Value Distribution in Ultrametric Analysis and
Applications

Alain Escassut

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Alain Escassut. Value Distribution in Ultrametric Analysis and Applications. WSPC. WSPC, In press. hal-04799547

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About the Author

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**Alain Escassut** after studies in Licence and Master at the University of Bordeaux obtained a PhD in 1970 and a Doctorat d'Etat in Mathematics in 1972, working in this university as Maître-Assistant and Maître de conférence until 1987 (he was also Visiting-Assistant Professor at Princeton University in Spring 1981). Next, he was Professor at Université Blaise Pascal (from Clermont-Ferrand) until 2012 and has been Emeritus Professor since 2012 at this university which become now Université Clermont Auvergne.

Alain Escassut is a specialist in ultrametric analysis, particularly *p*-adic analysis and first studied algebras of analytic elements (long ago defined by Marc Krasner). He first examined the problem of analytical extension where the rôle of *T*-filters first appeared, explaining also algebraic properties of algebras, and he characterized Krasner-Tate algebras. He constructed a holomorphic functional calculus where *T*-filters are essential, again. He then examined the rôle of the multiplicative spectrum of an ultrametric Banach algebra.

With Abdelbaki Boutabaa and Labib Haddad, he obtained applications of the *p*-adic Nevanlinna theory to problems of unicity. Joint works with C.C. Yang and Ta Thi Hoai An let them obtain new results. He obtained results on small functions and solved the problem of the Hayman conjecture in a *p*-adic field. He also examined
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Value Distribution in Ultrametric Analysis and Application

Valiron-like results on the growth of entire functions in a *p*-adic field. He examined the transcendence type in $C_p$ and gave a new proof of the Hermite–Lindemann theorem in an ultrametric field, including the case of a field of residue characteristic 0. All these studies are detailed in this book.
# Contents

<table><tr><td>About the Author</td><td>v</td></tr><tr><td>Introduction</td><td>xi</td></tr><tr><td>Chapter 1. Affinely Rigid Sets</td><td>1</td></tr><tr><td>Chapter 2. Properties of Ultrametric Fields</td><td>9</td></tr><tr><td>Chapter 3. Monotonous and Circular Filters</td><td>21</td></tr><tr><td>Chapter 4. Ultrametric Absolute Values for Rational Functions</td><td>35</td></tr><tr><td>Chapter 5. Hensel Lemma</td><td>47</td></tr><tr><td>Chapter 6. Extensions of Ultrametric Fields: The Field C<sub>p</sub></td><td>53</td></tr><tr><td>Chapter 7. Normal Extensions of Q<sub>p</sub> Inside C<sub>p</sub></td><td>63</td></tr><tr><td>Chapter 8. Spherically Complete Extensions</td><td>67</td></tr><tr><td>Chapter 9. Transcendence Order and Transcendence Type</td><td>73</td></tr><tr><td>Chapter 10. Algebras R(D)</td><td>81</td></tr><tr><td>Chapter 11. Analytic Elements</td><td>89</td></tr><tr><td>Chapter 12. Composition of Analytic Elements</td><td>97</td></tr></table>
viii Value Distribution in Ultrametric Analysis and Application

<table><tr><td>Chapter 13.</td><td>Multiplicative Spectrum of H(D)</td><td>103</td></tr><tr><td>Chapter 14.</td><td>Power and Laurent Series</td><td>109</td></tr><tr><td>Chapter 15.</td><td>Krasner Mittag-Leffler Theorem</td><td>119</td></tr><tr><td>Chapter 16.</td><td>Factorization of Analytic Elements</td><td>133</td></tr><tr><td>Chapter 17.</td><td>Algebras H(D)</td><td>139</td></tr><tr><td>Chapter 18.</td><td>Derivative of Analytic Elements</td><td>147</td></tr><tr><td>Chapter 19.</td><td>Properties of the Function Ψ for Analytic Elements</td><td>161</td></tr><tr><td>Chapter 20.</td><td>Vanishing Along a Monotonous Filter</td><td>167</td></tr><tr><td>Chapter 21.</td><td>Quasi-Minorated Elements</td><td>175</td></tr><tr><td>Chapter 22.</td><td>Zeros of Power Series</td><td>183</td></tr><tr><td>Chapter 23.</td><td>Image of a Disk</td><td>195</td></tr><tr><td>Chapter 24.</td><td>Quasi-Invertible Analytic Elements</td><td>209</td></tr><tr><td>Chapter 25.</td><td>Logarithm and Exponential in a p-adic Field</td><td>219</td></tr><tr><td>Chapter 26.</td><td>Problems on p-adic Exponentials</td><td>225</td></tr><tr><td>Chapter 27.</td><td>Divisors of Analytic Functions</td><td>241</td></tr><tr><td>Chapter 28.</td><td>Michel Lazard's Problem</td><td>255</td></tr><tr><td>Chapter 29.</td><td>Motzkin Factorization and Roots of Analytic Functions</td><td>263</td></tr><tr><td>Chapter 30.</td><td>Order of Growth for Entire Functions</td><td>281</td></tr><tr><td>Chapter 31.</td><td>Type of Growth for Entire Functions</td><td>293</td></tr><tr><td>Chapter 32.</td><td>Growth of the Derivative of an Entire Function</td><td>305</td></tr><tr><td>Chapter 33.</td><td>Growth of an Analytic Function in an Open Disk</td><td>311</td></tr></table>
Contents

ix

<table><tr><td>Chapter 34.</td><td>The Set Mult($A_b(d(0, R^-)$)</td><td>321</td></tr><tr><td>Chapter 35.</td><td>The Corona Problem on $A_b(d(0, 1^-)$)</td><td>329</td></tr><tr><td>Chapter 36.</td><td>Meromorphic Functions in $K$</td><td>345</td></tr><tr><td>Chapter 37.</td><td>Residues of Meromorphic Functions</td><td>355</td></tr><tr><td>Chapter 38.</td><td>Bezout Algebras of Analytic Functions</td><td>363</td></tr><tr><td>Chapter 39.</td><td>Meromorphic Functions Out of a Hole</td><td>367</td></tr><tr><td>Chapter 40.</td><td>Shilov Boundary for Algebras $H(D)$</td><td>371</td></tr><tr><td>Chapter 41.</td><td>Mappings from $\Phi(D)$ to the Tree $\Phi(K)$</td><td>385</td></tr><tr><td>Chapter 42.</td><td>Injective Analytic Elements</td><td>395</td></tr><tr><td>Chapter 43.</td><td>Counting Functions and Nevanlinna Theory</td><td>411</td></tr><tr><td>Chapter 44.</td><td>A Non-Clean Entire Function</td><td>421</td></tr><tr><td>Chapter 45.</td><td>Nevanlinna Theory in $K$ and Inside a Disk</td><td>427</td></tr><tr><td>Chapter 46.</td><td>Nevanlinna Theory Out of a Hole</td><td>433</td></tr><tr><td>Chapter 47.</td><td>Immediate Applications of the Nevanlinna Theory</td><td>443</td></tr><tr><td>Chapter 48.</td><td>Applications to Curves</td><td>451</td></tr><tr><td>Chapter 49.</td><td>Branched Values</td><td>461</td></tr><tr><td>Chapter 50.</td><td>Exceptional Values of Functions and Derivatives</td><td>471</td></tr><tr><td>Chapter 51.</td><td>Small Functions</td><td>485</td></tr><tr><td>Chapter 52.</td><td>Meromorphic Functions Sharing Some Small Ones</td><td>501</td></tr><tr><td>Chapter 53.</td><td>The $p$-adic Hayman Conjecture</td><td>519</td></tr></table>
x Value Distribution in Ultrametric Analysis and Application

<table><tr><td>Chapter 54. Composition of Meromorphic Functions</td><td>533</td></tr><tr><td>Chapter 55. Functions of Uniqueness</td><td>541</td></tr><tr><td>Chapter 56. Urscm and Ursim</td><td>557</td></tr><tr><td>Chapter 57. Other Urscm, Usim, and Non-Urscm</td><td>573</td></tr><tr><td>Chapter 58. Nevanlinna Theory in Characteristic p</td><td>585</td></tr><tr><td>Chapter 59. Strong Uniqueness and URSCM in Characteristic p</td><td>595</td></tr><tr><td>Chapter 60. The Functional Equation P(f) = Q(g)</td><td>603</td></tr><tr><td>Chapter 61. Yoshida's Equation in the Field K</td><td>613</td></tr><tr><td>Chapter 62. Yoshida's Equation Inside a Disk</td><td>619</td></tr><tr><td>Bibliography</td><td>629</td></tr><tr><td>Definitions</td><td>637</td></tr><tr><td>Notations</td><td>645</td></tr></table>
# Introduction

Value distribution for complex meromorphic functions is a topic that knew an impressive development during the 20th century and the early 2000, particularly from the eighties until now. The Nevanlinna theory has been very fruitful and has many applications, for instance, in algebraic geometry, problems of uniqueness, and value sharing values and small functions. On the other hand, during the second part of the 20th century and later on, ultrametric analysis became a new domain of research in value distribution and many problems once considered on the complex field were similarly considered on an ultrametric field. That needed the use of a complete algebraically closed field. That was done with the field $\mathbb{C}_p$ which is the completion of the algebraic closure of $\mathbb{Q}_p$, the completion of $\mathbb{Q}$ with respect to the $p$-adic absolute value. Thus, $\mathbb{C}_p$ is a field having certain properties similar to those of $\mathbb{C}$ but many other properties are much different, particularly, it has no connected subset except singletons. However, there exist other complete ultrametric fields, particularly the Levi-Civita field $\mathcal{R}$ [88] and its algebraic closure that is complete.

A field is said to be spherically complete when every decreasing sequence of disks has a not empty intersection. A field may be complete for an ultrametric absolute value but not spherically complete. For instance, $\mathbb{C}_p$ is not spherically complete and several serious problems happen because of that, particularly the well-known Lazard's problem. However, following, for example, a method proposed by Bertin Diarra, we show how to construct a spherically complete extension of a field. This technique enables us to solve certain problems when a spherically complete field is necessary.
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Value Distribution in Ultrametric Analysis and Application

This book is aimed at gathering classical results on analytic
elements and meromorphic functions known for almost 70 or 80 years
but also many recent works in *p*-adic value distribution and also some
results in number theory, concerning problems of transcendence and
exponential polynomials. Overall, it is aimed at giving correct proofs
for all properties of the functions we consider, while several times we
can find in the literature approximative citations or proofs putting
in doubt certain statements. Particularly, here as in [44] and [53] and
in [48] and [49], the role of circular filters is emphasized because it is
the best way to obtain rigorous explanations. Concerning the famous
Corona problem, here we give a description of all maximal ideals and
of most of continuous multiplicative semi-norms, even though it is
not known whether there exist continuous multiplicative semi-norms
other than those we describe.

Most of the results presented in [49] are also presented here
with sometimes some small corrections but several new results are
given, particularly on *p*-adic transcendence, exponential polynomials,
growth order of analytic functions, Picard's values of meromorphic
functions, meromorphic functions sharing small functions, and every
application of the Nevanlinna theory.

In the first chapters, we define affinely rigid sets, a basic notion
indispensable in problems of uniqueness [22], and we present the main
properties of *p*-adic fields, defining fields such as $\mathbb{Q}_p$, $\mathbb{C}_p$ and exam-
ining some classical properties. Thanks to the help of Bertin Diarra,
finite extensions of $\mathbb{Q}_p$ are examined, particularly normal extensions
and ramification index, in order to study the notion of transcendence
order over $\mathbb{Q}_p$: this transcendence order is stable through an algebraic
extension and we construct *p*-adic numbers of order $\le 1 + \epsilon \ \forall \epsilon > 0$.
The help of Bertin Diarra was very useful again. Similar to the tran-
scendence type of complex numbers, the transcendence type of a
*p*-adic transcendental number over $\mathbb{Q}$ is also defined and examined:
we show that it is also stable through an algebraic extension and
I am very grateful to Michel Waldschmidt for his help in that study
[55]. The numbers of zeros of exponential polynomials have a good
estimation through a computation that is not too complicated [57].
The ultrametric versions of the Hermite-Lindemann theorem, the
Gelfond-Schneider theorem, and the exponentials theorem are given
in residue characteristic *p* and 0. The case of residue characteristic 0
may apply to Levi-Civita fields [88].
## Introduction

xiii

Explaining correctly proofs requires to know perfectly all properties of analytic elements because all analytic or meromorphic functions are at least locally, analytic elements, as defined by Marc Krasner [69]. This is why we first recall them in order to make an autocontained book. Particularly, any student in master of mathematics, having a knowledge in general topology, can understand definitions and proofs.

Defining holomorphic functions in a *p*-adic field was not easy, except if we restrict ourselves to disks or sets with finitely many holes. The foundation of a veritable theory of holomorphic functions is due to Marc Krasner who defined analytic elements as limits of sequences of rational functions with respect to the topology of uniform convergence. Since the only connected sets are singletons, this requires the use of sets enjoying a property that can replace connectivity. Marc Krasner defined quasi-connected sets [69]. But most of the useful properties of quasi-connected sets actually hold in a larger class of sets: the infraconnected sets [36], [44], and [45].

Thus, we must place our study in an algebraically closed field $K$, complete with respect to an ultrametric absolute value. For a study of Krasner's analytic functions or the generalization of that theory to infraconnected analytic sets made by Philippe Robba, readers can refer [83]. On the other hand, another theory of analytic functions was made by John Tate with the use of a sheaf, considering Banach algebras, usually called affinoid algebras [89]. The comparison between the two theories through their algebras of functions is made in [37] and [45]. Here, these theories won't be developed again because they are not necessary to examine properties of meromorphic functions and problems of value distribution in *p*-adic analysis.

Mittag-Leffler's theorem due to Marc Krasner is another important tool to describe the behavior of analytic elements. The derivative of an analytic element $f$ is examined with its Mittag-Leffler series relatively to those of $f$. Overall, the maximum principle satisfied by an analytic element comes from its behavior on circular filters.

A key problem for analytic elements is whether it can vanish along a circular filter. This was thoroughly explained in [44] and [53]. Since it is a quite substantial problem involving the characterization of *T*-filters [38], [53], we do not describe that in this book, but one can find it in [44] and [53]: here, we only recall sufficient conditions, on infraconnected sets, to avoid that any analytic element vanishes
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Value Distribution in Ultrametric Analysis and Application

along a large circular filter, which is sufficient for our further studies. However, the description of that problem of filters is essential to understand the behavior of analytic and meromorphic functions in any context. Moreover, we give a very simple example of *T*-filter which is indispensable for the solution of the Corona problem.

An important particularity of the theory of analytic functions on $K$ holds in the role played by multiplicative semi-norms, characterized by circular filters on $K$. Their role in ultrametric Banach algebras was discovered by Bernard Guennebaud in the seventies [62] and they particularly apply to all kinds of analytic functions. Several properties of the spectral semi-norms are examined in [40], [45], and [53] with the use of multiplicative semi-norms and circular filters. On the other hand, Berkovich's theory [9], made in the late eighties, is entirely based on continuous multiplicative semi-norms. Here we recall their description and their intervention to determine, for instance, what we denote by $|f|(r)$, a kind of "maximum modulus principle".

We characterize the continuous multiplicative norms on the algebra $A$ of bounded analytic functions inside an “open” disk. A major problem posed in ultrametric analysis is the Corona problem, through a kind of analogy with the well-known Kakutani problem in $C$, solved by L. Carleson [28]. Concerning the algebra $A$, we explain why it cannot be considered in a similar way. However, by considering continuous multiplicative semi-norms, we can define an almost similar problem and show that the set of continuous multiplicative semi-norms whose kernel is a maximal ideal of codimension 1 is dense inside the whole set of all continuous multiplicative semi-norms whose kernel is a maximal ideal. That result, first proved when $K$ is spherically complete [48], is generalized here. The characterization of all continuous multiplicative semi-norms of that algebra is far from over, but we present examples of continuous multiplicative semi-norms called *Araujo's semi-norms*, whose kernel is neither a maximal ideal nor the null ideal, and I have been grateful to Jesus Araujo for communicating me his results.

The Shilov boundary of an algebra of analytic elements is described with the help of circular filters, in connection with the maximum principle [15]. Mappings between the multiplicative spectra of such algebras are studied by using the hyperbolic distance in order to characterize injective analytic elements, a property also
## Introduction

xv

stated by J. Rivera-Letelier in [82] in connection with previous works [42] and [44]. The idea consists of showing that an injective meromorphic function has an absolute value that looks like this of a Moebius function (also called a linear fractional function). A correct proof requires first to show that an analytic element $f$ maps the Shilov boundary of a set $D$ into that of $f(D)$.

The divisor of an analytic function is introduced and Lazard's problem for analytic functions inside an "open" disk is described. The very complicated construction made by Michel Lazard in a spherically complete field is explained in a way that is more comprehensible [72]. In a field that is not spherically complete, a counter-example due to Michel Lazard shows that the theorem may not be generalized and this is a gap for certain problems. However, we can provide a result that is less accurate but sufficient for certain problems. Yet, sometimes we can place ourselves in a spherically complete extension. All closed ideals of algebras of entire functions or algebras of analytic functions inside an open disk are principal, we show that the algebra of entire functions is a Bezout ring, and when $K$ is spherically complete, so is the algebra of analytic functions inside an open disk.

Similar to definitions well known in complex analysis, we have notions of order of growth and type of growth for an entire function in $K$. But here, another notion appears essential: the *cotype of growth* [16]. Thus, we can obtain similar formulas for the growth order and the growth type. Let $f$ be an entire function in $K$ whose order of growth is finite. We give bounds for the cotype of growth and also for the lower cotype of growth. We show that the type of growth of $f$ is equal to its lower type if and only if its cotype is equal to its lower cotype, and when these are realized, then the cotype is the product of the type by the order of growth. Such entire functions are called *clean functions*. But we construct an example of non-clean entire function $f$ where the cotype is strictly superior to the product of the type by the order, which answers negatively a question asked long ago.

Suppose that $K$ is of characteristic 0. Given a meromorphic function $f = \frac{g}{h}$, if the type or the cotype of $h$ is finite, then $f$ assumes all values infinitely many times. A counter-example is constructed showing that if the lower order of growth is equal to the order, that does not imply that the lower type of growth is equal to the type of
xvi

growth, although such a claim was made for complex meromorphic functions [32] and we contest the proof.

Next, we examine notions of order of growth and type of growth for analytic functions inside an open disk [19].

Meromorphic functions in $K$ and inside a disk are introduced and their basic properties are examined. Next, we examine meromorphic functions out of a hole, using Motzkin products. A meromorphic function admits primitives if and only if all residues are null. This property seems obvious but actually requires the use of Mittag-Leffler theorem to be correctly proved and does not seem to appear in other work. We study exceptional values of a function or its derivative, particularly we recall why a meromorphic function in $K$ or an “unbounded” meromorphic function in an open disk admits at most one exceptional value, which suggests a conjecture.

In a field such as $K$, the first fundamental theorem is almost trivial, but the *p*-adic Nevanlinna theory entirely holds in the so-called second fundamental theorem that we call here the main fundamental theorem. It was completely established in [21] by Abdelbaki Boutabaa for meromorphic functions in the whole field $K$, after the first study by Ha Huy Khoai [63]. The generalization for meromorphic functions in an open disk was made further [23], taking into account Lazard's problem: this theorem is useful when it applies to functions which are not a quotient of two bounded analytic functions in that disk. Next, a similar statement was established for meromorphic functions out of a hole [54] which lets us obtain applications for these functions.

This is the beginning of many applications to the behaviour of meromorphic functions. Using the main fundamental theorem, we can show that many functional equations have no solution, particularly, elliptic and hyperelliptic curves admit no parametrization by meromorphic functions (more generally, Berkovich theory shows that no curve of genus $\ge 1$ admits any parametrization by non-constant meromorphic functions). Similarly, equations of the form $f^m+g^n = 1$ admit no non-constant meromorphic solutions in the whole field, provided $\min(m,n) \ge 2$, $\max(m,n) \ge 3$ (results inside a disk are also given).

As it was done for complex functions, we define small functions. Generalizing Picard's values for meromorphic functions, given a meromorphic function $f$ in $K$ or an unbounded meromorphic function
inside an open disk, there exists at most one small function $w$ such that $f - w$ has finitely many zeros. One can easily obtain the main fundamental theorem on three small functions, as in $\mathbb{C}$, which was first noticed by Peichu Hu and C.C. Yang [67]. This is a useful tool in certain problems of uniqueness. Unfortunately, contrary to complex analysis, we know no main fundamental theorem on $n$ small functions ($n > 3$) as it was made in $\mathbb{C}$ by K. Yamanoi [94]. However, a result has just been obtained by Ta Thi Hoai An and An-Nguyen Phuong, which we explain here.

When two meromorphic functions share (ignoring multiplicity) seven small functions, in a work with C.C. Yang, we showed that they are equal [56]. As corollary, if two entire functions share three small functions, then they are equal. But a much better result has been obtained by Ta Thi Hoai An and An-Nguyen Phuong, showing, through big computations, that it is sufficient to assume the two meromorphic functions to share five small ones [5]. All results in that domain hold in $\mathbb{K}$ like in an open disk and out of a hole.

As in complex analysis [29], we can also define and study perfectly and totally branched values, showing that in a $p$-adic field, a meromorphic function in $\mathbb{K}$ has at most three totally branched values and four perfectly branched values. But we can also prove that a meromorphic function having finitely many poles has at most one perfectly branched value, which is obtained without the use of the Nevanlinna main theorem but with just the use of square roots of certain analytic functions [17]. This last result has a generalization to branched rational functions.

Concerning functions inside a disk, we obtain results that look like those obtained for complex functions in the whole field [33]. We can also show that a meromorphic function whose numerator and denominator have different growth order admits at most two perfectly branched values. Moreover, a meromorphic function whose numerator and denominator have same finite growth order but different growth type admits at most three perfectly branched values [17].

We examine the Hayman conjecture in the $p$-adic context: given $f \in \mathcal{M}(\mathbb{K})$, can we state that $f'f^n$ takes every value $b \neq 0$ infinitely many times? We first give the proof by Jacqueline Ojeda obtained for every $n \ge 3$ [80]. Next we give the proof for $n = 2$ [52]. Moreover, all results given for $n \ge 3$ also apply to meromorphic functions inside
xviii

Value Distribution in Ultrametric Analysis and Application

an open disk and to meromorphic functions out of a hole. Many particular cases are also examined around such problems.

Concerning entire functions, a link then appears between growth order and small functions. Let $f, h \in \mathcal{A}(\mathbb{K})$ be such that $0 < \rho(f) < +\infty$ and $\rho(h) < \tilde{\rho}(f)$. Then, $h$ is a small function with respect to $f$. Let $f, h \in \mathcal{A}(\mathbb{K})$ be such that $\rho(h) = \rho(f)$ and such that $0 < \rho(f) < +\infty$. If $f$ is clean and if $\sigma(h) = 0 < \sigma((f))$, then $h$ is a small function with respect to $f$.

Let $f, g \in \mathcal{A}(\mathbb{K})$ be transcendental, regular and share three distinct functions, ignoring multiplicity, $\omega_j \in \mathcal{A}(\mathbb{K})$, $j = 1, 2, 3$, such that $\max_{1\le j\le 3} \rho(\omega_j) < \min(\rho(f), \rho(g))$. Then, $f = g$.

Let $f, g \in \mathcal{A}(\mathbb{K})$ be clean and share $h_1, h_2, h_3 \in \mathcal{A}(\mathbb{K})$, ignoring multiplicity, such that $\rho(h_j) = \rho(f) = \rho(g)$, $j = 1, 2, 3$, $\sigma(h_j) = 0$, $j = 1, 2, 3$, and $0 < \min(\sigma(f), \sigma(g))$. Then, $f = g$.

We show that a meromorphic function admitting finitely many poles of order $\ge 3$ and admitting primitives must have infinitely many zeros. That suggests the conjecture: *a meromorphic function admitting primitives must have infinitely many zeros*. Thanks to an idea due to J.P. Bézivin [10], the conjecture is proved for all meromorphic functions $f$ having a total number of poles $t(r, f)$ in the disk of center 0 and radius $r$ satisfying: $\text{Log}(t(r, f)) \le O(\text{Log}(r))$, which makes the conjecture much likely. That suggests an interpretation by using type and cotype of the denominator of a meromorphic function: *given a meromorphic function $f = \frac{q}{h}$ admitting primitives, if the type of growth of $h$ is finite, then $f$ should have infinitely many zeros*.

We examine functions of uniqueness and unique range sets, counting or ignoring multiplicity, for $p$-adic analytic and meromorphic functions in the whole field and inside a disk. The Nevanlinna theory is the main tool for most of the results. Unique range sets counting multiplicity for entire functions in $\mathbb{K}$ and for polynomials are characterized as the affinely rigid subsets of $\mathbb{K}$ [22] and [45] (such a characterization has no equivalent for complex functions). The result, first proved for polynomials and for entire functions and sets of three points [22], was generalized to entire functions on $\mathbb{K}$ by W. Cherry and C.C. Yang [31].

It is shown that the situation is much more complicated concerning meromorphic functions. Examples of sets which are or are not
Introduction

xix

unique range sets counting multiplicity for meromorphic function
are presented. The situation is examined in the whole field and inside
a disk.

The problem of unique range sets counting multiplicity for entire
functions in $\mathbb{K}$ is closely linked to this of polynomials of uniqueness.
That suggests a more general study of meromorphic functions of
uniqueness for various kinds of functions (in the whole field and inside
a disk) [47], in connection with zeros of the derivative, a notion first
defined by H. Fujimoto [60].

A Nevanlinna theory in characteristic $p \neq 0$ is made and appli-
cations are given to problems of the form $P(f) = Q(g)$, with $P, Q$
polynomials [25] and [26]. Finally, the Malmquist–Yoshida equation
is studied in characteristic $p \ge 0$ and all solutions in the whole field
are characterized [24]. The problem is finally examined in character-
istic $0$ for functions in an open disk following a work by Abdelbaki
Boutabaa [27].

I would also like to thank many colleagues for many exchanges,
particularly Ta Thi Hoai An, Kamal Boussaf, Jesus Araujo, J.P.
Bézivin, and C.C. Yang. Abdelbaki Boutabaa passed away in 2022.
He made the *p*-adic Nevanlinna theory and took part in most of our
work, then.
# Chapter 1

## Affinely Rigid Sets

Affinely rigid sets have proven to play a crucial role in studying problems of uniqueness. They were introduced in [22] and were first called affinely rigid sets. Next, fearing a confusion with rigid geometry and affinoid sets, they were called *stiff sets* in [22]. However, the term affinely rigid was already known and popular. This is why it has become the one currently used in all further papers [3], [4], and [93]. It is a very basic notion that requires to be thoroughly examined.

**Definition.** In this chapter, we denote by $\mathbb{F}$ an algebraically closed field. We call *centered similarity in the field $\mathbb{F}$* a similarity in $\mathbb{F}$ of the form $\phi(x) = a + \alpha(x-a)$, and if $\phi$ is not the identity, the point $a$ is called the *center of $\phi$*.

We call *translation* a similarity of the form $\phi(x) = x + b$ with $b \neq 0$.

A subset $S$ of $\mathbb{F}$ is said to be *affinely rigid* if there exists no affine mapping $\phi$ from $\mathbb{F}$ to $\mathbb{F}$, other than the identity, such that $\phi(S) = S$.

Let $S$ be a non-affinely rigid set. $S$ is said to be a *centered non-affinely rigid set* (respectively *non-centered non-affinely rigid set*) with respect to a *centered similarity* $\phi$ if it is preserved by a centered similarity $\phi$ and if the center of $\phi$ belongs to $S$ (respectively does not belong to $S$).

Propositions 1.1 and 1.2 are immediate:

**Proposition 1.1.** A similarity $\phi$ is not centered if and only if it is a translation.
**Proposition 1.2.** If a finite subset $S$ of $\mathbb{F}$ is preserved by a similar-
ity $\phi$, that $\phi$ is a centered similarity.

**Proof.** Let $\phi$ be a non-centered similarity preserving a finite set $S$. Then, $\phi^n$ also preserves $S$. By Proposition 1.1, $\phi(x)$ is of the form $x+b$ with $b \neq 0$. Then, $\phi^n(x) = x+nb$. Since $\mathbb{F}$ is of characteristic 0, the sequence $(x+nb)_{n \in \mathbb{N}}$ is injective. Particularly, taking $u \in S$, we have $u+nb \in S \forall n \in \mathbb{N}$, hence $S$ has infinitely many elements, a contradiction. $\square$

Lemma 1.3 is immediate:

**Lemma 1.3.** Let $\phi$ be a similarity and let $S$ be a non-affinely rigid set. Then $\phi(S)$ is a non-affinely rigid set. Moreover, if $S$ is centered (respectively non-centered), so is $\phi(S)$.

**Theorem 1.4.** Let $P \in \mathbb{F}[x]$ be of degree $n$ and have all its zeros of order 1 and let $S$ be the set of zeros of $P$. Then, $S$ is not affinely rigid if and only if there exists a centered similarity of center $a$: $\phi(x) = a + \alpha(x-a)$ such that $P(\phi(x)) = \alpha^n P(x)$, with $\alpha \neq 1$. Let $S$ be non-affinely rigid and let $\phi$ be such a centered similarity $\phi(x) = a + \alpha(x-a)$ preserving $S$. Then putting $u = x-a$, $Q(u) = P(x)$, $Q$ is of one of the following two forms:

(i) $Q(u) = \sum_{k=0}^{q} a_{kd} u^{kd}$, with $\alpha^d = 1$ and $d \ge 2$ and then $S$ is non-centered.

(ii) $Q(u) = \sum_{k=0}^{q} a_{kd+1} u^{kd+1}$, with $\alpha^d = 1$ and $d \ge 2$ and then $S$ is centered.

Moreover, if two centered similarities preserve $S$, they have the same center. Further, $S$ is never both centered and non-centered.

**Proof.** Considering the coefficients of degree $n$, it is obvious that $S$ is not affinely rigid if and only if there exists a centered similarity of center $a$: $\phi(x) = a + \alpha(x-a)$ such that $P(\phi(x)) = \alpha^n P(x)$, with $\alpha \neq 1$.

If (i) or (ii) is satisfied, it is obviously seen that $S$ is preserved by the centered similarity $\phi$ of center $a$. Moreover, in Case (i), $a$ does not lie in $S$, hence $S$ is non-centered. In Case (ii), $a$ lies in $S$, hence $S$ is centered.

Now, suppose $S$ is not affinely rigid and let $\phi(x) = a + \alpha(x-a)$ be a centered similarity preserving $S$. Then, $P(\phi(x)) = P(a + \alpha u) = Q(\alpha u)$ and $\alpha^n P(x) = \alpha^n Q(u)$, hence $Q(\alpha u) = \alpha^n Q(u)$.
Let $Q(u) = \sum_{j=0}^{n} a_j u^j$. We have $\alpha^j a_j = \alpha^n a_j \quad \forall j = 0, \dots, n$. Consequently, $(1) \quad a_j(\alpha^{n-j} - 1) = 0 \quad \forall j = 0, \dots, n$.

Suppose first $a_0 \neq 0$. Clearly, we have $\alpha^n = 1$. Let $d$ be the order of $\alpha$ as a $n$th root of 1. Since $\phi$ is not the identity, $\alpha$ is $\neq 1$. Consequently, by (1), we note that $d \geq 2$ because $\alpha^{n-1} \neq 1$. Then, $n$ is of the form $qd$ ($q \in \mathbb{N}^*$), and then, by (1), we have $a_j = 0$ for every $j$ which is not multiple of $d$. Consequently, we have obtained $Q(u) = \sum_{k=0}^{q} a_{kd} u^{kd}$, with $\alpha^d = 1$ and $d \geq 2$.

Suppose now $a_0 = 0$. Since the zeros of $P$ are not multiple, neither are those of $Q$. Consequently, $a_1 \neq 0$, hence, by (1), we have $\alpha^{n-1} = 1$. Let $d$ be the order of $\alpha$ as an $(n-1)$th root of 1. Since $\alpha \neq 1$, by (1), we have $d \geq 2$ because $\alpha^{n-2} \neq 1$. Then, $n$ is of the form $qd+1$ ($q \in \mathbb{N}^*$), and by (1), we have $a_j = 0$ for every $j$ which is not of the form $kd+1$. Consequently, we obtain $Q(u) = \sum_{k=0}^{q} a_{kd+1} u^{kd+1}$, with $\alpha^d = 1$ and $d \geq 2$.

Suppose $S$ is non-centered with respect to $\phi$ and consider another centered similarity $\psi$ preserving $S$. Let $Q$ be a polynomial defined as in (i) with respects to $\phi$, admitting $S$ for set of zeros, all of order 1. Then, we have $\psi(u) = \lambda u + \mu$ and $Q(\psi(u)) = \lambda^n Q(u)$. Consequently,

$$ \sum_{k=0}^{q} a_{kd+1}(\lambda u + \mu)^{kd+1} = (\lambda)^{qd+1} \sum_{k=0}^{q} a_{kd+1} u^{kd+1}. $$

Examining terms of degree $qd$, we can check that $a_{qd+1}(qd+1)\lambda^{qd} = 0$ because $d \geq 2$. Consequently, $\mu = 0$. Therefore, $Q(\psi(u)) = \sum_{k=0}^{q} a_{kd}(\lambda u)^{kd}$ and then, the polynomial associated with $Q$ with respect to $\psi$, as defined in (i), is $Q$ itself. Consequently, there exists no centered similarity $\psi$ preserving $S$ such that the polynomial associated with $Q$ with respect to $\psi$ is of the form (ii). This proves that $S$ is non-centered with respect to $\psi$. That ends the proof of Theorem 1.4. $\square$

**Theorem 1.5.** Let $S$ be a finite centered (respectively non-centered) non-affinely rigid set of $n$ elements and let $\phi(x) = a + \alpha(x - a)$ be a similarity that preserves $S$. The order of $\phi$ divides $n-1$ (respectively divides $n$) and $\alpha$ is some $(n-1)$th root of $n-1$ (respectively $\alpha$ is some $(n-1)$th root of $n$).

**Proof.** Let $\phi(x) = a + \alpha(x - a)$. Without loss of generality, we can assume that $a = 0$ and hence $\phi(x) = \alpha x$.
We first note that the restriction $\bar{\phi}$ of $\phi$ to $S$ is a bijection from $S$ to $S$ and hence is a permutation of $S$.

Suppose first that $S$ is centered. So, 0 is the center of $\phi$. Then, $\bar{\phi}$ preserves $S\setminus\{0\}$ and hence its order divides $n-1$. Consequently, $\alpha$ is an $(n-1)$th root of 1.

Suppose now that $S$ is non-centered. Then, $\bar{\phi}$ preserves $S$ but does not preserve any point of $S$. Consequently, its order divides $n$ and hence $\alpha$ is an $n$th root of 1. $\square$

**Corollary 1.6.** Let $S$ be a finite centered non-affinely rigid set of $n$ elements. If $n-1$ is prime, then $S$ is of the form $\{a, a+u^k h \mid k = 0, \dots, n-2\}$ with $a, h \in \mathbb{F}$ and $u$ some $(n-1)$th root of 1.

Let $S$ be a finite non-centered non-affinely rigid set of $n$ elements. If $n$ is prime, then $S$ is of the form $\{a+u^k h \mid k = 0, \dots, n-1\}$ with $a, h \in \mathbb{F}$ and $u$ is some $n$th root of 1.

**Corollary 1.7.** Let $S$ be a finite centered non-affinely rigid set of three elements. Then, $S$ is of the form $\{a, a-h, a+h\}$ with $a, h \in \mathbb{F}$.

Let $S$ be a finite non-centered non-affinely rigid set of three elements. Then, $S$ is of the form $\{a+u^k h \mid k = 0, 1, 2\}$ with $a, h \in \mathbb{F}$ and $u$ is a primitive third root of 1.

**Theorem 1.8.** Let $S$ be a subset of $\mathbb{F}$ of four elements. The following two conditions are equivalent. Then:

(1) $S$ is a centered non-affinely rigid set if and only if it is of the form $\{a, a+h, a+uh, a+u^2h\}$, with $h \in \mathbb{F}$ and $u^3 = 1$, $u \neq 1$.

(2) $S$ is a non-centered non-affinely rigid set if and only if it is of the form $\{a-h, a+h, a-k, a+k\}$, with $h, k \in \mathbb{F}^*$.

**Proof.** If $S$ is of the form $\{a, a+h, a+uh, a+u^2h\}$, with $h \in \mathbb{F}$ and $u^3 = 1$, $u \neq 1$, it is obvious that the similarity $\phi(x) = a+u(x-a)$ preserves $S$. Conversely, suppose that $S$ is a centered non-affinely rigid set and let $\phi$ be a similarity of center $a$ preserving $S\setminus\{a\}$ and different from the identity. Then, $u$ is a third- root of 1 different from 1.

If $S$ is of the form $\{a-h, a+h, a-k, a+k\}$, with $h, k \in \mathbb{F}^*$, then the similarity $\phi(x) = a - (x-a)$ preserves $S$ and does not preserve any point of $S$, hence $S$ is a non-centered non-affinely rigid set. Conversely, suppose that $S$ is a non-centered non-affinely rigid
set and let $\phi(x) = a + u(x-a)$. For convenience, suppose $a=0$. Then, $u$ is a root of 1 of order 2 or 4. Suppose first $u$ is a root of 1 of order 2 i.e. $u = -1$. Then, $\phi(x) = -x$. Let $S = \{b_1, b_2, b_3, b_4\}$. Then, up to a reordering, we have $\phi(b_1) = b_3$, $\phi(b_2) = b_4$, hence $b_3 = -b_1$, $b_4 = -b_2$ which proves the claim.

Suppose now that $u$ is root of 1 of order 4. Then, up to a reordering, we must have $\phi(b_j) = b_{j+1} \forall j = 1, 2, 3$ and of course, $\phi(b_4) = b_1$. But then we note that $b_3 = -b_1$, $b_4 = -b_2$ again, which finishes proving the claim. □

The following condition (F) was introduced by C.C. Yang and Hirotaka Fujimoto.

**Definition.** A polynomial $P \in \mathbb{F}[x]$ is said to satisfy Condition (F) [60] if for any two distinct zeros $a, b$ of $P'$, we have $P(a) \neq P(b)$ (i.e. the restriction of $P$ to the set of zeros of $P'$ is injective).

**Lemma 1.9.** Let $P \in \mathbb{F}[x]$ be of degree 4 and assume it does not satisfy Condition (F). Then, $P'$ admits three distinct zeros.

**Proof.** Of course, $P'$ can't have a unique triple zero. Suppose $P'$ only has two distinct zeros. Up to an affine change of variable, $P'$ is of the form $\lambda(x^3 - bx^2)$. Since $P$ does not satisfy Condition (F), we can conclude $b=0$, hence $P'$ has a triple zero, a contradiction. □

**Theorem 1.10.** Let $P \in \mathbb{F}[x]$ be monic of degree 4. The following three statements are equivalent:

(i) $P$ does not satisfy Condition (F).

(ii) $P$ is of the form $[(x-a+l)(x-a-l)]^2 + B$ with $B \in \mathbb{F}$, $l \in \mathbb{F}^*$.

(iii) There exists an affine change of variable transforming $P$ into an even function.

Moreover, if those conditions are satisfied, assuming $P(c_1) = P(c_2)$, then we have $P(c_3) = -P(c_1)$ if and only if $B = -\frac{l^4}{2}$.

**Proof.** By putting $t = x-a$, it is obvious that (ii) implies (iii) and that (iii) implies (i). Suppose that (i) is satisfied. By Lemma 1.9, $P'$ has three distinct zeros. Suppose first $P(c_3) = P(c_2) = P(c_1)$. Then, $P - P(c_1)$ has three distinct zeros of order $\ge 2$, which contradicts the hypothesis $\deg(P) = 4$. Hence, we can assume $P(c_3) \neq P(c_1)$.
Let $a = \frac{c_1+c_2}{2}$, $c_1 = a+l$, $c_2 = a-l$. Thus, $P$ is of the form $[(x-a+l)(\bar{x}-a-l)]^2 + P(c_1)$, with $P(c_1) \neq 0$. Moreover, since $P'$ has three distinct zeros, we see that $l \neq 0$, which also proves that (ii) is satisfied.

Now, suppose that the above three conditions are satisfied. By (ii), we have $P(c_1) = P(c_2) = B$ and $P'(x) = 4(x-a)[(x-a)^2 - l^2]$. So the zeros of $P'$ are $a, a+l, a-l$. Since $P(c_1) = P(c_2)$, the pair $\{c_1, c_2\}$ is $\{a-l, a+l\}$. Therefore, $c_3 = a$ and hence $P(c_3) = l^4 + B = l^4 + P(c_1)$. Consequently, $P(c_3) = -P(c_1)$ holds if and only if $l^4 = -2P(c_1) = -2B$, which ends the proof. $\square$

**Corollary 1.11.** Let $P \in \mathbb{F}[x]$ be of degree 4 such that its set of zeros is affinely rigid. Then, $P$ satisfies Condition (F).

**Proof.** Indeed, suppose that $P$ does not satisfy Condition (F). By Theorem 1.10, through a change of variable, $P$ is an even function, hence for each zero, $h$ and $-h$ are other zeros. Consequently, the set of zeros is not affinely rigid. $\square$

**Theorem 1.12.** Let $P(x) = \sum_{k=0}^{q} a_{kd+1}x^{kd+1}$, with $a_1 \neq 0$, $d \ge 2$, and let $Z$ be the set of the zeros of $P'$. For all $c \in \mathbb{F}^*\setminus P(Z)$, $P-c$ admits $qd+1$ distinct zeros and its set of zeros is affinely rigid.

**Proof.** Let $c \in \mathbb{F}^*\setminus P(Z)$, let $G(x) = P(x) - c$, and let $T$ be the set of zeros of $G$. Suppose that $T$ is not affinely rigid. By Theorem 1.3, there exists a centered similarity $\phi$ defined as $\phi(x) = \lambda x + \mu$ such that $G(\phi(x)) = \lambda^n G(x)$. So, we have $\sum_{k=0}^{q} a_{kd+1}(\lambda x + \mu)^{kd+1} - c = (\lambda)^{qd+1}(\sum_{k=0}^{q} a_{kd+1}x^{kd+1} - c)$ with $d \ge 2$. Examining terms of degree $qd$, we can check that $a_{qd+1}(qd+1)\lambda^{qd}\mu = 0$ because $d \ge 2$. Consequently, $\mu = 0$. Therefore, $\phi$ has center 0. Next, examining terms of degree 0 shows that $\lambda^n = 1$ because $c \neq 0$. Hence, $G(x) = \sum_{k=0}^{q} a_{kd+1}x^{kd+1} - c$. But since the center of $\phi$ is 0, by Theorem 1.3, $G(x)$ is either of the form $\sum_{j=0}^{s} b_j x^{jt}$ or of the form $\sum_{j=0}^{s} b_j x^{jt+1}$ with $t \ge 2$. Now, since $c \neq 0$, the second case is excluded. Consequently, we have $G(x) = \sum_{j=0}^{s} b_j x^{jt}$ and therefore $\sum_{j=0}^{s} b_j x^{jt} = \sum_{k=0}^{q} a_{kd+1}x^{kd+1} - c$.
In particular, since $t \ge 2$, we note that the coefficient of $x$ on the right-hand side is null, a contradiction to $a_1 \ne 0$. Therefore, the assumption “$T$ is not affinely rigid” is not true, which ends the proof. $\square$

**Theorem 1.13.** $P \in \mathbb{F}[x]$ be such that $P'$ admits exactly two distinct zeros $c_1, c_2$. Then, $P(c_1) \ne P(c_2)$. Assume that all zeros of $P$ are distinct, of order 1. If the set of zeros $S$ of $P$ is not affinely rigid, then $P'$ is of the form $B((x - c_1)(x - c_2))^m$, $B \in \mathbb{F}$, and the unique similarity preserving $S$, other than the identity, is the mapping $\phi(x) = -x + c_1 + c_2$.

**Proof.** Let $P'(x) = B(x - c_1)^{m_1}(x - c_2)^{m_2}$. Of course, $\deg(P) = m_1 + m_2 + 1$. Without loss of generality, we may obviously assume $B = 1$. Suppose first that $P(c_1) = P(c_2)$. Then, $P - P(c_1)$ admits each $c_j$ as a zero of order $m_j + 1$ ($j = 1, 2$) and therefore $\deg(P) = m_1 + m_2 + 2$, a contradiction. Hence, $P(c_1) \ne P(c_2)$.

Suppose that $S$ is not affinely rigid and let $\phi(x) = ax+b$ preserve $S$, $\phi$ being different from the identity. Since $\phi$ preserves $S$ and since all zeros of $P$ are of order 1, $P \circ \phi$ is a polynomial of same degree as $P$, admitting the same zeros, all of order 1 and therefore $P \circ \phi$ is of the form $\lambda P$ with $\lambda \in \mathbb{F}^*$. Consequently, $P'(x) = \lambda a(ax + b - c_1)^{m_1}(ax + b - c_2)^{m_2}$.

Suppose first that $m_1 \ne m_2$. Then, we can identify $c_1$ with $\frac{c_1-b}{a}$ and $c_2$ with $\frac{c_2-b}{a}$. Consequently, $b=0$, $a=1$, a contradiction since $\phi$ is not the identity. Thus we are led to assume that $m_1 = m_2$. Put $m = m_1 = m_2$. Thus, $P'(x) = [(x - c_1)(x - c_2)]^m$. We may now write $P'(x) = a[(ax + b - c_1)(ax + b - c_2)]^m$ and we see that either $\frac{c_1-b}{a} = c_1$, $\frac{c_2-b}{a} = c_2$ which yields $a=1, b=0$ again or $\frac{c_1-b}{a} = c_2$, $\frac{c_2-b}{a} = c_1$. And since $\phi$ is not the identity, the second conclusion is the only possible. Then $a = -1$ and $b = c_1 + c_2$. $\square$
# Chapter 2

## Properties of Ultrametric Fields

In this chapter, we recall basic definitions and properties on ultra-metric fields: ultrametric absolute values, valuation rings, and residue fields. We must define holes of a subset and infraconnected subsets that are essential for the behavior of analytic functions (certain authors improperly call such sets “connected sets” which makes no sense in topology since there are no connected sets except singletons in an ultrametric field). A major interest of the class of infraconnected sets is that it is the biggest class of sets in an ultrametric complete algebraically closed field where the famous Krasner Mittag-Leffler theorem applies.

**Definition and notation.** Throughout this book, we denote by $\mathbb{N}$ the set of integers $\ge 0$, by $\mathbb{Z}$ the ring of relative integers, by $\mathbb{Q}$ the field of rational numbers, by $\mathbb{R}$ the field of real numbers, and by $\mathbb{C}$ the field of complex numbers.

Given a topological space $T$ and a subset $S$ of $T$, we denote by $\bar{S}$ its closure (also called adherence) and by $\hat{S}$ its interior (also called opening).

Let $E$ be a field provided with an absolute value $|\cdot|$ and let $\log$ be a real logarithm function of basis $\theta > 1$. We call *valuation associated with that absolute value* the mapping $v$ from $E$ to $\mathbb{R}$ defined as $v(x) = -\log|x|$ and here, we set $\Psi(x) = \log|x|$ and $|E| = \{|x| \mid x \in E\}$. If a set $F$ contains the zero of a ring, we denote by $F^*$ the set $F\setminus\{0\}$. An absolute value is said to be *trivial* if $|x| = 1 \ \forall x \in E\setminus\{0\}$.
Throughout this chapter, we denote by $\mathbb{L}$ a field complete with
respect to a non-trivial ultrametric absolute value, and throughout
this book, we denote by $\mathbb{K}$ an algebraically closed field complete
with respect to a non-trivial ultrametric absolute value. We denote
by $|\cdot|_{\infty}$ the archimedean absolute value defined on $\mathbb{R}$.

**Lemma 2.1.** Let $\mathbb{E}$ be a field provided with an ultrametric absolute value $|\cdot|$. The completion of $\mathbb{E}$ with respect to that absolute value is provided with an ultrametric absolute value which continues that of $\mathbb{E}$. The set $\{|x| \mid x \in \mathbb{E}^*\}$ is a subgroup of the multiplicative group $\mathbb{R}_+^*$.

**Definition.** Given a field $\mathbb{E}$ provided with an ultrametric absolute value $|\cdot|$, the multiplicative group $\{|x| \mid x \in \mathbb{E}^*\}$ is called the *value group of* $\mathbb{E}$ and the additive group $\{v(x) \mid x \in \mathbb{E}\}$ is called *valuation group of* $\mathbb{E}$.

Similarly, the set $\{\Psi(x) \mid x \in \mathbb{E}^*\}$ is a subgroup of $\mathbb{R}$ called *valuation group of* $\mathbb{E}$.

The field $\mathbb{E}$ is said to have *discrete valuation* or to have *discrete absolute value* if its valuation group is a discrete subgroup of $\mathbb{R}$ and hence is isomorphic to $\mathbb{Z}$. Else, the valuation group is dense in $\mathbb{R}$ and $\mathbb{E}$ is said to have *dense valuation* or to have a *dense absolute value*.

Lemma 2.2 is classical and proven in the same way no matter
what the absolute value of E.

**Lemma 2.2.** Let $\mathbb{E}$ be a field provided with two absolute values whose associated valuations are $v$ and $w$, respectively. They are equivalent if and only if there exists $r > 0$ such that $w(x) = rv(x)$ whenever $x \in \mathbb{E}$.

**Proof.** If such an $r$ exists, the two absolute values are seen to be equivalent. Reciprocally, we assume them to be equivalent and take $a \in \mathbb{E}$ such that $v(a) \ge 0$. It is seen that $w(a) \ge 0$. On the other hand, for all $x \in \mathbb{E}$ and for all $m, n \in \mathbb{N}$, we have $v(x/a^n) > 0$ if and only if $w(x/a^n) > 0$. Therefore, we see that $v(x)/(v(a)^m) > n/m$ is equivalent to $w(x)/(w(a)^m) > m/n$. Then, since $\mathbb{Q}$ is dense in $\mathbb{R}$, we have $v(x)/(v(a)) = w(x)/(w(a))$ whenever $x \in \mathbb{E}$, and therefore, $w(x)/(v(a)) = w(a)/(v(a))$. $\square$

**Notation.** The set of the $x \in E$ such that $|x| \le 1$ will be denoted by $U_E$ and the set of the $x \in E$ such that $|x| < 1$ will be denoted by $M_E$.
Then Lemma 2.3 is immediate:

**Lemma 2.3.** $U_E$ is a local subring of $E$ whose maximal ideal is $M_E$.

**Definition and notation.** Henceforth, $U_E$ is called the valuation ring of $E$. The maximal ideal $M_E$ of $U_E$ is called the valuation ideal and the field $\mathcal{E} = \frac{U_E}{M_E}$ is called the residue class field of $E$. For any $a \in E$, the residue class of $a$ is denoted by $\bar{a}$.

The characteristic of $\mathcal{E}$ is named the residue characteristic of $E$ and is denoted by $p$.

**Lemma 2.4.** Let $F$ be a subfield of $E$, let $\mathcal{E}$ (respectively $\mathcal{F}$) be the residue class field of $E$ (respectively of $F$). Then, $\mathcal{F}$ is a subfield of $\mathcal{E}$. If $E$ is algebraically closed and if its valuation is not trivial, it is dense.

**Proof.** The first statement is immediate. Next, given $\alpha \in E$ such that $0 < |\alpha| < 1$ and $\beta \in E$ such that $\beta^q = \alpha^s$, we have $v(\beta) = \frac{q}{s}v(\alpha)$ whenever $s \in \mathbb{N}^*$ and $q \in \mathbb{Z}$. $\square$

**Lemma 2.5.** Let $V$ be an $\mathbb{L}$-vector space of finite dimension provided with two norms. Then the two norms are equivalent.

**Proof.** Let $\| \cdot \|$ and $\| \cdot \|'$ be the two norms on $V$. We proceed by induction on the dimension of $V$ and assume the equivalence true for subspaces of dimension $n < q$. Let $V$ have dimension $q$. Let $e_1, \dots, e_q$ be a base of $V$. Let us suppose that the two norms are not equivalent on $V$. Then there exists a sequence $(u_n)_{n \in \mathbb{N}}$ of the form $u_n = \sum_{j=1}^q a_{j,n}e_j$, with $\|u_n\| \ge 1$, such that $\lim_{n \to \infty} \|u_n\|' = 0$. Let $S$ be the subspace of $V$ generated by $\{e_1, \dots, e_{q-1}\}$. For every $n \in \mathbb{N}$, we put $v_n = \sum_{j=1}^{q-1} a_{j,n}e_j$.

First, we suppose that

$$ (1) \quad \lim_{n \to \infty} |a_{q,n}| = 0. $$

Since $\lim_{n \to \infty} \|u_n\|' = 0$, we have $\lim_{n \to \infty} \|v_n\|' = 0$. By hypothesis, the restrictions of the two norms to $S$ are equivalent, hence we have $\lim_{n \to \infty} \|v_n\| = 0$. But since $\|u_n\| \ge 1$ for all $n \in \mathbb{N}$, this contradicts (1).

Now, since (1) is not true, there exists a subsequence of the sequence $(|a_{q,n}|)_{n \in \mathbb{N}}$ that admits a strictly positive lower bound and therefore, without loss of generality, we can clearly assume that there
exists $r > 0$ such that $|a_{q,n}| \ge r$ for all $n \in \mathbb{N}$. Let $(x_n)_{n \in \mathbb{N}}$ be the sequence defined as $x_n = \frac{u_n}{a_{q,n}}$. It is seen that

$$ (2) \quad \lim_{n \to \infty} \|x_n\|' = 0. $$

The two norms $\| \cdot \|$ and $\| \cdot \|'$ are equivalent on $S$ and they both are equivalent to the product norm $\| \cdot \|''$ defined as $\|\sum_{j=1}^{q-1} b_j e_j\|'' = \max_{1 \le j \le q-1} |b_j|$. Since $\mathbb{L}$ is complete, $S$ is complete with respect to $\| \cdot \|''$, hence $S$ is closed in $V$ with respect to the two norms $\| \cdot \|$ and $\| \cdot \|'$. Hence, by (2), $e_q$ belongs to $S$, which is absurd and finishes the proof. $\square$

**Theorem 2.6.** Let $F$ be an algebraic extension of $\mathbb{L}$, provided with two absolute values extending the one $\mathbb{L}$. These absolute values are equal.

**Proof.** Let $v, w$ be the valuations associated with these absolute values. Let $a \in F$. By Lemma 2.5, the two absolute values are equivalent on $\mathbb{L}[a]$. Hence, by Lemma 2.2, there exists $r > 0$ such that $w(x) = rv(x)$ whenever $x \in \mathbb{L}[a]$. But since $v(x) = w(x)$ whenever $x \in \mathbb{L}$ and since there exists $u \in \mathbb{L}$ such that $v(u) \ne 0$, we have $r = 1$. $\square$

**Lemma 2.7.** Let $A$ be an $\mathbb{L}$-algebra. Let $\phi$ be a semi-norm of $\mathbb{L}$-algebra satisfying $\phi(x^n) = (\phi(x))^n \forall x \in A$. Then, $\phi$ is ultrametric.

**Proof.** Let $a, b \in A$ satisfy $\phi(a) \ge \phi(b)$. We just have to show that $\phi(a + b) \le \phi(a)$. Obviously, we have $\phi((a+b)^n) = \phi(\sum_{k=0}^n C_n^k a^k b^{n-k})$. For each $k = 0, \dots, n$, we have $\phi(C_n^k a^k b^{n-k}) = |C_n^k| \phi(a^k b^{n-k}) \le \phi(a)^k \phi(b)^{n-k} \le \phi(a)^n$, hence $\phi((a+b)^n) \le (n+1)\phi(a)^n$, and therefore, $\phi(a+b) \le \sqrt[n]{n+1} \phi(a)$ for all $n \in \mathbb{N}^*$. Finally, we obtain $\phi(a+b) \le \phi(a)$. $\square$

The most classical example of an ultrametric complete algebraically closed field is the field $\mathbb{C}_p$ that is described later.

**Definition and notation.** Consider the field $\mathbb{E}$ provided with an ultrametric absolute value. Let $a \in \mathbb{E}$ and let $r \in \mathbb{R}_+$. We denote by $d(a, r)$ the disk $\{x \in \mathbb{E} | |x - a| \le r\}$, by $d(a, r^-)$ the disk $\{x \in \mathbb{E} | |x - a| < r\}$, and we call *circle of center a, of radius r* the set $C(a, r) = d(a, r)\backslash d(a, r^-)$.

Given $r_1$ and $r_2$ such that $0 < r_1 < r_2$, we denote by $\Gamma(a, r_1, r_2)$ the annulus $\{x \in \mathbb{E} | r_1 < |x - a| < r_2\}$ and by $\Delta(a, r_1, r_2)$ the annulus $\{x \in \mathbb{E} | r_1 \le |x - a| \le r_2\}$.
We know that if $b \in d(a, r)$, then $d(b, r) = d(a, r)$. In the same way, if $b \in d(a, r^{-})$, then $d(b, r^{-}) = d(a, r^{-})$. Moreover, given two disks $T$ and $T'$ such that $T \cap T' \neq \emptyset$, then either $T \subset T'$ or $T' \subset T$.

We denote by $\delta$ the distance defined on $\mathbb{E}$ by $\delta(a, b) = |a-b|$. Given $a \in \mathbb{E}$ and a subset $D$ of $\mathbb{E}$, we set $\delta(a, D) = \inf\{|x-a| \mid x \in D\}$, and given two subsets $D, F$ of $\mathbb{E}$, we set $\delta(D, F) = \inf\{|x-y| \mid x \in D, y \in F\}$.

We set $\text{diam}(D) = \sup\{|x-y| \mid |x \in D, y \in D\}$ and $\text{diam}(D)$ is named the *diameter* of $D$.

Similarly, we set $\text{codiam}(D) = \sup\{|x-y| \mid |x \in D, y \notin D\}$ and $\text{codiam}(D)$ is named the *codiameter* of $D$.

Of course, the following three statements are seen to be equivalent:

(i) $d(a, r) = d(a, r^{-})$.

(ii) $C(a, r) = \emptyset$.

(iii) $r \notin |E|$.

Further, the disks $d(b, r^{-})$ included in $C(a, r)$ (respectively in $d(a, r)$) are the disks $d(b, r^{-})$ such that $b \in C(a, r)$ (respectively in $d(a, r)$). They are called the *classes of $C(a, r)$* (respectively of $d(a, r)$).

Henceforth, $D$ will denote a subset of the field $\mathbb{L}$.

The closure of $D$ (also called *adherence of $D$*) is denoted by $\bar{D}$ and the interior of $D$ (also called *opening of $D$*) is denoted by $\tilde{D}$.

Given a point $a \in L$, we put $\delta(a, D) = \inf\{|x-a| \mid x \in D\}$. Then, $\delta(a, D)$ is named the *distance of a to D*.

Given two subsets $D, D'$ of $\mathbb{L}$, we put $\delta(D, D') = \inf\{|x-y| \mid x \in D, y \in D'\}$. Then, $\delta(D, D')$ is called *the distance between D and D'*.

Given a bounded subset $D$ of $\mathbb{L}$ of diameter $R$, we denote by $\tilde{D}$ the disk $d(a, R)$, whenever $a \in D$. Given an unbounded subset $D$ of $\mathbb{L}$, we put $\tilde{D} = \mathbb{L}$.

We denote by $\tilde{\mathbb{L}}$ an extension of $\mathbb{L}$ provided with an absolute value that extends that of $\mathbb{L}$. Given $a \in \tilde{\mathbb{L}}$, $r > 0$, $\tilde{d}(a, r)$ (respectively $\tilde{d}(a, r^{-})$) denotes the disk $\{x \in \tilde{\mathbb{L}} \mid |x-a| \le r\}$ (respectively $\{x \in \tilde{\mathbb{L}} \mid |x-a| < r\}$).

**Lemma 2.8.** Let $d(a, r)$, $d(b, s)$ be disks such that $d(a, r) \cap d(b, s) \neq \emptyset$ with $r \le s$. Then, $d(a, r) \subset d(b, s)$.
Let us also note the following basic lemma:

**Lemma 2.9.** Suppose that the residue class field $\mathcal{E}$ of the field $\mathbb{E}$ is finite, of cardinal $q$. Then, for every disk $d(a, r)$ with $a \in \mathbb{E}$ and $r \in |\mathbb{E}|$ admits only $q$ classes.

**Lemma 2.10.** $\tilde{D}\backslash\overline{D}$ admits a unique partition of the form $(T_i)_{i \in I}$, whereas each $T_i$ is a disk of the form $d(a_i, r_i^{-})$ with $r_i = \delta(a_i, D)$.

**Proof.** For every $a \in \tilde{D}\backslash\overline{D}$, let $r(a) = \delta(a, D)$. Let $\alpha$ and $\beta$ be two points in $\tilde{D}\backslash\overline{D}$ such that $|\beta - \alpha| < r(\alpha)$. It is easily seen that for every $x \in D$, we have $|x - \beta| = |x - \alpha|$, and then the family of the disks $T(\alpha) = d(\alpha, r(\alpha))^{-}$ ($\alpha \in \tilde{D}\backslash\overline{D}$) makes a partition of $\tilde{D}\backslash D$ because given $\alpha$ and $\beta \in \tilde{D}\backslash\overline{D}$, either $|\alpha - \beta| < r(\alpha)$ and then $T(\alpha) = T(\beta)$, or $|\alpha - \beta| \ge r(\alpha)$ and then $|\alpha - \beta| \ge r(\beta)$, hence $T(\alpha) \cap T(\beta) = \emptyset$. $\square$

**Definition and notation.** Such disks $d(a_i, r_i^{-})$ are called the holes of $D$.

**Example 1.** The holes of a disk $d(a, r^{-})$, with $r \in |\mathbb{I}|$, are the classes of $C(a, r)$.

**Example 2.** The only one hole of $\mathbb{I}\backslash d(0, 1^{-})$ is $d(0, 1^{-})$.

**Example 3.** The holes of $\mathbb{I}\backslash d(0, 1)$ are the disks $d(a, 1^{-})$ with $a \in d(0, 1)$.

**Definition.** $D$ is said to be *infraconnected* [35], [36], [44], [49] if for every $a \in D$, the mapping $I_a$ from $D$ to $\mathbb{R}_+$ defined by $I_a(x) = |x - a|$ has an image whose closure in $\mathbb{R}_+$ is an interval. In other words, $D$ is not infraconnected if and only if there exist $a$ and $b \in D$ and an annulus $\Gamma(a, r_1, r_2)$ with $0 < r_1 < r_2 < |a - b|$ such that $\Gamma(a, r_1, r_2) \cap D = \emptyset$.

Lemma 2.11 is immediate:

**Lemma 2.11.** If $D$ is infraconnected of diameter $R \in \mathbb{R}$ (respectively $+\infty$), then $\bar{I}_a(D) = [0, R]$ (respectively $\bar{I}_a(D) = [0, +\infty[$).

Lemma 2.12 gives a point of view from a hole of $D$.

**Lemma 2.12.** Let $D$ be infraconnected and let $\alpha$ belong to a hole $T$ of diameter $\rho$. The closure of the set $\{|x - \alpha| \mid x \in D\}$ is an interval whose lower bound is $\rho$.
**Proof.** We just have to show that for every $r$ and $r'$ such that $\rho < r < r' < \text{diam}(D)$, there exists $\beta \in D$ such that $r < |\beta - \alpha| < r'$. By definition of the holes, there exists $b \in D$ such that $|\alpha - b| < r$ and then, since $D$ is infraconnected, there exists $\beta \in D$ such that $r < |b - \beta| < r'$. But it is seen that $|\beta - \alpha| = |b - \beta|$. $\square$

Given two infraconnected sets $A$ and $B$, we may prove $A \cup B$ to be infraconnected in the following two hypothesis (Theorems 2.13 and 2.15).

**Theorem 2.13.** Let $A$ and $B$ be two infraconnected sets such that $A \cap B \neq \emptyset$. Then, $A \cup B$ is infraconnected.

**Proof.** If $A$ and $B$ are not bounded, the statement is obvious because for every $a \in A$, $\overline{I_a(A)} = \mathbb{R}_+$ and for every $a \in B$, we have $\overline{I_a(B)} = \mathbb{R}_+$. Now, we may assume $A$ to be bounded, of diameter $R$, while $B$ has diameter $R' \ge R$ (respectively is not bounded). Then, $A \cup B$ has diameter $R'$ (respectively is not bounded). Let $c \in A \cap B$, let $a \in A \cup B$, and let us show that $\overline{I_a(A \cup B)} = [0, R']$ (resp. $[0, +\infty[$).

For convenience, we first assume $B$ to be bounded. Since $c \in A \cap B$, we see that $|x-a| \le \max(|x-c|, |c-a|) \le R'$ whenever $x \in A \cup B$, hence $\overline{I_a(A \cup B)} \subset [0, R']$. Hence, we just have to show that $\overline{I_a(A \cup B)} \supset [0, R']$. Obviously, $\overline{I_a(A \cup B)} = \overline{I_a(A)} \cup \overline{I_a(B)} = [0, R] \cup \overline{I_a(B)}$. Hence, we have to show that $\overline{I_a(B)} \supset [R, R']$. But when $x \in B$ with $|x-a| > R$, we see that $|x-a| = |x-c|$ (because $|c-a| \le R$), hence $\overline{I_a(B) \cap R, R'} = \overline{I_c(B) \cap R, R'}$, and finally, $\overline{I_a(B)} \supset [R, R']$ because $\overline{I_c(B)} \supset [R, R']$.

When $B$ is not bounded, in the same way it is seen that $\overline{I_a(A \cup B)} = [0, +\infty[$. This finishes showing that $A \cup B$ is infraconnected. $\square$

**Corollary 2.14.** The relation $\mathcal{R}$ defined by $x\mathcal{R}y$ if there exists an infraconnected subset of $D$ that contains $x$ and $y$ is an equivalence relation.

**Proof.** $\mathcal{R}$ is obviously reflexive and symmetric. It is transitive by Theorem 2.13. $\square$

**Definition.** The equivalence classes with respect to this relation are called the *infraconnected components*.
**Examples.**

(1) $d(0, 1^{-}) \cup d(1, 1^{-})$ is infraconnected. Its holes are the disks $d(\alpha, 1^{-})$ with $|\alpha| = |\alpha - 1| = 1$.

(2) Let $r \in ]0, 1[$ and let $D = d(0, 1^{-}) \cup d(1, r)$. Then, $D$ is not infraconnected; its infraconnected components are $d(0, 1^{-})$ and $d(1, r)$. The holes of $D$ are the disks $d(\alpha, 1^{-})$ with $|\alpha| = |\alpha - 1| = 1$ and the disks $d(\alpha, |\alpha - 1|^{-})$ with $r < |\alpha - 1| < 1$.

**Theorem 2.15.** Let $A$ and $B$ be infraconnected sets such that $\tilde{A} = \tilde{B}$. Then, $A \cup B$ is infraconnected.

**Proof.** Obviously, $\tilde{A} \cup \tilde{B} = \tilde{A}$. If $A$ is bounded, let $\tilde{A} = d(\alpha, R)$, and otherwise, let $\tilde{A} = L$. First, let us assume $A$ to be bounded. For $a \in A$, the set $\{|x - a| \mid x \in A\}$ is dense in $[0, R]$, hence so is the set $\{|x - a| \mid x \in A \cup B\}$. In the same way, $B$ plays the same role, hence this still holds for $a \in B$. Finally, if $A$ is not bounded, we just replace $[0, R]$ with $[0, +\infty[$. That finishes proving Theorem 2.15. $\square$

**Definition.** An infraconnected subset $D$ of $\mathbb{L}$ is said to be *affinoid* if it is of the form $d(a, R) \setminus \bigcup_{k=1}^{q} d(b_k, r_k^{-})$ with $R$ and $r_k \in |\mathbb{L}| \forall k$. A subset $D$ of $\mathbb{L}$ is said to be *affinoid* if it is a finite union of infra-connected affinoid subsets.

**Proposition 2.16.** Let $D_1, D_2$ be two infraconnected affinoid subsets of $\mathbb{L}$ such that $D_1 \cap D_2 \neq \emptyset$ and set $D = D_1 \cup D_2$ and $E = D_1 \cap D_2$. Then both $D$ and $E$ are infraconnected affinoid. Moreover, $\tilde{D}$ is either $\tilde{D}_1$ or $\tilde{D}_2$ and each hole of $D$ is either a hole of $D_1$ or a hole of $D_2$.

**Proof.** By Theorem 2.13, $D$ is infraconnected. Consider now $D_1$ of the form $d(a, r) \setminus (\bigcup_{i=1}^{m} d(a_i, r_i^{-}))$ and $D_2$ of the form $d(b, s) \setminus (\bigcup_{i=1}^{n} d(b_i, s_i^{-}))$. Suppose, for instance, $r \leq s$ and let $c \in D_1 \cap D_2$. Then, we can check that

$$E = d(c, r) \setminus ( ( \bigcup_{i=1}^{d} (a_i, r_i^{-}) ) \cup ( \bigcup_{i=1}^{n} d(b_i, s_i^{-}) ) ),$$

which is an infraconnected affinoid again. Since $D_1 \cap D_2 \neq \emptyset$, we have $\tilde{D}_1 \cap \tilde{D}_2 \neq \emptyset$, hence $\tilde{D}$ is either $\tilde{D}_1$ or $\tilde{D}_2$, hence $\text{diam}(D) \in |\mathbb{L}|$. Next, since the holes of both sets are in finite number, each hole of
$D$ is either a hole of $D_1$ or a hole of $D_2$, so each hole of $D$ has a diameter in $|\mathbb{I}|$ and of course they are in finite number. $\square$

**Definition.** We call empty annulus of $D$ an annulus $\Gamma(a, r_1, r_2)$ such that

(i) $r_1 = \sup\{|x - a| \mid x \in D, |x - a| \le r_2\}$,

(ii) $r_2 = \inf\{|x - a| \mid x \in D, |x - a| \ge r_1\}$.

The set $d(a, r_1) \cap D$ is denoted by $\mathcal{I}_D(\Gamma(a, r_1, r_2))$, while the set $(\mathbb{I} \setminus d(a, r_2^-)) \cap D$ is denoted by $\mathcal{E}_D(\Gamma(a, r_1, r_2))$. When there is no risk of confusion about the set $D$, we just write $\mathcal{I}(\Gamma(a, r_1, r_2))$ (respectively $\mathcal{E}(\Gamma(a, r_1, r_2))$) instead of $\mathcal{I}_D(\Gamma(a, r_1, r_2))$ (respectively $\mathcal{E}_D(\Gamma(a, r_1, r_2))$).

**Remark 1.** By definition, $D$ is not infraconnected if and only if it admits an empty annulus.

**Remark 2.** By definition, $\{\mathcal{I}(\Gamma(a, r_1, r_2)), \mathcal{E}(\Gamma(a, r_1, r_2))\}$ is a partition of $D$.

**Example.** Let $r \in ]0, 1[$, let $D = d(0, r) \cup d(1, 1^{-})$, and let $D' = d(0, r^{-}) \cup d(1, r)$. Then, $\Gamma(0, r, 1)$ is an empty annulus of $D$ and also of $D'$. In the same way, $\Gamma(1, r, 1)$ is also an empty annulus of $D'$.

**Notation.** Let $\mathcal{X}(D)$ be the set of the empty annuli of $D$. Given $\Lambda_1$ and $\Lambda_2 \in \mathcal{X}(D)$, it is easily seen that $\mathcal{I}(\Lambda_1) \subset \mathcal{I}(\Lambda_2)$ is equivalent to $\mathcal{E}(\Lambda_1) \supset \mathcal{E}(\Lambda_2)$. We denote by $\le$ the relation defined on $\mathcal{X}(D)$ by $\Lambda_1 \le \Lambda_2$ if $\mathcal{I}(\Lambda_1) \subset \mathcal{I}(\Lambda_2)$ and we set $\Lambda_1 < \Lambda_2$ if $\Lambda_1 \le \Lambda_2$ and $\Lambda_1 \ne \Lambda_2$.

Lemmas 2.17 and 2.18 are easily seen.

**Lemma 2.17.** The relation $\le$ is a relation of order on $\mathcal{X}(D)$. Let $\Lambda_1$ and $\Lambda_2$ be two empty annuli of $D$. The following assertions are equivalent:

(i) $\Lambda_1$ and $\Lambda_2$ are not comparable with respect to the order $\le$.

(ii) $\mathcal{I}(\Lambda_1) \subset \mathcal{E}(\Lambda_2)$.

(iii) $\mathcal{I}(\Lambda_2) \subset \mathcal{E}(\Lambda_1)$.

(iv) $\mathcal{I}(\Lambda_1) \cap \mathcal{I}(\Lambda_2) = \emptyset$.

**Lemma 2.18.** Let $\Lambda \in \mathcal{X}(D)$ and let $x \in \mathcal{I}(\Lambda)$ (respectively $x \in \mathcal{E}(\Lambda)$). The infraconnected component of $x$ is included in $\mathcal{I}(\Lambda)$
(respectively in $\mathcal{E}(\Lambda)$). If $\Lambda' \in \mathcal{X}(D)$ is such that $\Lambda < \Lambda'$, then
$\mathcal{I}(\Lambda') \cap \mathcal{E}(\Lambda) \neq \emptyset$.

Lemma 2.19 is a direct consequence of Lemmas 2.17 and 2.18.

**Lemma 2.19.** Let $\Theta$ be an empty annulus of $D$. The family of the empty annuli $\Lambda \ge \Theta$ is totally ordered.

**Proof.** Let $\Lambda_1$ and $\Lambda_2 \in \mathcal{X}(D)$ satisfy $\Lambda_1 \ge \Theta, \Lambda_2 \ge \Theta$. Then, $\mathcal{I}(\Lambda_1) \cap \mathcal{I}(\Lambda_2) \supset \mathcal{I}(\Theta) \ne \emptyset$, hence $\mathcal{I}(\Lambda_1)$ is not included in $\mathcal{E}(\Lambda_2)$, hence $\Lambda_1$ and $\Lambda_2$ are comparable. $\square$

**Lemma 2.20.** Let $\Theta$ be a minimal element of $\mathcal{X}(D)$ for the order $\le$. Then, $\mathcal{I}(\Theta)$ is an infraconnected component of $D$.

**Proof.** Suppose that $\mathcal{I}(\Theta)$ is not infraconnected. By definition, $\mathcal{I}(\Theta)$ is of the form $d(a, R) \cap D$, hence there exists an empty annulus $\Lambda = \Gamma(\alpha, r_1, r_2)$ of $\mathcal{I}(\Theta)$ with $\alpha \in d(a, R), r_1 < r_2 \le R$ and some $\beta \in \mathcal{I}(\Theta)$ such that $r_2 \le |\alpha - \beta| \le R$. Since $\Lambda \subset d(a, R)$, we see that $\Lambda \cap D = \emptyset$, hence $\Lambda$ is an empty annulus of $D$, and therefore, $\Lambda < \Theta$. This ends the proof of Lemma 2.20. $\square$

**Theorem 2.21.** *D* has finitely many infraconnected components if and only if it has finitely many empty annuli. Moreover, if so does *D*, then one of the infraconnected components is $A_0 = \bigcap_{\Theta \in \mathcal{X}(D)} \mathcal{E}(\Theta)$, while the others are of the form $A_i = \mathcal{I}(\Lambda_i) \cap (\bigcap_{\Theta < \Lambda_i} \mathcal{E}(\Theta))$, with $\Lambda_i \in \mathcal{X}(D)$.

**Proof.** We first assume $\mathcal{X}(D)$ to be finite and we prove that the infraconnected components are in the form $A_i$, above, so that there will be finitely many ones.

Let $\Lambda_1, \dots, \Lambda_n$ be these empty annuli of $D$, and for every $i = 0, \dots, n$, let $A_i$ be the subsets of $D$ defined from $\Lambda_1, \dots, \Lambda_n$ as above. For every $x \in D$, for every $i = 1, \dots, n$, either $x \in \mathcal{I}(\Lambda_i)$ or $x \in \mathcal{E}(\Lambda_i)$, hence it is easily seen that $x$ belongs to one of the $A_i$, hence $D = \bigcup_{i=0}^n A_i$. We check that $A_i \cap A_j = \emptyset$ whenever $i \ne j$. First, we assume $i = 0, j > 0$. Hence, $A_0 \subset \mathcal{E}(\Lambda_j)$, while $A_j \subset \mathcal{I}(\Lambda_j)$, hence $A_0 \cap A_j = \emptyset$. Now, we suppose $i > 0, j > 0$. If $\Lambda_i < \Lambda_j$, then $a_j \subset \mathcal{E}(\Lambda_i)$, while $A_i \subset \mathcal{I}(\Lambda_i)$ and then $A_i \cap A_j = \emptyset$. Hence, we may assume that $\Lambda_1$ and $\Lambda_2$ are not comparable, and then, by Lemma 2.17, we
have $\mathcal{I}(\Lambda_i) \cap \mathcal{I}(\Lambda_j) = \emptyset$, hence $A_i \cap A_j = \emptyset$. Consequently, the family $(A_i)_{0 \le i \le n}$ makes a partition of $D$.

Now, we show that each $A_i$ is infraconnected. Suppose that a certain $A_h$ is not infraconnected for some $h > 0$ (respectively $h = 0$). Then, it admits an empty annulus $\Lambda = \Gamma(a, r_1, r_2)$. First, we note that if $h = 0$, then $\Lambda_h > \Lambda$ because both $a, b$ are centers of $\Lambda_h$. Now, if $h = 0$ (respectively $h > 0$), let $\Theta \in \mathcal{X}(D)$ (respectively let $\Theta \in \mathcal{X}(D)$ be such that $\Theta < \Lambda_h$). Since both $a, b$ belong to $\mathcal{E}(\Theta)$, it is seen that all $\Lambda$ is included in $\mathcal{E}(\Theta)$ and therefore is included in $A_h$. This contradicts the hypothesis and finishes proving that $A_h$ is infraconnected.

Next, we check that each $A_j$ is maximal in the set of the infraconnected subsets of $D$. Indeed, let $B$ be a subset of $D$ that strictly contains a certain $A_h$ and let $a \in B \setminus A_h$. If $h = 0$, there exists $\Theta \in \mathcal{X}(D)$ such that $a \in \mathcal{I}(\Theta)$, but $A_h \subset \mathcal{E}(\Theta)$, and therefore, $\Theta$ is included in an empty annulus of $B$. If $h > 0$, either $a$ belongs to $\mathcal{E}(\Lambda_h)$ whereas $A_h \subset \mathcal{I}(\Lambda_h)$ and then $\Lambda_h$ is included in an empty annulus of $B$, or there exists $\Theta \in \mathcal{X}(D)$ satisfying $\Theta < \Lambda_h$ and $a \in \mathcal{I}(\Theta)$, but then $A_h \subset \mathcal{I}(\Theta)$ and therefore $\Theta$ is included in an empty annulus of $B$. Thus, in each case, $B$ is not infraconnected and this finishes showing that each $A_i$ is maximal in the set of the infraconnected subsets of $D$. As a consequence, the infraconnected components of $D$ are the $A_i$.

Now, conversely, we assume $D$ to have infinitely many empty annuli. First, let us suppose that $D$ has a sequence of empty annuli $(\Lambda_n)_{n \in \mathbb{N}}$ such that $\Lambda_n < \Lambda_{n+1}$ (respectively $\Lambda_n > \Lambda_{n+1}$) for all $n \in \mathbb{N}$. By Theorem 2.15, for every $n \in \mathbb{N}$, there exists $x_n \in \mathcal{E}(\Lambda_n) \cap \mathcal{I}(\Lambda_{n+1})$ (respectively $x_n \in \mathcal{I}(\Lambda_n) \cap \mathcal{E}(\Lambda_{n+1}))$ and then the infraconnected component $X_n$ of $x_n$ satisfies $X_n \subset \mathcal{E}(\Lambda_n) \cap \mathcal{I}(\Lambda_{n+1})$ (respectively $X_n \subset \mathcal{I}(\Lambda_n) \cap \mathcal{E}(\Lambda_{n+1}))$, hence $X_n \cap X_m = \emptyset$ for all $n \ne m$, hence $D$ has infinitely many infraconnected components.

Finally, we may assume that every totally ordered set of empty annuli is finite. Hence, there exists a sequence of empty annuli $\Lambda_n$ that are minimal elements for the order $\le$ on $\mathcal{X}(D)$ and then $\mathcal{I}(\Lambda_n) \cap \mathcal{I}(\Lambda_m) = \emptyset$ whenever $n \ne m$. By Lemma 2.19, $\mathcal{I}(\Lambda_n)$ is an infraconnected component $D_n$ of $D$ such that $D_n \cap D_m = \emptyset$ whenever $n \ne m$. This finishes proving that $D$ has infinitely many infraconnected components and this ends the proof. $\square$
# Chapter 3

## Monotonous and Circular Filters

Monotonous and circular filters are essential on an ultrametric field, mainly because for any rational function, its absolute value admits a limit along each circular filter [36], [44], [45], [49], [53] and circular filters are the least thin filters having this property. Most of properties of analytic functions of all kinds are derived from that property of circular filters. Certain authors call “generic disk” a notion which is not clearly defined but actually represents a circular filter. We see that, given a bounded sequence, there exists a subsequence thinner than a circular filter.

For certain problems, we can reduce ourselves to consider monotonous filters instead of circular filters. Monotonous filters are linked to sequences $(a_n)$ such that $|a_{n+1} - a_n|$ is strictly monotonous. Moreover, decreasing filters let us define spherically complete fields.

**Definition.** Let $J$ be set. A filter $\mathcal{F}$ on $J$ is said to be *thinner* than a filter $\mathcal{G}$ if every element of $\mathcal{G}$ belongs to $\mathcal{F}$. In such a case, $\mathcal{G}$ is said to be *less thin than* $\mathcal{F}$. Two filters $\mathcal{F}, \mathcal{G}$ are said to be *secant* if for all $A \in \mathcal{F}, B \in \mathcal{G}$ we have $A \cap B \neq \emptyset$.

A filter $\mathcal{F}$ is said to be *secant* to a subset $B \subset J$ if $\{F \cap B \mid F \in \mathcal{F}\}$ is a filter.

A sequence $(u_n)_{n \in \mathbb{N}}$ in $J$ is said to be *thinner* than a filter $\mathcal{G}$ if so is the filter defined by the sets $A_q = \{u_n|n \ge q\}$ ($q \in \mathbb{N}$). In such a case, $\mathcal{G}$ is said to be *less thin than* the sequence $(u_n)_{n \in \mathbb{N}}$.

A sequence $(u_n)_{n \in \mathbb{N}}$ in $\mathbb{L}$ will be said to be an *increasing distances sequence* (respectively a *decreasing distances sequence*) if the
sequence $|u_{n+1} - u_n|$ is strictly increasing (respectively decreasing) and has a limit $\ell \in \mathbb{R}_+^*$.

The sequence $(u_n)_{n \in \mathbb{N}}$ will be said to be a monotonous distances sequence if it is either an increasing distances sequence or a decreasing distances sequence.

A sequence $(u_n)_{n \in \mathbb{N}}$ in $\mathbb{L}$ will be said to be an equal distances sequence if $|u_n - u_m| = |u_m - u_q|$ whenever $n, m, q \in \mathbb{N}$ such that $n \neq m \neq q$.

**Theorem 3.1.** Let $\mathbb{E}$ be a field provided with an ultrametric absolute value. Let $(u_n)_{n \in \mathbb{N}}$ be a bounded sequence in $\mathbb{E}$. Either we may extract a Cauchy subsequence or we may extract a monotonous distances subsequence or we may extract an equal distances subsequence from the sequence $(u_n)_{n \in \mathbb{N}}$. Further, if the absolute value of $\mathbb{E}$ is discrete, there is no monotonous distances sequence in $\mathbb{E}$. And if the residue class field of $\mathbb{E}$ is finite, there is no equal distances sequence in $\mathbb{E}$.

**Proof.** Suppose Theorem 3.1 to be false. For every $q \in \mathbb{N}$, the set of the circles $C(u_q, r)$ that contain some $u_n$ is then finite.

Suppose that we have already defined integers $n_q$ for $q \le t$ satisfying

(1) $|u_{n_q} - u_{n_{q-1}}| < |u_{n_{q-1}} - u_{n_{q-2}}|$ for $2 \le q \le t$ and such that $d(u_{n_q}, |u_{n_q} - u_{n_{q-1}}|^{-})$ contains infinitely many terms of the sequence $(u_n)$. For every $q = 2, \dots, t$, let $r_q = |u_{n_q} - n_{q-1}|$. Obviously, at least one of the circles $C(u_{n_t}, r)$, with $r < r_t$, contains infinitely many terms of the sequence $(u_n)_{n \in \mathbb{N}}$. Let $C(u_{n_t}, r_{t+1})$ be such a circle. It is seen that at least one class $\Lambda$ of this circle contains infinitely many terms of the sequence because otherwise we would have a sequence of classes $(\Lambda_j)$ each one containing at least one term $u_{\tau(j)}$ and then they should satisfy $|u_{\tau(j)} - u_{\tau(i)}| = r_{t+1}$ whenever $i \ne j$. Hence, the sequence $(u_n)_{n \in \mathbb{N}}$ should admit an equal distances subsequence. Then we may pick up one term $u_{n_{t+1}}$ in $\Lambda$ and we have constructed the finite subsequence up to the rank $t+1$, satisfying the properties mentioned above. In the same way, we may initiate the induction by defining $n_2$ from arbitrary $n_0$, $n_1$. The sequence $(u_{n_t})_{t \in \mathbb{N}}$ is then defined for every $t \in \mathbb{N}$ and satisfies (1) for $t > 1$. Let $\ell = \lim_{t \to \infty} |u_{n_t} - u_{n_{t+1}}|$. If $\ell = 0$, the subsequence $(u_{n_t})_{t \in \mathbb{N}}$ is a Cauchy subsequence. If $\ell > 0$, this is a decreasing distances subsequence. Thus, we have proven that we can extract a sequence which is either a convergent sequence, or a monotonous distances sequence, or an equal distances sequence.
Now, suppose that the absolute value is discrete and suppose that we have extracted monotonous distances sequences $(b_m)_{m \in \mathbb{N}}$ from the sequence $(u_n)$. Then the strictly monotonous sequence $|b_{m+1} - b_m|$ must tend to 0, a contradiction. Finally, suppose that the residue class field $\mathcal{E}$ of $\mathbb{E}$ is finite and suppose that we have extracted an equal distances sequences $(b_m)_{m \in \mathbb{N}}$. So, $|b_0 - b_m| = |b_0 - b_n| = |b_m - b_n| \quad \forall m \neq n, m \neq 0, n \neq 0$. Let $q$ be the cardinal of $\mathcal{E}$. Then, by Lemma 2.9, the set of terms $b_m$ is at most $q$, a contradiction. $\square$

Henceforth, throughout this chapter, the field $\mathbb{L}$ is supposed to have a dense valuation and $D$ is an infraconnected subset of $\mathbb{L}$.

**Definition and notation.** Let $a \in \tilde{D}$ and $R \in \mathbb{R}_+^*$ be such that $\Gamma(a, r, R) \cap D \neq \emptyset$ whenever $r \in [0, R]$ (respectively $\Gamma(a, R, r) \cap D \neq \emptyset$ whenever $r > R$). We call an increasing (respectively a decreasing) filter of center $a$ and diameter $R$, on $D$ the filter $\mathcal{F}$ on $D$ that admits for basis the family of sets $\Gamma(a, r, R) \cap D$ (respectively $\Gamma(a, R, r) \cap D$). For every sequence $(r_n)_{n \in \mathbb{N}}$ such that $r_n < r_{n+1}$ (respectively $r_n > r_{n+1}$) and $\lim_{n \to \infty} r_n = R$, it is seen that the sequence $\Gamma(a, r_n, R) \cap D$ (respectively $\Gamma(a, R, r_n) \cap D$) is a basis of $\mathcal{F}$, and such a basis is called a *canonical basis*. We call a decreasing filter with no center of canonical basis $(D_n)_{n \in \mathbb{N}}$ and diameter $R > 0$, on $D$ a filter $\mathcal{F}$ on $D$ that admits for basis a sequence $(D_n)_{n \in \mathbb{N}}$ in the form $D_n = d(a_n, r_n) \cap D$ with $D_{n+1} \subset D_n$, $r_{n+1} < r_n$, $\lim_{n \to \infty} r_n = R$, and $\bigcap_{n \in \mathbb{N}} d(a_n, r_n) = \emptyset$.

Given an increasing (respectively a decreasing) filter $\mathcal{F}$ on $D$ of center $a$ and diameter $r$, we denote by $\mathcal{B}_D(\mathcal{F})$ the set $\{x \in D \mid |x - a| \ge r\}$ (respectively the set $\{x \in D \mid |x - a| \le r\}$ and by $\mathcal{C}_D(\mathcal{F})$ the set $\{x \in D \mid |x - a| < r\}$ (respectively the set $\{x \in D \mid |x - a| > r\}$). When there is no risk of confusion, we only write $\mathcal{B}(\mathcal{F})$ instead of $\mathcal{B}_D(\mathcal{F})$ and $\mathcal{C}(\mathcal{F})$ instead of $\mathcal{C}_D(\mathcal{F})$. Next, $\mathcal{C}_D(\mathcal{F})$ is named the body of $\mathcal{F}$ and $\mathcal{B}_D(\mathcal{F})$ is named the beach of $\mathcal{F}$.

We call a *monotonous filter on D* a filter which is either an increasing filter or a decreasing filter (with or without a center). Given a monotonous filter $\mathcal{F}$, we denote by $\text{diam}(\mathcal{F})$ its diameter.

The field $\mathbb{L}$ is said to be *spherically complete* if every decreasing filter on $\mathbb{L}$ has a center in $\mathbb{L}$. The field $\mathbb{C}_p$, for example, is not spherically complete (see Chapter 6). However, every algebraically closed complete ultrametric field admits a spherically complete algebraically closed extension and this is recalled in Chapter 8.
**Lemma 3.2.** Let $(a_n)_n \in \mathbb{N}$ be an increasing distances (respectively a decreasing distances) sequence in $D$. There exists a unique increasing (respectively decreasing) filter $\mathcal{F}$ on $D$ such that the sequence $(a_n)_n \in \mathbb{N}$ is thinner than $\mathcal{F}$.

**Proof.** Let $r_n = |a_{n+1} - a_n|$ and let $R = \lim_{n \to \infty} r_n$.

We first suppose $(a_n)_{n \in \mathbb{N}}$ to be an increasing distances sequence. The increasing filter $\mathcal{F}$ of center $a_0$, of diameter $R$ is obviously less thin than the sequence $(a_n)_{n \in \mathbb{N}}$. We show that $\mathcal{F}$ is unique. Let $\mathcal{G}$ be an increasing filter of center $a$, of diameter $R'$, less thin than the sequence $(a_n)_{n \in \mathbb{N}}$. For every $r < R'$, there exists $q \in \mathbb{N}$ such that $a_n \in \Gamma(a, r, R')$ whenever $n \ge q$. If $a \in d(a_0, R^{-})$, this clearly requires that $R = R'$ and then $\mathcal{G} = \mathcal{F}$. Let us suppose that $a \notin d(a_0, R^{-})$. Then we have $|a - a_n| = |a_n - a_m| = C$ whenever $n \ne m$ so $R' > R$ and then $\Gamma(a, r, R')$ does not contain the $a_n$ whenever $r > R$. Finally, $\mathcal{G} = \mathcal{F}$.

We now suppose the sequence $(a_n)_{n \in \mathbb{N}}$ to be a decreasing distances sequence with a point $a$ such that $|a - a_n| = |a_{n+1} - a_n|$ whenever $n \in \mathbb{N}$. Then the decreasing filter of center $a$, of diameter $R$ is a decreasing filter less thin than the sequence $(a_n)_{n \in \mathbb{N}}$. We show it to be the only decreasing filter less thin than the sequence $(a_n)_{n \in \mathbb{N}}$. Indeed, given a decreasing filter $\mathcal{G}$ less thin than the sequence $(a_n)_{n \in \mathbb{N}}$, it must have a center because if it had no center, the sequence $d(a_{n+1}, |a_{n+1} - a_n|)$ would be one of its canonical basis, but by definition, it has an intersection that contains $a$. Then, symmetrical to the case when $\mathcal{F}$ is increasing, it is easily seen that $\mathcal{F}$ is unique.

Now, we suppose that the sequence $(a_n)_{n \in \mathbb{N}}$ is a decreasing distances sequence and that there does not exist $a \in L$ such that $|a - a_n| = |a_{n+1} - a_n|$ whenever $n \in \mathbb{N}$. We put $|a_{n+1} - a_n| = r_n$. Hence, the sequence of disks $d(a_{n+1}, r_n)$ has empty intersection and then the filter $\mathcal{F}$, a basis of which is the sequence $(D_n)_{n \in \mathbb{N}}$ with $D_n = d(a_{n+1}, r_n) \cap D$, is a decreasing filter with no center, of diameter $R$. There is no decreasing filter with center $a \in \mathbb{I}$, less thin than the sequence $(a_n)$ because we should have $|a - a_n| = r_n$ whenever $n \in \mathbb{N}$. Hence, it just remains to show that $\mathcal{F}$ is the only decreasing filter with no center less thin than the sequence $(a_n)$. Let us suppose that there exists another decreasing filter $\mathcal{G}$ of diameter $R'$ with no center, of canonical basis $(D'_m)_{m \in \mathbb{N}}$ less thin than the sequence
$(a_n)$. If $R' > R$, since every $D'_m$ contains points $a_n$, it is seen that all the $a_n$ lie in $D \cap d(a_0, R) \subset D'_m$ whenever $m \in \mathbb{N}$ and this contradicts that $\mathcal{G}$ has no center. Hence, we have $R' \le R$. But symmetrically, we have $R \le R'$. Hence, $R = R'$. We show that $\mathcal{G} = \mathcal{F}$. For every $m \in \mathbb{N}$, let $\rho_m$ be the diameter of $D'_m$ and let $a_q \in D'_m$ be such that $r_q \le \rho_m$. Clearly, $a_n \in D'_m$ whenever $n \ge q$, hence $D_n \subset D'_m$ whenever $n > q$. In the same way, let $n \in \mathbb{N}$ and $t \in \mathbb{N}$ be such that $\rho_m < r_n$ whenever $m \ge t$. Then, $D'_m$ contains some $a_s$ which belongs to $d(a_{n+1}, r_n) \cap D = D_n$, hence $D'_m \subset D_n$ whenever $m \ge t$. That finishes showing that $\mathcal{G} = \mathcal{F}$ and that ends the proof of Lemma 3.2. $\square$

**Lemma 3.3.** Let $\mathcal{F}$ be an increasing filter (respectively a decreasing filter) on $\mathbb{I}$, of center $a \in \tilde{D}$ and diameter $R \le \text{diam}(D)$ (respectively $R < \text{diam}(D)$) such that a does not belong to a hole of diameter $\rho \ge R$ (respectively $\rho > R$). Then, $\mathcal{F}$ is secant with $D$ and induces on $D$ an increasing filter (respectively a decreasing filter) of center $a$ and diameter $R$, on $D$.

**Proof.** We just have to check that $\Gamma(a, r, R) \cap D \ne \emptyset$ whenever $r \in ]0, R[$ (respectively $\Gamma(a, R, r) \cap D \ne \emptyset$ whenever $r > R$) and this is obvious when $a \in \tilde{D}$ because $D$ is infraconnected and $R \le \text{diam}(D)$ (respectively $R < \text{diam}(D)$). Now, let us assume $a$ to belong to a hole $T$ of diameter $\rho < R$ (respectively $\rho \le R$). Since $|\mathbb{I}|$ is dense in $[0, +\infty[$, for every $r < R$ (respectively $r > R$), $D$ has points $\alpha$ such that $r < |a - \alpha| < R$ (respectively $R < |a - \alpha| < r$) and this ends the proof. $\square$

**Definition.** Let $\mathcal{F}$ be an increasing (respectively a decreasing) filter of center $a$ and diameter $R$ on $D$. $\mathcal{F}$ is said to be pierced if for every $r \in ]0, R[$ (respectively $r > R$), $\Gamma(a, r, R)$ (respectively $\Gamma(a, R, r)$) contains some hole $T_m$ of $D$.

A decreasing filter with no center $\mathcal{F}$ of canonical basis $(D_m)_{m \in \mathbb{N}}$ on $D$ is said to be pierced if for every $m \in \mathbb{N}$, $\tilde{D}_m \setminus \tilde{D}_{m+1}$ contains some hole $T_m$ of $D$.

**Remarks.** The definition of a pierced filter with no center also applies to a decreasing filter with a center and then is equivalent to that given just above for such a filter.
If $\mathcal{F}$ is an increasing (respectively a decreasing) filter of center $a$, of diameter $R$, $\mathcal{F}$ is pierced if and only if there exists a sequence of holes $(T_n)_{n \in \mathbb{N}}$ of $D$ such that $\delta(a, T_n) < \delta(a, T_{n+1})$ (respectively $\delta(a, T_n) > \delta(a, T_{n+1})$), $\lim_{n \to \infty} \delta(a, T_n) = R$.

Given a Cauchy filter $\mathcal{F}$ on $D$, of limit $a$ in $|L$, we call a canonical basis of $\mathcal{F}$ a sequence $D_m$ in the form $d(a, r_m) \cap D$ with $0 < r_m < r_{m+1}$ and $\lim_{m \to \infty} r_m = 0$. The filter $\mathcal{F}$ is said to be pierced if for every $m \in \mathbb{N}$, $\tilde{D}_m$ contains some hole of $D$.

Let $a \in \tilde{D}$. Let $(T_{m,i})_{1 \le i \le s(m)}$ be a sequence of holes of $D$ which satisfies $\delta(a, T_{m,i}) = d_m$ ($1 \le i \le h_m$), $d_m < d_{m+1}$ (respectively $d_m > d_{m+1}$), $\lim_{m \to \infty} d_m = S > 0$.

The sequence $(T_{m,i})_{1 \le i \le s(m)}$ is called an increasing (respectively a decreasing) distances holes sequence that runs the increasing (respectively decreasing) filter of center $a$, of diameter $R$.

Now, let $(T_{m,i})_{1 \le i \le s(m)}$ be a sequence of holes of $D$ that satisfies $\delta(a_m, T_{m,i}) = d_m$ ($1 \le i \le s(m)$), $d_m > d_{m+1}$, $\lim_{m \to \infty} d_m = R > 0$, where the filter $\mathcal{F}$ of basis $D_m = d(a_m, d_m) \cap D$ is a decreasing filter with no center. The sequence $(T_{m,i})_{1 \le i \le s(m)}$ is called a decreasing distances holes sequence that runs $\mathcal{F}$.

Summarizing these definitions, an increasing (respectively decreasing) distances holes sequence that runs an increasing (respectively decreasing) filter $\mathcal{F}$ is just named an increasing (respectively decreasing) distances holes sequence and the filter $\mathcal{F}$ is named the increasing (respectively decreasing) filter associated with the sequence $(T_{m,i})_{1 \le i \le s(m)}$. The diameter of $\mathcal{F}$ is called the diameter of the sequence $(T_{m,i})_{1 \le i \le s(m)}$. If $\mathcal{F}$ has a center $a$, $a$ is named the center of the sequence $(T_{m,i})_{1 \le i \le s(m)}$. If $\mathcal{F}$ has no center, the sequence $(T_{m,i})$ is called a decreasing distances holes sequence with no center.

Finally, an increasing (respectively decreasing) distances holes sequence is called a monotonous distances holes sequence and the sequence $(d_m)_{m \in \mathbb{N}}$ is called the monotony of the monotonous distances holes sequence.
Let $(T_{m,i})_{\substack{1\le i\le s(m) \\ m\in\mathbb{N}}}$ be a monotonous distances holes sequence and for every $(m,i)_{\substack{1\le i\le s(m) \\ m\in\mathbb{N}}}$ let $\rho_{m,i} = \text{diam}(T_{m,i})$. The number $\inf_{\substack{1\le i\le s(m) \\ m\in\mathbb{N}}} \rho_{m,i}$ is called *piercing of the sequence* $(T_{m,i})_{\substack{1\le i\le s(m) \\ m\in\mathbb{N}}}$.

If a monotonous holes sequence has a piercing $\rho > 0$, it is said to be *well pierced*. If a monotonous filter $\mathcal{F}$ is run by a well-pierced monotonous holes sequence, $\mathcal{F}$ is said to be well pierced.

In each case, the sequence of circles $C(a, d_m)$ when $\mathcal{F}$ has center $a$ (respectively $C(a_{m+1}, d_m)$ when $\mathcal{F}$ has no center) is said to run the filter $\mathcal{F}$ and to carry the monotonous distances holes sequence $(T_{m,i})_{\substack{1\le i\le s(m) \\ m\in\mathbb{N}}}$.

A monotonous distances holes sequence $(T_{m,i})_{\substack{1\le i\le s(m) \\ m\in\mathbb{N}}}$ is said to be *simple* if $s(m) = 1$ for all $m \in \mathbb{N}$.

Next, a sequence of holes $(T_m)_{m\in\mathbb{N}}$ of $D$ is called a *Cauchy sequence of holes of limit* $a \in \mathbb{I}$ if $\lim_{m\to\infty} \delta(a, T_m) = 0$. Such a sequence is said to run the Cauchy filter of basis $\{d(a, r) \cap D | r > 0\}$.

**Notation.** In all the propositions, theorems, corollaries, and Lemmas 3.4–3.10, $\gamma$ is the Moebius function $b + \frac{1}{x-a}$ with $a, b \in \mathbb{I}$.

**Proposition 3.4.** Let $\alpha \in D, r > 0$ be such that $|a - \alpha| < t$. Then, $\gamma(C(\alpha, r)) = C(b, \frac{1}{r})$.

**Proof.** We may assume $b = 0$ and then the proof is immediate. $\square$

**Corollary 3.5.** Let $\alpha \in \mathbb{I}, r_1, r_2 \in ]0, +\infty[$ with $|a - \alpha| < r_1 < r_2$. Then, $\gamma(\Gamma(\alpha, r_1, r_2)) = \Gamma(b, \frac{1}{r_2}, \frac{1}{r_1})$.

**Corollary 3.6.** Let $\mathcal{F}$ be the increasing (respectively decreasing) filter of center $\alpha$ and diameter $R > |a - \alpha|$, on $\mathbb{I} \setminus \{a\}$. Then, $\gamma(\mathcal{F})$ is the decreasing (respectively increasing) filter of center $b$ and diameter $\frac{1}{R}$.

**Lemma 3.7.** Let $\alpha \in \mathbb{I}$ be such that $|\alpha - a| \neq r$. Then,

$$ \gamma(C(\alpha, r)) = C\left(\gamma(\alpha), \frac{r}{|a - \alpha|^2}\right). $$
**Proof.** When $x$ belongs to $C(\alpha, r)$, we have

$$\gamma(x) - \gamma(\alpha) = \left| \frac{\alpha - x}{(x - \alpha)(a - \alpha)} \right| = \frac{r}{|a - \alpha|^2},$$

hence $\gamma(C(\alpha, r)) \subset C(\gamma(\alpha), \frac{r}{|a-\alpha|^2})$. Now, let $\xi(u) = \gamma^{-1}(u) = a + \frac{1}{u-b}$. We see that $C(\gamma(\alpha), \frac{r}{|a-\alpha|^2}) \subset C(\alpha, r)$. Since $\gamma$ and $\xi$ are injective, we see that $\gamma$ must be a surjection onto $C(\gamma(\alpha), \frac{r}{|a-\alpha|^2})$. $\square$

**Corollary 3.8.** Let $\alpha \in \mathbb{I}$ and $r, r' \in ]0, +\infty[$ be such that $0 < r < r' < |a - \alpha|$. Then, we have $\gamma(\Gamma(\alpha, r, r')) = \Gamma(\gamma(\alpha), \frac{r}{|a-\alpha|^2}, \frac{r'}{|a-\alpha|^2})$, $\gamma(d(\alpha, r)) = d(\gamma(\alpha), \frac{r}{|a-\alpha|^2})$, $\gamma(d(\alpha, r^-)) = d(\gamma(\alpha), \frac{r}{|a-\alpha|^2}^-)$.

**Corollary 3.9.** Let $\mathcal{F}$ be the increasing (respectively decreasing) filter of center $\alpha$ and diameter $R$ on $\mathbb{I} \setminus \{a\}$ with $|a - \alpha| > R$. Then, $\gamma(\mathcal{F})$ is an increasing (respectively a decreasing) filter of center $\gamma(\alpha)$, of diameter $\frac{R}{|a-\alpha|^2}$ on $\mathbb{I} \setminus \{b\}$.

**Corollary 3.10.** Let $\mathcal{F}$ be a decreasing filter with no center, of basis $(D_n)_{n \in \mathbb{N}}$ on $\mathbb{I} \setminus \{a\}$ such that $a \notin D_0$. Then, $\gamma(\mathcal{F})$ is a decreasing filter with no center, of canonical basis $(\gamma(D_n))_{n \in \mathbb{N}}$ on $\mathbb{I} \setminus \{b\}$.

**Theorem 3.11.** We suppose $a \in D$. Let $D' = \gamma(D)$. Let $\mathcal{F}$ be a filter on $D$ which is either a monotonous filter or a Cauchy filter. Then, $\mathcal{F}$ is pierced if and only if $\gamma(\mathcal{F})$ is a pierced filter on $D'$.

**Proof.** For example, we suppose first $\mathcal{F}$ to be a monotonous filter. By definition, $\mathcal{F}$ is the intersection with $D$ of a monotonous filter $\mathcal{G}$ of $\mathbb{I} \setminus \{a\}$. Hence, $\mathcal{F}$ is pierced if and only if $\mathcal{G}$ is secant with $(\mathbb{I} \setminus \{a\}) \setminus \overline{D}$. Since $\gamma$ is bicontinuous in $\mathbb{I} \setminus \{a\}$, we see that $\gamma(\mathcal{G})$ is secant with $(\mathbb{I} \setminus \{b\}) \setminus \overline{D'}$ if and only if $\mathcal{G}$ is secant with $(\mathbb{I} \setminus \{a\}) \setminus \overline{D}$ because $\gamma(\overline{D}) \setminus \{a\} = \overline{D'} \setminus \{b\}$. Hence, the conclusion is clear. In the same way if $\mathcal{F}$ is a Cauchy filter of limit $\alpha \in \mathbb{I}$, we consider the filter $\mathcal{G}'$ of the neighborhoods of $\gamma(\alpha)$ in $\mathbb{I} \setminus \{b\}$ and we
see that $\mathcal{G}$ is secant with $(L\setminus\{a\})\setminus\overline{D}$ if and only if $\mathcal{G}'$ is secant with $(\mathbb{L}\setminus\{b\})\setminus\overline{D'}$. $\square$

We are now going to define circular filters, which roughly characterize the absolute values on $\mathbb{L}(x)$ when $\mathbb{L}$ is algebraically closed.

**Definition.** Let $a \in \mathbb{L}$ and let $R \in ]0, +\infty[$. We call circular filter of center $a$ and diameter $R$ on $L$ the filter $\mathcal{F}$ which admits as a generating system the family of sets $\Gamma(\alpha, r', r'') \cap D$ with $\alpha \in d(a, R)$, $r' < R < r''$, i.e. $\mathcal{F}$ is the filter which admits for basis the family of sets of the form $(\bigcap_{i=1}^q \Gamma(\alpha_i, r'_i, r''_i))$ with $\alpha_i \in d(a, R)$, $r'_i < R < r''_i$ ($1 \le i \le q$, $q \in \mathbb{N}$).

For reasons that appear when characterizing the absolute values of $\mathbb{L}(x)$ when $\mathbb{L}$ is algebraically closed, a decreasing filter with no center on $\mathbb{L}$, of canonical basis $(D_n)_{n \in \mathbb{N}}$ is also called circular filter on $\mathbb{L}$ with no center, of canonical basis $(D_n)_{n \in \mathbb{N}}$.

Finally, the filter of the neighborhoods of a point $a \in \mathbb{L}$ is called circular filter of the neighborhoods of a on $\mathbb{L}$. It is also named circular filter of center $a$ and diameter $0$. A circular filter on $\mathbb{L}$ is said to be large if it has a diameter different from $0$.

A circular filter on $\mathbb{L}$ secant with $D$ is called circular filter on $D$. Given a circular filter $\mathcal{F}$ on $\mathbb{L}$, its diameter is denoted by $\text{diam}(\mathcal{F})$ and we call $\mathcal{F}$-affinoid any infraconnected affinoid subset of $\mathbb{L}$ lying in $\mathcal{F}$.

**Lemma 3.12** lets us describe circular filters on an infraconnected subset of $\mathbb{L}$.

**Lemma 3.12.** Let $a \in \tilde{D}$, let $\rho$ be the distance from $a$ to $D$, and let $R$ be such that $\rho \le R \le \text{diam}(D)$. For $j = 1, \dots, q$, let $\alpha_j \in d(a, R)$ and let $r'_j, r''_j \in \mathbb{R}_+$ be such that $r'_j < R < r''_j$. Then, $\bigcap_{j=0}^q (\Gamma(\alpha_j, r'_j, r''_j) \cap D) \ne \emptyset$.

**Proof.** If $\rho < R$, we put $r' = \max_{1 \le j \le q} r'_j$ and we see that $\Gamma(\alpha, r', R) \cap D$ is not empty (because $D$ is infraconnected) and is included in every set $\Gamma(\alpha_j, r'_j, r''_j) \cap D$. If $R < \text{diam}(D)$, we put $r'' = \min_{1 \le j \le q} r''_j$, and in the same way, $\Gamma(\alpha, R, r'') \cap D$ is not empty and is included in every set $\Gamma(\alpha_j, r'_j, r''_j) \cap D$.

Now, if $\rho = R = \text{diam}(D)$, let $b \in D$ and let $r' = \max_{1 \le j \le q} r'_j$. Then, $\Gamma(b, r', R) \cap D$ is not empty and is included in every set $\Gamma(\alpha_j, r'_j, r''_j) \cap D$. $\square$
**Corollary 3.13.** Let $a \in \tilde{D}$, let $\rho$ be the distance from $a$ to $D$, and let $R$ be such that $\rho \le R \le \text{diam}(D)$. The circular filter on $\mathbb{L}$ of center $a$ and diameter $R$ is secant with $D$.

Proposition 3.14 is immediate according to definitions:

**Proposition 3.14.** Let $\mathcal{F}$ be an increasing filter (respectively a decreasing filter) of center $a$ and diameter $R$, on $D$. Then, the circular filter of center $a$ and diameter $R$ on $\mathbb{L}$ is secant with $D$ and is the only circular filter on $D$ less thin than $\mathcal{F}$.

Conversely, let $\mathcal{F}$ be a circular filter of center $a$ and diameter $R$, on $D$ secant with $d(a, R^{-})$ (respectively $\mathbb{L}\setminus d(a, R)$). Then, the increasing filter (respectively decreasing filter) of center $a$ and diameter $R$ on $\mathbb{L}$ is secant with $D$ and thinner than $\mathcal{F}$.

**Lemma 3.15.** Let $\mathcal{F}$ be a circular filter. Then, $\mathcal{F}$ admits a basis consisting of the family of all $\mathcal{F}$-affinoids. If $\mathcal{F}$ does not admit a countable basis, it has a center and its diameter belongs to $|\mathbb{L}|$. If $\mathcal{F}$ has no center and is secant with an infraconnected affinoid subset $B$ of $\mathbb{L}$, then $B$ lies in $\mathcal{F}$. If $\mathcal{F}$ has center $a$ and diameter $r$, then an infraconnected affinoid set $B$ lies in $\mathcal{F}$ if and only if satisfy $E \cap (\mathbb{L}\setminus d(a,r)) \neq \emptyset$, $B \cap d(b,r^{-}) \neq \emptyset \forall b \in d(a,r)$.

**Proof.** By definition, a circular filter with no center has a countable basis and of course so does a Cauchy circular filter. In both cases, it admits a basis consisting of a family of disks which are $\mathcal{F}$-affinoid sets.

Now, consider a circular filter of center $a$ and diameter $r > 0$. Then, $\mathcal{F}$ admits for basis the family of sets of the form $d(a, r + \frac{1}{n}) \setminus (\cup_{i=1}^q d(a_i, (r - \frac{1}{n})^{-}))$, where the $a_i$ are centers of $\mathcal{F}$ satisfying $|a_i - a_j| = r$. In particular, if $r \notin |L|$, we have $q = 1$ and we obtain a basis of the form $\Gamma(a, r - \frac{1}{n}, r + \frac{1}{n})$ which is countable.

Now, suppose that $\mathcal{F}$ is secant with an infraconnected affinoid subset $B$ of $\mathbb{L}$. Suppose first that $\mathcal{F}$ has no center. Let $(A_n)_{n \in \mathbb{N}}$ be a canonical basis of $\mathcal{F}$. Since each $A_n$ admits common points with $B$, each is included in $\tilde{B}$, and therefore, it is included in $B$ if and only if it contains no hole of $B$. But since $\mathcal{F}$ has no center, $\bigcap_{n=0}^{\infty} A_n = \emptyset$, hence there exists $q \in \mathbb{N}$ such that $A_n \subset B \ \forall n \ge q$, and therefore, $B \in \mathcal{F}$.
Now, suppose that $\mathcal{F}$ has center $a$ and diameter $r$. If $B \in \mathcal{F}$, it obviously satisfies $B \cap (\mathbb{I} \setminus d(a,r)) \neq \emptyset$, $B \cap d(b,r^{-}) \neq \emptyset$ $\forall b \in d(a,r)$. Since $B$ has finitely many holes, on one hand, there exists $s > r$ such that $\Gamma(a,r,s) \subset E$, and on the other hand, all classes of $d(a,r)$ are included in $B$, except finitely many: $d(b_j, r^{-})$, $1 \le j \le n$. And for each $j = 1, \dots, n$, there exists $r_j < r$ such that $\Gamma(b_j, r_j, r) \subset B$. Finally, $B$ contains the set $d(a,s) \setminus \bigcup_{j=1}^n d(b_j, r_j)$ which obviously lies in $\mathcal{F}$. $\square$

**Corollary 3.16.** Let $\mathcal{F}$ be a circular filter of diameter $r$. For every $s \in ]0, r[$, the family of $\mathcal{F}$-affinoid of codiameter $\rho > s$ is a basis of $\mathcal{F}$. If two disks $d(a,r)$ and $d(b,s)$ have no common points and if $\mathcal{F}$ is secant with $d(a,r)$, it is not secant with $d(b,s)$.

**Proposition 3.17.** Let $a \in \tilde{D}$ and let $S$ be the closure of $\{|x-a| \mid x \in D\}$ in $\mathbb{R}$. For every $r \in S$, the circular filter $\mathcal{F}$ of center $a$ and diameter $r$ is secant with $D$.

**Proof.** Let $a \in \tilde{D}$. We first suppose that $C(a,r) \cap D = \emptyset$. Then, either there exists a sequence $(x_n)_{n \in \mathbb{N}}$ in $D$ such that $r < |x_{n+1}-a| < |x_n - a|$, $\lim_{n \to \infty} |x_n - a| = r$, or there exists a sequence $(x_n)_{n \in \mathbb{N}}$ in $D$ such that $|x_n - a| < |x_{n+1} - a| < r$, $\lim_{n \to \infty} |x_n - a| = r$. In both cases, the circular filter $\mathcal{F}$ of center $a$ and diameter $r$ is clearly secant with $D$.

Now, we may suppose that $C(a,r) \cap D \neq \emptyset$. Let $b \in C(a,r) \cap D$. We see that $b$ is also a center of $\mathcal{F}$. Since $D$ is infraconnected and since $|a-b| = r \le \text{diam}(D)$, there does exist a sequence $(x_n)_{n \in \mathbb{N}}$ in $D$ such that $\lim_{n \to \infty} |x_n - b| = r$. Hence, $\mathcal{F}$ (that has center $b$ and diameter $r$) is secant with $D$. $\square$

**Proposition 3.18.** Let $(a_n)_{n \in \mathbb{N}}$ be a sequence in $D$ that is either a monotonous distances sequence or a constant distances sequence. Then, there exists a unique circular filter on $D$ less thin than the sequence $(a_n)$.

**Proof.** First, we suppose that the sequence $(a_n)_{n \in \mathbb{N}}$ is an increasing (respectively a decreasing) distances sequence. By Lemma 3.2, there exists a unique increasing (respectively decreasing) filter $\mathcal{F}$ on $D$ less thin than the sequence $(a_n)_{n \in \mathbb{N}}$. If $\mathcal{F}$ has center $a$ and diameter $R$, by Proposition 3.14, $\mathcal{F}$ is less thin than the circular filter of center $a$,
of diameter $R$ on $D$. If $\mathcal{F}$ is decreasing with no center, $\mathcal{F}$ is a circular filter.

Now, we suppose that $(a_n)_{n \in \mathbb{N}}$ is a constant distances sequence. We put $R = |a_n - a_m|$ for $n \neq m$ and $a = a_0$. The circular filter $\mathcal{F}$ of center $a$ of diameter $R$ on $\mathbb{I}$ is clearly secant with $D$ because each set $\Lambda_n = \Gamma(a_n, r', r'')$ with $r' < R < r''$ belongs to a generating system of $\mathcal{F}$ and contains $a_m$ for every $m > n$, hence its intersection with $D$ is a circular filter $\mathcal{C}$ on $D$ less thin than the sequence $(a_n)_{n \in \mathbb{N}}$. That ends the proof. $\square$

**Corollary 3.19.** Let $\mathcal{F}$ and $\mathcal{G}$ be two circular filters that are secant. Then, they are equal.

**Proof.** We can find a monotonous sequence thinner than $\mathcal{F}$. Then, the sequence is thinner than $\mathcal{G}$ and hence $\mathcal{G} = \mathcal{F}$. $\square$

**Notation.** Let $\mathbb{I}'$ be an extension of $\mathbb{I}$ provided with an absolute value that extends the one of $\mathbb{I}$. Let $D$ be a set in $\mathbb{I}$, let $\mathcal{F}$ be a monotonous filter on $D$, and let $D'$ be a set in $\mathbb{I}'$ that contains $D$. Let $(a_n)_{n \in \mathbb{N}}$ be a monotonous distances sequence that runs $\mathcal{F}$. In $\hat{D}$, there is a unique monotonous filter less thin than the sequence $(a_n)_{n \in \mathbb{N}}$. This filter is denoted by $\hat{\mathcal{F}}$.

In the same way, let $\mathcal{G}$ be a circular filter of center $a$ and diameter $r$ on $D$. We denote by $\hat{\mathcal{G}}$ the filter of center $a$ and diameter $r$ on $D'$. Finally, let $\mathcal{G}$ be a circular filter with no center. Then, it is a decreasing filter, hence we have already previously defined $\hat{\mathcal{G}}$.

**Corollary 3.20.** Let $(a_n)_{n \in \mathbb{N}}$ be a bounded sequence in $\mathbb{I}$. Then, there exists a subsequence $(a_{n_l})_{l \in \mathbb{N}}$ and a unique circular filter $\mathcal{F}$ on $\mathbb{I}$ less thin than the subsequence $(a_{n_l})_{l \in \mathbb{N}}$.

**Proof.** Since the sequence $(a_n)_{n \in \mathbb{N}}$ is bounded, by Theorem 3.1, we can extract either a monotonous distances subsequence, or a constant distances subsequence, or a converging subsequence. In all cases, once such a subsequence is chosen, there exists a unique circular filter $\mathcal{F}$ on $\mathbb{I}$ less thin than the subsequence. $\square$

**Theorem 3.21.** Let $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$ be two sequences such that $|a_n - b_n| \le t < r \forall n \in \mathbb{N}$. Suppose that the sequence $(a_n)_{n \in \mathbb{N}}$ is thinner than a circular filter $\mathcal{F}$ of diameter $r$. Then, the sequence $(b_n)_{n \in \mathbb{N}}$ also is thinner than $\mathcal{F}$.
**Proof.** By Corollary 3.16, $\mathcal{F}$ admits a basis consisting of $\mathcal{F}$-affinoids $S$ of codiameter $\rho > t$. Consider such a $\mathcal{F}$-affinoid $S$. Then, if $a_n$ belongs to $S$, so does $b_n$. Now, when $n$ is big enough, all $a_n$ belong to $S$ and hence so do all $b_n$. And since $\mathcal{F}$ admits a basis of $\mathcal{F}$-affinoids with a codiameter $s > t$, we see that the sequence $(b_n)_{n \in \mathbb{N}}$ is thinner than $\mathcal{F}$. $\square$
# Chapter 4

## Ultrametric Absolute Values for Rational Functions

**Notation.** As mentioned in Chapter 1, log denotes a real logarithm function of basis $\theta > 1$ (eventually, we can take for $\theta$ an integer $p$ that is the residue characteristic of $K$). When a function $f$ from an interval $I$ to $\mathbb{R}$ admits a right side (respectively a left side) derivative at a point $a \in I$, we denote it by $f^{r}(a)$ (respectively $f^{l}(a)$). If the variable is $\mu$, we also denote it by $\frac{d_r f}{d\mu}$ (respectively $\frac{d_l f}{d\mu}$).

Moreover, though out this chapter, we denote by $\mathbb{I}$ be a field provided with an ultrametric absolute value $|\cdot|$.

The set of circular filters on $K$ secant with a subset $D$ of $K$ is denoted by $\Phi(D)$ and the subset of large circular filters on $K$ secant with $D$ is denoted by $\Phi^{\circ}(D)$. We show the absolute values on the field $K(x)$ of rational functions to be characterized by the circular filters on $K$. Actually, the most important property of such absolute values comes from the fact that the logarithm of an absolute value is a piecewise affine function of the logarithm of the absolute value of the variable. And next, a valuation function is then defined for any $h \in K(x)$ in the following way: let $r \in ]0, +\infty[$ be such that $\mu = -\log r$ and let $\mathcal{F}$ be the circular filter of center 0 and diameter $r$. Following classical notations [1], [14], one sets

$$v(h, \mu) = -\log(\lim_{\mathcal{F}} |h(x)|).$$
This function $v(h, \mu)$, called the valuation function of $h$, is convenient mainly because it is piecewise affine. However, in order to avoid many changes of sign, here we consider $\Psi(h, \log r) = \log(\lim_F |h(x)|)$ and we show that when $\frac{d_r \Psi(h, \mu)}{d\mu} \neq \frac{d_l \Psi(h, \mu)}{d\mu}$ then $\frac{d_r \Psi(h, \mu)}{d\mu} - \frac{d_l \Psi(h, \mu)}{d\mu}$ is equal to the difference between the number of zeros and the number of poles of $h$ (taking multiplicity into account) on the circle $C(0, r)$ such that $\log r = \mu$. This translates properties of $|h(x)|$ into terms of piecewise affine functions.

However, this kind of definition presents the inconvenient of changing the sense of monotony for both $|x|$ and $|h(x)|$. Moreover, its sign is opposite to this of the counting function of zeros for entire functions in the Nevanlinna theory. This is why, here we adopt another set of notation and put $\Psi(x) = \log|x| \quad \forall x \in \mathbb{K}$. First, we have to state several basic properties that work not only in an algebraically closed field such as $\mathbb{K}$ but more generally in a field $\mathbb{E}$ that is just provided with an ultrametric absolute value.

Let $\mathbb{L}[x_1, \dots, x_q]$ be an algebra of polynomials in $q$ indeterminates, with coefficients in $\mathbb{L}$. For each $P(x_1, \dots, x_q) = \sum_{j_1+\dots+j_q \le t} a_{i_1, \dots, i_q} x_1^{j_1} \dots x_q^{j_q}$ we set

$$ \overline{P(x_1, \dots, x_q)} := \sum_{j_1+\dots+j_q \le t} \overline{a_{i_1, \dots, i_q}} x_1^{j_1} \dots x_q^{j_q}. $$

$$ \text{On } \mathbb{L}[x_1, \dots, x_q], \text{ we put } \|P\| := \sup_{j_1+\dots+j_q \le t} |a_{i_1, \dots, i_q}|_0. $$

However, when there is no risk of confusion, we just write $\| \cdot \|$ instead of $\| \cdot \|_0$.

**Lemma 4.1.** $\| \cdot \|$ is a multiplicative norm of $\mathbb{L}$-algebra.

**Proof.** Let $B = \mathbb{L}[x_1, \dots, x_q]$. Clearly, $\| \cdot \|$ is an ultrametric norm of $\mathbb{L}$-vector space on $B$. We check that $\|PQ\| = \|P\|\|Q\|$ whenever $P, Q \in B$. Both $\|P\|$, $\|Q\|$ belong to $|\mathbb{L}|$. Hence, without loss of generality, we may clearly assume $\|P\| = \|Q\| = 1$. Thus, we have $\bar{P} = \bar{Q} = \bar{1}$. Let $\mathcal{L}$ be the residue field of $\mathbb{L}$. Since $\mathcal{L}[x_1, \dots, x_q]$ is a ring without divisors of zeros, we have $\overline{PQ} \ne \overline{0}$, and therefore, $\|PQ\| = 1$. This ends the proof. $\square$

**Definition and notation.** The norm $\| \cdot \|$ on $\mathbb{L}[x_1, \dots, x_q]_0$ is called the Gauss norm. Given a polynomial $P(x) = \sum_{j=1}^q a_j x^j \in \mathbb{L}[x]$, for any $r > 0$, we set $|P|(r) = \max_{0 \le j \le q} (|a_j|r^j)$.
By ultrametricity, Lemma 4.2 is then immediate:

**Lemma 4.2.** Let $P(x) \in \mathbb{I}[x]$. For all $x \in \mathbb{I}$, one has $|P(x)| \le |P|(|x|)$.

**Lemma 4.3.** Suppose $\mathbb{I}$ is algebraically closed. Let $P(x) = \sum_{j=0}^n a_j x^j \in \mathbb{I}[x]\setminus\{0\}$ and let $r \in \mathbb{R}_+$. Then, $|P(x)|$ admits a limit equal to $|P(r)|$ when $|x|$ approaches $r$ but remains different from $r$. Let $x \in d(0, r)$. Then, $|P(x)| \le |P|(r)$. If $P$ has no zero in the class of $x$ in $d(0, r)$, then $|P(x)| = |P|(r)$. If $P$ has at least one zero in that class, then $|P(x)| < |P|(r)$.

**Proof.** Let $r' < r$ and $r'' > r$ be such that $P$ has no zero in $\Gamma(0, r', r) \cup \Gamma(0, r, r'')$. We may obviously assume $P$ to be monic. Let $P(x) = \prod_{i=1}^n (x - \alpha_i)$ be the factorization of $P$ in an algebraic closure of $\mathbb{I}$, with

$$|\alpha_i| < r' \text{ for } i \le h,$$

$$|\alpha_i| > r'' \text{ for } i \ge \ell,$$

$$|\alpha_i| = r \text{ for } i = h, \dots, \ell.$$

Now, let $x \in \Gamma(0, r', r)$. Clearly, $|x - \alpha_i| = |x|$ whenever $i \le h$, while $|x - \alpha_i| = |\alpha_i|$ whenever $i > h$, hence $|P(x)| = |x|^h \prod_{i=h+1}^n |\alpha_i|$, hence $\lim_{|x| \to r^{-}} |P(x)| = r^h \prod_{i=h+1}^n |\alpha_i|$.

Symmetrically, we show that $\lim_{|x| \to r^+} |P(x)| = r^{\ell-1} \prod_{i=\ell}^n |\alpha_i| = r^h \prod_{i=h+1}^n |\alpha_i|$. But the terms $|a_j x^j|$ are all different for every $|x|$ except for finitely many values so that there exist $\rho' \in [r', r[$ and $\rho'' \in ]r, r'[$] such that the $|a_j x^j|$ are all different when $x \in \Gamma(0, \rho', r) \cup \Gamma(0, r, \rho'')$. Then, we have $|P(x)| = \max_{0 \le j \le n} |a_j| |x|^j$ and hence $\lim_{|x| \to r} |P(x)| = \max_{0 \le j \le n} |a_j| r^j$.

Now, let $x \in d(0, r)$ and let us assume $P$ to have no zero in the class $\Lambda$ of $x$ in $d(0, r)$. This means that $|x - \alpha_i| = r$ whenever $i = 0, \dots, \ell - 1$ and $|x - \alpha_i| = |\alpha_i|$ whenever $i \ge \ell$. Thus, $|P(x)| = r^{\ell-1} \prod_{i=\ell}^n |\alpha_i|$. If $P$ has at least one zero $\alpha_{h+1}$ in the class of $x$, we see that $|x - \alpha_{h+1}| < r$, while $|x - \alpha_i| \le r$ for every $i = h+2, \dots, \ell-1$, hence $|P(x)| < r^{\ell-1} \prod_{i=\ell}^n |\alpha_i|$, and finally, $|P(x)| < |P|(r)$. $\square$

**Theorem 4.4.** Let $P \in \mathbb{I}[X]$ and let $U$ be the unit disk $\{x \in |x| \le 1\}$ of $\mathbb{K}$. Then, $||P|| = \sup_{x \in U} (|P(x)|)$.
**Proof.** On one hand, $|P(x)| \le \|P\| \forall x \in U$. On the other hand, by Theorem 4.3, $\lim_{|x| \to 1, x \in U, |x| \ne 1} |P(x)| = \|P\|$. Consequently, the equality holds. □

**Corollary 4.5.** Let $P \in \mathbb{I}[X]$ and let $t \in \mathbb{I}$ be such that $|t| \le 1$. Let $Q(X) = P(X + t)$. Then, $\|P\| = \|Q\|$. If $\|P\| \le 1$, then $P$ is 1-Lipschitzian.

**Theorem 4.6.** For every $r > 0$, the mapping from $\mathbb{I}[x]$ to $\mathbb{R}_+$ defined by $P \to |P|(r)$ is an absolute value on $\mathbb{I}[x]$ such that $|P(a)| \le |P|(|a|) \forall a \in \mathbb{I}$.

**Proof.** Due to the definition of $|P|(r)$, it is easily checked that $|P|(r) = 0$ if and only if $P = 0$ and that

$$|P + Q|(r) \le \max(|P|(r), |Q|(r)).$$

We can also check that $|P(a)| \le |P|(|a|) \forall a \in \mathbb{I}$. Now, set $P(x) = \sum_{j=1}^m a_j x^j$, $Q(x) = \sum_{j=1}^n b_j x^j$, and let $P(x)Q(x) = \sum_{j=1}^{m+n} c_j x^j$. Let $s$ (respectively $t$) be the biggest of the integers such that $|P|(r) = |a_s|r^s$ (respectively $|Q|(r) = |b_t|r^t$). Then, $|P|(r)|Q|(r) = |a_s b_t|r^{s+t}$. On the one hand, we can check that, obviously, $|c_j|r^j \le |a_s b_t|r^{s+t} \forall j = 0, \dots, m+n$, hence $|PQ|(r) \le |P|(r)|Q|(r)$. On the other hand, since $|a_j|r^j < |a_s|r^s \forall j > s$ and $|b_j|r^j < |b_t|r^t \forall j > t$, we have $|c_{s+t}|r^{s+t} = |a_s b_t|r^{s+t}$, which proves that $|PQ|(r) \ge |P|(r)|Q|(r)$ and hence ends the proof. □

Now, Lemma 4.7 shows that we can change the origin, inside the disk $d(0, r)$.

**Lemma 4.7.** Suppose $\mathbb{I}$ is algebraically closed. Let $r \in \mathbb{R}_+$ and let $a \in \mathbb{I}$ be such that $|a| \le r$. Then, $|P(x)|$ has a limit when $|x - a|$ approaches $r$ but remains different from $r$. Further, that limit does not depend on $a \in d(0, r)$ and it belongs to $|\mathbb{I}|$ if and only if so does $r$.

**Proof.** We set $x = a + u$ and $P_a(u) = P(a + u)$. For every $P \in \mathbb{I}[x]$, we have $\lim_{|u| \to r} |P_a(u)| = |P_a|(r)$. In particular, $|P_a|(r) = \lim_{|u| \ne r} |P_a(u)|$. But for every $\rho > r$, $C(0, \rho) = C(a, \rho)$, hence $\lim_{|u| \to r^+} |P_a(u)|$.
$\lim_{|x| \to r^+} |P(x)| = \lim_{|u| \to r^+} |P(a+u)| = \lim_{|u| \to r^+} |P_a(u)|$. That ends the proof of Lemma 4.7. □

**Theorem 4.8.** Suppose $\mathbb{I}$ is algebraically closed. Let $P(x) = \sum_{j=0}^{q} a_j x^j \in \mathbb{I}[x]$ be a monic polynomial such that $a_j \in d(0, 1)$ whenever $j = 0, \dots, q$. Then, the $q$ zeros of $P$ belong to $d(0, 1)$.

**Proof.** Let $\psi$ be the absolute value defined on $\mathbb{I}[x]$ by $\psi(P) = \lim_{|u| \to 1, |u| \neq 1} |P(u)|$ and let $P(x) = \prod_{j=1}^{q} (x - c_j)$. By Lemma 4.3, for each $j = 1, \dots, q$, it is seen that $\psi(x - c_j) \ge 1$, while $\psi(P) = 1$. Hence, $\psi(x - c_j) = 1$ for every $j = 1, \dots, q$, and therefore, by Lemma 4.3 again, we have $c_j \le 1$ whenever $j = 1, \dots, q$. □

**Notation.** These ultrametric absolute values defined on $\mathbb{I}[x]$ are immediately extended to rational functions by $\left|\frac{P}{Q}\right|(r) := \frac{|P|(r)}{|Q|(r)}$. Then, Lemma 4.9 is immediate:

**Lemma 4.9.** Suppose $\mathbb{I}$ is algebraically closed. Let $h \in \mathbb{I}(x)$ and let $r \in \mathbb{R}_+$. For every $a \in d(0, r)$, we have $\lim_{\substack{|x-a| \to r \\ |x-a| \neq 0}} |h(x)| = |h|(r)$. Let $x \in C(0, r)$. If $h$ has no zero and no pole in the class of $x$ in $C(0, r)$, then $|h(x)| = |h|(r)$. Further, $|h|(r)$ belongs to $|\mathbb{K}|$ if and only if so does $r$.

Circular filters characterize the multiplicative norms defined on $\mathbb{K}(x)$ [61], [36], [49].

**Theorem 4.10 (B. Guennebaud).** For every circular filter $\mathcal{F}$ on $\mathbb{K}$, for every polynomial $P(x) \in \mathbb{K}[x]$, $|P(x)|$ has a limit $\varphi_{\mathcal{F}}(P)$ along the filter $\mathcal{F}$. The mapping $\mathcal{F} \to \varphi_{\mathcal{F}}$ from $\Phi(\mathbb{K})$ into the set of the multiplicative semi-norms on $\mathbb{K}[x]$ is a bijection. Moreover, for every large circular filter on $\mathbb{K}$, $\varphi_{\mathcal{F}}$ has continuation to $\mathbb{K}(x)$ and the mapping $\mathcal{F} \to \varphi_{\mathcal{F}}$ from $\Phi^{\circ}(\mathbb{K})$ into the set of the multiplicative norms on $\mathbb{K}(x)$ is a bijection.

If $\mathcal{F}$ has center 0 and diameter $r$, then $\varphi_{\mathcal{F}}(h) = |h|(r)$.

**Proof.** We first suppose that $\mathcal{F}$ has center $a \in \mathbb{K}$ and diameter $r > 0$. With no loss of generality, we may obviously assume $a = 0$ by means of the change of variable $x = a + u$. Then, by Lemma 4.3,
$|h(x)| = |h|(r)$ holds in every class of $C(0, r)$ but finitely many ones $\Lambda_1, \dots, \Lambda_q$. For every $j = 1, \dots, q$, we take $\alpha_j \in \Lambda_j$ and set $\alpha_0 = 0$. Let $\epsilon$ be $> 0$. By Lemma 4.3, there exist $\rho' \in ]0, r[$ and $\rho'' > r$ such that $|h(x)| - |h|(r)|_\infty \le \epsilon$ for every $x \in \bigcap_{j=0}^q \Gamma(\alpha_j, \rho', \rho'')$, so $\lim_{\mathcal{F}} |h(x)| = |h|(r)$.

Now, we suppose that $\mathcal{F}$ has no center in $\mathbb{K}$. It admits a canonical basis $(D_n)_{n \in \mathbb{N}}$, and then given $h \in \mathbb{K}(x)$, there exists $q \in \mathbb{N}$ such that $h$ has neither any zero nor any pole inside $D_q$. Hence, $|h(x)|$ is equal to a constant $l$ in $D_q$, and therefore, we have $\lim_{\mathcal{F}} |h(x)| = l$. By the same kind of reasoning as in Theorem 4.6, it is easily seen that $\varphi_{\mathcal{F}}$ is an absolute value on $\mathbb{K}(x)$.

Now, we check that the mapping $\mathcal{F} \to \varphi_{\mathcal{F}}$ is injective. Indeed, let $\mathcal{F}_1, \mathcal{F}_2$ be two different circular filters and let $r_1$ (respectively $r_2 \ge r_1$) be the diameter of $\mathcal{F}_1$ (respectively $\mathcal{F}_2$). We first suppose that we may find disks $\Lambda_1 = d(a_1, \rho_1)$ and $\Lambda_2 = d(a_2, \rho_2)$ such that $\Lambda_1 \cap \Lambda_2 = \emptyset$ and such that $\mathcal{F}_1$ (respectively $\mathcal{F}_2$) is secant with $\Lambda_1$ (respectively $\Lambda_2$). Then, it is seen that $|a_1 - a_2| > \rho_1 \ge r_1$, hence $\varphi_{\mathcal{F}_1}(x - a_1) \le r_1$, while $\varphi_{\mathcal{F}_2}(x - a_1) = |a_1 - a_2| > r_1$, and therefore, $\varphi_{\mathcal{F}_1} \ne \varphi_{\mathcal{F}_2}$.

We now suppose that we cannot find disks $\Lambda_1, \Lambda_2$ defined as above. Since $r_1 \le r_2$, any disk $\Lambda$ which belongs to $\mathcal{F}_1$ is included in any disk that belongs to $\mathcal{F}_2$, and therefore, any point of $\Lambda_1$ is a center of $\mathcal{F}_2$. Thus, $\mathcal{F}_2$ admits a center $a \in \Lambda$, and then, $\mathcal{F}_1$ is secant with $d(a, r_2)$. Hence, we have $r_1 < r_2$ because otherwise $\mathcal{F}_1$ would be equal to $\mathcal{F}_2$. In particular, $\mathcal{F}_2$ is secant with one class $d(\alpha, r_2^{-})$ of $d(a, r_2)$. Then, we have $\varphi_{\mathcal{F}_1}(x-\alpha) \le r_1$, while $\varphi_{\mathcal{F}_2}(x-\alpha) = r_2$. This finishes showing that the mapping $\mathcal{F} \to \varphi_{\mathcal{F}}$ is injective.

Now, we show that this mapping defined on $\Phi^\circ(\mathbb{K})$ is also surjective onto the set of multiplicative norms i.e. the absolute values on $\mathbb{K}[x]$ continuing these of $\mathbb{K}$. Indeed, let $\psi$ be such an absolute value on $\mathbb{K}[x]$ and let $r = \inf_{\lambda \in \mathbb{K}} \psi(x - \lambda)$.

We first suppose that there exists $a \in \mathbb{K}$ such that $\psi(x-a) = r$. Since $\psi$ is an absolute value, we check that $r > 0$ because if $r = 0$, we have $\psi(h) = h(a)$ for every $h \in \mathbb{K}[x]$ and then $\psi$ is not an absolute value. Hence, we can assume $r > 0$. Let $\mathcal{F}$ be the circular filter of center $a$, of diameter $r$. By Lemma 2.7, we know that $\psi$ is ultrametric, and then, for every $b \in \mathbb{K}$, we have $\psi(x-b) \le \max(\psi(x-a), |a-b|) = \max(|a-b|, r)$. But, by definition, we have $\psi(x-b) \ge r$, hence $(1) r \le \psi(x-b) \le \max(r, |a-b|)$.
If $|a-b| > r$, then both $\psi(x-b)$, $\varphi_F(x-b)$ are equal to $|a-b|$. If $|a-b| \le r$, then $b$ is another center of $\mathcal{F}$ and we have $\varphi_F(x-a) = \varphi_F(x-b) = r$. But, by (1), we see that $\psi(x-b) = r$. So, we have shown that $\varphi_F(x-b) = \psi(x-b)$ for all $b \in \mathbb{K}$, and since $\mathbb{K}$ is algebraically closed, this finishes proving that $\psi = \varphi_F$.

We now suppose that there does not exist $a \in \mathbb{K}$ such that $r = \psi(x-a)$. There exists $\alpha_n \in \mathbb{K}$ such that $r < \psi(x - \alpha_n) < r + \frac{1}{n}$. Let $\rho_n = \psi(x - \alpha_n)$. For $b \in \mathbb{K} \setminus d(\alpha_n, \rho_n)$, clearly we have

$$ (2) \quad \psi(x-b) = |b - \alpha_n| > \rho_n $$

because $\psi(x - \alpha_n) < |\alpha_n - b|$. Further, if $\mathcal{F}_n$ is the circular filter of center $\alpha_n$ and diameter $\rho_n$, we have

$$ (3) \quad \psi(x-b) = |b - \alpha_n| = \varphi_{\mathcal{F}_n}(x-b). $$

However, there exists $\alpha_{n+1} \in \mathbb{K}$ such that $r < \psi(x - \alpha_{n+1}) < \min(\rho_n, r + \frac{1}{n+1})$. Hence, by (2), we see that $\alpha_{n+1} \in d(\alpha_n, \rho_n)$. That way, we may define a decreasing sequence of disks $D_n = d(\alpha_n, \rho_n)$ such that $r < \rho_n < r + \frac{1}{n}$ and $\psi(x - \alpha_n) = \rho_n$. Let $D'_n = D_n \cap D$. Then the decreasing filter $\mathcal{F}$ of basis $(D'_n)_{n \in \mathbb{N}}$ satisfies $\lim_{\mathcal{F}} \psi(x - \alpha_n) = r$. It is easily seen that $\mathcal{F}$ has no center because if $\alpha$ is a center of $\mathcal{F}$ then $\psi(x - \alpha) \le \max(|\psi - \alpha_n|, |\alpha_n - \alpha|)$ hence $\psi(x - \alpha) = r$. We show that $\psi = \varphi_{\mathcal{F}}$. Let $b \in \mathbb{K}$ and let $q \in \mathbb{N}$ be such that $b \notin D_q$. Then, by (3), for $n \ge q$, we have $\psi(x-b) = \varphi_{\mathcal{F}_n}(x-b)$. On the other hand, it is easily seen that $\varphi_{\mathcal{F}_n}(x-b) = \varphi_{\mathcal{F}}(x-b)$. Thus, $\psi(x-b) = \varphi_{\mathcal{F}}(x-b)$ whenever $b \in \mathbb{K}$ and then $\psi = \varphi_{\mathcal{F}}$.

Finally, let $\psi$ be a multiplicative semi-norm that is not a norm: there exists a polynomial $P$ such that $\psi(P) = 0$, and hence, there exists $a \in \mathbb{K}$ such that $\psi(x-a) = 0$. Let $b \in \mathbb{K}$. Then, $\psi(x-b) = \psi((x-a) + (a-b)) \le \max \psi(x-a), \psi(a-b))$. But since $\psi(t) = |t| \quad \forall t \in \mathbb{K}$, we have $\psi(x-b) = |a-b|$, hence, putting $h(x) = x-b$, we have $\psi(h) = |h(a)|$, which shows that the equality $\psi(h) = |h(a)|$ holds for every polynomial of degree 1, and therefore, it holds in all $\mathbb{K}[x]$. This ends the proof of Theorem 4.10. $\square$

**Theorem 4.11.** Let $\mathcal{H}$ be a filter on $\mathbb{K}$ and let $\mathcal{F}, \mathcal{G} \in \Phi(\mathbb{K})$ be less thin than $\mathcal{H}$. Then, $\mathcal{F} = \mathcal{G}$.

**Proof.** Since $\mathcal{H}$ is thinner than $\mathcal{F}$ and $\mathcal{G}$, we have $\lim_{\mathcal{H}} |P(x)| = \varphi_{\mathcal{F}}(P) = \varphi_{\mathcal{G}}(P) \quad \forall P \in \mathbb{K}[x]$, hence $\varphi_{\mathcal{F}} = \varphi_{\mathcal{G}}$. But, by Theorem 4.10, the mapping that associates with each circular filter $\mathcal{F}$ the multiplicative semi-norm $\varphi_{\mathcal{F}}$ is injective and hence $\mathcal{F} = \mathcal{G}$. $\square$
**Notation.** When $\mathcal{F}$ is the circular filter of center $a$, of diameter $r$, we also denote by $\varphi_{a,r}$ the absolute value $\varphi_{\mathcal{F}}$. Hence, by definition, we have $\varphi_{a,r}(h) = \lim_{|x-a|\to r} |h(x)|$. In particular, we note that

$$ \varphi_{0,r}(h) = |h|(r). $$

Finally, we denote by $\varphi_a$ the multiplicative semi-norm defined on rational functions with no pole at $a$ as $\varphi_a(h) = |h(a)|$.

Now, let us go back to the field $\mathbb{L}$. For $\mu \in \mathbb{R}$, we set $\Psi(h, \mu) = \log(|h|(\theta^\mu))$, for simplicity, we set $\Psi(h) = \Psi(h, 0)$. Thus, comparative to the valuation function $v(h, \mu)$ defined and used in previous works [1], [14], [49], we have $\Psi(h, \mu) = -v(h, -\mu)$. The advantage of the function $\Psi$ is to respect the sense of variation of $|h|(r)$.

The translation of statements 4.1, 4.2, 4.3, 4.6, 4.7, 4.8, and 4.9 into terms of valuation allows us to obtain Lemmas 4.12 and 4.13.

**Lemma 4.12.** Let $P(x) = \sum_{j=0}^{n} a_j x^j \in \mathbb{K}[x]\setminus\{0\}$. For every $\mu \in \mathbb{R}$, we have $\Psi(P, \mu) = \max_{0 \le j \le n} \Psi(a_j) + j\mu$. Moreover, $\Psi(P(a)) \le \Psi(P, \Psi(a)) \forall a \in L$.

Suppose $P \in \mathbb{K}[x]$. The equality $\Psi(P(x)) = \Psi(P, \Psi(x))$ holds if and only if $P$ has no zero $\alpha$ such that $\Psi(x - \alpha) < \Psi(x)$.

**Lemma 4.13.** Let $h \in \mathbb{K}(x)\setminus\{0\}$. We have $\Psi(h(x)) = \Psi(h, \Psi(x))$ for every $x \in \mathbb{K}$ such that $h$ has no zero $\alpha$ satisfying $\Psi(x - \alpha) < \Psi(x)$ and no pole $\beta$ satisfying $\Psi(x - \beta) < \Psi(x)$.

**Lemma 4.14.** Let $h_1, h_2 \in \mathbb{K}(x)\setminus\{0\}$. Then, we have

$$ \Psi(h_1 + h_2, \mu) \le \max(\Psi(h_1, \mu), \Psi(h_2, \mu)). $$

When $\Psi(h_1, \mu) > \Psi(h_2, \mu)$, then we have $\Psi(h_1 + h_2, \mu) = \Psi(h_1, \mu)$. Moreover, $\Psi(h_1.h_2, \mu) = \Psi(h_1, \mu) + \Psi(h_2, \mu)$.

**Notation.** In order to perform easily any change of origin, for every $a \in \mathbb{K}$ and $h \in \mathbb{K}(x)\setminus\{0\}$, we put $\Psi_a(h, \mu) = \Psi(h_a, \mu)$ with $h_a(u) = h(a+u)$. Thus, if $\mathcal{F}$ denotes the circular filter of center $a$ and diameter $\theta^\mu$, then $\Psi_a(h, \mu) = \log(\varphi_{\mathcal{F}}(h))$.

We now consider again a polynomial $P(x) = \sum_{j=0}^{n} a_j x^j \ne 0$. We denote by $\nu^+(P, \mu)$ (respectively $\nu^-(P, \mu)$) the biggest (respectively the smallest) index $j$ such that $\Psi(a_j) + j\mu = \Psi(P, \mu)$.
Lemma 4.15 is a consequence of Lemma 4.9:

**Lemma 4.15.** Let $h \in K(x)\setminus\{0\}$ and let $a, b \in K$. For every $\mu \ge \Psi(a-b)$, we have $\Psi_a(h, \mu) = \Psi_b(h, \mu)$.

**Theorem 4.16.** Let $P(x) = \sum_{j=0}^{n} a_j x^j \in K[x]$. For every $\mu \in \mathbb{R}$, $\nu^+(P, \mu) - \nu^-(P, \mu)$ is equal to the number of zeros admitted by $P$ in the circle $C(0, \theta^\mu)$ in $K$. The function $\nu^+(P, \cdot)$ (respectively $\nu^-(P, \cdot)$) is increasing and continuous on the right (respectively on the left). Moreover, given $Q \in K[x]$, then $\nu^+$, $\nu^-$ satisfy $\nu^+(PQ, \mu) = \nu^+(P, \mu) + \nu^+(Q, \mu)$, $\nu^-(PQ, \mu) = \nu^-(P, \mu) + \nu^-(Q, \mu)$. Further, if $\nu^+(P, \mu) = \nu^-(P, \mu)$, then both are constant in a neighborhood of $\mu$.

The function $\Psi(P, \cdot)$ is continuous, piecewise affine, increasing, and convex and has a right-side derivative (respectively a left-side derivative) equal to $\nu^+(P, \mu)$ (respectively $\nu^-(P, \mu)$).

**Proof.** It is easily seen that the equality (1) $\nu^+(P, \mu) = \nu^-(P, \mu)$ holds for every $\mu$ but finitely many values, at most $n$. It is also clear that the functions $\nu^+(P, \cdot)$ and $\nu^-(P, \cdot)$ are increasing. By continuity, we see that the function $\nu^+(P, \mu)$ is continuous on the right at each point, while $\nu^-(P, \mu)$ is continuous on the left at each point. Finally, if (1) holds in an interval $]\mu', \mu''[$, then the functions $\nu^+(P, \cdot)$ and $\nu^-(P, \cdot)$ are constant and equal. Consider an interval $]\mu', \mu''[$ such that $\nu^+(P, \mu) = \nu^-(P, \mu)$ for all $\mu \in]\mu', \mu''[$ and let $j = \nu^+(P, \mu)$ whenever $\mu \in]\mu', \mu''[$. Then, $\Psi(P, \mu) = \Psi(a_j) + j\mu$ so that the function $\Psi(P, \cdot)$ is in the form $A + i\mu$ in this interval.

Now, let $\mu$ be such that $\nu^+(P, \mu) < \nu^-(P, \mu)$. We see that $\Psi(P, \cdot)$ is still continuous at $\mu$ and has a left-side derivative equal to $\nu^-(P, \mu)$ and a right-side derivative equal to $\nu^+(P, \mu)$. Finally, the function $\Psi(P, \cdot)$ is continuous, piecewise affine, convex, and largely increasing.

If $P$ and $Q \in K(x)\setminus\{0\}$, then $\nu^+(PQ, \mu)$ is the right-side derivative of the function $\Psi(PQ, \cdot)$. But $\Psi(PQ, \cdot) = \Psi(P, \cdot) + \Psi(Q, \cdot)$, hence its right-side derivative at $\mu$ is just $\nu^+(P, \mu) + \nu^+(Q, \mu)$. In the same way, we have $\nu^-(PQ, \mu) = \nu^-(P, \mu) + \nu^-(Q, \mu)$ by considering left-side derivatives.

Then, to prove that $\nu^+(P, \mu) - \nu^-(P, \mu)$ is the number of zeros of $P$ in $C(0, \theta^\mu)$, it is sufficient to show this when $P$ is a binomial $x-a$. But then, this is obvious because $\nu^+(P, \mu) = \nu^-(P, \mu) = 0$ whenever $\mu < \Psi(\alpha)$, $\nu^+(P, \mu) = \nu^-(P, \mu) = 1$ whenever $\mu > \Psi(\alpha)$,
while $ν^+(P, \Psi(\alpha)) = 1, ν^-(P, \Psi(\alpha)) = 0$. So all the statements of Theorem 4.16 have been proven. □

Applying Lemma 4.12 and Theorem 4.16 to the numerator and the denominator of a rational function, we obtain Corollary 4.17.

**Corollary 4.17.** Let $P(x) = \sum_{j=0}^{n} a_j x^j \in \mathbb{K}[x]$. For every $\mu \in \mathbb{R}$, $ν^+(P, \log(r))$ is equal to the number of zeros admitted by $P$ in $d(0, r)$.

**Corollary 4.18.** Let $h \in \mathbb{K}(x)\setminus\{0\}$. The function in $\mu \Psi(h, \mu)$ is continuous and piecewise affine.

If $\mu$ is such that $d(0, \theta^\mu)$ contains $s$ zeros and $t$ poles of $h$ (taking multiplicity into account), but neither any zero nor any pole in $C(0, \theta^\mu)$, then $\Psi(h, \cdot)$ has a derivative equal to $s-t$ at $\mu$.

If $\mu$ is such that $C(0, \theta^\mu)$ contains $s$ zeros and $t$ poles of $h$ (taking multiplicity into account), then we have $\frac{d_j \Psi(h, \mu)}{d\mu} - \frac{d_r \Psi(h, \mu)}{d\mu} = s-t$. Further, if the function $\Psi(f, \mu)$ is not derivable at $\mu$, then $\mu$ lies in $\Psi(\mathbb{K})$.

**Corollary 4.19.** Let $h \in \mathbb{K}(x)\setminus\{0\}$ have no pole (respectively no zero) in an annulus $\Gamma(0, r', r'')$. Then, $\Psi(h, \mu)$ is convex (respectively concave) in $[\log r', \log r'']$.

**Corollary 4.20.** Let $h \in \mathbb{K}(x)\setminus\{0\}$ have $s$ zeros and $t$ poles in $d(0, r')$ and have neither any zero nor any pole in an annulus $\Gamma(0, r', r'')$. Then in $\Gamma(0, r', r'')$, $\Psi(h, \log|x|)$ is of the form $A + (s-t)\log|x|$. Moreover, $ν^+(h, \mu) - ν^-(h, \mu) = s-t \quad \forall \mu \in \mathbb{R}$ log $r'$, log $r''$ [and $ν^+(h, \mu)$ (respectively $ν^-(h, \mu)$) is continuous on the right (respectively on the left). Finally, given $g \in \mathbb{K}(x)$, we have $ν^+(gh, \mu) = ν^+(g, \mu) + ν^+(h, \mu)$, $ν^-(gh, \mu) = ν^-(g, \mu) + ν^-(h, \mu)$.

**Theorem 4.21.** Let $G \in \Phi(\mathbb{K})$. Let $f \in \mathbb{K}(x)$ and take $\epsilon > 0$. There exists a $G$-affinoid $E$ such that $|f(x)| - \varphi_G(f)|_\infty \le \epsilon, \forall x \in E$.

**Proof.** Let $r = \text{diam}(G)$ and let $l > r$. If $G$ has no center, there exists a disk $d(a, l) \in G$, with $r \in |\mathbb{K}|$, containing neither zeros nor poles of $f$, therefore, by Lemma 4.9, $|f(x)|$ is a constant equal to $\varphi_{\mathcal{F}}(f)$ in $d(a, r)$, so our claim is obvious. Now, suppose that $G = \mathcal{F}_{a,r}$. Let $\Lambda_1, \dots, \Lambda_q$ be the classes of $d(a, r)$ containing at least one zero or one pole of $f$. By Lemma 4.9, $|f(x)|$ is a constant equal to $\varphi_G(f)$ in $d(a, r)\setminus(\bigcup_{j=1}^q \Lambda_j)$. Consider a class $\Lambda_j = d(a_j, r^-)$
and let $s_j$ (respectively $t_j$) be the number of zeros (respectively poles) of $f$ inside $\Lambda_j$ and let $s_0$ (respectively $t_0$) be the number of zeros (respectively poles) of $f$ in all $d(a, r)$. Let $\rho \in ]0, r[ \cap \mathbb{K}$ be such that $|(\frac{r}{\rho})^{s_j-t_j} - 1| \varphi_G(f) \le \epsilon \ \forall j = 0, \dots, q$. Let $l = \frac{r^2}{\rho}$ and let $E = d(a, l) \setminus \bigcup_{j=1}^q d(a_j, \rho^-)$. By Corollary 4.19, we can check that the inequality $|f(x)| - \varphi_G(f)|_\infty \le \epsilon$ holds in all $E$. Since $\rho < r$, $E$ is an infraconnected affinoid set which belongs to $G$. Moreover, by definition, $l > r$. $\square$

**Theorem 4.22.** Let $\mathcal{F}$ be a filter on $\mathbb{K}$ such that for every $h \in \mathbb{K}(x)$, $|h(x)|$ admits a limit along $\mathcal{F}$. Then there exists a circular filter $\mathcal{H}$ less thin than $\mathcal{F}$.

**Proof.** For every $h \in \mathbb{K}(x)$, set $\phi(h) = \lim_{\mathcal{F}} |h(x)|$. Then, $\phi$ belongs to Mult($|K(x)|$), and hence, by Theorem 4.10, there exists a unique circular filter $\mathcal{H}$ such that $\phi = \varphi_{\mathcal{H}}$. Suppose that $\mathcal{F}$ is not thinner than $\mathcal{H}$. There exists a subset $B$ of $\mathbb{K}$ such that $\mathcal{F}$ is secant with $B$ but $\mathcal{H}$ is not. Since $\mathcal{H}$ admits a basis consisting of affinoid subsets, there exists an $\mathcal{H}$-affinoid $D$ such that $D \cap B = \emptyset$. Since $D$ is affinoid, it is of the form $d(a, R) \setminus (\bigcup_{k=1}^q d(a_i, r_i^-))$ and $\mathcal{H}$ also admits an $\mathcal{H}$-affinoid $E$ of the form $d(a, S) \setminus (\bigcup_{k=1}^q d(a_i, s_i^-))$ with $S < R$ and $r_i < s_i \ \forall i = 1, \dots, q$. Let $b \in \Gamma(a, S, R)$ and let $h(x) = \frac{\prod_{i=1}^q (x-a_i)^{m_i}}{(x-b)^n}$. Then with integers $m_i$ and $n$ big enough, we can get

$$ \inf\{|h(x)| \mid x \in E\} > \sup\{|h(x)| \mid x \in B\}, $$

a contradiction to the hypothesis: $|h(x)|$ admits a limit along $\mathcal{F}$. $\square$
# Chapter 5

## Hensel Lemma

The Hensel lemma is a classical tool for studying the factorization of analytic functions on a circle [1], [44] and is indispensable in Chapter 6. It is a strong result that roughly says. “In a complete field $\mathbb{L}$, if $\bar{P}$ splits in the form $\gamma\eta$ with $(\gamma, \eta) = \bar{1}$, then $P$ also splits in $\mathbb{L}[x]$ in the form $gh$ with $\bar{g} = \gamma$, $\bar{h} = h$, $\deg(\bar{g}) = \deg(\gamma)$”. The proof is not very easy and requires a serious preparation. Here, we roughly follow the same process as in [1], [44].

**Notation.** According to Chapter 2, we denote by $U_{\mathbb{L}}$ the unit ball of $\mathbb{L}$ and by $M_{\mathbb{L}}$ its maximal ideal. We denote by $\mathcal{L}$ the residue class field of $\mathbb{L}$: $\frac{U_{\mathbb{L}}[x]}{M_{\mathbb{L}}[x]}$.

Given $g, h \in \mathbb{L}[x]$, here $(g, h)$ denotes the monic greatest common divisor of $g$ and $h$. Given $Q(x) = \sum_{j=0}^{n} b_j x^j \in U_{\mathbb{L}}[x]$, as in Chapter 2, we denote by $\overline{Q}$ the polynomial $\sum_{j=0}^{q} \overline{a_j} x^j \in \mathcal{L}[x]$. In this chapter, $P(x) = \sum_{j=0}^{q} a_j x^j \in \mathbb{L}[x]$ denotes a polynomial of degree $q$.

**Lemma 5.1.** For all $\alpha \in \mathbb{L}$, we have $\Psi(P(\alpha)) \le \Psi(P, 0) + \max(0, q\Psi(\alpha))$.

**Proof.** By Lemma 4.12, we have $\Psi(P(\alpha)) \le \max_{0 \le j \le q}(\Psi(a_j) + j\Psi(\alpha)) \le \max_{0 \le j \le q} \Psi(a_j) + \max_{0 \le j \le q} j\Psi(\alpha)$. But $\max_{0 \le j \le q} \Psi(a_j) = \Psi(P, 0)$ and $\max_{0 \le j \le q} j\Psi(\alpha) = \max(0, q\Psi(\alpha))$. $\square$

**Definition.** A polynomial $\sum_{j=0}^{q} a_j x^j \in L[x]$ is said to be *quasi-monic* if $|a_q| = 1$.
**Lemma 5.2.** Let $F, D \in U_{\mathbb{L}}[x]$ with $D$ quasi-monic. Let $Q, R \in U_{\mathbb{L}}[x]$ satisfy $F = DQ + R$ and $\deg(R) < \deg(D)$. Then, we have $\Psi(Q, 0) \le \Psi(F, 0)$ and $\Psi(R, 0) \le \Psi(F, 0)$.

**Proof.** We can clearly assume $F \ne 0$. Then, by multiplying $F$ by a suitable constant $\lambda$, we can also assume $\Psi(F, 0) = 0$. Since $D$ is quasi-monic, the Euclidean division of $F$ by $D$ is clearly possible in $U_{\mathbb{L}}[x]$, and therefore, $Q$ is the quotient and $R$ is the rest of this division, due to the fact that $\deg(R) < \deg(D)$. So we have $\Psi(Q, 0) \le 0$, $\Psi(R, 0) \le 0$ because both $Q, R$ belong to $U_{\mathbb{L}}[x]$. $\square$

**Corollary 5.3.** Let $F, D \in \mathbb{K}[x]$ with $D$ having all its zeros in $d(0, 1)$. Let $Q, R \in U_{\mathbb{K}}[x]$ satisfy $F = DQ+R$ and $\deg(R) < \deg(D)$. Then we have $\Psi(Q, 0) \le \Psi(F, 0) - \Psi(D, 0)$ and $\Psi(R, 0) \le \Psi(F, 0)$. Moreover, if $F$ has all its zeros in $d(0, 1)$, then $\Psi(Q, 0) = \Psi(F, 0) - \Psi(D, 0)$.

**Proof.** The first statement is just an application of Lemma 5.2. Next, if $F$ has all its zeros in $d(0, 1)$, we can assume that both $F, D$ are monic and satisfy $\|F\| = \|D\| = 1$. Consequently, $Q$ also must be monic and hence $\|Q\| = 1$, which ends the proof. $\square$

**Lemma 5.4.** Let $g, h \in U_{\mathbb{L}}[x]$ be quasi-monic such that $(\bar{g}, \bar{h}) = \overline{1}$ and $\deg(P) < \deg(g) + \deg(h)$. There exist $V, W \in \mathbb{L}[x]$ satisfying $\Psi(Vg + Wh - P, 0) < \Psi(P, 0), \Psi(V, 0) \le \Psi(P, 0), \Psi(W, 0) \le \Psi(P, 0), \deg(V) < \deg(h), \deg(W) < \deg(g)$.

**Proof.** Since $(\bar{g}, \bar{h}) = \overline{1}$, by Bezout's theorem, there exists $v$ and $\tau \in \mathcal{L}[x]$ such that $v\bar{g} + \tau\bar{h} = \overline{1}$, $\deg(v) < \deg(\bar{h})$, $\deg(\tau) < \deg(\bar{g})$. Let $S, T \in U_{\mathbb{L}}[x]$ satisfy $\overline{S} = v$, $\overline{T} = \tau$, $\deg(S) = \deg(v)$, $\deg(T) = \deg(\tau)$. Thus, we have $Sg + Th - \overline{1} = \overline{0}$ i.e. (1) $\Psi(Sg + Th - \overline{1}) < 0$.

We now consider the Euclidean division of $SP$ by $h$ and $TP$ by $g$, respectively. We obtain $SP = S_0h+V$ and $TP = T_0g+W$. By Lemma 5.1, it is seen that $\max(\Psi(V, 0), \Psi(S_0, 0)) \le \Psi(SP, 0) \le \Psi(P, 0)$. Moreover, by hypothesis, we have

(1) $\deg(V) < \deg(h)$ and

(2) $\deg(W) < \deg(g)$.

Let $M = Sg + Th - \overline{1}$. Then, we have $MP = (S_0 + T_0)gh + Vg + Wh - P$. Since $\deg(P) < \deg(g) + \deg(h)$, by (1) and (2), we see that
$deg(Vg+Wh-P) < deg(g)+deg(h)$, and therefore, $Vg+Wh-P$ is just the remainder of the Euclidean division of BP by gh. But then, by Lemma 5.2, we have $\Psi(Vg+Wh-P, 0) \le \Psi(MP, 0) = \Psi(M, 0)+\Psi(P, 0)$, and therefore, by (1) and by definition of M, it is seen that $\Psi(M, 0) < 0$. This finishes proving that $\Psi(Vg+Wh-P, 0) < \Psi(P, 0)$ and this ends the proof of Lemma 5.4. □

**Notation.** Let $g, h \in U_{\mathbb{L}}[x]$ be monic and satisfy $(\overline{g}, \overline{h}) = \overline{1}$. We denote by $B(f, g)$ the set of constants $c \in \mathbb{R}_+$ such that, for every polynomial $Q \in \mathbb{L}[x]$ satisfying $deg(Q) < deg(g)+deg(h)$, there exist $V, W \in \mathbb{L}[x]$ satisfying $\Psi(Vg+Wh-Q, 0) \le \Psi(Q, 0)+c$, $\Psi(V, 0) \le \Psi(Q, 0)$, $\Psi(W, 0) \le \Psi(Q, 0)$, $deg(V) < deg(h)$, $deg(W) < deg(g)$.

**Lemma 5.5.** Let $g, h \in U_{\mathbb{L}}[x]$ be quasi-monic and satisfy $(\overline{g}, \overline{h}) = \overline{1}$ and let $d = deg(g) + deg(h)$. Then, $B(f, g)$ is a not empty interval whose lower bound is 0. Moreover, given $\lambda \in B(f, g)$ and monic polynomials $s, t \in U_{\mathbb{L}}[x]$ such that $\Psi(g-s, 0) \le \lambda$, $\Psi(h-t, 0) \le \lambda$, then $B(s, t) = B(g, h)$.

**Proof.** Let $d = deg(g) + deg(h)$. We can apply Lemma 5.4 to each polynomial $Q_n = x^n$ for every $n = 0, \dots, d-1$. Thus, we have polynomials $V_n, W_n$ satisfying $\Psi(V ng + W nh - x^n, 0) < 0$, $\Psi(V_n, 0) \le 0$, $\Psi(W_n, 0) \le 0$, $deg(V_n) < deg(h)$, $deg(W_n) < deg(g)$. We put $\lambda_n = \Psi(V ng + W nh - x^n, 0)$, $(0 \le n \le d-1)$. Now, let $Q = \sum_{n=0}^{d-1} a_n x^n$, let $V = \sum_{n=0}^{d-1} a_n V_n$, $W_n = \sum_{n=0}^{d-1} a_n W_n$, and let $\lambda(g, h) = \max_{0 \le n \le d-1} \lambda_n$. Clearly, we have $\Psi(Vg + Wh - Q, 0) \le \max_{0 \le n \le d-1} (\Psi(a_n) + \lambda_n) \le \max_{0 \le n \le d-1} \Psi(a_n) + \max_{0 \le n \le d-1} \lambda_n = \Psi(Q, 0) + \lambda(g, h)$.

But trivially,

$$ \begin{aligned} \Psi(V, 0) &\le \max_{0 \le n \le d-1} \Psi(a_n), \quad \Psi(W, 0) \le \Psi(Q, 0), \\ \deg(V) &\le \max_{0 \le n \le d-1} (\deg(V_n)) \\ < \deg(h), \quad \deg(W) &\le \max_{0 \le n \le d-1} (\deg(W_n)) < \deg(h). \end{aligned} $$

So, $\lambda(g, h)$ lies in $B(f, g)$, and hence, it is obviously seen that $B(f, g)$ is a not empty interval and that its lower bound is 0.

Now, let $c \in B(f, g)$ and let $s, t \in U_{\mathbb{L}}[x]$ be monic and satisfy $\Psi(g-s, 0) < c$, $\Psi(h-t, 0) < c$. Since $\Psi(V, 0) \le \Psi(Q, 0)$, $\Psi(W, 0) \le \Psi(P, 0)$
$\Psi(Q, 0)$, it is easily seen that $\Psi(V(g-s)+W(h-t), 0) < c+\Psi(Q, 0)$,
and therefore, $\Psi(Vs + Wt - Q, 0) < c + \Psi(Q, 0)$. This shows that
$\lambda(s,t) \le c$, and therefore, $B(f,g) \subset B(s,t)$. But similarly we have
$B(s,t) \subset B(g,h)$ and this ends the proof of Lemma 5.5.
□

**Lemma 5.6.** Let $Q \in L[x]$ and let $g, h \in U_{\mathbb{L}}[x]$ be quasi-monic and satisfy $(\bar{g}, \bar{h}) = \bar{T}$. Let $c \in B(g, h)$. There exist monic polynomials $V, W \in L[x]$ satisfying

$$
\begin{gather*}
\Psi(Vg + Wh - Q, 0) \le c + \Psi(Q, 0), \\
\deg(W) < \deg(g), \quad \deg(V) \le \max(\deg(h), \deg(Q) - \deg(g)), \\
\Psi(V, 0) \le \Psi(Q, 0), \quad \Psi(W, 0) \le \Psi(Q, 0).
\end{gather*}
$$

**Proof.** We consider the Euclidean division of $Q$ by $gh : Q = \ell gh + Q_1$. Hence, $\deg(Q_1) < \deg(g) + \deg(h)$. By Lemma 5.2, we have

$$
(1) \Psi(Q_1, 0) \leq \Psi(Q, 0),
$$

$$
(2) \Psi(\ell, 0) \leq \Psi(Q, 0).
$$

By Lemma 5.5, there exist $V_1, W_1 \in \mathbb{L}[x]$ satisfying

$$
(3) \Psi(V_1 g + W_1 h - Q_1, 0) \le \Psi(Q_1, 0) + c,
$$

$$
(4) \Psi(V_1, 0) \leq \Psi(Q_1, 0),
$$

$$
(5) \Psi(W_1, 0) \leq \Psi(Q_1, 0),
$$

$$
(6) \deg(V_1) < \deg(h),
$$

(7) $\deg(W_1) < \deg(g)$,

Now, we put $V = V_1 + \ell h$, $W = W_1$. So, we have $Vg + Wh - Q = V_1 + \ell gh + W_1h - \ell gh - Q_1$, and therefore, by (3), we obtain $\Psi(Vg + Wh - Q, 0) \le \Psi(Q_1, 0) + c$. Hence, by (1), we obtain

$$
(8) \quad v(Vg + Wh - Q) \ge v(Q) + c.
\text{Now, by (1), (5), it is seen that}
$$

$$
(9) \quad \Psi(W, 0) \le \Psi(Q, 0).
$$

We check

$$
(10) \quad \Psi(V, 0) \le \Psi(Q, 0).
$$

Indeed, we have $\Psi(h, 0) = 0$, hence, by (2), we see

$$
(11) \quad \Psi(\ell h, 0) \le \Psi(Q, 0).
$$
But, by (4), we have $\Psi(V_1, 0) \le \Psi(Q, 0)$, and therefore, by (11), we obtain (10). Finally, by definition, we have $\deg(\ell) = \deg(Q) - \deg(gh)$, and therefore,

(12) $\deg(V) \le \max(\deg(V_1), \deg(\ell h)) \le \max(\deg(h), \deg(Q) - \deg(g))$. Thanks to (7), (8), (9), (10), and (12), Lemma 5.6 is now proven. $\square$

**Theorem 5.7 (Hensel Lemma).** $\mathbb{L}$ is supposed to be complete. Let $P \in U_{\mathbb{L}}[x]$ be such that $\bar{P}$ splits in $\mathcal{L}[x]$ in the form $\gamma\eta$ with $\gamma, \eta$ relatively prime. There exists $g, h \in U_{\mathbb{L}}[x]$ such that $P = gh, \bar{g} = \gamma, \bar{h} = \eta, \deg(g) = \deg(\gamma)$.

**Proof.** We can obviously take quasi-monic polynomials $g_0, h_0 \in U_{\mathbb{L}}[x]$ such that $\overline{g_0} = \gamma, \overline{h_0} = \eta$. We put $\xi = \Psi(P - g_0 h_0, 0)$ and take $\zeta \in B(g_0, h_0)$ satisfying $\zeta \le \xi$. We construct sequences $(g_n)_{n \in \mathbb{N}}, (h_n)_{n \in \mathbb{N}}$ in $\mathbb{L}[x]$ satisfying for all $n \ge 0$:

$$ 
\begin{align*}
(i_n) \quad & \Psi(P - g_n h_n, 0) \le (n+1)\zeta, \\
(ii_n) \quad & \Psi(g_n - g_{n-1}, 0) \le n\zeta, \quad \Psi(h_n - h_{n-1}, 0) \le n\zeta, \\
(iii_n) \quad & \deg(h_n) \le \deg(P) - \deg(g_0), \quad \deg(g_n) = \deg(g_0), \\
(iv_n) \quad & \overline{g_n} = \gamma, \quad \overline{h_n} = \eta, \\
(v_n) \quad & \zeta \in B(g_n, h_n).
\end{align*}
$$

First, we put $P_1 = P - g_0 h_0$. We note that $\deg(P_1) = \deg(P)$. We now apply Lemma 5.6 to the case when $(Q, g, h) = (P_1, g_0, h_0)$: there exist $V_1, W_1 \in \mathbb{L}[x]$ satisfying (1) $\deg(W_1) < \deg(g_0)$, (2) $\deg(V_1) < \deg(P) - \deg(g_0)$, (3) $\Psi(V_1, 0) \le \zeta$, (4) $\Psi(W_1, 0) \le \zeta$, (5) $\Psi(V_1 g_0 + W_1 h_0 - P_1, 0) \le \zeta + \Psi(P_1, 0)$. Next, we put $g_1 = g_0 + W_1, h_1 = h_0 + V_1$. We check that $(i_1), (ii_1), (iii_1)$, and $(iv_1)$ are satisfied. Moreover, by (3) and (4) and by Lemma 5.5, $\zeta$ lies in $B(g_1, h_1)$, hence $v_1$ is satisfied.

Now, we suppose we have already constructed the pairs $(g_m, h_m)$ satisfying $(i_m), (ii_m), (iii_m), (iv_m)$, and $(v_m)$ for every $m = 0, \dots, n$. Then, we put $P_{n+1} = P - g_n h_n$.

We can apply Lemma 5.6 to the case when $(Q, g, h)$ is equal to $(P_{n+1}, g_n, h_n)$. So, we can obtain $V_{n+1}, W_{n+1} \in L[x]$ satisfying (6) $\Psi(W_{n+1}g_n + V_{n+1}h_n - P_{n+1}, 0) \le \zeta + \Psi(P_{n+1}, 0)$, (7) $\deg(W_{n+1}) < \deg(g_n)$, (8) $\deg(V_{n+1}) \le \max(\deg(h_n), \deg(P_{n+1}) - \deg(g_n))$, (9) $\Psi(V_{n+1}, 0) \le \Psi(P_{n+1}, 0)$, $\Psi(W_{n+1}, 0) \le \Psi(P_{n+1}, 0)$. By (6) and by $(v_n)$, we obtain (10) $\Psi(W_{n+1}g_n + V_{n+1}h_n - P_{n+1}, 0) \le (n+2)\zeta$.
Now we put $g_{n+1} = g_n + W_{n+1}$, $h_{n+1} = h_n + W_{n+1}$. We check that $P - g_{n+1}h_{n+1} = (P_{n+1} - h_n W_{n+1} - g_n V_{n+1}) - V_{n+1} W_{n+1} = P_{n+1} - h W_{n+1} - g_{n+1} V_{n+1} + (h_{n+1} - h_n) W_{n+1} + (g_{n+1} - g_n) V_{n+1} + V_{n+1} W_{n+1}$.
By $(ii_m)$ true for $m \le n$, we note that (11) $\Psi(g_n - g_{n+1}, 0) \le (n+1)\zeta$, $\Psi(h_n - h_{n+1}, 0) \le (n+1)\zeta$, and then, by (9) and (10), we obtain $(i)_{n+1} \quad \Psi(P - g_{n+1}h_{n+1}, 0) \le (n+2)\zeta$.

Relation $(i_{n+1})$ is true by definition and $(iii_{n+1})$ and $(iv_{n+1})$ are
easily checked. By (11) and by Lemma 5.5, Relation $(v_{n+1})$ is also
clear.

Therefore, the sequences $(g_n)_{n \in \mathbb{N}}$, $(h_n)_{n \in \mathbb{N}}$ satisfying $(i_n)$, $(ii_n)$, $(iii_n)$, $(iv_n)$, and $(v_n)$ are now constructed. Since $\mathbb{L}$ is complete, the vector space $\mathbb{L}_q[x]$ of polynomial of degree $m \le q$ is obviously complete with respect to the Gauss norm $\| \cdot \|$ which is characterized by $\log \|Q\| = \Psi(Q, 0)$.

Then, by Relation $(ii_n)$, the sequences $(g_n)_{n \in \mathbb{N}}$, $(h_n)_{n \in \mathbb{N}}$ converge
in $\mathbb{L}_q[x]$. We put $g = \lim_{n \to \infty} g_n$, $h = \lim_{n \to \infty} h_n$. By $(iii_n)$, we have
$\deg(g) = \deg(g_0) = \deg(\gamma)$. By $(iv_n)$, we have $\bar{g} = \gamma$, $\bar{h} = \eta$, and
finally, by $(i_n)$, we have $\Psi(P - gh, 0) = +\infty$, hence $P = gh$. That
ends the proof of Theorem 5.7. $\square$
# Chapter 6

## Extensions of Ultrametric Fields: The Field $C_p$

All considerations on analytic and meromorphic functions require considering a complete ultrametric algebraically closed field $\mathbb{K}$. Here we construct the field $C_p$ and study finite extensions of $\mathbb{Q}_p$ [70]. And we show that $C_p$ is not spherically complete.

**Notation.** As in the previous chapters, $\mathbb{I}$ denotes a complete ultrametric field whose absolute value is not trivial and whose residue class field is $\mathcal{L}$. We denote by $F$ an algebraically closed ultrametric field whose absolute value is not trivial.

Let $\mathbb{E}$ be a field, let $\mathbb{B}$ be a finite algebraic extension of $\mathbb{E}$, and let $q = [\mathbb{B} : \mathbb{E}]$. We denote by $\mathcal{N}$ the algebraic norm of $\mathbb{B}$ over $\mathbb{E}$. Given $a \in \mathbb{B}$, we denote by $\text{irr}(a, \mathbb{E})$ the minimal polynomial of $a$ over $\mathbb{E}$.

Lemma 6.1 is classical in algebra:

**Lemma 6.1.** Let $q = [\mathbb{B} : \mathbb{E}]$, and let $\mathcal{N}$ be the norm of $\mathbb{B}$ over $\mathbb{E}$. Let $a \in \mathbb{B}$, let $P_a = \text{irr}(a, \mathbb{E})$, and let $d = \text{deg}(P_a)$. Then $\mathcal{N}$ satisfies
$$ \mathcal{N}(a) = ((-1)^d P_a(0))^\frac{q}{d} \quad \text{and} \quad \mathcal{N}(ab) = \mathcal{N}(a)\mathcal{N}(b), \forall b \in \mathbb{B}. $$

**Theorem 6.2.** Let $\mathbb{E}$ be an algebraic extension of $\mathbb{I}$, let $a \in \mathbb{E}$, and let $P = \text{irr}(a, \mathbb{I})$. Then $a$ is integral over $U_{\mathbb{I}}$ if and only if $|P(0)| \le 1$. Moreover, if $|P(0)| \le 1$, then $\|P\| = 1$. Finally, if $|P(0)| = 1$, then $\text{irr}(\bar{a}, \mathcal{L}) = \bar{P}$.

**Proof.** First, we assume $a$ to be integral over $U_{\mathbb{I}}$. Then there exists a monic polynomial $Q \in U_{\mathbb{I}}[x]$ such that $Q(a) = 0$. Therefore,
$P$ divides $Q$ in $\mathbb{I}[x]$. Let $Q(x) = P(x)T(x)$. Since both $P, Q$ are monic, so is $T$. Therefore, $\Psi(P(, 0) \ge 0)$, $\Psi(T(, 0) \ge 0)$. But since $\Psi(Q, 0) = 0$ and since $\Psi(Q, 0) = \Psi(P, 0) + \Psi(T, 0)$, $P, T$ must satisfy $\Psi(P, 0) = \Psi(T, 0) = 0$, and therefore, $|P(0)| \le 1$.

Now, we assume $|P(0)| \le 1$. Suppose $\Psi(P, 0) > 0$. There exists $b \in M_{\mathbb{I}}$ such that $\Psi(bP, 0) = 0$ and then $|bP(0)| < 1$, hence $\bar{0}$ is a zero of $\overline{bP}$. Further, we note
(1) $\deg(\overline{bP}) < \deg(P)$.

Let $\overline{bP} = x^d\phi$ with $\phi(\bar{0}) \ne \bar{0}$. Then, $x^d$ and $\phi$ are relatively prime in $\mathcal{L}[x]$. Therefore, by Theorem 5.7, there exist $g, h \in U_{\mathbb{I}}[x]$ such that $\overline{g} = x^d, \overline{h} = \phi, \deg(g) = d$, and $P = gh$. But since $P$ is irreducible in $\mathbb{I}[x]$ and since $d > 0, h$ must be a constant, and therefore, $\deg(P) = d$, a contradiction to (1). Consequently, $\Psi(P, 0) = 0$. Now, suppose $|P(0)| = 1$. Since $P$ is irreducible in $\mathbb{I}[X]$, by Theorem 5.7, so is $\overline{P}$ in $\mathcal{L}[X]$, hence $\text{irr}(\overline{\alpha}, \mathcal{L}) = \overline{P}$. $\square$

**Corollary 6.3.** Let $\mathbb{E}$ be an algebraic extension of $\mathbb{I}$ equipped with the unique extension of the absolute value of $\mathbb{I}$. Let $a \in \mathbb{E}$ be such that $|a| = 1$, of degree $l$ over $\mathbb{I}$. Then the residue class $\overline{a}$ of $a$ in the residue class field of $\mathbb{E}$ is algebraic, of degree $l$ over $\mathcal{L}$.

**Proof.** Let $P(x) = \sum_{j=0}^{m} a_j x^j = \text{irr}(a, \mathbb{I})$. Since $|a| = 1$, we have $|a_0| = 1$, hence, by Theorem 6.2, we have $\|P\| = 1$ and $\overline{a}$ satisfies $\text{irr}(\overline{a}, \mathcal{L}) = \overline{P}$, which ends the proof. $\square$

**Theorem 6.4.** Let $\mathbb{E}$ be an algebraic extension of $\mathbb{I}$. There exists a unique absolute value $\varphi$ on $\mathbb{E}$ that extends the one of $\mathbb{I}$. Further, this absolute value is ultrametric and defined as follows: given $a \in \mathbb{E}$, $Q = \text{irr}(a, \mathbb{I})$, and $t = \text{deg}(Q)$, then $\varphi(a) = \sqrt[t]{|Q(0)|}$.

**Proof.** We first note that $\varphi(a) = |a|$ whenever $a \in \mathbb{I}$. We show that $\varphi$ is an ultrametric absolute value on $\mathbb{E}$. Clearly, we have $\varphi(a) \ne 0$ whenever $a \in \mathbb{E}$. By Lemma 6.1, it is easily seen that we have $\varphi(ab) = \varphi(a)\varphi(b)$ whenever $a, b \in \mathbb{E}$, and therefore, $\varphi(a)^{-1} = \varphi(a^{-1})$. So it remains to show the ultrametric inequality. For this, we show $\varphi(1+z) \le 1$ for every $z \in U_{\mathbb{E}}$. For convenience, we put again $P_z = \text{irr}(z, \mathbb{E})$ whenever $z \in \mathbb{E}$. Let $z \in U_{\mathbb{E}}$. So, we have $|P_z(0)| \le 1$, and then by Theorem 6.2, $z$ is integral over $U_{\mathbb{I}}$, hence so is $1+z$. Hence, by Theorem 6.2, we have $|P_{1+z}(0)| \le 1$, and therefore, $\varphi(1+z) \le 1$. Now, the ultrametric
inequality is easily derived. Let $a, b \in \mathbb{E}$ satisfy $0 < |a| \le |b|$. We have $\varphi(a+b) = \varphi(b(1+\frac{a}{b})) = \varphi(b)\varphi(1+\frac{a}{b})$. But $\varphi(1+\frac{a}{b}) \le 1$, hence finally, $\varphi(a+b) \le \varphi(b)$. Thus, we have now proven $\varphi$ to be an ultra-metric absolute value that extends that of $\mathbb{I}$. Then, by Theorem 2.6, this absolute value on $\mathbb{E}$ is unique, which ends the proof. $\square$

**Corollary 6.5.** Let $\Omega$ be an algebraic closure of $\mathbb{I}$. There exists a unique absolute value $\varphi$ on $\Omega$ that extends the one of $\mathbb{I}$. Further, this absolute value is ultrametric and defined as follows: given $a \in \Omega$, $Q = \text{irr}(a, \mathbb{I})$ and $t = \text{deg}(Q)$, then $\varphi(a) = \sqrt[t]{|Q(0)|}$.

**Corollary 6.6.** Let $P(x) \in \mathbb{I}[x]$ be irreducible over $\mathbb{I}$, let $\Omega$ be an algebraic closure of $\mathbb{I}$ provided with the absolute value extending that of $\mathbb{I}$, and let $b_1, \dots, b_q$ be the zeros of $P$ in $\Omega$. Then, $|b_i| = |b_j| \forall i, j \le q$.

**Corollary 6.7.** Let $\Omega$ be an algebraic closure of $\mathbb{I}$ provided with the unique absolute value $|\cdot|$ that extends the one of $\mathbb{I}$. Then, $U_\Omega$ is equal to the integral closure of $U_{\mathbb{I}}$. Moreover $|\Omega| = \{\sqrt[r]{r} | r \in |\mathbb{I}|, n \in \mathbb{N}^*\}$.

**Corollary 6.8.** Suppose that the value group of $\mathbb{I}$ is $\mathbb{Z}$. Let $\mathbb{E}$ be a finite algebraic extension of $\mathbb{I}$ of degree $t$ provided with the unique absolute value $|\cdot|$ that extends the one of $\mathbb{I}$. There exists a rational $r$ of the form $\frac{s}{t}$ such that the value group of $\mathbb{E}$ is $r\mathbb{Z}$.

**Lemma 6.9.** Let $\mathbb{B}$ be an algebraic extension of $\mathbb{E}$ provided with an absolute value extending that of $\mathbb{E}$. Then the residue class field of $\mathbb{B}$ is algebraic over the residue class field of $\mathbb{E}$.

**Proof.** Let $t \in U_{\mathbb{B}}$ and let $\hat{\mathbb{E}}$ be the completion of $\mathbb{E}$ with respect to the absolute value of $\mathbb{E}$. Since $t$ is algebraic over $\mathbb{E}$, so much the more it is algebraic over $\hat{\mathbb{E}}$. Then, by Corollary 6.5, the residue class $\bar{t}$ of $t$ is algebraic over the residue class field of $\hat{\mathbb{E}}$. But $\hat{\mathbb{E}}$ obviously has the same residue class field as $\mathbb{E}$. $\square$

**Corollary 6.10.** Let $\mathbb{B}$ be an algebraic extension of $\mathbb{I}$ provided with the unique absolute value that extends the one of $\mathbb{I}$. Then the residue class field of $\mathbb{B}$ is an algebraic extension of the residue class field of $\mathbb{I}$. Moreover, if $\mathbb{B}$ is finite over $\mathbb{I}$, then residue class field of $\mathbb{B}$ is finite over the residue class field of $\mathbb{I}$.
**Proof.** Suppose first that $\mathbb{B}$ is of the form $\mathbb{I}[u]$. Without loss of generality, we may assume that $|u| = 1$. Then the residue class field of $\mathbb{B}$ is $\mathcal{L}[\bar{u}]$. Next, we can generalize by induction. $\square$

**Theorem 6.11.** Let $\Omega$ be an algebraic closure of $\mathbb{I}$ provided with the unique absolute value extending the one of $\mathbb{I}$. Then the residue class field of $\Omega$ is an algebraic closure of $\mathcal{L}$.

**Proof.** Let $\mathcal{T}$ be the residue class field of $\Omega$. Let $u \in U_{\Omega}$ and let $P = \text{irr}(u, \mathbb{I})$. By Corollary 6.7, $P$ belongs to $U_{\mathbb{I}}[x]$ and obviously satisfies $\bar{P}(\bar{u}) = \bar{0}$, hence $\bar{u}$ is algebraic over $\mathcal{L}$. So, $\mathcal{T}$ is an algebraic extension of $\mathcal{L}$. Now, let $q \in \mathcal{L}[x]$, and let $Q \in U_{\mathbb{I}}[x]$ be a polynomial such that $\bar{Q} = q$. Then, $Q$ factorizes in $\Omega[x]$ in the form $\prod_{j=1}^{s}(x - a_j)$ with $|a_j| \le 1 \quad \forall j = 1, \dots, s$, hence $\bar{a}_j$ belongs to $\mathcal{T}$ and $q(x) = \bar{a}_j = \prod_{j=1}^{s}(x - \bar{a}_j)$. So, $\mathcal{T}$ contains the algebraic closure of $\mathcal{L}$. And since it is an algebraic extension of $\mathcal{L}$, then it is the algebraic closure of $\mathcal{L}$. $\square$

**Corollary 6.12.** The residue class field $\mathcal{K}$ of $\mathbb{K}$ is algebraically closed.

**Lemma 6.13.** Let $P(x) = \sum_{j=0}^{t} a_j x^j$, $Q(x) = \sum_{j=0}^{t} b_j x^j$ be monic, belong to $\mathbb{B}[x]$ and satisfy $||\bar{P}||||Q|| = 1$. For each zero $\alpha$ of $P$, $Q$ admits at least one zero $\beta$ such that $|\alpha - \beta|^t \le \max_{0 \le j \le t} |a_j - b_j|$.

**Proof.** Let $s = \max_{0 \le j \le t} |a_j - b_j|$ and let $\alpha$ be a zero of $P$. By Theorem 4.6, we have $|(P - Q)(x)| \le s$ whenever $x \in d(0, 1)$, hence, in particular, $|Q(\alpha)| \le s$. Let $\beta_1, \dots, \beta_t$ be the zeros of $Q$ (taking multiplicities into account). So, we have $\prod_{j=1}^{t} |\beta_j - \alpha| \le s$ and then that at least one of the $\beta_j$ satisfies $|\beta_j - \alpha| \le s^{\frac{1}{t}}$. $\square$

**Theorem 6.14.** Let $\mathbb{B}$ be an algebraically closed extension of $\mathbb{I}$ provided with the unique absolute value that extends the one of $\mathbb{I}$. The completion of $\mathbb{B}$ also is algebraically closed.

**Proof.** Let $\tilde{\mathbb{B}}$ be the completion of $\mathbb{B}$ and let $P(x) = \sum_{j=0}^{t} a_j x^j \in \tilde{\mathbb{B}}[x]$ be monic. Let $\Omega$ be an algebraic closure of $\tilde{\mathbb{B}}$, provided with the unique absolute value extending that of $\tilde{\mathbb{B}}$ and let $\alpha_1, \dots, \alpha_t$ be the zeros of $P$ in $\Omega$. Up to a change of variable, we may assume that $|\alpha_j| \le 1 \quad \forall j = 1, \dots, t$. Let $\varepsilon \in ]0, 1[$ and let $Q(x) = \sum_{j=0}^{t} b_j x^j \in \tilde{\mathbb{B}}[x]$ be such that $\max_{0 \le j \le t} |a_j - b_j| \le \varepsilon^t$. For each $j = 1, \dots, t$,
by Lemma 6.13, $Q$ admits a zero $\beta$ such that $|\alpha_j - \beta|^t \le \varepsilon$. Since $Q \in \mathbb{B}[x]$, obviously $\beta$ belongs to $\mathbb{B}$, and therefore, we see that $\alpha_j$ belongs to $\tilde{\mathbb{B}}$. That ends the proof. $\square$

The following theorem is due to Marc Krasner [70].

**Theorem 6.15 (M. Krasner).** Let $\mathbb{L}$ have characteristic zero. Let $\Omega$ be an algebraic closure of $\mathbb{L}$ provided with the unique absolute value extending the one of $\mathbb{L}$. Let $a \in \Omega$, let $a_2, \dots, a_n$ be the conjugates of $a$ in $\Omega$, and let $b \in \Omega$ satisfy $|b-a| < |b-a_j|$ for every $j=2, \dots, n$. Then, we have $\mathbb{L}[a] \subset \mathbb{L}[b]$.

**Proof.** Let $a_1 = a$ and let $P(x) = \text{irr}(a, \mathbb{L})$. In $\Omega[x]$, the polynomial $P$ splits in the form $\prod_{j=1}^n (x - a_j)$. Let $Q(x) = \text{irr}(a, \mathbb{L}[b])$. Then, $Q$ divides $P$. Let $t = \text{deg}(Q)$ and suppose the $a_j$ ranged in such a way that $Q(x) = \prod_{j=1}^t (x - a_j)$. Let $R(y) = Q(b+y)$. Then $R(y)$ is seen to be irreducible in $\mathbb{L}[b][y]$ like $Q(x)$ in $\mathbb{L}[b][x]$. Moreover, the zeros of $R$ are just the $a_j - b$, with $1 \le j \le t$. Thus, we have $R = \text{irr}(a - b, \mathbb{L}[b])$. But since $\mathbb{L}$ is complete, by Corollary 6.5, we have $|a_j - b| = \sqrt[q]{|R(0)|}$ for every $j = 1, \dots, t$. In particular, for $j = 2, \dots, t$, we have $|a_j - b| = |a - b|$ and this contradicts the hypothesis. Finally, we have $t = 1$, and therefore, $a$ lies in $\mathbb{L}[b]$. $\square$

**Corollary 6.16.** Let $\mathbb{L}$ have characteristic zero. Let $\Omega$ be an algebraic closure of $\mathbb{L}$ provided with the unique absolute value extending the one of $\mathbb{L}$. Let $a \in \Omega$, let $a_2, \dots, a_n$ be the conjugates of $a$ in $\Omega$, and let $b \in \Omega$ satisfy $|b-a| < |b-a_j|$ for every $j = 2, \dots, n$ and $[\mathbb{L}[b] : \mathbb{L}] \le n$. Then, we have $\mathbb{L}[a] = \mathbb{L}[b]$.

We can now recall the construction of *p*-adic fields.

**Definition and notation.** Let $p$ be a prime number. On $\mathbb{Z}$, the *p*-adic absolute value is defined as follows: given $n \in \mathbb{Z}^*$, it factorizes in a unique way in the form $p^s q$, with $q \in \mathbb{Z}^*$, prime to $p$. So, here we take $\theta = p$ and set $|n|_p = p^{-s}$.

Lemma 6.17 is immediate.

**Lemma 6.17.** $|\cdot|_p$ is an ultrametric absolute value on $\mathbb{Z}$ that has continuation to $\mathbb{Q}$ and defines an ultrametric absolute value on $\mathbb{Q}$.
and $\mathbb{N}$ is dense in $\mathbb{Z}$. Then, $|n| \ge \frac{1}{n}$, $\forall n \in \mathbb{N}^*$,

$$U_{\mathbb{Q}} = \{p^n\left(\frac{a}{b}\right) \mid n \in \mathbb{N}, a \in \mathbb{Z}, b \in \mathbb{Z}^*, \text{ g.c.d.}(a, p) = \text{g.c.d.}(b, p) = 1\},$$

$$M_{\mathbb{Q}} = \{p^n\left(\frac{a}{b}\right) \mid n \in \mathbb{N}^*, a \in \mathbb{Z}, b \in \mathbb{Z}^*, \text{ g.c.d.}(a, p) = \text{g.c.d.}(b, p) = 1\},$$

and the residue characteristic of $\mathbb{Q}$ is $p$. The residue class field of $\mathbb{Q}$ is the field of $p$ elements $\mathbb{F}_p$. The valuation group of $\mathbb{Q}$ is isomorphic to the additive group $\mathbb{Z}$.

**Remark and notation.** Now, $\mathbb{Q}$ admits a completion with respect to the $p$-adic absolute value and its completion is denoted by $\mathbb{Q}_p$. The closure of $\mathbb{Z}$ in $\mathbb{Q}_p$ is denoted by $\mathbb{Z}_p$.

On $\mathbb{Q}_p$, we extend the valuation and the absolute value $| \cdot |_p$ defined on $\mathbb{Q}$ and we set again $\Psi_p(x) = -v_p(x)$.

An algebraic closure $\Omega_p$ of $\mathbb{Q}_p$ is equipped with the unique extension of the $p$-adic absolute value defined on $\mathbb{Q}_p$ and we again denote it by $| \cdot |_p$. The valuation group of $\mathbb{Q}_p$ is obviously equal to the one of $\mathbb{Q}$. Next, the valuation group of $\Omega_p$ is easily seen to be isomorphic to $(\mathbb{Q}, +)$. In Chapter 8, we see that $\Omega_p$ is not complete.

By Theorem 6.14, $\Omega_p$ has a completion denoted by $\mathbb{C}_p$ that is algebraically closed. The valuation group of $\mathbb{C}_p$ is then isomorphic to $(\mathbb{Q}, +)$ like the one of $\Omega_p$. Moreover, by Theorem 6.11, the residue class field of $\Omega_p$ is an algebraic closure of $\mathbb{F}_p$ and the one of $\mathbb{C}_p$ is seen to be the same. The absolute value $| \cdot |_p$ defined on $\Omega_p$ has a natural extension to $\mathbb{C}_p$ and the associated valuation is denoted by $v_p$ again, and we set again $\Psi_p(x) = -v_p(x)$, $\forall x \in \mathbb{C}_p$. However, when there is no risk of confusion, we just write $\Psi$ instead of $\Psi_p$.

**Theorem 6.18.** Let a be integral over $\mathbb{Z}$ and let $a_2, \dots, a_q$ be the conjugates of a over $\mathbb{Q}$. Then, $|a| \le 1$ and $a_j| \le 1$, $\forall j = 2, \dots, q$.

**Proof.** Since $a$ is integral over $\mathbb{Z}$, it is integral over $\mathbb{Z}_p$. Let $P(X) = \text{irr}(a, \mathbb{Q})$ and let $B(X) = \text{irr}(a, \mathbb{Q}_p)$. Let $a_1, \dots, a_h$ be the conjugates of $a$ over $\mathbb{Q}_p$ (with $a_1 = a$). Then, $P(0) = \prod_{j=1}^q a_j$. Then, $B$ divides $P$ in $\mathbb{Q}_p[X]$. Moreover, $\|P\| = 1$ and $\|B\| = 1$. Next, $B(0) = \prod_{j=1}^h a_j$. By Corollary 6.6, we have $|a_1| =, \dots, = |a_h|$, while $|B(0)| \le 1$, hence $|a_j| \le 1$, i.e. $|a| \le 1$. Next, what is true for $a_1$ also holds for every $a_j, \dots, j = 2, \dots, q$, hence $|a_j| \le 1$. $\square$
In the future, we use Lemma 6.19:

**Lemma 6.19.** Let $a \in \mathbb{C}_p$ be algebraic over $\mathbb{Q}_p$ such that $\log_p(|a|)$ is of the form $\frac{\lambda}{t}$, with $\lambda \in \mathbb{Z}$ and $t$ in $\mathbb{N}^*$. Take $m, n \in \mathbb{N}$ and $b \in \mathbb{C}_p$ such that $\log_p(|b|)$ is of the form $\frac{u}{w}$ with $u \in \mathbb{N}$ and $w \in \mathbb{N}$ prime, prime with $u$ and such that $w > \max(m, n, t)$.

Let $f, g \in \mathbb{Q}_p[a]$ be such that $|fb^m|_p = |gb^n|_p$. Then, $m = n$.

**Proof.** We note that for every $x \in \mathbb{Q}_p[a]$, $\Psi_p(|x|)$ is of the form $\frac{\ell}{t}$ with $\ell \in \mathbb{Z}$. Consequently, $\Psi_p(|f|)$ is of the form $\frac{h}{t}$ and $\log_p(|g|)$ is of the form $\frac{k}{t}$ with $h$ and $k \in \mathbb{Z}$. Consequently, $\Psi_p(|fb^m|) = \frac{h}{t} + \frac{mu}{w}$ and $\Psi_p(|gb^n|) = \frac{k}{t} + \frac{nu}{w}$, and therefore, due to the equality $|fb^m|_p = |gb^n|_p$, we have $(h-k)w = ut(n-m)$. But since $w > t$, it is prime with $ut$, hence it must divide $n-m$, which is impossible because $\max(m,n) < w$, except if $m=n$. $\square$

**Lemma 6.20.** Let $|\cdot|.$ be an ultrametric absolute value on $\mathbb{Q}$. If this absolute value is trivial, the residue characteristic is zero. If the absolute value is not trivial, there exists a prime number $q$ such that $|\cdot|.$ is equivalent to $|\cdot|_q$.

**Proof.** If this absolute value is trivial, it is clear that the residue characteristic is zero. So, we suppose that the absolute value is not trivial. For every $n \in \mathbb{N}^*$, we have $|n| \le 1$. Since $|\cdot|$ is not trivial, there certainly exists $s \in \mathbb{N}^*$ such that $|s| < 1$. Let $q$ be the smallest $s \in \mathbb{N}^*$ such that $|s| < 1$. It is easily checked that $q$ is prime. Since $M_{\mathbb{Q}}$ is a principal ideal of $\mathbb{Z}$, we have $M_{\mathbb{Q}} = q\mathbb{Z}$.

Let $t = \frac{1}{|q|}$. It is easily checked that given $m \in \mathbb{Z}^*$, of the form $q^s n$, with $n \in \mathbb{Z}^*$, prime to $q$, we have $|m| = t^{-s}$. Let $w$ be the valuation associated with this absolute value. Then, $w$ is clearly proportional to $v_q$ and by Lemma 2.2 is equivalent to $v_q$. This ends the proof. $\square$

Lemma 6.21 is easily seen.

**Lemma 6.21.** $\mathbb{N}$ is dense in $\mathbb{Z}_p$, the invertible elements in $\mathbb{Z}_p$ are the ones whose absolute value is 1, $\mathbb{Z}_p$ is compact, equal to $U_{\mathbb{Q}_p}$, and $p\mathbb{Z}_p$ is equal to $M_{\mathbb{Q}_p}$. $\mathbb{Q}_p$ is locally compact. The residue class field of $\mathbb{Q}_p$ is equal to the field of $p$ elements $\mathbb{F}_p$. Finally, $U_p$ is the union of $p$ disks $d(u, \frac{1}{p})$.
60

**Proof.** All statements are immediate except the compacity of $Z_p$. Consider a sequence $(a_n)$ in $Z_p$. Since it is bounded, by Theorem 3.1, we can extract either a monotonous distances sequence or an equal distances sequence, or a converging sequence. But since each circle $d(a, r)$ with $r \in |Q_p|$ only has $p$ classes, there are no equal distances sequences in $Q_p$. And since the absolute value is discrete, there is no monotonous distances sequence in $Q_p$. Hence, we can extract a converging sequence from the sequence $(a_n)$. $\square$

**Corollary 6.22.** For each $m \in \mathbb{N}$, $U_p$ is the union of $p^m$ distinct disks $d(u, \frac{1}{p^m})$. Let $E$ be a finite algebraic extension of $Q_p$. There exists a constant $B > 0$ such that for all $m \in \mathbb{N}$ the number of distinct $d(u, \frac{1}{p^m})$ is inferior or equal to $Bp^m$.

By definition and construction of $C_p$, we have this corollary:

**Corollary 6.23.** The field of algebraic numbers is dense in $C_p$. Hence, $C_p$ contains a dense countable subset.

In Theorem 6.24, we follow the method of [84].

**Theorem 6.24.** $C_p$ is not spherically complete.

**Proof.** Let $(r_n)_{n \in \mathbb{N}}$ be a real sequence such that $0 < r < r_{n+1} < r_n < 1$, $\forall n \in \mathbb{N}$. Let $S$ be the set of sequences of the set $\{0,1\}$. Suppose that for each $n = 0, \dots, q$ we have defined $2^n$ disks $d(a_{n,k}, r_n)$, $k = 0,1$, such that for each $n = 1, \dots, q$ and for each $k = 0,1$ the disks $d(a_{n,0}, r_n)$ and $d(a_{n,1}, r_n)$ are included in some $d(a_{n-1,k}, r_{n-1})$ (with $k=0$ or $k=1$) and have an empty intersection. It is then immediate to define in each disk $d(a_{q,k}r_q)$ two disks $d(a_{q+1,0}, r_{q+1})$ and $d(a_{q+1,1}, r_{q+1})$ having an empty intersection. So, the family is defined for every $n \in \mathbb{N}$.

Now, let $(u_n) \in S$ and let $(V_n)$ be a decreasing sequence of disks defined as follows: suppose defined $(V_n)$ for $n \le q$. If $u_{q+1} = 0$, we set $V_{q+1} = d(a_{q+1,0}, r_{q+1})$, and if $u_{q+1} = 1$, we set $V_{q+1} = d(a_{q+1,1}, r_{q+1})$. That mapping which associates with each sequence $(u_n)$ the decreasing sequence of disks $(V_n)$ is clearly injective. Now, consider two distinct such sequences $(u_n)$ and $(u'_n)$. Let $q$ be the smallest integer such that $u_q \ne u'_q$. The distance from $V_q$ to $V'_q$ is at least $r_q$. Consequently, for every $n \ge q$, the distance between $V_n$ and $V'_n$ is at least $r_q \ge r$.
Suppose now that $\mathbb{C}_p$ is spherically complete. For each sequence $(u_n)$ the intersection of the decreasing sequence of disks $(V_n)$ contains a point $\alpha((u_n))$ and hence, by the last conclusion, if $(u_n)$ and $(u'_n)$ are two different sequences, we have $|\alpha((u_n)) - \alpha((u'_n))| \ge r$. But we know that the set of sequences $(u_n)$ is not countable and hence the set of the $\alpha((u_n))$, $((u_n) \in S)$ is not countable. Consequently, $\mathbb{C}_p$ contains an uncountable subset $\Sigma$ such that $|x - y| \ge r, \forall x \ne y, x, y \in \Sigma$. This contradicts the fact that $\mathbb{C}_p$ contains a dense countable subset. $\square$
# Chapter 7

## Normal Extensions of $Q_p$ Inside $C_p$

**Notation.** Recall that $\mathbb{L}$ is a complete field with respect to an ultra-metric absolute value. For every $s \in \mathbb{N}^*$, we put $u_s = \frac{1}{p^{s-1}(p-1)}$ and $r_s = p^{-u_s}$. We study the $p^s$th roots of 1 and we show that they lie in circles of center 1 and radius $r_s$. We examine normal extensions of $Q_p$ and totally ramified extensions and show the role of Eisenstein polynomials.

We need certain technical lemmas.

**Lemma 7.1.** Let $s \in \mathbb{N}^*$. For every $n \in \mathbb{N}^*$ such that $n < p^s$, we have $\left|\binom{p^s}{n}\right|_p = \frac{1}{p^s|n|_p}$.

**Proof.** We note that for any $n \in \mathbb{N}^*$ such that $n < p^s$, $n$ is a multiple of $p^h$ for some $h < s$, if and only if so is $p^s - n$. Now, let $B$ be the bijection from $\{1, \dots, n-1\}$ onto $\{(p^s - n + 1), \dots, p^s - 1\}$, defined as $B(j) = p^s - j$. Thus, for every $j = 1, \dots, n-1$, we have $|j|_p = |B(j)|_p$. Now, obviously, $\left|\binom{p^s}{n}\right|_p = \left|\frac{p^s}{n}\right|_p \cdot \left|\prod_{j=1}^{n-1} \frac{B(j)}{j}\right|_p$. But as we just saw, each factor $\left|\frac{B(j)}{j}\right|_p$ is equal to 1, and therefore, the conclusion is clear. $\square$

**Notation.** Let $s \in \mathbb{N}$. We denote by $W_s$ the group of the $p^s$th roots of 1 in $C_p$ i.e. the $\zeta \in C_p$ such that $\zeta^{p^s} = 1$ and we denote by $B_s$ the set $W_s \setminus W_{s-1}$ and we set $W = \bigcup_{s \in \mathbb{N}} W_s$.
$F_s$ denotes the polynomial $\sum_{j=0}^{p-1} x^{jp^{s-1}}$ and we put $G_s(x) = F_s(1+x)$.

**Definition.** A monic polynomial $P(x) = \sum_{j=0}^{q} a_j x^j \in \mathbb{I}[x]$ is called an Eisenstein polynomial if it satisfies $a_j \in M_{\mathbb{I}}$ whenever $j = 0, \dots, q-1$ and $a_0 \notin (M_{\mathbb{I}})^2$.

**Theorem 7.2 (Eisenstein).** Let $\mathbb{I}$ have a discrete valuation. Let $P(x) = \sum_{j=0}^{q} a_j x^j \in \mathbb{I}[x]$ be an Eisenstein polynomial. Then, $P$ is irreducible in $\mathbb{I}[x]$.

**Proof.** We suppose $P$ not irreducible. Then, $P$ splits in $\mathbb{I}[x]$ in the form $S(x)T(x)$ with $S(x) = \sum_{j=0}^{m} \alpha_j x^j$, $T(x) = \sum_{j=0}^{n} \beta_j x^j$, and $\alpha_m = \beta_n = 1$. Since $S, T$ are monic, we have $\|S\| \ge 1$, $\|T\| \ge 1$, and since $\|S\|\|T\| = \|ST\| = \|P\| = 1$, we have $\|S\| = \|T\| = 1$. Hence, both $S, T$ belong to $U_{\mathbb{I}}[x]$. First, we note that if $\alpha_0$ belongs to $M_{\mathbb{I}}$, then $\beta_0$ does not because $a_0 \notin (M_{\mathbb{I}})^2$. Hence, we may assume $\alpha_0 \in M_{\mathbb{I}}$ and $\beta_0 \notin M_{\mathbb{I}}$. Further, we have $\alpha_j \in M_{\mathbb{I}}$ for every $j = 0, \dots, m-1$. Indeed, let $\ell$ be the smallest of the integers $h$ such that $|\alpha_h| = 1$. Then we have $|a_\ell| = |\beta_0\alpha_\ell + \sum_{j=1}^\ell \beta_j\alpha_{\ell-j}| = 1$ because $|\beta_0\alpha_\ell| = 1$ and $\sum_{j=1}^\ell \beta_j\alpha_{\ell-j} \in M_{\mathbb{I}}$. Consequently, $\ell = q$, therefore $P$ is irreducible. $\square$

The following lemmas are useful in the sequel.

**Lemma 7.3.** Let $\mathcal{G}$ be a subgroup of the multiplicative group $(\mathbb{K}^*, \cdot)$ included in $C(0, 1)$ and let $u \in \mathcal{G}$. The bijection $\gamma$ from $\mathcal{G}$ onto $\mathcal{G}$ defined as $\gamma(x) = ux$ is isometric.

**Lemma 7.4.** Let $(j, n) \in \mathbb{N} \times \mathbb{N}^*$ be such that $j < n$. Then, we have

$$ \binom{n}{j+1} = \sum_{h=j}^{n-1} \binom{h}{j}. $$

**Lemma 7.5.** For every $s \in \mathbb{N}$, $G_s$ is an Eisenstein polynomial.

**Proof.** First, we suppose $s = 1$. We have

$$ G_1(x) = \sum_{h=0}^{p-1} \left( \sum_{j=0}^{h} \binom{h}{j} x^j \right) = \sum_{j=0}^{p-1} \left( \sum_{h=j}^{p-1} \binom{h}{j} \right) x^j. $$
Hence, by Lemma 7.4, we have $G_1(x) = \sum_{j=0}^{p-1} \binom{p}{j+1} x^j$. Moreover, by Lemma 7.1, we have $\left|\binom{p}{j+1}\right|_p = \frac{1}{p}$ for every $j = 0, \dots, p-2$, and therefore, $G_1$ is an Eisenstein polynomial.

Now we consider the general case $s \in \mathbb{N}^*$. First, we put $T_s(x) = (1+x)^{p^s}$. By Lemma 7.1, it is seen that $T_1(x)$ is of the form $1+x^p + \gamma_1(x)$ with $\gamma_1(x) \in px\mathbb{Z}_p[x]$ and $\deg(\gamma_1) = p-1$. Then by an immediate induction, we see that $T_s(x)$ is of the form $1+x^{p^s} + \gamma_s(x)$ with $\gamma_s(x) \in px\mathbb{Z}_p[x]$ and $\deg(\gamma_s) = p^s - 1$.

As a consequence, it is easily seen that $G_s$ is an Eisenstein polynomial if and only if so is the polynomial $g_s(x) = \sum_{j=0}^{p-1} (1+x^{p^s})^j$. But we have $g_s(x) = G_1(x^{p^s})$. Since $G_1$ is an Eisenstein polynomial, so is $g_s$. This ends the proof. $\square$

**Theorem 7.6.** For each $s \in \mathbb{N}^*$, $B_s$ consists of $p^s - p^{s-1}$ roots of $1$ of order $p^s$ that lie in $C(1, r_s)$. For every $t \in B_s$, $\text{irr}(t, \mathbb{Q}_p)$ is equal to $F_s$ and $B_s \cap d(t, r_s^-)$ is equal to $tW_{s-1}$.

**Proof.** Let $t \in B_s$ and let $F(x) = \text{irr}(t, \mathbb{Q}_p)$. Then, $F$ divides $x^{p^s} - 1$ and has a degree $d > p^{s-1}$. Since $x^{p^s} - 1 = (x^{p^{s-1}} - 1)F_s$, $F$ divides $F_s$. But by Lemma 7.5 and Theorem 6.1, $G_s$ is irreducible over $\mathbb{Q}_p$, hence so is $F_s$, and therefore, $F = F_s$. Then, by Corollary 6.5, we have $\Psi_p(t-1) = \frac{\Psi_p(G_s(0))}{p^s-p^{s-1}} = -u_s$. Hence, $B_s$ is included in $C(1, r_s)$ and obviously consists of $p^s - p^{s-1}$ different points in this circle. Let $\phi$ be the mapping defined in $W_{s-1}$ as $\phi(\xi) = t\xi$. Since $t$ is of order $p^s$ and any element of $W_{s-1}$ is of order $< p^s$, one sees that $\phi$ is an injection from $W_{s-1}$ into $B_s$, and then, by Lemma 7.3, we have $\phi(W_{s-1}) \subset B_s \cap d(t, r_s^-)$. Conversely, given $u \in B_s \cap d(t, r_s^-)$, then $t^{-1}u$ lies in $W_{s-1}$ which shows that $\phi$ is a bijection from $W_{s-1}$ onto $B_s \cap d(t, r_s^-)$. $\square$

**Corollary 7.7.** For every $\xi$, $\zeta \in W_1$ of $\xi \neq \zeta$, we have $\Psi_p(\xi - \zeta) = -\frac{1}{p-1}$.

**Theorem 7.8.** Let $n \in \mathbb{N}^*$ and let $\zeta$ be an $n$th root of $1$ of order $n$. Then $\zeta$ belongs to $d(1, 1^-)$ if and only if $n$ is of the form $p^s$ ($s \in \mathbb{N}$).

**Proof.** By Theorem 7.6, we know that if $n$ is of the form $p^s$, then $\zeta$ belongs to $d(1, 1^-)$. Now we suppose $\zeta \in d(1, 1^-)$ and put $n = q p^s$.
with $q$ prime to $p$. Let $\xi = \zeta^{(p^s)}$. It is seen that $\xi$ also belongs to $d(1, 1^{-})$ because $|\xi - 1|_p = |\zeta - 1|_p \left| \sum_{j=0}^{p^s-1} \zeta^j \right|_p$. Let $P(x) = x^q - 1$. If $\xi \neq 1$, then $\bar{P}(x)$ admits 1 as a zero of order $t \ge 2$. But $\bar{P}'(\bar{1}) = \bar{q} \neq \bar{0}$, hence $\xi = 1$. Therefore, we have $q = 1$ and this ends the proof. $\square$

**Definition.** Let $\mathbb{E}$ be a finite extension of $\mathbb{Q}_p$. Recall that the residue class field of $\mathbb{Q}_p$ is $\mathbb{F}_p$. Let $\mathcal{E}$ be the residue class field of $\mathbb{E}$. Since $\mathbb{E}$ is finite over $\mathbb{Q}_p$, it is locally compact, hence $\sup\{|x| \mid x \in \mathbb{E} \mid x| < 1\} < 1$. So, we can choose an element $s \in \mathbb{E}$ such that $|s| < 1$ and such that $|s| = \sup\{|x| \mid x \in \mathbb{E} \mid x| < 1\}$. Such an element $s$ is called a *uniformizer* of $\mathbb{E}$. Since $s$ is algebraic over $\mathbb{Q}_p$, $|p|$ is of the form $|s|^c$ with $c \in \mathbb{N}^*$. The number $c$ is called *ramification index of $\mathbb{E}$*.

Next, by Corollary 6.3, we know that if $a \in \mathbb{E}$ is algebraic over $\mathbb{Q}_p$ of degree $q$ and such that $|a| = 1$, then its residue class $\bar{a}$ is algebraic over $\mathbb{F}_p$, of degree $\le q$. Consequently, if $\mathbb{E}$ is finite over $\mathbb{Q}_p$, then $\mathcal{E}$ is finite over $\mathbb{F}_p$. The number $[\mathcal{E} : \mathbb{F}_p]$ is called residual degree of $\mathbb{E}$ and is denoted by $f$.

The extension $\mathbb{E}$ is said to be *ramified* if $c > 1$ and *unramified* if $c = 1$.

Lemma 7.9 is just a remark:

**Lemma 7.9.** Let $T$ be an unramified extension of $\mathbb{Q}_p$ and let $\mathbb{E}$ be an extension of $\mathbb{Q}_p$ such that $\mathbb{E} \subset T$. Then, $\mathbb{E}$ is unramified.

**Theorem 7.10.** Let $n \in \mathbb{N}^*$ be prime to $p$ and let $u, w$ be two distinct roots of $1$ in $\Omega_p$ of order $n$. Then, in $\Omega_p$, we have $|u - w| = 1$.

**Proof.** Let $h = \frac{u}{w}$. Then, $h^n = 1$. Suppose $|h - 1| < 1$. By Theorem 7.8, $n$ is of the form $p^s$, a contradiction. $\square$
# Chapter 8

## Spherically Complete Extensions

Several problems on *p*-adic analytic functions require one to consider an ultrametric algebraically closed extension of $K$ which is spherically complete, in order to give every circular filter a center. Others require to have a complete algebraically closed extension which admits a non-countable residue class field. Proving the existence of a spherically complete algebraically closed extension of the ground field $K$ isn't easy, most of the ways involving basic considerations in logic. Here we follow the method proposed by Bertin Diarra that is only based on the notion of ultraproducts [33].

**Definition and notation.** Here we denote by $(\mathbb{E}_i)_{i \in I}$ an infinite family of field extensions of $\mathbb{L}$, provided each with an ultrametric absolute value $|\cdot|_i$ extending that of $\mathbb{L}$. Next, $\mathcal{U}$ denotes an ultrafilter on $I$. We remember that $\mathcal{U}$ is said to be *principal* if there exists $\alpha \in I$ such that $\mathcal{U}$ is the set of the subsets of $I$ that contain $\alpha$. Then, $\mathcal{U}$ is said to be *incomplete* if there exists a decreasing sequence $(X_n)_{n \in \mathbb{N}}$ of elements of $\mathcal{U}$ such that $\bigcap_{n \in \mathbb{N}} X_n = \emptyset$. Since $I$ is infinite, there obviously exist incomplete ultrafilters on $I$. In particular, any incomplete ultrafilter is not principal. $\mathcal{R}$ denotes the subring of $\prod_{i \in I} \mathbb{E}_i$ that consists of the set $(a_i)_{i \in I} \in \prod_{i \in I} \mathbb{E}_i$ such that $\sup_{i \in I} |a_i|_i < +\infty$. Of course, $\mathcal{R}$ is an $\mathbb{L}$-algebra.

We denote by $\varphi$ the mapping from $\mathcal{R}$ into $\mathbb{R}_+$ defined as $\varphi((a_i)_{i \in I}) = \lim_{\mathcal{U}} |a_i|_i$. Then, $\varphi$ is seen to be a multiplicative semi-norm of the $\mathbb{L}$-algebra $\mathcal{R}$.
We put $\mathcal{J} = \text{Ker}(\varphi)$ and $\mathcal{S} = \mathcal{R}/\mathcal{J}$ and we denote by $\psi$ the canonical surjection from $\mathcal{R}$ onto $\mathcal{S}$. Then, $\mathcal{S}$ is obviously provided with an absolute value $|\cdot|$ defined as $|\psi(a)| = \varphi(a)$, $(a \in \mathcal{R})$.

On the other hand, $\mathcal{R}$ is seen to be provided with a norm of $\mathbb{L}$-algebra $\|\cdot\|$ defined as $\|(a_i)_{i \in I}\| = \sup_{i \in I} |a_i|$. Next, we denote by $|||\cdot|||$ the semi-norm quotient of the norm of $\mathbb{L}$-algebra by the ideal $\mathcal{J}$, defined on $\mathcal{R}$ as $|||a||| = \inf_{t \in \mathcal{J}} \|a - t\|$.

**Theorem 8.1 (B. Diarra).** $\mathcal{S}$ is a field extension of $\mathbb{L}$ and its absolute value $|\cdot|$ extends the one of $\mathbb{L}$. Moreover, if $\mathcal{U}$ is non-principal and if each $\mathbb{E}_i$ has a dense valuation group, then $\mathcal{S}$ has a valuation group equal to $(\mathbb{R}, +)$. Further, if each $\mathbb{E}_i$ is algebraically closed, then so is $\mathcal{S}$.

**Proof.** Let $\alpha \in \mathcal{S}\setminus\{0\}$ and let $a = (a_i)_{i \in I} \in \mathcal{R}$ be such that $\psi(a) = \alpha$. By definition, we have $\lim_{\mathcal{U}} |a_i|_i \neq 0$. Hence, there exists $J \in \mathcal{U}$ such that for every $i \in J$ we have $\frac{\varphi(a)}{2} < |a_i|_i < \frac{3\varphi(a)}{2}$, hence $a_i \neq 0$, and therefore,

$$ (1) \quad \frac{2}{3\varphi(a)} < |a_i^{-1}|_i < \frac{2}{\varphi(a)} \quad \text{whenever } i \in J. $$

Now, let $b = (b_i)_{i \in I} \in \prod_{i \in I} \mathbb{E}_i$ be defined as $b_i = a_i^{-1}$ whenever $i \in J$ and $b_i = 1$ whenever $i \in I \setminus J$. By (1), it is seen that $b$ does belong to $\mathcal{R}$. But now, $ab - 1$ is an element $(c_i)_{i \in I}$ of $\mathcal{R}$ that satisfies $c_i = 0$ whenever $i \in J$, hence $\lim_{\mathcal{U}} |c_i| = 0$. Therefore, $ab - 1$ belongs to $\mathcal{J}$, and finally, in $\mathcal{S}$ we have $\psi(a)\psi(b) = 1$. This shows $\mathcal{S}$ to be a field.

Next, we suppose that for each $i \in I$ the valuation group of $\mathbb{E}_i$ is dense. Let $r \in ]0, +\infty[$. We can obviously find a family $(\varepsilon_i)_{i \in I}$ in $]0, +\infty[$ such that $\lim_{\mathcal{U}} \varepsilon_i = 0$. For every $i \in I$, let $a_i \in \mathbb{E}_i$ satisfy $r - \varepsilon_i < |a_i| < r$ and let $a = (a_i)_{i \in I}$. Of course, $a$ belongs to $\mathcal{R}$ and satisfies $\lim_{\mathcal{U}} |a_i|_i = r$, hence $r$ belongs to $|\mathcal{S}|$. This shows that the valuation group of $\mathcal{S}$ is equal to $\mathbb{R}$.

Finally, we suppose that each field $\mathbb{E}_i$ is algebraically closed. Let

$$ P(x) = \sum_{n=0}^{q} \lambda_n x^n \in \mathcal{S}[x] $$

with $\lambda_q = 1$, and $q > 0$. We show that $P$ admits at least one zero in $\mathcal{S}$. For every $n = 0, 1, \dots, q-1$, let $(a_{i,n})_{i \in I} \in \mathcal{R}$ satisfy $\psi((a_{i,n})_{i \in I}) = \lambda_n$ and let $a_{i,q} = 1$ whenever $i \in I$. So, we have $\psi((a_{i,n})_{i \in I}) = \lambda_n$ whenever $n = 0, \dots, q$. For every $i \in I$, we put $T_i(x) = \sum_{n=0}^{q} a_{i,n} x^n$.
Since $\mathbb{E}_i$ is algebraically closed, and since $a_{i,q} = 1$ for every $i \in I$,
at least one of the zeros $\alpha_i$ of $T_i$ in $\mathbb{E}_i$ satisfies $|\alpha_i|^q \le |a_{i,0}|$. But by
hypothesis, $(a_{i,0})_{i \in I}$ belongs to $\mathcal{R}$, hence so does $(\alpha_i)_{i \in I}$. Hence, we
can put $\alpha = \psi((\alpha_i)_{i \in I})$ and then we have $P(\alpha) = 0$. This finishes
showing that $\mathcal{S}$ is algebraically closed and this ends the proof of
Theorem 8.1.
□

**Lemma 8.2 (B. Diarra).** Let $a = (a_i)_{i \in I} \in \mathcal{R}$. Then we have
$$ \|a\| = \varphi(a) $$.

**Proof.** Let $J \in \mathcal{U}$ and let $e = (e_i)_{i \in I} \in \mathcal{R}$ be defined as $e_i = 0$ whenever $i \in J$ and $e_i = 1$ whenever $i \notin J$. For convenience, we put $b = ae$. Clearly, $b$ belongs to $\mathcal{J}$, hence, we have $\|\|a\|\|$ = $\inf_{t \in \mathcal{J}} \|a-t\| \le \|a-b\|$. But now, we check that $\|a-b\| = \sup_{i \in J} |a_i|$. Further, this is true for every $J \in \mathcal{U}$. Hence, we obtain $\|\|a\|\| \le \inf_{J \in \mathcal{U}} (\sup_{i \in J} |a_i|) = \lim_{\mathcal{U}} |a_i| = \varphi(a)$. On the other hand, for all $t \in \mathcal{J}$, we have $\varphi(a-t) = \varphi(a) \le \|a-t\|$, hence $\varphi(a) \le \|\|a\|\|$. This ends the proof of Lemma 8.2.
□

**Theorem 8.3 (B. Diarra).** If $\mathcal{U}$ is incomplete, then $\mathcal{S}$ is spherically complete.

**Proof.** Let $(\alpha_n)_{n \in \mathbb{N}}$ be a decreasing distances sequence in $\mathcal{S}$ and for every $n \in \mathbb{N}$, let $a_n \in \mathcal{R}$ be such that $\psi(a_n) = \alpha_n$. By induction, we can easily construct another sequence $(b_n)_{n \in \mathbb{N}}$ in $\mathcal{R}$ satisfying

$(\mathcal{V}_n) \quad \psi(b_n) = \alpha_n \text{ for every } n \in \mathbb{N},$

$(W_n) \qquad \|b_n - b_{n-1}\| < r_{n-2} \text{ whenever } n > 1.$

Indeed, let $q \in \mathbb{N}^*$ suppose that we have defined $b_0, \dots, b_q$ satisfying
$(\mathcal{V}_n)$ for every $n = 0, \dots, q$. Of course, we have

$$
(1) \quad \varphi(b_q - a_{q+1}) = \varphi(a_q - a_{q+1}) = r_q < r_{q-1}.
$$

By Lemma 8.2, we have $\varphi(b_q - a_{q+1}) = \|\|b_q - a_{q+1}\|\|$, hence by (1), there exists $c \in \mathcal{J}$ such that $\|b_q - a_{q+1} - c\| < r_{q-1}$. So, we put $b_{q+1} = a_{q+1} + c$ and then $(\mathcal{V}_{q+1})$, $(\mathcal{W}_{q+1})$ are satisfied. In order to begin the induction, we put $b_0 = a_0$, $b_1 = a_1$ and then, we can define the sequence $(b_n)_{n \in \mathbb{N}}$ for every $n \in \mathbb{N}$, satisfying $(\mathcal{V}_n)$ and $(\mathcal{W}_n)$.

Now, for each $n \in \mathbb{N}$, we put $b_n = (b_{i,n})_{i \in I}$. Since $\mathcal{U}$ is incomplete,
we can take a decreasing sequence $(X_n)_{n \in \mathbb{N}}$ of elements of $\mathcal{U}$ such
that $\bigcap_{n \in \mathbb{N}} X_n = \emptyset$. We put $I_0 = I \setminus X_0$ and for each $n \in \mathbb{N}^*$,
$I_{n+1} =
X_n \setminus X_{n+1}$. Thus, the family $(I_n)_{n \in \mathbb{N}}$ makes a partition of $I$. Further,
for each $q \in \mathbb{N}$, $(I_n)_{n \ge q}$ makes a partition of $X_q$. Hence, we can define a surjective mapping $g$ from $I$ onto $\mathbb{N}$ as $g(i) = n$ whenever $i \in I_n$. Now, for every $i \in I$, we put $h_i = b_{i,g(i)}$. By $(W_n)$, we have $\|b_n\| \le r_0$ whenever $n \in \mathbb{N}$, hence $|h_i|_i \le r_0$ for each $i \in I$, and therefore, $(h_i)_{i \in I}$ belongs to $\mathcal{R}$. We put $h = (h_i)_{i \in I}$ and $w = \psi(h)$. We show

(2) $|w - \alpha_n| \le r_{n-1}$ whenever $n \in \mathbb{N}^*$.

Let $n \in \mathbb{N}^*$ be fixed. It is seen that for every $m > n$, we have $\|b_n - b_m\| < r_{n-1}$, hence for every $i \in I$, we have $|b_{i,n} - b_{i,m}|_i < r_{n-1}$. Moreover, since $(I_m)_{m>n}$ makes a partition of $X_{n+1}$, for every $i \in X_{n+1}$, there exists $m > n$ such that $i \in I_m$ and then we have $|b_{i,n} - h_i|_i = |b_{i,n} - b_{i,g(i)}|_i = |b_{i,n} - b_{i,m}|_i < r_{n-1}$ whenever $i \in X_{n+1}$. But $X_{n+1}$ belongs to $\mathcal{U}$, and therefore, in $\mathcal{S}$ we have $|b_n - h| = \lim_{\mathcal{U}} |b_{i,n} - h_i|_i \le r_{n-1}$. This is true for every $n \in \mathbb{N}^*$ and finally this shows (2). Hence, $w$ belongs to $\bigcap_{n \in \mathbb{N}} d_{\mathcal{S}}(\alpha_{n+1}, r_n)$ and this finishes proving that $\mathcal{S}$ is spherically complete. $\square$

**Theorem 8.4.** $\mathbb{K}$ admits a spherically complete algebraically closed extension whose residue class field is not countable and whose valuation group is equal to $\mathbb{R}$.

**Proof.** First, we construct a complete algebraically closed extension of $\mathbb{K}$ whose residue class field is not countable. Let $T$ be a transcendental extension of the form $\mathbb{K}((x_j)_{j \in \mathbb{R}})$ provided with the absolute value $| \cdot |$ defined on $\mathbb{K}[(x_j)_{j \in \mathbb{R}}]$ by

$$ \left| \sum_{j_1, \dots, j_q \le N} a_{j_1, \dots, j_q} x_{j_1}^{t_1} \cdots x_{j_q}^{t_q} \right| = \max_{j_1, \dots, j_q \le N} |a_{j_1, \dots, j_q}|. $$

It is seen that $|x_j - x_h| = 1$ whenever $j, h \in \mathbb{R}$ such that $j \ne h$, and therefore, the residue class field of $T$ is not countable. Let $T'$ be the completion of $T$ and let $\mathbb{E}$ be an algebraic closure of $T'$, provided with the unique absolute value that extends the one of $T'$. Let $\mathbb{E}'$ be the completion of $\mathbb{E}$. By Theorem 6.14, $\mathbb{E}'$ is algebraically closed. Obviously, its residue class field contains one of $T$ and therefore is not countable.

Now, we can construct $\mathcal{S}$ by taking $I = \mathbb{N}$ and $\mathbb{E}_i = \mathbb{E}'$ for every $i \in \mathbb{N}$. Since $\mathbb{E}'$ is algebraically closed, by Theorem 8.1, so is $\mathcal{S}$. Moreover, the valuation group of $\mathbb{E}'$ obviously is dense and therefore, by Theorem 7.1, $\mathcal{S}$ has a valuation group equal to $\mathbb{R}$. Finally, by Theorem 8.3, $\mathcal{S}$ is spherically complete. That ends the proof of Theorem 8.4. $\square$
Thanks to Theorem 8.3, we can generalize Corollary 5.3:

**Theorem 8.5.** Let $P, F \in \mathbb{K}[x]$ with $F$ having all its zeros in $d(0, r)$. Let $Q, R \in \mathbb{K}[x]$ satisfy $P = FQ + R$ and $\deg(R) < \deg(F)$. Then we have $\Psi(Q, \log r) \le \Psi(P, \log r) - \Psi(F, \log r)$ and $\Psi(R, \log r) \le \Psi(P, \log r)$. Moreover, if $P$ has all its zeros in $d(0, r)$, then $\Psi(Q, \log r) = \Psi(P, \log r) - \Psi(F, \log r)$.

**Proof.** Without loss of generality, we can assume that the valuation group of $\mathbb{K}$ is $\mathbb{R}$. Consequently, up to a change of variable, we can suppose that $r = 1$. We can also assume that $F$ is monic. Now, since $F$ has all zeros in $d(0, r)$, this means that $\Psi(F, 0) = 0$ and hence Theorem 8.5 is reduced to Corollary 5.3. $\square$

**Notation.** Henceforth, $\hat{\mathbb{K}}$ denotes an algebraically closed spherically complete extension of $\mathbb{K}$.

For every disk $d(a, r^{-})$ (respectively $d(a, r)$) in $\mathbb{K}$, we denote by $\hat{d}(a, r^{-})$ (respectively $\hat{d}(a, r)$) the disk of same center and diameter in $\hat{\mathbb{K}}$. Similarly, we denote by $\hat{C}(a, r)$ the circle $\{x \in \hat{\mathbb{K}} \mid |x - a| = r\}$.

**Remark.** There exists another way to construct a spherically complete extension, due to Irving Kaplansky [68].

**Definition.** Let $\mathbb{E}$ be an extension of $\mathbb{I}$ provided with an ultrametric absolute value that extends that of $\mathbb{I}$. The extension $\mathbb{E}$ is said to be *immediate* if its residue class field is identical to that of $\mathbb{I}$ and its value group also is identical to that of $\mathbb{I}$.
The following theorem is due to I. Kaplansky [68]:

**Theorem 8.6.** $\mathbb{I}$ admits an immediate extension that is maximal with respect to the inclusion.

An immediate extension of $\mathbb{I}$ is spherically complete if and only if it is maximal with respect to the inclusion.

A maximal immediate extension of $\mathbb{I}$ is unique up to an $\mathbb{I}$-isomorphism.

The proof of this theorem represents a very big work. In what follows, we do not need that theorem.
# Chapter 9

## Transcendence Order and
Transcendence Type

In $C_p$, we can define a notion of transcendence order stating that if $a$ is transcendental over $Q_p$ and has a transcendence order $\le t$ and if $b$ is transcendental over $Q_p$ but algebraic over $Q_p[a]$, then $b$ also has a transcendence order $\le t$. We show the existence of numbers of order less than $1 + \epsilon$ for every $\epsilon > 0$ [41], [55].

**Definition.** Let $\tau \in ]0, +\infty[$. Let $F$ be a transcendental extension of $Q_p$ provided with an absolute value $|\cdot|.$ extending that of $Q_p$. An element $a \in F$ is said to have transcendence order $\le \tau$ or order $\le \tau$ in brief if there exists a constant $C_a \in ]0, +\infty[$ such that every polynomial $P \in Q_p[x]$ satisfies $\log_p(|P(a)|) \ge \log(\|P\|) - C_a(\deg(P))^{\tau}$. Moreover, $a$ is said to have weak transcendence order $\le \tau^+$ or weak order $\le \tau^+$ in brief if $a$ has transcendence order $\le \tau + \epsilon$ for every $\epsilon > 0$.

**Notation.** We denote by $S(\tau)$ the set of numbers $x \in C_p$ having transcendence order $\le \tau$ and by $S(\tau^+)$ the set of numbers $x \in C_p$ having weak transcendence order $\le \tau^+$.

Finally, we say that a number $x \in C_p$ is of infinite order if it does not belong to $S(\tau)$ for all $\tau \in \mathbb{R}_+^*$.

**Remark.** By definition, an element $a \in C_p$ having transcendence order $\le \tau$ or weak transcendence order $\le \tau^+$ is transcendental over $Q_p$.
**Theorem 9.1.** Let $\tau \in ]0, +\infty[$. If $S(\tau) \neq \emptyset$ then $\tau \ge 1$.

**Proof.** Let $a \in \mathbb{C}_p$, $a \neq 0$ be transcendental over $\mathbb{Q}_p$ and have transcendence order $\le \tau$. We can find $b \in \Omega_p$ ($b \neq 0$) such that $|a - b|_p < 1$. Consider the minimal polynomial $Q$ of $b$ over $\mathbb{Q}_p$. Let $b_2, \dots, b_q$ be the conjugates of $b$ over $\mathbb{Q}_p$ and set $b_1 = b$. We note that by Corollary 6.6 all conjugates $b_j$ of $b$ over $\mathbb{Q}_p$ satisfy $|b_j|_p = |b|_p$.

Suppose first that $|a|_p \le 1$. Since $|b_j|_p = |b|_p = |a|_p \le 1$, all coefficients of $Q$ belong to $\mathbb{Z}_p$. Obviously, $Q$ is monic, hence $\|Q\| = 1$. By hypothesis, there exists $C_a \in ]0, +\infty[$ such that $\Psi_p(P(a)) \ge \log_p(\|P\|) - C_a(\deg(P))^{\tau}$, $\forall P \in \mathbb{Q}_p[x]$. Consequently, $-n\Psi_p(Q(a)) = -\Psi_p((Q(a))^n) \le C_a(n \deg(Q))^{\tau}$, $\forall n \in \mathbb{N}^*$. Since $Q(b) = 0$ and since, by Corollary 4.5, $Q$ is 1-Lipschitzian in $U$, we have $-\Psi_p(Q(a)) > 0$, and therefore, if $\tau < 1$, the inequality $-n\Psi_p(Q(a)) \le C_a(n \deg(Q))^{\tau} \forall n \in \mathbb{N}^*$ is impossible when $n$ tends to $+\infty$.

Suppose now $|a|_p > 1$. Set $Q(X) = \sum_{k=0}^q c_k X^k$. Since the $b_j$ satisfy $|b_j|_p = |a|_p$ ($1 \le j \le q$), we have $|c_k|_p \le (|a|_p)^{q-k}$ and particularly $|c_0|_p = \prod_{j=1}^q |b_j|_p = (|a|_p)^q$. Consequently, $\|Q\| = (|a|_p)^q$, and therefore, considering the sequence $(Q^n)_{n \in \mathbb{N}}$, for every $n \in \mathbb{N}^*$, we have

$$ (1) \quad -n\Psi_p(Q(a)) \le -nq\Psi_p(a) + C_a(nq)^{\tau}. $$

On the other hand, $Q(a) = Q(a) - Q(b) = (a-b) \sum_{k=1}^q c_k \sum_{j=0}^{k-1} a_j b^{k-j-1}$ and hence $|Q(a)|_p \le |a-b|_p (|a|_p)^{q-1}$. Consequently, we obtain $-n\Psi_p(a-b) - n(q-1)\Psi_p(a) \le \Psi_p(Q(a)$, and hence, by (1), $-n\Psi_p(a-b) - n(q-1)\Psi_p(a) \le -nq\Psi_p(a) + C_a(nq)^{\tau}$. Finally, $n(\Psi_p(a) - \Psi_p(a-b)) \le C_a(nq)^{\tau}$. Since $|a|_p > 1$ and $|a-b|_p < 1$, this inequality is impossible again when $n$ tends to $+\infty$, which ends the proof. $\square$

**Theorem 9.2.** There exists $b \in \mathbb{C}_p$ transcendental over $\mathbb{Q}_p$, of order $\le 1 + \epsilon$, for every $\epsilon > 0$.

**Proof.** Consider first a strictly decreasing sequence $(\epsilon_n)_{n \in \mathbb{N}}$ such that $\lim_{n \to +\infty} \epsilon_n = 0$ and $\lim_{n \to +\infty} \epsilon_n \log(n) = +\infty$.

We can always divide any polynomial $P \in \mathbb{Q}_p[x]$ by some $\lambda \in \mathbb{Q}_p$ such that $|\lambda|_p = \|P\|$ and hence we go back to the hypothesis $\|P\| = 1$. So, if we can find some $b \in \mathbb{C}_p$ and, for every $\omega > 0$, a constant $C(\omega) > 0$ and show that for every $P \in \mathbb{Q}[X]$ such that
$\|P\| = 1$, we have $-\log(|P(b)|_p) \le C(\omega)(\deg(P))^{1+\omega}$, Theorem 9.2 will be proven.

By induction, we can define a strictly increasing sequence $(r_n)_{n \in \mathbb{N}}$ of $\mathbb{Q}$ and a sequence $(a_n)_{n \in \mathbb{N}}$ of $\mathbb{C}_p$ with $r_n = \frac{u_n}{v_n}$, irreducible and $(v_n)_{n \in \mathbb{N}}$ a strictly increasing sequence of prime numbers satisfying further the following properties:

(i) $\lim_{n \to +\infty} r_n = +\infty$,

(ii) for every $n \in \mathbb{N}$, $n^{\epsilon_n} < r_n < (n+1)^{\epsilon_n}$,

(iii) $v_n > \prod_{j=1}^{n-1} v_j$,

(iv) $(a_n)^{v_n} = p^{u_n}$.

By construction, the sequence $(|a_n|_p)_{n \in \mathbb{N}}$ is strictly decreasing and tends to 0 and all terms belong to $U$. Set $b = \sum_{n=1}^{\infty} a_n$. Now, let us fix $\epsilon > 0$. We show that $b$ is transcendental over $\mathbb{Q}_p$ and has a transcendence order $\le 1 + \epsilon$.

Since the sequence $(\epsilon_n)$ tends to 0, we can find an integer $t(\epsilon)$ such that $\epsilon_m < \epsilon \ \forall m \ge t(\epsilon)$. Thus, as the first step, let us take $q \ge t(\epsilon)$ and let us find a constant $C(\epsilon) > 0$, not depending on $b$, such that for every $P \in \mathbb{Q}[X]$ satisfying $\|P\| = 1$ and $\deg(P) = q$, we have $-\log_p(|P(b)|_p) \le C(\epsilon)q^{1+\epsilon}$.

For each $n \in \mathbb{N}^*$, set $b_n = \sum_{m=1}^n a_m$. Since the sequence $(|a_m|_p)_{m \in \mathbb{N}}$ is strictly decreasing, we have $|b - b_n|_p = |a_{n+1}|_p$, and since $P$ is obviously 1-Lipschitzian in the disk $U$, we have $|P(b) - P(b_n)|_p \le |a_{n+1}|_p$, hence

$$ (1) \quad \log_p(|P(b) - P(b_n)|_p) \le \log_p(|a_{n+1}|_p) = -r_{n+1}. $$

Now, since the sequence $\epsilon_n \log_p(n)$ tends to $+\infty$, we can choose $n(q)$ such that $(n(q) + 1)^{\epsilon_n(q)+1} > (q+1)^{1+\epsilon}$. Then, by (1), we have

$$ (2) \quad \log_p(|P(b) - P(b_{n(q)})|_p) < \log_p(|a_{n(q)+1}|_p) = -(r_{n(q)+1}) < -(n(q)+1)^{\epsilon_n(q)+1} < -(q+1)^{1+\epsilon}. $$

We show the following inequality:

$$ (3) \quad -\log_p(|P(b_{n(q)})|_p) \le (q+1)^{1+\epsilon}. $$

Thus, suppose (3) is wrong. Set $h_q = \sum_{m=q}^{n(q)} a_m$. Then, $b_{n(q)} = b_{q-1} + h_q$. Now, developing $P$ at the point $b_{q-1}$, we have

$$ (4) \quad \log_p(|P(b_{n(q)})|_p) = \log_p\left(\left|\sum_{m=0}^q \frac{P^{(m)}(b_{q-1})}{m!} (h_q)^m\right|_p\right) < -(q+1)^{1+\epsilon}. $$