}\to
A_{i}[(s_{1}s_{2})^{-1}]$.

We require these data to satisfy the following conditions:

  * (a)

$\alpha_{j,i}^{*}\circ\alpha_{i,j}^{*}=\operatorname{id}$ for every $i,j\in
I;$

  * (b)

for every $i,j,k\in I$, the following diagram is commutative

$\textstyle{A_{j,i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{i,j}^{*}}$$\textstyle{A_{i,j}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A_{j,i,k}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{i,j,k}^{*}}$$\textstyle{A_{i,j,k}}$

where the vertical morphisms are the natural ones induced by localizations;

  * (c)

for every $i,j,k\in I$,
$\alpha_{i,j,k}^{*}\circ\alpha_{j,k,i}^{*}=\alpha_{i,k,j}^{*}$;

  * (d)

for every $i\in I$, $\sum_{s\in S_{i}}sh_{s}=1\in A_{i}$;

  * (e)

for every $i\in I$, and every $s_{1},s_{2}\in S_{i}$, the following diagram is
commutative

$\textstyle{A_{\sigma_{i}(s_{1})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta_{i,s_{1}}^{*}}$$\textstyle{A_{i}[s_{1}^{-1}]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A_{\sigma_{i}(s_{1}),\sigma_{i}(s_{2})}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta_{i,s_{1},s_{2}}^{*}}$$\textstyle{A_{i}[(s_{1}s_{2})^{-1}]}$

where the vertical morphisms are the natural ones induced by localizations;

  * (f)

for every $i\in I$, and every $s_{1},s_{2}\in S_{i}$,
$\theta^{*}_{i,s_{2},s_{1}}\circ\alpha^{*}_{\sigma_{i}(s_{2}),\sigma_{i}(s_{1})}=\theta^{*}_{i,s_{1},s_{2}}$.

Assume that $\mathcal{G}$ is a (GD) over $R$. We explain how to define a
compact topological space $X(\mathcal{G})$ , and a continuous self-map
$f_{\mathcal{G}}\colon X(\mathcal{G})\to X(\mathcal{G})$.

First, we construct $X(\mathcal{G})$ as the disjoint union of the sets
$\operatorname{\bar{P}}(A_{i})$ modulo the identification
$x\in\operatorname{\bar{P}}(A_{i})\sim y\in\operatorname{\bar{P}}(A_{j})$ iff
$h_{i,j}\notin x$, $h_{j,i}\notin y$ and $\alpha_{i,j}^{*}(x)=y$. Conditions
(a), (b), (c) imply this to be an equivalence relation. Note that we have
natural maps $\phi_{i}\colon\operatorname{\bar{P}}(A_{i})\to X(\mathcal{G})$
and that these maps are injective. Write
$U_{i}:=\phi_{i}(\operatorname{\bar{P}}(A_{i}))\subset X$. The topology on
$X(\mathcal{G})$ is then defined by declaring that $U$ is open iff for all
$i$, the set $\phi_{i}^{-1}(U\cap U_{i})$ is open for the constructible
topology on $\operatorname{\bar{P}}(A_{i})$. It follows that $U_{i}$ (which is
compact and totally disconnected) forms a finite open cover of
$X(\mathcal{G})$, hence $X(\mathcal{G})$ is compact and totally disconnected.

Note that $\phi_{i}(\operatorname{\bar{P}}(A_{i,j}))=U_{i}\cap U_{j}$ and
$\phi_{j}^{-1}\circ\phi_{i}|_{\operatorname{\bar{P}}(A_{i,j})}\colon\operatorname{\bar{P}}(A_{i,j})\to\operatorname{\bar{P}}(A_{j,i})$
is induced by $\alpha_{i,j}^{*}.$ Also
$\phi_{j}^{-1}\circ\phi_{i}|_{\operatorname{\bar{P}}(A_{i,j,k})}:=\operatorname{\bar{P}}(A_{i,j,k})\to\operatorname{\bar{P}}(A_{j,i,k})$
is induced by $\alpha_{i,j,k}^{*}.$

######  Remark 5.10.

The space $X(\mathcal{G})$ has a natural structure of $\imath$-scheme in the
sense of Remark 5.7. It is also possible to show that $X(\mathcal{G})$ is
defined by a suitable universal property, as in the case of schemes, see
[Stack, Lemma 26.14.1].

Observe that for each $i\in I$, condition (d) implies that
$\\{\operatorname{\bar{P}}(A_{i}[s^{-1}])\\}_{s\in S_{i}}$ forms an open cover
of $\operatorname{Id}(A_{i}).$ For any $i\in I$, and any $s\in S_{i}$, set
$U_{i,s}:=\phi_{i}(\operatorname{\bar{P}}(A_{i}[s^{-1}]))\subset
X(\mathcal{G})$. By conditions (e) and (f), there is a unique continuous map
$f_{\mathcal{G}}\colon X(\mathcal{G})\to X(\mathcal{G})$ such that:

  * •

for any $i\in I$ and any $s\in S_{i}$ we have
$f_{\mathcal{G}}(U_{i,s})\subseteq U_{\sigma_{i}(s)}$;

  * •

and the morphism

$\phi_{\sigma_{i}(s)}^{-1}\circ
f_{\mathcal{G}}\circ\phi_{i}|_{\operatorname{\bar{P}}(A_{i}[s^{-1}])}\colon\operatorname{\bar{P}}(A_{i}[s^{-1}])\to\operatorname{\bar{P}}(A_{\sigma_{i}(s)})$

is induced by $\theta_{i,s}^{*}.$

######  Remark 5.11.

As above, the map $f_{\mathcal{G}}$ is a morphism of $\imath$-schemes, and it
can be characterized by a suitable universal property.

######  Definition 5.2.

We say that a (GD) over $R$ is finitely generated if the $R$-algebras $A_{i}$
are finitely generated over $R$ for all $i\in I$.

######  Remark 5.12.

If $R^{\prime}$ is a subring of $R$, then any (GD) over $R$ is a (GD) over
$R^{\prime}$.

####  5.5.2. Gluing data for $(\mathfrak{X}[\varepsilon],f_{\varepsilon})$

Fix any $0<\varepsilon\leq 1$. Our objective in this section is to construct a
(GD) $\mathcal{G}_{\varepsilon}$ that is finitely generated over
$k_{\varepsilon}$, and such that $X(\mathcal{G}_{\varepsilon})$ is
homeomorphic to $\mathfrak{X}[\varepsilon]$ and
$f_{\mathcal{G}_{\varepsilon}}$ corresponds to $f_{\varepsilon}$ under that
homeomorphism.

We first collect data from the construction of $\mathfrak{X}[\varepsilon]$
done in §5.4 . Since $\mathfrak{X}_{s}$ is projective, we may choose any very
ample line bundle $L_{s}\to\mathfrak{X}_{s}$, pick a (finite) basis
$\\{\sigma_{i}\\}$ of $H^{0}(\mathfrak{X}_{s},L_{s})$, and set
$\tilde{V}_{i}:=\\{\sigma_{i}\neq 0\\}$. We obtain a finite affine open cover
$\\{\tilde{V}_{i}\\}_{i\in I}$ of $\mathfrak{X}_{s}$. Observe that
$\tilde{h}_{i,j}:=\sigma_{j}/\sigma_{i}\in\mathcal{O}_{\mathfrak{X}_{s}}(\tilde{V}_{i})$,
and $\tilde{V}_{i}\cap\tilde{V}_{j}=\\{\tilde{h}_{i,j}\neq 0\\}$ in
$\tilde{V}_{i}$. Recall that
$V_{i}:=\operatorname{red}_{\mathfrak{X}}^{-1}(\tilde{V}_{i})$ forms a cover
of $X^{\operatorname{an}}$ by affinoid domains (see Lemma 5.9) so that
$\mathcal{O}(V_{i})$ is an affinoid algebra, and we have a canonical finite
morphism
$\mathcal{O}_{\mathfrak{X}_{s}}(\tilde{V}_{i})\to\widetilde{\mathcal{O}(V_{i})}$.

For each $i$, we set $A_{i}:=\mathcal{O}(V_{i})_{\varepsilon}$. By
construction, $U_{i}:=\operatorname{\bar{P}}(A_{i})$ forms an open cover of
$\mathfrak{X}[\varepsilon]$. Note that $A_{i}$ is a finitely generated
$k_{\varepsilon}$-algebra by Lemma 5.1.

Since the reduction map
$\mathcal{O}(V_{i})^{\circ}\to\widetilde{\mathcal{O}(V_{i})}$ is surjective,
we may choose a family of lifts $\hat{h}_{i,j}\in\mathcal{O}(V_{i})^{\circ}$
such that $\hat{h}_{i,i}=1$ for all $i$, and the image of $\hat{h}_{i,j}$
equals $\tilde{h}_{i,j}$ in $\mathcal{O}_{\mathfrak{X}_{s}}(\tilde{V}_{i})$.
Let $h_{i,j}$ be the $\varepsilon$-reduction of $\hat{h}_{i,j}$ in $A_{i}$.
Write $A_{i,j}=A_{i}[h_{i,j}^{-1}]$.

The equality $V_{i}\cap\\{h_{i,j}\neq 0\\}=V_{j}\cap\\{h_{j,i}\neq 0\\}$ in
$\mathfrak{X}_{s}$ gives an isomorphism at the level of affinoid algebras
$\mathcal{O}(V_{i})\langle\hat{h}_{i,j}^{-1}\rangle\to\mathcal{O}(V_{j})\langle\hat{h}_{j,i}^{-1}\rangle$,
and induces a ring isomorphism $\alpha_{i,j}^{*}\colon A_{j,i}\to A_{i,j}$. In
the same way, the equality $V_{i}\cap\\{h_{i,j}h_{i,k}\neq
0\\}=V_{j}\cap\\{h_{j,i}h_{j,k}\neq 0\\}$ in $\mathfrak{X}_{s}$ gives
isomorphisms $\alpha_{i,j,k}^{*}\colon A_{j,i,k}\to A_{i,j,k}$ with
$A_{i,j,k}:=A_{i}[(h_{i,j}h_{i,k})^{-1}]$. The first set of conditions (a),
(b) and (c) are easily satisfied.

Now we collect the data defining the map
$f_{\varepsilon}\colon\mathfrak{X}[\varepsilon]\to\mathfrak{X}[\varepsilon]$.
Recall that we are given morphisms $f\colon X^{\operatorname{an}}\to
X^{\operatorname{an}}$, and
$\mathfrak{f}_{s}\colon\mathfrak{X}_{s}\to\mathfrak{X}_{s}$ such that the
diagram (16) commutes.

It follows that for each $i\in I$, there is a finite subset $\tilde{S}_{i}$ of
$\mathcal{O}_{\mathfrak{X}_{s}}(\tilde{V}_{i})$, and a map
$\tilde{\sigma}_{i}\colon\tilde{S}_{i}\to I$ such that:

  * •

the family $\\{\tilde{s}\neq 0\\}_{\tilde{s}\in\tilde{S}_{i}}$ forms an open
cover of $\tilde{V}_{i}$;

  * •

$\mathfrak{f}_{s}(\tilde{V}_{i}\setminus\\{\tilde{s}=0\\})\subseteq\tilde{V}_{\tilde{\sigma}(\tilde{s})}$
for every $\tilde{s}\in\tilde{S}_{i}.$

Now lift each $\tilde{s}\in\tilde{S}_{i}$ to some $\hat{s}$ in
$\mathcal{O}(V_{i})^{\circ},$ and write $\hat{S}_{i}=\\{\hat{s}\\}$. Since the
ideal generated by $\tilde{S}_{i}$ in
$\mathcal{O}_{\mathfrak{X}_{s}}(\tilde{V}_{i})$ is the unit ideal,
$\hat{S}_{i}$ also generates the unit ideal in $\mathcal{O}(V_{i})^{\circ}$.
Fix a set $\hat{h}_{\hat{s}}\in\mathcal{O}(V_{i})^{\circ}$ such that
$\sum\hat{s}\hat{h}_{\hat{s}}=1$. Let $s$ (resp. $h_{s}$) be the image of
$\hat{s}$ (resp. of $\hat{h}_{\hat{s}}$) in $A_{i}$. Write
$S_{i}=\\{s\\}\subset A_{i}$, and observe that $\sum_{s\in S_{i}}sh_{s}=1$ so
that condition (d) holds. The map $\tilde{\sigma}_{i}$ descends to a map
$\sigma_{i}\colon S_{i}\to I$.

Since
$f(\operatorname{red}_{\mathfrak{X}}^{-1}(\tilde{V}_{i}\setminus\\{\tilde{s}=0\\}))\subseteq\operatorname{red}_{\mathfrak{X}}^{-1}(\tilde{V}_{\tilde{\sigma}(\tilde{s})})$
for every $\tilde{s}\in\tilde{S}_{i}$, the $\varepsilon$-reduction of this
morphism induces a morphism $\theta_{i,s}^{*}\colon A_{\sigma_{i}(s)}\to
A_{i}[s^{-1}]$, and conditions (e) and (f) are easily satisfied.

It follows from the very definitions that $\mathfrak{X}[\varepsilon]$ is
homeomorphic to $X(\mathcal{G}_{\varepsilon})$ and that $f_{\varepsilon}$ is
conjugate to $f_{\mathcal{G}_{\varepsilon}}$ under that homeomorphism.

####  5.5.3. Descent argument

Let us fix a (GD) $\mathcal{G}$ over some ring $R$ as in Section 5.5.1 (so
that it is determined by a finite set $I$, a collection of rings
$\\{A_{i}\\}_{i\in I}$, elements $h_{i,j}\in A_{i}$, etc.). We say that
$\mathcal{G}^{\prime}$ is a sub-(GD) and we write
$\mathcal{G}^{\prime}\subseteq\mathcal{G}$ if the following conditions hold:

  * •

$\mathcal{G}^{\prime}$ is determined by a family of sub-rings $B_{i}\subset
A_{i}$ indexed by the set $I$;

  * •

$B_{i}$ contains $h_{i,j}$ for all $j$, the set $S_{i}$, and $h_{s}$ for all
$s\in S_{i}$;

  * •

the set $S_{i}(\mathcal{G}^{\prime})$ is equal to $S_{i}$, the map
$\sigma_{i}(\mathcal{G}^{\prime})$ is equal to $\sigma_{i}$, and the maps
$S_{i}\to B_{i}$ are induced by the maps $S_{i}\to A_{i}$ (i.e. the image of
$S_{i}\to A_{i}$ belongs to $B_{i}$);

  * •

all morphisms $\alpha_{i,j}^{*}(\mathcal{G}^{\prime})$,
$\alpha_{i,j,k}^{*}(\mathcal{G}^{\prime})$,
$\alpha^{*}_{i,s}(\mathcal{G}^{\prime})$,
$\alpha^{*}_{i,s_{1},s_{2}}(\mathcal{G}^{\prime})$ are induced by the
corresponding ones in $\mathcal{G}$, that is, we impose:

$\displaystyle\alpha_{i,j}^{*}(B_{j,i})\subseteq B_{i,j};\,\,$
$\displaystyle\alpha_{i,j,k}^{*}(B_{j,i,k})\subseteq B_{i,j,k}$
$\displaystyle\alpha_{i,s}^{*}(B_{\sigma_{i}(s)})\subseteq B_{i}[s^{-1}];\,\,$
$\displaystyle\alpha^{*}_{i,s_{1},s_{2}}(B_{\sigma_{i}(s_{1}),\sigma_{i}(s_{2})})\subseteq
B_{i}[(s_{1}s_{2})^{-1}]~{}.$

When $\mathcal{G}^{\prime}$ is a sub-(GD) of $\mathcal{G}$, the inclusion
morphisms $B_{i}\subset A_{i}$ induce a canonical continuous map
$\pi^{\mathcal{G}^{\prime}/\mathcal{G}}\colon X(\mathcal{G})\to
X(\mathcal{G}^{\prime})$ such that
$\pi^{\mathcal{G}/\mathcal{G}^{\prime}}\circ
f_{\mathcal{G}}=f_{\mathcal{G}^{\prime}}\circ\pi^{\mathcal{G}/\mathcal{G}^{\prime}}$.

######  Lemma 5.12.

Suppose $\mathcal{G}$ is a (GD) that is finitely generated over $R$. Then
there exists an inductive set $\mathcal{N}$, a family of Noetherian rings
$\\{R_{t}\\}_{t\in\mathcal{N}}$ and a family of sub-(GD)
$\mathcal{G}_{t}\subset\mathcal{G}$, each defined over $R_{t}$ such that:

  1. (1)

$\mathcal{G}_{t}\subset\mathcal{G}_{t^{\prime}}\subset\mathcal{G}$ for any
$t\leq t^{\prime}$;

  2. (2)

$\mathcal{G}_{t}$ is finitely generated over $R_{t}$ for all
$t\in\mathcal{N}$;

  3. (3)

the canonical map $X(\mathcal{G})\to\projlim_{\mathcal{N}}X(\mathcal{G}_{t})$
is a homeomorphism.

###### Proof.

The set $\mathcal{N}$ will be defined as the poset of all subrings
$R^{\prime}$ of $R$ ordered by inclusion subject to the conditions that
$R^{\prime}$ is finitely generated (over $\mathbf{Z}$) and contains a finite
set $O(\mathcal{G})\subset R$ that we define now.

For each $i$, let us fix $T_{i}=\\{t_{i,\cdot}\\}$ a finite set of generators
of $A_{i}$ as an $R$-algebra that contains $S_{i}$, $h_{i,j}$ for all $j$, and
$h_{s}$ for all $s\in S_{i}$. We express all morphisms
$\alpha_{i,j},\alpha_{i,j,k},\theta_{i,s}$, and $\theta_{i,s_{1},s_{2}}$ in
these basis. For instance, we write

(17)
$\alpha_{i,j}(t_{j,\beta})=\sum_{n\in{\mathbb{Z}}}\sum_{\alpha}r_{\alpha}(n,j,\beta)t_{i}^{\alpha}h_{i,j}^{n}$

where $r(n,j,\beta)\in R$, and the sum is finite. In this way, we collect a
finite set of elements of $R$, and denote by $O(\mathcal{G})$ its union.

Now take any finitely generated subring $R^{\prime}$ of $R$ containing
$O(\mathcal{G})$. We let $\mathcal{G}^{\prime}$ be the (GD) over $R^{\prime}$
determined by the collection of subrings $A^{\prime}_{i}$ of $A_{i}$ generated
by $T_{i}$ as an $R^{\prime}$ algebra.

Define $\alpha_{i,j}^{\prime}\colon A_{j}^{\prime}\to A_{i}$ as the
restriction of the morphism $\alpha_{i,j}\colon A_{j}\to A_{i}$ to
$A_{j}^{\prime}$. By (17), $\alpha_{i,j}^{\prime}(A_{j}^{\prime})$ is
contained in $A_{i}^{\prime}$, so that we get a morphism
$\alpha_{i,j}^{\prime}\colon A_{j}^{\prime}\to A_{i}^{\prime}$. We proceed in
the same way to construct $\alpha^{\prime}_{i,j,k},\theta^{\prime}_{i,s}$, and
$\theta^{\prime}_{i,s_{1},s_{2}}$.

It is clear that $\mathcal{G}^{\prime}$ is a sub-(GD) of $\mathcal{G}$, that
$\mathcal{G}^{\prime}\subset\mathcal{G}^{\prime\prime}$ whenever
$R^{\prime}\subset R^{\prime\prime}$ and $\mathcal{G}^{\prime}$ is finitely
generated over $R^{\prime}$ by construction.

Finally the injective limit of all finitely subrings $R^{\prime}$ of $R$
containing $O(\mathcal{G})$ is equal to $R$, hence for any $i$,
$\operatorname{Id}(A_{i})$ is homeomorphic to
$\projlim\operatorname{Id}(A^{\prime}_{i})$ by Lemma 4.4 which implies (3). ∎

###### Proof of Proposition 5.11.

Let $\mathcal{G}_{\varepsilon}$ be the (GD) associated to
$(\mathfrak{X}[\varepsilon],f_{\varepsilon})$ as defined in Section 5.5.2 so
that $X(\mathcal{G}_{\varepsilon})=\mathfrak{X}[\varepsilon]$ and
$f_{\varepsilon}=f_{\mathcal{G}_{\varepsilon}}$.

Pick any inductive family of sub-(GD)
$\mathcal{G}_{t}\subseteq\mathcal{G}_{\varepsilon}$ as in Lemma 5.12. For each
$t\in\mathcal{N}$, $\mathcal{G}_{t}$ is finitely generated over a Noetherian
ring $R_{t}$. The space $X(\mathcal{G}_{t})$ is a finite union of compact
subsets of Noetherian Priestley spaces of the form $\operatorname{Id}(A)$ for
some Noetherian ring $A$, hence any Radon measure on $X(\mathcal{G}_{t})$ is
atomic by Theorem 3.3. Since $X(\mathcal{G}_{t})$ is compact, the variationnal
principle (Theorem 1.2) implies that
$\operatorname{h_{top}}(f_{\mathcal{G}_{t}})=0$ for all $t$. As
$X(\mathcal{G})$ is homeomorphic to the projective limit
$\projlim_{\mathcal{N}}X(\mathcal{G}_{t})$ and the following diagrams commute

$\textstyle{X(\mathcal{G}_{\varepsilon})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{\mathcal{G}_{\varepsilon}}}$$\scriptstyle{\pi^{\mathcal{G}_{\varepsilon}/\mathcal{G}_{t}}}$$\textstyle{X(\mathcal{G}_{\varepsilon})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{\mathcal{G}_{\varepsilon}/\mathcal{G}_{t}}}$$\textstyle{X(\mathcal{G}_{t})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f_{\mathcal{G}_{t}}}$$\textstyle{X(\mathcal{G}_{t})}$

Lemma 4.3 shows
$\operatorname{h_{top}}(f_{\varepsilon})=\operatorname{h_{top}}(f_{\mathcal{G}_{\varepsilon}})\leq\limsup_{t}\operatorname{h_{top}}(f_{\mathcal{G}_{t}})=0$
as required. ∎

##  Appendix A Topological entropy and field extensions

Let $k$ be any complete metrized field, and $X$ a projective variety defined
over $k$. Let $f\colon X\dashrightarrow X$ be any dominant rational map.
Recall that $\operatorname{h_{top}}(f)$ is defined in §2.4, and coincides with
the topological entropy of the action of $f$ on the Berkovich analytification
of $X$ when $f$ is a regular map.

We propose the following problem.

######  Conjecture 2.

Let $K/k$ be any complete metrized field extension, and let $f_{K}\colon
X_{K}\dashrightarrow X_{K}$ be the rational self-map induced by base change.
Then we have $\operatorname{h_{top}}(f_{K})=\operatorname{h_{top}}(f)$.

Let us prove the following easy bound (thereby proving Proposition 2.4).

######  Proposition A.1.

Let $K/k$ be any complete metrized field extension, and let $f_{K}\colon
X_{K}\dashrightarrow X_{K}$ be the rational self-map induced by base change.
Then $\operatorname{h_{top}}(f_{K})\geq\operatorname{h_{top}}(f)$.

###### Proof.

Base change induces a canonical continuous map $\pi_{K/k}\colon
X^{\operatorname{an}}_{K}\to X^{\operatorname{an}}_{k}$. This map is
surjective since the scheme morphism $X_{K}\to X_{k}$ is surjective, and any
norm on a field $L$ admits at least one extension to any of its field
extension $L^{\prime}/L$, see [ZS60, §VI.6].999This fact also follows from
Lemma 4.1. Observe now that
$\operatorname{Ind}(f_{K})=\operatorname{Ind}(f)_{K}$. We conclude the proof
using Corollary 1.12. ∎

We now prove a partial converse to this proposition.

######  Proposition A.2.

Let $\widehat{k^{\operatorname{alg}}}$ be the completion of an algebraic
closure of $k$. For any regular map $f\colon X\to X$, we have
$\operatorname{h_{top}}(f_{\widehat{k^{\operatorname{alg}}}})=\operatorname{h_{top}}(f)$.

Let $p$ be the characteristic of $k$.

###### Proof.

Let us consider the separable closure $k^{\operatorname{sep}}\subset
k^{\operatorname{alg}}$ which consists of the union of all separable finite
extensions of $k$. Note that the algebraic extension
$k^{\operatorname{alg}}/k^{\operatorname{sep}}$ is purely inseparable (i.e.,
for any $a\in k^{\operatorname{alg}}$ we have $a^{p^{n}}\in
k^{\operatorname{sep}}$ for some $n$), and $k^{\operatorname{sep}}/k$ is an
algebraic Galois extension. It follows that the group
$G:=\operatorname{Gal}(k^{\operatorname{sep}}/k)$ is the projective limit of
$\operatorname{Gal}(K/k)$ where $K$ ranges over the set $\mathfrak{G}$ of all
finite Galois subextensions of $k$ inside $k^{\operatorname{alg}}$, see e.g.
[Lan, VI.§14].

Recall the norm on $k$ extends in a unique way to any finite extension as a
multiplicative norm, so that any element of
$G:=\operatorname{Gal}(k^{\operatorname{sep}}/k)$ induces an isometry on
$k^{\operatorname{sep}}$. In particular $G$ acts isometrically on its
completion $\widehat{k^{\operatorname{sep}}}$.

By [Ber90, Proposition 1.3.5] the analytic space $X_{k}^{\operatorname{an}}$
is homeomorphic to the quotient of
$X_{\widehat{k^{\operatorname{sep}}}}^{\operatorname{an}}$ by the action of
$G$, and the base change morphism
$X_{\widehat{k^{\operatorname{alg}}}}^{\operatorname{an}}\to
X_{\widehat{k^{\operatorname{sep}}}}^{\operatorname{an}}$ is a homeomorphism,
so that

$X_{\widehat{k^{\operatorname{alg}}}}^{\operatorname{an}}\simeq
X_{\widehat{k^{\operatorname{sep}}}}^{\operatorname{an}}\simeq\projlim_{K\in\mathfrak{G}}X_{K}^{\operatorname{an}}~{}.$

Since $f$ lifts to a continuous map to $X_{K}^{\operatorname{an}}$ for any
complete field extension $K/k$, we conclude that
$\operatorname{h_{top}}(f_{\widehat{k^{\operatorname{alg}}}})=\sup_{K\in\mathfrak{G}}\operatorname{h_{top}}(f_{K})$
by Lemma 4.3.

When $K$ is a finite extension of $k$, then the map
$X^{\operatorname{an}}_{K}\to X^{\operatorname{an}}_{k}$ has finite fibers. It
follows from the Abramov-Rokhlin’s formula (see e.g., [AdM79, Proposition
2.15]) that the measure entropy is preserved under finite extension. We
conclude by the variational principle that
$\operatorname{h_{top}}(f)=\operatorname{h_{top}}(f_{K})$, and therefore
$\operatorname{h_{top}}(f_{\widehat{k^{\operatorname{alg}}}})=\operatorname{h_{top}}(f)$.
∎

We conclude this Appendix with the following solution to the conjecture in the
case $\dim(X)=1$.

######  Proposition A.3.

Let $f\colon X\to X$ be any rational self-map of a smooth projective curve $X$
defined over $k$. Then for any complete metrized field extension $K/k$, we
have $\operatorname{h_{top}}(f_{K})=\operatorname{h_{top}}(f)$.

We rely on the next lemma whose proof is left to the reader (see [FR10,
Théorème C]).

######  Lemma A.4.

Suppose $f\colon\mathbb{D}(0,1)\to\mathbb{D}(0,1)$ is an analytic map of the
open unit disk. Then any $f$-invariant probability measure has zero metric
entropy.

###### Proof of Proposition A.3.

We may assume that the genus of $X$ is either $0$ or $1$ since otherwise $f$
has finite order.

By Proposition A.1, it is sufficient to prove
$\operatorname{h_{top}}(f_{K})\leq\operatorname{h_{top}}(f)$. By Proposition
A.2, we may assume that $K$ and $k$ are both algebraically closed. By [Po13,
Corollaires 3.7 et 3.14], there exists a canonical injective continuous map
$\sigma_{K/k}\colon X\to X_{K}$ such that
$\pi_{K/k}\circ\sigma_{K/k}=\operatorname{id}_{\mathbb{P}^{1}_{k}}$. Since
this lift is canonical, we have $\sigma_{K/k}\circ f=f_{K}\circ\sigma_{K/k}$
so that $f_{K}|_{\sigma_{K/k}(X_{k})}=\operatorname{h_{top}}(f)$.

Suppose first that $X$ has genus $1$. By the semi-stable reduction theorem
[Ber90, Theorem 4.3.1], either $X$ admits a smooth model $\mathfrak{X}$ over
$k^{\circ}$, or $X$ retracts by deformation on a unique topological circle
$\operatorname{Sk}(X)\subset X$.

In the former case, the unique point $x$ corresponding to the special fiber of
$\mathfrak{X}$ is totally invariant by $f$ and $X\setminus\\{x\\}$ is a union
of open balls. It follows from Lemma A.4 that $\operatorname{h_{top}}(f)=0$ By
base change, $X_{K}$ also admits a smooth model over $K^{\circ}$, hence
$\operatorname{h_{top}}(f_{K})=0$.

In the latter case, the complement $X\setminus\operatorname{Sk}(X)$ is a union
of open balls, hence
$\operatorname{h_{top}}(f)=\operatorname{h_{top}}(f|_{\operatorname{Sk}(X)})$
by Lemma A.4. Now note that $\sigma_{K/k}(\operatorname{Sk}(X))$ is then a
topological circle, hence by uniqueness we must have
$\operatorname{Sk}(X_{K})=\sigma_{K/k}(\operatorname{Sk}(X))$. We conclude
that

$\operatorname{h_{top}}(f_{K})=\operatorname{h_{top}}(f_{K}|_{\sigma_{K/k}(\operatorname{Sk}(X))})=\operatorname{h_{top}}(f|_{\operatorname{Sk}(X)})=\operatorname{h_{top}}(f)~{}.$

Now suppose that $X=\mathbb{P}^{1}_{k}$. In that case,
$\mathbb{P}^{1}_{K}\setminus\sigma_{K/k}(\mathbb{P}^{1}_{k})$ is a union of
open balls, hence again by Lemma A.4 we have

$\operatorname{h_{top}}(f_{K})=\operatorname{h_{top}}(f_{K}|_{\sigma_{K/k}(\mathbb{P}^{1}_{k})})=\operatorname{h_{top}}(f)$

as required. ∎

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