# On Preimages Question

Yohsuke Matsuzawa  and  Kaoru Sano  Department of Mathematics, Graduate School
of Science, Osaka Metropolitan University, 3-3-138, Sugimoto, Sumiyoshi,
Osaka, 558-8585, Japan  NTT Institute for Fundamental Mathematics, NTT
Communication Science Laboratories, NTT Corporation, 2-4, Hikaridai, Seika-
cho, Soraku-gun, Kyoto 619-0237, Japan<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

For a surjective self-morphism on a projective variety defined over a number
field, we study the preimages question, which asks if the set of rational
points on the iterated preimages of an invariant closed subscheme eventually
stabilize. We prove the answer is positive for self-morphisms on
${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$. We also prove that the answer is
positive even if we allow finite extension of the number field by a bounded
degree if the morphism is étale or the invariant subscheme is $0$-dimensional.
We prove certain uniformity of the stabilization for self-morphisms on abelian
varieties. When the subscheme is not invariant, we provide counterexamples and
a result in a positive direction for products of polynomial maps on
${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$.

###### Key words and phrases:

Self-morphisms on projective varieties, rational points, preimages of
subvarieties.

###### 1991 Mathematics Subject Classification:

Primary: 37P55; Secondary: 37P15, 14G05.

###### Contents

  1. 1 Introduction
  2. 2 Equivalence of two Preimages Questions
  3. 3 Preimages Question for a fixed point
  4. 4 Preimages Question when the subvariety is not invariant
  5. 5 Étale morphisms
  6. 6 Uniformity of $s_{0}$ for abelian varieties
  7. 7 Preliminaries on polydisc algebras
  8. 8 $p$-adic uniformization
  9. 9 Preimages Question on ${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$

##  1\. Introduction

Let $f\colon X\longrightarrow X$ be a surjective morphism on a projective
variety $X$ defined over a number field $K$. For a closed subscheme $Y\subset
X$, the iterated preimages $f^{-n}(Y)$ tend to become complicated as $n$
approaches $\infty$, and so it is reasonable to expect that the $K$-rational
points on $f^{-n}(Y)$ might be sparse in a certain sense. Such a kind of
question is investigated in [2], motivated by a question proposed in [10]. The
central problem is the following ([10, Question 8.4(1)],[2, Question 1.1]).

######  Question 1.1 (Preimages Question (PIQ)).

Let $K$ be a number field. Let $X$ be a projective variety over $K$ (i.e.
integral projective scheme over $K$) and $f\colon X\longrightarrow X$ a
surjective morphism over $K$. Let $Y\subset X$ be an $f$-invariant closed
subscheme (i.e. $f|_{Y}$ factors through $Y$). Then is there
$s_{0}\in{\mathbb{Z}}_{\geq 0}$ such that for all $s\geq s_{0}$, we have

(1.1) $\displaystyle\left(f^{-s-1}(Y)\smallsetminus
f^{-s}(Y)\right)(K)=\emptyset\ ?$

Let us give some comments on the assumptions. We cannot remove the
projectivity of $X$ as there is a counterexample [2, section 6.1]. The
$f$-invariance of $Y$ is more mysterious. We can pose a similar question for
$Y$ that is not necessarily $f$-invariant (see Question 4.1). While there are
counterexamples (cf. Section 4), it seems plausible to assume that certain
additional conditions on $f$ or $Y$ might yield a positive answer. The
$f$-invariance of $Y$ can be seen as one such condition, but we anticipate
that other types of conditions exist. Currently, we do not comprehend what
causes the set of rational points on $f^{-n}(Y)$ to be sparse.

Let us also give a comment on other ground fields. Obviously, the question is
nonsense if $K$ is algebraically closed. A slightly non-trivial fact is that
it has a counterexample over $p$-adic fields in general (see [2, section
6.2]). However, for a specific type of morphisms, PIQ is true over $p$-adic
fields or finitely generated fields over ${\mathbb{Q}}$ (cf. [2, Proof of
Theorem 3.1 Step 2], Theorem 9.1).

######  Remark 1.2.

The statement of Question 1.1 makes sense for arbitrary self-map $f\colon
X\longrightarrow X$ on a set $X$ and its $f$-invariant subset $Y\subset X$. We
use the terminology Preimages Question or PIQ for such situations flexibly.
For example, PIQ is true for $(X,f,Y)$ means there is
$s_{0}\in{\mathbb{Z}}_{\geq 0}$ such that for all $s\geq s$, we have

(1.2) $\displaystyle f^{-s-1}(Y)\smallsetminus f^{-s}(Y)=\emptyset.$

In this terminology, Question 1.1 is equivalent to PIQ for $f\colon
X(K)\longrightarrow X(K)$ and subset $Y(K)\subset X(K)$.

######  Remark 1.3.

For a map $f\colon X\longrightarrow X$ on a set $X$ and an $f$-invariant
subset $Y\subset X$, the following are equivalent.

  1. (1)

PIQ is true for $(X,f,Y)$.

  2. (2)

PIQ is true for $(X,f^{n},Y)$ for some $n\geq 1$.

  3. (3)

There is $s_{0}\geq 0$ with the following property: for every $x\in X$, if
$f^{s}(x)\in Y$ for some $s\geq 0$, then $f^{s_{0}}(x)\in Y$.

Back to Question 1.1, the original [10, Question 8.4(1)] was the following
slightly stronger question.

######  Question 1.4.

Let $K$ be a number field. Let $X$ be a projective variety over $K$ and
$f\colon X\longrightarrow X$ a surjective morphism over $K$. Let $Y\subset X$
be an $f$-invariant closed subscheme. Let $d\in{\mathbb{Z}}_{\geq 0}$. Then is
there a $s_{0}\in{\mathbb{Z}}_{\geq 0}$ such that for all $s\geq s_{0}$, we
have

(1.3) $\displaystyle\left(f^{-s-1}(Y)\smallsetminus
f^{-s}(Y)\right)(L)=\emptyset$

for all field extensions $K\subset L$ with $[L:K]\leq d$.

It turns out:

######  Proposition A (Proposition 2.1).

Question 1.1 for all $K,X,f$, and $Y$ is equivalent to Question 1.4 for all
$K,X,f,Y$, and $d$.

We verify Question 1.4 for the following cases.

######  Theorem B (Theorems 5.1 and 3.1).

The answer to Question 1.4 is affirmative if either $\dim Y=0$, or $f$ is
étale.

Note that Question 1.1 was proven for étale morphisms in [2].

In [2], the authors proved that Question 1.1 is true for self-morphisms on
${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ of the form $f\times f$ and the
diagonal subvariety. We generalized this theorem to arbitrary self-morphisms
on ${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$ and any invariant closed subscheme
(as well as to finitely generated fields).

######  Theorem C (Theorem 9.1).

Let $K$ be a finitely generated field over ${\mathbb{Q}}$. Let
$\varphi\colon{\mathbb{P}}^{1}_{K}\times_{K}{\mathbb{P}}^{1}_{K}\longrightarrow{\mathbb{P}}^{1}_{K}\times_{K}{\mathbb{P}}^{1}_{K}$
be a surjective morphism over $K$. Let
$Y\subset{\mathbb{P}}^{1}_{K}\times_{K}{\mathbb{P}}^{1}_{K}$ be a
$\varphi$-invariant closed subscheme. Then there is $s_{0}\geq 0$ such that
for all $s\geq s_{0}$, we have

(1.4)
$\displaystyle\left(\varphi^{-s-1}(Y)\smallsetminus\varphi^{-s}(Y)\right)(K)=\emptyset.$

We have mentioned that the invariance of $Y$ in Question 1.1 cannot be removed
in general. But we have the following particular case that the answer is
positive.

######  Theorem D (Theorem 4.5).

Let $K$ be a number field. Let
$f,g\colon{\mathbb{P}}^{1}_{K}\longrightarrow{\mathbb{P}}^{1}_{K}$ be
polynomial maps of the same degree $d\geq 2$. Consider the product morphism
$f\times
g\colon{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}\longrightarrow{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}$
and the diagonal subvariety
$\Delta\subset{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}$. There is
$s_{0}\in{\mathbb{Z}}_{\geq 0}$ such that for all $s\geq s_{0}$, we have

(1.5) $\displaystyle\biggl{(}(f\times
g)^{-s-1}(\Delta)\smallsetminus\bigcup_{0\leq i\leq s}(f\times
g)^{-i}(\Delta)\biggr{)}(K)=\emptyset.$

We have seen that the answer to Question 1.1 is positive for some classes of
morphisms and varieties. It is then natural to ask how much extent the
(minimum of the) number $s_{0}$ in Question 1.1 depends on $f$, $Y$, and $X$.
When $X$ is an abelian variety, we prove the following uniformity on $s_{0}$.

######  Theorem E (Theorem 6.4).

Let $X$ be an abelian variety over a number field $K$. Then there exists
$s_{0}\in{\mathbb{Z}}_{\geq 0}$ depending only on $X$ with the following
property. For any surjective morphism $f\colon X\longrightarrow X$ and any
reduced closed subscheme $Y\subset X$ with $f(Y)\subset Y$, we have

(1.6) $\displaystyle(f^{-s_{0}-1}(Y)\smallsetminus
f^{-s_{0}}(Y))(K)=\emptyset.$

######  Remark 1.5.

In the course of the proof of Theorem E, we prove a uniform bound of the
periods of periodic abelian subvarieties under isogenies Lemma 6.2. While our
proof heavily relies on several special properties of abelian varieties, it is
worth noting the similarity of this and the problem raised in [10, Question
8.4 (2)].

######  Remark 1.6.

Under the assumption of general uniform boundedness conjecture on the torsion
points of abelian varieties, we obtain a stronger version of Theorem E,
subject to the condition that $Y$ is geometrically irreducible. Refer to
Proposition 6.6 for details.

Notation.

  1. (1)

A variety over a field $k$ is an irreducible reduced separated scheme of
finite type over $k$. For a separated scheme $X$ of finite type over $k$, a
subvariety of $X$ is an irreducible reduced closed subscheme of $X$.

  2. (2)

For a self-map $f\colon X\longrightarrow X$ on a set $X$,
$\operatorname{Fix}(f,X)$ is the set of fixed point, i.e.
$\operatorname{Fix}(f,X)=\\{x\in X\mid f(x)=x\\}$.

  3. (3)

For a rational number $a$, we set $|a|_{p}=p^{-v_{p}(a)}$, where $v_{p}$ is
the $p$-adic valuation. In particular, $|p^{n}|_{p}=p^{-n}$ for
$n\in{\mathbb{Z}}$.

  4. (4)

A valued field $(k,|\ |)$ is a field equipped with a multiplicative absolute
value $|\ |$.

  5. (5)

For a non-archimedean valued field $k$, the classical (closed) disc of radius
$r>0$ is denoted by $\overline{\mathbf{D}}_{k}(r)$:
$\overline{\mathbf{D}}_{k}(r)=\\{a\in k\mid|a|\leq r\\}$.

Organization of the paper.

In Section 2, we prove Proposition A, i.e., the two preimages questions are
equivalent. In Section 3, we prove that Question 1.4 is true if the invariant
subscheme is $0$-dimensional. In Section 5, we prove it for étale morphisms.
In Section 4, we introduce a variant of Question 1.1 where the subscheme is
not assumed to be invariant. We prove Theorem D, which gives an affirmative
answer. We also note that a certain dynamical structure produces
counterexamples to the generalized question. In Section 6, we prove Theorem E.
In Sections 7 and 8, we recall some facts and prepare several lemmas that are
used in the proof of Theorem C. In Section 9, we prove Theorem C.

######  Acknowledgements.

The authors would like to thank Yusuke Okuyama, Joseph Silverman, and Yu
Yasufuku for helpful discussions. The first author is supported by JSPS
KAKENHI Grant Number JP22K13903. The second author is supported by JSPS
KAKENHI Grant Number JP20K14300.

##  2\. Equivalence of two Preimages Questions

In this section, we prove the following.

######  Proposition 2.1.

Question 1.1 for all $K,X,f$, and $Y$ is equivalent to Question 1.4 for all
$K,X,f,Y$, and $d$.

######  Remark 2.2.

We do not claim the equivalence of two questions for individual dynamical
systems $(X,f,Y)$.

###### Proof.

It is clear that Question 1.4 implies Question 1.1. Assume Question 1.1 is
true for all $K$, $X$, $f$, and $Y$. Let $K$, $X$, $f$, $Y$, and $d$ as in
Question 1.4. By replacing $K$ with a finite extension and replacing $f$ with
iterates, we may assume $X$ is geometrically irreducible. It is enough to show
the statement for $L$ that is Galois over $K$ with $[L:K]=d$. Now let
$X^{(d)}:=X^{d}/\mathfrak{S}_{d}$ be the symmetric product of $X$. Note that
$X^{(d)}$ is projective over $K$ (cf. Remark 2.3).

Let us consider the following commutative diagram

(2.1)
${Y^{d}}$${X^{d}}$${X^{d}}$${Y^{(d)}}$${X^{(d)}}$${X^{(d)}}$${Z}$${Z}$$\scriptstyle{f\times\cdots\times
f}$$\scriptstyle{\pi}$$\scriptstyle{\pi}$$\scriptstyle{g}$${\subset}$${\subset}$

where $\pi$ is the quotient morphism, $Z$ is the scheme theoretic image of
$Y^{(d)}\longrightarrow X^{(d)}$, and $g$ is the induced surjective morphism.
Let $L$ be any Galois extension of $K$ of degree $d$ with Galois group
$G=\\{\sigma_{1},\dots,\sigma_{d}\\}$. For any point $x\in X(L)$, we define
$\widetilde{x}\in X^{d}(L)$ by

(2.2)
${\operatorname{Spec}L}$${X^{d}}$${\operatorname{Spec}L}$${X}$$\scriptstyle{\widetilde{x}}$$\scriptstyle{\sigma_{i}^{*}}$$\scriptstyle{\operatorname{pr}_{i}}$$\scriptstyle{x}$

for all $i=1,\dots,d$. Then we have $\pi\circ\widetilde{x}$ factors through
$\operatorname{Spec}K$ and defines a $K$-point $\overline{x}\in X^{(d)}(K)$.
It is easy to see that

(2.3) $\displaystyle x\in Y(L)\iff\overline{x}\in Z(K).$

Now let $s_{0}\geq 0$ be as in Question 1.1 for $(X^{(d)},g,Z)$. If $x\in
X(L)$ satisfies $f^{s}(x)\in Y(L)$ for some $s\geq 0$, then
$\overline{f^{s}(x)}\in Z(K)$. Since
$\overline{f^{s}(x)}=g^{s}(\overline{x})$, we have $g^{s_{0}}(\overline{x})\in
Z(K)$ and therefore $f^{s_{0}}(x)\in Y(L)$. Since $s_{0}$ depends only on
$X^{(d)},g,Z$, and $K$, we are done.

∎

######  Remark 2.3.

To see $X^{(d)}$ is projective, it is enough to prove that
$X^{(d)}_{\overline{K}}=X_{\overline{K}}^{d}/\mathfrak{S}_{d}$ is projective.
In general, finite group action on a projective variety over an algebraically
closed field is linearizable by a very ample line bundle (cf. [4, Proposition
3.4.5]). Thus, the standard construction of quotient in GIT shows that the
quotient $X_{\overline{K}}^{d}/\mathfrak{S}_{d}$ is projective (cf. [5,
Proposition 8.1, Exercise 8.5]).

##  3\. Preimages Question for a fixed point

The simplest case of Question 1.1 is when the invariant subset $Y$ consists of
a single rational point $y\in X(K)$. If, for example, $f$ is polarized, it is
easy to see that the set $\bigcup_{s\geq 0}f^{-s}(y)(\overline{K})$ is a set
of bounded height and hence the answer to Question 1.1, as well as Question
1.4, is affirmative. In general, we cannot use such a height argument, but the
answer is still affirmative.

######  Proposition 3.1.

Let $K$ be a number field. Let $X$ be a projective variety over $K$ and
$f\colon X\longrightarrow X$ a surjective morphism. Let $Y\subset X$ be a
$0$-dimensional $f$-invariant closed subscheme. Then the answer to Question
1.4 is affirmative, i.e. for $d\in{\mathbb{Z}}_{\geq 1}$, there is
$s_{0}\in{\mathbb{Z}}_{\geq 0}$ such that for all $s\geq s_{0}$, we have

(3.1) $\displaystyle\left(f^{-s-1}(Y)\smallsetminus
f^{-s}(Y)\right)(L)=\emptyset$

for all field extensions $K\subset L$ with $[L:K]\leq d$.

###### Proof.

By taking base change to a finite extension of $K$, we may assume $Y$ is a
finite set of $K$-rational points. Then, replacing $f$ with an iterate, we may
further assume that $Y=\\{y\\}$ where $y\in X(K)$ is a fixed point of $f$.

Suppose for any $s_{0}\geq 0$, there are $s\geq s_{0}$ and $K\subset L$ with
$[L:K]\leq d$ such that

(3.2) $\displaystyle\left(f^{-s-1}(y)\smallsetminus
f^{-s}(y)\right)(L)\neq\emptyset.$

Then for all $s\geq 0$,

(3.3) $\displaystyle A_{s+1}:=\\{z\in f^{-s-1}(y)\smallsetminus
f^{-s}(y)\mid[\kappa(z):K]\leq d\\}\neq\emptyset$

where $\kappa(z)$ is the residue field of $z$. We set $A_{0}=\\{y\\}$. If
$z\in A_{s+1}$, then $f(z)\in f^{-s}(y)\smallsetminus f^{-(s-1)}(y)$ and
$[\kappa(f(z)):K]\leq[\kappa(z):K]\leq d$. Thus we have $f(A_{s+1})\subset
A_{s}$. Thus we can take a $(x_{s})_{s\geq 0}\in\varprojlim A_{s}$. By
replacing $K$ with a finite extension, we may regard $x_{s}$ as $K$-rational
points of $X$ for all $s$. Note that $x_{0}=y$.

Let us take a model ${\mathcal{X}}$ and $\widetilde{f}$ of $X$ and $f$ over
the ring of $S$-integers ${\mathcal{O}}_{K,S}$ of $K$. For a finite place
$v\not\in S$, ${\mathbb{F}}_{v}$ denotes the residue field of
${\mathcal{O}}_{K,S}$ at $v$, and we write
$X_{{\mathbb{F}}_{v}}={\mathcal{X}}\times_{{\mathcal{O}}_{K,S}}{\mathbb{F}}_{v}$,
$f_{{\mathbb{F}}_{v}}=\widetilde{f}\times_{{\mathcal{O}}_{K,S}}{\mathbb{F}}_{v}$.
Every $x_{s}$ corresponds to
$\widetilde{x_{s}}\in{\mathcal{X}}({\mathcal{O}}_{K,S})$. Its reduction to
${\mathbb{F}}_{v}$ is denoted by $x_{s}(v)$.

(3.4)
${\widetilde{f}}$${f}$${X}$${{\mathcal{X}}}$${X_{{\mathbb{F}}_{v}}}$${f_{{\mathbb{F}}_{v}}}$${\operatorname{Spec}K}$${\operatorname{Spec}{\mathcal{O}}_{K,S}}$${\operatorname{Spec}{\mathbb{F}}_{v}}$

$\circlearrowright$

$\circlearrowright$

$\circlearrowleft$

$\scriptstyle{x_{s}}$$\scriptstyle{\widetilde{x_{s}}}$$\scriptstyle{x_{s}(v)}$

Since $X_{{\mathbb{F}}_{v}}({\mathbb{F}}_{v})$ is a finite set and
$f_{{\mathbb{F}}_{v}}(x_{0}(v))=x_{0}(v)$, there is $n_{v}\geq 0$ with the
following property: any point $a\in X_{{\mathbb{F}}_{v}}({\mathbb{F}}_{v})$
with $f_{{\mathbb{F}}_{v}}^{n}(a)=x_{0}(v)$ for some $n\geq 0$, we have
$f_{{\mathbb{F}}_{v}}^{n_{v}}(a)=x_{0}(v)$. Since
$f_{{\mathbb{F}}_{v}}^{n}(x_{n}(v))=x_{0}(v)$ for all $n\geq 0$, we have for
all $m\geq 0$

(3.5) $\displaystyle
x_{m}(v)=f_{{\mathbb{F}}_{v}}^{n_{v}}(x_{m+n_{v}}(v))=x_{0}(v).$

Since this holds for all $v\notin S$, we get $x_{m}=x_{0}$ for all $m\geq 0$.
This contradicts to $x_{m}\notin f^{-(m-1)}(y)$. ∎

##  4\. Preimages Question when the subvariety is not invariant

Let $f\colon X\longrightarrow X$ be a surjective morphism on a projective
variety. Consider an arbitrary subvariety $Y\subset X$, not necessarily
$f$-invariant. One might expect that the preimages $f^{-s}(Y)$ are typically
geometrically complicated and admit few rational points. Stemming from this
intuition, the following question (PIQ without the constraint of $Y$ being
invariant) naturally arises.

######  Question 4.1.

Let $K$ be a number field. Let $X$ be a projective variety over $K$ and
$f\colon X\longrightarrow X$ a surjective morphism over $K$. Let $Y\subset X$
be a closed subscheme. Then is there $s_{0}\in{\mathbb{Z}}_{\geq 0}$ such that
for all $s\geq s_{0}$, we have

(4.1) $\displaystyle\biggl{(}f^{-s-1}(Y)\smallsetminus\bigcup_{0\leq i\leq
s}f^{-i}(Y)\biggr{)}(K)=\emptyset\ ?$

There are situations where the answer is negative by trivial reason, e.g., $f$
is an automorphism and $Y$ is a single rational point with infinite $f$-orbit.
It was somewhat surprising that the question has a negative answer even if $f$
is polarized. The following is a recipe for constructing counterexamples.

######  Proposition 4.2.

Let $K$ be a number field and $X$ a projective variety over $K$. Let
$f,g\colon X\longrightarrow X$ be surjective morphisms over $K$ and $Y\subset
X$ be a closed subvariety. Suppose

  1. (1)

$f\circ g=g\circ f$;

  2. (2)

$f(g(Y))=Y$;

  3. (3)

$Y\cap g^{n}(Y)\subsetneq Y$ for all $n\geq 1$;

  4. (4)

$Y(K)$ is Zariski dense in $Y$.

Then the answer to Question 4.1 for $X,f,Y$ is negative.

###### Proof.

For all integers $n\geq m\geq 0$, we have

(4.2) $\displaystyle f^{m}(g^{n}(Y))=g^{n-m}\left((f\circ
g)^{m}(Y)\right)=g^{n-m}(Y).$

Thus for all $s\geq 0$, we have

(4.3) $\displaystyle\biggl{(}f^{-s-1}(Y)\smallsetminus\bigcup_{0\leq i\leq
s}f^{-i}(Y)\biggr{)}(K)$ (4.4)
$\displaystyle\supset\biggl{(}g^{s+1}(Y)\smallsetminus\bigcup_{0\leq i\leq
s}f^{-i}(Y\cap g^{s+1-i}(Y))\biggr{)}(K).$

By assumption, $Y\cap g^{s+1-i}(Y)$ is a proper closed subset of $Y$. Since
$Y$ is irreducible and $f^{i},g^{s+1}$ are finite, $f^{-i}(Y\cap
g^{s+1-i}(Y))$ does not contain $g^{s+1}(Y)$. Since $g^{s+1}(Y)(K)$ is Zariski
dense in $g^{s+1}(Y)$, the right hand side of eq. 4.3 is non-empty. ∎

######  Example 4.3.

There is a polarized endomorphism for which Question 4.1 is not true. Let $E$
be an elliptic curve over ${\mathbb{Q}}(i)$ such that
$E_{{\mathbb{C}}}\simeq{\mathbb{C}}/({\mathbb{Z}}+i{\mathbb{Z}})$ (cf. [16,
II, Example 1.3.1] or [16, I, Proposition 4.5]). The endomorphisms defined by
multiplication by $1+2i$ and $1-2i$ on $E_{{\mathbb{C}}}$ is defined over
${\mathbb{Q}}(i)$ (cf. [16, II, Theorem 2.2 (b)]). Let $\varphi$, $\psi$
denote those endomorphisms on $E$ respectively. Then we have

(4.5) $\displaystyle\varphi\circ\psi=\psi\circ\varphi=[5]$ (4.6)
$\displaystyle\deg\varphi=\deg\psi=5$

where $[5]$ stands for the multiplication by $5$. We are going to apply
Proposition 4.2. Set $X=E\times E$, $f=\varphi\times\psi$,
$g=\psi\times\varphi$, and $Y=\Delta\subset X$, the diagonal subvariety. Then,
it is easy to see that

(4.7) $\displaystyle f\circ g=g\circ f$ (4.8) $\displaystyle f(g(Y))=Y.$

There is a finite extension $K$ of ${\mathbb{Q}}(i)$ such that $Y(K)=E(K)$ is
infinite. We then replace everything with the base change to $K$. So far, we
confirmed conditions (1)(2)(4). To see (3), it is enough to show that
$\psi^{n}\neq\varphi^{n}$ for all $n\geq 1$. This follows from the fact that

(4.9) $\displaystyle\frac{1+2i}{1-2i}$

is not a root of $1$ as the unit group of ${\mathbb{Q}}(i)$ is generated by
$i$. We have proven that the morphism $f\colon E\times E\longrightarrow
E\times E$ and the diagonal $\Delta\subset E\times E$ give a counterexample to
Question 4.1. Note that $f$ is polarized and étale.

######  Example 4.4.

In Example 4.3, $\varphi$ and $\psi$ commute with $[-1]\colon E\longrightarrow
E$. Hence they induce endomorphisms on
${\mathbb{P}}^{1}_{K}=E/\langle[-1]\rangle$. Let
$\widetilde{\varphi},\widetilde{\psi}\colon{\mathbb{P}}^{1}_{K}\longrightarrow{\mathbb{P}}^{1}_{K}$
be the induced endomorphisms. They commute and
$\deg\widetilde{\varphi}=\deg\widetilde{\psi}=5$. Set
$X={\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}$,
$f=\widetilde{\varphi}\times\widetilde{\psi}$,
$g=\widetilde{\psi}\times\widetilde{\varphi}$, and $Y=\Delta\subset X$, the
diagonal subvariety. Then we have

(4.10) $\displaystyle f\circ g=g\circ f,\ f(g(Y))=Y,\ \text{$Y(K)$ is Zariski
dense in $Y$}.$

Moreover,

(4.11) $\displaystyle g^{n}(Y)\subset
Y\iff\varphi^{n}=\pm\psi^{n}\iff\left(\frac{1+2i}{1-2i}\right)^{n}=\pm 1$

and hence $g^{n}(Y)\not\subset Y$ for $n\geq 1$. Therefore, by Proposition
4.2,
$f\colon{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}\longrightarrow{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}$
and $\Delta\subset{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}$ give a
counterexample to Question 4.1. Note that $f$ is polarized.

The above construction of counterexamples uses two mutually commutative self-
morphisms. Such a situation is somewhat special, and it may be reasonable to
expect Question 4.1 holds for “general” self-morphisms. The following is one
attempt in this direction.

######  Theorem 4.5.

Let $K$ be a number field. Let
$f,g\colon{\mathbb{P}}^{1}_{K}\longrightarrow{\mathbb{P}}^{1}_{K}$ be
polynomial maps of the same degree $d\geq 2$. Then the answer to Question 4.1
for $f\times
g\colon{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}\longrightarrow{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}$
and the diagonal subvariety
$\Delta\subset{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}$ is affirmative,
i.e. there is $s_{0}\in{\mathbb{Z}}_{\geq 0}$ such that for all $s\geq s_{0}$,
we have

(4.12) $\displaystyle\biggl{(}(f\times
g)^{-s-1}(\Delta)\smallsetminus\bigcup_{0\leq i\leq s}(f\times
g)^{-i}(\Delta)\biggr{)}(K)=\emptyset.$

The proof of this theorem is similar to the proof of [2, Theorem 1.6]. We use
the following straightforward generalization of [2, Lemma 4.3].

######  Lemma 4.6.

Let $k$ be an algebraically closed field of characteristic zero. Let $P,Q\in
k[x]$ be non-constant polynomials such that $\deg Q\geq 2$ and $\deg Q\mid\deg
P$. We use the same symbols $P,Q$ to denote the self-morphisms on
${\mathbb{P}}^{1}_{k}$ defined by $P,Q$. Consider the following diagram

(4.13)
${\widetilde{X}}$${X}$${\Delta}$${{\mathbb{P}}^{1}_{k}\times{\mathbb{P}}^{1}_{k}}$${{\mathbb{P}}^{1}_{k}\times{\mathbb{P}}^{1}_{k}}$${{\mathbb{P}}^{1}_{k}}$${{\mathbb{P}}^{1}_{k}}$$\scriptstyle{\nu}$$\scriptstyle{\pi_{1}}$${\subset}$${\subset}$$\scriptstyle{P\times
Q}$$\scriptstyle{\operatorname{pr}_{1}}$$\scriptstyle{\operatorname{pr}_{1}}$$\scriptstyle{P}$

where $\Delta$ is the diagonal subvariety, $X$ is an arbitrary irreducible
component of $(P\times Q)^{-1}(\Delta)$ with reduced structure, $\nu$ is the
normalization of $X$, and $\pi_{1}$ is the composite of $\nu$ and the first
projection $\operatorname{pr}_{1}$. Then $\pi_{1}$ is étale over $\infty$.

###### Proof.

The proof is a straightforward modification of the proof of [2, Lemma 4.3].
Let $\pi_{2}=\operatorname{pr}_{2}\circ\nu$ and
$y=1/x\in{\mathcal{O}}_{{\mathbb{P}}^{1}_{k},\infty}\subset
k({\mathbb{P}}^{1}_{k})$. Suppose there is a closed point
$\alpha\in\widetilde{X}$ such that $\pi_{1}(\alpha)=\infty$ and $\pi_{1}$ is
ramified at $\alpha$. Let $e=e_{\pi_{1},\alpha}(\geq 2)$ be the ramification
index at $\alpha$. (We use this notation $e_{{\rm map},{\rm point}}$ for the
ramification index in the following.) Since $P\circ\pi_{1}=Q\circ\pi_{2}$, we
have $Q(\pi_{2}(\alpha))=P(\pi_{1}(\alpha))=P(\infty)=\infty$. Thus
$\pi_{2}(\alpha)=\infty$. Moreover, we have
$e_{P,\infty}e_{\pi_{1},\alpha}=e_{Q,\infty}e_{\pi_{2},\alpha}$ and hence

(4.14) $\displaystyle e_{\pi_{2},\alpha}=\frac{\deg P}{\deg Q}e=de$

where $d:=\deg P/\deg Q$.

Let

(4.15) $\displaystyle\varphi:=\pi_{1}^{*}y,\ \psi:=\pi_{2}^{*}y.$

Note that we have $\varphi,\psi\in{\mathcal{O}}_{\widetilde{X},\alpha}$ and
$v_{\alpha}(\varphi)=e$, $v_{\alpha}(\psi)=de$. Here $v_{\alpha}$ is the
discrete valuation of ${\mathcal{O}}_{\widetilde{X},\alpha}$. Note also that
$\varphi$ and $\psi$ generate $k(\widetilde{X})$ over $k$ as a field. By the
relation $P\circ\pi_{1}=Q\circ\pi_{2}$, we have

(4.16) $\displaystyle\pi_{1}^{*}(P^{*}y)=\pi_{2}^{*}(Q^{*}y),\ \text{i.e.}$
(4.17) $\displaystyle\pi_{1}^{*}(1/P(1/y))=\pi_{2}^{*}(1/Q(1/y)),\
\text{i.e.}$ (4.18) $\displaystyle P(1/\varphi)=Q(1/\psi).$

Now, by Hensel, there is a uniformizer
$u\in\widehat{{\mathcal{O}}}_{\widetilde{X},\alpha}$ such that
$\varphi=u^{e}$. We identify
$\widehat{{\mathcal{O}}}_{\widetilde{X},\alpha}=k\llbracket u\rrbracket$ and
write

(4.19) $\displaystyle\frac{1}{\psi}=\sum_{\begin{subarray}{c}n\geq-de\\\ e\mid
n\end{subarray}}a_{n}u^{n}+\sum_{\begin{subarray}{c}n\geq-de\\\ e\not\mid
n\end{subarray}}a_{n}u^{n}\quad(a_{n}\in k,\ a_{-de}\neq 0).$

We set

(4.20) $\displaystyle f=\sum_{\begin{subarray}{c}n\geq-de\\\ e\mid
n\end{subarray}}a_{n}u^{n},\quad g=\sum_{\begin{subarray}{c}n\geq-de\\\
e\not\mid n\end{subarray}}a_{n}u^{n}.$

We claim $g=0$. Indeed, assume $g\neq 0$ and let $m_{0}$ be the degree of the
lowest degree term of $g$. By the definition of $g$, we have $m_{0}\geq-de+1$
and $e\not\mid m_{0}$. By eq. 4.18, we have

(4.21) $\displaystyle P(1/u^{e})=Q(f+g)=Q(f)+\sum_{i=1}^{\deg
Q}\frac{Q^{(i)}(f)}{i!}g^{i}.$

The lowest degree of $\frac{Q^{(i)}(f)}{i!}g^{i}$ is $(-de)(\deg Q-i)+im_{0}$.
Since $m_{0}\geq-de+1$, we have

(4.22) $\displaystyle(-de)(\deg Q-i)+im_{0}>(-de)(\deg Q-1)+m_{0}$

for all $i\geq 2$. Thus, the lowest degree of

(4.23) $\displaystyle\sum_{i=1}^{\deg Q}\frac{Q^{(i)}(f)}{i!}g^{i}$

is $(-de)(\deg Q-1)+m_{0}$, which is not divisible by $e$. But this is a
contradiction because

(4.24) $\displaystyle\sum_{i=1}^{\deg
Q}\frac{Q^{(i)}(f)}{i!}g^{i}=P(1/u^{e})-Q(f)$

is a Laurent series of $u^{e}$. This proves $g=0$ and, in particular, $\psi$
is a power series of $u^{e}$.

Finally, since $\varphi$ and $\psi$ generate $k(\widetilde{X})$, there is a
rational function of $F(\varphi,\psi)$ of $\varphi$ and $\psi$ with
coefficient in $k$ such that $F(\varphi,\psi)$ is a uniformizer of
${\mathcal{O}}_{\widetilde{X},\alpha}$. But $F(\varphi,\psi)$ is a Lauren
series of $u^{e}$ in
$\widehat{{\mathcal{O}}}_{\widetilde{X},\alpha}=k\llbracket u\rrbracket$, and
hence, its valuation is divisible by $e$. This is a contradiction, and we
complete the proof. ∎

###### Proof of Theorem 4.5.

Let $\varphi=f\times g$. By [6, Proposition 2.1], there is
$\delta\in{\mathbb{Z}}_{\geq 1}$ such that the closed immersion
$i\colon{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}\hookrightarrow{\mathbb{P}}^{N}_{K}$
defined by
${\mathcal{O}}_{{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}}(\delta,\delta)$
is equivariant, i.e. there is
$F\colon{\mathbb{P}}^{N}_{K}\longrightarrow{\mathbb{P}}^{N}_{K}$ such that
$F\circ i=i\circ\varphi$. The height $h$ of subvarieties of
${\mathbb{P}}^{N}_{K}$ and the canonical height $\hat{h}_{F}$ are defined as
in [8, section 4]. The restriction of $h$ and $\hat{h}_{F}$ to subvarieties of
${\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}$ are denoted by $h$ (abuse of
notation) and $\hat{h}_{\varphi}$. We refer [8, section 4] or [2, Lemma 5.1]
for the properties of $h$ and $\hat{h}_{\varphi}$ that we need for this proof.

Now suppose Question 4.1 does not hold for $\varphi$ and $\Delta$. Then for
all $m\in{\mathbb{Z}}_{\geq 0}$, the set

(4.25)
$\displaystyle\Sigma_{m+1}:=\biggl{(}\varphi^{-m-1}(\Delta)\smallsetminus\bigcup_{0\leq
i\leq m}\varphi^{-i}(\Delta)\biggr{)}(K)$

are infinite. Indeed, if there is $m\in{\mathbb{Z}}_{\geq 0}$ such that
$\Sigma_{m+1}$ is finite, then the set

(4.26)
$\displaystyle\Sigma:=\\{x\in({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})(K)\mid\text{$\varphi^{l}(x)\in\Sigma_{m+1}$
for some $l\in{\mathbb{Z}}_{\geq 0}$}\\}$

is finite because it is a set of bounded heights. But we have

(4.27) $\displaystyle\bigcup_{n\geq m+1}\Sigma_{n}\subset\Sigma$

and the left-hand side is an infinite set by our hypothesis. This is a
contradiction.

For all $m\in{\mathbb{Z}}_{\geq 0}$, let

(4.28) $\displaystyle S_{m+1}:=\left\\{C\ \middle|\ \parbox{200.0003pt}{ $C$
is an irreducible component of $\varphi^{-m-1}(\Delta)$ which is not contained
in $\bigcup_{0\leq i\leq m}\varphi^{-i}(\Delta)$ and $C(K)$ is infinite
}\right\\}.$

We set $S_{0}=\\{\Delta\\}$. By the above argument, $S_{m}\neq\emptyset$ for
all $m\geq 0$. Also $\varphi$ maps members of $S_{m}$ to members of $S_{m-1}$.
Thus, they form an inverse system of non-empty finite sets. Let us take an
element $(C_{m})_{m\geq 0}\in\varprojlim S_{m}$. By the construction,
$C_{0}=\Delta$, $C_{m+1}\not\subset\bigcup_{0\leq i\leq
m}\varphi^{-i}(\Delta)$, $\\#C_{m}(K)=\infty$ for all $m\geq 0$, and we have
the following commutative diagram:

(4.29)
${\cdots}$${C_{m+1}}$${C_{m}}$${\quad\cdots\quad}$${\Delta}$${\cdots}$${{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}}$${{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}}$${\quad\cdots\quad}$${{\mathbb{P}}^{1}_{K}\times{\mathbb{P}}^{1}_{K}}$${\subset}$${\subset}$${\subset}$$\scriptstyle{\varphi}$$\scriptstyle{\varphi}$$\scriptstyle{\varphi}$$\scriptstyle{\varphi}$

We note that $C_{m}$’s are geometrically integral because $C_{m}(K)$ is dense
in $C_{m}$.

Let $\deg
C_{m}:=\deg{\mathcal{O}}_{{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}}(1,1)|_{C_{m}}$,
i.e. the degree of $C_{m}$ with respect to the ample line bundle
${\mathcal{O}}_{{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}}(1,1)$. We claim that
$\lim_{m\to\infty}\deg C_{m}=\infty$. Indeed, by [8, Theorem 4.10] (cf. [2,
Lemma 5.1]), we have

(4.30)
$\displaystyle\frac{\hat{h}_{\varphi}(\Delta)}{\deg{\mathcal{O}}(\delta,\delta)|_{\Delta}}=\frac{\hat{h}_{\varphi}(\varphi^{m}(C_{m}))}{\deg{\mathcal{O}}(\delta,\delta)|_{\varphi^{m}(C_{m})}}=d^{m}\frac{\hat{h}_{\varphi}(C_{m})}{\deg{\mathcal{O}}(\delta,\delta)|_{C_{m}}}.$

Here $\varphi^{m}(C_{m})$ is equipped with the reduced structure. This implies

(4.31)
$\displaystyle\hat{h}_{\varphi}(C_{m})=\frac{\hat{h}_{\varphi}(\Delta)}{\deg{\mathcal{O}}(\delta,\delta)|_{\Delta}}\frac{\delta\deg
C_{m}}{d^{m}}.$

In particular, if there is a sequence $0\leq m_{1}<m_{2}<\cdots$ such that
$\deg C_{m_{i}}$ is bounded, then $\hat{h}_{\varphi}(C_{m_{i}})$ is also
bounded. This implies the set of subvarieties $\\{C_{m_{i}}\mid i\geq 1\\}$ is
a finite set (cf. [8] or [2, Lemma 5.1]). But this contradicts our
construction of $C_{m}$.

To end the proof, we prove that the genus of the normalization of
$C_{m}\times_{K}\overline{K}$ is at least two if $m$ is large enough. We get a
contradiction with the fact that $\\#C_{m}(K)=\infty$ and the Faltings’
theorem.

We replace everything with the base change to $\overline{K}$ and we work over
$\overline{K}$. Let us decompose

(4.32) $\displaystyle f=\alpha_{1}\circ\beta_{1},\ g=\alpha_{2}\circ\beta_{2}$

where $\alpha_{1}$ and $\alpha_{2}$ are indecomposable polynomial maps of
degree at least two and $\beta_{1}$ and $\beta_{2}$ are polynomial maps. Here
a surjective morphism ${\mathbb{P}}^{1}\longrightarrow{\mathbb{P}}^{1}$ is
said to be indecomposable if the induced function field extension admits no
non-trivial intermediate fields. Consider the following commutative diagram.

(4.33)
${C_{m+1}}$${X}$${C_{m}}$${\quad\cdots\quad}$${\Delta}$${{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}}$${{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}}$${\quad\cdots\quad}$${{\mathbb{P}}^{1}\times{\mathbb{P}}^{1}}$${{\mathbb{P}}^{1}}$${{\mathbb{P}}^{1}}$${{\mathbb{P}}^{1}}$${\subset}$${\subset}$$\scriptstyle{\nu_{i}}$${\subset}$$\scriptstyle{\operatorname{pr}_{i}}$$\scriptstyle{\operatorname{pr}_{i}}$$\scriptstyle{\varphi}$$\scriptstyle{\varphi}$$\scriptstyle{\beta_{i}}$$\scriptstyle{\alpha_{i}}$

Here $i$ is $1$ or $2$, $\nu_{i}$ is the composition of the inclusion and the
projection, and $X$ is the reduced scheme of the fiber product of
$C_{m}\xlongrightarrow{\nu_{i}}{\mathbb{P}}^{1}$ and
${\mathbb{P}}^{1}\xlongrightarrow{\alpha_{i}}{\mathbb{P}}^{1}$. The fiber
product’s universal property induces the morphism $C_{m+1}\longrightarrow X$
is induced by the fiber product’s universal property. Let
$\mu\colon\widetilde{C}_{m}\longrightarrow C_{m}$ be the normalization. By
Lemma 4.6, $\nu_{i}\circ\mu$ is étale over $\infty$. Thus by [2, Proposition
4.1], $X$ is irreducible and

(4.34) $\displaystyle
2g_{X}-2\geq\deg\alpha_{i}(2g_{C_{m}}-2)+(\deg\alpha_{i}-1)\deg\nu_{i}$

where $g_{X}$ and $g_{C_{m}}$ are the genus of the normalization of $X$ and
$C_{m}$ respectively. Now note that $\deg C_{m}=\deg\nu_{1}+\deg\nu_{2}$.
Since $\deg C_{m}\to\infty$, we may take $m$ large enough so that
$\deg\nu_{i}\geq 5$ for $i=1$ or $2$. Then by eq. 4.34,

(4.35) $\displaystyle
2g_{X}-2\geq-2\deg\alpha_{i}+5(\deg\alpha_{i}-1)=3\deg\alpha_{i}-5\geq 1.$

Therefore $g_{C_{m+1}}\geq g_{X}\geq 2$ and we are done. ∎

##  5\. Étale morphisms

######  Theorem 5.1.

Let $K$ be a number field. Let $X$ be a projective variety over $K$ and
$f\colon X\longrightarrow X$ be an étale morphism over $K$. Let $Y\subset X$
be a closed subscheme such that $f|_{Y}$ factors through $Y\subset X$. Let
$d\in{\mathbb{Z}}_{\geq 1}$. Then there exists a non-negative integer $s_{0}$
such that

$(f^{-s-1}(Y)\smallsetminus f^{-s}(Y))(L)=\emptyset$

for all field extension $K\subset L$ with $[L:K]\leq d$ and $s\geq s_{0}$.

######  Remark 5.2.

Let us remark that Theorem 5.1 does not automatically follow from (the proof
of) Proposition 2.1 and [2, Theorem 2.1], which is $d=1$ version of Theorem
5.1. Indeed, in the proof of Proposition 2.1, we apply PIQ (with fixed number
field) to the self-morphism induced on the symmetric product $X^{(d)}$. But
this induced morphism could be non-étale.

The following lemma is the key in the proof of Theorem 5.1.

######  Lemma 5.3.

Let $f\colon X\longrightarrow X$ be an étale morphism of a projective variety
$X$ over a number field $K$. Let $Y\subset X$ be a normal closed subscheme.
Let $d\in{\mathbb{Z}}_{\geq 1}$. Then there is a non-negative integer $t$ such
that for all $s\geq 0$, we have

(5.1) $\displaystyle\\#\left\\{Z\ \middle|\ \parbox{200.0003pt}{$Z$ is an
irreducible component of $f^{-s}(Y)$ such that $Z(L)\neq\emptyset$ for some
$L$ with $[L:K]\leq d$.}\right\\}\leq t$

###### Proof.

As in [2, Lemma 2.2], let $S$ be a sufficiently large finite set of primes of
$K$ so that we have models ${\mathcal{X}},F$, and ${\mathcal{Y}}$ over
${\mathcal{O}}_{K,S}$ of $X,f$, and $Y$ respectively with the following
properties:

  1. (1)

${\mathcal{X}}$ is an integral projective scheme over ${\mathcal{O}}_{K,S}$;

  2. (2)

$F\colon{\mathcal{X}}\longrightarrow{\mathcal{X}}$ is an étale surjective
morphism over ${\mathcal{O}}_{K,S}$;

  3. (3)

${\mathcal{Y}}$ is a closed subscheme of ${\mathcal{X}}$ and normal.

Since $F$ is étale and ${\mathcal{Y}}$ is normal, $F^{-s}({\mathcal{Y}})$ is a
disjoint union of normal integral schemes for any $s\geq 0$:

(5.2) $\displaystyle
F^{-s}({\mathcal{Y}})=\coprod_{i=1}^{r_{s}}\mathcal{Z}_{i}^{(s)}.$

Let $Z_{i}^{(s)}$ be the generic fiber of $\mathcal{Z}_{i}^{(s)}$. Then
$Z_{i}^{(s)}$’s are the irreducible components of $f^{-s}(Y)$:

(5.3) $\displaystyle f^{-s}(Y)=\coprod_{i=1}^{r_{s}}Z_{i}^{(s)}.$

By relabeling if necessary, we may assume that the first $l$ components admit
rational points of degree at most $d$ over $K$. That is, for $i=1,\dots,l$,
there are finite extensions $K\subset L_{i}$ with $[L_{i}:K]\leq d$ and

(5.4) $\displaystyle Z_{i}^{(s)}(L_{i})\neq\emptyset.$

Let $R_{i}$ be the integral closure of ${\mathcal{O}}_{K,S}$ in $L_{i}$. Then
any point of $Z_{i}^{(s)}(L_{i})$ extends to a point of
$\mathcal{Z}_{i}^{(s)}(R_{i})$ because $\mathcal{Z}_{i}^{(s)}$ is proper over
${\mathcal{O}}_{K,S}$. Let us take $x_{i}\in\mathcal{Z}_{i}^{(s)}(R_{i})$ for
each $i=1,\dots,l$. Pick a maximal ideal
${\mathfrak{p}}\subset{\mathcal{O}}_{K,S}$ and take maximal ideals
${\mathfrak{q}}_{i}\subset R_{i}$ containing ${\mathfrak{p}}R_{i}$. Then we
get the following diagram:

(5.5)
${\operatorname{Spec}R_{i}}$${\operatorname{Spec}R_{i}/{\mathfrak{p}}R_{i}}$${\operatorname{Spec}R_{i}/{\mathfrak{q}}_{i}}$${{\mathcal{X}}}$${{\mathcal{X}}\times_{{\mathcal{O}}_{K,S}}{\mathcal{O}}_{K,S}/{\mathfrak{p}}\mathrlap{=:{\mathcal{X}}_{{\mathfrak{p}}}}}$${\operatorname{Spec}{\mathcal{O}}_{K,S}}$${\operatorname{Spec}{\mathcal{O}}_{K,S}/{\mathfrak{p}}}$$\scriptstyle{x_{i}}$$\scriptstyle{\xi_{i}}$

where the squares are fiber products and $\xi_{i}$ is the induced morphism. We
have

(5.6)
$\displaystyle[R_{i}/{\mathfrak{q}}_{i}:{\mathcal{O}}_{K,S}/{\mathfrak{p}}]\leq[L_{i}:K]\leq
d.$

In particular, if $\\#{\mathcal{O}}_{K,S}/{\mathfrak{p}}=q$, then
$\\#R_{i}/{\mathfrak{q}}_{i}=q^{j}$ for some $1\leq j\leq d$. Also the images
of $\xi_{i}$’s are disjoint because $\mathcal{Z}_{i}^{(s)}$ are disjoint.
Therefore, we get

(5.7) $\displaystyle
l\leq\\#\bigcup_{j=1}^{d}\mathop{\mathrm{Hom}}\nolimits(\operatorname{Spec}{\mathbb{F}}_{q^{j}},{\mathcal{X}}_{{\mathfrak{p}}})$

where the $\mathop{\mathrm{Hom}}\nolimits$ in the right-hand side stands for
the set of morphisms of schemes. We are done since the right-hand side is
finite and independent of $s$. ∎

###### Proof of Theorem 5.1.

The proof is the same as that of [2, Theorem 2.1]. First, we may assume $Y$ is
reduced. If $Y$ is normal, the statement follows from Lemma 5.3.

For the general case, we proceed by induction on $\dim Y$. If $\dim Y=0$, then
$Y$ is normal, and we are done.

Suppose $\dim Y>0$. Let us write $Y=Y_{1}\cup Y_{2}$ where $Y_{1}$ is the
union of all maximal dimensional irreducible components and $Y_{2}$ is the
union of other components. We endow $Y_{1},Y_{2}$ with the reduced scheme
structure. Then as in the proof of [2, Theorem 2.1], we have

(5.8) $\displaystyle f(Y_{1})\subset Y_{1},\quad
f(\operatorname{Sing}Y_{1})\subset\operatorname{Sing}Y_{1}$

where $\operatorname{Sing}Y_{1}$ is the non-regular locus of $Y_{1}$.

Let $\nu\colon\widetilde{Y_{1}}\longrightarrow Y_{1}$ be the normalization,
i.e. $\widetilde{Y_{1}}$ is the union of the normalization of irreducible
components of $Y_{1}$. Take a closed immersion
$i\colon\widetilde{Y_{1}}\longrightarrow{\mathbb{P}}^{N}_{K}$ and we regard
$\widetilde{Y_{1}}$ as a closed subscheme of ${\mathbb{P}}^{N}\times X$ via

$\widetilde{Y_{1}}\xrightarrow{(i,\nu)}{\mathbb{P}}^{N}\times
Y_{1}\subset{\mathbb{P}}^{N}\times X.$

Set $g=\operatorname{id}\times f\colon{\mathbb{P}}^{N}\times
X\longrightarrow{\mathbb{P}}^{N}\times X$. Since $g$ is étale and
$\widetilde{Y_{1}}$ is normal, by Lemma 5.3, there exists a positive integer
$t$ such for all $s$, at most $t$ irreducible components of
$g^{-s}(\widetilde{Y_{1}})$ admit $L$-points for some $L$ with $[L:K]\leq d$.

Note that the diagram

(5.9)
${g^{-s-1}(\widetilde{Y_{1}})}$${\widetilde{Y_{1}}}$${f^{-s-1}(Y_{1})}$${Y_{1}}$$\scriptstyle{g^{s+1}}$$\scriptstyle{\nu}$$\scriptstyle{f^{s+1}}$

is cartesian and $\nu$ is an isomorphism over
$Y_{1}\smallsetminus\operatorname{Sing}Y_{1}$.

We apply the induction hypothesis to $f$ and $\operatorname{Sing}Y_{1}$. Then
there exists a non-negative integer $s_{0}$ such that

(5.10) $\displaystyle(f^{-s-1}(\operatorname{Sing}Y_{1})\smallsetminus
f^{-s}(\operatorname{Sing}Y_{1}))(L)=\emptyset$

for all $L$ with $[L:K]\leq d$ and $s\geq s_{0}$. We claim that

(5.11) $\displaystyle(f^{-s_{0}-t-1}(Y_{1})\smallsetminus
f^{-s_{0}-t}(Y_{1}))(L)=\emptyset$

for all $L$ with $[L:K]\leq d$. Indeed, suppose there is $L$ with $[L:K]\leq
d$ and $x\in(f^{-s_{0}-t-1}(Y_{1})\smallsetminus f^{-s_{0}-t}(Y_{1}))(L)$.
Then $x,f(x),\dots,f^{t}(x)$ are contained in different irreducible components
of $f^{-s_{0}-t-1}(Y_{1})$. Moreover, they are not mapped into
$\operatorname{Sing}Y_{1}$ by $f^{s_{0}+t+1}$ because $x\not\in
f^{-s_{0}-t}(Y_{1})$ and eq. 5.10. Thus $x,f(x),\dots,f^{t}(x)$ are contained
in the open subset over which
$g^{-s_{0}-t-1}(\widetilde{Y}_{1})\longrightarrow f^{-s_{0}-t-1}(Y_{1})$ is
isomorphic. Thus $x,f(x),\dots,f^{t}(x)$ produce $L$-points of
$g^{-s_{0}-t-1}(\widetilde{Y}_{1})$ that are on distinct irreducible
components. This contradicts the definition of $t$.

Now we have proven that

$(f^{-s-1}(Y_{1})\smallsetminus f^{-s}(Y_{1}))(L)=\emptyset$

for all $L$ with $[L:K]\leq d$ and $s\geq s_{0}+t$.

Let $Y_{3}\subset Y_{2}$ be the union of all irreducible components, which are
eventually mapped into $Y_{1}$ and let $Y_{4}\subset Y_{2}$ be the union of
all other components. Then we have $f(Y_{4})\subset Y_{4}$. Take $r>0$ so that
$f^{r}(Y_{3})\subset Y_{1}$. Then for $s\geq s_{0}+t$ and for all $L$ with
$[L:K]\leq d$,

(5.12) $\displaystyle\hphantom{=}(f^{-s-1}(Y)\smallsetminus f^{-s}(Y))(L)$
(5.13) $\displaystyle\subset(f^{-s-1}(Y_{1})\smallsetminus
f^{-s}(Y_{1}))(L)\cup(f^{-s-1}(Y_{3})\smallsetminus f^{-s}(Y))(L)$ (5.14)
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\cup(f^{-s-1}(Y_{4})\smallsetminus
f^{-s}(Y_{4}))(L)$ (5.15) $\displaystyle=(f^{-s-1}(Y_{3})\smallsetminus
f^{-s}(Y))(L)\cup(f^{-s-1}(Y_{4})\smallsetminus f^{-s}(Y_{4}))(L)$ (5.16)
$\displaystyle\subset(f^{-s-1-r}(Y_{1})\smallsetminus
f^{-s}(Y))(L)\cup(f^{-s-1}(Y_{4})\smallsetminus f^{-s}(Y_{4}))(L)$ (5.17)
$\displaystyle=(f^{-s-1}(Y_{4})\smallsetminus f^{-s}(Y_{4}))(L).$

Applying the induction hypothesis to $Y_{4}$ and $f$, we are done. ∎

##  6\. Uniformity of $s_{0}$ for abelian varieties

In this section, we prove Theorem E. At its core, the uniformity arises from
the finiteness of torsion points defined over a number field, and a finiteness
results on abelian subvarieties due to Lenstra and Oort [9]. Let us start with
a linear algebra lemma.

######  Lemma 6.1.

Let $V$ be a finite-dimensional ${\mathbb{Q}}$-vector space. Then there exists
$n_{0}\in{\mathbb{Z}}_{\geq 1}$ that depends only on $\dim V$ with the
following property. For any ${\mathbb{Q}}$-linear automorphism $f\colon
V\longrightarrow V$ and any $f$-periodic ${\mathbb{Q}}$-subspace $W\subset V$,
we have

(6.1) $\displaystyle f^{n_{0}}(W)=W.$

Here a subspace $W$ is said to be $f$-periodic if $f^{n}(W)=W$ for some
$n\in{\mathbb{Z}}_{\geq 1}$.

###### Proof.

Let $W$ be an $f$-periodic ${\mathbb{Q}}$-subspace of $V$ and let $d=\dim W$.
If $d=0$, then $f(W)=W$. Suppose $d\geq 1$. Consider the ${\mathbb{Q}}$-linear
automorphism

(6.2)
$\displaystyle\scalebox{1.0}[1.5]{$\wedge$}^{d}\,f\colon\scalebox{1.0}[1.5]{$\wedge$}^{d}\,V\longrightarrow\scalebox{1.0}[1.5]{$\wedge$}^{d}\,V.$

Then, it is easy to see that

(6.3) $\displaystyle
f^{n}(W)=W\iff(\scalebox{1.0}[1.5]{$\wedge$}^{d}\,f)^{n}(\scalebox{1.0}[1.5]{$\wedge$}^{d}\,W)=\scalebox{1.0}[1.5]{$\wedge$}^{d}\,W$

for all $n\in{\mathbb{Z}}_{\geq 1}$. Therefore, it is enough to show the
statement for one-dimensional subspaces $W$.

Now, suppose $W\subset V$ is an $f$-periodic one-dimensional subspace. We will
show that there is $n_{0}\in{\mathbb{Z}}_{\geq 1}$ depending only on $\dim V$
such that $f^{n_{0}}(W)=W$. Take $n\in{\mathbb{Z}}_{\geq 1}$ such that
$f^{n}(W)=W$. Let $w\in W\smallsetminus\\{0\\}$ and set

(6.4) $\displaystyle U=\langle w,f(w),\dots,f^{n-1}(w)\rangle_{{\mathbb{Q}}},$

the subspace of $V$ generated by $w,\dots,f^{n-1}(w)$. We have $f(U)=U$ and
there is $a\in{\mathbb{Q}}^{\times}$ such that $f|_{U}^{n}$ is multiplication
by $a$. In particular, $f|_{U}$ is diagonalizable over ${\mathbb{C}}$.
Replacing $n$ with $2n$ if necessary, we may assume $a>0$. Take
$b\in{\mathbb{R}}_{>0}$ such that $b^{n}=a$. Let
$\lambda_{1},\dots,\lambda_{r}$ be all the roots of the characteristic
polynomial of $f|_{U}$ (possibly repeated). Note $r=\dim U\leq\dim V$. Then we
have $\lambda_{i}^{n}=a$, and hence we can write

(6.5) $\displaystyle\lambda_{i}=\zeta_{i}b$

where $\zeta_{i}\in{\mathbb{C}}$ with $\zeta_{i}^{n}=1$. Then

(6.6)
$\displaystyle{\mathbb{Q}}\ni\lambda_{1}\cdots\lambda_{r}=\zeta_{1}\cdots\zeta_{r}b^{r}$

implies $b^{r}\in{\mathbb{Q}}$. Thus

(6.7)
$\displaystyle[{\mathbb{Q}}(\zeta_{i}):{\mathbb{Q}}]=[{\mathbb{Q}}(b^{-1}\lambda_{i}):{\mathbb{Q}}]\leq
r^{2}\leq(\dim V)^{2}.$

Therefore, there is $n_{0}\in{\mathbb{Z}}_{\geq 1}$ depending only on $\dim V$
such that

(6.8)
$\displaystyle\lambda_{1}^{n_{0}}=\cdots=\lambda_{r}^{n_{0}}\in{\mathbb{Q}}.$

Remembering $f|_{U}$ is diagonalizable, we can deduce that $f|_{U}^{n_{0}}$ is
a multiplication by a non-zero rational number, and hence $f^{n_{0}}(W)=W$. ∎

We can uniformly bound the period of periodic abelian subvarieties.

######  Lemma 6.2.

Let $X$ be an abelian variety over a number field $K$. Then, there exists
$n_{0}\in{\mathbb{Z}}_{\geq 1}$ that depends only on $X$ with the following
property. For any isogeny $\varphi\colon X\longrightarrow X$ and any
$\varphi$-periodic abelian subvariety $A\subset X$, we have

(6.9) $\displaystyle\varphi^{n_{0}}(A)=A.$

Here $A$ being $\varphi$-periodic means that $\varphi^{n}(A)=A$ for some
$n\in{\mathbb{Z}}_{\geq 1}$.

###### Proof.

By [9], the action of group automorphisms of $X$ defined over $K$ on the set
of abelian subvarieties of $X$ has only finitely many orbits. Let
$A_{1},\dots,A_{n}\subset X$ be abelian subvarieties that represent these
finitely many orbits. Then there is a finite extension $L$ of $K$ such that
$A_{i}(L)$ is Zariski dense in $(A_{i})_{L}$ for all $i=1,\dots,n$. Since
every abelian subvarieties of $X$ are isomorphic to one of $A_{i}$ over $K$,
it follows that every abelian subvariety has densely many $L$-rational points.
Replacing $K$ with $L$ and $X$ with $X_{L}$, it is enough to prove the
statement under the additional assumption that $A(K)$ is Zariski dense in $A$.

Let $V=X(K)\otimes_{{\mathbb{Z}}}{\mathbb{Q}}$. Then, $V$ is a finite-
dimensional ${\mathbb{Q}}$-vector space. Let $\varphi$ and $A$ be as in the
statement. Then $\varphi$ induces an ${\mathbb{Q}}$-linear automorphism
$\varphi_{\mathbb{Q}}\colon V\longrightarrow V$, and
$A(K)\otimes_{{\mathbb{Z}}}{\mathbb{Q}}\subset V$ is a
$\varphi_{\mathbb{Q}}$-periodic subspace. By Lemma 6.1, there is
$n_{0}\in{\mathbb{Z}}_{\geq 1}$ depending only on $\dim V$ such that

(6.10)
$\displaystyle\varphi_{{\mathbb{Q}}}^{n_{0}}(A(K)\otimes_{{\mathbb{Z}}}{\mathbb{Q}})=A(K)\otimes_{{\mathbb{Z}}}{\mathbb{Q}}.$

This implies there is $m\in{\mathbb{Z}}_{\geq 1}$ such that

(6.11) $\displaystyle m\varphi^{n_{0}}(A(K))\subset A(K),$

and therefore $\varphi^{n_{0}}(mA(K))\subset A(K)$. Since $A(K)$ is Zariski
dense in $A$, $mA(K)$ is also Zariski dense in $[m](A)=A$. (Here $[m]$ stands
for the multiplication by $m$ map acting on the scheme $A$.) Taking Zariski
closure, we get

(6.12)
$\displaystyle\varphi^{n_{0}}(A)=\overline{\varphi^{n_{0}}(mA(K))}\subset A.$

Looking at the dimension, we get $\varphi^{n_{0}}(A)=A$ and we are done. ∎

We summarize some facts on minimal polynomials of endomorphisms on abelian
varieties that we use later. These are well-known, but we give proofs for
completeness.

######  Lemma 6.3.

Let $k$ be a field and $X$ be an abelian variety over $k$. For a group
endomorphism $\varphi\colon X\longrightarrow X$, the minimal polynomial of
$\varphi$ is denoted by $P_{\varphi}$. ($P_{\varphi}$ is a monic polynomial
with integer coefficients.)

  1. (1)

For a group endomorphism $\varphi\colon X\longrightarrow X$ and
$n\in{\mathbb{Z}}_{\geq 1}$, $P_{\varphi}$ has a cyclotomic polynomial as a
factor if and only if so does $P_{\varphi^{n}}$.

  2. (2)

There are $n_{0},m_{0}\in{\mathbb{Z}}_{\geq 1}$ depending only on $\dim X$
such that for every group endomorphism $\varphi\colon X\longrightarrow X$
whose minimal polynomial is a product of cyclotomic polynomials, we have
$(\varphi^{n_{0}}-\operatorname{id})^{m_{0}}=0$.

  3. (3)

There is $n_{0}\in{\mathbb{Z}}_{\geq 1}$ depending only on $\dim X$ such that
for every group endomorphism $\varphi\colon X\longrightarrow X$, the minimal
polynomial of $\varphi^{n_{0}}$ is of the form

(6.13) $\displaystyle P_{\varphi^{n_{0}}}(t)=(t-1)^{m}Q(t)$

for some $m\in{\mathbb{Z}}_{\geq 0}$ and $Q(t)\in{\mathbb{Z}}[t]$ such that
$Q(t)$ does not have cyclotomic polynomials as a factor.

###### Proof.

(1) Note that $P_{\varphi}$ has a cyclotomic factor if and only if there is
$m\in{\mathbb{Z}}_{\geq 1}$ such that $\varphi^{m}-\operatorname{id}$ is not
surjective. Indeed, let us write $P_{\varphi}=Q(t)R(t)$ where
$Q,R\in{\mathbb{Z}}[t]$ are monic and $Q$ has no cyclotomic factor and $R$ is
a product of cyclotomic polynomials (possibly repeated). Suppose $R\neq 1$.
Take $l,m\in{\mathbb{Z}}_{\geq 1}$ such that $R(t)\mid(t^{m}-1)^{l}$. Since
$R\neq 1$, $R(\varphi)$ is not surjective and hence
$(\varphi^{m}-\operatorname{id})^{l}$ is not surjective. This implies
$\varphi^{m}-\operatorname{id}$ is not surjective. Conversely, suppose
$\varphi^{m}-\operatorname{id}$ is not surjective for some
$m\in{\mathbb{Z}}_{\geq 1}$. If $R=1$, then $P_{\varphi}(t)$ and $t^{m}-1$ are
coprime. This implies $\varphi^{m}-\operatorname{id}$ is surjective and
contradiction.

Now, the statement easily follows.

(2) Since

(6.14)
$\displaystyle{\mathbb{Q}}[t]/(P_{\varphi}(t))\longrightarrow\operatorname{End}(X)_{{\mathbb{Q}}}$

is an injection of ${\mathbb{Q}}$-vector spaces, we have

(6.15) $\displaystyle\deg
P_{\varphi}\leq\dim\operatorname{End}(X)_{{\mathbb{Q}}}\leq 4(\dim X)^{2}.$

Thus there exists $n_{0}\in{\mathbb{Z}}_{\geq 1}$ that is determined by $\dim
X$ such that all the roots of $P_{\varphi}$ are $n_{0}$-th root of $1$. Hence
$P_{\varphi}(t)\mid(t^{n_{0}}-1)^{4(\dim X)^{2}}$ and we are done.

(3) Let us write $P_{\varphi}=QR$ as in the proof of $(1)$. Let
$A=Q(\varphi)(X)$ and $B=R(\varphi)(X)$. Then $A,B\subset X$ are
$\varphi$-invariant abelian subvarieties. Moreover, we have the commutative
diagram

(6.16) ${A\times B}$${A\times
B}$${X}$${X}$$\scriptstyle{\varphi_{A}\times\varphi_{B}}$$\scriptstyle{\varphi}$

where the vertical arrows are additions and $\varphi_{A}$ and $\varphi_{B}$
are restrictions of $\varphi$ on $A$ and $B$ respectively. By construction,
the vertical arrow is an isogeny. (cf. [17, section 3].)

The minimal polynomial of $\varphi_{A}$ is $R(t)$. By (2), there are
$n_{0},m_{0}$ depending only on $\dim X$ such that
$(\varphi_{A}^{n_{0}}-1)^{m_{0}}=0$. Thus
$P_{\varphi_{A}^{n_{0}}}(t)\mid(t-1)^{m_{0}}$ and hence
$P_{\varphi_{A}^{n_{0}}}(t)=(t-1)^{m}$ for some $m\in{\mathbb{Z}}_{\geq 1}$.
On the other hand, the minimal polynomial of $\varphi_{B}$ is $Q$, which does
not have cyclotomic factors. By (1), $P_{\varphi_{B}^{n_{0}}}$ has no
cyclotomic factor as well. Therefore $P_{\varphi_{A}^{n_{0}}}$ and
$P_{\varphi_{B}^{n_{0}}}$ are coprime. Thus

(6.17) $\displaystyle
P_{\varphi^{n_{0}}}=P_{(\varphi_{A}\times\varphi_{B})^{n_{0}}}=P_{\varphi_{A}^{n_{0}}}P_{\varphi_{B}^{n_{0}}}=(t-1)^{m}P_{\varphi_{B}^{n_{0}}}$

and we are done. ∎

Now we are ready to prove Theorem E.

######  Theorem 6.4 (Theorem E).

Let $X$ be an abelian variety over a number field $K$. Then there exists
$s_{0}\in{\mathbb{Z}}_{\geq 0}$ depending only on $X$ with the following
property. For any surjective morphism $f\colon X\longrightarrow X$ and any
reduced closed subscheme $Y\subset X$ with $f(Y)\subset Y$, we have

(6.18) $\displaystyle(f^{-s_{0}-1}(Y)\smallsetminus
f^{-s_{0}}(Y))(K)=\emptyset.$

###### Proof.

We prove the statement by induction on $\dim X$. When $\dim X=0$, there is
nothing to prove. Suppose $\dim X\geq 1$.

Let $f$ and $Y$ be as in the statement. If $Y(K)=\emptyset$, then eq. 6.18
holds for all $s_{0}\in{\mathbb{Z}}_{\geq 0}$. Suppose $Y(K)\neq\emptyset$.
Let $Z$ be the Zariski closure of $Y(K)$. Then we have $f(Z)\subset Z$ and

(6.19) $\displaystyle\left(f^{-s-1}(Y)\smallsetminus
f^{-s}(Y)\right)(K)=\left(f^{-s-1}(Z)\smallsetminus f^{-s}(Z)\right)(K)$

for all $s\in{\mathbb{Z}}_{\geq 0}$. Thus replacing $Y$ with $Z$, we may
assume $Y(K)$ is Zariski dense in $Y$. Note that in this case, every
irreducible component of $Y$ has Zariski dense set of $K$-rational points and
in particular geometrically irreducible. Let $Y=Y_{1}\cup\cdots\cup Y_{r}$ be
the irreducible decomposition so that $Y_{1},\dots,Y_{r^{\prime}}$ are
$f$-periodic and $Y_{r^{\prime}+1},\dots,Y_{r}$ are mapped into periodic ones
by some iterates of $f$. Take $m\in{\mathbb{Z}}_{\geq 1}$ so that for all
$j\in\\{r^{\prime}+1,\dots,r\\}$, $f^{m}(Y_{j})\subset Y_{i}$ for some
$i\in\\{1,\dots,r^{\prime}\\}$. Let $Y^{\prime}=Y_{1}\cup\cdots\cup
Y_{r^{\prime}}$. Then we have

(6.20) $\displaystyle f^{-s-1}(Y)(K)\subset f^{-s-1-m}(Y^{\prime})(K)$

for all $s\in{\mathbb{Z}}_{\geq 0}$. Thus, we may replace $Y$ with
$Y^{\prime}$ to prove the statement. In particular, we may assume every
irreducible component of $Y$ is $f$-periodic. Moreover, since it is enough to
prove the existence of $s_{0}$ for each $f$-cycles of irreducible components,
we may further assume that irreducible components of $Y$ form a cycle under
$f$, say, $f(Y_{i})=Y_{i+1}$ for $i=1,\dots,r-1$ and $f(Y_{r})=Y_{1}$. Since
$Y_{1}(K)\neq\emptyset$, picking a point $y\in Y_{1}(K)$ and taking conjugate
by $T_{y}\colon X\longrightarrow X$, we may assume $0\in Y_{1}$. Finally, let
us write $f=T_{a}\circ\varphi$ with $a\in X(K)$ and a surjective group
endomorphism $\varphi$. By Lemma 6.3(3), replacing $f$ with iterates depending
only on $\dim X$, we may assume the cyclotomic factor of the minimal
polynomial of $\varphi$ is of the form $(t-1)^{d}$ for some
$d\in{\mathbb{Z}}_{\geq 0}$.

In summary, it is enough to prove the existence of $s_{0}$ in the following
setting:

  * •

$f=T_{a}\circ\varphi$ where $P_{\varphi}(t)=(t-1)^{d}Q(t)$ for some
$d\in{\mathbb{Z}}_{\geq 0}$ and $Q$ having no cyclotomic factors;

  * •

$Y(K)\subset Y$ is Zariski dense;

  * •

We have an irreducible decomposition $Y=Y_{1}\cup\cdots\cup Y_{r}$ such that
$0\in Y_{1}$ and $f(Y_{i})=Y_{i+1}$ for $i=1,\dots,r-1$ and $f(Y_{r})=Y_{1}$.

Let $A:=\operatorname{Stab}(Y_{1})^{\circ}$ be the identity component of the
stabilizer of $Y_{1}$ in $X$. This is an abelian subvariety. Applying $f$ to
the equality $Y_{1}+A=Y_{1}$, we get $Y_{2}+\varphi(A)=Y_{2}$. Repeating this,
we get $Y_{1}+\varphi^{r}(A)=Y_{1}$. In particular, we get $\varphi^{r}(A)=A$.
By Lemma 6.2, there exists $n_{0}\in{\mathbb{Z}}_{\geq 1}$ depending only on
$X$ such that $\varphi^{n_{0}}(A)=A$.

We claim that we may replace $f$ with $f^{n_{0}}$ and $Y$ with the
$f^{n_{0}}$-orbit of $Y_{1}$. Indeed, let $W$ be the $f^{n_{0}}$-orbit of
$Y_{1}$ and suppose that there is $s_{0}\in{\mathbb{Z}}_{\geq 0}$ depending
only on $X$ such that

(6.21)
$\displaystyle((f^{n_{0}})^{-s_{0}-1}(W)\smallsetminus(f^{n_{0}})^{-s_{0}}(W))(K)=\emptyset.$

Take $x\in X(K)$ such that $f^{s}(x)\in Y$ for some $s\in{\mathbb{Z}}_{\geq
0}$. Then $f^{s}(x)\in W$ for some $s\in{\mathbb{Z}}_{\geq 0}$ as well. Write
$s=qn_{0}+l$ with $q\in{\mathbb{Z}}_{\geq 0}$ and $0\leq l\leq n_{0}-1$. Then
$(f^{n_{0}})^{q}(f^{l}(x))\in W$ implies $(f^{n_{0}})^{s_{0}}(f^{l}(x))\in W$
and therefore $f^{n_{0}s_{0}+n_{0}-1}(x)\in Y$.

Now replacing $f$ with $f^{n_{0}}$ and $Y$ with $W$, we may assume
$\varphi(A)=A$. Note that in this case, we have
$\operatorname{Stab}(Y_{i})^{\circ}=A$ for all $i=1,\dots,r$. Consider

(6.22)
${X}$${X}$${X/A}$${X/A}$$\scriptstyle{f}$$\scriptstyle{\pi}$$\scriptstyle{\pi}$$\scriptstyle{\overline{f}:=T_{\pi(a)}\circ\overline{\varphi}}$

where $\pi$ is the quotient morphism and $\overline{\varphi}$ is the group
endomorphism induced by $\varphi$. Since $\pi(Y_{1})(K)$ is Zariski dense in
$\pi(Y_{1})$, $\pi(Y_{1})$ is geometrically irreducible (cf. [14, Proposition
2.3.26]).

We also have
$\operatorname{Stab}(\pi(Y_{1})_{\overline{K}})^{\circ}=\operatorname{Stab}(\pi(Y_{1}))^{\circ}_{\overline{K}}=0$.
Thus by [12, (3.7) Theorem], $\pi(Y_{1})_{\overline{K}}$ is a variety of
general type (possibly a single point). As $\pi(Y_{1})$ is
$\overline{f}$-periodic, there is $m\in{\mathbb{Z}}_{\geq 1}$ such that
$\overline{f}^{m}|_{\pi(Y_{1})}=\operatorname{id}_{\pi(Y_{1})}$. Applying this
to $0\in\pi(Y_{1})$, we get

(6.23) $\displaystyle\sum_{i=0}^{m-1}\overline{\varphi}^{i}(\pi(a))=0.$

In particular, we get
$\overline{\varphi}^{m}|_{\pi(Y_{1})}=\operatorname{id}_{\pi(Y_{1})}$. Let
$H\subset X/A$ be the abelian subvariety generated by the
$\overline{\varphi}$-orbit of $\pi(Y_{1})$. Then $H$ is
$\overline{\varphi}$-invariant and
$\overline{\varphi}|_{H}^{m}=\operatorname{id}_{H}$. Thus
$P_{\overline{\varphi}|_{H}}(t)\mid t^{m}-1$ and we also have
$P_{\overline{\varphi}|_{H}}\mid P_{\overline{\varphi}}\mid P_{\varphi}$.
Since the cyclotomic factor of $P_{\varphi}$ is $(t-1)^{d}$, we get
$P_{\overline{\varphi}|_{H}}(t)=t-1$, in other words,
$\overline{\varphi}|_{H}=\operatorname{id}_{H}$.

Now let $B=(X/A)/(\overline{\varphi}-\operatorname{id})(X/A)$.

Case 1. Suppose $\dim B\geq 1$. Let $\nu\colon X/A\longrightarrow B$ be the
quotient morphism. The morphism $\overline{\varphi}$ descends to
$\operatorname{id}$ on $B$ and thus by eq. 6.23, $\nu(\pi(a))\in B(K)_{\rm
tors}$. Note that we have $B=X/C$ where $C=A+(\varphi-\operatorname{id})(X)$
is an abelian subvariety. By [9], the isomorphic classes of the quotient of
$X$ by an abelian subvariety are finite. Thus there is
$m_{0}\in{\mathbb{Z}}_{\geq 1}$ depending only on $X$ such that
$m_{0}\nu(\pi(a))=0$. Thus, we have the commutative diagram

(6.24)
${X}$${X}$${B}$$\scriptstyle{f^{m_{0}}}$$\scriptstyle{\mu}$$\scriptstyle{\nu\circ\pi=:\mu}$

where $\mu$ is the quotient morphism of $X$ by $C$. Now let $x\in X(K)$ be
such that $f^{s}(x)\in Y$ for some $s\in{\mathbb{Z}}_{\geq 0}$. Then
$(f^{m_{0}})^{s}(x)\in Y$ for some $s\in{\mathbb{Z}}_{\geq 0}$ as well. Note
that $f^{m_{0}}$ induces

(6.25) $\displaystyle f^{m_{0}}\colon\mu^{-1}(\mu(x))=x+C\longrightarrow x+C.$

Note also that $x+C$ can be equipped with an abelian variety structure so that
$x$ is the identity element so that it is isomorphic to $C$ as abelian
varieties. As $\dim B\geq 1$, we have $\dim C<\dim X$, we can apply the
induction hypothesis to $f^{m_{0}}|_{x+C}$ and $Y\cap(x+C)$. Then there is
$s_{0}$ depending only on $C$ such that $(f^{m_{0}})^{s_{0}}(x)\in Y$. Since
$C$ is an abelian subvariety of $X$, by [9] again, we can take $s_{0}$
depending only on $X$.

Case 2. Suppose $B=0$. Then $\overline{\varphi}-\operatorname{id}$ is an
isogeny and hence $H=0$. In particular, we have $\pi(Y_{1})=0$. Moreover,
$P_{\overline{\varphi}}$ has no cyclotomic factor because if it has, it must
be the power of $t-1$.

By eq. 6.23, $\overline{f}^{m}=\overline{\varphi}^{m}$. Noting
$\pi(Y_{2})=\overline{f}(\pi(Y_{1}))=\pi(a)$, we have
$\overline{\varphi}^{m}(\pi(a))=\pi(a)$. Hence $\pi(a)\in(X/A)(K)_{\rm tors}$.

Now let $x\in X(K)$ be such that $f^{s}(x)\in Y$ for some
$s\in{\mathbb{Z}}_{\geq 0}$. Then $f^{s}(x)\in Y_{1}$ for some
$s\in{\mathbb{Z}}_{\geq 1}$ as well. Then $\overline{f}^{s}(\pi(x))=0$ and
this implies

(6.26)
$\displaystyle\overline{\varphi}^{s}(\pi(x))=-\sum_{i=0}^{s-1}\overline{\varphi}^{i}(\pi(a)).$

The right-hand side is a torsion point, and thus $\pi(x)\in(X/A)(K)_{\rm
tors}$. Since $\pi(a)$ is a torsion point, we have
$O_{\overline{f}}(\pi(x))\subset(X/A)(K)_{\rm tors}$. Again by [9], there is
$m_{0}\in{\mathbb{Z}}_{\geq 1}$ depending only on $X$ such that
$\\#X/A)(K)_{\rm tors}\leq m_{0}$. Then we have
$\overline{f}^{m_{0}}(\pi(x))\in\pi(Y)$ and therefore
$f^{m_{0}}(x)\in\pi^{-1}(\pi(Y))=Y+A=Y$. This completes the proof. ∎

In the proof of Theorem 6.4, we use several finiteness properties of abelian
varieties, namely finiteness of the set of torsion points, finiteness of
Mordell-Weil rank, and finiteness of abelian subvarieties [9]. As for the
first one, there is a following strong uniformity conjecture.

######  Conjecture 6.5.

Let $g\in{\mathbb{Z}}_{\geq 1}$ and $d\in{\mathbb{Z}}_{\geq 1}$. Then there
exists $C\geq 1$ depending only on $g$ and $d$ such that for all abelian
variety $X$ of dimension $g$ defined over a number field $K$ with
$[K:{\mathbb{Q}}]\leq d$, we have

(6.27) $\displaystyle\\#X(K)_{\rm tors}\leq C.$

If we assume this conjecture, we have the following.

######  Proposition 6.6.

Assume Conjecture 6.5. Let $g\in{\mathbb{Z}}_{\geq 1}$ and
$d\in{\mathbb{Z}}_{\geq 1}$. Then there exists $s_{0}\geq 0$ depending only on
$g$ and $d$ with the following property. Let $X$ be an abelian variety of
dimension $g$ defined over a number field $K$ with $[K:{\mathbb{Q}}]\leq d$.
Let $f\colon X\longrightarrow X$ be a surjective morphism, and let $Y\subset
X$ be an $f$-invariant geometrically irreducible closed subvariety. Then

(6.28) $\displaystyle(f^{-s_{0}-1}(Y)\smallsetminus
f^{-s_{0}}(Y))(K)=\emptyset.$

###### Proof.

Let $X,f,Y$ be as in the statement. We may assume there exists at least one
$K$-rational point $y\in Y(K)$. Let $A$ be the identity component of the
stabilizer $\operatorname{Stab}(Y)$ of $Y$ in $X$. Let us write
$f=T_{a}\circ\varphi$ with $a\in X(K)$ and a group endomorphism $\varphi\colon
X\longrightarrow X$. Then we have $\varphi(A)=A$. Indeed,

(6.29) $\displaystyle
Y+\varphi(A)=(a+\varphi(Y))+\varphi(A)=a+\varphi(Y+A)=a+\varphi(Y)=Y$

and this implies $\varphi(A)=A$.

Consider the following commutative diagram:

(6.30)
${X}$${X}$${X/A}$${X/A}$$\scriptstyle{f}$$\scriptstyle{\pi}$$\scriptstyle{\pi}$$\scriptstyle{\overline{f}:=T_{\pi(a)}\circ\overline{\varphi}}$

where $\pi$ is the quotient morphism and $\overline{\varphi}$ is the group
morphism induced by $\varphi$. Then
$\operatorname{Stab}(\pi(Y)_{\overline{K}})=\operatorname{Stab}(\pi(Y))_{\overline{K}}$
is finite. Since $\pi(Y)$ is geometrically irreducible, by [12, (3.7)
Theorem], $\pi(Y)_{\overline{K}}$ is a variety of general type. Thus there is
$n\in{\mathbb{Z}}_{\geq 1}$ such that
$\overline{f}|_{\pi(Y)}^{n}=\operatorname{id}_{\pi(Y)}$.

Let $P_{\overline{\varphi}}$ be the minimal polynomial of
$\overline{\varphi}$. Let us factor $P_{\overline{\varphi}}(t)=Q(t)R(t)$ such
that $Q$ is a product of cyclotomic polynomials and $R$ has no cyclotomic
factors. Then set $B=R(\overline{\varphi})(X/A)$. Since $B$ is
$\overline{\varphi}$-invariant and thus we can form the following commutative
diagram:

(6.31)
${X/A}$${X/A}$${(X/A)/B}$${(X/A)/B}$$\scriptstyle{\overline{f}}$$\scriptstyle{\nu}$$\scriptstyle{\nu}$$\scriptstyle{h:=T_{\nu(\pi(a))}\circ\psi}$

where $\nu$ is the quotient morphism and $\psi$ is the morphism induced by
$\overline{\varphi}$. Then, first note that the minimal polynomial
$P_{\psi}(t)$ of $\psi$ is equal to $R(t)$. In particular,
$\psi^{n}-\operatorname{id}$ is an isogeny. As
$\overline{f}|_{\pi(Y)}^{n}=\operatorname{id}_{\pi(Y)}$, we have
$h^{n}=\operatorname{id}$ on $\nu(\pi(Y))$. This implies

(6.32)
$\displaystyle(\psi^{n}-\operatorname{id})(\nu(\pi(Y)))=-\sum_{i=0}^{n-1}\psi^{i}(\nu(\pi(a))).$

Thus we get $\dim\nu(\pi(Y))=0$. Since $Y$ is irreducible, so is $\nu(\pi(Y))$
and hence $\nu(\pi(Y))=\\{\nu(\pi(y))\\}$. In particular, we have

(6.33) $\displaystyle h(\nu(\pi(y)))=\nu(\pi(y))$ (6.34)
$\displaystyle\pi(Y)\subset\pi(y)+B.$

Moreover $\pi(y)+B$ is invariant under $\overline{f}$ and the induced self-
morphism on $\pi(y)+B$ is an isomorphism. Indeed, we have the following
commutative diagram

(6.35)
${\pi(y)+B}$${\pi(y)+B}$${B}$${B}$$\scriptstyle{\overline{f}}$$\scriptstyle{T_{\pi(y)}}$$\scriptstyle{\overline{\varphi}}$$\scriptstyle{T_{\overline{\varphi}(\pi(y))+\pi(a)}}$

and the bottom arrow $\overline{\varphi}$ action on $B$ is an isomorphism
because its minimal polynomial is $Q$.

Now let $x\in X(K)$ and suppose $f^{s}(x)\in Y$ for some
$s\in{\mathbb{Z}}_{\geq 0}$. Then we have $h^{s}(\nu(\pi(x)))=\nu(\pi(y))$.
Let us consider the following set

(6.36) $\displaystyle S=\\{z\in((X/A)/B)(K)\mid\text{$h^{t}(z)=\nu(\pi(y))$
for some $t\in{\mathbb{Z}}_{\geq 0}$}\\}.$

For any $z_{1},z_{2}\in S$, we have $\psi^{t}(z_{1}-z_{2})=0$. Since $\psi$ is
an isogeny, $z_{1}-z_{2}$ is a torsion point and hence
$\\#S\leq((X/A)/B)(K)_{\rm tors}$. By Conjecture 6.5, there is
$s_{0}\in{\mathbb{Z}}_{\geq 1}$ depending only on $g,d$ such that $\\#S\leq
s_{0}$. Then we have $h^{s_{0}}(\nu(\pi(x)))=\nu(\pi(y))$. This implies
$\overline{f}^{s_{0}}(\pi(x))\in\pi(y)+B$. Since
$\overline{f}^{s}(\pi(x))\in\pi(Y)$ and $\overline{f}|_{\pi(y)+B}$ is an
isomorphism (as well as $\overline{f}(\pi(Y))=\pi(Y)$), we must have
$\overline{f}^{s_{0}}(\pi(x))\in\pi(Y)$. Therefore we get

(6.37) $\displaystyle f^{s_{0}}(x)\in\pi^{-1}(\pi(Y))=Y+A=Y$

and we are done. ∎

######  Remark 6.7.

We do not know if we can get rid of the geometric irreducibility of $Y$. In
the proof of Theorem 6.4, we use uniform bound of the period of periodic
abelian subvarieties to deal with possibly non-irreducible $Y$. Our bound of
the periods depends on the Mordell-Weil rank of $X(L)$ where $L$ is a finite
extension of $K$, which also depends on $X$.

##  7\. Preliminaries on polydisc algebras

We prepare some preliminary results on polydisc algebras, especially on prime
factorization in the polydisc algebras. The contents of this section might be
well-known to experts, but we include them here for completeness. For basic
materials on Tate algebras or strictly affinoid algebras, see for example [3,
11].

Let $(K,|\ |)$ be a complete Non-archimedean valued field. For
$r=(r_{1},\dots,r_{n})\in{\mathbb{R}}_{>0}^{n}$, the polydisc algebra with
radii $r$ is

(7.1) $\displaystyle K\\{r^{-1}T\\}$
$\displaystyle=K\\{r_{1}^{-1}T_{1},\dots,r_{n}^{-1}T_{n}\\}$ (7.2)
$\displaystyle=\biggl{\\{}\sum_{I\in{\mathbb{Z}}_{\geq 0}^{n}}a_{I}T^{I}\in
K\llbracket
T\rrbracket\mathrel{\bigg{|}}\lim_{I\to\infty}|a_{I}|r^{I}=0\biggr{\\}}.$

Here we use the standard multi index notation: for $I=(i_{1},\dots,i_{n})$,
$T^{I}:=T_{1}^{i_{1}}\cdots T_{n}^{i_{n}}$ and $r^{I}=r_{1}^{i_{1}}\cdots
r_{n}^{i_{n}}$. The Gauss norm on $K\\{r^{-1}T\\}$ is

(7.3) $\displaystyle\biggl{\|}\sum_{I\in{\mathbb{Z}}_{\geq
0}^{n}}a_{I}T^{I}\biggr{\|}_{K\\{r^{-1}T\\}}=\max\left\\{|a_{I}|r^{I}\
\middle|\ I\in{\mathbb{Z}}_{\geq 0}^{n}\right\\}.$

We drop the subscript and simply write $\|\ \|$ when there should be no
confusion. For $I=(i_{1},\dots,i_{n})\in{\mathbb{Z}}_{\geq 0}^{n}$, we set
$|I|=i_{1}+\cdots+i_{n}$.

######  Definition 7.1.

For $f\in K\\{r^{-1}T\\}\smallsetminus\\{0\\}$, we define

(7.4)
$\displaystyle\operatorname{ord}(f)=\operatorname{ord}_{K\\{r^{-1}T\\}}(f):=\max\left\\{|I|\
\middle|\ |a_{I}|r^{I}=\|f\|\right\\}.$

When $f=0$, we set $\operatorname{ord}_{K\\{r^{-1}T\\}}(0)=\infty$.

######  Lemma 7.2.

Let $K$ be a complete Non-archimedean valued field and
$r=(r_{1},\dots,r_{n})\in{\mathbb{R}}_{>0}^{n}$.

  1. (1)

For $f,g\in K\\{r^{-1}T\\}$, we have
$\operatorname{ord}(fg)=\operatorname{ord}(f)+\operatorname{ord}(g)$.

  2. (2)

For $f\in K\\{r^{-1}T\\}$, $f$ is unit if and only if
$\operatorname{ord}(f)=0$.

  3. (3)

For any finite extension $L$ of $K$ equipped with the uniquely extended
absolute value and $f\in K\\{r^{-1}T\\}$, we have

(7.5)
$\displaystyle\operatorname{ord}_{L\\{r^{-1}T\\}}(f)=\operatorname{ord}_{K\\{r^{-1}T\\}}(f).$

  4. (4)

For $f\in K\\{r^{-1}T\\}\smallsetminus\\{0\\}$, the following set is finite:

(7.6) $\displaystyle{\mathcal{I}}(f):=\left\\{I\in{\mathbb{Z}}_{\geq 0}^{n}\
\middle|\ \exists
s\in\prod_{i=1}^{n}(0,r_{i}],|a_{I}|s^{I}=\|f\|_{K\\{s^{-1}T\\}}\right\\}.$

In particular, the set of orders

(7.7) $\displaystyle\left\\{\operatorname{ord}_{K\\{s^{-1}T\\}}(f)\ \middle|\
s\in\prod_{i=1}^{n}(0,r_{i}]\right\\}$

is finite.

###### Proof.

(1) Let us write $f=\sum_{k\geq 0}f_{k}$, $g=\sum_{l\geq 0}g_{l}$ where
$f_{k},g_{l}$ are homogeneous part of degree $k,l$ respectively. Let
$d=\operatorname{ord}(f),e=\operatorname{ord}(g)$. Then, by the standard
argument, we have

(7.8) $\displaystyle\left\|\sum_{i=0}^{d+e}f_{i}g_{d+e-i}\right\|$
$\displaystyle=\left\|f_{d}g_{e}\right\|=\left\|fg\right\|,$ (7.9)
$\displaystyle\left\|\sum_{i=0}^{m}f_{i}g_{m-i}\right\|$
$\displaystyle<\left\|f_{d}g_{e}\right\|\quad(\text{for }m>d+e)$

and thus $\operatorname{ord}(fg)=d+e$.

(2) If $f$ is a unit, then
$0=\operatorname{ord}(1)=\operatorname{ord}(ff^{-1})=\operatorname{ord}(f)+\operatorname{ord}(f^{-1})$.
This implies $\operatorname{ord}(f)=0$. Suppose $\operatorname{ord}(f)=0$.
Then $f(0)\neq 0$ and we can write

(7.10) $\displaystyle f=f(0)\left(1+g\right)$

with $g$ such that $\|g\|<1$. Thus, $f$ is a unit.

(3) This is obvious from the definition.

(4) For $I=(i_{1},\dots,i_{n})\in{\mathbb{Z}}_{\geq 0}^{n}$, let

(7.11) $\displaystyle C(I)=\\{(j_{1},\dots,j_{n})\in{\mathbb{Z}}_{\geq
0}^{n}\mid\text{$j_{k}\geq i_{k}$ for $k=1,\dots,n$}\\}.$

Note that for any subset ${\mathcal{I}}\subset{\mathbb{Z}}_{\geq 0}^{n}$,
there are finitely many elements $I_{1},\dots,I_{t}$ such that

(7.12)
$\displaystyle\bigcup_{I\in{\mathcal{I}}}C(I)=\bigcup_{k=1}^{t}C(I_{k}).$

To see this, note that $C(I)$ is exactly the set of exponents of monomials
contained in the ideal generated by $x_{1}^{i_{1}}\cdots x_{n}^{i_{n}}$ in the
polynomial ring ${\mathbb{Q}}[x_{1},\dots,x_{n}]$. The left-hand side is the
set of exponents of monomials contained in the ideal generated by
$(x^{I})_{I\in{\mathcal{I}}}$, but the Hilbert basis theorem says this ideal
is generated by finitely many $x^{I}$’s.

Now, we go back to our situation. We prove (4) by induction on $n$. The
statement is clear when $n=1$. Suppose $n\geq 2$ and
$\\#{\mathcal{I}}(f)=\infty$. By the above discussion, there are
$I_{1},\dots,I_{t}\in{\mathcal{I}}(f)$ such that

(7.13)
$\displaystyle{\mathcal{I}}(f)\subset\bigcup_{I\in{\mathcal{I}}(f)}C(I)=\bigcup_{k=1}^{t}C(I_{k}).$

Since $\\#{\mathcal{I}}(f)=\infty$, there is $I_{k}$ such that

(7.14) $\displaystyle\\#{\mathcal{I}}(f)\cap C(I_{k})=\infty.$

Define $J_{1}=I_{k}$. Next,

(7.15) $\displaystyle\bigcup_{I\in{\mathcal{I}}(f)\cap
C(J_{1})\smallsetminus\\{J_{1}\\}}C(I)$

is also a finite union, and thus there is $J_{2}\in{\mathcal{I}}(f)\cap
C(J_{1})\smallsetminus\\{J_{1}\\}$ such that $\\#\left({\mathcal{I}}(f)\cap
C(J_{1})\smallsetminus\\{J_{1}\\}\right)\cap C(J_{2})=\infty$. Repeat this and
we get a sequence $(J_{k})_{k=1}^{\infty}$ such that

(7.16) $\displaystyle J_{k}\in{\mathcal{I}}(f)\ \text{for $k=1,2,\dots$}$
(7.17) $\displaystyle C(J_{1})\supsetneq C(J_{2})\supsetneq\cdots.$

Now let us write $f=\sum_{I}a_{I}T^{I}$ and take $I_{0}\in{\mathbb{Z}}_{\geq
0}^{n}$ such that

(7.18) $\displaystyle|a_{I_{0}}|r^{I_{0}}=\|f\|_{K\\{r^{-1}T\\}}\ \text{and}\
|I_{0}|=\operatorname{ord}_{K\\{r^{-1}T\\}}(f).$

Then we have ${\mathcal{I}}(f)\cap C(I_{0})=\\{I_{0}\\}$. Indeed, if $J\in
C(I_{0})\smallsetminus\\{I_{0}\\}$ and $s\in\prod_{i=1}^{n}(0,r_{i}]$, then

(7.19)
$\displaystyle|a_{J}|s^{J}=|a_{J}|r^{J}\frac{s^{J}}{r^{J}}<|a_{I_{0}}|r^{I_{0}}\frac{s^{J}}{r^{J}}=|a_{I_{0}}|s^{I_{0}}\frac{r^{I_{0}}}{s^{I_{0}}}\frac{s^{J}}{r^{J}}\leq|a_{I_{0}}|s^{I_{0}}.$

By eq. 7.16, we have $J_{k}\not\in C(I_{0})$ for all $k$. By replacing it with
a subsequence, we may assume that one of the coordinates of $J_{k}$ are
constant. By changing coordinates, we may assume the last coordinates are
constant. Let us write $J_{k}=(J_{k}^{\prime},j)$ where
$J_{k}^{\prime}\in{\mathbb{Z}}_{\geq 0}^{n-1}$ and $j\in{\mathbb{Z}}_{\geq
0}$. Consider

(7.20) $\displaystyle
g=\sum_{k=1}^{\infty}a_{J_{k}}T^{J_{k}}\frac{1}{T_{n}^{j}}\in
K\\{r_{1}^{-1}T_{1},\dots,r_{n-1}^{-1}T_{n-1}\\}.$

Since $J_{k}\in{\mathcal{I}}(f)$, we have $J_{k}^{\prime}\in{\mathcal{I}}(g)$.
This implies that ${\mathcal{I}}(g)$ is infinite, which contradicts the
induction hypothesis. ∎

######  Corollary 7.3.

Let $K$ be a complete Non-archimedean valued field and
$r=(r_{1},\dots,r_{n})\in{\mathbb{R}}_{>0}^{n}$. Let $f\in
K\\{r^{-1}T\\}\smallsetminus\\{0\\}$. Then there is $C>0$ with the following
property: for all finite extension $L$ of $K$ with the uniquely extended
absolute value and for all $s=(s_{1},\dots,s_{n})\in\prod_{i=1}^{n}(0,r_{i}]$
with $s_{i}\in|L|$ for $i=1,\dots n$, we have

(7.23) $\displaystyle\left(\hbox{\hbox{\kern
0.0pt\raise-9.44444pt\vbox{\halign{\relax\hfil\txtline@@{#}\hfil\cr\hbox{{\ignorespaces\leavevmode
the number of prime factors counted with multiplicity
}\\\\\hbox{{\ignorespaces of $f$ as an element of
$L\\{s^{-1}T\\}$}\crcr}}}}\right)\leq C.}}$

Here note that $L\\{s^{-1}T\\}$ is UFD, because $s_{i}\in|L|$ and hence
$L\\{s^{-1}T\\}\simeq L\\{T\\}$. (cf. [3, (5.2.6) Theorem 1].)

###### Proof.

Let $L$ and $s$ as in the statement. Let $f=P_{1}\dots P_{l}$ be the prime
factorization in $L\\{s^{-1}T\\}$. Then, by Lemma 7.2 (1), (2), and (3), we
obtain

(7.24)
$\displaystyle\operatorname{ord}_{K\\{s^{-1}T\\}}(f)=\operatorname{ord}_{L\\{s^{-1}T\\}}(f)=\sum_{i=1}^{l}\operatorname{ord}_{L\\{s^{-1}T\\}}(P_{i})\geq
l.$

By Lemma 7.2 (4), $\operatorname{ord}_{K\\{s^{-1}T\\}}(f)$ is bounded when $s$
varies, we are done. ∎

######  Corollary 7.4.

Let $K$ be a complete Non-archimedean valued field. Let
$r=(r_{1},\dots,r_{n})\in{\mathbb{R}}_{>0}^{n}$ with $r_{i}\in|K|$ for
$i=1,\dots,n$. Let $f\in K\\{r^{-1}T\\}\smallsetminus\\{0\\}$. Then there is a
finite extension $L$ of $K$ such that for all finite extensions $L^{\prime}$
of $L$, the prime factorization of $f$ in $L\\{r^{-1}T\\}$ is the prime
factorization of $f$ in $L^{\prime}\\{r^{-1}T\\}$ as well.

###### Proof.

By Corollary 7.3, the number of prime factors of $f$ as an element of
$L\\{r^{-1}T\\}$ are bounded when $L$ varies finite extensions of $K$. Take
$L$ such that the number of prime factors attains the maximum. Let
$f=P_{1}\dots P_{l}$ be the prime factorization in $L\\{r^{-1}T\\}$. For any
finite extension $L^{\prime}$ of $L$, every $P_{i}$ is not a unit in
$L^{\prime}\\{r^{-1}T\\}$ because
$\operatorname{ord}_{L^{\prime}\\{r^{-1}T\\}}(P_{i})=\operatorname{ord}_{L\\{r^{-1}T\\}}(P_{i})>0$.
Therefore, $f=P_{1}\dots P_{l}$ must be the prime factorization of $f$ as an
element of $L^{\prime}\\{r^{-1}T\\}$. ∎

Asymptotic behavior of prime factorization

Let $K$ be a complete Non-archimedean valued field as before. Let
$r=(r_{1},\dots,r_{n})\in{\mathbb{R}}_{>0}^{n}$ and suppose $r_{i}\in|K|$ for
$i=1,\dots n$. Fix a sequence of finite extensions of fields

(7.25) $\displaystyle K=K_{0}\subset K_{1}\subset K_{2}\subset\cdots$

and equip $K_{m}$ with the uniquely extended absolute values for
$m\in{\mathbb{Z}}_{\geq 1}$. Fix also a sequence of radii

(7.26)
$\displaystyle\\{s^{(m)}=(s_{1}^{(m)},\dots,s_{n}^{(m)})\\}_{m=0}^{\infty}$

in ${\mathbb{Z}}_{>0}^{n}$ such that

(7.27) $\displaystyle r_{j}=s_{j}^{(0)}\geq s_{j}^{(1)}\geq
s_{j}^{(2)}\geq\cdots\quad(j=1,2,\ldots,n)$ (7.28) $\displaystyle
s_{j}^{(m)}\in|K_{m}|\quad(\text{$j=1,2,\ldots,n$ and $m\geq 0$}).$

We set

(7.29) $\displaystyle
R_{m}=K_{m}\\{(s_{1}^{(m)})^{-1}T_{1},\dots,(s_{n}^{(m)})^{-1}T_{n}\\}.$

Note that $R_{m}$ are UFD. We have a sequence of inclusions

(7.30) $\displaystyle
R_{0}\xlongrightarrow{i_{1}}R_{1}\xlongrightarrow{i_{2}}R_{2}\xlongrightarrow{i_{3}}\cdots.$

An element $u\in R_{m}$ for some $m$ is said to be eventually unit if there is
some $m^{\prime}\geq m$ such that $u$ is unit in $R_{m^{\prime}}$. In this
case, we say $u$ is eventually unit with respect to $(R_{m})_{m}$.

######  Lemma 7.5.

Under the setting above, let $f\in R_{0}\setminus{0}$. Let $f_{m}$ be the
image of $f$ in $R_{m}$. Then there is $m_{0}\geq 0$ such that

(7.31) $\displaystyle f_{m_{0}}=ug_{1}\cdots g_{l}$

for some $u,g_{1},\dots,g_{l}\in R_{m_{0}}$ with the following properties:

  * •

$u$ is eventually unit:

  * •

for any $m\geq m_{0}$, the image of $g_{j}$ in $R_{m}$ is of the form

(7.32) $\displaystyle vh$

where $v\in R_{m}$ is eventually unit and $h\in R_{m}$ is a prime element.

###### Proof.

Define an inverse system

(7.33) $\displaystyle\cdots
S_{m}\xlongrightarrow{\varphi_{m-1}}S_{m-1}\longrightarrow\cdots\longrightarrow
S_{1}\xlongrightarrow{\varphi_{0}}S_{0}$

as follows. Label the prime factors of $f$ as an element of $R_{0}$. If the
same prime factor appears multiple times, we attach distinct labels to them.
Let $S_{0}$ be the set of the labels. Note that if $f$ is unit, then
$S_{0}=\emptyset$. Next, for each prime factor $P$ of $f$ in $R_{0}$, we
similarly label the prime factors of $P$ as an element of $R_{1}$. Then, set
$S_{1}$ as their disjoint union varying $P$. Define $\varphi_{0}\colon
S_{1}\longrightarrow S_{0}$ so that the labels corresponding to factors of $P$
are mapped to the label corresponding to $P$. Repeat this, and we get the
inverse system. For each $i\in S_{m}$, let us write the corresponding prime
factor of $f$ in $R_{m}$ as $P_{m,i}$. In particular we have
$f_{m}=u\prod_{i\in S_{m}}P_{m,i}$ for some $u\in R_{m}^{\times}$.

First note that $S_{m}=\emptyset$ for some $m$ if and only if $f$ is
eventually unit. In this case, the statement is clear. Suppose
$S_{m}\neq\emptyset$ for all $m\geq 0$. Then

(7.34) $\displaystyle S:=\varprojlim_{m\geq 0}S_{m}\neq\emptyset.$

By Corollary 7.3, there is $C>0$ such that $\\#S_{m}\leq C$ for all $m$. This
implies $\\#S\leq C$. Let $\\#S=l$. Let us write

(7.35) $\displaystyle S=\left\\{(i_{1}(m))_{m},\dots,(i_{l}(m))_{m}\right\\}.$

Then there is $m_{0}$ such that

(7.36) $\displaystyle i_{1}(m_{0}),\dots,i_{l}(m_{0})$

are distinct. Then, we can write

(7.37) $\displaystyle f_{m_{0}}=u\prod_{i\in
S_{m_{0}}\smallsetminus\\{i_{1}(m_{0}),\dots,i_{l}(m_{0})\\}}P_{m_{0},i}\prod_{j=1}^{l}P_{m_{0},i_{j}(m_{0})}$

for some $u\in R_{m_{0}}^{\times}$.

By eq. 7.35, for each $i\in
S_{m_{0}}\smallsetminus\\{i_{1}(m_{0}),\dots,i_{l}(m_{0})\\}$ there is $m\geq
m_{0}$ such that

(7.38) $\displaystyle(S_{m}\longrightarrow S_{m_{0}})^{-1}(i)=\emptyset.$

Indeed, if there is no such $m$, an element of

(7.39) $\displaystyle\varprojlim_{m\geq m_{0}}(S_{m}\longrightarrow
S_{m_{0}})^{-1}(i)$

produces an element of $S$ different from any of $(i_{j}(m))_{m}$. In
particular, we proved that $P_{m_{0},i}$ is eventually a unit.

Finally, for every $m\geq m_{0}$, we can write

(7.40) $\displaystyle
P_{m_{0},i_{j}(m_{0})}=vP_{m,i_{j}(m)}\prod_{\begin{subarray}{c}k\in S_{m}\\\
k\neq i_{j}(m)\\\ (S_{m}\longrightarrow
S_{m_{0}})(k)=i_{j}(m_{0})\end{subarray}}P_{m,k}$

in $R_{m}$ for some $v\in R_{m}^{\times}$. As before, the last factor is
eventually a unit. Thus, we are done. ∎

Question. Is it possible to get rid of “eventually”? That is, is there such
$m_{0}$ with $u$ being unit in $R_{m_{0}}$ and $g_{j}$ being prime in $R_{m}$
for all $m\geq m_{0}$?

##  8\. $p$-adic uniformization

In this section, we recall Rivera-Letelier’s $p$-adic uniformization theorem
and some basic facts on formal power series.

The following is well-known, and the proof is elementary.

######  Lemma 8.1.

Let $R$ be a ring. Let $f=\sum_{n\geq 1}a_{n}z^{n}\in R\llbracket z\rrbracket$
be a formal power series. If $a_{1}\in R^{\times}$, then there is unique $g\in
zR\llbracket z\rrbracket$ such that $f(g(z))=g(f(z))=z$.

Convergent power series that admit compositional inverse induces isomorphism
between small discs around $0$. The following is an algebraic expression of
this fact.

######  Lemma 8.2.

Let $(K,|\ |)$ be a complete non-archimedean field and $r_{0},s_{0}>0$. Let
$u\in K\\{r_{0}^{-1}z\\}$ and $v\in K\\{s_{0}^{-1}z\\}$ with $u(0)=v(0)=0$.
Suppose $u(v(z))=v(u(z))=z$ in $K\llbracket z\rrbracket$. Then there is
$r_{1}\in(0,r_{0}]$ with the following properties. For all $r\in(0,r_{1}]$,
the morphism

(8.1) $\displaystyle K\\{(|u^{\prime}(0)|r)^{-1}z\\}\longrightarrow
K\\{r^{-1}z\\}\ ;z\mapsto u(z)$

is an isometric isomorphism. Moreover, $\|v\|_{|u^{\prime}(0)|r}=r$ and

(8.2) $\displaystyle K\\{r^{-1}z\\}\longrightarrow
K\\{(|u^{\prime}(0)|r)^{-1}z\\}\ ;z\mapsto v(z)$

is the inverse of the above morphism. Here $\|\ \|_{r}$ denotes the Gauss norm
on $K\\{r^{-1}z\\}$.

###### Proof.

Let us write

(8.3) $\displaystyle u(z)=az+\widetilde{u}(z)$ (8.4) $\displaystyle
v(z)=bz+\widetilde{v}(z)$

for some $a,b\in K\smallsetminus\\{0\\}$ and
$\widetilde{u}(z),\widetilde{v}(z)\in z^{2}K\llbracket z\rrbracket$. Then
$u^{\prime}(0)=a$. We have $ab=1$ because $u(v(z))=z$.

Now, there is $r_{1}\in(0,r_{0}]$ such that $\|u\|_{r}=|a|r$ for all
$r\in(0,r_{1}]$. If we further make $r_{1}$ small, then we also have
$\|v\|_{\|u\|_{r}}=\|v\|_{|a|r}=|b||a|r=r$ for all $r\in(0,r_{1}]$. These
imply the morphisms in the statements are well-defined and contractive. Since
$u(v(z))=v(u(z))=z$, they are inverse maps of each other. ∎

Now, we state the $p$-adic uniformization in the form we use later.

######  Theorem 8.3.

Let $r>0$ and $K$ be a finite extension of ${\mathbb{Q}}_{p}$ equipped with
the multiplicative absolute value $|\ |$ extending $|\ |_{p}$. Let
$f=\sum_{n\geq 1}a_{n}z^{n}\in K\\{r^{-1}z\\}$

  1. (1)

Suppose $0<|a_{1}|<1$. Then there is $r^{\prime}\in(0,r]$, $s>0$, and
isometric isomorphism $\varphi\colon K\\{s^{-1}z\\}\longrightarrow
K\\{{r^{\prime}}^{-1}z\\}$ such that

     * •

$\varphi(z)(0)=0$

     * •

$\|f\|_{K\\{{r^{\prime}}^{-1}z\\}}\leq r^{\prime}$

     * •

the following diagram commutes.

${z}$${f(z)}$${K\\{{r^{\prime}}^{-1}z\\}}$${K\\{{r^{\prime}}^{-1}z\\}}$${K\\{s^{-1}z\\}}$${K\\{s^{-1}z\\}}$${z}$${a_{1}z}$$\scriptstyle{\varphi}$$\scriptstyle{\varphi}$

  2. (2)

Suppose $a_{1}=\cdots=a_{d-1}=0$ and $a_{d}\neq 0$ for some $d\geq 2$. Let $L$
be a finite extension of $K$ containing a $(d-1)$-th root of $a_{d}$. Then
there is $r^{\prime}\in(0,r]$, $s>0$, and isometric isomorphism $\varphi\colon
L\\{s^{-1}z\\}\longrightarrow L\\{{r^{\prime}}^{-1}z\\}$ such that

     * •

$\varphi(z)(0)=0$

     * •

$\|f\|_{L\\{{r^{\prime}}^{-1}z\\}}\leq r^{\prime}$

     * •

the following diagram commutes.

${z}$${f(z)}$${L\\{{r^{\prime}}^{-1}z\\}}$${L\\{{r^{\prime}}^{-1}z\\}}$${L\\{s^{-1}z\\}}$${L\\{s^{-1}z\\}}$${z}$${z^{d}}$$\scriptstyle{\varphi}$$\scriptstyle{\varphi}$

###### Proof.

See the proof of [15, Proposition 3.3]. See also [7, Theorem 1]. ∎

##  9\. Preimages Question on ${\mathbb{P}}^{1}\times{\mathbb{P}}^{1}$

In this section, the $p$-adic absolute value $|\ |_{p}$ on ${\mathbb{Q}}_{p}$
is simply denoted by $|\ |$. All algebraic extensions of ${\mathbb{Q}}_{p}$ as
well as ${\mathbb{C}}_{p}$ are equipped with the multiplicative absolute
values extending $|\ |_{p}$ and they are denoted by $|\ |$.

The following is the main theorem in this section.

######  Theorem 9.1.

Let $K$ be a finitely generated field over ${\mathbb{Q}}$. Let
$\varphi\colon{\mathbb{P}}^{1}_{K}\times_{K}{\mathbb{P}}^{1}_{K}\longrightarrow{\mathbb{P}}^{1}_{K}\times_{K}{\mathbb{P}}^{1}_{K}$
be a surjective morphism over $K$. Let
$Y\subset{\mathbb{P}}^{1}_{K}\times_{K}{\mathbb{P}}^{1}_{K}$ be a
$\varphi$-invariant closed subscheme. Then there is $s_{0}\geq 0$ such that
for all $s\geq s_{0}$, we have

(9.1)
$\displaystyle\left(\varphi^{-s-1}(Y)\smallsetminus\varphi^{-s}(Y)\right)(K)=\emptyset.$

###### Proof.

By replacing $\varphi$ with $\varphi^{2}$, we may write $\varphi=f\times g$
for some finite surjective morphisms
$f,g\colon{\mathbb{P}}^{1}_{K}\longrightarrow{\mathbb{P}}^{1}_{K}$. Take a
subring $R\subset K$ which is finitely generated over ${\mathbb{Z}}$ so that
$K=\operatorname{Frac}R$ and there are

  * •

finite surjective morphisms
$f_{R},g_{R}\colon{\mathbb{P}}^{1}_{R}\longrightarrow{\mathbb{P}}^{1}_{R}$;

  * •

closed subscheme
$Y_{R}\subset{\mathbb{P}}^{1}_{R}\times_{R}{\mathbb{P}}^{1}_{R}$

with the following properties. The morphisms $f,g$ are base change of $f_{R}$
and $g_{R}$, $Y$ is the base change of $Y_{R}$, and $Y_{R}$ is $f_{R}\times
g_{R}$-invariant. By [1, Proposition 2.5.3.1] (cf. [2, Proposition 3.6]),
there is a prime number $p$ and an embedding $\sigma\colon
K\longrightarrow{\mathbb{Q}}_{p}$ such that

  * •

$\sigma(R)\subset{\mathbb{Z}}_{p}$

  * •

$p$ does not divide any ramification indices of $f$ and $g$ considered as
self-morphisms on ${\mathbb{P}}^{1}_{\overline{K}}$.

Then let us write the base change of $f_{R},g_{R}$, and $Y_{R}$ to
${\mathbb{Z}}_{p}$ (resp. ${\mathbb{Q}}_{p}$) via $\sigma$ as
$f_{{\mathbb{Z}}_{p}},g_{{\mathbb{Z}}_{p}}$, and $Y_{{\mathbb{Z}}_{p}}$ (resp.
$f_{{\mathbb{Q}}_{p}},g_{{\mathbb{Q}}_{p}}$, and $Y_{{\mathbb{Q}}_{p}}$), cf.
the following diagram.

${\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}\times_{{\mathbb{Z}}_{p}}{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}$P${\mathbb{P}}^{1}_{{\mathbb{Q}}_{p}}\times_{{\mathbb{Q}}_{p}}{\mathbb{P}}^{1}_{{\mathbb{Q}}_{p}}$P${\mathbb{P}}^{1}_{R}\times_{R}{\mathbb{P}}^{1}_{R}$P${\mathbb{P}}^{1}_{K}\times_{K}{\mathbb{P}}^{1}_{K}$P$\operatorname{Spec}{\mathbb{Z}}_{p}$$\operatorname{Spec}{\mathbb{Q}}_{p}$$\operatorname{Spec}R$$\operatorname{Spec}K$$f_{R}\times
g_{R}$$f_{{\mathbb{Z}}_{p}}\times
g_{{\mathbb{Z}}_{p}}$$f_{{\mathbb{Q}}_{p}}\times
g_{{\mathbb{Q}}_{p}}$$f_{K}\times g_{K}$

Then, we have the following commutative diagram.

(9.2)
${({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})(K)}$${({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})({\mathbb{Q}}_{p})}$${({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})({\mathbb{Z}}_{p})}$${({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})(K)}$${({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})({\mathbb{Q}}_{p})}$${({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})({\mathbb{Z}}_{p})}$$\scriptstyle{f\times
g}$$\scriptstyle{\sim}$$\scriptstyle{f_{{\mathbb{Q}}_{p}}\times
g_{{\mathbb{Q}}_{p}}}$$\scriptstyle{f_{{\mathbb{Z}}_{p}}\times
g_{{\mathbb{Z}}_{p}}}$$\scriptstyle{\sim}$

We also have

(9.3) $\displaystyle
Y_{{\mathbb{Q}}_{p}}({\mathbb{Q}}_{p})\cap({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})(K)=Y(K)$
(9.4) $\displaystyle
Y_{{\mathbb{Q}}_{p}}({\mathbb{Q}}_{p})=Y_{{\mathbb{Z}}_{p}}({\mathbb{Z}}_{p})\
\text{via the above identification.}$

Therefore, PIQ for $f\times g$ and $Y$ follows from that of
$f_{{\mathbb{Z}}_{p}}\times
g_{{\mathbb{Z}}_{p}}\colon({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})({\mathbb{Z}}_{p})\longrightarrow({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})({\mathbb{Z}}_{p})$
and $Y_{{\mathbb{Z}}_{p}}({\mathbb{Z}}_{p})$. By the choice of $p$, $p$ does
not divide the ramification indices of $f_{{\mathbb{Q}}_{p}}$ and
$g_{{\mathbb{Q}}_{p}}$ considered as self-morphisms on
${\mathbb{P}}^{1}_{\overline{{\mathbb{Q}}}_{p}}$. Thus, the theorem follows
from the following Proposition 9.2. ∎

######  Proposition 9.2.

Let
$f,g\colon{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}\longrightarrow{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}$
be finite surjective morphism over ${\mathbb{Z}}_{p}$. Suppose $p$ is coprime
with every ramification index of $f$ and $g$ considered as self-morphisms on
${\mathbb{P}}^{1}_{\overline{{\mathbb{Q}}}_{p}}$. Let
$Y\subset{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}\times_{{\mathbb{Z}}_{p}}{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}$
be an $f\times g$-invariant closed subscheme. Then there is $s_{0}\geq 0$ such
that if $x\in({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})({\mathbb{Z}}_{p})$
satisfies $f^{s}(x)\in Y({\mathbb{Z}}_{p})$ for some $s\geq 0$, then we have
$f^{s_{0}}(x)\in Y({\mathbb{Z}}_{p})$.

###### Proof.

Since we are interested in ${\mathbb{Z}}_{p}$-points, we may assume $Y$ is
reduced. By replacing $f,g$ with some iterates, we may assume $Y$ is
irreducible. If either
$Y={\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}\times_{{\mathbb{Z}}_{p}}{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}$
or $Y({\mathbb{Z}}_{p})=\emptyset$, there is nothing to prove. Thus, we may
assume $\dim Y=1$ or $2$. When $\dim Y=1$, by replacing $f,g$ with some
iterates again, we may assume $Y$ is a single ${\mathbb{Z}}_{p}$-point
$\operatorname{Spec}{\mathbb{Z}}_{p}\hookrightarrow{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}\times_{{\mathbb{Z}}_{p}}{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}$.
When $\dim Y=2$, $Y$ is locally defined by a principal ideal.

In this proof, we use the following terminology. For a triple $(X,f,Y)$
consisting of a set $X$, self-map $f\colon X\longrightarrow X$, and subset
$Y\subset X$ with $f(Y)\subset Y$, we say PIQ is true for $(X,f,Y)$ if the
following holds: there is $s_{0}$ such that if $x\in X$ satisfies $f^{s}(x)\in
Y$ for some $s\geq 0$, then $f^{s_{0}}(x)\in Y$.

Step 1. For a morphism $f$ over ${\mathbb{Z}}_{p}$ or a
${\mathbb{Z}}_{p}$-point $x$, their reduction mod $p$ are denoted by
$\overline{f}$ and $\overline{x}$.

Since
${\mathbb{P}}^{1}({\mathbb{F}}_{p})\times{\mathbb{P}}^{1}({\mathbb{F}}_{p})$
is a finite set, by replacing $f,g$ with iterates, we may assume
$\overline{(f\times g)}=\overline{(f\times g)}^{2}$ as self-maps on
${\mathbb{P}}^{1}({\mathbb{F}}_{p})\times{\mathbb{P}}^{1}({\mathbb{F}}_{p})$.
Let

(9.5)
$\displaystyle\operatorname{Fix}(\overline{f},{\mathbb{P}}^{1}({\mathbb{F}}_{p}))$
$\displaystyle=\\{\xi_{1},\dots,\xi_{r_{1}}\\},$ (9.6)
$\displaystyle\operatorname{Fix}(\overline{g},{\mathbb{P}}^{1}({\mathbb{F}}_{p}))$
$\displaystyle=\\{\eta_{1},\dots,\eta_{r_{2}}\\}$

and set

(9.7) $\displaystyle U_{\xi_{i}}$
$\displaystyle:=\\{\alpha\in{\mathbb{P}}^{1}({\mathbb{Z}}_{p})\mid\overline{\alpha}=\xi_{i}\\},$
(9.8) $\displaystyle V_{\eta_{j}}$
$\displaystyle:=\\{\alpha\in{\mathbb{P}}^{1}({\mathbb{Z}}_{p})\mid\overline{\alpha}=\eta_{j}\\}.$

Then we have $f(U_{\xi_{i}})\subset U_{\xi_{i}}$ and $g(V_{\eta_{j}})\subset
V_{\eta_{j}}$. Since $\overline{(f\times g)}=\overline{(f\times g)}^{2}$ on
${\mathbb{P}}^{1}({\mathbb{F}}_{p})\times{\mathbb{P}}^{1}({\mathbb{F}}_{p})$,
for any point $x\in({\mathbb{P}}^{1}\times{\mathbb{P}}^{1})({\mathbb{Z}}_{p})$
we have $(f\times g)(x)\in U_{\xi_{i}}\times V_{\eta_{j}}$ for some $i$ and
$j$. Therefore, it is enough to show that PIQ is true for

(9.9) $\displaystyle(f\times g)|_{U_{\xi_{i}}\times V_{\eta_{j}}}\colon
U_{\xi_{i}}\times V_{\eta_{j}}\longrightarrow U_{\xi_{i}}\times V_{\eta_{j}}$

and $Y({\mathbb{Z}}_{p})\cap(U_{\xi_{i}}\times V_{\eta_{j}})$ for every $i$
and $j$.

Let us fix $i$ and $j$ and simply write $\xi_{i}$ and $\eta_{j}$ by $\xi$ and
$\eta$, respectively. Consider $\xi$ and $\eta$ as closed points of
${\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}$ and fix the following isomorphisms of
${\mathbb{Z}}_{p}$-algebras:

(9.10)
$\displaystyle\widehat{{\mathcal{O}}}_{{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}},\xi}\simeq{\mathbb{Z}}_{p}\llbracket
z\rrbracket,\
\widehat{{\mathcal{O}}}_{{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}},\eta}\simeq{\mathbb{Z}}_{p}\llbracket
w\rrbracket.$

By these isomorphisms, we get bijections

(9.11) $\displaystyle U_{\xi}\simeq p{\mathbb{Z}}_{p},\ V_{\eta}\simeq
p{\mathbb{Z}}_{p}$

where $a\in p{\mathbb{Z}}_{p}$ corresponds to ${\mathbb{Z}}_{p}$-algebra
homomorphism ${\mathbb{Z}}_{p}\llbracket
z\rrbracket\longrightarrow{\mathbb{Z}}_{p};\ z\mapsto a$ for example. Under
these bijections, we get

(9.16)

where $F=({\mathbb{Z}}_{p}\llbracket
z\rrbracket\xlongrightarrow[]{f^{*}}{\mathbb{Z}}_{p}\llbracket
z\rrbracket)(z)$ and $G=({\mathbb{Z}}_{p}\llbracket
w\rrbracket\xlongrightarrow[]{g^{*}}{\mathbb{Z}}_{p}\llbracket
w\rrbracket)(w)$. The image $Z$ of $Y({\mathbb{Z}}_{p})\cap(U_{\xi}\times
V_{\eta})$ to $(p{\mathbb{Z}}_{p})^{2}$ is either empty set, singleton, or

(9.17) $\displaystyle\\{(a,b)\in(p{\mathbb{Z}}_{p})^{2}\mid\Phi(a,b)=0\\}$

where $\Phi\in{\mathbb{Z}}_{p}\llbracket z,w\rrbracket$ is a local defining
equation of $Y$ at
$(\xi,\eta)\in{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}\times_{{\mathbb{Z}}_{p}}{\mathbb{P}}^{1}_{{\mathbb{Z}}_{p}}$.
When $Z=\emptyset$, there is nothing to prove. When $Z$ is a singleton, it is
a fixed point of $F\times G$. When $Z$ is the zero set of $\Phi$, we have

(9.18) $\displaystyle\Phi(z,w)\mid\Phi(F(z),G(w))\ \text{in
${\mathbb{Z}}_{p}\llbracket z,w\rrbracket$}.$

Note also that we may assume the constant term of $\Phi$ is zero modulo $p$
since otherwise the zero set of $\Phi$ in $(p{\mathbb{Z}}_{p})^{2}$ is empty.

In summary, we reduced the problem to show PIQ for the map

(9.19) $\displaystyle F\times
G\colon(p{\mathbb{Z}}_{p})^{2}\longrightarrow(p{\mathbb{Z}}_{p})^{2}$

and the set $Z\subset(p{\mathbb{Z}}_{p})^{2}$ where

  * •

$F\in{\mathbb{Z}}_{p}\llbracket z\rrbracket$, $G\in{\mathbb{Z}}_{p}\llbracket
w\rrbracket$;

  * •

$F,G\neq 0(\mathrm{mod}\ p)$;

  * •

constant terms of $F,G$ are zero modulo $p$;

  * •

$Z$ is a fixed point of $F\times G$, or the zero set of
$\Phi\in{\mathbb{Z}}_{p}\llbracket z,w\rrbracket\smallsetminus\\{0\\}$. In the
latter case, the constant term of $\Phi$ is zero modulo $p$ and
$\Phi(z,w)\mid\Phi(F(z),G(w))$ in ${\mathbb{Z}}_{p}\llbracket z,w\rrbracket$.

Step 2. Let $F,G$, and $Z$ as above. Then $F,G$ define morphisms of strictly
${\mathbb{Q}}_{p}$-affinoid algebras

(9.20)
$\displaystyle{\mathbb{Q}}_{p}\\{pz\\}\longrightarrow{\mathbb{Q}}_{p}\\{pz\\};z\mapsto
F(z)$ (9.21)
$\displaystyle{\mathbb{Q}}_{p}\\{pw\\}\longrightarrow{\mathbb{Q}}_{p}\\{pw\\};w\mapsto
G(w)$

because $\|F\|_{{\mathbb{Q}}_{p}\\{pz\\}}\leq p^{-1}$ and
$\|G\|_{{\mathbb{Q}}_{p}\\{pw\\}}\leq p^{-1}$. Their completed tensor product
$\varphi\colon{\mathbb{Q}}_{p}\\{pz,pw\\}\longrightarrow{\mathbb{Q}}_{p}\\{pz,pw\\}$
induces continuous map
$\varphi^{*}\colon{\mathcal{M}}({\mathbb{Q}}_{p}\\{pz,pw\\})\longrightarrow{\mathcal{M}}({\mathbb{Q}}_{p}\\{pz,pw\\})$
on the Berkovich spectrum. The map eq. 9.19 is embedded in this map
$\varphi^{*}$ as maps between type I points:

(9.27)

We also remark that when $Z$ is the zero locus of $\Phi$, we have
$\Phi\mid\varphi(\Phi)$ in ${\mathbb{Q}}_{p}\\{pz,pw\\}$.

Let us write

(9.28) $\displaystyle F(z)=a_{0}+a_{1}z+a_{2}z^{2}+\cdots$ (9.29)
$\displaystyle G(w)=b_{0}+b_{1}w+b_{2}w^{2}+\cdots$

where $a_{n},b_{m}\in{\mathbb{Z}}_{p}$. We assumed $|a_{0}|,|b_{0}|<1$.
(Recall that $|\ |$ is the $p$-adic absolute value normalized as $|p|=p^{-1}$.
) We treat the following cases separately.

Case 1 |  $|a_{1}|=|b_{1}|=1$.
---|---
Case 2 |  $|a_{1}|=1$ and $|b_{1}|<1$.
Case 3 |  $|a_{1}|<1$ and $|b_{1}|<1$.

Case 1. Suppose $|a_{1}|=|b_{1}|=1$. In this case, $F(z)-a_{0}$ and
$G(w)-b_{0}$ have compositional inverse in $z{\mathbb{Z}}_{p}\llbracket
z\rrbracket$ and $w{\mathbb{Z}}_{p}\llbracket w\rrbracket$ respectively. This
implies the morphism

(9.30)
$\displaystyle\varphi\colon{\mathbb{Q}}_{p}\\{pz,pw\\}\longrightarrow{\mathbb{Q}}_{p}\\{pz,pw\\}\
;z\mapsto F(z),w\mapsto G(w)$

is an isometric isomorphism. In particular, $F\times
G\colon(p{\mathbb{Z}}_{p})^{2}\longrightarrow(p{\mathbb{Z}}_{p})^{2}$ is a
bijection. This proves PIQ for the case where $Z$ is a fixed point of $F\times
G$. When $Z=(\Phi=0)$, recall that we have $\Phi\mid\varphi(\Phi)$ in
${\mathbb{Q}}_{p}\\{pz,pw\\}$. Since $\varphi$ is a ring automorphism, the
number of prime factors of $\varphi(\Phi)$ counted with multiplicity is equal
to that of $\Phi$. Hence we get $\varphi(\Phi)=\gamma\Phi$ for some
$\gamma\in{\mathbb{Q}}_{p}\\{pz,pw\\}^{\times}$. Thus for
$(a,b)\in(p{\mathbb{Z}}_{p})^{2}$, we have

(9.31) $\displaystyle(F\times G)(a,b)\in
Z\iff\Phi(F(a),G(b))=0\iff\varphi(\Phi)(a,b)=0$ (9.32)
$\displaystyle\iff\gamma(a,b)\Phi(a,b)=0\iff\Phi(a,b)=0\iff(a,b)\in Z.$

This proves PIQ.

Case 2. Suppose $|a_{1}|=1$ and $|b_{1}|<1$. In this case, $F(z)-a_{0}$ has
compositional inverse in $z{\mathbb{Z}}_{p}\llbracket z\rrbracket$ and hence

(9.33)
$\displaystyle{\mathbb{Q}}_{p}\\{pz\\}\longrightarrow{\mathbb{Q}}_{p}\\{pz\\}\
;z\mapsto F(z)$

is an isometric isomorphism. We claim that $G\colon
p{\mathbb{Z}}_{p}\longrightarrow p{\mathbb{Z}}_{p}$ has a unique fixed point.
Indeed, consider
$G(w)-w-b_{0}=(b_{1}-1)w+b_{2}w^{2}+\cdots\in{\mathbb{Z}}_{p}\llbracket
w\rrbracket$. Since $|b_{1}-1|=1$, this has a compositional inverse in
$w{\mathbb{Z}}_{p}\llbracket w\rrbracket$. Therefore,
$p{\mathbb{Z}}_{p}\longrightarrow p{\mathbb{Z}}_{p}\ ;w\mapsto G(w)-w-b_{0}$
is a bijection and thus $G$ has a unique fixed point in $p{\mathbb{Z}}_{p}$.

Let $c\in p{\mathbb{Z}}_{p}$ be the fixed point. Let
$\widetilde{G}(w)=G(w+c)-c\in{\mathbb{Z}}_{p}\llbracket w\rrbracket$. Then we
have the commutative diagram

(9.34)
${w}$${G(w)}$${w}$${{\mathbb{Q}}_{p}\\{pw\\}}$${{\mathbb{Q}}_{p}\\{pw\\}}$${w}$${w+c}$${{\mathbb{Q}}_{p}\\{pw\\}}$${{\mathbb{Q}}_{p}\\{pw\\}}$${w+c}$${w}$${\widetilde{G}(w)}$

where the vertical arrows are isometric isomorphisms. We also have

(9.35) $\displaystyle\widetilde{G}(0)=0$ (9.36)
$\displaystyle\widetilde{G}^{\prime}(0)=G^{\prime}(c)=b_{1}+2b_{2}c+\cdots,\
\text{and hence}\ |\widetilde{G}^{\prime}(0)|\leq p^{-1}.$

Note that for arbitrary $b\in{\mathbb{C}}_{p}$ with $|b|\leq p^{-1}$ (we fix a
${\mathbb{C}}_{p}$, completion of the algebraic closure of ${\mathbb{Q}}_{p}$,
and will work inside it), we have

(9.37)
$\displaystyle|\widetilde{G}(b)|\leq\max\\{|\widetilde{G}^{\prime}(0)|,|b|\\}|b|\leq
p^{-1}|b|$

and hence

(9.38) $\displaystyle|\widetilde{G}^{\circ n}(b)|\leq p^{-n-1}$

for all $n\in{\mathbb{Z}}_{\geq 0}$. To prove PIQ for $F\times G$, it is
enough to prove it for $F\times\widetilde{G}$ acting on
$(p{\mathbb{Z}}_{p})^{2}$, and so we replace $G$ with $\widetilde{G}$.

Case 2-1. First we treat the case where $G^{\prime}(0)=:\lambda\neq 0$. Let us
write $|\lambda|=p^{-r}$ for some $r\in{\mathbb{Z}}_{\geq 1}$.

By Theorems 8.3 and 8.2, we have the following commutative diagram

(9.39)
${w}$${G(w)}$${{\mathbb{Q}}_{p}\\{pw\\}}$${{\mathbb{Q}}_{p}\\{pw\\}}$${{\mathbb{Q}}_{p}\\{p^{n}w\\}}$${{\mathbb{Q}}_{p}\\{p^{n}w\\}}$${{\mathbb{Q}}_{p}\\{p^{N}w\\}}$${{\mathbb{Q}}_{p}\\{p^{N}w\\}}$${w}$${\lambda
w}$$\scriptstyle{\iota}$$\scriptstyle{\iota}$

where $n,N\in{\mathbb{Z}}_{\geq 1}$, the top vertical arrows are inclusions,
and $\iota$ is an isometric isomorphism such that $\iota(w)(0)=0$. Taking
completed tensor product with ${\mathbb{Q}}_{p}\\{pz\\}$, we get

(9.40)
${{\mathbb{Q}}_{p}\\{pz,pw\\}}$${{\mathbb{Q}}_{p}\\{pz,pw\\}}$${{\mathbb{Q}}_{p}\\{pz,p^{n}w\\}}$${{\mathbb{Q}}_{p}\\{pz,p^{n}w\\}}$${{\mathbb{Q}}_{p}\\{pz,p^{N}w\\}}$${{\mathbb{Q}}_{p}\\{pz,p^{N}w\\}}$$\scriptstyle{\chi}$$\scriptstyle{\begin{subarray}{c}z\mapsto
F(z)\\\ w\mapsto
G(w)\end{subarray}}$$\scriptstyle{\sim}$$\operatorname{id}\widehat{\otimes}\iota$$\scriptstyle{\sim}$$\operatorname{id}\widehat{\otimes}\iota$$\scriptstyle{\alpha}$$\scriptstyle{\begin{subarray}{c}z\mapsto
F(z)\\\ w\mapsto\lambda w\end{subarray}}$

where $\chi$ is the composite of the vertical arrows and $\alpha$ is the
${\mathbb{Q}}_{p}$-algebra morphism defined by sending $z$ to $F(z)$ and $w$
to $\lambda w$. The map $\chi$ induces

(9.41) $\displaystyle(p{\mathbb{Z}}_{p})^{2}\supset p{\mathbb{Z}}_{p}\times
p^{n}{\mathbb{Z}}_{p}\xleftarrow[(\operatorname{id}\widehat{\otimes}\iota)^{*}]{\sim}p{\mathbb{Z}}_{p}\times
p^{N}{\mathbb{Z}}_{p}.$

Let

(9.42) $\displaystyle
Z^{\prime}=\left((\operatorname{id}\widehat{\otimes}\iota)^{*}\right)^{-1}(Z\cap(p{\mathbb{Z}}_{p}\times
p^{n}{\mathbb{Z}}_{p}))\subset p{\mathbb{Z}}_{p}\times p^{N}{\mathbb{Z}}_{p}.$

By eq. 9.38, we have

(9.43) $\displaystyle(F\times G)^{\circ(n-1)}((p{\mathbb{Z}}_{p})^{2})\subset
p{\mathbb{Z}}_{p}\times p^{n}{\mathbb{Z}}_{p}.$

Therefore, it is enough to prove PIQ for the map $F(z)\times\lambda w$ acting
on $p{\mathbb{Z}}_{p}\times p^{N}{\mathbb{Z}}_{p}$ and $Z^{\prime}$. When
$Z^{\prime}$ consists of a single fixed point, PIQ is clear because $F\colon
p{\mathbb{Z}}_{p}\longrightarrow p{\mathbb{Z}}_{p}$ is a bijection and the
only fixed point of $\lambda w$ is $0$. Consider the case $Z=(\Phi=0)$. In
this case, set $\Psi=\chi(\Phi)$. By Corollary 7.3, there is arbitrary large
$N^{\prime}\geq N$ such that for all $m\geq N^{\prime}$, we have

(9.44) number of prime factors of $\Psi$ in ${\mathbb{Q}}_{p}\\{pz,p^{m}w\\}$
counted with multiplicity (9.45) $\displaystyle\leq\text{that of $\Psi$ in
${\mathbb{Q}}_{p}\\{pz,p^{N^{\prime}}w\\}$}.$

We replace $N$ with such large $N^{\prime}$ (cf. Lemma 8.2). Then, consider
the decomposition

(9.46)
${{\mathbb{Q}}_{p}\\{pz,p^{N}w\\}}$${{\mathbb{Q}}_{p}\\{pz,p^{N}w\\}}$${{\mathbb{Q}}_{p}\\{pz,p^{N+r}w\\}}$$\scriptstyle{\alpha}$$\scriptstyle{\begin{subarray}{c}z\mapsto
F(z)\\\ w\mapsto\lambda w\end{subarray}}$$\scriptstyle{\sim}$

where the first map is inclusion, and the second one is an isometric
isomorphism. By this diagram, the number of prime factors of $\alpha(\Psi)$ is
at most that of $\Psi$ by the choice of $N$. Then we have

(9.47) $\displaystyle\alpha(\Psi)=\gamma\Psi$

for some $\gamma\in{\mathbb{Q}}_{p}\\{pz,p^{N}w\\}^{\times}$ because
$\Psi\mid\alpha(\Psi)$. Therefore, for $(a,b)\in p{\mathbb{Z}}_{p}\times
p^{N}{\mathbb{Z}}_{p}$, we have

(9.48) $\displaystyle(F(a),\lambda b)\in Z^{\prime}\iff\Psi((F(a),\lambda
b))=0\iff\alpha(\Psi)(a,b)=0$ (9.49)
$\displaystyle\iff\gamma(a,b)\Psi(a,b)=0\iff\Psi(a,b)=0\iff(a,b)\in
Z^{\prime}$

and PIQ follows from this.

Case 2-2. Next suppose $G^{\prime}(0)=0$. Let

(9.50) $\displaystyle G(w)=b_{d}w^{d}+\text{higher order terms}$

with $b_{d}\neq 0$. Set $K={\mathbb{Q}}_{p}(b_{d}^{1/(d-1)})$ where
$b_{d}^{1/(d-1)}$ is a fixed $(d-1)$-th root of $b_{d}$. By Theorems 8.3 and
8.2, we have the following commutative diagram

(9.51)
${w}$${G(w)}$${K\\{pw\\}}$${K\\{pw\\}}$${K\\{p^{n}w\\}}$${K\\{p^{n}w\\}}$${K\\{p^{N}w\\}}$${K\\{p^{N}w\\}}$${w}$${w^{d}}$$\scriptstyle{\iota}$$\scriptstyle{\iota}$

where $n,N\in{\mathbb{Z}}_{\geq 1}$, the top vertical arrows are inclusions,
and $\iota$ is an isometric isomorphism such that $\iota(w)(0)=0$. Taking
completed tensor product with $K\\{pz\\}$, we get

(9.52)
${K\\{pz,pw\\}}$${K\\{pz,pw\\}}$${K\\{pz,p^{n}w\\}}$${K\\{pz,p^{n}w\\}}$${K\\{pz,p^{N}w\\}}$${K\\{pz,p^{N}w\\}}$$\scriptstyle{\chi}$$\scriptstyle{\begin{subarray}{c}z\mapsto
F(z)\\\ w\mapsto
G(w)\end{subarray}}$$\scriptstyle{\sim}$$\operatorname{id}\widehat{\otimes}\iota$$\scriptstyle{\sim}$$\operatorname{id}\widehat{\otimes}\iota$$\scriptstyle{\alpha}$$\scriptstyle{\begin{subarray}{c}z\mapsto
F(z)\\\ w\mapsto w^{d}\end{subarray}}$

where $\chi$ is the composite of the vertical arrows and $\alpha$ is the
$K$-algebra morphism defined by sending $z$ to $F(z)$ and $w$ to $w^{d}$. Set
$\Psi=\chi(\Phi)$. As before, we have

(9.53) $\displaystyle(F\times
G)^{\circ(n-1)}(\overline{\mathbf{D}}_{K}(p^{-1})^{2})\subset\overline{\mathbf{D}}_{K}(p^{-1})\times\overline{\mathbf{D}}_{K}(p^{-n}).$

Thus, it is enough to prove PIQ for $F(z)\times w^{d}$ acting on
$\overline{\mathbf{D}}_{K}(p^{-1})\times\overline{\mathbf{D}}_{K}(p^{-N})$ and
the invariant subset $Z^{\prime}$ where

(9.54) $\displaystyle Z^{\prime}=\begin{cases}\text{a fixed point, or}\\\
\\{(a,b)\in\overline{\mathbf{D}}_{K}(p^{-1})\times\overline{\mathbf{D}}_{K}(p^{-N})\mid\Psi(a,b)=0\\}.\end{cases}$

When $Z^{\prime}$ is a fixed point, PIQ holds because
$F\colon\overline{\mathbf{D}}_{K}(p^{-1})\longrightarrow\overline{\mathbf{D}}_{K}(p^{-1})$
is bijective and $w^{d}$ has only one fixed point, $0$, in
$\overline{\mathbf{D}}_{K}(p^{-N})$. Suppose
$Z^{\prime}=\\{(a,b)\in\overline{\mathbf{D}}_{K}(p^{-1})\times\overline{\mathbf{D}}_{K}(p^{-N})\mid\Psi(a,b)=0\\}$.

We first note that we may assume $\Psi$ is prime. Indeed, we claim that
$\alpha$ acts on the set of minimal prime ideals containing $\Psi$ by pull-
back. Let ${\mathfrak{p}}\subset K\\{pz,p^{N}w\\}$ be a minimal prime ideal
containing $\Psi$. Then we have $\Psi\in\alpha^{-1}({\mathfrak{p}})$ and thus
it is enough to show that $\alpha^{-1}({\mathfrak{p}})$ is not a maximal
ideal. Consider the following decomposition.

(9.55)
${K\\{pz,p^{N}w\\}}$${K\\{pz,p^{N}w\\}}$${K\\{pz,p^{Nd}w\\}}$$\scriptstyle{i}$$\scriptstyle{\alpha}$$\scriptstyle{\begin{subarray}{c}z\mapsto
F(z)\\\ w\mapsto w^{d}\end{subarray}}$$\scriptstyle{j}$

The second map $j$ is finite injective and thus $j^{-1}({\mathfrak{p}})$ is a
height one prime ideal. Then
$i^{-1}(j^{-1}{\mathfrak{p}})=\alpha^{-1}({\mathfrak{p}})$ is not a maximal
ideal because $K\\{pz,p^{Nd}w\\}$ is a Jacobson ring and Nullstellensatz
holds. Finally, we replace $\alpha$ with some iterates so that every minimal
prime of $\Psi$ is invariant or mapped to an invariant one. Hence, it is
enough to prove PIQ for invariant minimal primes, and this means we may assume
$\Psi$ is a prime.

Now we consider the case $w\mid\Psi$. In this case we can write $\Psi=\gamma
w$ for some $\gamma\in K\\{pz,p^{N}w\\}^{\times}$. Thus for
$(a,b)\in\overline{\mathbf{D}}_{K}(p^{-1})\times\overline{\mathbf{D}}_{K}(p^{-N})$,

(9.56) $\displaystyle(F(a),b^{d})\in
Z^{\prime}\iff\Psi(F(a),b^{d})=0\iff\gamma(F(a),b^{d})b^{d}=0$ (9.57)
$\displaystyle\iff b=0\iff(a,b)\in Z^{\prime}$

and we are done.

Suppose $w\not\mid\Psi$. Let us write

(9.58) $\displaystyle\Psi(z,w)$ $\displaystyle=\Psi_{0}(z)+w\Psi_{1}(z,w)$
(9.59) $\displaystyle\qquad\text{for some}\ \Psi_{0}\in
K\\{pz\\}\smallsetminus\\{0\\}\ \text{and}\ \Psi_{1}\in K\\{pz,p^{N}w\\},$
(9.60) $\displaystyle\alpha(\Psi)$ $\displaystyle=\gamma\Psi\ \text{for some}\
\gamma\in K\\{pz,p^{N}w\\}.$

Let us also write $\gamma=\gamma_{0}+w\gamma_{1}$ with $\gamma_{0}\in
K\\{pz\\}$ and $\gamma_{1}\in K\\{pz,p^{N}w\\}$. Then we have

(9.61) $\displaystyle\alpha(\Psi)(z,w)$
$\displaystyle=\Psi_{0}(F(z))+w^{d}\Psi(F(z),w^{d})$ (9.62)
$\displaystyle=\left(\gamma_{0}(z)+w\gamma_{1}(z,w)\right)\left(\Psi_{0}(z)+w\Psi_{1}(z,w)\right).$

Thus we have

(9.63) $\displaystyle\Psi_{0}(F(z))=\gamma_{0}(z)\Psi_{0}(z).$

Since $K\\{pz\\}\longrightarrow K\\{pz\\};z\mapsto F(z)$ is an isometric
isomorphism, the number of prime factors of $\Psi_{0}(F(z))$ is equal to that
of $\Psi_{0}(z)$. Thus $\gamma_{0}\in K\\{pz\\}^{\times}$. If we take
$N^{\prime}\geq N$ so that
$\|\gamma_{0}\|_{K\\{pz\\}}>\|w\gamma_{1}\|_{K\\{pz,p^{N^{\prime}}w\\}}$, then
$\gamma\in K\\{pz,p^{N^{\prime}}w\\}^{\times}$. Therefore, for
$(a,b)\in\overline{\mathbf{D}}_{K}(p^{-1})\times\overline{\mathbf{D}}_{K}(p^{-N})$,

(9.64) $\displaystyle(F^{\circ n}(a),b^{d^{n}})\in Z^{\prime}\iff\Psi(F^{\circ
n}(a),b^{d^{n}})=0$ (9.65)
$\displaystyle\iff\gamma(F^{\circ(n-1)}(a),b^{d^{n-1}})\Psi(F^{\circ(n-1)}(a),b^{d^{n-1}})=0$
(9.66) $\displaystyle\iff\Psi(F^{\circ(n-1)}(a),b^{d^{n-1}})=0$ (9.67)
$\displaystyle\iff(F^{\circ(n-1)}(a),b^{d^{n-1}})\in Z^{\prime}$

if $|b^{d^{n-1}}|\leq p^{-N^{\prime}}$, which is true when $p^{-Nd^{n-1}}\leq
p^{-N^{\prime}}$. This proves PIQ for this case.

Case 3. Finally, we treat the case where $|a_{1}|,|b_{1}|<1$. As at the
beginning of Case 2, we may assume $F(0)=G(0)=0$ by taking conjugate by
translation. Moreover, we have

(9.68) $\displaystyle|F^{\prime}(0)|\leq p^{-1},\ |G^{\prime}(0)|\leq p^{-1}$
(9.69) $\displaystyle(F\times G)^{\circ
n}\left(\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(p^{-1})^{2}\right)\subset\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(p^{-n-1})^{2}.$

Case 3-1. Let us first treat the case $F^{\prime}(0)=:\lambda\neq 0$. By
Theorems 8.3 and 8.2, there is a finite extension $K$ of ${\mathbb{Q}}_{p}$
and the following commutative diagram.

(9.70)
${K\\{pz,pw\\}}$${K\\{pz,pw\\}}$${K\\{p^{n}z,p^{n}w\\}}$${K\\{p^{n}z,p^{n}w\\}}$${K\\{p^{N}z,p^{N}w\\}}$${K\\{p^{N}z,p^{N}w\\}}$$\scriptstyle{\chi}$$\scriptstyle{\begin{subarray}{c}z\mapsto
F(z)\\\ w\mapsto
G(w)\end{subarray}}$$\scriptstyle{\sim}$$\iota$$\scriptstyle{\sim}$$\iota$$\scriptstyle{\alpha}$$\scriptstyle{\begin{subarray}{c}z\mapsto\lambda
z\\\ w\mapsto g(w)\end{subarray}}$

Here, $n,N\in{\mathbb{Z}}_{\geq 1}$, the top vertical arrows are inclusion,
$\iota$ is an isometric isomorphism, and $\alpha$ is the $K$-algebra morphism
defined by sending $z$ to $\lambda z$ and $w$ to $g(w)$, where $g(w)=\mu w$
for some $\mu\in p{\mathbb{Z}}_{p}\smallsetminus\\{0\\}$ or $g(w)=w^{d}$ for
some $d\in{\mathbb{Z}}_{\geq 2}$. Let $\Psi=\chi(\Phi)$. By eq. 9.69, it is
enough to prove PIQ for the map $\lambda z\times g$ acting on
$\overline{\mathbf{D}}_{K}(p^{-N})^{2}$ and the invariant subset

(9.71) $\displaystyle Z^{\prime}=\begin{cases}\text{a fixed point, or}\\\
\\{(a,b)\in\overline{\mathbf{D}}_{K}(p^{-N})^{2}\mid\Psi(a,b)=0\\}.\end{cases}$

When $Z^{\prime}$ is a fixed point, PIQ is clear because the only fixed point
is $(0,0)$ and its only preimage is itself. Suppose $Z^{\prime}=(\Psi=0)$. As
in Case 2-2, by replacing $\alpha$ with iterates, we may assume $\Psi$ is
prime.

Consider the case $w\mid\Psi$. In this case we can write $\Psi=\gamma w$ for
some $\gamma\in K\\{p^{N}z,p^{N}w\\}^{\times}$. Thus for
$(a,b)\in\overline{\mathbf{D}}_{K}(p^{-N})^{2}$,

(9.72) $\displaystyle(\lambda a,g(b))\in Z^{\prime}\iff\Psi(\lambda
a,g(b))=0\iff\gamma(\lambda a,g(b))g(b)=0$ (9.73) $\displaystyle\iff
b=0\iff(a,b)\in Z^{\prime}$

and we are done.

Suppose $w\not\mid\Psi$. Let us write

(9.74) $\displaystyle\Psi(z,w)=\Psi_{0}(z)+w\Psi_{1}(z,w)$ (9.75)
$\displaystyle\qquad\text{for some}\ \Psi_{0}\in
K\\{p^{N}z\\}\smallsetminus\\{0\\}\ \text{and}\ \Psi_{1}\in
K\\{p^{N}z,p^{N}w\\},$ (9.76) $\displaystyle\alpha(\Psi)=\gamma\Psi\ \text{for
some}\ \gamma\in K\\{p^{N}z,p^{N}w\\}.$

Let us also write $\gamma=\gamma_{0}+w\gamma_{1}$ for some $\gamma_{0}\in
K\\{p^{N}z\\}$ and $\gamma_{1}\in K\\{p^{N}z,p^{N}w\\}$. Then we have

(9.77) $\displaystyle\alpha(\Psi)(z,w)$ $\displaystyle=\Psi_{0}(\lambda
z)+g(w)\Psi_{1}(\lambda z,g(w))$ (9.78)
$\displaystyle=\left(\gamma_{0}(z)+w\gamma_{1}(z,w)\right)\left(\Psi_{0}(z)+w\Psi_{1}(z,w)\right)$

and thus

(9.79) $\displaystyle\Psi_{0}(\lambda z)=\gamma_{0}(z)\Psi_{0}(z).$

Since $\operatorname{ord}_{K\\{p^{N}z\\}}\Psi_{0}(\lambda
z)\leq\operatorname{ord}_{K\\{p^{N}z\\}}\Psi_{0}(z)$ by $|\lambda|<1$, we get
$\operatorname{ord}_{K\\{p^{N}z\\}}\gamma_{0}(z)=0$, i.e. $\gamma_{0}\in
K\\{p^{N}z\\}^{\times}$. Then there is $N^{\prime}\geq N$ such that $\gamma\in
K\\{p^{N}z,p^{N^{\prime}}w\\}^{\times}$. Now for
$(a,b)\in\overline{\mathbf{D}}_{K}(p^{-N})^{2}$, we have

(9.80) $\displaystyle(\lambda^{n}a,g^{\circ n}(b))\in
Z^{\prime}\iff\Psi(\lambda^{n}a,g^{\circ n}(b))=0$ (9.81)
$\displaystyle\iff\gamma(\lambda^{n-1}a,g^{\circ(n-1)}(b))\Psi(\lambda^{n-1}a,g^{\circ(n-1)}(b))=0$
(9.82) $\displaystyle\iff\Psi(\lambda^{n-1}a,g^{\circ(n-1)}(b))=0$

if $|g^{\circ(n-1)}(b)|\leq p^{-N^{\prime}}$, which is true when
$p^{-N-(n-1)}\leq p^{-N^{\prime}}$. This proves PIQ for this case.

Case 3-2. Finally we treat the case $F^{\prime}(0)=G^{\prime}(0)=0$. Let

(9.83) $\displaystyle F(z)=a_{d}z^{d}+\text{higher order terms}$ (9.84)
$\displaystyle G(w)=b_{e}w^{e}+\text{higher order terms}$

where $d,e\in{\mathbb{Z}}_{\geq 2}$ and $a_{d},b_{e}\neq 0$.

By Theorems 8.3 and 8.2, there is a finite extension $K$ of ${\mathbb{Q}}_{p}$
and the following commutative diagram.

(9.85)
${K\\{pz,pw\\}}$${K\\{pz,pw\\}}$${K\\{p^{n}z,p^{n}w\\}}$${K\\{p^{n}z,p^{n}w\\}}$${K\\{p^{N}z,p^{N}w\\}}$${K\\{p^{N}z,p^{N}w\\}}$$\scriptstyle{\chi}$$\scriptstyle{\begin{subarray}{c}z\mapsto
F(z)\\\ w\mapsto
G(w)\end{subarray}}$$\scriptstyle{\sim}$$\iota$$\scriptstyle{\sim}$$\iota$$\scriptstyle{\alpha}$$\scriptstyle{\begin{subarray}{c}z\mapsto
z^{d}\\\ w\mapsto w^{e}\end{subarray}}$

Here, $n,N\in{\mathbb{Z}}_{\geq 1}$, the top vertical arrows are inclusion,
$\iota$ is an isometric isomorphism, and $\alpha$ is the $K$-algebra morphism
defined by sending $z$ to $z^{e}$ and $w$ to $w^{e}$. Let $\Psi=\chi(\Phi)$.
By eq. 9.69, it is enough to prove PIQ for the map $z^{d}\times w^{e}$ acting
on $\overline{\mathbf{D}}_{K}(p^{-N})^{2}$ and the invariant subset

(9.86) $\displaystyle Z^{\prime}=\begin{cases}\text{a fixed point, or}\\\
\\{(a,b)\in\overline{\mathbf{D}}_{K}(p^{-N})^{2}\mid\Psi(a,b)=0\\}.\end{cases}$

When $Z^{\prime}$ is a fixed point, PIQ is clear because the only fixed point
is $(0,0)$ and its only preimage is itself. Suppose $Z^{\prime}=(\Psi=0)$.

Let us define a sequence of finite extensions of $K$ as follows. Let $K_{0}$
be a finite extension of $K$ such that the prime factorization of $\Psi\in
K_{0}\\{p^{N}z,p^{N}w\\}$ stabilizes along ground field extensions (such
$K_{0}$ exists by Corollary 7.4). Next, let $K_{1}$ be a finite extension of
$K_{0}$ such that the prime factorization of $\Psi\in
K_{1}\\{p^{Nd}z,p^{Ne}w\\}$ stabilizes. Having defined
$K_{0},K_{1},\dots,K_{m}$, the next finite extension $K_{m+1}/K_{m}$ is
defined such that the prime factorization of $\Psi\in
K_{m+1}\\{p^{Nd^{m+1}}z,p^{Ne^{m+1}}w\\}$ stabilizes. Inductively we get
$K_{0}\subset K_{1}\subset\cdots$. We set
$R_{m}=K_{m}\\{p^{Nd^{m}}z,p^{Ne^{m}}w\\}$. Then we have

(9.87) $\displaystyle R_{0}\subset R_{1}\subset R_{2}\subset\cdots.$

By Lemma 7.5, prime factorization of $\Psi$ in $R_{m}$ eventually stabilizes
in the following sense: There is $m_{0}\geq 0$ such that

(9.88) $\displaystyle\Psi=\gamma P_{1}\cdots P_{l}\ \text{in $R_{m_{0}}$}$

where $\gamma\in R_{m_{0}}$ is eventually unit with respect to eq. 9.87,
$P_{i}\in R_{m_{0}}$ is prime, and for any $m\geq m_{0}$,
$P_{i}=vP_{i}^{\prime}$ in $R_{m}$ for some prime $P_{i}^{\prime}\in R_{m}$
and eventual unit $v\in R_{m}$.

To prove PIQ, we may replace $\alpha\colon K\\{p^{N}z,p^{N}w\\}\longrightarrow
K\\{p^{N}z,p^{N}w\\}$ with

(9.89) $\displaystyle
K_{m_{0}}\\{p^{Nd^{m_{0}}}z,p^{Ne^{m_{0}}}w\\}\longrightarrow
K_{m_{0}}\\{p^{Nd^{m_{0}}}z,p^{Ne^{m_{0}}}w\\}\ ;z\mapsto z^{d},w\mapsto
w^{e}.$

Then, as in Case 2-2, it is enough to prove PIQ for an iterate $\alpha^{t}$ of
$\alpha$, and each prime factor $Q$ of $\Psi$ in $R_{m_{0}}$ that is invariant
under $\alpha^{t}$ i.e. $Q\mid\alpha^{t}(Q)$. Consider
$L=K_{m_{0}+t}(\zeta_{d^{t}},\zeta_{e^{t}})$, where $\zeta_{l}$ stands for an
$l$-th primitive root of $1$ in ${\mathbb{C}}_{p}$. (We add
$\zeta_{d^{t}},\zeta_{e^{t}}$ for the later use.) Then, form the following
commutative diagram

(9.90)
${L\\{p^{Nd^{m_{0}}}z,p^{Ne^{m_{0}}}w\\}}$${L\\{p^{Nd^{m_{0}}}z,p^{Ne^{m_{0}}}w\\}}$${L\\{p^{Nd^{m_{0}+t}}z,p^{Ne^{m_{0}+t}}w\\}}$$\scriptstyle{\alpha^{t}}$$\scriptstyle{\begin{subarray}{c}z\mapsto
z^{d^{t}}\\\ w\mapsto w^{e^{t}}\end{subarray}}$

where the first slant arrow is the inclusion. By the choice of $K_{m}$ and
$L$, the element $Q$ is still prime in
$L\\{p^{Nd^{m_{0}}}z,p^{Ne^{m_{0}}}w\\}$. Moreover, either

  * •

$Q$ is eventually unit with respect to $(L\\{p^{Nd^{m}}z,p^{Ne^{m}}w\\})_{m}$,
or

  * •

$Q=vQ^{\prime}$ in $L\\{p^{Nd^{m_{0}+t}}z,p^{Ne^{m_{0}+t}}w\\}$, where
$Q^{\prime}$ is a prime element of
$L\\{p^{Nd^{m_{0}+t}}z,p^{Ne^{m_{0}+t}}w\\}$ and $v$ is eventually unit with
respect to
$(L\\{p^{Nd^{m}}z,p^{Ne^{m}}w\\})_{m}$.

Here, we claim that the first case does not happen. Indeed, if $Q$ is
eventually a unit, then it is a unit in
$L\\{p^{Nd^{m_{0}+kt}}z,p^{Ne^{m_{0}+kt}}w\\}$ for some $k\geq 1$. We have
$Q\mid\alpha^{t}(Q)$ and this implies $Q\mid\alpha^{tk}(Q)$ in
$L\\{p^{Nd^{m_{0}}}z,p^{Ne^{m_{0}}}w\\}$. Since $\alpha^{tk}$ factors through
$L\\{p^{Nd^{m_{0}+kt}}z,p^{Ne^{m_{0}+kt}}w\\}$, $\alpha^{tk}(Q)$ is unit and
hence so is $Q$. This contradicts to $Q$ being prime.

Let us set $r=p^{-Nd^{m_{0}}}$ and $s=p^{-Ne^{m_{0}}}$. Then, it is enough to
prove PIQ for

(9.91) $\displaystyle\alpha^{t}\colon L\\{r^{-1}z,s^{-1}w\\}\longrightarrow
L\\{r^{-1}z,s^{-1}w\\}$

and $Q$.

We reset the notation: we write $K$ in place of $L$, $\alpha$ in place of
$\alpha^{t}$, $\Psi$ in place of $Q$, $\Psi^{\prime}$ in place of
$Q^{\prime}$, and $\gamma$ in place of $v$. We remark that by this
replacement, the original degrees $d,e$ are replaced with their power
$d^{t},e^{t}$.

The situation is summarized as follows.

  * •

$K$ is a finite extension of ${\mathbb{Q}}_{p}$ such that
$\zeta_{d},\zeta_{e}\in K$.

  * •

$r,s\in|K^{\times}|$, $r,s\leq p^{-1}$.

  * •

$\Psi\in K\\{r^{-1}z,s^{-1}w\\}$ is a prime element that is not eventually
unit with respect to $(K\\{r^{-d^{m}}z,s^{-e^{m}}w\\})_{m}$.

  * •

$\Psi=\gamma\Psi^{\prime}$ in $K\\{r^{-d}z,s^{-e}w\\}$, where $\Psi^{\prime}$
is a prime in $K\\{r^{-d}z,s^{-e}w\\}$ and $\gamma$ is eventually unit with
respect to $(K\\{r^{-d^{m}}z,s^{-e^{m}}w\\})_{m}$.

We have the following commutative diagram.

(9.92)
${\Psi}$${K\\{r^{-1}z,s^{-1}w\\}}$${K\\{r^{-1}z,s^{-1}w\\}}$${\gamma\Psi^{\prime}}$${K\\{r^{-d}z,s^{-e}w\\}}$${\in}$$\scriptstyle{\alpha}$${\in}$$\scriptstyle{\begin{subarray}{c}z\mapsto
z^{d}\\\ w\mapsto w^{e}\end{subarray}}$$\scriptstyle{\widetilde{\alpha}}$

Here, we call the second slant arrow $\widetilde{\alpha}$.

If $z\mid\Psi$, then we have $\Psi=uz$ for some $u\in K\\{r^{-1}z,s^{-1}w\\}$
because $\Psi$ is prime. Then for
$(a,b)\in\overline{\mathbf{D}}_{K}(r)\times\overline{\mathbf{D}}_{K}(s)$, we
have

(9.93) $\displaystyle\Psi(a^{d},b^{e})=0\iff u(a^{d},b^{e})a^{d}=0\iff
a=0\iff\Psi(a,b)=0$

and PIQ follows. By the same reason, we are done if $w\mid\Psi$.

Now we assume $z\not\mid\Psi$ and $w\not\mid\Psi$. We claim that $d=e$. First
note that have $\Psi(0,0)=0$ because $\Psi$ is not eventually unit. Recall
that we have
$\Psi\mid\alpha(\Psi)=\widetilde{\alpha}(\gamma)\widetilde{\alpha}(\Psi^{\prime})$.
Since $\gamma$ is eventually a unit, it is easy to see that
$\widetilde{\alpha}(\gamma)$ is also eventually unit with respect to
$\left(K\\{r^{-d^{m}}z,s^{-e^{m}}w\\}\right)_{m}$. Thus $\Psi$ cannot divide
$\widetilde{\alpha}(\gamma)$ and hence
$\Psi\mid\widetilde{\alpha}(\Psi^{\prime})$.

Let us consider the action of
$G:=\langle\zeta_{d}\rangle\times\langle\zeta_{e}\rangle$ on
$K\\{r^{-1}z,s^{-1}w\\}$ defined by

(9.94) $\displaystyle(\zeta_{d}^{i},\zeta_{e}^{j})\cdot
f(z,w)=f(\zeta_{d}^{i}z,\zeta_{e}^{j}w).$

We have
$K\\{r^{-1}z,s^{-1}w\\}^{G}=\widetilde{\alpha}\left(K\\{r^{-d}z,s^{-e}w\\}\right)$.
Thus, $G$ acts transitively on the set of prime ideals of
$K\\{r^{-1}z,s^{-1}w\\}$ that lie over $\widetilde{\alpha}(\Psi^{\prime})$.
Thus all the prime factors of $\widetilde{\alpha}(\Psi^{\prime})$ are of the
form $\Psi(\zeta_{d}^{i}z,\zeta_{e}^{j}w)$. Thus we have

(9.95)
$\displaystyle\alpha(\Psi)=\Psi(z^{d},w^{e})=\widetilde{\alpha}(\gamma)u\prod_{i=1}^{k}\Psi(\zeta_{i}z,\eta_{i}w)$

for some $u\in K\\{r^{-1}z,s^{-1}w\\}^{\times}$, $k\in{\mathbb{Z}}_{\geq 1}$,
and $(\zeta_{i},\eta_{i})\in G$. Let us write

(9.96) $\displaystyle\Psi(z,w)$
$\displaystyle=\Psi_{0}(z)+\Psi_{1}(w)+zw\Psi_{2}(z,w)$ (9.97)
$\displaystyle\widetilde{\alpha}(\gamma)u$
$\displaystyle=c+\delta_{0}(z)+\delta_{1}(w)+zw\delta_{2}(z,w)$

where $\Psi_{0},\delta_{0}\in zK\\{r^{-1}z\\}$, $\Psi_{1},\delta_{1}\in
wK\\{s^{-1}w\\}$, $\Psi_{2},\delta_{2}\in K\\{r^{-1}z,s^{-1}w\\}$, and $c\in
K$. Note that $c\neq 0$ since $\widetilde{\alpha}(\gamma)u$ is eventually
unit. By substituting into eq. 9.95, we have

(9.98)
$\displaystyle\hphantom{=}\Psi_{0}(z^{d})+\Psi_{1}(w^{e})+z^{d}w^{e}\Psi_{2}(z^{d},w^{e})$
(9.99)
$\displaystyle=\left(c+\delta_{0}(z)+\delta_{1}(w)+zw\delta_{2}(z,w)\right)$
(9.100)
$\displaystyle\quad\times\prod_{i=1}^{k}\left(\Psi_{0}(\zeta_{i}z)+\Psi_{1}(\eta_{i}w)+\zeta_{i}\eta_{i}zw\Psi_{2}(\zeta_{i}z,\eta_{i}w)\right).$

Therefore we get

(9.101) $\displaystyle\Psi_{0}(z^{d})$
$\displaystyle=\left(c+\delta_{0}(z)\right)\prod_{i=1}^{k}\Psi_{0}(\zeta_{i}z)$
(9.102) $\displaystyle\Psi_{1}(w^{e})$
$\displaystyle=\left(c+\delta_{1}(w)\right)\prod_{i=1}^{k}\Psi_{1}(\eta_{i}w).$

We get $d=k=e$ by comparing the lowest degree terms.

Next, we will detect the shape of the zero set of $\Psi$. In the following
claim, we use the fact $p\not\mid d$. This can be seen as follows. By our
assumption, $p$ does not divide any ramification indices of original maps
$f,g\colon{\mathbb{P}}^{1}\longrightarrow{\mathbb{P}}^{1}$. In the course of
reduction in Step 1 and Step 2 so far, we replaced $f,g$ with iterates and $d$
with its power. Therefore, $d$ is a ramification index of iterated $f$ (and
$g$), and by the chain rule, we see that $d$ is a product of ramification
indices of $f$. Thus, $p$ does not divide $d$.

######  Claim 9.3.

The set
$\left\\{(a,b)\in\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(r)\times\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(s)\
\middle|\ \Psi(a,b)=0\right\\}$ is contained in

(9.103)
$\displaystyle\hphantom{\cup}\left(\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(r)\times\\{0\\}\right)\cup\left(\\{0\\}\times\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(s)\right)$
(9.104) $\displaystyle\cup\bigcup_{k,l\geq
1}\left\\{(a,b)\in\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(r)\times\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(s)\
\middle|\ a^{k}-b^{l}=0\right\\}$

###### Proof of Claim 9.3.

Let us write

(9.105) $\displaystyle\Psi=\sum_{i,j\geq 0}c_{ij}z^{i}w^{j}.$

Let $a,b\in{\mathbb{C}}_{p}$ be such that $|a|\leq r$, $|b|\leq s$, and
$\Psi(a,b)=0$. It is enough to show that $a^{k}-b^{l}=0$ for some $k,l\geq 1$
under the assumption $ab\neq 0$. Let

(9.106) $\displaystyle M$ $\displaystyle=\max\\{|a^{i}b^{j}|\mid c_{ij}\neq
0\\}\text{ and}$ (9.107) $\displaystyle S$
$\displaystyle=\\{(i,j)\in{\mathbb{Z}}_{\geq
0}^{2}\mid|a^{i}b^{j}|=M,c_{ij}\neq 0\\}.$

Since $|a|,|b|<1$, $M$ exists and $S$ is a finite set. Let us take
$(i_{0},j_{0})\in S$ such that the first coordinate is the smallest among
elements of $S$.

Since $\Psi\mid\alpha(\Psi)=\Psi(z^{d},w^{d})$, we have
$\Psi(a^{d^{n}},b^{d^{n}})=0$ for all $n\geq 1$. Thus

(9.108) $\displaystyle 0$
$\displaystyle=\frac{\Psi(a^{d^{n}},b^{d^{n}})}{(a^{i_{0}}b^{j_{0}})^{d^{n}}}=\frac{1}{(a^{i_{0}}b^{j_{0}})^{d^{n}}}\sum_{i,j\geq
0}c_{ij}(a^{i}b^{j})^{d^{n}}$ (9.109) $\displaystyle=\sum_{(i,j)\in
S}c_{ij}\frac{(a^{i}b^{j})^{d^{n}}}{(a^{i_{0}}b^{j_{0}})^{d^{n}}}+\sum_{(i,j)\notin
S}c_{ij}\frac{(a^{i}b^{j})^{d^{n}}}{(a^{i_{0}}b^{j_{0}})^{d^{n}}}$

for all $n\geq 1$. If $(i,j)\notin S$, then we have

(9.110)
$\displaystyle\left|\frac{c_{ij}(a^{i}b^{j})^{d^{n}}}{(a^{i_{0}}b^{j_{0}})^{d^{n}}}\right|=\left|\frac{c_{ij}a^{i}b^{j}}{a^{i_{0}}b^{j_{0}}}\right|\left|\frac{(a^{i}b^{j})^{d^{n}-1}}{(a^{i_{0}}b^{j_{0}})^{d^{n}-1}}\right|$
(9.111)
$\displaystyle\leq\frac{\|\Psi\|_{K\\{r^{-1}z,s^{-1}w\\}}}{|a^{i_{0}}b^{j_{0}}|}\left|\frac{a^{i}b^{j}}{a^{i_{0}}b^{j_{0}}}\right|^{d^{n}-1}\leq\frac{\|\Psi\|_{K\\{r^{-1}z,s^{-1}w\\}}}{|a^{i_{0}}b^{j_{0}}|}\left(\frac{1}{p^{1/\delta}}\right)^{d^{n}-1}$

where $\delta\in{\mathbb{Z}}_{\geq 1}$ is independent of $i,j$. By this, we
get

(9.112) $\displaystyle\left|\sum_{(i,j)\notin
S}c_{ij}\frac{(a^{i}b^{j})^{d^{n}}}{(a^{i_{0}}b^{j_{0}})^{d^{n}}}\right|\leq\frac{\|\Psi\|_{K\\{r^{-1}z,s^{-1}w\\}}}{|a^{i_{0}}b^{j_{0}}|}\left(\frac{1}{p^{1/\delta}}\right)^{d^{n}-1}\xrightarrow{n\to\infty}0.$

Thus we have

(9.113) $\displaystyle\lim_{n\to\infty}\sum_{(i,j)\in
S}c_{ij}\frac{(a^{i}b^{j})^{d^{n}}}{(a^{i_{0}}b^{j_{0}})^{d^{n}}}=0.$

Now, since $|a^{i-i_{0}}b^{j-j_{0}}|=1$ for $(i,j)\in S$, we have

(9.114) $\displaystyle j-j_{0}=-\frac{\log|a|}{\log|b|}(i-i_{0}).$

Let us write $-\log|a|/\log|b|=m/q$ by coprime integers $m,q$ with $m\leq-1$
and $q\geq 1$. Then fix a $q$-th root $b^{\prime}$ of $b$ in
${\mathbb{C}}_{p}$. Then we have

(9.115)
$\displaystyle\frac{a^{i}b^{j}}{a^{i_{0}}b^{j_{0}}}=a^{i-i_{0}}b^{m(i-i_{0})/q}=a^{i-i_{0}}{b^{\prime}}^{m(i-i_{0})}=(a{b^{\prime}}^{m})^{i-i_{0}}.$

Set $\xi=a{b^{\prime}}^{m}$. Then we have $|\xi|=1$ and

(9.116) $\displaystyle\lim_{n\to\infty}\sum_{(i,j)\in
S}c_{ij}\xi^{i-i_{0}}=0.$

By Lemma 9.5 below, $\xi$ is a root of $1$. This implies $a^{q}b^{m}$ is a
root of $1$, and we are done.

∎

The above claim implies the following.

######  Claim 9.4.

There are $k,l\in{\mathbb{Z}}_{\geq 1}$ such that

(9.117) $\displaystyle\Psi\mid z^{k}-w^{l}$

in $K\\{r^{-1}z,s^{-1}w\\}$.

###### Proof of Claim 9.4.

Let $\Psi=u\Psi_{1}^{e_{1}}\cdots\Psi_{n}^{e_{n}}$ be the prime factorization
in ${\mathbb{C}}_{p}\\{r^{-1}z,s^{-1}w\\}$ with $u$ being a unit and
$\Psi_{i}$ being primes. Suppose $\Psi_{i}$ does not divide any of $z,w$, and
$z^{k}-w^{l}$ $(k,l\in{\mathbb{Z}}_{\geq 1})$ in
${\mathbb{C}}_{p}\\{r^{-1}z,s^{-1}w\\}$. Let $f$ be any of $z,w$, and
$z^{k}-w^{l}$. Then the Krull dimension of
${\mathbb{C}}_{p}\\{r^{-1}z,s^{-1}w\\}/(\Psi_{i},f)$ is zero and hence, it is
an Artinian ring. Thus, it has only finitely many maximal ideals, and thus

(9.118)
$\displaystyle\left\\{(a,b)\in\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(r)\times\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(s)\
\middle|\ \Psi_{i}(a,b)=f(a,b)=0\right\\}$

is a finite set. By Claim 9.3, this implies

(9.119)
$\displaystyle\left\\{(a,b)\in\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(r)\times\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(s)\
\middle|\ \Psi_{i}(a,b)=0\right\\}$

is countable. This is a contradiction because this set is uncountable. Indeed,
first, take an isometric isomorphism
$\sigma_{1}\colon{\mathbb{C}}_{p}\\{r^{-1}z,s^{-1}w\\}\xrightarrow{\sim}{\mathbb{C}}_{p}\\{z,w\\}$
by scaling $z$ and $w$ (recall $r,s\in|{\mathbb{C}}_{p}^{\times}|$). Then,
take isometric isomorphism
$\sigma_{2}\colon{\mathbb{C}}_{p}\\{z,w\\}\xlongrightarrow{\sim}{\mathbb{C}}_{p}\\{z,w\\}$
so that $\sigma_{2}(\sigma_{1}(\Psi_{i}))=uW$, where
$u\in{\mathbb{C}}_{p}\\{z,w\\}^{\times}$ and $W\in{\mathbb{C}}_{p}\\{z\\}[w]$
is a Weierstrass polynomial. Note that the degree of $W$ in $w$ is at least
one since $\Psi_{i}$ is not a unit. Then we get

(9.120)
$\displaystyle{\mathbb{C}}_{p}\\{r^{-1}z,s^{-1}w\\}/(\Psi_{i})\simeq{\mathbb{C}}_{p}\\{z\\}[w]/(W)$

and this is finite as a ${\mathbb{C}}_{p}\\{z\\}$-module. (cf. for example [3,
section 5.2] or [11, section 2.3 and 2.4].) Hence eq. 9.119 has a surjection
onto $\overline{\mathbf{D}}_{{\mathbb{C}}_{p}}(1)$, which is uncountable.

Thus we proved every $\Psi_{i}$ divides one of $z,w$, and $z^{k}-w^{l}$
$(k,l\in{\mathbb{Z}}_{\geq 1})$ in ${\mathbb{C}}_{p}\\{r^{-1}z,s^{-1}w\\}$.
Thus there is a finite product $\widetilde{\Psi}$ of some of $z,w$, and
$z^{k}-w^{l}$ for $k,l\in{\mathbb{Z}}_{\geq 1}$ such that
$\Psi\mid\widetilde{\Psi}$ in ${\mathbb{C}}_{p}\\{r^{-1}z,s^{-1}w\\}$. Since
$\Psi,\widetilde{\Psi}\in K\\{r^{-1}z,s^{-1}w\\}$, by Lemma 9.6 below, $\Psi$
divides $\widetilde{\Psi}$ in $K\\{r^{-1}z,s^{-1}w\\}$. Finally, since $\Psi$
is prime and we assumed $z\not\mid\Psi$ and $w\not\mid\Psi$, there is some
$k,l\in{\mathbb{Z}}_{\geq 1}$ such that $\Psi\mid z^{k}-w^{l}$. ∎

Now note that $z^{k}-w^{l}$ is invariant under the map $(z^{d},w^{d})$. Thus,
replacing $\alpha$ with iterates, we may assume every minimal prime ideal of
$K\\{r^{-1}z,s^{-1}w\\}$ containing $z^{k}-w^{l}$ is either fixed or mapped to
a fixed one by the pullback by $\alpha$ (cf. Case 2-2). Let
$z^{k}-w^{l}=uP_{1}^{e_{1}}\cdots P_{m}^{e_{m}}$ be the prime factorization in
$K\\{r^{-1}z,s^{-1}w\\}$ where $u$ is a unit, $P_{i}$ are distinct primes with
$P_{1}=\Psi$, and $e_{i}\in{\mathbb{Z}}_{\geq 1}$. Now suppose
$(a,b)\in\overline{\mathbf{D}}_{K}(r)\times\overline{\mathbf{D}}_{K}(s)$
satisfies $\Psi(a^{d^{n}},b^{d^{n}})=0$ for some $n\in{\mathbb{Z}}_{\geq 0}$.
We want to show that there is $n_{0}\geq 0$ independent of $(a,b)$ such that
$\Psi(a^{d^{n_{0}}},b^{d^{n_{0}}})=0$. First note that we have
$(a^{d^{n}})^{k}-(b^{d^{n}})^{l}=0$. If $ab=0$, we have $a=b=0$, and we are
done. Suppose $ab\neq 0$. Then

(9.121) $\displaystyle\left(\frac{a^{k}}{b^{l}}\right)^{d^{n}}=1.$

Since $a^{k}/b^{l}\in K$ and $K$ is a finite extension of ${\mathbb{Q}}_{p}$,
there is $n_{1}\geq 0$ depending only on $K$ and $d$ such that

(9.122) $\displaystyle\left(\frac{a^{k}}{b^{l}}\right)^{d^{n_{1}}}=1$

or equivalently

(9.123) $\displaystyle(a^{d^{n_{1}}})^{k}-(b^{d^{n_{1}}})^{l}=0.$

(cf. [13, II (7.12),(7.13)], [2, Proposition 3.6].) Thus there is $P_{i}$
whose zero locus is invariant under $(z^{d},w^{d})$ and such that
$P_{i}(a^{d^{n_{1}+1}},b^{d^{n_{1}+1}})=0$. If $i=1$, then we are done.
Suppose $i\geq 2$. Then by the $(z^{d},w^{d})$-invariance of $(P_{i}=0)$ and
$(\Psi=0)$ as well as our assumption, we have

(9.124) $\displaystyle