# Ramification theory from homotopical point of view, II

Tomoyuki Abe

## Introduction

This paper is a continuation from [Part I], and we will prove a theorem which
we used in [Part I]. However, the method is almost independent from [Part I],
and, for example, we do not use the language of $\infty$-categories at all. In
the last two sections, we use the terminologies of [Part I, §4], but this
section can be read independently from the other sections of [Part I]. The
goal of this paper is to show the following theorem which we promised in [Part
I, 5.7]. We keep the notations from [Part I, §4], some of which we will recall
in §4.

######  Theorem (Corollary 5.9). —

Assume that $X$ and $S$ are noetherian affine $\mathbb{F}_{p}$-schemes (not
necessarily of finite type), and assume we are given a morphism $X\rightarrow
S$ of finite type. For any
$F\in\mathscr{S}_{d}(S)(:=\mathrm{Fl\acute{E}t}_{d}(X/S))$ and a finite
ordered set $\mathbf{h}$ of
$\mathcal{O}_{X\times\Delta^{m}}(X\times\Delta^{m})$, we can find a finite
ordered set $\mathbf{h}^{\prime}$ of $\mathcal{O}_{X}(X)$ such that for a
surjective $\mathrm{Mdf}$-sequence $\\{\mathbf{V}_{i}\\}$ adapted to
$(F;\mathbf{h}\vee\mathbf{h}^{\prime})$, $F[\\{\mathbf{V}_{i}\\}]_{\infty}$ is
in $\mathscr{S}_{d-1}$.

By limit argument, we may and do assume that $S$ is of finite type over a
field $k$ of characteristic $p$. We may also reduce to the case where
$X\rightarrow S$ is smooth. If we are given a morphism $h\colon
Y\rightarrow\mathbb{A}^{1}$, we denote $h^{*}\mathcal{L}$, where $\mathcal{L}$
is the Artin-Schreier sheaf on $\mathbb{A}^{1}$ associated with a non-trivial
additive character, by $\mathcal{L}(h)$. The function $h$ is said to be
$S$-separable if the smooth locus in $X$ of the induced morphism
$X\rightarrow\mathbb{A}^{1}\times S$ is fiberwise dense over $S$. A key to
show the above theorem is a weak structure theorem of the Artin-Schreier sheaf
$\mathcal{L}(h)$. This states that after a suitable alteration of $S$,
$\mathcal{L}(h)$ can be written, generically with respect to each fiber of
$X\rightarrow S$, as $\mathcal{L}(f/\sigma)\otimes\mathcal{G}$ where $f$ is a
separable function on $X$, $\sigma$ is a function on $S$, and $\mathcal{G}$ is
a locally constant constructible module on $X$. A benefit of writing in this
way is that we can compute the nearby cycles of $\mathcal{L}(h)$ very
explicitly, at least generically with respect to each fiber. In this
introduction, we will focus on how to prove the weak structure theorem. For a
proof of this key result, we use the argument invented by Kedlaya used to
prove semistable reduction theorem [Ke4] and the existence of good formal
model [Ke5].

From now on, for simplicity, we fix a smooth morphism $X\rightarrow S$ between
integral schemes of finite type over a field $k$ of characteristic $p$. Since,
for a function $f$, we have $\mathcal{L}(f^{p}-f)\cong\Lambda$, the weak
structure theorem follows readily from the following (purely geometric)
result:

######  Theorem (Theorem Theorem). —

For any $h\in k(X)$, we can find an alteration $S^{\prime}\rightarrow S$ over
which $h$ is admissible. Here, $h$ is said to be admissible if, Zariski
locally on $S$, there exists a fiberwise dense open subscheme $U\subset X$, an
$S$-separable function $f$ on $U$, a regular function
$\sigma\in\mathcal{O}_{S}\cap k(S)^{\times}$, and $g\in k(X)$ such that
$h|_{U}+(g^{p}-g)\in f/\sigma+\mathcal{O}_{X}$.

Let us outline the strategy of the proof. Since $X\rightarrow S$ is assumed to
be smooth, we may assume that there exists a local coordinate $\\{x_{i}\\}$ of
$X$ over $S$. It is not hard to reduce to the case where $h=h^{\prime}/\tau$
where $h^{\prime}\in\mathcal{O}_{X}$, $\tau\in\mathcal{O}_{S}$. For
simplicity, we assume $S=\mathrm{Spec}(k[s,t])$ in the following. Formally, we
may write $h=\sum_{\underline{k}\geq
0}a_{\underline{k}}\underline{x}^{\underline{k}}$ where $a_{\underline{k}}\in
k(S)$.

Let us see how we take $g$ in some examples. First consider $h=\frac{s}{t}x$.
In this case, $h$ is not of the form $f/\sigma$ at $(s,t)=(0,0)$. What we
should do is to take the blowup $S^{\prime}\rightarrow S$ at $(0,0)$. Then we
may write $h=ux$ or $h=x/u$ using the coordinate around the exceptional fiber,
and these are the forms that we need.

Now, let us consider $h=\frac{1}{s}x^{p}+\frac{t}{s}x$. In this case, $h$ is
not separable over $t=0$. What we will do in this case is to “mollify” $h$ by
adding functions of the form $g-g^{p}$. In this case, we take
$g=\frac{1}{s^{1/p}}x$ (by extending $S$). Then we have
$h+(g-g^{p})=\frac{t+s^{p-1/p}}{s}x$. Now, this function can be handled as in
the first example, by taking a suitable blowup to transform it into the
desired form. As this example shows, a basic strategy is to “eliminate” terms
$a_{\underline{k}}\underline{x}^{\underline{k}}$ such that
$p\mid\underline{k}$ (i.e. $p\mid k_{i}$ for any $i$) by adding $g-g^{p}$.
Such a $g$ is called a mollifier.

Finally, let us consider $h=\frac{1}{s}+\frac{1}{s}x+\frac{1}{s}x^{2}+\dots$.
Since $h$ is equal to $\frac{1}{s}\frac{1}{1-x}$, this is already admissible.
In other words, we have nothing to do. However, if we take the philosophy of
the 2nd example seriously, we could remove the factors of the form
$\frac{1}{s}x^{pl}$ by adding some $g$. Thus, we need to distinguish the case
where we need further mollification and the case where we do not.

Keeping these basic examples in mind, our strategy, after Kedlaya, is to
attain the admissibility “locally” in the Zariski-Riemann space of $S$, and
globalize by using the compactness of Zariski-Riemann space. By writing
valuation of height greater than $1$ by composition of valuation of height
$1$, it is not too hard to reduce the argument to the height $1$ case, so we
only consider height $1$ valuation in this introduction. To explain the
strategy, let us fix a height $1$ valuation $v$ of $k(S)$, trivial on $k$. Let
$f=\sum a_{\underline{k}}\underline{x}^{\underline{k}}$. Since $v$ is of
height $1$, $v(k(S)^{\times})\subset\mathbb{R}$. We may consider a number

$\mathrm{VP}(f)(v):=\min_{\underline{k}\in\mathbb{N}^{d}\setminus\\{\underline{0}\\}}\bigl{\\{}0,v(a_{\underline{k}})\bigr{\\}}\leq
0.$

We view this as a function on $v$ later. Now, we will perform mollifications,
namely adding function of the form $g^{p}-g$, formally. For this, put
$f^{(m)}_{\underline{k}}:=\sum_{k=0}^{m}(a_{p^{k}\underline{k}})^{1/p^{k}}$.
Using this, we may define $\mathrm{VP}^{(\infty)}(f)(v)$. This is, roughly
saying, the limit
$\lim_{m\rightarrow\infty}\mathrm{VP}(\sum_{\underline{k}}f^{(m)}_{\underline{k}}\underline{x}^{\underline{k}})$.
It is not hard to show the existence of the limit. With this limit value, we
may distinguish the 2nd and the 3rd example: if
$\mathrm{VP}^{(\infty)}(f)(v)=\mathrm{VP}(f)(v)$, then we do not need further
mollification. In the 2nd example, consider a valuation $v$ such that
$v(s)>v(t)>0$, $v(s)-v(t)>v(s)/p$. Then
$\mathrm{VP}^{(\infty)}(f)(v)\neq\mathrm{VP}(f)(v)$, which implies that we do
need mollifications around these valuations. Thus, the question will be to
find $g$ such that $\mathrm{VP}(f+g-g^{p})(v)=\mathrm{VP}^{(\infty)}(f)(v)$.
This is not possible if we take $f$ to be an arbitrary formal function.
However, if $f$ comes from a function on $X$, we will show that we may find
such a function $g$ if we take an alteration “locally around $v$”. Once this
is shown, we find a global alteration dominating all the local alterations.
This is done by using the fact that the Riemann-Zariski space is compact. From
now on, we focus on the local arguments.

Generally, valuations are hard to describe explicitly. However, this
difficulty has a hierarchy, called the transcendence defect after Kedlaya. The
transcendence defect is a non-negative integer for each valuation, and when
the transcendence defect is $0$, the valuation is relatively easy to describe.
For example, consider the case $S=\mathrm{Spec}(k[s,t])$ and $v$ has
transcendence defect $0$ and the dimension of the center is of dimension $0$.
This kind of valuation is usually called an Abhyankar valuation. Then $v(s)$,
$v(t)$ are $\mathbb{Q}$-linearly independent, and we have
$v\bigl{(}\sum\alpha_{m,n}s^{m}t^{n}\bigr{)}=\min\bigl{\\{}mv(s)+nv(t)\mid\alpha_{m,n}\neq
0\bigr{\\}}$. In this case, we may concretely construct a mollifier $g$ that
attains local admissibility around $v$. See Lemma Lemma.

In order to treat valuations with positive transcendence defect, we use the
induction. Let $v_{0}$ be a valuation with transcendence defect $n>0$ that we
wish to show the local admissibility. We assume that the local admissibility
is known for valuations with transcendence defect less than $n$. In the
Riemann-Zariski space of $S$, valuations with transcendence defect less than
$n$ clusters around $v_{0}$. The main step is to find a valuation $v_{1}$
close enough to $v_{0}$ so that the transcendence defect is $n-1$ and
$\mathrm{VP}^{(\infty)}(h)(v_{0})=\mathrm{VP}^{(\infty)}(h)(v_{1})$. Once we
find such a $v_{1}$, we may transfer the local admissibility of $v_{1}$ to
$v_{0}$ using certain the monotonicity of valuation polygon. To show the
existence of $v_{1}$, we choose a suitable “Berkovich disc containing
$v_{0}$”. If we consider $\mathrm{VP}^{(\infty)}(f)(v)$ as a function of $v$
on this Berkovich disc, a polygon appears. This is the reason for the
notation, and $\mathrm{VP}$ stands for “valuation polygon”. In this disc,
$v_{0}$ becomes a terminal point, more precisely, a point of type either 1 or
4. In the type 1 case, it is not hard to find a valuation $v_{1}$ in the disc,
and the only issue is to treat the type 4 points. What we need to show now is
that as the valuation $v$ approaches to $v_{0}$ the valuation polygon
$\mathrm{VP}^{(\infty)}(f)(v)$ stabilizes. This is not true unless we use the
algebraicity of $f$, and we believe that this is the core of the proof. For
this stability, we use the fact that Hadamard product preserves algebraic
functions shown in [SW].

Let us overview the structure of this paper. In §1, we fix terminologies and
show basic facts in algebraic geometry. In §2, we perform an analysis around
type 4 point of a Berkovich disc. Especially, we prove the stability of the
valuation polygon around type 4 points using the algebraicity of Hadamard
product. In §3, we show the admissibility of the function using the stability
result of §2. This section is strongly inspired by Kedlaya’s argument. In §4,
we recall some notations from Part I, and prove some complimentary results. In
§5, we show the main result.

### Acknowledgments

The author wishes to thank Kiran Kedlaya for answering various questions on
his paper [Ke4]. He also thanks Takeshi Saito for encouragements and numerous
feedbacks. In particular, the author learned how to use reduced fiber theorem
from him. He is grateful to Yoichi Mieda, Kazuhiro Fujiwara, and Deepam Patel
for comments and answering questions.

This work is supported by JSPS KAKENHI Grant Numbers 16H05993, 18H03667,
20H01790.

### Notations and conventions

Let $\underline{n}=(n_{1},\dots,n_{d})\in\mathbb{N}^{d}$. For an integer $m$,
we denote by $m\mid\underline{n}$ if $m\mid n_{i}$ for any $i$, and by
$m\nmid\underline{n}$ otherwise. We put $\underline{0}:=(0,\dots,0)$, and
$\underline{n}\geq\underline{0}$ if and only if $n_{i}\geq 0$ for any $i$.

Let $A$ be an integral $\mathbb{F}_{p}$-algebra. We denote by
$A^{p}:=\\{a^{p}\mid a\in A\\}\subset A$, and $A\subset A^{1/p}$ the ring $A$
considered as an $A$-algebra by the Frobenius endomorphism $A\rightarrow
A=:A^{1/p}$.

Let $X\rightarrow S$ and $T\rightarrow S$ are morphisms of schemes. We often
denote $X\times_{S}T$ by $X_{T}$. In particular, for a morphism $X\rightarrow
S$ and a point $s\in S$, we put $X_{s}:=X\otimes_{S}\mathrm{Spec}(k(s))$.
Similarly, for a sheaf $\mathcal{F}$ on $X$, we denote by $\mathcal{F}_{T}$
the pullback on $X_{T}$.

For a scheme $X$, we put
$\mathrm{R}(X):=\mathop{\underrightarrow{\mathrm{lim}}}_{U}\Gamma(U,\mathcal{O}_{X})$
where $U\subset X$ runs over open dense subscheme, which is usually called the
ring of rational functions. If $X$ is an integral $k$-scheme, we sometimes
denote $\mathrm{R}(X)$ by $k(X)$, which is a field. We often denote by
$f\in\mathcal{O}_{X}$ for $f\in\Gamma(X,\mathcal{O}_{X})$.

We fix two different prime numbers $p$, $\ell$. We fix a finite commutative
ring $\Lambda$ whose residue characteristic is $\ell$. Throughout this paper,
we fix a field $k$ of characteristic $p\,(>0)$.

##  1 Preliminary — Algebraic geometry

### 1.1

Recall that a morphism $\pi\colon X\rightarrow Y$ between schemes is said to
be maximally dominant if any generic point of $X$ is sent to a generic point
of $Y$. Flat morphisms between schemes are examples of maximally dominant
morphisms (cf. [G, Exp. II, 1.1.3]). We say that $\pi$ is an alteration if it
is proper, surjective, generically finite, maximally dominant. We say that $f$
is a modification if it is an alteration and there exists a open dense
subscheme $V\subset Y$ such that $\pi^{-1}(V)\rightarrow V$ is an isomorphism.

######  1.2 Definition. —

Let $\pi\colon X\rightarrow S$ be a morphism of finite type, and
$g\in\mathcal{O}_{X}$.

  1. 1.

If $S=\mathrm{Spec}(k)$, the function $g$ is said to be non-constant if the
induced morphism $X\rightarrow\mathbb{A}^{1}_{k}$ is maximally dominant. If
the image of $X$ in $\mathbb{A}^{1}_{k}$ is not dense, we say that $g$ is
constant. For a general $S$, it is $S$-non-constant if for any $s\in S$, the
induced function $g_{s}$ on $X_{s}$ is non-constant.

  2. 2.

Assume $\pi$ is smooth. The function $g$ is said to be
$S$-separable(1)(1)(1)We caution that this terminology is not consistent with
[SGA I, Exp. X]. if there exists an open subscheme $U\subset X$ such that the
induced morphism $U\rightarrow\mathbb{A}^{1}\times S$ is smooth and
$U_{s}\subset X_{s}$ is dense for any $s\in S$.

  3. 3.

Assume $\pi$ is smooth. If $X$ is connected, the function $g$ is said to be
$S$-Frobenius separable(2)(2)(2) This terminology is suggested to the author
by T. Saito. if there exists an $S$-separable function $g^{\prime}$ such that
$g=g^{\prime q}$ with some power $q$ of $p$. If $X$ is not connected, $g$ is
said to be $S$-Frobenius separable if the restriction to each connected
component of $X$ is so.

If no confusion may arise, we often omit the prefix “$S$-”.

######  1.3 Lemma. —

Let $\pi\colon X\rightarrow S$ be a morphism of finite type, and
$g\in\mathcal{O}_{X}$.

  1. 1.

Non-constancy is stable under base change. Moreover, for a surjective morphism
$S^{\prime}\rightarrow S$, if the pullback of $g$ to $X\times_{S}S^{\prime}$
is $S^{\prime}$-non-constant, then $g$ is $S$-non-constant.

  2. 2.

If $\pi$ is flat, then $g$ is $S$-non-constant if and only if the induced
morphism $X\xrightarrow{\pi\times g}S\times\mathbb{A}^{1}$ is flat.

  3. 3.

Assume $\pi$ is smooth, then $g$ is $S$-separable if and only if for any point
$s\in S$ the base change $g_{s}$ is separable. This is equivalent to saying
that $Z_{s}\subset X_{s}$ is nowhere dense for any $s\in S$, where $Z$ is the
zero-locus of $\mathrm{d}g\in\Omega^{1}_{X/S}$.

  4. 4.

Assume $\pi$ is smooth, $S$ is excellent, and $g$ is $S$-non-constant. Then
there exists a quasi-finite dominant morphism $F\colon S^{\prime}\rightarrow
S$ such that $F^{*}(g)$ in $\mathcal{O}_{X_{S^{\prime}}}$ is
$S^{\prime}$-Frobenius separable. In fact, $F$ can be taken as an iterated
Frobenius endomorphism followed by an immersion $U\subset
S_{\mathrm{red}}\hookrightarrow S$, where $U$ is open inside
$S_{\mathrm{red}}$.

###### Proof.

To show 1, we may easily reduce to the case where $S$ is the spectrum of a
field. In this case, all the irreducible components of $X$ that appear after
changing the base lie over the generic points of $X$ by [EGA IV, 4.5.11], and
the claim follows. To show 2, if $S$ is $\mathrm{Spec}(k)$, the claim follows
since any maximally dominant morphism
$X\rightarrow\mathbb{A}^{1}=\mathrm{Spec}(k[t])$ from a reduced scheme $X$ is
flat, because $\mathcal{O}_{X}$ is $t$-torsion free. When $S$ is general, use
[EGA IV, 11.3.10]. To show 3, since $f$ is smooth, the induced morphism
$X_{s}\rightarrow\mathbb{A}^{1}_{k(s)}$ is flat for any $s\in S$ by 2. Now,
use [EGA IV, 17.8.2].

Let us check 4. Since the claim is local around the generic points of $S$, we
may shrink $S$ and assume that $S$ is irreducible, so replacing by its reduced
scheme, we may even assume that $S$ is integral with the generic point $\eta$.
Shrinking $S$ further, we may assume that $S$ is regular since $S$ is assumed
excellent. Thus, $X$ is a disjoint union of irreducible components, and we may
assume that $X$ is integral as well.

Now, let $K/k$ be an extension of fields of finite type. We claim that
$\bigcap_{n}k(K^{p^{n}})\subset l$, where $l$ is the algebraic closure of $k$
in $K$. Indeed, we may assume that $k$ is perfect. If
$x\in\bigcap_{n}k(K^{p^{n}})$, then
$k(x^{1/p^{\infty}})\subset\bigcap_{n}k(K^{p^{n}})\subset K$. Since $K$ is
finitely generated over $k$, so is $k(x^{1/p^{\infty}})$. This is possible
only when $x$ is algebraic over $k$, and the claim follows. Now, put
$K:=\mathrm{R}(X)$, $k:=\mathrm{R}(S)$ which are fields by assumption, and
apply this observation. Since any function in $l$ is constant, the non-
constancy assumption implies that there exists an integer $n$ such that $g\in
k(K^{p^{n}})$ but $g\not\in k(K^{p^{n+1}})$. Thus, there exists $g^{\prime}\in
k^{1/p^{n}}(K)$ such that $g^{\prime p^{n}}=g$. Then since $g^{\prime}\not\in
k^{1/p^{n}}(K^{p})$, by [EGA $0_{\mathrm{IV}}$, 21.4.6], $g^{\prime}$ is a
separable function defined generically. Now, since $X$ is regular, the
morphism $\mathcal{O}_{X}^{q}\rightarrow\mathcal{O}_{X}$ is faithfully flat
for any power $q$ of $p$. Moreover, we have
$\mathrm{R}(X)\cong\mathrm{R}(X)^{q}\otimes_{\mathcal{O}_{X}^{q}}\mathcal{O}_{X}$
because the Frobenius endomorphism is homeomorphic. Thus, by [EGA
$0_{\mathrm{I}}$, 6.6.3], we have
$\mathrm{R}(X)^{q}\cap\mathcal{O}_{X}=\mathcal{O}_{X}^{q}$ in $\mathrm{R}(X)$.
Since $g\in\mathrm{R}(X)^{p^{n}}\cap\mathcal{O}_{X}$, we have
$g^{\prime}\in\mathcal{O}_{X}$, and the claim follows. ∎

######  1.4 Lemma. —

Let $Y$ be an integral smooth scheme over $k$, and $\\{y_{i}\\}$ be a
coordinate system on $Y$ at a $k$-rational point, which determines an
embedding $\mathcal{O}_{Y}\hookrightarrow k[\\![\underline{y}]\\!]$ by [EGA
IV, 17.6.3]. Then for a power $q$ of $p$, we have

$\mathcal{O}_{Y}\cap
k[\\![\underline{y}^{q}]\\!]=\mathcal{O}^{q}_{Y}\otimes_{k^{q}}k.$

In particular, if $k$ is perfect, $f=\sum
a_{\underline{k}}\underline{y}^{\underline{k}}\in\mathcal{O}_{Y}$ is separable
if and only if there exists $p\nmid\underline{n}$ such that
$a_{\underline{n}}\neq 0$.

###### Proof.

Since $Y$ is smooth over a field $k$, the homomorphisms
$\mathcal{O}_{Y}^{q}\rightarrow\mathcal{O}_{Y}^{q}\otimes_{k^{q}}k\rightarrow\mathcal{O}_{Y}$
are faithfully flat, which can be checked easily by taking local coordinates.
We have the canonical homomorphism
$k[\\![\underline{y}^{q}]\\!]\otimes_{(\mathcal{O}_{Y}^{q}\otimes_{k^{q}}k)}\mathcal{O}_{Y}\rightarrow
k[\\![\underline{y}]\\!]$. This is a surjective homomorphism between finite
flat $k[\\![\underline{y}^{q}]\\!]$-modules of the same rank, and the
homomorphism is an isomorphism. Thus, [EGA $0_{\mathrm{I}}$, 6.6.3] implies
the formula. Let us show the second claim. Let $K_{Y}:=k(Y)$. The smoothness
implies $K_{Y}^{q}\cap\mathcal{O}_{Y}=\mathcal{O}_{Y}^{q}$ in $K_{Y}$, and by
taking $\otimes_{k^{q}}k$, we get
$k(K_{Y}^{q})\cap\mathcal{O}_{Y}=\mathcal{O}_{Y}^{q}\otimes_{k^{q}}k$ in
$K_{Y}\otimes_{k^{q}}k\cong K_{Y}$. By [EGA $0_{\mathrm{IV}}$, 21.4.6], the
claim follows. ∎

######  1.5 Lemma. —

Let $\pi\colon X\rightarrow S$ be a flat morphism of finite type between
noetherian schemes, and $x\in\mathcal{O}_{X}$ be an $S$-non-constant function.
Let $g$ be a function on an open subscheme $U\subset X$ such that $g|_{U_{s}}$
is not nilpotent around each generic point of $U_{s}$ for any $s\in S$. Then
there exists an integer $N$ such that $(x^{n}g)|_{U}$ is $S$-non-constant for
any $n>N$.

###### Proof.

We show the lemma using the noetherian induction. It suffices to check that
there exists an open dense subscheme $W\subset S$ such that the lemma holds
for $S$ replaced by $W$. We may assume $S$ irreducible. It suffices to show
the existence of $W$ after changing the base by a dominant morphism
$\alpha\colon S^{\prime}\rightarrow S$ of finite type. Indeed, if the lemma
holds over an open subscheme $W^{\prime}\subset S^{\prime}$, then the lemma
holds over the set $\alpha(W^{\prime})$. Since $\alpha(W^{\prime})$ is a
constructible set by [EGA IV, 1.8.4] which contains the generic point of $S$,
we may take an open subscheme $W\subset\alpha(W^{\prime})$, and the claim
follows. Moreover, it suffices to show the claim for each irreducible
component of $X$. Thus, we may assume that for the generic point $\eta\in S$,
the fiber $X_{\eta}$ is geometrically irreducible. Now, take a
compactification $X\hookrightarrow\overline{X}$ over $S\times\mathbb{A}^{1}$.
Replacing $\overline{X}$ by the closure of $X_{\eta}$ in $\overline{X}$, we
may assume that the the open immersion $X_{\eta}\subset\overline{X}_{\eta}$ is
dense, and thus, $\overline{X}_{\eta}$ is geometrically irreducible. Since $x$
is assumed flat and $S$ is irreducible, the morphism
$\overline{x}\colon\overline{X}\rightarrow S\times\mathbb{A}^{1}$ is
surjective. Thus
$V(\overline{x})\cong\overline{X}\times_{\overline{x},\mathbb{A}^{1}}\\{0\\}\subset\overline{X}\rightarrow
S$ is surjective. Shrinking $S$, we may assume that any fiber of
$\overline{\pi}\colon\overline{X}\rightarrow S$ is geometrically irreducible
by [EGA IV, 9.7.7]. Shrinking $S$ further, we may take an open affine
subscheme $V\subset\overline{X}$ such that $V(\overline{x})\cap V\rightarrow
S$ is surjective. By the fiberwise irreducibility, $V_{s}\cap U_{s}\subset
U_{s}$ is dense. It suffices to show the lemma after replacing $X$ by $V$ and
$x$ by $\overline{x}$. Thus, it suffices to show the lemma assuming that $X$,
$S$ are irreducible affine, $V(x)\rightarrow S$ is surjective, $V(x)\subsetneq
X$, and $X\rightarrow S$ is fiberwise irreducible, but $x\colon X\rightarrow
S\times\mathbb{A}^{1}$ is not anymore flat.

Let $Y$ be a noetherian scheme and $\mathcal{I}\subset\mathcal{O}_{Y}$ be an
ideal. Consider the $\mathcal{I}$-adic filtration on $\mathcal{O}_{Y}$. For
$f\in\mathcal{O}_{Y}$, $\mathrm{ord}_{\mathcal{I}}(f)=:n$ denotes the largest
integer $k$ or $\infty$ such that $f\in\mathcal{I}^{k}$. For such $f$, the
image in $\mathrm{gr}_{\mathcal{I}}^{n}\mathcal{O}_{Y}$ is denoted by
$\sigma(f)$ if $n$ is an integer and $0$ if $n=\infty$. This defines a map
$\sigma_{Y}\colon\mathcal{O}_{Y}\rightarrow\mathrm{gr}_{\mathcal{I}}(\mathcal{O}_{Y})$,
which is merely a map of sets. By [EGA $0_{\mathrm{I}}$, 7.3.7], the
$\mathcal{I}$-adic filtration is separated if $Y$ is integral and
$\mathcal{I}\neq\mathcal{O}_{Y}$ since $1+\mathcal{I}$ does not contain $0$.
In this case, this in particular implies that $\sigma(f)\neq 0$ unless $f=0$.
The following claims are easy to check:

  * •

Let $Y\rightarrow\mathrm{Spec}(k)$ be a morphism of finite type between
irreducible schemes, and $y$ a non-constant function on $Y$. If
$h\in\mathcal{O}_{Y}$ is a constant function which is non-nilpotent, then
$\mathrm{ord}_{y\mathcal{O}_{Y}}(hg)=\mathrm{ord}_{y\mathcal{O}_{Y}}(g)$ for
any function $g$. This is because $h$ is invertible.

  * •

Let $\mathcal{I}\subset\mathcal{O}_{X}$ be an ideal, and
$S^{\prime}\rightarrow S$ be a morphism. Put $\rho\colon
X^{\prime}:=X\times_{S}S^{\prime}\rightarrow X$. If
$\mathrm{gr}_{\mathcal{I}}\mathcal{O}_{X}$ is $\mathcal{O}_{S}$-flat, then we
have a canonical isomorphism
$\mathrm{gr}_{\mathcal{I}\mathcal{O}_{X^{\prime}}}\mathcal{O}_{X^{\prime}}\cong\mathrm{gr}_{\mathcal{I}}\mathcal{O}_{X}\otimes_{\mathcal{O}_{S}}\mathcal{O}_{S^{\prime}}$.
Indeed, the direct summand $\mathcal{I}^{n-1}/\mathcal{I}^{n}$ is
$\mathcal{O}_{S}$-flat, thus $\mathcal{O}_{X}/\mathcal{I}^{n}$ is flat as
well, and
$\mathcal{I}^{n}\otimes_{\mathcal{O}_{S}}\mathcal{O}_{S^{\prime}}\rightarrow\mathcal{I}^{n}\mathcal{O}_{X^{\prime}}$
is an isomorphism.

  * •

We keep the notation. Let $g$ be a function on $X$. If
$\rho^{*}\sigma_{X}(g)\in\mathrm{gr}_{\mathcal{I}}\mathcal{O}_{X}\otimes_{\mathcal{O}_{S}}\mathcal{O}_{S^{\prime}}$
does not vanish, then $\rho^{*}\sigma_{X}(g)=\sigma_{X^{\prime}}\rho^{*}(g)$
via the previous isomorphism.

Put $\mathcal{I}:=x\mathcal{O}_{X}$. We may replace $X$ and $S$ by its
underlying reduced schemes, and assume $X$ and $S$ to be integral. Invoking
[EGA IV, 9.7.7], by shrink $S$ further, we may assume that each fiber of $\pi$
is geometrically integral. Since $X$ is assumed integral and $V(x)$ is non-
empty, the $\mathcal{I}$-adic filtration on $\mathcal{O}_{X}$ is separated.
This implies that the symbol map $\sigma$ is injective. Moreover, since
$V(x)\neq X$, $x$ is non-zero. Thus, $\mathrm{Ann}(x)=0$, and the
multiplication by $\sigma(x)$ on $\mathrm{gr}_{\mathcal{I}}\mathcal{O}_{X}$ is
injective. Shrinking $S$, we may assume that
$\mathrm{Spec}_{X}(\mathrm{gr}_{\mathcal{I}}\mathcal{O}_{X})\rightarrow S$ is
flat by [EGA IV, 6.9.1]. Note that, since $V(x)\rightarrow S$ is assumed
surjective, the $\mathcal{I}$-adic filtration on $\mathcal{O}_{X}$ remains
separated even after shrinking $S$.

Now, write $g=g_{1}/g_{2}$ with $g_{i}\in\mathcal{O}_{X}$. Since $g$ is not
$0$, we have $\sigma(g_{i})\neq 0$ by the injectivity of $\sigma$. By
shrinking $S$ further, we may assume that $\bigl{(}\sigma(g_{i})\bigr{)}_{s}$
does not vanish for any $s\in S$. Take
$N:=\max\bigl{\\{}0,\mathrm{ord}_{\mathcal{I}}(g_{2})-\mathrm{ord}_{\mathcal{I}}(g_{1})\bigr{\\}}$.
Since $(x^{n}g_{1}/g_{2})_{s}$ is not $0$ by the assumption that $\pi$ is
fiberwise geometrically integral, if $(x^{n}g_{1}/g_{2})_{s}$ is constant, we
have

($\star$)
$\mathrm{ord}_{\mathcal{I}_{s}}\sigma\bigl{(}(x^{n}g_{1})_{s}\bigr{)}=\mathrm{ord}_{\mathcal{I}_{s}}\sigma(g_{2,s}).$

On the other hand,
$\mathrm{ord}_{\mathcal{I}}(x^{n}g_{1})=n+\mathrm{ord}_{\mathcal{I}}(g_{1})$.
Since $\bigl{(}\sigma(g_{i})\bigr{)}_{s}$ does not vanish for any $s\in S$, we
have
$\mathrm{ord}_{\mathcal{I}}(g_{i})=\mathrm{ord}_{\mathcal{I}_{s}}(g_{i,s})$.
Thus, ($\star$ ‣ 1.1) cannot happen for $n>N$, and we have the lemma. ∎

### 1.6

Let $\pi\colon X\rightarrow S$ be a morphism. A set of functions
$\mathfrak{h}:=\\{h_{i}\\}_{i\in I}$ in $\mathcal{O}_{X}$ is said to be a non-
constant family on $X$ over $S$ if for any point $s\in S$, there exists $i\in
I$ such that $h_{i}|_{Y}$ is non-constant for any irreducible component $Y$ of
$X_{s}$ of dimension $>0$, and $h\in\mathfrak{h}$ implies
$h^{n}\in\mathfrak{h}$ for any integer $n>0$.

######  Remark. —

In the definition, let $X^{\prime}_{s}\subset X_{s}$ be the union of
irreducible components of dimension $>0$. Caution that
$h_{i}|_{X^{\prime}_{s}}$ is a non-constant function (cf. Definition 1.2.1),
but $h_{i}|_{X_{s}}$ is not if $X^{\prime}_{s}\neq X_{s}$. For example, if
$\pi$ is a finite morphism, then any family of functions is non-constant
family, but no function is non-constant.

######  1.7 Lemma. —

  1. 1.

Let $S$ be a scheme of finite type over $k$, and let $\pi\colon X\rightarrow
Y$ be a finite surjective morphism between flat $S$-schemes of finite type
which are equidimensional over $k$. If $\mathfrak{h}$ is a non-constant family
on $Y$, so is $\pi^{*}\mathfrak{h}$.

  2. 2.

Let $X$ be an affine scheme of finite type over $k$. Let $\Omega/k$ be a field
extension, and assume we are given a closed subscheme $Z\subset X_{\Omega}$
such that each irreducible component is of dimension $>0$. Then there exists a
function $f\in\mathcal{O}_{X}$ such that its pullback to $Z$ is non-constant.

  3. 3.

Let $X\rightarrow S$ be a morphism of finite type between affine schemes. For
any morphism $S^{\prime}\rightarrow S$, and any closed subscheme $Z\subset
X\times_{S}S^{\prime}$, the pullback of $\Gamma(X,\mathcal{O}_{X})$ is a non-
constant family on $Z$.

###### Proof.

Let us show 1. Take $s\in S$. Since $X$ and $Y$ are flat over $S$,
$X_{s}\rightarrow Y_{s}$ is a finite morphism between equidimensional
$k(s)$-schemes with the same dimension by [EGA IV, 14.2.4]. Thus, this is
maximally dominant. Since the pullback of a non-constant function by a
maximally dominant morphism remains to be non-constant by Definition 1.2.1,
the claim follows.

Let us show 2. By limit argument, we may assume that $\mathrm{Spec}(\Omega)$
is the generic point of an integral affine scheme $W$ of finite type over $k$
and there exists a closed subscheme $\widetilde{Z}\subset X_{W}$ such that
$\widetilde{Z}_{\Omega}=Z$ and $\widetilde{Z}$ is maximally dominant over $W$.
Let $\\{\widetilde{Z}_{i}\\}$ be the set of irreducible components of
$\widetilde{Z}$. By shrinking $W$, we may assume that
$\widetilde{Z}_{i}\cap\widetilde{Z}_{j}$ is either faithfully flat over $W$ or
empty for any $i,j$ by [EGA IV, 6.9.1]. For any $w\in W$, any irreducible
component of $\widetilde{Z}_{i,w}$ is an irreducible component of
$\widetilde{Z}_{w}$. Indeed, let $Y\subset\widetilde{Z}_{i,w}$ be an
irreducible component. If $Y$ were not an irreducible component of
$\widetilde{Z}_{w}$, we could take an irreducible component $Y^{\prime}$ of
$\widetilde{Z}_{w}$ which contains $Y$. There exists $j$ such that
$Y^{\prime}$ is contained in $\widetilde{Z}_{j}$. Since $Y^{\prime}\neq Y$,
$i$ and $j$ are not the same. By using [EGA IV, 14.2.4], we have
$\dim(\widetilde{Z}_{i,\eta})=\dim(Y)\leq\dim(\widetilde{Z}_{i}\cap\widetilde{Z}_{j})_{\eta}<\dim(\widetilde{Z}_{i,\eta})$,
which is a contradiction, and $Y$ is an irreducible component of
$\widetilde{Z}_{w}$.

Now, fix a closed point $w\in W$, and assume that there exists a function
$f\in\mathcal{O}_{X_{W}}$ such that $f|_{\widetilde{Z}_{w}}$ is non-constant.
For each $i$, we take a point $z_{i}\in\widetilde{Z}_{i,w}$ which is a generic
point of $\widetilde{Z}_{w}$. Since
$f|_{\widetilde{Z}_{w}}\colon\widetilde{Z}_{w}\rightarrow\mathbb{A}^{1}_{k(w)}$
is maximally dominant, this is flat around $z_{i}$. Then, since
$\widetilde{Z}\rightarrow W$ is assumed flat,
$\widetilde{Z}\rightarrow\mathbb{A}^{1}_{W}$ is flat around $z_{i}$ by [EGA
IV, 11.3.10]. Thus $f$ is flat around the generic point of $\widetilde{Z}_{i}$
by [EGA IV, 11.1.1], and $f$ is non-constant over $\Omega$. Since $X_{W}$ is
affine, any function on $X_{w}$ extends to a function on $X_{W}$, and thus, it
suffices to find a function $g$ on $X_{w}$ such that $g|_{\widetilde{Z}_{w}}$
is non-constant. In particular, we may assume that $\Omega$ is an algebraic
finite extension of $k$. By taking the image of $Z$ to $X$, we may assume that
$\Omega=k$. Since $X$ is affine, any function on $Z$ extends to $X$, and thus,
we may even assume that $Z=X$. Now, let $X=\bigcup_{1\leq i\leq N}X_{i}$ be
the decomposition into irreducible components, and take a closed immersion
$X\hookrightarrow\mathbb{A}^{n}_{k}$ such that $n>N$. For each $i$, let
$l_{i}\subset\mathbb{A}^{n}_{k}$ be a line (i.e. $1$-dimensional irreducible
closed subscheme defined by polynomials of degree $1$ with coefficients in
$k$) contained in $X_{i}$ if it exists, and $\emptyset$ it does not. Let
$l_{i,0}$ be the parallel transform of $l_{i}$ which contains the origin.
Since $n>N$, we may take a hyperplane $H\subset\mathbb{A}^{n}$ passing through
the origin which does not contain any of $l_{i,0}$. The function defining $H$
is a function we are looking for.

Let us show the last claim. We may assume that
$S^{\prime}=\mathrm{Spec}(\Omega)$ is a point for some field $\Omega$. Since
$S$ is assumed affine, for any closed subscheme $W\subset S$, the map
$\Gamma(S,\mathcal{O}_{S})\rightarrow\Gamma(W,\mathcal{O}_{W})$ is surjective.
Thus, we may assume that $S$ is integral and the image of
$\mathrm{Spec}(\Omega)$ is the generic point $\eta$ of $S$. By 2, we may find
a function $f\in\Gamma(X,\mathcal{O}_{X})\otimes\mathcal{O}_{S,\eta}$ such
that $f$ is non-constant on each irreducible component of positive dimension
of $Z$. There exists $g\in\Gamma(S,\mathcal{O}_{S})^{\times}$ such that $gf$
is defined on $\Gamma(X,\mathcal{O}_{X})$. By construction,
$gf\in\Gamma(X,\mathcal{O}_{X})$ is non-constant on each irreducible component
of positive dimension of $Z$, and we conclude the proof. ∎

######  1.8 Lemma. —

Let $X\rightarrow S$ be a smooth morphism between excellent (noetherian)
schemes of finite dimension. Then there exists a finite surjective morphism
$T\rightarrow S$ such that

> (*) for any $x\in X_{T}:=X\times_{S}T$, there exists an open neighborhood
> $U$ such that each fiber of $U\rightarrow T$ is geometrically irreducible.

###### Proof.

First, we note that the following holds:

  * •

If (*) holds for $T$, then it also holds for any $T^{\prime}\rightarrow T$.

  * •

Let $T^{\prime}=\bigcup T^{\prime}_{i}$ where $T^{\prime}_{i}$ is an
irreducible component. If (*) holds for $T=T^{\prime}_{i}$, then (*) holds for
$T^{\prime}$.

For the second claim, we note the following: Let $Y$ be a scheme and $Y_{i}$
be an irreducible component. Take $y\in\bigcap_{i\in I}Y_{i}$ for some finite
set $I$. Assume we are given an open neighborhood $V_{i}\subset Y_{i}$ of $y$
for each $i\in I$. Put $V:=Y\setminus\bigcup_{i\in I}(Y_{i}\setminus V_{i})$.
Then $V$ is a neighborhood of $y$ in $Y$ and $V\cap Y_{i}\subset V_{i}$ for
each $i\in I$.

Since the dimension of $S$ is finite, we show the claim by the induction on
the dimension of $S$. Assume the claim holds for $\dim(S)<n$, and assume now
that $\dim(S)=n$. Consider the finite morphism $\coprod
S_{i,\mathrm{red}}\rightarrow S$, where $\\{S_{i}\\}$ is the set of
irreducible components, and replacing $S$ by $\coprod S_{i}$, we may assume
that $S$ is integral with generic point $\eta$. There exists a finite
extension $L$ of $k(\eta)$ such that the number of irreducible components of
$X_{L}$ is the same as that of geometric irreducible components. We replace
$S$ by the normalization of $S$ in $L$, we may assume that $S$ is normal and
each irreducible component of the generic fiber is geometrically irreducible.
By considering the connected component, we may assume that $X$ is connected.

In this situation, let us show that the generic fiber is geometrically
irreducible. Let $\\{X_{\eta,i}\\}$ be the set of irreducible components of
$X_{\eta}$. Let $X_{i}$ be the closure of $X_{\eta,i}$, and put
$X_{ij}:=X_{i}\cap X_{j}$. Assume $X_{ij}\cap X_{t}\neq\emptyset$ for some
$t\in S$. Take a point $x$ in the intersection. Then $\mathcal{O}_{X,x}$ has
more than one minimal primes, so it cannot be geometrically unibranch.
However, since $\mathcal{O}_{S,t}$ is assumed to be normal, it is
geometrically unibranch (cf. [EGA $0_{\mathrm{IV}}$, 23.2.1]) this contradicts
with [EGA IV, 17.5.7]. Thus, $X_{ij}$ does not intersect with $X_{t}$ for any
$t\in S$. This implies that $X_{i}$ and $X_{j}$ are disjoint. Since $X$ is
assumed connected, this can happen only when $X_{i}=X_{j}$, thus $X_{\eta}$ is
irreducible.

Since $X_{\eta}$ is geometrically irreducible, there exists an open dense
subscheme $U\subset S$ such that for any $s\in U$, $X_{s}$ is geometrically
irreducible by [EGA IV, 9.7.8]. Since the dimension of $Z:=S\setminus U$ is
$<n$, we may use the induction hypothesis, and we have a finite dominant
morphism $W\rightarrow Z$ over which (*) is satisfied. Use Gabber’s lemma [O2,
1.6.2] to produce a finite dominant morphism $S^{\prime}\rightarrow S$ such
that each irreducible component of $Z^{\prime}:=S^{\prime}\times_{S}Z$ maps to
$W$. Thus (*) holds for $Z^{\prime}$ by using the remarks at the beginning of
this proof. Take any point $x\in X^{\prime}:=X\times_{S}S^{\prime}$. If $x$ is
not over $Z^{\prime}$, we can just take the complement of
$X^{\prime}\times_{S^{\prime}}Z^{\prime}\subset X^{\prime}$ since each fiber
of $X_{U}\rightarrow U$ is geometrically irreducible. Assume $x$ is mapped to
$Z^{\prime}$. Then there exists an open neighborhood $V$ of $x$ in
$X^{\prime}\times_{S^{\prime}}Z^{\prime}$ which is fiberwise geometrically
irreducible over $Z^{\prime}$. Take any extension of $V$ to an open subset in
$X^{\prime}$. Since $X^{\prime}$ is already fiberwise geometrically
irreducible outside of $Z^{\prime}$, this neighborhood is what we need, and
(*) is satisfied for $X^{\prime}\rightarrow S^{\prime}$. ∎

### 1.9

For terminologies concerning valuations, we follow [V]. Let $v$ be a valuation
on an integral ring $A$ over a finite field $\mathbb{F}_{p}$. Then $v$ extends
uniquely to the perfection
$A^{\mathrm{perf}}:=A^{1/p^{\infty}}:=\mathop{\underrightarrow{\mathrm{lim}}}A^{1/p^{n}}$.
Abusing notations, we denote the extension also by $v$.

Let $A$ be an integral noetherian ring, and $v$ be a valuation on it. Then it
is known that the value semigroup is well-ordered by [V, Prop 9.1]. We
moreover have the following result which can be found in Zariski-Samuel’s
book. For the reader’s convenience, we recall also the proof.

######  Lemma. —

Let $A$ be an integral noetherian ring with its field of fractions $K$, and
let $v$ be a height $1$ valuation on $K$ centered on $A$. Then the value
semigroup $\Gamma:=v(A\setminus\\{0\\})\subset\mathbb{R}_{\geq 0}$ is
discrete.

###### Proof.

Let $\mathfrak{p}\subset A$ be the center of $A$. Since the value semigroup of
$A$ is contained in that of $A_{\mathfrak{p}}$, we may assume that
$\mathfrak{p}$ is a maximal ideal of $A$. For $r\in\mathbb{R}_{\geq 0}$, let
$I_{r}:=\bigl{\\{}a\in A\mid v(a)\geq r\bigr{\\}}$, which is an ideal of $A$.
The semigroup $\Gamma$ is well-ordered as we remarked above. Take the minimal
value $\gamma\in\Gamma\setminus\\{0\\}$. Then $\mathfrak{p}=I_{\gamma}$. Take
any $\rho\in\mathbb{R}_{\geq 0}$ greater than $\gamma$. We may find an integer
$N$ such that $\rho<N\gamma$. Then we have $\mathfrak{p}^{N}\subset I_{\rho}$.
Since $\mathfrak{p}$ is assumed to be maximal, the quotient
$A/\mathfrak{p}^{N}$ is an Artinian ring. This implies that there can only be
finitely many distinct ideals in the set $\\{I_{r}\\}_{\gamma\leq r\leq\rho}$,
and thus $\Gamma\cap[\gamma,\rho]$ is a finite set. ∎

######  Corollary. —

Let $K$ be a field, $v$ be a valuation on $K$ of height $1$, and $R$ be a
noetherian subring of $K$ such that $v(R)\geq 0$. Let $M\subset K$ be a
finitely generated $R$-submodule. Then $v(M\setminus\\{0\\})\subset\mathbb{R}$
is a discrete subset.

###### Proof.

Write $M=\sum_{i\in I}R\cdot\alpha_{i}$ where $\alpha_{i}\in K$. Since $M$ is
finitely generated, we may take $I$ to be a finite set. Thus, there exists
$\beta\in K^{\times}$ such that $v(\beta\cdot\alpha_{i})\geq 0$ for any $i\in
I$. Consider $R^{\prime}:=R\left[\\{\beta\cdot\alpha_{i}\\}_{i\in
I}\right]\subset K$. This is a noetherian ring such that
$v(R^{\prime}\setminus\\{0\\})\geq 0$ and contains $\beta\cdot M$. Invoking
the lemma above, $v(R^{\prime}\setminus\\{0\\})$ is a discrete subset of
$\mathbb{R}_{\geq 0}$. Thus, $v(M\setminus\\{0\\})=v(\beta\cdot
M\setminus\\{0\\})-v(\beta)\subset v(R^{\prime}\setminus\\{0\\})-v(\beta)$ is
discrete. ∎

##  2 Analysis around type 4 points

### 2.1

In this section, we fix an algebraically closed complete non-archimedean
valuation field $\ell$ of characteristic $p>0$ with non-trivial valuation $v$.
We denote by $\mathfrak{o}_{\ell}$ the ring of integers of $\ell$. We,
moreover, fix a subring $R\subset\mathfrak{o}_{\ell}$, which is assumed
noetherian. This implies that the value semigroup $v(R)$ is a discrete subset
of $\mathbb{R}_{\geq 0}$ by Lemma Lemma. The Tate algebra over $\ell$ is
denoted by $\ell\\!\left<t\right>$, and
$R\\!\left<t\right>:=R[\\![t]\\!]\cap\mathfrak{o}_{\ell}\\!\left<t\right>$.

### 2.2

A Berkovich point $\alpha$ of the Tate algebra $\ell\\!\left<t\right>$ is a
multiplicative semi-norm $|\cdot|_{\alpha}$ such that $\leq\|\cdot\|$, where
$\|\cdot\|$ is the $0$-Gauss norm. However, in order to keep accordance with
the notations for Riemann-Zariski space, we use valuation theoretic notation,
and put $v_{\alpha}(f):=\log(|f|_{\alpha})$. This is possible since there is
no archimedean semi-norm on $\ell\\!\left<t\right>$. Beware that, since points
of Berkovich space are merely semi-norms, the valuation
$v_{\alpha}\colon\ell\\!\left<t\right>\rightarrow\mathbb{R}_{\infty}$
corresponding to a point $\alpha$, where
$\mathbb{R}_{\infty}:=\mathbb{R}\sqcup\\{\infty\\}$, may have non-trivial
preimage of $\infty$.

For $\omega\in\ell$ and $r\in\left[0,\infty\right]$, we define a semi-norm
$v_{\omega,r}$ by

$v_{\omega,r}\bigl{(}\sum
a_{i}(t-\omega)^{i}\bigr{)}:=\min\bigl{\\{}v(a_{i})+ir\bigr{\\}}.$

This semi-norm is multiplicative, and defines a point of $\mathbb{D}_{\ell}$.
This point is denoted by $\alpha_{\omega,e^{-r}}$ in [Ke4]. Let $\alpha$ be a
Berkovich point on $\ell\\!\left<t\right>$. We put
$r_{\alpha}:=-\log(r(\alpha))$ in $\mathbb{R}_{\infty}$, where $r(\alpha)$ is
the radius of $\alpha$ (cf. [Ke4, 2.2.11]). For $s\in[0,r_{\alpha}]$, we have
the valuation $v_{\alpha,s}$ corresponding to the point $\alpha_{e^{-s}}$.
Note that $v_{\alpha,0}$ corresponds to the Gauss point for any $\alpha$.
These points can be depicted in the following figure.

$v_{\alpha,r_{\alpha}}$$v_{\alpha,0}=v_{0,0}$Gauss$\alpha$

For $f\in\ell\\!\left<t\right>$, we consider the function

$\mathrm{VP}_{\alpha}(f)\colon\left[0,r_{\alpha}\right]\rightarrow\mathbb{R};\quad
s\mapsto v_{\alpha,s}(f).$

This function $\mathrm{VP}_{\alpha}(f)$ is continuous piecewise affine,
convex, and there are only fintely many breaks. Indeed, we may assume that
$\ell$ is spherically complete. In this case, we may assume that
$\alpha=v_{\omega,r}$ for some $\omega\in\ell$ and $r$. Then
$\mathrm{VP}_{\alpha}(f)$ is nothing but (a truncation of) the valuation
polygon of $f(z+\omega)$ around $v_{0,\infty}$. Now, let $g=\sum
a_{i}t^{i}\in\ell\\!\left<t\right>$. Then there exists $n$ such that
$v(a_{i})>v(a_{n})$ for any $i>n$. Then we have
$\mathrm{VP}_{v_{0,\infty}}(g)=\mathrm{VP}_{v_{0,\infty}}(\sum_{i\leq
n}a_{i}t^{i})$. The latter function obviously satisfies the claim, and so is
$\mathrm{VP}_{\alpha}(f)$. This function is called the valuation polygon at
$\alpha$.

######  2.3 Lemma. —

Let $R^{\prime}\subset\mathfrak{o}_{\ell}$ be a noetherian ring, and
$\\{f_{i}\\}_{i\in I}$ be a set of functions in $R^{\prime}\\!\left<t\right>$.
Then the function $V(s):=\inf_{i\in
I}\bigl{\\{}\mathrm{VP}_{\alpha}(f_{i})(s)\bigr{\\}}$ is continuous piecewise
affine, convex, and there are only fintely many breaks.

###### Proof.

Since the slopes of $\mathrm{VP}(f_{i})(s)$ are non-negative integers and the
slope decreases as $s$ goes to the infinity, the lemma holds over
$\left[\varepsilon,\infty\right[$ for any $\varepsilon>0$. It suffices to
check that the infimum has “initial slope” around $0$, namely there exists
$\epsilon>0$ and $d$, $a$ such that $V(s)=ds+a$ for any $s\in[0,\epsilon]$.
Because $R^{\prime}$ is assumed notherian,
$D:=v(R^{\prime})\subset\left[0,\infty\right[$ is discrete by Lemma Lemma.
Thus, the set $\\{v_{0}(f_{i})\\}$ has the minimum. Let $a$ be the minimum.
Then $V(0)=a$. Let $J$ be the set of $i\in I$ such that
$\mathrm{VP}(f_{i})(0)=a$. Since the slopes are integer, there exists the
minimum in the set of initial slopes of $\mathrm{VP}(f_{i})$ for $i\in J$. Let
$d$ be the slope. Let $a^{\prime}$ be the element in $D$ next to $a$. Then we
may check that $V(s)=ds+a$ for $s\in\left[0,(a^{\prime}-a)/s\right]$. Thus the
lemma follows. ∎

### 2.4

Now, we fix a Berkovich point $\alpha$, and we omit the index $\alpha$ from
$\mathrm{VP}_{\alpha}(f)$. Let us fix an element $\pi\in R$, and let
$\mathbf{a}:=\\{a_{k}\\}_{k\geq 0}$ be a sequence of elements in
$R\\!\left<t\right>$. Valuations on $\ell\\!\left<t\right>$ extends uniquely
to that of $\ell\\!\left<t\right>^{\mathrm{perf}}$, and we identify the
Berkovich space of $\ell\\!\left<t\right>$ and
$\ell\\!\left<t\right>^{\mathrm{perf}}$. Now, for $n\geq 0$, we put

$g^{(n)}(\mathbf{a}):=\sum_{k=0}^{n}(a_{k}/\pi)^{1/p^{k}},\quad\mathrm{VP}^{(n)}(\mathbf{a}):=\min_{n}\bigl{\\{}0,\mathrm{VP}(g^{(n)}(\mathbf{a}))\bigr{\\}}.$

For any $\varepsilon>0$, there exists $N$ such that
$-v(\pi)/p^{N}>-\varepsilon$. This implies that

(2.4.1)
$\min\bigl{\\{}-\varepsilon,\mathrm{VP}^{(n)}(\mathbf{a})\bigr{\\}}=\min\bigl{\\{}-\varepsilon,\mathrm{VP}^{(N)}(\mathbf{a})\bigr{\\}}\leq
0$

for any $n\geq N$. Thus,

$\mathrm{VP}^{(\infty)}(\mathbf{a}):=\lim_{n\rightarrow\infty}\mathrm{VP}^{(n)}(\mathbf{a})$

converges. Now, let $I$ be a set, and for $i\in I$, assume we are given a
sequence $\mathbf{a}_{i}:=\\{a_{i,k}\\}_{k\geq 0}$ in $R\\!\left<t\right>$. We
put

$\mathrm{VP}^{(\infty)}(\\{\mathbf{a}_{i}\\}_{i\in I}):=\inf_{i\in
I}\big{\\{}\mathrm{VP}^{(\infty)}(\mathbf{a}_{i})\\}.$

The infimum is in fact the minimum. Indeed, for any $N>0$, we have
$v\bigl{(}\sum_{N\geq n\geq 0}(R/\pi)^{1/p^{n}}\bigr{)}\subset
v(R^{1/p^{N}}/\pi)$. Since $R^{1/p^{N}}$ is noetherian, this is a discrete set
by Lemma Lemma. Thus, by (2.4.1), $v\bigl{(}\sum_{n\geq
0}(R/\pi)^{1/p^{n}}\bigr{)}\cap\left]-\infty,0\right[$ is a discrete subset,
and the minimum exists.

######  2.5 Lemma. —

The function $\mathrm{VP}^{(\infty)}(\\{\mathbf{a}_{i}\\}_{i\in I})(s)$ is
continuous piecewise affine and convex. Furthermore, there are only fintely
many breaks. If $\alpha$ is a type 1 point, then the terminal slope is $0$, in
other words the function is constant for $s\gg 0$.

###### Proof.

We may assume that $\ell$ is spherically complete. In this case, we may write
$\alpha=v_{\omega,r}$. Since $\mathrm{VP}(f)$ is the same as the valuation
polygon of $f(t+\omega)$ around $v_{0,\infty}$ (and truncate to $[0,r]$) for
any $f\in\\!\ell\left<t\right>^{1/q}$ where $q$ is some power of $p$, by
replacing $R$ by $R[\omega]\subset\mathfrak{o}_{\ell}$, which is indeed
noetherian, we may assume that $\alpha=v_{0,\infty}$. Note that if $\alpha$ is
a point of type 1 equal to $v_{\omega,\infty}$, then
$\mathrm{VP}_{\alpha}(f(t))=\mathrm{VP}_{v_{0,\infty}}(f(t+\omega))$ and we do
not need truncation. This implies that the claim for type 1 point is also
reduced to the case $\alpha=v_{0,\infty}$. By (2.4.1) together with Lemma 2.3
applying for $R^{\prime}=R^{1/p^{N}}$,
$\bigl{(}\mathrm{VP}^{(\infty)}(\mathbf{a})\bigr{)}^{-1}(\left]-\infty,-\varepsilon\right[)$
is continuous piecewise affine and convex for any $\varepsilon>0$. Thus, we
only need to check that the breaks are finite and the terminal slope is $0$.

Let
$C_{i}:=\lim_{n\rightarrow\infty}\min\bigl{\\{}0,v(\sum_{k=0}^{n}(a_{i,k}(0)/\pi)^{1/p^{k}})\bigr{\\}}$.
This limit converges since we have
$\lim_{k\rightarrow\infty}\min\bigl{\\{}0,v(a_{i,k}(0)/\pi)/p^{k}\bigr{\\}}=0$.
Write $a_{i,k}=a^{\prime}_{i,k}+a_{i,k}(0)$, and put
$\mathbf{a}^{\prime}_{i}:=\\{a^{\prime}_{i,k}\\}_{k}$. Since
$\mathrm{VP}(a^{\prime}_{i,k})$ does not contain segments of slope $0$ for any
$i,k$, we have

$\mathrm{VP}^{(\infty)}(\mathbf{a}_{i})=\min\bigl{\\{}\mathrm{VP}^{(\infty)}(\mathbf{a}^{\prime}_{i}),C_{i}\bigr{\\}}.$

Thus, we may assume that $a_{i,k}(0)=0$ for any $k$, namely $a_{i,k}$ does not
have the contant term.

Fix a point $\gamma\in\left]0,\infty\right[$ such that
$\mathrm{VP}^{(\infty)}(\mathbf{a})(\gamma)<0$. If we do not have such a
point, $\mathrm{VP}^{(\infty)}(\mathbf{a})=0$, and there is nothing to prove.
On the interval $[0,\gamma]$, the function
$\mathrm{VP}^{(\infty)}(\mathbf{a})$ has only finitely many breaks because of
(2.4.1) and Lemma 2.3 (applied for $R^{\prime}=R^{1/p^{N}}$). Thus, we only
need to check the finiteness for $\left[\gamma,\infty\right[$. There exists an
integer $N>0$ such that $(N+1)\gamma>v(\pi)$. For $g=\sum_{i\geq
0}g_{i}t^{i}\in R\\!\left<t\right>$, denoted by $g^{\leq N}:=\sum_{N\geq i\geq
0}g_{i}t^{i}\in R[t]$, the $N$-truncated polynomial. Since $g_{i}\in R$ and
thus $v(g_{i})\geq 0$ and $\alpha=v_{0,\infty}$, we have
$\min\\{0,v_{s}(g/\pi)\\}=\min\\{0,v_{s}(g^{\leq N}/\pi)\\}$ for any
$s\in\left[\gamma,\infty\right[$. Thus, by putting $\mathbf{a}^{\leq
N}_{i}:=\\{a_{i,k}^{\leq N}\\}_{k}$, we have
$\mathrm{VP}^{(\infty)}(\\{\mathbf{a}_{i}\\})=\mathrm{VP}^{(\infty)}(\\{\mathbf{a}_{i}^{\leq
N}\\})$ on $\left[\gamma,\infty\right[$. Thus, from now on, we may and do
assume that $a_{i,k}$ is a polynomial without constant term of degree $\leq N$
for any $i\in I$ and $k\geq 0$.

Now, put $R^{\prime}:=R^{1/p^{\lceil\log_{p}(N)\rceil}}$, and let
$\mathbb{N}^{\prime}$ be the set of natural numbers $0<n\leq N$ such that
$p\nmid n$. Denote by $[1,N]_{\mathbb{N}}:=[1,N]\cap\mathbb{N}$. Let $P$ be
the (non-empty) set of monomials $\alpha t^{i}$ ($i\in\mathbb{Q}$) in
$\ell[t]^{\mathrm{perf}}$ such that

  1. 1.

$i$ is either in $p^{-k}\mathbb{N}^{\prime}$ for some positive integer $k$, or
in $[1,N]_{\mathbb{N}}$. In the latter case, put $k=0$.

  2. 2.

$\alpha\in(\pi^{-1}R^{\prime})^{1/p^{k}}$ and $v(\alpha)+i\gamma<0$.

Now, let $\\{a_{m}\\}_{m}$ be a sequence of $a_{m}\in R[t]$ of degree $\leq N$
and $a_{m}(0)=0$. Let $n\geq 0$, and write $g^{(n)}(\\{a_{m}\\})=\sum_{i\in
p^{-n}\mathbb{N}}g_{i}t^{i}$ with $g\in\ell^{\mathrm{perf}}$. We claim that
either $g_{i}t^{i}\in P$ or $v(g_{i})+i\gamma\geq 0$. Indeed, we may assume
$v(g_{i})+i\gamma<0$, and we must show that $g_{i}t^{i}$ belongs to $P$. By
construction of $g^{(n)}$ and the assumption that there is no constant term in
$a_{m}$, the coefficient $g_{i}$ is non-zero only if

$i\in p^{-n}\mathbb{N}^{\prime}\sqcup p^{-n+1}\mathbb{N}^{\prime}\dots
p^{-1}\mathbb{N}^{\prime}\sqcup[1,N]_{\mathbb{N}}.$

We need to check that for $i\in p^{-k}\mathbb{N}^{\prime}$ (resp.
$i\in[1,N]_{\mathbb{N}}$), $g_{i}\in(\pi^{-1}R^{\prime})^{1/p^{k}}$ (resp.
$g_{i}\in\pi^{-1}R^{\prime}$). Indeed, for $i\in[1,N]_{\mathbb{N}}$, since the
degree of $a_{m}$ is bounded by $N$, there is no contribution to $g_{i}t^{i}$
from $(a_{m}/\pi)^{1/p^{m}}$ if $m>\lceil\log_{p}(N)\rceil$. This implies that

$g_{i}\in\pi^{-1}R+(\pi^{-1}R)+\dots+(\pi^{-1}R)^{1/p^{\lceil\log_{p}(N)\rceil}}\subset\pi^{-1}R^{\prime}.$

We can check similarly for $i\in p^{-k}\mathbb{N}^{\prime}$.

Now, we claim that there exists a (non-empty) finite set $L$ of (half) lines
with positive integral slope in $\left[\gamma,\infty\right[\times\mathbb{R}$
such that $\mathrm{VP}(\alpha t^{i})\in\bigcup_{i\geq 0}p^{-i}L$ for any
$\alpha t^{i}\in P$. Indeed, let $\alpha
t^{i}\in(\pi^{-1}R^{\prime})^{1/p^{k}}t^{i}$ in $P$. Then we have

$\mathrm{VP}(\alpha t^{i})(s)=p^{-k}\times\mathrm{VP}\bigl{(}(\alpha
t^{i})^{p^{k}}\bigr{)}(s),$

so we only need to construct $L$ such that $\mathrm{VP}(\alpha t^{i})\in L$
for any $\alpha t^{i}\in P$ and $i\in\mathbb{N}$. We show that the set

$L:=\Bigl{\\{}\mathrm{VP}(\alpha
t^{i})\subset\left[\gamma,\infty\right[\times\mathbb{R}\,\big{|}\,\mbox{(*)\,$\alpha
t^{i}\in P$ and $i\in\mathbb{N}$}\Bigr{\\}}$

is finite. Since $R^{\prime}$ is noetherian, the set
$v(R^{\prime})-v(\pi)+\mathbb{N}_{\leq N}\cdot\gamma$ is discrete. We have

$\mathrm{VP}(\alpha
t^{i})(\gamma)\in\bigl{\\{}v(R^{\prime})-v(\pi)+\mathbb{N}_{\leq
N}\gamma\bigr{\\}}\cap\left]-\infty,0\right]=:D$

for any $\alpha t^{i}$ satisfying (*), and the discreteness implies that $D$
is actually a finite set. Any line in $L$ has the endpoint in the set
$\bigl{\\{}(\gamma,d)\mid d\in D\bigr{\\}}$ and the slope is in
$\mathbb{N}_{\leq N}$. There are only $N\times(\\#D)$ such lines, and the
finiteness of $L$ follows.

What we have shown implies that there is a set $S\subset\bigcup_{i\geq
0}p^{-i}L$ such that $\mathrm{VP}^{(\infty)}(\mathbf{a})=\inf_{l\in
S}\\{l,\mathbf{0}\\}$, where
$\mathbf{0}:=\left[\gamma,\infty\right[\times\\{0\\}\subset\left[\gamma,\infty\right[\times\mathbb{R}$.
Since $L$ is a finite set, the set

$Z:=\bigl{\\{}l\cap\mathbf{0}\mid l\in\bigcup
p^{-i}L\bigr{\\}}=\bigl{\\{}l\cap\mathbf{0}\mid l\in
L\bigr{\\}}\subset\mathbb{R}^{2}$

is finite (and non-empty). Let $\gamma_{\max}:=\max\\{\mathrm{pr}_{1}(Z)\\}$,
where $\mathrm{pr}_{1}\colon\mathbb{R}^{2}\rightarrow\mathbb{R}$ is the first
projection. Then for $s\geq\gamma_{\max}$, we have
$\mathrm{VP}^{(\infty)}(\mathbf{a})(s)=0$, thus the terminal slope is $0$.
Since $\mathrm{VP}^{(\infty)}(\mathbf{a})(\gamma)<0$, there exists
$\gamma_{0}\in Z$ such that
$\mathrm{VP}^{(\infty)}(\mathbf{a})(\gamma_{0})=0$. Take
$\gamma_{1}<\gamma_{0}$ such that
$\left(\left[\gamma_{1},\gamma_{0}\right[\times\\{0\\}\right)\cap
Z=(\gamma_{1},0)$ if it exists and $\gamma_{1}=\gamma$ if not. Then
$\mathrm{VP}^{(\infty)}(\mathbf{a})(\left[\gamma_{1},\gamma_{0}\right])$ is a
segment. Indeed, if there were a break on
$\left]\gamma_{1},\gamma_{0}\right[$, then there would be a line in $\bigcup
p^{-i}L$ which has intersection with
$\left]\gamma_{1},\gamma_{0}\right[\times\\{0\\}$, which is impossible by the
choice. Thus there are finitely many breaks on $[\gamma,\gamma_{0}]$, and the
lemma follows. ∎

### 2.6

Let
$f=\sum_{\underline{k}}a_{\underline{k}}\underline{x}^{\underline{k}}\in\ell\\!\left<t\right>[\\![x_{1},\dots,x_{n}]\\!]$,
and let $\alpha\in\mathbb{D}_{\ell}$. Let $\Lambda$ be the set
$\bigl{\\{}\underline{n}\in\mathbb{N}^{n}\setminus\\{\underline{0}\\}\mid
p\nmid\underline{n}\bigr{\\}}$. For $\underline{n}\in\Lambda$, we put
$A_{\underline{n}}:=\\{a_{p^{k}\underline{n}}\\}_{k\geq 0}$, and

$\mathrm{VP}^{(\infty)}_{\alpha}(f/\pi):=\mathrm{VP}^{(\infty)}_{\alpha}(\\{A_{\underline{n}}\\}_{\underline{n}\in\Lambda}).$

We wish to study the behavior of this polygon around terminal points (i.e.
$\alpha$ is either of type 1 or 4) when $f$ is algebraic.

######  Proposition. —

Let $f=\sum a_{\underline{k}}x^{\underline{k}}$ be a formal power series in
$R\\!\left<t\right>[\\![x_{1},\dots,x_{n}]\\!]$ which is algebraic over
$\mathrm{Frac}\left(\ell\\!\left<t\right>[x_{1},\dots,x_{n}]\right)$. Assume
that $\alpha$ is a terminal point. Then there exists $\varepsilon>0$ such that
the function $\mathrm{VP}^{(\infty)}_{\alpha}(f/\pi)(s)$ is constant on
$s\in[r_{\alpha}-\varepsilon,r_{\alpha}]$.

The rest of this section is devoted to proving this proposition.

### 2.7

We start by recalling some basic facts. Till 2.13, the noetherian ring $R$
plays no role. Let $K$ be a field of characteristic $p>0$. A sequence
$\\{x_{k}\\}_{k\geq 0}$ in $K$ is said to satisfy a linear recurrence relation
(resp. reversed linear recurrence relation), abbreviated as LRR (resp. rLRR),
if there exist $c_{0},\dots,c_{n}\in\overline{K}$ such that

$c_{0}x_{k}+c_{1}x_{k+1}^{p}+\dots+c_{n}x_{k+n}^{p^{n}}=0\qquad(\mbox{resp.\
}c_{0}x_{k}^{p^{n}}+c_{1}x_{k+1}^{p^{n-1}}+\dots+c_{n}x_{k+n}=0)$

for any $k\geq 0$. We believe that the following lemma is well-known to
experts. A proof can be found, for example, in [Ke1, Lemma 4].

######  Lemma. —

Let $c_{0},\dots,c_{n}$ be elements in $\overline{K}$ such that
$c_{0},c_{n}\neq 0$.

(i) The roots of the “characteristic polynomial”
$P(X):=c_{0}X+c_{1}X^{p}+\dots+c_{n}X^{p^{n}}$ (resp.
$P(X):=c_{0}X^{p^{n}}+c_{1}X^{p^{n-1}}+\dots+c_{n}X$) form a
$\mathbb{F}_{p}$-vector subspace of $\overline{K}$ of dimension $n$.

(ii) Let $z_{1},\dots,z_{n}$ be a basis of the vector space of (i). Then a
sequence $\\{x_{k}\\}_{k\geq 0}$ in $K$ satisfies LRR (resp. rLRR) with
respect to $c_{0},\dots,c_{n}$ if and only if there exist
$\lambda_{1},\dots,\lambda_{n}$ in $\overline{K}$ such that $x_{k}$ is of the
form

$x_{k}=z_{1}\lambda_{1}^{1/p^{k}}+\dots+z_{n}\lambda_{n}^{1/p^{k}},\qquad\mbox{{(}resp.\
$x_{k}=z_{1}\lambda_{1}^{p^{k}}+\dots+z_{n}\lambda_{n}^{p^{k}}${)}}.$

for any $k\geq 0$.

######  2.8 Lemma. —

Let $K$ be a field of characteristic $p>0$. Let $f=\sum
a_{\underline{k}}\underline{x}^{\underline{k}}$ be a formal power series in
$K[\\![x_{1},\dots,x_{n}]\\!]$ which is algebraic over $K(x_{1},\dots,x_{n})$.
Then for any $\underline{n}\neq\underline{0}$, the series

$h(X):=\sum_{k\geq 0}a_{p^{k}\cdot\underline{n}}X^{p^{k}}$

in $K[\\![X]\\!]$ is algebraic over $K(X)$.

###### Proof.

The argument here is a variant of [SW, Theorem 7.1]. Without loss of
generality, we may assume $n_{1}\neq 0$. Let $g=\sum_{k\geq
0}\underline{x}^{p^{k}\cdot\underline{n}}$. This is algebraic over
$K(x_{1},\dots,x_{n})$ since $g-g^{p}=\underline{x}^{\underline{n}}$. Then

$f*g=\sum_{k\geq
0}a_{p^{k}\cdot\underline{n}}\underline{x}^{p^{k}\cdot\underline{n}}=h(\underline{x}^{\underline{n}}),$

where $*$ denotes the Hadamard product (cf. [SW, Definition 2.1]). Since
algebraicity is preserved by taking Hadamard product by [SW], $f*g$ is
algebraic over $K(x_{1},\dots,x_{n})$. This implies that
$h(\underline{x}^{\underline{n}})=(f*g)(x_{1},\dots,x_{n})$ is algebraic over
$K(X,x_{2},\dots,x_{n})$ where $X=\underline{x}^{\underline{n}}$, since
$K(x_{1},\dots,x_{n})\supset K(X,x_{2},\dots,x_{n})$ is a finite algebraic
extension by the assumption $n_{1}\neq 0$. Note that $X,x_{2},\dots,x_{n}$ do
not have algebraic relation over $K$ in $K(x_{1},\dots,x_{n})$, thus
$K(X,x_{2},\dots,x_{n})$ is the polynomial field with $n$-indeterminates. Then
there exist an integer $N>0$ and polynomials $b_{i}\in K[X,x_{2},\dots,x_{n}]$
for $0\leq i\leq N$ such that $b_{N}\neq 0$ and $h(X)$ is a solution of the
equation

($\star$) $b_{N}t^{N}+b_{N-1}t^{N-1}+\dots+b_{0}=0.$

We write the left side of the equation as
$\sum_{\underline{k}=(k_{2},\dots,k_{n})}c_{\underline{k}}\cdot(x_{2}^{k_{2}}\dots
x_{n}^{k_{n}})$ where $c_{\underline{k}}\in K[X,t]$, a polynomial in
$x_{2},\dots,x_{n}$ with coefficients in $K[X,t]$. We have
$c_{\underline{k}}(X,h(X))=0$ for any $\underline{k}$ in $K(X)$. Since the
equation ($\star$ ‣ 2.7) is non-trivial, there exists $\underline{k}_{0}$ such
that $c_{\underline{k}_{0}}\neq 0$, and then, $h(X)$ is a solution of
$c_{\underline{k}_{0}}(X,t)=0$, as required. ∎

######  2.9 Lemma. —

Let $f:=\sum a_{i}\,X^{p^{i}}\in K[\\![X]\\!]$ be algebraic over $K(X)$. Then
there exists $N$ such that $\\{a_{i+N}\\}_{i\geq 0}$ is rLRR.

###### Proof.

Since $f$ is algebraic, the set $\\{1,f,f^{p},f^{p^{2}},\dots\\}$ is linearly
dependent over $K(X)$. Thus, there exists $b,b_{0},\dots,b_{n}\in K[X]$ such
that

($\star$) $b=b_{0}f+b_{1}f^{p}+\dots+b_{n}f^{p^{n}}$

in $K[X]$. Let $M:=\max_{0\leq i\leq n}\\{\deg(b),\deg(b_{i})\\}$. Take $N$
such that $p^{n+N-1}>M$. Let $c_{i}$ be the constant term of $b_{i}$. For
$i\geq N$, comparing the $X^{p^{i+n}}$-terms of ($\star$ ‣ 2.7), we have

$0=c_{0}a_{i+n}+c_{1}a_{i+n-1}^{p}+\dots+c_{n}a_{i}^{p^{n}}.$

This is what we wanted. ∎

### 2.10

Let $K$ be a field equipped with a valuation $v$. Let $P(X)=\sum a_{i}X^{i}\in
K[X]$, where $a_{i}\in K$. For $a\in\mathbb{R}$, we define
$\overline{\mathrm{VP}}(P)(a):=\min_{i}\bigl{\\{}v(a_{i})+ia\bigr{\\}}$. This
function $\overline{\mathrm{VP}}(P)$ is called the complete valuation polygon
of $P$. Evidently, this function is piecewise affine and convex.

######  Lemma. —

Assume that the valuation polygon $\overline{\mathrm{VP}}(P)$ has a break at
$r\in\mathbb{R}$ and the slope changes by $n$.

  1. 1.

Let $\overline{K}$ be an algebraic closure of $K$ equipped with an extension
$\widetilde{v}$ of $v$. There exist $n$ roots $\alpha_{i}\in\overline{K}$ of
the equation $P(t)=0$, counted with multiplicities, such that
$\widetilde{v}(\alpha_{i})=r$.

  2. 2.

Assume $K$ is complete. Let $\alpha\in K$ such that $v(\alpha)\geq r$. Then
there is an extension $\widetilde{v}$ of $v$ on
$K\\!\left<\alpha^{-1}X\right>/(P(X))$ so that $\widetilde{v}(X)=r$.

###### Proof.

Let us check the first claim. We may assume $P$ to be monic, and let
$P(X)=\prod(X-\alpha)$ in $\overline{K}$. Since the valuation polygon of
$X-\alpha$ has the break at $r=\widetilde{v}(\alpha)$ and the slope changes by
$1$, the lemma follows since
$\overline{\mathrm{VP}}(P)=\sum\overline{\mathrm{VP}}(X-\alpha)$.

Let us check the second one. Let $L$ be the completion of an algebraic closure
of $K$. Note that $v$ extends (uniquely) on the algebraic closure by invoking
[Ke3, 1.4.9]. It suffices to construct an extension $w$ of $v$ on
$L\\!\left<\alpha^{-1}X\right>/(P(X))$ so that $w(X)=r$. We write
$P(X)=\prod_{i\in I}(X-\alpha_{i})^{r_{i}}$ where $\alpha_{i}\in L$ and
$r_{i}\in\mathbb{N}$ such that $\alpha_{i}\neq\alpha_{j}$ for $i\neq j$. Then
we have $L\\!\left<\alpha^{-1}X\right>/(P(X))\cong\prod_{i\in
I^{\prime}}L\\!\left<\alpha^{-1}X\right>/(X-\alpha_{i})^{r_{i}}$, where
$I^{\prime}$ is the subset of $i\in I$ such that $v(\alpha_{i})\geq
v(\alpha)$. There exists $i_{0}\in I^{\prime}$ such that $v(\alpha_{i_{0}})=r$
by assumption. We may check easily that there exists a unique valuation on
$L\\!\left<\alpha^{-1}X\right>/(X-\alpha_{i_{0}})^{r_{i_{0}}}$ extending $v$
by using [Ke3, 2.3.1]. ∎

### 2.11

Let $\alpha$ be a point of the Berkovich disk
$\mathbb{D}_{\ell}:=\mathrm{Sp}(\ell\\!\left<t\right>)$, and put
$r:=-\log(r(\alpha))$ as 2.2. Denote by $v_{0}$ the Gauss point of
$\mathbb{D}_{\ell}$.

######  Lemma. —

Let $\\{a_{i}\\}_{i\geq 0}$ be an rLRR sequence in $\ell\\!\left<t\right>$ and
$a$ be an element of $\ell\\!\left<t\right>^{\mathrm{perf}}$. Assume that
there exists $\lambda>0$ such that $v_{0}(a_{i})>-\lambda$ for any $i$. For
$n\geq 0$, consider the partial sum

$S_{n}:=a+\sum_{n\geq i\geq 0}a_{i}^{1/p^{i}}.$

Fix a type 4 point $\alpha\in\mathbb{D}_{\ell}$, and put $r:=r_{\alpha}$. Then
there exists a morphism of $\ell$-affinoid spaces
$\psi\colon\mathbb{X}=\mathrm{Sp}(A)\rightarrow\mathbb{D}_{\ell}$, a
continuous map $\varphi\colon\left[r-\gamma,r\right]\hookrightarrow\mathbb{X}$
such that $\gamma>0$ and $\psi\circ\varphi(r)=\alpha$, and an element $S$ in
$A$ such that the following holds: for any $\varepsilon>0$, there exists an
integer $N$ such that for any $n>N$ and $\rho\in\left[r-\gamma,r\right]$, we
have

$v_{\varphi(\rho)}(S-S_{n})>-\varepsilon.$

###### Proof.

Let $K:=\mathrm{Frac}(\ell\\!\left<t\right>)$, and fix an algebraic closure
$\overline{K}$ of $K$. We may assume $a=0$. Indeed, let $S^{\prime}$ be the
element in $\overline{K}$ which satisfies the condition for $a=0$. Then
$S:=S^{\prime}+a$ satisfies the condition. From now on, we assume that $a=0$.

Since $\\{a_{i}\\}$ is rLRR, there exist $\alpha_{j}$, $\beta_{j}$ for $1\leq
j\leq m^{\prime}$ in $\overline{K}$ such that

$a_{i}=\alpha_{1}\beta_{1}^{p^{i}}+\dots+\alpha_{m^{\prime}}\beta_{m^{\prime}}^{p^{i}}$

by Lemma 2.7. Thus, we have
$a_{i}^{1/p^{i}}=\sum_{j=1}^{m^{\prime}}\alpha_{j}^{1/p^{i}}\beta_{j}$, which
implies that $\\{b_{i}:=a_{i}^{1/p^{i}}\\}_{i\geq 0}$ satisfies LRR. Now, let
$V\subset\overline{K}$ be the $\mathbb{F}_{p}$-subspace spanned by
$\\{\beta_{j}\\}$. We note that $m:=\dim(V)\leq m^{\prime}$. We put

$P(X):=\prod_{\beta\in
V}(X-\beta)=d_{0}X+d_{1}X^{p}+\dots+d_{m-1}X^{p^{m-1}}+d_{m}X^{p^{m}}.$

Of course, $d_{m}=1$. By Lemma 2.7 again, we have

$d_{0}b_{i}+d_{1}b_{i+1}^{p}+\dots+d_{m-1}b_{i+m-1}^{p^{m-1}}+d_{m}b_{i+m}^{p^{m}}=0$

for any $i\geq 0$. We put

$f(x_{1},\dots,x_{m}):=\sum_{0\leq i\leq m}d_{i}\,\bigl{(}\sum_{1\leq j\leq
i}x_{j}\bigr{)}^{p^{i}},\qquad g(x_{1},\dots,x_{m}):=\sum_{0\leq i\leq
m}d_{i}\,\bigl{(}\sum_{i+1\leq j\leq m}x_{j}\bigr{)}^{p^{i}}.$

Since $d_{i}(x_{j}+y_{j})^{p^{i}}=d_{i}x_{j}^{p^{i}}+d_{i}y_{j}^{p^{i}}$, we
have

$f(x_{1}+y_{1},x_{2}+y_{2},\dots,x_{m}+y_{m})=f(x_{1},\dots,x_{m})+f(y_{1},\dots,y_{m}).$

Furthermore, we have

$\displaystyle f(b_{n+1},$ $\displaystyle
b_{n+2},\dots,b_{n+m})+g(b_{n-m+1},b_{n-m+2},\dots,b_{n})$
$\displaystyle=\sum_{0\leq i\leq m}d_{i}\,\bigl{(}\sum_{1\leq j\leq
i}b_{n+j}\bigr{)}^{p^{i}}+\sum_{0\leq i\leq m}d_{i}\,\bigl{(}\sum_{i+1\leq
j\leq m}b_{n-m+j}\bigr{)}^{p^{i}}$ $\displaystyle=\sum_{0\leq i\leq
m}d_{i}\,\bigl{(}\sum_{m+1\leq j\leq m+i}b_{n-m+j}\bigr{)}^{p^{i}}+\sum_{0\leq
i\leq m}d_{i}\,\bigl{(}\sum_{i+1\leq j\leq m}b_{n-m+j}\bigr{)}^{p^{i}}$
$\displaystyle=\sum_{0\leq i\leq m}d_{i}\,\bigl{(}\sum_{i+1\leq j\leq
m+i}b_{n-m+j}\bigr{)}^{p^{i}}=\sum_{0\leq i\leq m}d_{i}\,\bigl{(}\sum_{1\leq
j\leq m}b_{n-m+i+j}\bigr{)}^{p^{i}}$ $\displaystyle=\sum_{1\leq j\leq
m}\sum_{0\leq i\leq m}d_{i}b_{n-m+i+j}^{p^{i}}=0.$

For $n^{\prime}>n>0$, put
$S_{n,n^{\prime}}:=S_{n^{\prime}}-S_{n-1}=\sum_{n\leq i\leq n^{\prime}}b_{i}$.
We put $f(\mathbf{b}_{n}):=f(b_{n},\dots,b_{n+m-1})$. Now, the summands of
$P(S_{n,n^{\prime}})$ can be seen as follows.

$\begin{array}[]{cccccccccc}&&&d_{0}b_{n}&d_{0}b_{n+1}&\cdots&d_{0}b_{n^{\prime}-m}&d_{0}b_{n^{\prime}-m+1}&\cdots&d_{0}b_{n^{\prime}}\\\
&&d_{1}b_{n}^{p}&d_{1}b^{p}_{n+1}&d_{1}b^{p}_{n+2}&\cdots&d_{1}b^{p}_{n^{\prime}-m+1}&d_{1}b^{p}_{n^{\prime}-m+2}&&\\\
&&\vdots&\vdots&\vdots&&\vdots&\vdots\\\
&&\cdots&\cdots&\cdots&&\cdots&d_{m-1}b_{n^{\prime}}^{p^{m-1}}\\\
d_{m}b_{n}^{p^{m}}&\cdots&d_{m}b_{n+m-1}^{p^{m}}&d_{m}b_{n+m}^{p^{m}}&d_{m}b_{n+m+1}^{p^{m}}&\cdots&d_{m}b_{n^{\prime}}^{p^{m}}&&&\\\
\lx@intercol\hfil\underbrace{\hskip
113.81102pt}_{f(b_{n},\dots,b_{n+m-1})}\hfil\lx@intercol&\lx@intercol\hfil\underbrace{\hskip
184.9429pt}_{0}\hfil\lx@intercol&\lx@intercol\hfil\underbrace{\hskip
99.58464pt}_{g(b_{n^{\prime}-m+1},\dots,b_{n^{\prime}})}\hfil\lx@intercol\end{array}$

Thus, we have

($\star$) $\displaystyle
P(S_{n,n^{\prime}})=f(b_{n},\dots,b_{n+m-1})+g(b_{n^{\prime}-m+1},\dots,b_{n^{\prime}})=f(\mathbf{b}_{n})-f(\mathbf{b}_{n^{\prime}+1}),$

where the second equality holds by the relation above. For the convenience of
the reader, let us outline the rest of the argument. We take a “sufficiently
large” $n_{0}$, and define $S$ to be a (suitable) solution of the equation
$P(X)=f(\mathbf{b}_{n_{0}})$. For $n\geq n_{0}$, ($\star$ ‣ 2.11) implies that
$P(S-S_{n_{0},n})=f(\mathbf{b}_{n+1})$. To estimate $S-S_{n_{0},n}$ at some
valuation $v$, we may use Lemma Lemma by considering the valuation polygon of
$P(X)=f(\mathbf{b}_{n+1})$.

Now, let $\mathbb{X}^{\prime}:=\mathrm{Sp}(A^{\prime})$ be the normalization
of $\mathrm{Sp}\bigl{(}\ell\\!\left<t\right>[d_{i}]\bigr{)}$. Since $d_{i}$
are algebraic over $\ell\\!\left<t\right>$, the morphism
$\psi^{\prime}\colon\mathbb{X}^{\prime}\rightarrow\mathbb{D}_{\ell}$ is
finite, and we may find an extension $\widetilde{\alpha}$ of $\alpha$. Since
$\alpha$ is of type 4, so is $\widetilde{\alpha}$. Take a continuous embedding
$\varphi^{\prime}\colon\left[r-\gamma,r\right]\hookrightarrow\mathbb{X}^{\prime}$
with $\varphi^{\prime}(r)=\widetilde{\alpha}$. By [BPR, 5.12], by decreasing
$\gamma$, we may assume that $\psi^{\prime}\circ\varphi^{\prime}$ is injective
and the image in $\mathbb{D}_{\ell}$ is contained in the generic path of
$\alpha$ (cf. [Ke4, 2.2.13]). By decreasing $\gamma$ further, we may assume
that the function $\rho\mapsto v_{\varphi^{\prime}(\rho)}(d_{i})$ is constant
on $\left[r-\gamma,r\right]$ for any $i$, since $\widetilde{\alpha}$ is of
type 4 and we may take a neighborhood of $\widetilde{\alpha}$ which is
isomorphic to an open ball by [BPR, 4.27]. Let us consider the valuation
polygon of $P(X)$. Namely, let $w_{a}$ ($a\in\mathbb{R}$) be the valuation on
$\mathrm{Frac}(A^{\prime})[X]$ such that $w_{a}(\sum
c_{i}X^{i})=\min_{i}\\{v_{\varphi^{\prime}(\rho)}(c_{i})+ia\\}$. Recall that
the complete valuation polygon of $P(X)$ is the function sending
$a\in\mathbb{R}$ to $\overline{\mathrm{VP}}(a):=w_{a}(P)$. By the choice of
$\gamma$, $w_{a}$ does not depend on the choice of $\rho\in[r-\gamma,r]$. Note
that all the slopes of $\overline{\mathrm{VP}}$ are positive integers since
$P(0)=0$.

$R$$0$$M$$M^{\prime}$$a$$\overline{\mathrm{VP}}(a)$

Let
$M:=\overline{\mathrm{VP}}(0)=\min_{i}\\{v_{\varphi^{\prime}(\rho)}(d_{i})\\}$.
Let $R<0$ such that $(R,M^{\prime}:=\overline{\mathrm{VP}}(R))$ is a break
point of $\overline{\mathrm{VP}}(a)$ and there is no break point in
$\left]R,0\right[$. If there is no such a break point, we take any $R<0$. Note
that $M^{\prime}<M$. Recall that, since $a_{n}$ is in $\ell\\!\left<t\right>$
and $\varphi^{\prime}$ maps to a sub-segment of the generic path of $\alpha$,
we have $v_{\varphi^{\prime}(r-\gamma)}(b_{n})\leq
v_{\varphi^{\prime}(\rho)}(b_{n})$.

Let $\varepsilon^{\prime}>0$, and recall that we are given $\lambda>0$ in the
statement of the lemma. Take $N$ such that
$\varepsilon^{\prime}>p^{m-N}\lambda$. Then for any $n>N$ and
$\rho\in\left[r-\gamma,r\right]$, we have

$p^{m}v_{\varphi^{\prime}(\rho)}(b_{n})=p^{m}v_{\psi^{\prime}\varphi^{\prime}(\rho)}(b_{n})\geq
p^{m}v_{0}(b_{n})>-p^{m-n}\lambda>-\varepsilon^{\prime},$

where the first inequality holds since $a_{n}\in\ell\\!\left<t\right>$. Thus,

$v_{\varphi^{\prime}(\rho)}(f(\mathbf{b}_{n}))=v\Bigl{(}\sum_{0\leq i\leq
m}d_{i}\,\bigl{(}\sum_{1\leq j\leq
i}b_{n+j}\bigr{)}^{p^{i}}\Bigr{)}>M-\varepsilon^{\prime}.$

This implies that we may take large enough $N$ such that
$v_{\varphi^{\prime}(\rho)}(f(\mathbf{b}_{n}))>M^{\prime}$ and
$v_{\varphi^{\prime}(\rho)}(b_{n})>R$ for any $n\geq N$ and $\rho$. Since the
characteristic polynomial $P(X)$ remains the same even if we consider the
sequence $\\{a_{i+N}\\}_{i\geq 0}$ instead of $\\{a_{i}\\}_{i\geq 0}$, we may
and do assume that

($\star\star$)
$v_{\varphi^{\prime}(\rho)}(f(\mathbf{b}_{n}))>M^{\prime},\qquad
v_{\varphi^{\prime}(\rho)}(S_{n})>R$

for any $n\geq 0$.

Now, take $\alpha\in\ell$ such that $v(\alpha)\geq-R>0$, and put
$A:=A^{\prime}\\!\left<\alpha X\right>/(P(X)-f(\mathbf{b}_{0}))$ and
$\mathbb{X}:=\mathrm{Sp}(A)$. Since
$A\widehat{\otimes}_{A^{\prime}}\mathscr{H}(\varphi^{\prime}(r))\cong\mathscr{H}(\varphi^{\prime}(r))\\!\left<\alpha
X\right>/(P(X)-f(\mathbf{b}_{0}))$ and since
$v_{\varphi^{\prime}(r)}(f(\mathbf{b}_{0}))>M^{\prime}$ by ($\star\star$ ‣
2.11), using Lemma Lemma.2, there exists a valuation $w$ on
$A\widehat{\otimes}_{A^{\prime}}\mathscr{H}(\varphi^{\prime}(r))$ extending
that of $\mathscr{H}(\varphi^{\prime}(r))$ such that $w(X)>R$. This defines a
point $\beta$ of $\mathbb{X}$. Take any continuous map
$\varphi\colon[r-\gamma,r]\rightarrow\mathbb{X}$ such that $\varphi(r)=\beta$.
By [BPR, 5.12], decreasing $\gamma$ if necessary, $\varphi$ becomes a lifting
of $\varphi^{\prime}$ and further $v_{\varphi(\rho)}(X)>R$ for any
$\rho\in[r-\gamma,r]$. We take $S=X\,(=\alpha^{-1}\cdot(\alpha X)\in A)$,
which is a solution of $P(X)=f(\mathbf{b}_{0})$ such that
$v_{\varphi(\rho)}(S)>R$. Let us check that $S$ satisfies the condition of the
lemma. Let $\varepsilon>0$ as in the claim of the lemma. By ($\star$ ‣ 2.11),
we have

$\displaystyle P(S-S_{n})$ $\displaystyle=P(S-S_{0,n})=P(S)-P(S_{0,n})$
$\displaystyle=f(\mathbf{b}_{0})-\bigl{(}f(\mathbf{b}_{0})-f(\mathbf{b}_{n+1})\bigr{)}$
$\displaystyle=f(\mathbf{b}_{n+1}).$

We have checked that, for $n\gg 0$,
$v_{\varphi(\rho)}(f(\mathbf{b}_{n+1}))=v_{\varphi^{\prime}(\rho)}(f(\mathbf{b}_{n+1}))>\overline{\mathrm{VP}}(-\varepsilon)$
for any $\rho$. This implies that any solution $s$ of the equation
$P(X)=f(\mathbf{b}_{n+1})$ such that $v_{\varphi(\rho)}(s)>R$ have the
property that $v_{\varphi(\rho)}(s)>-\varepsilon$ by Lemma Lemma.1. On the
other hand, we have
$v_{\varphi(\rho)}(S-S_{n})\geq\min\\{v_{\varphi(\rho)}(S),v_{\varphi(\rho)}(S_{n})\\}>R$
for any $\rho$ by ($\star\star$ ‣ 2.11). Thus
$v_{\varphi(\rho)}(S-S_{n})>-\varepsilon$ as required. ∎

######  2.12 Corollary. —

Let $\alpha$ be a type 4 point on $\mathbb{D}_{\ell}$, and $\\{a_{i}\\}$ and
$S_{n}$ be as in the lemma. Then there exists $\gamma>0$ such that one of the
following conditions holds:

  * •

for any $\varepsilon>0$, there exists $N$ such that
$v_{\alpha,\rho}(S_{n})>-\varepsilon$ for any $n>N$ and
$\rho\in\left[r-\gamma,r\right]$.

  * •

there exists $\mu<0$ and $N>0$ such that $v_{\alpha,\rho}(S_{n})=\mu$ for any
$n>N$ and $\rho\in[r-\gamma,r]$.

###### Proof.

Let $\varphi\colon[r-\gamma,r]\rightarrow\mathbb{X}$ be as in the lemma. We
have

$v_{\alpha,\rho}(S_{n})=v_{\varphi(\rho)}(S_{n})=v_{\varphi(\rho)}(S_{n}-S+S)\geq\min\\{v_{\varphi(\rho)}(S_{n}-S),v_{\varphi(\rho)}(S)\\}.$

Since $\alpha$ is a type 4 point, by increasing $\gamma$, we may assume that
$v_{\varphi(\rho)}(S)$ is constant $\mu$ for any
$\rho\in\left[r+\gamma,r\right]$. The conclusion of the lemma implies that one
of the following conditions holds:

  * •

if $\mu\geq 0$ then for any $\varepsilon>0$, there exists $N$ such that
$v_{\alpha,\rho}(S_{n})>-\varepsilon$ for any $n>N$.

  * •

if $\mu<0$, then there exists $N$ such that
$v_{\alpha,\rho}(S_{n})=v_{\varphi(\rho)}(S)$ for any $n>N$.

Thus the corollary follows. ∎

###  2.13 Proof of Proposition 2.6

If $\alpha$ is of type 1, the claim follows by Lemma 2.5. Thus, we may assume
$\alpha$ to be of type 4. We use the notation of 2.6. By Lemma 2.5, there
exists the terminal slope of $\mathrm{VP}^{(\infty)}(f)$. Thus, it remains to
show that for any $\underline{n}\in\Lambda$, the terminal slope of the
valuation polygon $\mathrm{VP}^{(\infty)}(A_{\underline{n}})$ is $0$. By Lemma
2.8, the function $\sum_{k\geq 0}a_{p^{k}\underline{n}}X^{p^{k}}$ is
algebraic, so $\sum_{k\geq 0}(a_{p^{k}\underline{n}}/\pi)X^{p^{k}}$ is
algebraic as well. Thus, by putting $b_{i}:=a_{p^{i}\underline{n}}/\pi$, the
sequence $\\{b_{i}\\}_{i\geq N}$ satisfies rLRR for $N\gg 0$ by Lemma 2.9.
Apply Corollary 2.12 above for $a:=\sum_{k<N}b^{1/p^{k}}_{k}$ and
$a_{i}:=b^{1/p^{N}}_{i+N}$ to get the desired result. Indeed, with these
choices of $a$ and $a_{i}$, we have
$\mathrm{VP}(S_{n})=\mathrm{VP}^{(n+N)}(A_{\underline{n}})$. In the first case
of the corollary, for any $\rho\in[r-\gamma,r]$, we have
$\mathrm{VP}^{(\infty)}(A_{\underline{n}})(\rho)=\lim_{n\rightarrow\infty}\mathrm{VP}(S_{n})(\rho)=0$.
Thus, the terminal slope is $0$. In the second case where $\mu<0$, the
corollary implies that $\lim_{n\rightarrow\infty}\mathrm{VP}(S_{n})(\rho)=\mu$
for any $\rho\in[r-\gamma,r]$, and the terminal slope is also $0$, as
required. ∎

##  3 Existence of admissible alteration

This section is devoted to proving Theorem Theorem, which says that given a
smooth morphism $X\rightarrow S$ and a rational function $f$ on $X$, we may
take an alteration $S^{\prime}$ of $S$ so that $f$ is “admissible”. We follow
[Ke4] for a proof: We define “local admissibility” around each point of the
Riemann-Zariski space of $S$. It takes most of the section to prove that this
local admissibility can be attained after taking an alteration around each
valuation of $S$. At the very end, we deduce the theorem from the local
admissibility by using the quasi-compactness of Riemann-Zariski space.

######  3.1 Definition. —

Let $X\rightarrow S$ be a smooth morphism, and $h\in\mathrm{R}(X)$. We say
that $h$ is admissible if the following condition holds: there exists an open
finite covering $\\{V_{i}\\}_{i\in I}$ of $S$ and open subschemes
$\\{U_{i}\\}_{i\in I}$ of $X$ where $U_{i}\subset X\times_{S}V_{i}$ such that;
1\. $\bigcup_{i\in I}U_{i,s}\subset X_{s}$ is dense for any $s\in S$; 2\.
there exists a function $f_{i}\colon U_{i}\rightarrow\mathbb{A}^{1}_{S}$ which
is $S$-separable, a regular function
$\sigma_{i}\in\mathcal{O}_{S}(V_{i})\cap\mathrm{R}(S)^{\times}$, and
$g_{i}\in\mathrm{R}(X)$ such that

$h|_{U_{i}}+(g_{i}^{p}-g_{i})\in f_{i}/\sigma_{i}+\mathcal{O}_{X}(U_{i}).$

######  Remark. —

The 2nd condition that $f_{i}$ being $S$-separable can be replaced by $f_{i}$
being either $S$-separable or $0$. Indeed, assume that
$h|_{U_{i}}+(g_{i}^{p}-g_{i})\in\mathcal{O}_{X}(U_{i})$. By refining $U_{i}$
and $V_{i}$, we may assume that there exists an $S$-separable function
$\alpha_{i}\in\mathcal{O}_{U_{i}}$. Then we have
$h|_{U_{i}}+(g_{i}^{p}-g_{i})\in\alpha_{i}/1+\mathcal{O}_{X}(U_{i})$.

For a motivation of this definition, one can see Lemma 5.3.

### 3.2

Let $X\rightarrow S$ be a flat morphism and $S^{\prime}\rightarrow S$ be a
maximally dominant morphism. Then $X_{S^{\prime}}\rightarrow X$ is maximally
dominant by [G, Exp. II, 1.1.5]. Thus, the pullback homomorphism
$\mathrm{R}(X)\rightarrow\mathrm{R}(X_{S^{\prime}})$ is defined.

######  Theorem. —

Let $X\rightarrow S$ be a smooth morphism between $k$-schemes of finite type,
and take a rational function $h\in\mathrm{R}(X)$. Then there exists an
alteration $S^{\prime}\rightarrow S$ such that the pullback of $h$ to
$X_{S^{\prime}}$ is admissible.

### 3.3

From the next paragraph, we use Zariski-Riemann space. We recall briefly the
basic notions just to fix them. See [V, Ke2, Ke4] for more details. Let $S$ be
an integral scheme of finite type over $k$. We denote by $\left<S\right>$, or
$\left<S\right>_{/k}$ if we want to be more precise, inspired by the notation
of [FK, II, §E], the set of valuations $v$ on $k(S)$ such that $v|_{k}$ is
trivial and $v$ is centered on $S$. This space is called the Zariski-Riemann
space associated to $S$. When $S$ is proper, $\left<S\right>$ is often denoted
by $S_{k(S)/k}$ in the literature. As a set, we may write
$\left<S\right>=\mathop{\underleftarrow{\mathrm{lim}}}S^{\prime}$ where
$S^{\prime}$ runs over all the modifications of $S$. The topology induced by
the Zariski topology of $S^{\prime}$ is called the Zariski topology of
$\left<S\right>$. Let $U\subset S$ be an open subscheme. Define
$\left<S\right>_{U}:=\mathop{\underleftarrow{\mathrm{lim}}}S^{\prime}$ where
the projective limit runs over modifications $\rho\colon S^{\prime}\rightarrow
S$ such that $\rho^{-1}(U)\rightarrow U$ is an isomorphism. For
$v\in\left<S\right>$, denote by $k_{v}$ the residue field, $R_{v}$ the
valuation ring, and $\Gamma_{v}$ the value semigroup. The height (or rank) of
$v$ is denoted by $\mathrm{ht}(v)$. The rational rank of $v$ denoted by
$\mathrm{rat.rk}(v)$ is
$\mathrm{rk}_{\mathbb{Q}}(\Gamma_{v}\otimes\mathbb{Q})$. We say that $v$ is a
minimal valuation if $\mathrm{trdeg}_{k}(k_{v})=0$ and $\mathrm{ht}(v)=1$, an
Abhyankar valuation if $\dim(S)=\mathrm{rat.rk}(v)+\mathrm{trdeg}_{k}(k_{v})$,
a monomial valuation if it is Abhyankar and minimal, a divisorial valuation if
it is Abhyankar, $\mathrm{rat.rk}(v)=1$ and $\mathrm{ht}(v)=1$.

### 3.4

Until the beginning of 3.23, we further assume that the following condition
holds, and fix the data:

> (*) $X\rightarrow S$ is a smooth morphism between integral affine
> $k$-schemes of finite type with global coordinates $\\{x_{1},\dots,x_{d}\\}$
> over $S$ such that the fiber $X_{s}$ is geometrically irreducible over
> $k(s)$ for any $s\in S$, and it is endowed with a section $S\rightarrow X$
> whose image is the zero locus $\mathrm{Zero}:=V(x_{1},\dots,x_{d})$. We
> denote by $\eta$ the generic point of $S$.

Recall that we write $f\in\mathcal{O}_{T}$ in the sense of
$f\in\Gamma(T,\mathcal{O}_{T})$. For $v\in\left<S\right>$, a morphism of
finite type $S^{\prime}\rightarrow S$ is said to be a local alteration around
$v$ if it is generically finite, $S^{\prime}$ is affine integral, and $v$
extends to a valuation centered on $S^{\prime}$. Note that this condition (*)
is stable under changing the base $S$ by a local alteration around $v$. With
these fixed coordinates, by [EGA $0_{\mathrm{IV}}$, 19.5.4], the completion
$\widehat{X}:=\widehat{X}_{\mathrm{Zero}}$ is isomorphic canonically to
$\widehat{\mathbb{A}}^{d}_{S}$, where $S$ is considered as a formal scheme
with the discrete topology. Let $f\in\mathcal{O}_{X_{\eta}}$. We can write

$f=\sum_{\underline{k}\geq\underline{0}}a_{\underline{k}}\underline{x}^{\underline{k}}$

where $a_{\underline{k}}\in k(S)$ in
$\mathcal{O}_{S}[\\![x_{1},\dots,x_{d}]\\!]\otimes_{\mathcal{O}_{S}}k(S)\subset
k(S)[\\![x_{1},\dots,x_{d}]\\!]$. Put

$\Lambda:=\bigl{\\{}\underline{n}\in\mathbb{N}^{d}\mid\mbox{$\underline{n}\neq\underline{0}$
and $p\nmid\underline{n}$}\bigr{\\}}.$

For a point $v\in\left<S\right>$ of the Riemann-Zariski space, we define

$\mathrm{VP}(f)(v):=\min_{\underline{k}\in\mathbb{N}^{d}\setminus\\{\underline{0}\\}}\bigl{\\{}0,v(a_{\underline{k}})\bigr{\\}}=\min_{\underline{n}\in\Lambda}\,\min_{k}\bigl{\\{}0,v(a_{p^{k}\underline{n}})\bigr{\\}}$

in $\Gamma_{v}$. Note that the minimum exists. Indeed, since
$f\in\mathcal{O}_{X_{\eta}}$, we may write $f=g/\pi$ with
$g\in\mathcal{O}_{X}$ and $\pi\in\mathcal{O}_{S}$. Then
$a_{\underline{n}}\in\pi^{-1}\mathcal{O}_{S}$. Since $\mathcal{O}_{S}$ is
noetherian, the semigroup $v(\mathcal{O}_{S})\subset\Gamma_{v}$ is well-
ordered (cf. Lemma), so is
$v(\pi^{-1}\mathcal{O}_{S})=v(\mathcal{O}_{S})-v(\pi)$, and the minimum
exists.

Now, assume that $\mathrm{ht}(v)=1$. For $\underline{n}\in\Lambda$, let

$f^{(m)}_{\underline{n}}:=\sum^{m}_{k=0}(a_{p^{k}\underline{n}})^{1/p^{k}}.$

This sum is defined in $\pi^{-1}A:=\sum_{n\geq
0}(\pi^{-1}\mathcal{O}_{S})^{1/p^{n}}$. Note that $v$ extends uniquely to the
perfection $k(S)^{1/p^{\infty}}$. Since $v$ is of height $1$, for any
$\epsilon<0$ in $\Gamma_{v}\subset\mathbb{R}$, there exists an integer $N>0$
such that $\epsilon<-v(\pi)/p^{N}$. This implies that

(3.4.1) $\bigl{\\{}\lambda\in
v(\pi^{-1}A)\mid\lambda\leq\epsilon\bigr{\\}}=\bigl{\\{}\lambda\in
v\bigl{(}\mathop{\textstyle\sum}\nolimits_{n\leq
N}(\pi^{-1}\mathcal{O}_{S})^{1/p^{n}}\bigr{)}\mid\lambda\leq\epsilon\bigr{\\}}.$

The set on the right is well-ordered because $\mathcal{O}_{S}^{1/p^{N}}$ is
noetherian. This implies that the set $J:=\bigl{\\{}\lambda\in
v(\pi^{-1}A)\mid\lambda\leq 0\bigr{\\}}$ is well-ordered. For
$\underline{n}\in\Lambda$, assume
$\lim_{m\rightarrow\infty}\min\bigl{\\{}0,v(f^{(m)}_{\underline{n}})\bigr{\\}}\neq
0$. Then $\liminf_{m\rightarrow\infty}v(f^{(m)}_{\underline{n}})<0$. For any
$\liminf v(\dots)<\epsilon<0$, there exists $M>\log_{p}(-v(\pi)/\epsilon)$
such that $v(f^{(M)}_{\underline{n}})<\epsilon$. Then
$v(f^{(M)}_{\underline{n}})=v(f^{(n)}_{\underline{n}})$ for any $n\geq M$, and
$\lim_{m\rightarrow\infty}\min\bigl{\\{}0,v(f^{(m)}_{\underline{n}})\bigr{\\}}=v(f^{(M)}_{\underline{n}})\in
J$. Since $J$ is well-ordered, this implies that the following definition
makes sense:

$\mathrm{VP}^{(\infty)}(f)(v):=\min_{\underline{n}\in\Lambda}\Bigl{\\{}\lim_{m\rightarrow\infty}\min\bigl{\\{}0,v(f^{(m)}_{\underline{n}})\bigr{\\}}\Bigr{\\}}.$

Note that, by definition, $\mathrm{VP}(f)(v),\mathrm{VP}^{(\infty)}(f)(v)\leq
0$. For a dominant morphism $\alpha\colon S^{\prime}\rightarrow S$ of integral
schemes, $\mathrm{VP}$ and $\mathrm{VP}^{(\infty)}$ are invariant under base
change by $\alpha$. Namely, for any extension $v^{\prime}$ to $S^{\prime}$ of
$v\in\left<S\right>$, we have the equalities

(3.4.2)
$\mathrm{VP}(\alpha^{*}f)(v^{\prime})=\mathrm{VP}(f)(v),\qquad\mathrm{VP}^{(\infty)}(\alpha^{*}f)(v^{\prime})=\mathrm{VP}^{(\infty)}(f)(v).$

Let $L/k(S)$ be a finite extension, $w$ be an extension of $v$, and
$f^{\prime}\in\mathcal{O}_{X_{L}}$, where
$X_{L}:=X\times_{S}\mathrm{Spec}(L)$. Then we can take a local alteration
$S^{\prime}\rightarrow S$ such that $k(S^{\prime})\cong L$ and $w$ is centered
on $S^{\prime}$. Then
$f^{\prime}\in\mathcal{O}_{X_{\eta^{\prime}}}\cong\mathcal{O}_{X_{L}}$, where
$\eta^{\prime}$ is the generic point of $S^{\prime}$, and the values
$\mathrm{VP}(f^{\prime})(w)$, $\mathrm{VP}^{(\infty)}(f^{\prime})(w)$ are
defined. By (3.4.2), these values only depend on $L$ and $w$, and not on
specific “model” $S^{\prime}$. We use these values without referring to
$S^{\prime}$.

######  3.5 Definition. —

Take a point $v\in\left<S\right>$, and a function
$h\in\mathcal{O}_{X_{\eta}}$.

  * •

An ally of $h$ is a rational function of the form $h+(g^{p}-g)$ with
$g\in\mathcal{O}_{X_{\eta}}$. An ally $h^{\prime}$ of $h$ is said to be good
on $S$ at $v$ if it belongs to $f/\sigma+\mathcal{O}_{X,z}$, where $z$ is the
center of $v$ on $S$,
$\mathcal{O}_{X,z}:=\mathcal{O}_{X}\otimes_{\mathcal{O}_{S}}\mathcal{O}_{S,z}$,
$\sigma\in\mathcal{O}_{S}\setminus\\{0\\}$, and $f\in\mathcal{O}_{X}$ which is
either an $S$-separable function or in $\mathcal{O}_{S}$. Such a $g$ is called
a mollifier. For an open subscheme $U\subset S$, a mollifier $g$ is said to
provide a good ally of $h$ on $U$ if $h+(g^{p}-g)$ is a good ally of $h$ at
any valuation $v$ centered on $U$.

  * •

We say that $h$ is locally admissible at $v$ if there exists an integral
scheme $S^{\prime}$ such that $k(S^{\prime})=k(S)$, $v$ is centered on
$S^{\prime}$, and there exists a good ally of $h$ on $S^{\prime}$ at $v$.

  * •

Assume $\mathrm{ht}(v)=1$. We say that $h$ is weakly admissible at $v$ on
$k(S)$ if there exists a function
$g\in\mathcal{O}_{X_{\eta}}=\mathcal{O}_{X}\otimes_{\mathcal{O}_{S}}k(S)$ such
that $\mathrm{VP}^{(\infty)}(h)(v)=\mathrm{VP}(h+(g^{p}-g))(v)$. Such a $g$ is
called a weak mollifier.

Note that if $h$ is good at $v$ which is centered at $z\in S$, then there
exists an open neighborhood subscheme $U$ of $z$ over which $h$ is good.
Indeed, it suffices to show that if $f$ is separable at $z$, then $f$ is
separable on an open neighborhood of $z$. Since we are assuming $X\rightarrow
S$ to be geometrically irreducible, and thus equidimensional, this follows by
Lemma 1.3.3 and [EGA IV, 13.1.3]. Even though we do not use, we remark that if
$h$ admits a good ally, then it is locally weakly admissible. This can be seen
from Lemma 3.10 below. The converse is “almost” true by Lemma 3.13.

######  3.6 Lemma. —

For any $f,g\in\mathcal{O}_{X_{\eta}}$, we have

$\mathrm{VP}^{(\infty)}\bigl{(}f+(g^{p}-g)\bigr{)}(v)=\mathrm{VP}^{(\infty)}(f)(v),\qquad\mathrm{VP}(f)(v)\leq\mathrm{VP}^{(\infty)}(f)(v).$

###### Proof.

We may take $\pi\in\mathcal{O}_{S}\setminus\\{0\\}$ such that
$f,g\in\pi^{-1}\mathcal{O}_{S}$. Let $\mathbf{a}:=\\{a_{k}\\}_{k\geq 0}$ be a
sequence in $\pi^{-1}\mathcal{O}_{S}$, and put

$v(\mathbf{a}):=\min_{k}\bigl{\\{}0,v(a_{k})\bigr{\\}},\quad
v^{(\infty)}(\mathbf{a}):=\min\Bigl{\\{}0,\lim_{m\rightarrow\infty}v\Bigl{(}\sum_{k=0}^{m}a_{k}^{1/p^{k}}\Bigr{)}\Bigr{\\}}.$

Now, let $\mathbf{b}=\\{b_{k}\\}_{k\geq 0}$ be a sequence in
$\pi^{-1}\mathcal{O}_{S}$. Put $\mathbf{b}^{p}:=\\{b_{k-1}^{p}\\}_{k\geq 0}$
($b_{-1}:=0$). Then we claim that

$v^{(\infty)}(\mathbf{a}+(\mathbf{b}^{p}-\mathbf{b}))=v^{(\infty)}(\mathbf{a}),\qquad
v(\mathbf{a})\leq v^{(\infty)}(\mathbf{a}).$

If these claims hold, the lemma follows. Indeed, in order to show the lemma,
we take $\\{a_{k}\\}$ (resp. $\\{b_{k}\\}$) to be the coefficients of
$\underline{x}^{p^{k}\underline{n}}$ for each $\underline{n}\in\Lambda$ in the
expansion of $f$ (resp. $g$). Now, to check the first claim, we may compute

$\displaystyle(a_{0}-b_{0})+\sum_{k=1}^{m}\bigl{(}a_{k}+(b_{k-1}^{p}-b_{k})\bigr{)}^{1/p^{k}}=\bigl{(}\sum_{k=0}^{m}a_{k}^{1/p^{k}}\bigr{)}-b_{m}^{1/p^{m}}.$

Since $v(b_{m})\geq-v(\pi)$, we have
$\liminf_{m\rightarrow\infty}v(b_{m}^{1/p^{m}})\geq 0$, and the first equality
follows. The second inequality is easy to check. ∎

### 3.7

Let $f\in\mathcal{O}_{X}$ and $\pi\in\mathcal{O}_{S}$, and let
$v\in\left<S\right>$ whose center is $z\in S$. We assume that $k$ is perfect
and $k_{v}=k(z)$. We put

$A:=\pi\sum_{n\geq
0}(\pi^{-1}\mathcal{O}_{S})^{1/p^{n}}\supset\mathcal{O}_{S},\qquad
J:=\bigl{\\{}\lambda\in v(\pi^{-1}A)\mid\lambda\leq
0\bigr{\\}}\subset\Gamma_{v}.$

Then $A$ is an $\mathcal{O}_{S}$-module contained in
$\mathcal{O}_{S}^{1/p^{\infty}}$. Recall from 3.4 that $J$ is well-ordered.
The filtration defined by $v$ on $\mathcal{O}_{S}^{1/p^{\infty}}$ induces a
filtration on $A$. For a submodule $B\subset k(S)^{1/p^{\infty}}$ and
$\lambda\in\Gamma_{v}$, we put $\mathrm{Fil}^{\lambda}_{v}(B):=\bigl{\\{}f\in
B\mid v(f)\geq\lambda\bigr{\\}}$,
$\mathrm{Fil}^{\lambda+}_{v}(B):=\bigl{\\{}f\in B\mid v(f)>\lambda\bigr{\\}}$,
and $\mathrm{gr}^{\lambda}_{v}(B):=B^{\lambda}/B^{\lambda+}$. We often denote
$\mathrm{Fil}^{\lambda(+)}_{v}B$, $\mathrm{gr}^{\lambda}_{v}B$ by
$B^{\lambda(+)}$, $\mathrm{gr}^{\lambda}B$ if no confusion may arise. We have
inclusions

$\mathrm{gr}^{\lambda}(\mathcal{O}_{S})\subset\mathrm{gr}^{\lambda}(A)\subset\mathrm{gr}^{\lambda}(\mathcal{O}_{S}^{1/p^{\infty}}).$

By assumption on $v$, we have
$k_{v}:=\mathrm{gr}^{0}(k(S))=\mathrm{gr}^{0}(\mathcal{O}_{S})=:k(z)$. Since
$\mathcal{O}_{X}$ is flat over $\mathcal{O}_{S}$ we may define a filtration on
$\mathcal{O}_{X}\otimes\pi^{-1}A$ by
$(\mathcal{O}_{X}\otimes\pi^{-1}A)^{\lambda}:=\mathcal{O}_{X}\otimes(\pi^{-1}A)^{\lambda}$.
The flatness moreover implies that $\mathrm{gr}(\mathcal{O}_{X}\otimes
A)\cong\mathcal{O}_{X}\otimes\mathrm{gr}(A)$.

######  3.8 Lemma. —

Let $M$ be an $\mathcal{O}_{S}$-module, and $N\subset M$ be a submodule. Put
$\widehat{\mathcal{O}}_{X}:=\mathcal{O}_{\widehat{X}_{\mathrm{Zero}}}$. Then
we have the following equality in
$M\otimes_{\mathcal{O}_{S}}\widehat{\mathcal{O}}_{X}$:

$(N\otimes_{\mathcal{O}_{S}}\widehat{\mathcal{O}}_{X})\cap(M\otimes_{\mathcal{O}_{S}}\mathcal{O}_{X})=N\otimes_{\mathcal{O}_{S}}\mathcal{O}_{X}.$

###### Proof.

Let $J\subset\mathcal{O}_{X}$ be the ideal defining the closed subscheme
$Z:=\mathrm{Zero}$ in 3.4 (*). We put
$\mathcal{O}_{X,Z}:=(1+J)^{-1}\mathcal{O}_{X}$. The pair
$(\mathcal{O}_{X,Z},J\mathcal{O}_{X,Z})$ is a Zariskian pair [FK, I, B.1 (a)],
and thus, the completion homomorphism
$\mathcal{O}_{X,Z}\rightarrow\widehat{\mathcal{O}}_{X,Z}\cong\widehat{\mathcal{O}}_{X}$
is faithfully flat. The fully faithfulness implies that the following diagram
on the right is Cartesian by [EGA $0_{\mathrm{I}}$, 6.6.3]:

$\textstyle{N\otimes_{\mathcal{O}_{S}}\mathcal{O}_{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{N\otimes_{\mathcal{O}_{S}}\mathcal{O}_{X,Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\otimes\mathcal{O}_{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\otimes\mathcal{O}_{X,Z},}$
$\textstyle{N\otimes_{\mathcal{O}_{S}}\mathcal{O}_{X,Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\square}$$\textstyle{N\otimes_{\mathcal{O}_{S}}\widehat{\mathcal{O}}_{X,Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\otimes\mathcal{O}_{X,Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{M\otimes\widehat{\mathcal{O}}_{X,Z}.}$

Thus, it remains to show that the left diagram above is Cartesian. By the
flatness of $\mathcal{O}_{X}$ over $\mathcal{O}_{S}$, this is equivalent to
showing that the induced homomorphism $\alpha\colon
L:=(M/N)\otimes\mathcal{O}_{X}\rightarrow(M/N)\otimes\mathcal{O}_{X,Z}$ is
injective. Since $\alpha$ is a localization homomorphism by the multiplicative
system $1+J$, it suffices to show that, for any $f\in 1+J$, we have
$\mathrm{Ass}(L)\subset D(f)$ in $X=\mathrm{Spec}(\mathcal{O}_{X})$ by [EGA
IV, 3.1.9]. If there exist $f$, $\mathfrak{p}$ such that
$\mathfrak{p}\in\mathrm{Ass}(L)\cap V(f)$, then $V(J)\cap
V(\mathfrak{p})\subset V(f)\cap V(J)=\emptyset$. Thus, it suffices to check
that, for any $\mathfrak{p}\in\mathrm{Ass}(L)$, the intersection $V(J)\cap
V(\mathfrak{p})\neq\emptyset$. Since $\mathcal{O}_{X}$ is flat over
$\mathcal{O}_{S}$, the associated prime of $L$ is equal to
$\bigcup_{\mathfrak{p}\in\mathrm{Ass}(M/N)}\mathrm{Ass}\bigl{(}k(\mathfrak{p})\otimes\mathcal{O}_{X}\bigr{)}$
by [EGA IV, 3.3.1]. Since each fiber of $\mathcal{O}_{X}$ is geometrically
irreducible and is smooth over $\mathcal{O}_{S}$,
$k(\mathfrak{p})\otimes\mathcal{O}_{X}$ is integral, and thus
$\mathrm{Ass}\bigl{(}k(\mathfrak{p})\otimes\mathcal{O}_{X}\bigr{)}$ is equal
to the generic point of the fiber $X\otimes_{\mathcal{O}_{S}}k(\mathfrak{p})$.
Since the morphism $Z\rightarrow S$ is an isomorphism, the intersection is not
empty, and the claim follows. ∎

######  3.9 Corollary. —

Let $M$ be a finitely generated $\mathcal{O}_{S}$-module, and $N\subset M$ be
a submodule. Let $h\in M\otimes_{\mathcal{O}_{S}}\mathcal{O}_{X}$, and write
$h=\sum a_{\underline{k}}\underline{x}^{\underline{k}}$ in
$M\otimes\widehat{\mathcal{O}}_{X}\cong M[\\![x_{1},\dots,x_{d}]\\!]$. If
$a_{\underline{k}}\in N$ for any $\underline{k}$, then $h\in
N\otimes_{\mathcal{O}_{S}}\mathcal{O}_{X}$. In particular, if $k$ is perfect
and we are in the situation of 3.7, $h\in\mathcal{O}_{X}\otimes\pi^{-1}A$
belongs to $(\mathcal{O}_{X}\otimes\pi^{-1}A)^{\lambda}$ if and only if
$a_{\underline{k}}\in(\pi^{-1}A)^{\lambda}$ for any $\underline{k}$.

###### Proof.

Since $N$ is finitely generated, we have

$N\otimes_{\mathcal{O}_{S}}\widehat{\mathcal{O}}_{X}\cong\Bigl{\\{}\sum
a_{\underline{k}}\underline{x}^{\underline{k}}\mid a_{\underline{k}}\in
N\Bigr{\\}}\subset M[\\![x_{1},\dots,x_{d}]\\!],$

and the first claim follows by the lemma. To check the second claim, “if” part
is the only problem. The $\mathcal{O}_{S}$-module $M:=\pi^{-1}\sum_{n\leq
a}(\pi^{-1}\mathcal{O}_{S})^{1/p^{n}}$ is finitely generated for any $a$ since
$k$ is assumed to be perfect. Apply the first claim for
$N:=\bigl{(}\pi^{-1}\sum_{n\leq
a}(\pi^{-1}\mathcal{O}_{S})^{1/p^{n}}\bigr{)}^{\lambda}$, and take the
inductive limit over $a$ to conclude. ∎

### 3.10

Let $k$ be perfect and $v\in\left<S\right>$. This corollary implies that for
$h\in\mathcal{O}_{X}\otimes\pi^{-1}A$, $\mathrm{VP}(h)(v)$ is the largest
$\lambda\in J$ such that $h\in(\mathcal{O}_{X}\otimes\pi^{-1}A)^{\lambda}$.
The image of $h$ in
$\mathrm{gr}^{\mathrm{VP}(h)(v)}(\mathcal{O}_{X}\otimes\pi^{-1}A)$ is called
the symbol denoted by $\sigma_{v}(h)$ or simply by $\sigma(h)$. By
construction, $\sigma(h)\neq 0$ if $\lambda=\mathrm{VP}(h)(v)<0$. An element
in $\mathcal{O}_{X}\otimes\mathrm{gr}^{\lambda}(\pi^{-1}A)$ is said to be
separable if the image of the element by the homomorphism

$\mathcal{O}_{X}\otimes\mathrm{gr}^{\lambda}(\pi^{-1}A)\hookrightarrow\mathcal{O}_{X}\otimes\mathrm{gr}^{\lambda}\bigl{(}k(S)^{1/p^{\infty}}\bigr{)}\cong\mathcal{O}_{X}\otimes\mathrm{gr}^{0}\bigl{(}k(S)^{1/p^{\infty}}\bigr{)}\cong\mathcal{O}_{X}\otimes_{k}k_{v}^{1/p^{\infty}}$

is ($\mathrm{Spec}(k_{v}^{1/p^{\infty}})$-)separable. Note that the first
isomorphism between $k_{v}^{1/p^{\infty}}$-vector spaces
$\mathrm{gr}^{\lambda}$ and $\mathrm{gr}^{0}$ is given once we fix $f\in
k(S)^{1/p^{\infty}}$ such that $v(f)=\lambda$. In particular, it is not
canonical, but the separability does not depend on the choice.

######  Lemma. —

We keep assuming $k$ to be perfect. Let $h\in\mathcal{O}_{X}\otimes\pi^{-1}A$
and assume $\mathrm{VP}(h)(v)<0$.

  1. 1.

If $\sigma_{v}(h)$ is separable, then there exists a power $q$ of $p$ and a
local modification $S^{\prime}\rightarrow S^{1/q}$ such that $h$ can be
written as $h^{\prime}/\pi^{\prime}$ where $h^{\prime}$ is a separable
function on $\mathcal{O}_{X_{S^{\prime}}}$ and
$\pi^{\prime}\in\mathcal{O}_{S^{\prime}}$. If $h\in\pi^{-1}\mathcal{O}_{X}$,
then we may take $q=1$.

  2. 2.

Assume $\mathrm{ht}(v)=1$ and $\mathrm{VP}^{(\infty)}(h)(v)<0$. Then the
symbol $\sigma_{v}(h)$ is separable if and only if $h$ is weakly admissible at
$v$, namely $\mathrm{VP}^{(\infty)}(h)(v)=\mathrm{VP}(h)(v)$.

###### Proof.

Let us show the first one. Let $\lambda:=\mathrm{VP}(h)(v)<0$. We may take
$\pi^{\prime}\in k(S)^{1/p^{\infty}}$ such that $v(\pi^{\prime})=-\lambda>0$.
Take $S^{\prime}$ so that $\pi^{\prime}$ and $h^{\prime}:=\pi^{\prime}\cdot h$
belong to $\mathcal{O}_{S^{\prime}}$. Then $h^{\prime}:=\pi^{\prime}\cdot h$
is separable if we shrink $S^{\prime}$ around the center of $v$, thus the
claim follows.

Let us show the second claim. Write $h=\sum
a_{\underline{n}}\underline{x}^{\underline{n}}$. Let $\Sigma$ be the set of
$\underline{n}$ such that $v(a_{\underline{n}})=\mathrm{VP}(h)(v)$. The symbol
$\sigma(h)$ is separable if and only if there exists $\underline{n}\in\Sigma$
such that $p\nmid\underline{n}$ by Lemma 1.4. Assume there exists
$\underline{n}_{0}\in\Sigma$ such that $p\nmid\underline{n}_{0}$. Then for any
$i>0$,

$p^{-i}\cdot v(a_{p^{i}\underline{n}_{0}})>v(a_{\underline{n}_{0}}).$

Indeed, if $v(a_{p^{i}\underline{n}_{0}})\geq 0$, the claim follows since
$v(a_{\underline{n}_{0}})<0$, and otherwise, $p^{-i}\cdot
v(a_{p^{i}\underline{n}_{0}})>v(a_{p^{i}\underline{n}_{0}})\geq
v(a_{\underline{n}_{0}})$. This implies that
$v\bigl{(}\sum_{i=0}^{N}(a_{p^{i}\underline{n}_{0}})^{1/p^{i}}\bigr{)}=v(a_{\underline{n}_{0}})$
for any $N$. Thus, $\mathrm{VP}^{(\infty)}(h)(v)=\mathrm{VP}(h)(v)$.

Assume, on the other hand, that any element of $\Sigma$ is divisible by $p$.
Take $\underline{n}\in\Lambda$. If
$\bigl{\\{}p^{i}\underline{n}\bigr{\\}}_{i\geq 0}\cap\Sigma=\emptyset$, then
$v(a_{p^{i}\underline{n}})>\mathrm{VP}(h)(v)$ for any $i$. If
$p^{a}\underline{n}\in\bigl{\\{}p^{i}\underline{n}\bigr{\\}}_{i\geq
0}\cap\Sigma$, then $a>0$ by assumption. In this case, for any $k$, we have

$\displaystyle
v\bigl{(}\sum_{i=0}^{k}(a_{p^{i}\underline{n}})^{1/p^{i}}\bigr{)}>v(a_{p^{a}\underline{n}}).$

Since $\mathrm{VP}^{(\infty)}(h)(v)\in J$, which is well-ordered, we have
$\mathrm{VP}^{(\infty)}(h)(v)>\mathrm{VP}(h)(v)$. ∎

######  3.11 Lemma. —

Assume that $k$ is algebraically closed and $v$ is minimal. For any
$h\in\mathcal{O}_{X}\otimes\pi^{-1}A$ such that
$\mathrm{VP}(h)<\mathrm{VP}^{(\infty)}(h)$, there exists
$g\in\mathcal{O}_{X}\otimes(\pi^{-1}A)^{1/p}\subset\mathcal{O}_{X}\otimes\pi^{-1}A$
such that $\mathrm{VP}(h)<\mathrm{VP}\bigl{(}h+(g^{p}-g)\bigr{)}$.

###### Proof.

Since $\mathrm{trdeg}_{k}(k_{v})=0$ and $k$ is assumed to be algebraically
closed, $\mathrm{gr}^{0}(\mathcal{O}_{S})$ is equal to $k$. Moreover,

$\mathrm{gr}\bigl{(}k(S)^{1/p^{\infty}}\bigr{)}\cong k[\Gamma_{v}^{(p)}],$

where $\Gamma_{v}^{(p)}:=\bigcup_{i>0}p^{-i}\Gamma_{v}$. Caution that this is
a non-canonical isomorphism of $k$-modules. Since $\mathrm{gr}^{\lambda}(A)$
is a $\mathrm{gr}^{0}(\mathcal{O}_{S})\cong k$-vector subspace of
$\mathrm{gr}^{\lambda}(k(S)^{1/p^{\infty}})$, we get that
$\mathrm{gr}^{\lambda}(A)$ is either $k$-module of rank $0$ or $1$. Since the
map
$\mathrm{gr}^{\lambda}(\pi^{-1}A)\rightarrow\mathrm{gr}^{\lambda+v(\pi)}(A)$
sending $\alpha$ to $\pi\alpha$ is an isomorphism, the same is true for
$\mathrm{gr}^{\lambda}(\pi^{-1}A)$. In the same way,
$\mathrm{gr}^{\lambda}\bigl{(}(\pi^{-1}A)^{1/q}\bigr{)}$ is a $k$-vector space
of dimension either $0$ or $1$. Now, take $0>\lambda\geq-v(\pi)$ such that
$\mathrm{gr}^{\lambda}(\pi^{-1}A)$ is of rank $1$. Then there exists
$\alpha\in\pi^{-1}A$ such that $v(\alpha)=\lambda$. Let $q$ be a power of $p$.
Because valuations are multiplicative, the element $\sigma(\alpha^{1/q})$ does
not vanish in $\mathrm{gr}^{\lambda/q}\bigl{(}(\pi^{-1}A)^{1/q}\bigr{)}$. Thus
the latter $k$-vector space is also of dimension $1$. This implies that we
have the following commutative diagram whose vertical homomorphisms are
isomorphic:

($\star$)
$\textstyle{\mathcal{O}_{X}\otimes\mathrm{gr}^{\lambda/q}\bigl{(}(\pi^{-1}A)^{1/q}\bigr{)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(-)^{q}}$$\scriptstyle{\sim}$$\textstyle{\mathcal{O}_{X}\otimes\mathrm{gr}^{\lambda}(\pi^{-1}A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\sim}$$\textstyle{\mathcal{O}_{X}\otimes\mathrm{gr}^{\lambda/q}\bigl{(}k(S)^{1/p^{\infty}}\bigr{)}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{(-)^{q}}$$\textstyle{\mathcal{O}_{X}\otimes\mathrm{gr}^{\lambda}\bigl{(}k(S)^{1/p^{\infty}}\bigr{)}.}$

Here, the horizontal homomorphisms are $q$-th power map.

Now, let us go back to the proof. By Lemma 3.10.2, $\sigma(h)$ is not
separable. This implies that $\sigma(h)$ is in the image of the $q$-th power
map in $\mathcal{O}_{X}\otimes\mathrm{gr}(k(S)^{1/p^{\infty}})$ for some power
$q>1$ of $p$ by Lemma 1.3.4. Since the vertical homomorphisms of ($\star$ ‣
3.10) are isomorphisms, there exists a separable function
$g^{\prime}\in\mathrm{gr}\bigl{(}\mathcal{O}_{X}\otimes(\pi^{-1}A)^{1/q}\bigr{)}$
such that $g^{\prime q}=-\sigma(h)$ where $q>1$ is a power of $p$. Now, we can
take an extension $g\in\mathcal{O}_{X}\otimes(\pi^{-1}A)^{1/q}$ such that
$\sigma(g)=g^{\prime}$. By the choice of $g^{\prime}$, we have
$v(h+(g^{q}-g))>v(h)=0$. This implies that

$\mathrm{VP}\bigl{(}h+(g^{q}-g)\bigr{)}(v)>\mathrm{VP}(h)(v).$

Finally by replacing $g$ by $g+g^{p}+\dots+g^{q/p}$, we may take $q=p$, and
the lemma follows. ∎

######  3.12 Lemma. —

Let $f\in\mathcal{O}_{X}$, $\pi\in\mathcal{O}_{S}\setminus\\{0\\}$, and let
$v$ be a minimal valuation such that $\mathrm{VP}^{(\infty)}(f/\pi)(v)<0$.
Then there exists a finite extension $L$ of $k(S)$ such that $f/\pi$ is weakly
admissible at any extension of $v$ to $L$.

###### Proof.

We may assume that $k$ is algebraically closed. Since $v$ is of height $1$,
$\Gamma_{v}\subset\mathbb{R}$. For any $-v(\pi)<\varepsilon<0$, the set
$I_{\varepsilon}:=v(\pi^{-1}A)\cap\left]-\infty,\varepsilon\right]\subset J$
is finite by Lemma Lemma and (3.4.1). Set $I:=I_{\mathrm{VP}^{(\infty)}(v)}$.
By Lemma 3.6, we have
$\mathrm{VP}\bigl{(}f/\pi+(g^{p}-g)\bigr{)}(v)\leq\mathrm{VP}^{(\infty)}(v)$
for any $g\in\mathcal{O}_{X}\otimes_{\mathcal{O}_{S}}\pi^{-1}A$, we have
$\mathrm{VP}\bigl{(}f/\pi+(g^{p}-g)\bigr{)}(v)\in I$. Now, apply Lemma 3.11
with $h=f/\pi$, and we can find a function
$g\in\mathcal{O}_{X}\otimes(\pi^{-1}A)^{1/p}$ such that
$\mathrm{VP}\bigl{(}f/\pi+(g^{p}-g)\bigr{)}>\mathrm{VP}(f/\pi)$. Note that
$h+(g^{p}-g)$ is still in $\mathcal{O}_{X}\otimes\pi^{-1}A$, and thus
$\mathrm{VP}\bigl{(}f/\pi+(g^{p}-g)\bigr{)}\in I$. We again apply Lemma 3.11,
this time, with $h=f/\pi+(g^{p}-g)$. We repeat this procedure. Since $I$ is a
finite set, this procedure terminates after finitely many times, and we find
$g^{\prime}\in\mathcal{O}_{X}\otimes\pi^{-1}A$ such that
$\mathrm{VP}^{(\infty)}(f/\pi)=\mathrm{VP}\bigl{(}f/\pi+(g^{\prime
p}-g^{\prime})\bigr{)}$. We take $S^{\prime}:=S^{1/q}$ so that $\pi
g^{\prime}\in\mathcal{O}_{X_{S^{\prime}}}$ where $q$ is a power of $p$. ∎

######  3.13 Lemma. —

If $h\in\mathcal{O}_{X_{\eta}}$ is weakly admissible at a valuation $v$ of
height $1$, then there exists a finite extension $L$ of $k(S)$ so that $h$ is
admissible at any extension of $v$.

###### Proof.

Assume $\mathrm{VP}(h)(v)=0$. We may write $h=f/\pi$ where
$f\in\mathcal{O}_{X}$, $\pi\in\mathcal{O}_{S}$. Write $f=\sum
a_{\underline{k}}\underline{x}^{\underline{k}}$. By assumption,
$v(a_{\underline{k}})\geq v(\pi)$. Let $I\subset\mathcal{O}_{S}$ be the ideal
generated by $\\{a_{\underline{k}}\\}$. For any $b\in I$, we have $v(b)\geq
v(\pi)$. Since $\mathcal{O}_{S}$ is noetherian, there exist finitely many
generators $c_{i}$ of $I$. We take a modification $S^{\prime}$ of $S$ so that
$c_{i}/\pi$ are regular functions around the center of $v$, which is possible
since $v(c_{i}/\pi)\geq 0$. Then the claim follows since
$h\in\mathcal{O}_{X_{S^{\prime}}}$ by Corollary 3.9. Note that in this case,
we can take the local alteration to be a “local modification”. Assume, now,
that $\mathrm{VP}(h)(v)<0$. Let $l/k$ be an extension of fields. For a
$k$-scheme $Z$ and a morphism $a\colon Z\rightarrow\mathbb{A}^{1}$, if
$a\otimes_{k}l$ is smooth, then $a$ is smooth by [EGA IV, 17.7.3]. Thus, $h$
is a good ally if and only if $h\otimes_{k}l$ is a good ally, and we may
assume that $k$ is perfect. Then the claim follows by Lemma 3.10. ∎

### 3.14

By combining what we have proven so far, when $v$ is minimal and
$\mathrm{VP}^{(\infty)}(h)(v)<0$, there exists a local alteration over which
$h$ admits a good ally around an extension of $v$. The next step is to treat
the case where $v$ is minimal and $\mathrm{VP}^{(\infty)}(h)(v)=0$. This case
requires more serious analysis, and we use the induction on the transcendence
defect of $v$ following [Ke4]. The base of the induction is the case where $v$
is monomial. In fact, we can show the weakly admissibility for Abhyankar
valuation of height $1$ as follows:

######  Lemma. —

Let $v$ be an Abhyankar valuation of $\mathrm{ht}(v)=1$, and assume we are
given an étale morphism $S\rightarrow\mathbb{A}^{e}_{(s_{1},\dots,s_{e}),k}$
such that $v(s_{1}),\dots,v(s_{r})$ is $\mathbb{Q}$-linearly independent with
$r=\mathrm{rat.rk}(v)\leq e$, and $Z:=V(s_{1},\dots,s_{r})$ is irreducible.
Let $f\in\mathcal{O}_{X}$ and $\pi\in\mathcal{O}_{S}$. Assume that
$V(\pi)\subset V(s_{1}\cdots s_{r})$ and $\mathrm{VP}^{(\infty)}(f/\pi)(v)=0$.
Then $f/\pi$ is weakly admissible at $v$, and admits a weak mollifier so that
$g\in\pi^{-1}\mathcal{O}_{X}$. If $v$ is divisorial (i.e. $r=1$), then we can
take $g$ so that it provides a good ally of $f/\pi$ on $S$ at $v$.

###### Proof.

The closure of the center of $v$ on $S$ is equal to $Z$ since $Z$ is assumed
irreducible. Note that $k_{v}\supset\mathcal{O}_{Z}$. Invoking [EGA
$0_{\mathrm{IV}}$, 19.5.4], the coordinates $s_{i}$ together with the fixed
coordinates $x_{i}$ yield the embeddings

$\mathcal{O}_{X}\hookrightarrow\mathcal{O}_{X_{Z}}[\\![s_{1},\dots,s_{r}]\\!]\subset\mathcal{O}_{X_{v}}[\\![s_{1},\dots,s_{r}]\\!],$

where $X_{v}:=X\times_{S}\mathrm{Spec}(k_{v})$. For $g=\sum
a_{\underline{n}}\underline{s}^{\underline{n}}\in\mathcal{O}_{X_{v}}[\\![\underline{s}]\\!]$,
we put $\widetilde{v}(g):=\min\\{v(\underline{s}^{\underline{n}})\mid
a_{\underline{n}}\neq 0\\}$. Note that minimum is uniquely attained by the
assumption that $v(s_{1}),\dots,v(s_{r})$ are linearly independent, and
$\widetilde{v}$ is a valuation that extends $v$. If no confusion can arise, we
denote $\widetilde{v}$ abusively by $v$. Note that for any
$h\in\mathcal{O}_{X_{Z}}[\\![\underline{s}]\\!]$ and $\lambda>0$, there exists
$\widetilde{h}\in\mathcal{O}_{X}$ such that $v(\widetilde{h}-h)>\lambda$ in
$\mathcal{O}_{X_{v}}[\\![\underline{s}]\\!]$ because $v(s_{i})>0$ and
$\mathcal{O}_{X_{Z}}[\\![\underline{s}]\\!]$ is the completion of
$\mathcal{O}_{X}$ with respect to $(s_{1},\dots,s_{r})$-adic topology. Because
$S$ is regular and $V(\pi)\subset V(s_{1}\dots s_{r})$, using [EGA IV,
(21.1.3.3) and 21.6.9], we may write
$\pi=\underline{s}^{\underline{\alpha}}\pi^{\times}$ where
$\pi^{\times}\in\mathcal{O}^{\times}_{S}$ (thus $v(\pi^{\times})=0$). The set
$\Sigma$ of $\underline{n}\in\mathbb{N}^{r}$ such that
$v(\underline{s}^{\underline{n}})<v(\pi)=v(\underline{s}^{\underline{\alpha}})$
is finite. Then we may write uniquely
$(\pi^{\times})^{-1}f=\sum_{\underline{n}\in\Sigma}b_{\underline{n}-\underline{\alpha}}\underline{s}^{\underline{n}}+f_{\mathrm{res}}$
in $\mathcal{O}_{X_{Z}}[\\![\underline{s}]\\!]$ where $v(f_{\mathrm{res}})\geq
v(\pi)$ and $b_{\underline{n}}\in\mathcal{O}_{X_{Z}}$. Note that
$f_{\mathrm{res}}$ may not be in the image of $\mathcal{O}_{X}$. Then

$f/\pi=\bigl{(}\underline{s}^{-\underline{\alpha}}\sum_{\underline{n}\in\Sigma}b_{\underline{n}-\underline{\alpha}}\underline{s}^{\underline{n}}\bigr{)}+\underline{s}^{-\underline{\alpha}}\cdot
f_{\mathrm{res}}=\bigl{(}\sum_{\underline{n}\in\Sigma-\underline{\alpha}}b_{\underline{n}}\underline{s}^{\underline{n}}\bigr{)}+\underline{s}^{-\underline{\alpha}}\cdot
f_{\mathrm{res}}.$

Put
$\mathcal{O}_{X_{Z}}(\\!(\underline{s})\\!):=\mathop{\underrightarrow{\mathrm{lim}}}_{|\underline{n}|\rightarrow\infty}\underline{s}^{-\underline{n}}\cdot\mathcal{O}_{X_{Z}}[\\![\underline{s}]\\!]$.
Note that $\underline{s}^{-\underline{\alpha}}\cdot
f_{\mathrm{res}}\in\bigl{\\{}h\in\mathcal{O}_{X_{Z}}(\\!(\underline{s})\\!)\mid
v(h)\geq 0\bigr{\\}}=:\mathcal{O}_{X_{Z}}(\\!(\underline{s})\\!)_{\geq 0}$.
Now, put

$\Sigma^{\prime}:=\bigl{\\{}\underline{n}\in\Sigma-\underline{\alpha}\mid
p\underline{n}\not\in\Sigma-\underline{\alpha}\bigr{\\}}\subset\mathbb{Z}^{r},$

and for $\underline{n}\in\Sigma^{\prime}$, we put

$c_{\underline{n}}:=b_{\underline{n}}+b^{p}_{\underline{n}/p}+\dots+b^{p^{M}}_{\underline{n}/p^{M}}\quad\in\mathcal{O}_{X_{Z}},$

where $M$ is the largest number such that
$\underline{n}/p^{M}\in\mathbb{Z}^{r}$. Now, we have the following 2 cases:

  1. 1.

We have
$c_{\underline{n}}\in\bigl{(}\mathcal{O}_{X_{v}}\otimes_{k_{v}}k_{v}^{\mathrm{perf}}\bigr{)}^{p^{\infty}}\,(=\bigcap_{i}(\dots)^{p^{i}})$
for any $\underline{n}\in\Sigma^{\prime}$.

  2. 2.

There exists $\underline{n}_{0}\in\Sigma^{\prime}$ such that
$c_{\underline{n}_{0}}\not\in\bigl{(}\mathcal{O}_{X_{v}}\otimes_{k_{v}}k_{v}^{\mathrm{perf}}\bigr{)}^{p^{m+1}}$
for some integer $m$.

Put

$g:=\sum_{\underline{n}\in\Sigma^{\prime}}\sum_{i=1}^{d_{\underline{n}}}\bigl{(}\sum_{k=i}^{d_{\underline{n}}}\widetilde{b}_{\underline{n}/p^{k}}\underline{s}^{\underline{n}/p^{k}}\bigr{)}^{p^{i-1}},$

where
$d_{\underline{n}}:=\max\bigl{\\{}i\mid\underline{n}/p^{i}\in\mathbb{Z}^{r}\bigr{\\}}$
(this is the same as $M$ in the definition of $c_{\underline{n}}$), and
$\widetilde{b}_{\underline{n}}\in\mathcal{O}_{X}$ such that
$v\bigl{(}\widetilde{b}_{\underline{n}}-b_{\underline{n}}\bigr{)}>v(\pi)$.
Note that $\pi g\in\mathcal{O}_{X}$. Then we may compute

$\bigl{(}\sum_{\underline{n}\in\Sigma-\underline{\alpha}}b_{\underline{n}}\underline{s}^{\underline{n}}\bigr{)}+(g^{p}-g)\in\sum_{\underline{n}\in\Sigma^{\prime}}c_{\underline{n}}\underline{s}^{\underline{n}}+\mathcal{O}_{X_{Z}}(\\!(\underline{s})\\!)_{\geq
0}.$

Consider the case 1. The section of $X\rightarrow S$ yields a section of
$X_{v}\rightarrow\mathrm{Spec}(k_{v})$. Since $X_{v}$ is smooth over $k_{v}$,
we have the inclusion $\mathcal{O}_{X_{v}}\subset
k_{v}[\\![x_{1},\dots,x_{d}]\\!]$. We have

$k_{v}^{\mathrm{perf}}\subset\bigl{(}\mathcal{O}_{X_{v}}\otimes_{k_{v}}k_{v}^{\mathrm{perf}}\bigr{)}^{p^{\infty}}\subset\bigl{(}k_{v}^{\mathrm{perf}}[\\![x_{1},\dots,x_{d}]\\!]\bigr{)}^{p^{\infty}}=k_{v}^{\mathrm{perf}}.$

Thus, $c_{\underline{n}}\in k^{\mathrm{perf}}_{v}\cap\mathcal{O}_{X_{Z}}$ in
$(k_{v}[\\![x_{1},\dots,x_{d}]\\!])^{\mathrm{perf}}$ for any
$\underline{n}\in\Sigma^{\prime}$. On the other hand, we have

$k_{v}\subset k^{\mathrm{perf}}_{v}\cap\mathcal{O}_{X_{v}}\subset
k^{\mathrm{perf}}_{v}\cap k_{v}[\\![x_{1},\dots,x_{d}]\\!]=k_{v}$

and thus $k^{\mathrm{perf}}_{v}\cap\mathcal{O}_{X_{v}}=k_{v}$. This implies
that $c_{\underline{n}}\in k_{v}\cap\mathcal{O}_{X_{Z}}=\mathcal{O}_{Z}$ since
$X\rightarrow S$ is faithfully flat. Now, we can take
$\widetilde{c}_{\underline{n}}\in\mathcal{O}_{S}$ such that
$v\bigl{(}\widetilde{c}_{\underline{n}}-c_{\underline{n}}\bigr{)}>v(\pi)$.
Then we have
$f/\pi+(g^{p}-g)\in\sum_{\underline{n}\in\Sigma^{\prime}}\widetilde{c}_{\underline{n}}\underline{s}^{\underline{n}}+\mathcal{O}_{X_{Z}}(\\!(\underline{s})\\!)_{\geq
0}\subset\mathrm{R}(S)+\mathcal{O}_{X_{Z}}(\\!(\underline{s})\\!)_{\geq 0}$.
Thus, $\mathrm{VP}\bigl{(}f/\pi+(g^{p}-g)\bigr{)}(v)=0$, and $g$ is a weak
mollifier at $v$. If $v$ is divisorial, we have
$\mathcal{O}_{X_{Z}}(\\!(s_{1})\\!)_{\geq
0}=\mathcal{O}_{X_{Z}}[\\![s_{1}]\\!]$, and thus $f/\pi+(g^{p}-g)$ is already
a good ally since
$\mathcal{O}_{X_{Z}}[\\![s_{1}]\\!]\cap\mathcal{O}_{X_{\eta}}=\mathcal{O}_{X}$
by Corollary 3.9.

Finally, consider 2. Let us show that this occurs only when
$\mathrm{VP}^{(\infty)}(v)<0$ which is impossible from the assumption of the
lemma. Take $\widetilde{c}_{\underline{n}}\in\mathcal{O}_{X}$ (not in
$\mathcal{O}_{S}$ this time) such that
$v\bigl{(}\widetilde{c}_{\underline{n}}-c_{\underline{n}}\bigr{)}>v(\pi)$,
then
$f/\pi+(g^{p}-g)\in\sum_{\underline{n}\in\Sigma^{\prime}}\widetilde{c}_{\underline{n}}\underline{s}^{\underline{n}}+\mathcal{O}_{X_{v}}(\\!(\underline{s})\\!)_{\geq
0}$. Write $\widetilde{c}_{\underline{n}_{0}}=\sum
a_{\underline{k}}\underline{x}^{\underline{k}}$ in
$\mathcal{O}_{S}[\\![\underline{x}]\\!]$. We have the following commutative
diagram:

$\textstyle{\mathcal{O}_{X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{O}_{S}[\\![\underline{x}]\\!]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{O}_{X_{v}}[\\![\underline{s}]\\!]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{k_{v}[\\![\underline{s},\underline{x}]\\!].}$

Since
$c_{\underline{n}}\in\mathcal{O}_{X_{Z}}\subset\mathcal{O}_{X_{v}}[\\![\underline{s}]\\!]$,
we may write $c_{\underline{n}_{0}}=\sum
a^{\prime}_{\underline{k}}\underline{x}^{\underline{k}}$ with
$a^{\prime}_{\underline{k}}\in k_{v}$ in
$k_{v}[\\![\underline{s},\underline{x}]\\!]$. By the choice of
$\widetilde{c}_{\underline{n}}$, we have
$v(a_{\underline{k}}-a^{\prime}_{\underline{k}})>v(\pi)$ for any
$\underline{k}$, and in particular, $v(a_{\underline{k}})$ is either $0$ or
$>v(\pi)$. Because of Lemma 1.4 and the assumption that
$c_{\underline{n}_{0}}\not\in\mathcal{O}^{p^{m+1}}_{X_{v}}\otimes k_{v}$,
there exists $p^{m+1}\nmid\underline{k}_{0}$ such that
$v(a_{\underline{k}_{0}})=0$. Let $\underline{l}_{0}\in\Lambda$ such that
$p^{K}\underline{l}_{0}=\underline{k}_{0}$ with some $K\in\mathbb{N}$. Then,
for any integer $a\geq K$, we have

$\displaystyle v$
$\displaystyle\bigl{(}(\widetilde{c}_{\underline{n}_{0}}\underline{s}^{\underline{n}_{0}})^{(a)}_{\underline{l}_{0}}\bigr{)}=$
$\displaystyle
v\Bigl{(}\underline{s}^{\underline{n}_{0}}a_{\underline{l}_{0}}+(\underline{s}^{\underline{n}_{0}}a_{p\underline{l}_{0}})^{1/p}+\dots+(\underline{s}^{\underline{n}_{0}}a_{\underline{k}_{0}})^{1/p^{K}}+(\underline{s}^{\underline{n}_{0}}a_{p^{K+1}\underline{l}_{0}})^{1/p^{K+1}}+\dots+(\underline{s}^{\underline{n}_{0}}a_{p^{a}\underline{l}_{0}})^{1/p^{a}}\Bigr{)}$
$\displaystyle\hskip
230.00035pt\in\bigl{\\{}p^{-i}v(\underline{s}^{\underline{n}_{0}})\mid
i=0,\dots,K\bigr{\\}}=:D.$

For $\underline{n}\in\Sigma^{\prime}\setminus\\{\underline{n}_{0}\\}$, by a
similar computation,
$v\bigl{(}(\widetilde{c}_{\underline{n}}\underline{s}^{\underline{n}})^{(a)}_{\underline{l}_{0}}\bigr{)}$
is also a real number of the form $p^{-i}v(\underline{s}^{\underline{n}})$. By
the choice of $\Sigma^{\prime}$, this in particular implies that this number
cannot be in $D$. Thus, we have

$\mathrm{VP}^{(\infty)}(f/\pi)(v)=\mathrm{VP}^{(\infty)}\bigl{(}f/\pi+(g^{p}-g)\bigr{)}(v)\leq\lim_{a\rightarrow\infty}v\left(\bigl{(}{\textstyle\sum_{\underline{n}\in\Sigma^{\prime}}}\,\widetilde{c}_{\underline{n}}\underline{s}^{\underline{n}}\bigr{)}^{(a)}_{\underline{l}_{0}}\right)\leq
p^{-K}v(\underline{s}^{\underline{n}_{0}})<0,$

and the claim follows. ∎

### 3.15

We have the following purity type result.

######  Lemma. —

Let $v$ be a minimal valuation on $S$ centered on $z\in U\subset S$. Assume we
have a system of coordinates $\\{s_{1},\dots,s_{d}\\}$ on $U$ around $z$ such
that $\pi\in s_{1}^{a_{1}}\dots s_{k}^{a_{k}}\mathcal{O}^{\times}_{S}$. Let
$D_{i}:=V(s_{i})$, and denote by $v_{D_{i}}$ the associated divisorial
valuation. Then the following conditions are equivalent:

  1. 1.

$\mathrm{VP}^{(\infty)}(f/\pi)(v)=0$ and there exists a mollifier which
provides a good ally of $f/\pi$ on an open neighborhood of $z$ in $U$;

  2. 2.

for any divisorial valuation $v_{D}$ centered on $U$, we have
$\mathrm{VP}^{(\infty)}(f/\pi)(v_{D})=0$;

  3. 3.

$\mathrm{VP}^{(\infty)}(f/\pi)(v_{D_{i}})=0$ for any $i$.

###### Proof.

Assume the condition 1 holds. Take a mollifier $g$ on $U$ providing a good
ally around $z$. Then taking a suitable $\gamma\in\mathrm{R}(S)$,
$f/\pi+(g^{p}-g)+\gamma$ is a regular function around $v$ since it cannot be
of the form $f^{\prime}/\sigma+\alpha$ where $f^{\prime}$ is separable
function, $\alpha$ is a regular function, and $v(\sigma)>0$ by the assumption
that $\mathrm{VP}^{(\infty)}(f/\pi)(v)=0$. Thus, there exists a closed
subscheme $Z\subset U$ not containing $z$ such that $f/\pi+(g^{p}-g)+\gamma$
is in $\mathcal{O}_{X}\otimes\mathcal{O}_{U\setminus Z}$. This implies that
$\mathrm{VP}^{(\infty)}(f/\pi)(w)=0$ for any valuation $w$ centered on
$U\setminus Z$. This in particular implies that
$\mathrm{VP}^{(\infty)}(f/\pi)(v_{D_{i}})=0$, and condition 3 holds.

Now, let us check (3 $\Rightarrow$ 1). We show by the induction on $k$. Since
$v_{D_{k}}$ is a divisorial valuation, there exists a function
$g_{k}\in\mathcal{O}_{X}\otimes\pi^{-n}\mathcal{O}_{U}$ for some $n\geq 0$
such that $\mathrm{VP}\bigl{(}f/\pi+(g^{p}_{k}-g_{k})\bigr{)}(v_{D_{k}})=0$ by
applying Lemma Lemma to $U\setminus V(s_{1}\dots s_{k-1})$. Note that by
construction,
$f/\pi+(g^{p}_{k}-g_{k})\in\mathcal{O}_{X}\otimes\pi^{-pn}\mathcal{O}_{U}$.
However, since $\mathrm{VP}(f/\pi+(g^{p}_{k}-g_{k}))(v_{D_{k}})=0$, there
exists $\gamma_{k}\in\mathrm{R}(S)$ such that
$f/\pi+(g^{p}_{k}-g_{k})+\gamma_{k}\in\mathcal{O}_{X}\otimes(s_{1}^{a_{1}}\dots
s_{k-1}^{a_{k-1}})^{-pn}\mathcal{O}_{U}$. This construction also shows that
the mollifier can be taken in $\pi^{-n}\mathcal{O}_{X_{U}}$ for some $n$. Now,
use the induction hypothesis to conclude.

Finally, let us check that (1 $\Rightarrow$ 2) holds. Let $Z$ be as in the
proof of (1 $\Rightarrow$ 3). We only need to check the claim for valuations
centered on $Z$ above. If $w$ is a valuation centered on
$Z\setminus(D_{1}\cup\dots\cup D_{k})$, then $f/\pi$ already provides a good
ally, and we have nothing to show. Let $w$ be a valuation centered on
$Z\cap(D_{1}\cup\dots\cup D_{k})$. Then we can apply (3 $\Rightarrow$ 1) for
this $w$ to conclude. ∎

######  3.16 Corollary. —

Let $v\in\left<S\right>$ be of height $1$. Assume that
$\mathrm{VP}^{(\infty)}(f/\pi)(v)=0$ and there exists a finite extension $L$
of $k(S)$ and some extension $v^{\prime}$ of $v$ to $L$ such that $f/\pi$ is
locally admissible at $v^{\prime}$. Then there exists a Zariski open
neighborhood $W$ of $v$ in $\left<S\right>$ such that
$\mathrm{VP}^{(\infty)}(f/\pi)(w)=0$ for any $w\in W$ and $\mathrm{ht}(w)=1$.

###### Proof.

We may take a local alteration $S^{\prime}\rightarrow S$ around $v$ and a
coordinate system $\\{s_{1},\dots,s_{d}\\}$ on $S^{\prime}$ such that $V(\pi)$
is a union of $V(s_{i})$’s and there exists a mollifier of $v$ on $S^{\prime}$
which provides a good ally of $f/\pi$ at $v$ by Lemma 3.13. By shrinking
$S^{\prime}$ around the center of an extension of $v$ to $S^{\prime}$, we may
assume that $\mathrm{VP}^{(\infty)}(f/\pi)(w)=0$ for any valuation on
$k(S^{\prime})$ centered on $S^{\prime}$. Now, let $W^{\prime}$ be the set of
valuations in $\left<S\right>$ which extends to $S^{\prime}$. Lemma Lemma and
(3.4.2) imply that $\mathrm{VP}^{(\infty)}(f/\pi)(w)=0$ for any $w\in
W^{\prime}$ and $\mathrm{ht}(w)=1$. Thus, it remains to show that $W^{\prime}$
is a neighborhood of $v$.

Take a compactification $\overline{S}^{\prime}\rightarrow S$ of $S^{\prime}$
over $S$. By the Gruson-Raynaud flattening, we may find modifications
$T\rightarrow S$ and $\overline{T}^{\prime}\rightarrow\overline{S}^{\prime}$
and a finite flat morphism $\overline{T}^{\prime}\rightarrow T$. We may assume
that $\overline{T}^{\prime}$ is integral. Let $T^{\prime}$ be the pullback of
$S^{\prime}$ to $\overline{T}^{\prime}$. The situation can be depicted as
follows:

$\textstyle{T^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\square}$$\textstyle{\overline{T}^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{fin.\
flat}}$$\scriptstyle{\mathrm{modif}}$$\textstyle{T\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{modif}}$$\textstyle{S^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{open}}$$\textstyle{\overline{S}^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{prop}}$$\textstyle{S.}$

Since $g\colon T^{\prime}\rightarrow T$ is flat, the image
$g(T^{\prime})\subset T$ is open. Let $w$ be an element of $\left<S\right>$
whose center lies in $g(T^{\prime})$. Then we claim that $w$ extends on
$S^{\prime}$, and in particular belongs to $W^{\prime}$. To show this, it
suffices to show that $w$ extends on $T^{\prime}$. Let $z$ be the center of
$w$ in $g(T^{\prime})$. Take a point $z^{\prime}\in T^{\prime}$ over $z\in
g(T^{\prime})$. Let $\mathrm{Spec}(R_{v})\rightarrow T$ be the morphism from
the valuation ring associated with $v$, and consider
$P:=\mathrm{Spec}(R_{v})\times_{T}T^{\prime}$. This scheme $P$ is integral.
Indeed, by shrinking $T$ around $z$, we may assume that $T$ is affine. We have
inclusions $\mathcal{O}_{T}\subset R_{v}\subset k(T)$. Since
$\mathcal{O}_{T}\rightarrow\mathcal{O}_{T}^{\prime}$ is flat, this induces the
inclusions $\mathcal{O}_{T^{\prime}}\subset\mathcal{O}_{P}\subset
k(T)\otimes_{\mathcal{O}_{T}}\mathcal{O}_{T^{\prime}}$. Since $k(T)$ is a
localization of $\mathcal{O}_{T}$,
$k(T)\otimes_{\mathcal{O}_{T}}\mathcal{O}_{T^{\prime}}$ is a localization of
$\mathcal{O}_{T^{\prime}}$, and since $T^{\prime}$ is integral, we have an
inclusion $k(T)\otimes_{\mathcal{O}_{T}}\mathcal{O}_{T^{\prime}}\subset
k(T^{\prime})$. Thus, we have $\mathcal{O}_{P}\subset k(T^{\prime})$ and the
integrality follows. The point $z^{\prime}$ induces a point $z^{\prime\prime}$
of $P$. We may take a valuation of $k(P)=k(T^{\prime})$ centered on
$z^{\prime\prime}$, which is the desired valuation centered on $T^{\prime}$.
Thus, if we put $W$ to be the Zariski open neighborhood defined by
$g(T^{\prime})$ in $\left<S\right>$, we have $W^{\prime}\supset W$. ∎

### 3.17

We use the following lemma due to Kedlaya. Even though it is completely
contained in [Ke4, 5.4.4], it may not be easy to transform the statement into
the form we need, we decided to include the proof for the convenience of the
reader.

######  Lemma ([Ke4, 5.4.4]). —

Let $S\rightarrow S^{0}\times\mathbb{A}^{1}_{t}$ be an étale morphism,
$\pi\in\mathcal{O}_{S^{0}}$ and let $v\in\left<S\right>$. Put
$v^{0}:=v|_{S^{0}}$, and assume that $v(t)$ is not contained in the divisible
closure of $\Gamma_{v^{0}}$. Assume we are given a Zariski open neighborhood
$W\subset\left<S\right>$ of $v$. Then by replacing $S^{0}$ by a local
alteration around $v^{0}$, there exist

  * •

$x_{1},\dots,x_{r}\in\mathcal{O}_{S^{0}}$ such that the components of $V(\pi)$
are cut out by some of these functions;

  * •

rational numbers $a_{i},b_{i}\in\mathbb{Q}$ such that
$v^{0}(x_{i})/b_{i}<v(t)<v^{0}(x_{i})/a_{i}$;

such that
$S\bigl{(}w(x_{i})/b_{i}<w(t)<w(x_{i})/a_{i}\bigr{)}:=\bigl{\\{}w\in\left<S\right>\mid
w(x_{i})/b_{i}<w(t)<w(x_{i})/a_{i}\bigr{\\}}\subset W$.

###### Proof.

Put $Z:=V(\pi)$ and $r=\mathrm{rat.rk}(v^{0})$. Since $v(t)>0$, the center of
$v$ is contained in $V(t)$. Since $V(t)\rightarrow S^{0}$ is étale by
assumption, by replacing $S^{0}$ by $V(t)$, we may assume that $S\rightarrow
S^{0}$ has a section and fix one. By [Ke4, 2.3.5], by replacing $S^{0}$ by its
alteration around $v^{0}$, we may assume that there exists a coordinate system
$\\{x_{1},\dots,x_{d}\\}$ such that $\\{v^{0}(x_{i})\\}_{r=1,\dots,r}$ are
linearly independent over $\mathbb{Q}$, and $\\{x_{i}\\}_{i=1,\dots,r}$ cuts
out the components of $Z$. Let $h\in\mathcal{O}_{S^{0}}\setminus\\{0\\}$. Then
we may write that $h=(x_{1}^{e_{1}}\dots x_{r}^{e_{r}})\cdot a$ where
$e_{i}\in\mathbb{Q}$ and $v^{0}(a)=0$. Take a modification so that
$a\in\mathcal{O}_{S^{0}}^{\times}$ followed by an exposing alteration $g\colon
S^{\prime 0}\rightarrow S^{0}$ of $(S^{0},Z)$ (cf. [Ke4, 2.3.4] for the
terminology). Let $x^{\prime}_{i}$ be the coordinates of $S^{\prime 0}$. Then
since components of $Z^{\prime}:=g^{-1}(Z)$ are cut out by
$\\{x^{\prime}_{i}\\}_{i=1,\dots,r}$, by shrinking $S^{\prime 0}$ around the
center of $v^{0}$ if necessarily, we may write $g^{*}(x_{1}^{e_{1}}\dots
x_{r}^{e_{r}})=({x^{\prime}_{1}}^{e^{\prime}_{1}}\dots
x^{\prime}_{r}^{e^{\prime}_{r}})\cdot b^{\prime}$ where
$b^{\prime}\in\mathcal{O}_{S^{\prime 0}}^{\times}$. Thus, over $S^{\prime 0}$,
we can write $g^{*}h=({x^{\prime}_{1}}^{e^{\prime}_{1}}\dots
x^{\prime}_{r}^{e^{\prime}_{r}})\cdot u^{\prime}$ with
$u^{\prime}\in\mathcal{O}_{S^{\prime 0}}^{\times}$. Note that since
${x^{\prime}_{1}}^{e^{\prime}_{1}}\dots
x^{\prime}_{r}^{e^{\prime}_{r}}\in\mathcal{O}_{S^{0}}$, $e^{\prime}_{i}$ are
non-negative integers.

We can take $f^{(1)},\dots,f^{(2n)}\in\mathcal{O}_{S}$ such that
$W\supset\bigl{\\{}w\mid f^{(k)}/f^{(n+k)}(w)\geq
0\quad(k=0,\dots,n)\bigr{\\}}\ni v$. Let $f$ be one of $f^{(k)}$’s. Using the
fixed section of $S\rightarrow S^{0}$, we may write $f=\sum f_{i}t^{i}$ in
$\mathcal{O}_{S^{0}}[\\![t]\\!]$ with $f_{i}\in\mathcal{O}_{S^{0}}$. Since
$s:=v(t)$ is not in the divisible closure, there exists $h$ such that
$v^{0}(f_{i})+is>v^{0}(f_{h})+hs$ for any $i\neq h$. Note that this $h$ is
chosen so that $v(f_{i}t^{i})>v(f_{h}t^{h})$ for any $i\neq h$. Take an index
$N$ such that $(N-h)s>v^{0}(f_{h})$. Note that, with this choice of $N$, we
have $v(f_{i}t^{i})>v(f_{h}t^{h})$ for any $i\geq N$. By the argument of the
previous paragraph, by taking further alteration, we may assume that
$f_{i}=u_{i}\cdot(x_{1}^{e_{i1}}\dots x_{r}^{e_{ir}})$ where
$e_{ij}\in\mathbb{Z}$, $u_{i}\in\mathcal{O}_{S^{0}}^{\times}$ for
$i=0,\dots,N-1$.

Now, for $f=f^{(k)}$, we put superscript (k) to $e_{ij}$ etc. To ease the
notation, $e^{(k)}_{h^{(k)}i}$ is denoted by $e^{(k)}_{hi}$. We have
$v\bigl{(}(x_{1}^{e^{(k)}_{i1}}\dots x_{r}^{e^{(k)}_{ir}})\cdot
t^{i-h^{(k)}}\bigr{)}>0$ for $i=0,\dots,N^{(k)}-1$ and $i\neq h^{(k)}$,
$v\bigl{(}(x_{1}^{-e_{h1}^{(k)}}\dots x_{r}^{-e_{hr}^{(k)}})\cdot
t^{N^{(k)}-h^{(k)}}\bigr{)}>0$, and
$v\bigl{(}(x_{1}^{e^{(k)}_{h1}-e^{(n+k)}_{h1}}\dots
x_{r}^{e^{(k)}_{hr}-e^{(n+k)}_{hr}})t^{h^{(k)}-h^{(n+k)}}\bigr{)}\geq 0$. Note
that if the last $\geq 0$ is $=0$, then we have $h^{(k)}=h^{(n+k)}$ and
$e^{(k)}_{hi}=e^{(n+k)}_{hi}$ for any $i$ by the choice of $x_{i}$ and $t$. We
take $c_{i}$ such that $v(x_{i})=c_{i}s$. Then these conditions can be written
as

  * •

$c_{1}e^{(k)}_{i1}+\dots+c_{r}e^{(k)}_{ir}+(i-h^{(k)})>0$ for any $k$,
$i=0,\dots,N^{(k)}-1$, and $i\neq h^{(k)}$;

  * •

$-(c_{1}e^{(k)}_{h1}+\dots+c_{r}e^{(k)}_{hr})+(N^{(k)}-h^{(k)})>0$ for any
$k$;

  * •

$(c_{1}e^{(k)}_{h1}+\dots+c_{r}e^{(k)}_{hr})-(c_{1}e^{(n+k)}_{h1}+\dots+c_{r}e^{(n+k)}_{hr})+h^{(k)}-h^{(n+k)}\geq
0$ for any $1\leq k\leq n$, and $=0$ if and only if $h^{(k)}=h^{(n+k)}$ and
$e^{(k)}_{hi}=e^{(n+k)}_{hi}$ for any $i$.

Thus, we may take $a_{i}<c_{i}<b_{i}$ such that for any $a_{i}\leq d_{i}\leq
b_{i}$, the inequalities holds even if we replace $c_{i}$ by $d_{i}$. This is
what we need. ∎

### 3.18

Let $t,s_{2},\dots,s_{e}$ be coordinates on $S$ around $z\in S$. Let
$U:=(u,u_{2}\dots,u_{e})\in\mathbb{Q}_{>0}^{e}$. Then the toric valuation
$v_{U}$ associated to $U$ is defined to be

$v_{U}\bigl{(}\sum
a_{\underline{k}}\underline{s}^{\underline{k}}\bigr{)}=\min\Bigl{\\{}\sum
k_{i}u_{i}\mid\mbox{$\underline{k}=(k_{1},\dots,k_{e})$ such that
$a_{\underline{k}}\neq 0$}\Bigr{\\}}$

for $\sum a_{\underline{k}}\underline{s}^{\underline{k}}\in
k(z)[\\![t,s_{2}\dots,s_{e}]\\!]\supset\mathcal{O}_{S}$. This valuation is
divisorial. Let $U^{\prime}=(u^{\prime},u_{2}\dots,u_{e})$ such that
$u^{\prime}\geq u$. Then for $g\in\mathcal{O}_{S}$, we have
$v_{U^{\prime}}(g)\geq v_{U}(g)$. Assume we have an étale morphism
$S\rightarrow\mathbb{A}^{1}_{t}\times S^{0}$. Assume
$\pi\in\mathcal{O}_{S^{0}}$ (not just in $\mathcal{O}_{S}$!) and
$f\in\mathcal{O}_{X}$. Then we have

(3.18.1)
$\mathrm{VP}^{(\infty)}(f/\pi)(v_{U^{\prime}})\geq\mathrm{VP}^{(\infty)}(f/\pi)(v_{U}).$

### 3.19

Let us show that the weakly admissibility holds for any minimal valuation
after a local alteration. We use the induction on the transcendence defect.

######  Lemma. —

If $v$ is a minimal valuation, there then exists a finite extension $L$ of
$k(S)$ and some extension $v^{\prime}$ of $v$ to $L$ such that $f/\pi$ is
locally admissible at $v^{\prime}$.

###### Proof.

We may assume $k$ to be algebraically closed. When
$\mathrm{VP}^{(\infty)}(f/\pi)(v)<0$, then the lemma has already been verified
in Lemma 3.12, so we may assume $\mathrm{VP}^{(\infty)}(f/\pi)(v)=0$. We use
the induction on the transcendence defect (i.e.
$\dim(S)-\mathrm{rat.rk}(v)-\mathrm{trdeg}_{k}(k_{v})$) of $v$. Abhyankar’s
inequality implies that transcendence defect is non-negative (cf. [Ke4,
2.1.4]). When the transcendence defect is $0$, the valuation is monomial in