StrokeColor=black,TextRenderingMode=2,LineWidth=0.2pt

# On the ranks and implicit constant fields of valuations induced by pseudo
monotone sequences

Arpan Dutta  Department of Mathematics, IISER Mohali, Knowledge City, Sector
81, Manauli PO, SAS Nagar, Punjab, India, 140306<EMAIL_ADDRESS>
###### Abstract.

Given a valued field $(K,v)$ and a pseudo monotone sequence $E$ in $(K,v)$,
one has an induced valuation $v_{E}$ extending $v$ to $K(X)$. After fixing an
extension of $v_{E}$ to a fixed algebraic closure $\overline{K(X)}$ of $K(X)$,
we show that the implicit constant field of the extension $(K(X)|K,v_{E})$ is
simply the henselization of $(K,v)$. We consider the question: given a value
transcendental extension $w$ of $v$ to $K(X)$ and a pseudo monotone sequence
$E$ in $(K,v)$, under which precise conditions is $w$ induced by $E$? The dual
nature of pseudo convergent sequences of algebraic type and pseudo divergent
sequences is also explored. Further, we provide a complete description of the
various possibilities of the rank of the valuation $v_{E}$, provided that $v$
has finite rank.

###### Key words and phrases:

Valuation, minimal pairs, key polynomials, pseudo-Cauchy sequences, pseudo
monotone sequences, implicit constant fields

###### 2010 Mathematics Subject Classification:

12J20, 13A18, 12J25

##  1\. Introduction

A central problem in valuation theory is to study the set of all extensions of
a valuation $v$ from a field $K$ to a rational function field $K(X)$. On the
other hand, given such an extension $w$ of $v$ to $K(X)$, it is important to
give a complete description of the valuation $w$. Several different objects
have been introduced to tackle these interconnected problems. Minimal pairs of
definition were introduced by Alexandru, Popescu and Zaharescu [References,
References] to study residue transcendental extensions. The notion of key
polynomials was introduced by MacLane [References] which was later generalized
by Vaquié [References]. An alternative form of key polynomials have been
recently developed by Spivakovsky and Novacoski [References, References].
Ostrowski [References] introduced pseudo convergent sequences which were
succesfully used by Kaplansky [References] to answer maximality questions for
immediate extensions. Recently, this notion has been generalized to that of
pseudo monotone sequences which has found significant use in the works of
Peruginelli and Spirito [References]. The notion of implicit constant fields
has been introduced by Kuhlmann [References].

Our goal in this article is to study the extensions of a valued field $(K,v)$
induced by pseudo monotone sequences in $(K,v)$. In particular, given an
extension $v_{E}$ of $v$ to $K(X)$ induced by a pseudo monotone sequence $E$
in $(K,v)$ and an extension of $v_{E}$ to a fixed algebraic closure
$\overline{K(X)}$ of $K(X)$, we investigate the following:

  * •

the implicit constant field of the extension $(K(X)|K,v_{E})$,

  * •

the rank of the value group $v_{E}K(X)$.

It has recently been observed by Peruginelli and Spirito [References] that
every extension of a valuation $v$ from $K$ to $K(X)$ is induced by a pseudo
monotone sequence in $(K,v)$ if and only if the $v$-adic completion $\hat{K}$
is algebraically closed. As a consequence, our results hold for all extensions
$(K(X)|K,w)$ whenever $\hat{K}$ is algebraically closed.

Given a valued field extension $(K(X)|K,w)$ and an extension of $w$ to
$\overline{K(X)}$, the implicit constant field of the extension is defined to
be the relative algebraic closure of $K$ in the henselization of $K(X)$ and it
is denoted by $IC(K(X)|K,w)$. In particular, $IC(K(X)|K,w)$ is a separable-
algebraic extension of the henselization $K^{h}$ of $(K,v)$. It has been
observed in [References, Theorem 1.1 and Theorem 1.3] that explicit
computation of $IC(K(X)|K,w)$ can be quite difficult in general. In this
article we discover that

$IC(K(X)|K,v_{E})=K^{h}$

whenever $E$ is a pseudo monotone sequence in $(K,v)$ (see Theorem 6.2 and
Theorem 7.1).

The extension $(K(X)|K,v_{E})$ is value transcendental if and only if $E$ is
either pseudo convergent of algebraic type or pseudo divergent. We consider
the question:

######  Question 1.1.

Given a value transcendental extension $(K(X)|K,w)$ and a pseudo monotone
sequence $E$ in $(K,v)$, under which precise conditions is $w=v_{E}$?

Some related equivalences for this equality to hold are provided in Theorem
7.2 and Theorem 8.1. In particular, we observe the strong connections between
pairs of definition for $v_{E}$ over $K$ and the limits of $E$, which play a
crucial role throughout this paper. The duality of pseudo convergent sequences
of algebraic type and pseudo divergent sequences is made clear in Theorem
7.2$(iii)$ and Theorem 8.1$(iii)$. In the corollary to Theorem 7.2 we observe
that if $E$ is a Cauchy sequence and $(K(X)|K,w)$ is a value transcendental
extension such that $X$ is a limit of $E$, then $w=v_{E}$. Similarly, if $E$
diverges to infinity and $X$ is not a limit of $E$, then $w=v_{E}$.

We also provide a complete characterization of the rank of the value group
$v_{E}K(X)$, provided that $vK$ is of finite rank. It is well known that the
rank either remains same or increases by one. We give a constructive
description of the various possibilities (see Figures 1, 2, 3, 4). Since the
value group remains unchanged when $E$ is a pseudo convergent sequence of
transcendental type or a pseudo constant sequence, we only focus on the cases
when $E$ is pseudo convergent of algebraic type or pseudo divergent. The
duality observed in Theorem 7.2$(iii)$ and Theorem 8.1$(iii)$ comes to the
fore at this juncture.

After fixing an extension $\overline{v}$ of $v$ to $\overline{K}$, we
introduce the notion of supremum and infimum of a pseudo convergent sequence
of algebraic type or a pseudo divergent sequence, when $(K,v)$ has finite
rank. Thereafter, we observe in Theorem 9.3 that whenever $E$ is a Cauchy
sequence or a pseudo convergent sequence of algebraic type such that the
supremum of $E$ is contained in $\overline{v}\overline{K}$, we obtain
$\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1$. The analogous result holds
true when $E$ diverges to infinity or $E$ is pseudo divergent such that the
infimum of $E$ is contained in $\overline{v}\overline{K}$.

The paper is structured as follows: in Section 2 we give a quick tour of the
theories of minimal pairs of definition and key polynomials. The definitions
of pseudo monotone sequences and their limits are provided in Section 3. In
Section 4 we define valuations induced by pseudo monotone sequences and study
some of their basic properties. Section 5 deals with the extensions of $v_{E}$
to $\overline{K}(X)$. The definition of pure extensions is provided in Section
6, where we also observe that $(K(X)|K,v_{E})$ is a pure extension whenever
$E$ is not a pseudo convergent sequence without a limit in $K$. The remaining
case is considered in Section 7. Question 1.1 is also investigated in this
section, under the assumption that $E$ is pseudo convergent of algebraic type.
The extension $(K(X)|K,v_{E})$ when $E$ is pseudo divergent is studied in
greater detail in Section 8. Finally, we define the valuation theoretic
concept of rank and compute the rank of the valuation $v_{E}$ in Section 9.

## Acknowledgements

This work was supported by the Post-Doctoral Fellowship of the National Board
of Higher Mathematics, India. The author would like to thank Giulio
Peruginelli for directing his attention towards possible interrelation between
implicit constant fields and dominating degrees, and for pointing out a
mistake in an earlier version of this paper.

## Notations

A valued field $(K,v)$ is a field $K$ endowed with a valuation $v$. The value
group of $K$ is denoted by $vK$ and the residue field by $Kv$. The value of an
element $a\in K$ is denoted by $va$. By an extension of valued fields
$(L|K,v)$ we will mean that $L|K$ is an extension of fields, $v$ is a
valuation on $L$ and $K$ is equipped with the restricted valuation. If $L$ and
$K$ are subfields of a larger valued field $(F,v)$, then we will also mean
that both $L$ and $K$ are equipped with the restricted valuations.

##  2\. Minimal pairs and key polynomials

A valued field extension $(K(X)|K,v)$ satisfies the famous Abhyankar
inequality:

$\mbox{\rm rat rk}\,\frac{vK(X)}{vK}+\mbox{\rm trdeg}\,[K(X)v:Kv]\leq 1,$

where $\mbox{\rm rat rk}\,vK(X)/vK$ is the ${\mathbb{Q}}$-dimension of the
divisible hull ${\mathbb{Q}}\otimes_{{\mathbb{Z}}}(vK(X)/vK)$. This inequality
is a consequence of [References, Chapter VI, §10.3, Theorem 1]. The extension
$(K(X)|K,v)$ is said to be valuation transcendental if equality holds in the
above inequality. The extension is said to be value transcendental if
$\mbox{\rm rat rk}\,vK(X)/vK=1$ and is said to be residue transcendental if
$\mbox{\rm trdeg}\,[K(X)v:Kv]=1$.

Take a polynomial $f(X)\in K[X]$ and $a\in K$. Write
$f(X)=\sum_{i=0}^{n}c_{i}(X-a)^{i}$ where $c_{i}\in K$. Take an element
$\alpha$ in some ordered abelian group extending $vK$. We define the map
$v_{a,\alpha}:K[X]\longrightarrow vK+{\mathbb{Z}}\alpha$ by setting
$v_{a,\alpha}f:=\min\\{vc_{i}+i\alpha\\}$ and extend it canonically to $K(X)$.
Then $v_{a,\alpha}$ is a valuation transcendental extension of $v$ from $K$ to
$K(X)$ [References, Lemma 3.10]. Further, $v_{a,\alpha}$ is value
transcendental if and only if
$\alpha\notin{\mathbb{Q}}\otimes_{{\mathbb{Z}}}vK$ [References, Lemma 3.10].
The valuation $v_{a,\alpha}$ is said to be a monomial valuation. It is known
that the extension $(K(X)|K,v)$ is value (residue) transcendental if and only
if $(\overline{K}(X)|\overline{K},\overline{v})$ is also value (residue)
transcendental [References, Lemma 3.3], where $\overline{v}$ is some extension
of $v$ to $\overline{K}(X)$. Further,
$(\overline{K}(X)|\overline{K},\overline{v})$ is valuation transcendental if
and only if the valuation $\overline{v}$ on $\overline{K}(X)$ is a monomial
valuation [References, Theorem 3.11]. If we have
$\overline{v}=\overline{v}_{a,\alpha}$ on $\overline{K}(X)$ for some
$a\in\overline{K}$ and $\alpha\in\overline{v}\overline{K}(X)$, then we say
that $(a,\alpha)$ is a pair of definition for $v$ over $K$. It can be easily
checked that $(a,\alpha)$ is a pair of definition for $v$ over $K$ if and only
if

$\alpha=\overline{v}(X-a)=\max\overline{v}(X-\overline{K}),$

where $\overline{v}(X-\overline{K}):=\\{\overline{v}(X-c)\mid
c\in\overline{K}\\}$ (see [References, Section 2] for a proof). A valuation
transcendental extension can have multiple pairs of definition. The following
result, proved in [References, Proposition 3], gives us a relation between the
different pairs of definition:

######  Lemma 2.1.

Take $a,a^{\prime}$ in $\overline{K}$ and $\alpha,\alpha^{\prime}$ in some
ordered abelian group containing $\overline{v}\overline{K}$. Then,

$\overline{v}_{a,\alpha}=\overline{v}_{a^{\prime},\alpha^{\prime}}\text{ if
and only if }\alpha=\alpha^{\prime}\text{ and
}\overline{v}(a-a^{\prime})\geq\alpha.$

######  Definition 2.2.

A pair $(a,\alpha)\in\overline{K}\times\overline{v}\overline{K}(X)$ is said to
be a minimal pair of definition for $v$ over $K$ if the following conditions
are satisifed:

(MP1) $\alpha=\overline{v}(X-a)=\max\overline{v}(X-\overline{K})$,

(MP2) $\overline{v}(a-b)\geq\alpha\Longrightarrow[K(b):K]\geq[K(a):K]$ for all
$b\in\overline{K}$.

Take a polynomial $f(X)\in K[X]$. We define,

$\delta(f):=\max\\{\overline{v}(X-c)\mid c\text{ is a root of }f\\}.$

It has been observed in [References] that $\delta(f)$ is independent of the
choice of the extension of $v$ to $\overline{K}(X)$. A monic polynomial
$Q(X)\in K[X]$ is said to be a key polynomial for $v$ over $K$ if
$\delta(f)<\delta(Q)$ for all $f(X)\in K[X]$ such that $\deg f<\deg Q$.
Observe that a monic linear polynomial is a key polynomial by definition.

Given a polynomial $f(X)\in K[X]$ and a monic polynomial $Q(X)\in K[X]$, we
have a unique expansion $f=\sum_{i=0}^{n}f_{i}Q^{i}$, where $f_{i}(X)\in K[X]$
with $\deg f_{i}<\deg Q$. We define a map $v_{Q}:K[X]\longrightarrow vK(X)$ by
setting $v_{Q}(f):=\min\\{vf_{i}+ivQ\\}$, and extend it canonically to $K(X)$.
It is known that $v_{Q}$ is a valuation on $K(X)$ whenever $Q$ is a key
polynomial for $v$ over $K$ [References, Proposition 2.6]. A family $\Lambda$
of key polynomials for $v$ over $K$ is said to form a complete sequence of key
polynomials for $v$ over $K$ if the following conditions are satisfied:

(CSKP1) $\delta(Q)\neq\delta(Q^{\prime})$ for $Q,Q^{\prime}\in\Lambda$ with
$Q\neq Q^{\prime}$,

(CSKP2) $\Lambda$ is well ordered with respect to the ordering given by
$Q<Q^{\prime}$ if and only if $\delta(Q)<\delta(Q^{\prime})$,

(CSKP3) for any $f(X)\in K[X]$, there exists some $Q\in\Lambda$ such that
$vf=v_{Q}f$.

######  Lemma 2.3.

Let $(K(y)|K,v)$ be an extension of valued fields. Assume that $\alpha\in
v(y-K)$ such that $\alpha\notin vK$. Then $\alpha=\max v(y-K)$.

###### Proof.

Take $a\in K$ such that $\alpha:=v(y-a)\notin vK$. Suppose that there exists
$b\in K$ such that $v(y-b)>v(y-a)=\alpha$. It then follows from the triangle
inequality that $\alpha=v(a-b)\in vK$, which contradicts our assumption. ∎

##  3\. Pseudo monotone sequences

For the remainder of this article, we will assume that $(K,v)$ is a valued
field and $E:=\\{z_{\nu}\\}_{\nu<\lambda}$ is a sequence in $K$ where
$\lambda$ is some limit ordinal.

######  Definition 3.1.

The sequence $E$ is said to be a pseudo convergent sequence (pcs) if

$v(z_{\nu_{1}}-z_{\nu_{2}})<v(z_{\nu_{2}}-z_{\nu_{3}})\text{ for all
}\nu_{1}<\nu_{2}<\nu_{3}<\lambda.$

Observe that $v(z_{\nu}-z_{\mu})=v(z_{\nu}-z_{\nu+1})$ for all $\mu>\nu$. We
set $\delta_{\nu}:=v(z_{\nu}-z_{\nu+1})$. By definition,
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is a strictly increasing sequence.

######  Definition 3.2.

The sequence $E$ is said to be pseudo divergent sequence (pds) if

$v(z_{\nu_{1}}-z_{\nu_{2}})>v(z_{\nu_{2}}-z_{\nu_{3}})\text{ for all
}\nu_{1}<\nu_{2}<\nu_{3}<\lambda.$

Observe that $v(z_{\nu}-z_{\mu})=v(z_{\nu}-z_{\mu^{\prime}})$ for all
$\mu^{\prime},\mu<\nu$. We set $\delta_{\nu}:=v(z_{\nu}-z_{\mu})\text{ for all
}\mu<\nu$. By definition, $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is a strictly
decreasing sequence.

######  Definition 3.3.

The sequence $E$ is said to be a pseudo constant sequence (pcts) if

$v(z_{\nu}-z_{\mu})=v(z_{\nu^{\prime}}-z_{\mu^{\prime}})\text{ for all
}\nu\neq\mu,\,\nu^{\prime}\neq\mu^{\prime}.$

We define $\delta:=v(z_{\nu}-z_{\mu})\text{ for all }\nu\neq\mu$. If we are
making general statements about the sequence $E$ without differentiating
between the cases, we will write $\delta_{\nu}$ to mean $\delta$ in the pcts
case.

######  Definition 3.4.

A pcs $E$ is said to be a Cauchy sequence if
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is cofinal in $vK$. A pds $E$ will be said
to be diverging to infinity if $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is coinitial
in $vK$.

######  Definition 3.5.

The sequence $E$ is said to be pseudo monotone sequence (pms) if is either a
pcs, or pds, or pcts.

######  Definition 3.6.

An element $y\in K$ is said to be a limit of $E$ if

$v(y-z_{\nu})=\delta_{\nu}\text{ for all }\nu\text{ sufficiently large}.$

Hence $y$ is a limit if the sequence

$\\{v(y-z_{\nu})\\}_{\nu<\lambda}\text{ is }\begin{cases}\text{ultimately
strictly increasing,}&\quad\text{when }E\text{ is a pcs,}\\\ \text{ultimately
strictly decreasing,}&\quad\text{when }E\text{ is a pds,}\\\ \text{ultimately
constant,}&\quad\text{when }E\text{ is a pcts.}\end{cases}$

Observe that a limit of $E$ is not necessarily uniquely determined. Here are
some straightforward observations.

######  Remark 3.7.

Assume that $y$ is a limit of $E$. If $E$ is a pcs, then
$\\{v(y-z_{\nu})\\}_{\nu<\lambda}$ is strictly increasing. If $E$ is a pcts,
then $v(y-z_{\nu})=\delta$ for all but at most element of $E$. Both of these
are consequences of the triangle inequality. Further, every element of $E$ is
a limit of $E$ when $E$ is a pds or pcts. When $E$ is a pcs, then no element
of $E$ is a limit of $E$.

######  Lemma 3.8.

Let $(L|K,v)$ be an extension of valued fields and $E$ a pms in $(K,v)$.
Assume that $y$ is a limit of $E$. Take any $b\in L$. Then $b$ is also a limit
of $E$ if and only if $v(y-b)\geq\delta_{\nu}$ for all $\nu$ sufficiently
large.

###### Proof.

The fact that $y$ is a limit of $E$ implies that $v(y-z_{\nu})=\delta_{\nu}$
for all $\nu$ sufficiently large. By definition, $b$ is also a limit of $E$ if
and only if $v(b-z_{\nu})=\delta_{\nu}=v(y-z_{\nu})$ for all $\nu$
sufficiently large. The result now follows from the ultrametric inequality. ∎

######  Lemma 3.9.

Let $E$ be a pms in $(K,v)$ and take $y\in K$. Then either $y$ is a limit of
$E$ or the sequence $S:=\\{v(y-z_{\nu})\\}_{\mu<\lambda}$ is ultimately
constant.

###### Proof.

If $E$ is a pcts, then the sequence $S$ is ultimately constant either way.

We now assume that $E$ is a pds and $y$ is not a limit. Suppose that $S$ is
not ultimately constant. If $S$ is ultimately increasing, then we can find
$\nu_{1}<\nu_{2}<\nu_{3}$ such that

$v(y-z_{\nu_{1}})<v(y-z_{\nu_{2}})<v(y-z_{\nu_{3}}).$

By the triangle inequality we obtain

$\delta_{\nu_{2}}=v(y-z_{\nu_{1}})<v(y-z_{\nu_{2}})=\delta_{\nu_{3}}$

which contradicts the fact that the sequence
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is strictly decreasing. If $S$ is
ultimately decreasing, then the assumption that $y$ is not a limit implies
that $S$ is not ultimately strictly decreasing. So we obtain
$\nu_{1}<\nu_{2}<\nu_{3}$ such that

$v(y-z_{\nu_{1}})>v(y-z_{\nu_{2}})=v(y-z_{\nu_{3}}).$

It now follows from the triangle inequality that
$\delta_{\nu_{2}}=\delta_{\nu_{3}}$ which is again a contradiction. Thus $S$
is neither ultimately increasing nor ultimately decreasing. Now take any
$\nu<\lambda$. The fact that $S$ is not ultimately increasing implies that
there exists some $\mu_{1}>\nu$ such that $v(y-z_{\nu})>v(y-z_{\mu_{1}})$. The
fact that $S$ is not ultimately decreasing implies that there exists
$\mu_{2}>\mu_{1}$ such that $v(y-z_{\mu_{1}})<v(y-z_{\mu_{2}})$. Hence,

$v(y-z_{\nu})>v(y-z_{\mu_{1}})<v(y-z_{\mu_{2}}).$

We now obtain from the triangle inequality that
$\delta_{\mu_{1}}=v(y-z_{\mu_{1}})=\delta_{\mu_{2}}$ which again yields a
contradiction. Thus $S$ is ultimately constant.

We now assume that $E$ is a pcs and $y$ is not a limit. Suppose that $S$ is
not ultimately constant. If $S$ is ultimately increasing, then the assumption
that $y$ is not a limit implies that $S$ is not ultimately strictly
increasing. So we have a relation

$v(y-z_{\nu_{1}})=v(y-z_{\nu_{2}})<v(y-z_{\nu_{3}}).$

But this yields $\delta_{\nu_{1}}=\delta_{\nu_{2}}$ which is a contradiction.
If $S$ is ultimately decreasing, then we have a relation

$v(y-z_{\nu_{1}})>v(y-z_{\nu_{2}})>v(y-z_{\nu_{3}})$

which yields
$\delta_{\nu_{1}}=v(y-z_{\nu_{2}})>v(y-z_{\nu_{3}})=\delta_{\nu_{2}}$. But
this contradicts the fact that $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is strictly
increasing. Hence $S$ is neither ultimately increasing nor ultimately
decreasing. Thus we have a relation

$v(y-z_{\nu})>v(y-z_{\mu_{1}})<v(y-z_{\mu_{2}})$

where $\nu<\mu_{1}<\mu_{2}$. The triangle inequality again leads to the
contradiction $\delta_{\nu}=v(y-z_{\mu_{1}})=\delta_{\mu_{1}}$. Hence $S$ is
ultimately constant. ∎

Fix an extension $\overline{v}$ of $v$ to $\overline{K}$. Take a polynomial
$f(X)\in K[X]$. Write

$f(X)=c(X-a_{1})\dotsc(X-a_{j})\dotsc(X-a_{n}),$

where $c\in K$ and $a_{1},\dotsc,a_{j}$ are all the root-limits of $(f,E)$,
that is, they are all the roots of $f$ which are also limits of $E$. It
follows from Lemma 3.9 that $\\{\overline{v}(z_{\nu}-a_{i})\\}_{\nu<\lambda}$
is ultimately constant for all $i\geq j+1$. We denote the ultimate constant
value of $\\{\overline{v}(z_{\nu}-a_{i})\\}_{\nu<\lambda}$ by $\beta_{i}$.
Then,

$vf(z_{\nu})=j\delta_{\nu}+\beta\text{ for all }\nu\text{ sufficiently
large},$

where $\beta:=vc+\beta_{j+1}+\dotsc+\beta_{n}$. The fact that
$vf(z_{\nu}),\delta_{\nu}\in vK$ implies that

$\beta\in vK.$

The above observation naturally extends to rational functions. Take
$\varphi(X)\in K(X)$. Then there exist $d\in{\mathbb{Z}}$ and $\beta\in vK$
such that

$v\varphi(z_{\nu})=d\delta_{\nu}+\beta\text{ for all }\nu\text{ sufficiently
large}.$

The number $d$ is referred to as the dominating degree of $\varphi$ with
respect to $E$ in [References] and is denoted by
$\deg\mathrm{dom}\,_{E}(\varphi)$. Observe that if $\varphi=f/g$ where $f,g\in
K[X]$, then

$\deg\mathrm{dom}\,_{E}(\varphi)=d=\\#\\{\text{root-limits of
}(f,E)\\}-\\#\\{\text{root-limits of }(g,E)\\}.$

Finally, observe that
$v\varphi(z_{\nu})-v\varphi(z_{\nu+1})=d(\delta_{\nu}-\delta_{\nu+1})$ for all
$\nu$ sufficiently large. Since $v\varphi(z_{\nu})-v\varphi(z_{\nu+1})$ and
$\delta_{\nu}-\delta_{\nu+1}$ are solely dependent on $v$, we conclude that

$\text{the dominating degree $d$ is independent of the choice of the extension
of $v$ to }\overline{K}.$

If $E$ is a pcs or pds, then $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is either
strictly increasing or strictly decreasing. The following result is now
straightfoward.

######  Proposition 3.10.

Take $\varphi(X)\in K(X)$. Then $\\{v\varphi(z_{\nu})\\}_{\nu<\lambda}$ is
ultimately constant if and only if either $E$ is a pcts, or $E$ is a pcs or
pds such that $\deg\mathrm{dom}\,_{E}(\varphi)=0$.

It follows from our preceding discussions that when $E$ is a pcs, the sequence
$\\{vf(z_{\nu})\\}_{\nu<\lambda}$ is either ultimately constant or ultimately
strictly increasing, where $f(X)\in K[X]$. If the sequence is ultimately
constant for all polynomials $f$ over $K$, then we say that $E$ is a pcs of
transcendental type. In this case $E$ does not admit any algebraic limits.
Otherwise, $E$ is said to be a pcs of algebraic type. In this case, we can
find a polynomial $Q(X)\in K[X]$ of minimal degree such that
$\\{vQ(z_{\nu})\\}_{\nu<\lambda}$ is ultimately strictly increasing. Without
any loss of generality, we can choose $Q$ to be monic. It can be observed that
$Q(X)$ is irreducible over $K$. The polynomial $Q(X)$ is said to be an
associated minimal polynomial for $E$.

##  4\. Valuations induced by pseudo monotone sequences

Let $E:=\\{z_{\nu}\\}_{\nu<\lambda}$ be a pms in $(K,v)$. Define,

$\displaystyle V_{E}$ $\displaystyle:=\\{\varphi(X)\in K(X)\mid
v\varphi(z_{\nu})\geq 0\text{ for all }\nu\text{ sufficiently large}\\},$
$\displaystyle M_{E}$ $\displaystyle:=\\{\varphi(X)\in K(X)\mid
v\varphi(z_{\nu})>0\text{ for all }\nu\text{ sufficiently large}\\}.$

Then $(V_{E},M_{E})$ is a valuation domain of $K(X)$ [References, Theorem
3.4]. We denote the corresponding valuation by $v_{E}$. It follows from the
definition that $(V_{E},M_{E})$ dominates $(V,M)$, where $(V,M)$ is the
valuation domain of the valuation $v$. Hence $v_{E}$ is an extension of $v$ to
$K(X)$.

######  Proposition 4.1.

Take $\varphi(X)\in K(X)$. Then $\\{v\varphi(z_{\nu})\\}_{\nu<\lambda}$ is
ultimately constant if and only if $v_{E}(\varphi)\in vK$. Further, in this
case $v_{E}(\varphi)=\beta$ where $v\varphi(z_{\nu})=\beta$ for all $\nu$
sufficiently large.

###### Proof.

We first assume that $\\{v\varphi(z_{\nu})\\}_{\nu<\lambda}$ is ultimately
constant and write the constant value as $\beta$. Take $a\in K$ such that
$va=\beta$. Write $\varphi^{\prime}(X):=\frac{1}{a}\varphi(X)$. Then
$v\varphi^{\prime}(z_{\nu})=v(1/a)+v\varphi(z_{\nu})$. Thus
$\\{v\varphi^{\prime}(z_{\nu})\\}_{\nu<\lambda}$ is also ultimately constant
attaining the value $\beta-va=0$ ultimately. So without any loss of generality
we can assume that $\beta=0$, that is, $v\varphi(z_{\nu})=0$ for all $\nu$
sufficiently large. It now follows from the definition that $\varphi\in
V_{E}\setminus M_{E}$, that is, $v_{E}(\varphi)=0$.

We now assume that $v_{E}(\varphi)\in vK$. Take $a\in K$ such that
$va=v_{E}(\varphi)$. Write $\varphi^{\prime}(X):=\frac{1}{a}\varphi(X)$. Then
$v_{E}(\varphi^{\prime})=0$, that is, $\varphi^{\prime}\in V_{E}\setminus
M_{E}$. It follows from the definition that $v(\varphi^{\prime}(z_{\nu}))=0$
for all $\nu$ sufficiently large. Consequently, $v\varphi(z_{\nu})=va$ for all
$\nu$ sufficiently large and hence $\\{v\varphi(z_{\nu})\\}_{\nu<\lambda}$ is
ultimately constant. ∎

Recall that $v\varphi(z_{\nu})=d\delta_{\nu}+\beta$ for all $\nu$ sufficiently
large, where $\beta\in vK$ and $d=\deg\mathrm{dom}\,_{E}(\varphi)$. If $E$ is
a pcts, then $\delta_{\nu}=\delta$ for all $\nu$ and hence
$v_{E}(\varphi)=d\delta+\beta\in vK$. If $E$ is a pcs or pds then
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is either strictly increasing or strictly
decreasing. In this case $\\{v\varphi(z_{\nu})\\}_{\nu<\lambda}$ is ultimately
constant if and only if $d=0$. It follows that

$v_{E}(\varphi)\in vK\text{ if and only if }E\text{ is a pcts, or $E$ is a pcs
or pds and }\deg\mathrm{dom}\,_{E}(\varphi)=0.$

Conversely, take a polynomial $f(X)\in K[X]$. Then,

$v_{E}(f)\notin vK\text{ if and only if $E$ is a pcs or pds and }(f,E)\text{
admits root-limits}.$

Now assume that $v_{E}(f)\notin vK$. Suppose that $v_{E}(f)$ is a torsion
element modulo $vK$, that is, $nv_{E}(f)\in vK$ for some $n\in{\mathbb{N}}$.
Then $v_{E}(f^{n})\in vK$ and consequently, no root of $f^{n}$ is a limit of
$E$. It follows that no root of $f$ is a limit of $E$ which contradicts our
assumption. Thus,

$v_{E}(f)\notin vK\text{ if and only if
}v_{E}(f)\notin{\mathbb{Q}}\otimes_{{\mathbb{Z}}}vK.$

######  Remark 4.2.

($E$ is a pcs) Take any $\nu<\lambda$ and define
$\varphi(X):=\frac{X-z_{\nu+1}}{X-z_{\nu}}$. Observe that
$v\varphi(z_{\mu})=\delta_{\nu+1}-\delta_{\nu}>0$ for all $\mu>\nu+1$. By
definition, $v_{E}(\varphi)>0$ and hence
$v_{E}(X-z_{\nu+1})>v_{E}(X-z_{\nu})$. Thus the sequence
$\\{v_{E}(X-z_{\nu})\\}_{\nu<\lambda}$ is strictly increasing and hence $X$ is
a limit of $E$.

If $E$ is a pcs of transcendental type, then it follows from [References,
Theorem 2] that $(K(X)|K,v_{E})$ is an immediate extension.

######  Remark 4.3.

($E$ is a pds) Recall that every element of $E$ is a limit of $E$. It follows
from the preceding discussions that
$v_{E}(X-z_{\nu})\notin{\mathbb{Q}}\otimes_{{\mathbb{Z}}}vK$. Further,
$v_{E}(X-z_{\nu})\in v_{E}(X-K)$. It follows from Lemma 2.3 that

$v_{E}(X-z_{\nu})=\max
v_{E}(X-K)\notin{\mathbb{Q}}\otimes_{{\mathbb{Z}}}vK\text{ for all
}\nu<\lambda.$

As a consequence we obtain that $\\{v_{E}(X-z_{\nu})\\}_{\nu<\lambda}$ is a
constant sequence. In particular, $X$ is not a limit of $E$. Suppose that
$v_{E}(X-z_{\mu})\geq\delta_{\nu}$ for some $\mu,\nu<\lambda$. The fact that
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is strictly decreasing implies that
$v_{E}(X-z_{\mu})>\delta_{\nu}$ for all $\nu$ sufficiently large. In light of
Lemma 3.8, the observation that $z_{\mu}$ is a limit of $E$ then implies that
$X$ is a limit of $E$, contradicting our observations above. It follows that

$v_{E}(X-z_{\nu})<\delta_{\nu}\text{ for all }\nu<\lambda.$

######  Remark 4.4.

($E$ is a pcts) Observe that $v(z_{\mu}-z_{\nu})=\delta$ for all $\mu\neq\nu$.
It follows from Proposition 4.1 that $v_{E}(X-z_{\nu})=\delta$ for all
$\nu<\lambda$. As a consequence, we obtain that $X$ is a limit of $E$.

##  5\. Extending $v_{E}$ to $\overline{K}(X)$

Fix an extension $\overline{v}$ of $v$ to $\overline{K}$. Observe that $E$ is
a pms in $(\overline{K},\overline{v})$. Consider the valuation
$\overline{v}_{E}$ extending $\overline{v}$ to $\overline{K}(X)$ with the
corresponding valuation domain given as follows:

$\displaystyle\overline{V}_{E}$
$\displaystyle:=\\{\varphi(X)\in\overline{K}(X)\mid\overline{v}\varphi(z_{\nu})\geq
0\text{ for all }\nu\text{ sufficiently large}\\},$
$\displaystyle\overline{M}_{E}$
$\displaystyle:=\\{\varphi(X)\in\overline{K}(X)\mid\overline{v}\varphi(z_{\nu})>0\text{
for all }\nu\text{ sufficiently large}\\}.$

We can directly observe that $(\overline{V}_{E},\overline{M}_{E})$ dominates
$(V_{E},M_{E})$. Hence $\overline{v}_{E}$ is an extension of $v_{E}$ to
$\overline{K}(X)$ [References, Ch VI, §6, Lemma 1].

######  Proposition 5.1.

Let $E$ be a pms in $(K,v)$ and assume that $a\in\overline{K}$ is a limit of
$E$. Set $\alpha:=\overline{v}_{E}(X-a)$. Then

$\overline{v}_{E}=\overline{v}_{a,\alpha}.$

###### Proof.

We first assume that $E$ is a pcs or pds. The assumption that $a$ is a limit
of $E$ implies that $\overline{v}_{E}(X-a)\notin\overline{v}\overline{K}$.
Consider a polynomial $f(X)\in\overline{K}[X]$ and write
$f(X)=\sum_{i=0}^{n}c_{i}(X-a)^{i}$ where $c_{i}\in\overline{K}$. We observe
that all the monomials have distinct values. It follows that
$\overline{v}_{E}f=\min\\{\overline{v}c_{i}+i\alpha\\}$ and hence
$\overline{v}_{E}=\overline{v}_{a,\alpha}$.

We now assume that $E$ is a pcts. Observe that the fact that $E$ is a pcts
implies that $\alpha\in\overline{v}\overline{K}$. It follows from Proposition
4.1 that $\overline{v}_{E}(X-a)=\overline{v}(z_{\nu}-a)$ for all $\nu$
sufficiently large. The assumption that $a$ is a limit of $E$ then implies
that

$\alpha=\overline{v}_{E}(X-a)=\delta.$

Take any $b\in\overline{K}$. Then
$\overline{v}_{E}(X-b)=\overline{v}_{E}(X-a+a-b)\geq\min\\{\delta,\overline{v}(a-b)\\}=\overline{v}_{a,\delta}(X-b)$.
If $\delta\neq\overline{v}(a-b)$, then it follows from the triangle inequality
that $\overline{v}_{E}(X-b)=\overline{v}_{a,\delta}(X-b)$. Now assume that
$\overline{v}(a-b)=\delta$. It follows from Lemma 3.8 that $b$ is also a limit
of $E$. By definition, $\overline{v}(b-z_{\nu})=\delta$ for all $\nu$
sufficiently large. Consequently, it follows from Proposition 4.1 that
$\overline{v}_{E}(X-b)=\delta$. Thus,

$\overline{v}_{E}(X-b)=\overline{v}_{a,\delta}(X-b)\text{ for all
}b\in\overline{K}.$

Since any polynomial over $\overline{K}$ splits into linear factors over
$\overline{K}$, it follows that
$\overline{v}_{E}=\overline{v}_{a,\delta}=\overline{v}_{a,\alpha}$. ∎

######  Corollary 5.2.

Assume that $E$ is a pms in $(K,v)$ with a limit $a\in K$. Set
$\alpha:=v_{E}(X-a)$. Then $v_{E}=v_{a,\alpha}$.

######  Remark 5.3.

Assume that $E$ is a pcs or pds in $(K,v)$ and $a\neq b$ are elements in
$\overline{K}$ which are limits of $E$. It follows that
$\alpha:=\overline{v}_{E}(X-a)\notin\overline{v}\overline{K}$ and
$\overline{v}_{E}(X-b)\notin\overline{v}\overline{K}$. In light of Lemma 2.3
we conclude that
$\alpha=\overline{v}_{E}(X-a)=\max\overline{v}_{E}(X-\overline{K})=\overline{v}_{E}(X-b)$.
Consequently, $\overline{v}(a-b)>\alpha$.

We will write $\alpha>\overline{v}\overline{K}$ if $\alpha>\beta$ for all
$\beta\in\overline{v}\overline{K}$. Analogously we will write
$\alpha<\overline{v}\overline{K}$ to mean that $\alpha<\beta$ for all
$\beta\in\overline{v}\overline{K}$.

######  Proposition 5.4.

Let $E$ be a pms in $(K,v)$ and assume that $a\in\overline{K}$ is a limit of
$E$. Set $\alpha:=\overline{v}_{E}(X-a)$. Then

$\alpha>\overline{v}\overline{K}\text{ if and only if }E\text{ is a Cauchy
sequence of algebraic type}.$

In this case $a$ is purely inseparable over $K^{h}$. On the other hand,

$\alpha<\overline{v}\overline{K}\text{ if and only if }E\text{ diverges to
infinity}.$

###### Proof.

We first assume that $\alpha>\overline{v}\overline{K}$. The fact that
$\alpha\notin\overline{v}\overline{K}$ implies that $E$ is a pcs or pds.
Suppose that there exists some $b\in\overline{K}\setminus\\{a\\}$ such that
$b$ is also a limit of $E$. It follows from Remark 5.3 that
$\overline{v}(a-b)>\alpha$. However, this contradicts the assumption that
$\alpha>\overline{v}\overline{K}$. So $a$ is the unique algebraic limit of
$E$. As a consequence, $E$ is a pcs of algebraic type. Now suppose that
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is not cofinal in
$\overline{v}\overline{K}$. Then there exists some $c\in\overline{K}$ such
that $\overline{v}(a-c)>\delta_{\nu}$ for all $\nu<\lambda$. But this implies
that $c$ is a limit of $E$ which again yields a contradiction. Thus
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is cofinal in $\overline{v}\overline{K}$.
The fact that $vK$ is cofinal in $\overline{v}\overline{K}$ then implies that
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is cofinal in $vK$. Hence $E$ is a Cauchy
sequence in $(K,v)$ of algebraic type.

We now assume that $E$ is a Cauchy sequence of algebraic type, that is,
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is cofinal in $\overline{v}\overline{K}$.
Recall that $X$ is a limit of $E$. The fact that $a$ is also a limit of $E$
then implies that $\alpha:=\overline{v}_{E}(X-a)>\delta_{\nu}$ for all
$\nu<\lambda$. It follows that $\alpha>\overline{v}\overline{K}$.

We now assume that $E$ is a Cauchy sequence of algebraic type with the unique
algebraic limit $a\in\overline{K}$. Suppose that $a$ is not purely inseparable
over $K^{h}$. Then there exists some conjugate $\sigma a\neq a$ of $a$ over
$K^{h}$ where $\sigma\in\mathrm{Gal}\,(\overline{K}|K^{h})$. The fact that
$(K^{h},\overline{v})$ is henselian implies that
$\overline{v}\circ\sigma=\overline{v}$ on $\overline{K}$. It follows that

$\overline{v}(\sigma
a-z_{\nu})=(\overline{v}\circ\sigma)(a-z_{\nu})=\overline{v}(a-z_{\nu})=\delta_{\nu}\text{
for all }\nu<\lambda.$

Consequently, $\sigma a$ is also a limit of $E$, which contradicts the
uniqueness of $a$. Hence $a$ is purely inseparable over $K^{h}$.

We now assume that $\alpha<\overline{v}\overline{K}$. Hence either $E$ is a
pcs or $E$ is a pds. Take any $c\in\overline{K}$. The triangle inequality
implies that $\overline{v}_{E}(X-c)=\alpha\notin\overline{v}\overline{K}$ and
hence $c$ is a limit of $E$. In particular, every element of $E$ is a limit of
$E$. Hence $E$ is a pds. Further, we take any
$\beta\in\overline{v}\overline{K}$ and $b\in\overline{K}$ such that
$\overline{v}(a-b)=\beta$. The observation that $b$ is a limit of $E$ implies
that $\beta\geq\delta_{\nu}$ for all $\nu$ sufficiently large. Since this
condition holds for all $\beta\in\overline{v}\overline{K}$ we observe that
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is coinitial in $\overline{v}\overline{K}$,
that is, $E$ diverges to infinity.

Conversely, we assume that $E$ is a pds such that
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is coinitial in $\overline{v}\overline{K}$.
Suppose that $\alpha\geq\delta_{\nu}$ for some and hence all sufficiently
large $\nu$. It then follows from Lemma 3.8 that $X$ is a limit of $E$ which
yields a contradiction. Hence $\alpha<\delta_{\nu}$ for all $\nu<\lambda$,
that is, $\alpha<\overline{v}\overline{K}$. ∎

##  6\. Pure extensions

An extension $(K(X)|K,v)$ is said to be a pure extension if it satisfies one
of the following conditions:

(PE1) $v(X-a)$ is not a torsion element modulo $vK$ for some $a\in K$,

(PE2) $vb(X-a)=0$ and $b(X-a)v$ is transcendental over $Kv$ for some $a,b\in
K$,

(PE3) $X$ is the limit of some pcs $\\{z_{\nu}\\}_{\nu<\lambda}$ of
transcendental type in $(K,v)$.

Pure extensions are very well understood in terms of minimal pairs, key
polynomials and implicit constant fields.

######  Proposition 6.1.

Let $(\overline{K(X)}|K,v)$ be an extension of valued fields such that
$(K(X)|K,v)$ is pure. Then the following statements hold true:

(1) $IC(K(X)|K,v)=K^{h}$,

(2) if the extension $(K(X)|K,v)$ satisfies (PE1) or (PE2), then $(a,v(X-a))$
is a minimal pair of definition for $v$ over $K$ and $\\{X-a\\}$ forms a
complete sequence of key polynomials for $v$ over $K$,

(3) if the condition (PE3) is satisfied, then $(K(X)|K,v)$ is immediate and
$\\{X-z_{\nu}\\}_{\nu<\lambda}$ forms a complete sequence of key polynomials
for $v$ over $K$.

###### Proof.

The first assertion is proved in [References, Lemma 3.7].

We now assume that $v(X-a)$ is not a torsion element modulo $vK$, that is,
$v(X-a)\notin v\overline{K}$. Take a polynomial $f(X)\in K[X]$ and write
$f(X)=\sum_{i=0}^{n}c_{i}(X-a)^{i}$ where $c_{i}\in\overline{K}$. Observe that
each of the monomials in this expression have distinct values. It follows that
$vf=\min\\{vc_{i}+iv(X-a)\\}=v_{a,v(X-a)}f$, that is, $(a,v(X-a))$ is a pair
of definition for $v$ over $K$. The fact that $a\in K$ implies that
$(a,v(X-a))$ is a minimal pair of definition for $v$ over $K$. Further, the
observation $vf=v_{a,v(X-a)}f$ for all $f(X)\in K[X]$ implies that
$v|_{K(X)}=v_{Q}$, where $Q(X):=X-a$. By definition, $\\{X-a\\}$ forms a
complete sequence of key polynomials for $v$ over $K$.

We now assume that $vb(X-a)=0$ and $b(X-a)v$ is transcendental over $Kv$ for
some $a,b\in K$. It follows from [References, Lemma 2.10] that
$v|_{K(X)}=v_{Q}$ where $Q(X):=X-a$. Hence $\\{X-a\\}$ forms a complete
sequence of key polynomials for $v$ over $K$. Observe that $\delta(Q)=v(X-a)$,
since $Q$ is a linear polynomial. It now follows from [References, Theorem
1.1] that $(a,v(X-a))$ forms a minimal pair of definition for $v$ over $K$.

Finally, we assume that $(K(X)|K,v)$ satisfies the condition (PE3). In light
of [References, Theorem 2], we obtain that the extension $(K(X)|K,v)$ is
immediate. Set $Q_{\nu}:=X-z_{\nu}\in K[X]$. By definition,
$\\{\delta(Q_{\nu})\\}_{\nu<\lambda}$ is a strictly increasing sequence.
Further, for any $f(X)\in K[X]$, the sequence
$\\{vf(z_{\nu})\\}_{\nu<\lambda}$ is ultimately constant. It has been observed
in the proof of [References, Theorem 2] that $vf=v_{Q_{\nu}}f$ for all $\nu$
sufficeintly large. As a consequence, we obtain that
$\\{X-z_{\nu}\\}_{\nu<\lambda}$ forms a complete sequence of key polynomials
for $v$ over $K$. ∎

######  Theorem 6.2.

Assume that either $E$ is a pcs of transcendental type or $E$ has a limit
$a\in K$. Fix an extension $\overline{v}$ of $v$ to $\overline{K}$ and
consider the common extension $\overline{v}_{E}$ of $\overline{v}$ and $v_{E}$
to $\overline{K}(X)$. Fix an extension of $\overline{v}_{E}$ to
$\overline{K(X)}$. Then $(K(X)|K,v_{E})$ is a pure extension and
$IC(K(X)|K,v_{E})=K^{h}$.

###### Proof.

If $E$ is a pcs of transcendental type, then the extension $(K(X)|K,v_{E})$ is
pure by definition. Otherwise, we assume that there exists some $a\in K$ such
that $a$ is a limit of $E$. We have observed in Proposition 5.1 that
$(a,\alpha)$ is a pair of definition for $v_{E}$ over $K$, where
$\alpha:=v_{E}(X-a)$. The fact that $a\in K$ implies that $(a,\alpha)$ is a
minimal pair of definition. If $E$ is a pcs or pds, then the fact that $a$ is
a limit of $E$ implies that $\alpha\notin\overline{v}\overline{K}$.
Consequently, $(K(X)|K,v_{E})$ is pure by definition. If $E$ is a pcts, then
$\alpha\in vK$. Take some $b\in K$ such that $vb(X-a)=0$. The fact that
$(a,\alpha)$ is a minimal pair of definition for $v_{E}$ over $K$ where
$\alpha\in vK$ implies that $(K(X)|K,v_{E})$ is residue transcendental. It now
follows from [References, Theorem 2.1] that $b(X-a)v$ is transcendental over
$Kv$. We have thus proved the theorem. ∎

Recall that every element of $E$ is also a limit of $E$ whenever $E$ is a pds
or pcts. Thus the remaining case to consider is when $E$ is a pcs of algebraic
type without a limit in $(K,v)$.

##  7\. $E$ is a pcs of algebraic type

Throughout this section, we assume that $\overline{v}$ is a fixed extension of
$v$ to $\overline{K}$. Take the common extension $\overline{v}_{E}$ of $v_{E}$
and $\overline{v}$ to $\overline{K}(X)$. We fix an extension of
$\overline{v}_{E}$ to $\overline{K(X)}$. Our next result modifies [References,
Theorem 9.2] which illustrates that we can obtain stronger results in our
setting.

######  Theorem 7.1.

Assume that $E$ is a pcs of algebraic type without a limit in $K$. Take an
associated minimal polynomial $Q(X)\in K[X]$. Then the following statements
hold true:

(1) there exists some root $a$ of $Q$ such that $(a,\overline{v}_{E}(X-a))$
forms a minimal pair of definition for $v_{E}$ over $K$,

(2) $\\{Q_{\nu}\\}_{\nu<\lambda}\cup\\{Q\\}$ forms a complete sequence of key
polynomials for $v_{E}$ over $K$, where $Q_{\nu}:=X-z_{\nu}$ for all
$\nu<\lambda$,

(3) $IC(K(X)|K,v_{E})=K^{h}$.

###### Proof.

The assumption that $Q(X)$ is an associated minimal polynomial for $E$ implies
that $\\{vQ(z_{\nu})\\}_{\nu<\lambda}$ is ultimately strictly increasing.
Hence there exists some root $a$ of $Q$ such that $a$ is a limit of $E$
[Proposition 3.10]. It follows from Propostion 5.1 that $(a,\alpha)$ is a pair
of defninition for $v_{E}$ over $K$, where $\alpha:=\overline{v}_{E}(X-a)$.
Take another pair of definition $(a^{\prime},\alpha)$ for $v_{E}$ over $K$ and
consider the minimal polynomial $f(X)$ of $a^{\prime}$ over $K$. It follows
from [References, Theorem 9.2] that $a^{\prime}$ is also a limit of $E$.
Consequently, $\\{vf(z_{\nu})\\}_{\nu<\lambda}$ is ultimately strictly
increasing. The minimality of $Q$ then implies that $\deg f\geq\deg Q$, that
is, $[K(a^{\prime}):K]\geq[K(a):K]$. Hence $(a,\alpha)$ is a minimal pair of
definition for $v_{E}$ over $K$.

The observation $\alpha=\max\overline{v}_{E}(X-\overline{K})$ implies that
$\delta(Q)=\alpha$. It now follows from [References, Theorem 1.1] that
$v_{E}=v_{Q}$ and $Q(X)$ is a key polynomial for $v_{E}$ over $K$. Recall that
$X$ is a limit of $E$. Thus $\delta(Q_{\nu})=v_{E}(X-z_{\nu})=\delta_{\nu}$
for all $\nu<\lambda$. Hence $\\{\delta(Q_{\nu})\\}_{\nu<\lambda}$ is a
strictly increasing sequence and $\delta(Q)>\delta(Q_{\nu})$ for all
$\nu<\lambda$. Take any polynomial $g(X)\in K[X]$ with $\deg g<\deg Q$. The
minimality of $Q$ implies that $\\{vg(z_{\nu})\\}_{\nu<\lambda}$ is ultimately
constant. It follows from Proposition 4.1 that $v_{E}g=vg(z_{\nu})$ for all
$\nu$ sufficiently large. As a consequence, $vg=v_{Q_{\nu}}g$ for all $\nu$
sufficiently large. We have thus obtained that
$\\{Q_{\nu}\\}_{\nu<\lambda}\cup\\{Q\\}$ forms a complete sequence of key
polynomials for $v_{E}$ over $K$.

We set $k:=K^{h}$. Observe that $k(X)^{h}=K(X)^{h}$. As a consequence,

$IC(K(X)|K,v_{E})=\overline{K}\cap K(X)^{h}=\overline{k}\cap
k(X)^{h}=IC(k(X)|k,\overline{v}_{E}).$

Now, $E$ is a pcs of algebraic type in $(k,\overline{v})$ and $(a,\alpha)$ is
a pair of definition for $\overline{v}_{E}$ over $k$. Take a minimal pair of
definition $(b,\alpha)$ for $\overline{v}_{E}$ over $k$ such that $b$ is a
limit of $E$. It follows from [References, Theorem 1.1] that

$k\subseteq IC(k(X)|k,\overline{v}_{E})\subseteq k(b).$

If $b\in k$, then we have $IC(k(X)|k,\overline{v}_{E})=k$. If
$\alpha>\overline{v}\overline{K}$, then we obtain from Proposition 5.4 that
$b$ is purely inseparable over $k$. Consequently, it follows from [References,
Propostion 7.1] that $IC(k(X)|k,\overline{v}_{E})=k$. We now consider the
complementary case, that is, $\overline{v}k$ is cofinal in
$\overline{v}_{E}k(X)$ and $b\notin k$. In light of [References, Proposition
3.6] we can assume that $b$ is separable over $k$. Take any conjugate $\sigma
b\neq b$ of $b$ over $k$ where $\sigma\in\mathrm{Gal}\,(\overline{k}|k)$. The
fact that $(k,\overline{v})$ is henselian implies that
$\overline{v}\circ\sigma=\overline{v}$ on $\overline{k}$. Then,

$\overline{v}(\sigma
b-z_{\nu})=(\overline{v}\circ\sigma)(b-z_{\nu})=\overline{v}(b-z_{\nu})=\delta_{\nu}\text{
for all }\nu<\lambda.$

It follows that any conjugate of $b$ is also a limit of $E$. Thus
$\overline{v}_{E}(X-\sigma b)=\alpha$ for all conjuates $\sigma b$ of $b$ over
$k$. Take the minimal polynomial $q(X)\in k[X]$ of $b$ over $k$. Then

$\overline{v}_{E}(q)=n\alpha\text{ where }n:=\deg q=[k(b):k].$

It follows from [References, Remark 3.3] that
$\overline{v}_{E}k(X)=\overline{v}k(b)\oplus n{\mathbb{Z}}\alpha$. Further,
the extension $(k(b,X)|k(b),\overline{v}_{E})$ being pure implies that
$\overline{v}_{E}k(b,X)=\overline{v}k(b)\oplus{\mathbb{Z}}\alpha$. As a
consequence, we obtain that

$(\overline{v}_{E}k(b,X):\overline{v}_{E}k(X))=n=[k(b):k].$

Now, observe that $k(b,X)^{h}=k(X)^{h}(b)$. It follows that

$(\overline{v}_{E}k(b,X):\overline{v}_{E}k(X))=[k(b):k]\geq[k(b,X)^{h}:k(X)^{h}]\geq(\overline{v}_{E}k(b,X):\overline{v}_{E}k(X)).$

As a consequence,

$[k(b,X)^{h}:k(X)^{h}]=[k(b):k].$

It follows that $k(b)$ and $k(X)^{h}$ are linearly disjoint over $k$. Hence

$k(b)\cap k(X)^{h}=k.$

The fact that $IC(k(X)|k,\overline{v}_{E})\subseteq k(b)$ now implies that
$IC(k(X)|k,\overline{v}_{E})\subseteq k(b)\cap k(X)^{h}=k$. Further,
$IC(k(X)|k,\overline{v}_{E})$ is a separable-algebraic extension of $k$ by
definition. It then follows that $IC(k(X)|k,\overline{v}_{E})=k$. We have thus
obtained that

$IC(K(X)|K,v_{E})=K^{h}.$

∎

######  Theorem 7.2.

Let $(K(X)|K,w)$ be a value transcendental extension and
$E:=\\{z_{\nu}\\}_{\nu<\lambda}$ a pcs of algebraic type in $(K,v)$. Take
associated minimal polynomials $Q(X)\in K[X]$ and $Q^{h}(X)\in K^{h}[X]$ of
$E$ over $K$ and $K^{h}$ respectively. Assume that $X$ is a limit of $E$. Then

$\deg\mathrm{dom}\,_{E}(Q^{h})=\deg Q^{h}.$

Consider the following statements:

(i) $w=v_{E}$.

(ii) There exists some common extension $\overline{w}$ of $\overline{v}$ and
$w$ to $\overline{K}(X)$ satisfying the property that $a$ is a limit of $E$ if
and only if $(a,\overline{w}(X-a))$ is a pair of definition for $w$ over $K$.

(iii) Take a pair of definition $(a,\overline{w}(X-a))$ for $w$ over $K$,
where $\overline{w}$ is some common extension of $w$ and $\overline{v}$ to
$\overline{K}(X)$. Then for all $\beta\in\overline{v}\overline{K}$,
$\beta>\overline{w}(X-a)$ if and only if $\beta>\delta_{\nu}$ for all
$\nu<\lambda$.

(iv) $(a,\overline{w}(X-a))$ is a minimal pair of definition for $w$ over $K$
for some root $a$ of $Q$, where $\overline{w}$ is some common extension of $w$
and $\overline{v}$ to $\overline{K}(X)$.

(v) $(a,\overline{w}(X-a))$ is a pair of definition for $w$ over $K$ for some
root $a$ of $Q$, where $\overline{w}$ is some common extension of $w$ and
$\overline{v}$ to $\overline{K}(X)$.

(vi) $w(Q)\notin\overline{v}\overline{K}$.

(vii) $IC(K(X)|K,w)=K^{h}$, where we fix an extension of $w$ to
$\overline{K(X)}$.
Then we have the chains of relations:

$\displaystyle(i)\Longleftrightarrow(ii)\Longleftrightarrow(iii)\Longrightarrow(iv)\Longleftrightarrow(v)\Longleftrightarrow(vi).$
$\displaystyle(i)\Longleftrightarrow(ii)\Longleftrightarrow(iii)\Longrightarrow(vii).$

###### Proof.

If $E$ admits a limit $b^{\prime}\in K^{h}$, then $Q^{h}(X):=X-b^{\prime}\in
K^{h}[X]$ is a linear polynomial and hence we have
$\deg\mathrm{dom}\,_{E}(Q^{h})=1=\deg Q^{h}$. Otherwise, it follows from
[References, Theorem 9.2] that there exists some root $b$ of $Q^{h}$ which is
a limit of $E$. By definition, $\overline{v}(b-z_{\nu})=\delta_{\nu}$ for all
$\nu<\lambda$. The fact that $K^{h}$ is henselian implies that
$\overline{v}\circ\sigma=\overline{v}$ for all
$\sigma\in\mathrm{Gal}\,(\overline{K}|K^{h})$. It follows that

$\overline{v}(\sigma
b-z_{\nu})=(\overline{v}\circ\sigma)(b-z_{\nu})=\overline{v}(b-z_{\nu})=\delta_{\nu}.$

Consequently, every conjugate of $b$ over $K^{h}$ is also a limit of $E$, that
is, $\deg\mathrm{dom}\,_{E}(Q^{h})=\deg Q^{h}$.

We assume that $w=v_{E}$ and take the common extension $\overline{v}_{E}$ of
$\overline{v}$ and $v_{E}$ to $\overline{K}(X)$. We have observed in Section 4
that $a$ is a limit of $E$ if and only if
$\overline{v}_{E}(X-a)\notin\overline{v}\overline{K}$. It now follows from
Lemma 2.3 that $a$ is a limit of $E$ if and only if
$\overline{v}_{E}(X-a)=\max\overline{v}_{E}(X-\overline{K})$, that is,
$(a,\overline{v}_{E}(X-a))$ is a pair of definition for $v_{E}$ over $K$. We
thus have the assertion $(i)\Longrightarrow(ii)$.

We now assume that there exists some common extension $\overline{w}$ of
$\overline{v}$ and $w$ to $\overline{K}(X)$ such that $a$ is a limit of $E$ if
and only if $(a,\overline{w}(X-a))$ is a pair of definition for $w$ over $K$.
Take a limit $a$ of $E$ and set $\alpha:=\overline{w}(X-a)$. The fact that $w$
is a value transcendental extension implies that
$\alpha\notin\overline{v}\overline{K}$. Further, the fact that $X$ and $a$ are
limits of $E$ implies that $\alpha\geq\delta_{\nu}$ for all $\nu$ sufficiently
large [Lemma 3.8]. Since $E$ is a pcs, we have that
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is strictly increasing. Hence

$\alpha>\delta_{\nu}\text{ for all }\nu<\lambda.$

Take any $\beta\in\overline{v}\overline{K}$. If $\delta_{\nu}>\beta$ for some
and hence all $\nu$ sufficiently large, then it follows that $\alpha>\beta$.
Conversely, suppose that $\alpha>\beta$ but $\delta_{\nu}<\beta$ for all
$\nu<\lambda$. Take some $b\in\overline{K}$ such that
$\overline{v}(a-b)=\beta$. In light of Lemma 3.8, the observation that $a$ is
a limit of $E$ implies that $b$ is also a limit of $E$. However, the relation
$\overline{v}(a-b)<\alpha$ implies that $\overline{w}(X-b)=\beta<\alpha$ and
hence $(b,\overline{w}(X-b))$ is not a pair of definition for $w$ over $K$,
thereby contradicting our staring assumption. We have thus shown that for any
$\beta\in\overline{v}\overline{K}$,

$\alpha>\beta\text{ if and only if }\delta_{\nu}>\beta\text{ for all
}\nu\text{ sufficiently large}.$

As a consequence, we obtain that for fixed $d\in{\mathbb{Z}}$ and
$\beta\in\overline{v}\overline{K}$, we have that

$d\alpha+\beta\geq 0\text{ if and only if }d\delta_{\nu}+\beta\geq 0\text{ for
all }\nu\text{ sufficiently large}.$

Take $c\in\overline{K}$ such that $c$ is not a limit of $E$. It follows from
Lemma 3.8 that $\overline{w}(X-c)<\delta_{\nu}$ for all $\nu$ sufficiently
large. We then obtain from the triangle inequality that

$\overline{w}(X-c)=\overline{v}(z_{\nu}-c)\text{ for all }\nu\text{
sufficiently large}.$

We can then expand this observation to arbitrary polynomials over $K$. Take
$f(X)\in K[X]$ and write $f(X)=\prod_{i=1}^{n}(X-a_{i})$, where
$a_{1},\dotsc,a_{j}$ are all the root-limits of $(f,E)$. From the starting
assumptions we observe that $\overline{w}(X-a_{i})=\alpha$ for all
$i=1,\dotsc,j$. Hence

$wf=j\alpha+(\beta_{j+1}+\dotsc\beta_{n}),$

where $\beta_{r}:=\overline{v}(z_{\nu}-a_{r})$ for all $\nu$ sufficiently
large, for all $r=j+1,\dotsc,n$. Observe that $j$ and
$\beta_{j+1}+\dotsc\beta_{n}$ are uniquely determined by $f(X)$. The same
arguments imply that for any rational function $\varphi(X)\in K(X)$, we have
uniquely determined $d\in{\mathbb{Z}}$ and $\beta\in\overline{v}\overline{K}$
such that

$w\varphi=d\alpha+\beta.$

Further, observe that

$v\varphi(z_{\nu})=d\delta_{\nu}+\beta\text{ for all }\nu\text{ sufficiently
large}.$

From our observations earlier it now follows that

$w\varphi\geq 0\text{ if and only if }v\varphi(z_{\nu})\geq 0\text{ for all
}\nu\text{ sufficiently large}.$

We have thus shown that $W=V_{E}$, where $W$ is the valuation domain
corresponding to the valuation $w$. Hence we have proved the assertion
$(ii)\Longrightarrow(i)$.

We now take a pair of definition $(a,\overline{w}(X-a))$ for $w$ over $K$
where $\overline{w}$ is a common extension of $w$ and $\overline{v}$ to
$\overline{K}(X)$, and set $\alpha:=\overline{w}(X-a)$. The fact that $X$ is a
limit of $E$ implies that $w(X-z_{\nu})=\delta_{\nu}$ for all $\nu<\lambda$.
It now follows from Lemma 2.3 that $\alpha>\delta_{\nu}$ for all
$\nu<\lambda$, and as a consequence $a$ is also a limit of $E$. Take any
$\beta\in\overline{v}\overline{K}$ and $b\in\overline{K}$ such that
$\overline{v}(a-b)=\beta$. In light of Lemma 3.8 we observe that
$\beta>\delta_{\nu}$ for all $\nu<\lambda$ if and only if $b$ is also a limit
of $E$. On the other hand, $\beta>\alpha$ if and only if
$(b,\overline{w}(X-b))$ is a pair of definition for $w$ over $K$. We thus have
the equivalence $(iii)\Longleftrightarrow(ii)$.

If there exists some $a\in K$ which is a limit of $E$, then $Q(X)=X-a$ and
$(a,v_{E}(X-a))$ is a minimal pair of definition for $v_{E}$ over $K$. If $E$
does not have a limit in $K$ then the assertion $(i)\Longrightarrow(iv)$ is
proved in Theorem 7.1, where we consider the common extension
$\overline{v}_{E}$ of $\overline{v}$ and $v_{E}$ to $\overline{K}(X)$. The
assertion $(iv)\Longrightarrow(v)$ follows by definition. The assertion
$(v)\Longrightarrow(vi)$ is straightforward. Conversely, we assume that
$w(Q)\notin\overline{v}\overline{K}$. Thus
$\alpha:=\overline{w}(X-a)\notin\overline{v}\overline{K}$ for some root $a$ of
$Q$ where $\overline{w}$ is a common extension of $w$ and $\overline{v}$ to
$\overline{K}(X)$. It follows from Lemma 2.3 that
$\alpha=\max\overline{w}(X-\overline{K})$ and hence $(a,\alpha)$ is a pair of
definition for $w$ over $K$. Observe that $\delta_{\nu}=w(X-z_{\nu})\in
w(X-K)$. As a consequence, $\alpha>\delta_{\nu}\text{ for all }\nu<\lambda$.
Recall that $X$ is a limit of $E$. It now follows from Lemma 3.8 that $a$ is a
limit of $E$. We take another pair of definition $(b,\alpha)$ for $w$ over
$K$. Then $\overline{v}(a-b)>\alpha>\delta_{\nu}$ for all $\nu<\lambda$ and
hence $b$ is also a limit of $E$. Consequently, the sequence
$\\{vf(z_{\nu})\\}_{\nu<\lambda}$ is ultimately strictly increasing, where
$f(X)\in K[X]$ is the minimal polynomial of $b$ over $K$ [Proposition 3.10].
The minimality of $Q(X)$ then implies that $\deg f\geq\deg Q$, that is,
$[K(b):K]\geq[K(a):K]$. By definition, $(a,\alpha)$ is a minimal pair of
definition for $w$ over $K$, thereby proving the assertion
$(vi)\Longrightarrow(iv)$. We have thus obtained the equivalence
$(iv)\Longleftrightarrow(v)\Longleftrightarrow(vi)$.

The assertion $(i)\Longrightarrow(vii)$ follows from Theorem 7.1. ∎

The following result is now immediate:

######  Corollary 7.3.

Let $(K(X)|K,w)$ be a value transcendental extension and $E$ a Cauchy sequence
of algebraic type in $(K,v)$. Assume that $X$ is a limit of $E$. Then
$w=v_{E}$.

The assertions $(iv)\Longrightarrow(i)$ and $(vii)\Longrightarrow(i)$ in the
statement of Theorem 7.2 are not always true. An example illustrating this
fact is furnished below.

######  Example 7.4.

Take a prime number $p>0$. Consider the valued field
$({\mathbb{F}}_{p}(t),v_{t})$ where $v:=v_{t}$ is the $t$-adic valuation. Fix
an extension of $v$ to $\overline{{\mathbb{F}}_{p}(t)}$ which we again denote
by $v$. We define a sequence of elements
$\\{a_{i}\\}_{i\in{\mathbb{N}}}\subset\overline{{\mathbb{F}}_{p}(t)}$
recursively as follows:

$a_{0}:=-\frac{1}{t},\,\,a_{i+1}^{p}-a_{i+1}+a_{i}=0.$

Set $K:={\mathbb{F}}_{p}(t)^{h}(a_{i}\mid i\in{\mathbb{N}})$ and take
$\eta\in\overline{K}$ such that $\eta^{p}=\frac{1}{t}$. Then $(K,v)$ is a
henselian valued field with
$vK=\frac{1}{p^{\infty}}{\mathbb{Z}}:=\\{\frac{a}{p^{n}}\mid
a\in{\mathbb{Z}},n\in{\mathbb{N}}\\}$. It has been observed in [References,
Example 3.14] that $(K(\eta)|K,v)$ is an immediate extension and

$v(\eta-K)=\\{\gamma\in vK\mid\gamma<0\\}.$

It follows from [References, Theorem 1] that there exists a pcs
$E:=\\{z_{\nu}\\}_{\nu<\lambda}$ in $(K,v)$ without a limit in $K$ such that
$\eta$ is a limit of $E$. Hence $\delta_{\nu}=v(\eta-z_{\nu})<0$ for all
$\nu<\lambda$, where $\delta_{\nu}:=v(z_{\nu}-z_{\nu+1})$. The fact that
$\eta$ is algebraic over $K$ implies that $E$ is a pcs of algebraic type.
Further, it follows from Proposition 3.10 that
$\\{vQ(z_{\nu})_{\nu<\lambda}\\}$ is ultimately strictly increasing, where
$Q(X):=X^{p}-\frac{1}{t}\in K[X]$ is the minimal polynomial of $\eta$ over
$K$. The facts that $\deg Q$ is a prime number and $E$ does not have a limit
in $K$ imply that

$Q(X)\text{ is an associated minimal polynomial of $E$ over }K.$

Take an irrational number $\alpha>0$ and consider the value extension
$\overline{w}:=v_{\eta,\alpha}$ of $v$ from $\overline{K}$ to
$\overline{K}(X)$. Set $w:=\overline{w}|_{K(X)}$. In light of Lemma 3.8 we
observe that

$X\text{ is a limit of }E.$

Now, $w(Q)=w(X-\eta)^{p}=p\alpha\notin v\overline{K}$. From Theorem 7.2 we now
obtain that

$(\eta,\alpha)\text{ is a minimal pair of definition for $w$ over }K.$

As a consequence, we obtain from [References, Proposition 7.1] that

$IC(K(X)|K,w)=K.$

We obtain from [References, Remark 3.3] that
$wK(X)=vK(\eta)\oplus{\mathbb{Z}}w(Q)\subset{\mathbb{R}}$. Consequently,
$\mbox{\rm rk}\,wK(X)=1$. Observe that $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is
cofinal in $v(\eta-K)=\\{\gamma\in vK\mid\gamma<0\\}$. Further, given any
negative real number $r\in{\mathbb{R}}$, there exists $n\in{\mathbb{N}}$ such
that $r<-\frac{1}{p^{n}}<0$. It follows that $v(\eta-K)$ and hence
consequently $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is cofinal in
$0^{-}:=\\{r\in{\mathbb{R}}\mid r<0\\}$. Thus
$\sup\\{\delta_{\nu}\\}_{\nu<\lambda}=0$. It now follows from Theorem 9.3 that
$\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1=2$. As a consequence,

$w\neq v_{E}.$

We now provide a couple of examples to illustrate that the conditions $(iv)$
and $(vii)$ in the setting of Theorem 7.2 may not be interrelated.

######  Example 7.5.

Take the valued field $(K,v)$, $\eta\in\overline{K}$ and a pcs $E$ in $(K,v)$
with limit $\eta$ as in Example 7.4. Recall that $v(\eta-K)=\\{\gamma\in
vK\mid\gamma<0\\}$. Further, $Q(X):=X^{p}-\frac{1}{t}$ is an associated
minimal polynomial of $E$ over $K$. Take $a\in\overline{K}$ such that

$a^{p^{2}}=\frac{1}{t^{p}}+t.$

Then $K(a)|K$ is a purely inseparable extension and $[K(a):K]\leq p^{2}$.
Observe that $a^{p}=\frac{1}{t}+\frac{1}{\eta}$. The fact that $\eta\notin K$
implies that $a^{p}\notin K$. It follows that

$[K(a):K]=p^{2}.$

Applying Frobenius endomorphism, we obtain that $(a-\eta)^{p^{2}}=t$. As a
consequence,

$v(a-\eta)=\frac{1}{p^{2}}.$

It now follows from Lemma 3.8 that $a$ is a limit of $E$. Take the value
transcendental extension $\overline{w}:=v_{a,\alpha}$ of $v$ from
$\overline{K}$ to $\overline{K}(X)$, where $\alpha>\frac{1}{p^{2}}$ is an
irrational number. Set $w:=\overline{w}|_{K(X)}$. Then $(a,\alpha)$ is a pair
of definition for $w$ over $K$. Applying Lemma 3.8 again, we obtain that $X$
is also a limit of $E$. Further, it follows from [References, Lemma 5.1] that
$IC(K(X)|K,w)\subseteq K(a)$. The observation that $IC(K(X)|K,w)$ is a
separable-algebraic extension of $K$ while $K(a)|K$ is purely inseparable
implies that

$IC(K(X)|K,w)=K.$

From the triangle inequality we obtain that
$\overline{w}(X-\eta)=v(a-\eta)=\frac{1}{p^{2}}<\alpha$. Consequently,
$(\eta,\overline{w}(X-\eta))$ is not a pair of definition for $w$ over $K$.

######  Example 7.6.

Let $p>0$ be a prime number. Consider the valued field
$({\mathbb{F}}_{p}(t),v_{t})$ where $v:=v_{t}$ is the $t$-adic valuation. Fix
an extension of $v$ to $\overline{{\mathbb{F}}_{p}(t)}$ which we again denote
by $v$. Set $(K,v)$ to be the perfect hull of the henselization of
${\mathbb{F}}_{p}(t)$. Take $a\in\overline{K}$ such that
$a^{p}-a=\frac{1}{t}$. It has been observed in [References, Example 3.12] that
$(K(a)|K,v)$ is an immediate extension with

$v(a-K)=\\{\gamma\in vK\mid\gamma<0\\}.$

In light of [References, Thoerem 1] we can take a pcs $E$ in $(K,v)$ without
any limits in $K$ such that $a$ is a limit of $E$. The fact that
$a\in\overline{K}$ implies that $E$ is a pcs of algebraic type [References,
Theorem 2]. Take the minimal polynomial $Q(X):=X^{p}-X-\frac{1}{t}\in K[X]$ of
$a$ over $K$. It follows from Proposition 3.10 that
$\\{vQ(z_{\nu})\\}_{\nu<\lambda}$ is ultimately strictly increasing. Since
$\deg Q$ is a prime number, we conclude that

$Q(X)\text{ is an associated minimal polynomial of $E$ over }K.$

Take an irrational number $\alpha>0$ and consider the value transcendental
extension $\overline{w}:=v_{a,\alpha}$ of $v$ from $\overline{K}$ to
$\overline{K}(X)$. Set $w:=\overline{w}|_{K(X)}$. Applying Lemma 3.8, we
obtain that $X$ is also a limit of $E$. Further, $(a,\alpha)$ is a pair of
definition for $w$ over $K$. It then follows from Theorem 7.2 that
$(a,\alpha)$ is a minimal pair of definition for $w$ over $K$. Observe that
the set $\\{a+i\mid i\in{\mathbb{F}}_{p}\\}$ is the set of all the conjugates
of $a$ over $K$. Consequently, $\mathrm{kras}\,(a,K)=0<\alpha$, where
$\mathrm{kras}\,(a,K)$ is the Krasner constant of $a$ over $K$. It now follows
from [References, Theorem 1.3] that

$IC(K(X)|K,w)=K(a).$

##  8\. $E$ is a pds

Throughout this section we will assume that $E:=\\{z_{\nu}\\}_{\nu<\lambda}$
is a pds and we fix an extension $\overline{v}$ of $v$ to $\overline{K}$.
Recall that $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is a strictly decreasing
sequence.

######  Theorem 8.1.

Let $(K(X)|K,w)$ be a value transcendental extension and $E$ a pds in $(K,v)$.
Assume that $X$ is not a limit of $E$. Then the following are equivalent:

(i) $w=v_{E}$.

(ii) There exists some common extension $\overline{w}$ of $w$ and
$\overline{v}$ to $\overline{K}(X)$ satisfying the property that
$(a,\overline{w}(X-a))$ is a pair of definition for $w$ over $K$ if and only
if $a$ is a limit of $E$.

(iii) $(z_{\mu},w(X-z_{\mu}))$ is a pair of definition for $w$ over $K$ for
some $z_{\mu}\in E$. Further, for all $\beta\in\overline{v}\overline{K}$ we
have that $\beta<w(X-z_{\mu})$ if and only if $\beta<\delta_{\nu}$ for all
$\nu<\lambda$.

###### Proof.

The assertion $(i)\Longrightarrow(ii)$ follows from our discussions in Section
4.

We assume that $\overline{w}$ is a common extension such that
$(a,\overline{w}(X-a))$ is a pair of definition for $w$ over $K$ if and only
if $a$ is a limit of $E$. Take a limit $a$ of $E$ and set
$\alpha:=\overline{w}(X-a)$. Then $\alpha\notin\overline{v}\overline{K}$. The
fact that each element of $E$ is a limit of $E$ implies that
$\overline{w}(X-z_{\nu})=\alpha$ for each $\nu<\lambda$. Suppose that
$\alpha>\delta_{\nu}$ for some and hence all $\nu$ sufficiently large. It
follows consequently from Lemma 3.8 that $X$ is a limit of $E$ which
contradicts our starting assumption. Hence,

$\alpha<\delta_{\nu}\text{ for all }\nu<\lambda.$

Take any $\beta\in\overline{v}\overline{K}$ and $b\in\overline{K}$ such that
$\overline{v}(a-b)=\beta$. Suppose that $\alpha<\beta<\delta_{\nu}$ for all
$\nu$. It follows that $(b,\alpha)$ is a pair of definition for $w$ over $K$
but $b$ is not a limit of $E$, contradicting our starting assumptions. Hence,
for all $\beta\in\overline{v}\overline{K}$,

$\alpha>\beta\text{ if and only if }\delta_{\nu}>\beta\text{ for all
}\nu<\lambda.$

As a consequence, for fixed $d\in{\mathbb{Z}}$ and
$\beta\in\overline{v}\overline{K}$ we obtain that,

$d\alpha+\beta\geq 0\text{ if and only if }d\delta_{\nu}+\beta\geq 0\text{ for
all }\nu<\lambda.$

Take any $c\in\overline{K}$ which is not a limit of $E$. It follows that
$\overline{w}(X-c)<\alpha$ and hence

$\overline{w}(X-c)=\overline{v}(z_{\nu}-c)\text{ for all }\nu<\lambda.$

The above observations imply that for any rational function $\varphi(X)\in
K(X)$, we have uniquely determined $d\in{\mathbb{Z}}$ and
$\beta\in\overline{v}\overline{K}$ such that

$\displaystyle w\varphi$ $\displaystyle=d\alpha+\beta,$ $\displaystyle
v\varphi(z_{\nu})$ $\displaystyle=d\delta_{\nu}+\beta\text{ for all }\nu\text{
sufficiently large}.$

It now follows that $v\varphi(z_{\nu})\geq 0$ for all $\nu$ sufficiently large
whenever $w\varphi\geq 0$. Conversely, assume that $v\varphi(z_{\nu})\geq 0$
for all $\nu$ sufficiently large. Then $\\{\delta_{\nu}\\}_{\nu<\lambda}$
being a strictly decreasing sequence implies that $v\varphi(z_{\nu})\geq 0$
for all $\nu<\lambda$ and hence $w\varphi\geq 0$. As a consequence, we obtain
$W=V_{E}$ where $W$ is the valuation domain corresponding to the valuation
$w$. We have thus proved the assertion $(ii)\Longrightarrow(i)$.

We now assume that $(z_{\mu},w(X-z_{\mu}))$ is a pair of definition for $w$
over $K$ for some $z_{\mu}\in E$. Set $\alpha:=w(X-z_{\mu})$. Take a common
extension $\overline{w}$ of $w$ and $\overline{v}$ to $\overline{K}(X)$. Take
any $\beta\in\overline{v}\overline{K}$ and $b\in\overline{K}$ such that
$\overline{v}(z_{\mu}-b)=\beta$. The fact that
$\\{\delta_{\nu}\\}_{\nu<\lambda}$ is strictly decreasing implies that
$\beta\geq\delta_{\nu}$ for all $\nu$ sufficiently large if and only if
$\beta\geq\delta_{\nu}$ for some $\nu<\lambda$. It now follows from Lemma 3.8
that $b$ is not a limit of $E$ if and only if $\beta<\delta_{\nu}$ for all
$\nu<\lambda$. On the other hand, $\beta<\alpha$ if and only if
$(b,\overline{w}(X-b))$ is not a pair of definition for $w$ over $K$. We thus
have the equivalence $(iii)\Longleftrightarrow(ii)$. ∎

######  Corollary 8.2.

Let $(K(X)|K,w)$ be a value transcendental extension and $E$ a pds in $(K,v)$
diverging to infinity. Assume that $X$ is not a limit of $E$. Then $w=v_{E}$.

###### Proof.

Suppose that $w(X-z_{\mu})\geq\delta_{\nu}$ for some $z_{\mu}\in E$ and
$\nu<\lambda$. The fact that $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is a
decreasing sequence implies that $X$ is a limit of $E$, contradicting our
assumption. It follows that $w(X-z_{\mu})<\delta_{\nu}$ for all $\nu<\lambda$.
By definition, $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is coinitial in
$\overline{v}\overline{K}$. Hence $w(X-z_{\mu})\notin\overline{v}\overline{K}$
and $w(X-z_{\mu})<\overline{v}\overline{K}$. It now follows from Theorem 8.1
that $w=v_{E}$. ∎

##  9\. Rank computation

Let $\Gamma$ be an ordered abelian group. A subset $\Delta$ of $\Gamma$ is
said to be an isolated subgroup if it satisfies the following conditions:

  * •

$\Delta$ is a proper subgroup of $\Gamma$,

  * •

if $\alpha\in\Delta$, then $\beta\in\Delta$ for all $\beta\in\Gamma$ with
$-\alpha\leq\beta\leq\alpha$.

Thus the set of all isolated subgroups of $\Gamma$ are ordered by inclusion.
The rank of $\Gamma$ is defined to be the ordinal type of the set of all
isolated subgroups of $\Gamma$ and we write it as $\mbox{\rm rk}\,\Gamma$.

We now assume that $\Gamma_{1},\dotsc,\Gamma_{n}$ are ordered abelian groups
with $\mbox{\rm rk}\,\Gamma_{i}=r_{i}<\infty$ for all $i=1,\dotsc,n$. Let the
chain of isolated subgroups of $\Gamma_{i}$ be given by

$(0)=H_{i,1}\subsetneq H_{i,2}\subsetneq\dotsc\subsetneq H_{i,r_{i}}.$

Consider the ordered abelian group
$\Gamma:=(\Gamma_{1}\oplus\Gamma_{2}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$.
It is a straightforward check that the chain of isolated subgroups of $\Gamma$
is given by $H_{1,1}\oplus\dotsc\oplus H_{n,1}\subsetneq
H_{1,1}\oplus\dotsc\oplus H_{n-1,1}\oplus H_{n,2}\subsetneq\dotsc\subsetneq
H_{1,1}\oplus\dotsc\oplus H_{n-1,1}\oplus H_{n,r_{n}}\subsetneq
H_{1,1}\oplus\dotsc\oplus H_{n-1,1}\oplus\Gamma_{n}\subsetneq
H_{1,1}\oplus\dotsc\oplus H_{n-2,1}\oplus
H_{n-1,2}\oplus\Gamma_{n}\subsetneq\dotsc\subsetneq
H_{1,r_{1}}\oplus\Gamma_{2}\oplus\dotsc\oplus\Gamma_{n}$. It follows that

(1) $\mbox{\rm
rk}\,(\Gamma_{1}\oplus\Gamma_{2}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}=\mbox{\rm
rk}\,\Gamma_{n}+\dotsc+\mbox{\rm rk}\,\Gamma_{1}.$

For the rest of the section, we fix an extension $\overline{v}$ of $v$ to
$\overline{K}$. Given a pms $E$ in $(K,v)$ and assuming $\mbox{\rm rk}\,vK$ is
a finite positive integer, our goal in this section is to compute $\mbox{\rm
rk}\,v_{E}K(X)$. The extension $(K(X)|K,v_{E})$ is immediate when $E$ is a pcs
of transcendental type and is a pure residue transcendental extension when $E$
is a pcts [Theorem 6.2]. Hence $v_{E}K(X)=vK$ in either case. So we will
assume that $E$ is a pcs of algebraic type or $E$ is a pds. It is known that
that the rank of a valuation is invariant under algebraic extensions
[References, Chapter VI, §11, Lemma 2]. It follows that $\mbox{\rm
rk}\,\overline{v}_{E}\overline{K}(X)=\mbox{\rm rk}\,v_{E}K(X)$ and $\mbox{\rm
rk}\,\overline{v}\overline{K}=\mbox{\rm rk}\,vK$.

We first assume that $E$ is a pcs of algebraic type. Take $a\in\overline{K}$
such that

$a\text{ is a limit of }E.$

Let $\mbox{\rm rk}\,vK=n$ for some positive integer $n$. It has been observed
in [References, Proposition 2.10] that $\overline{v}\overline{K}$ is order
isomorphic to $(\Gamma_{1}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$,
where $\Gamma_{i}$ are subgroups of ${\mathbb{R}}$. From now on, we will
identify $\overline{v}\overline{K}$ with
$(\Gamma_{1}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$. Assume that under
this identification,

(2) $\delta_{\nu}\text{ corresponds to
}(\delta_{1,\nu},\dotsc,\delta_{n,\nu})\text{ for all }\nu<\lambda.$

Observe that the fact that $\overline{v}\overline{K}$ is a divisible group
implies that $\Gamma_{i}$ is a divisible group for each $i$. Hence, for any
$r_{i}\in{\mathbb{R}}$ we have that

$r_{i}\notin\Gamma_{i}\Longleftrightarrow r_{i}\text{ is torsion free over
}\Gamma_{i}.$

Recall that the Dedekind-MacNeille completion of ${\mathbb{Q}}$ is
$\tilde{{\mathbb{R}}}:=\\{-\infty\\}\cup{\mathbb{R}}\cup\\{\infty\\}$, where
$\tilde{{\mathbb{R}}}$ is endowed with an order extending the canonical order
on ${\mathbb{R}}$ such that $-\infty<r<\infty$ for all $r\in{\mathbb{R}}$
[References, Example 5.3.3]. By definition, $\tilde{{\mathbb{R}}}$ is also the
Dedekind-MacNeille completion of ${\mathbb{R}}$. Hence any subset
$A_{i}\subseteq\Gamma_{i}$ has a unique supremum and infimum in
$\tilde{{\mathbb{R}}}$, which we will denote as $\sup A_{i}$ and $\inf A_{i}$
respectively.

The fact that $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is a strictly increasing
sequence implies that $\\{\delta_{1,\nu}\\}_{\nu<\lambda}$ is an increasing
sequence. Further, $E$ is a Cauchy sequence if and only if
$\\{\delta_{1,\nu}\\}_{\nu<\lambda}$ is cofinal in $\Gamma_{1}$. Now the fact
that $\Gamma_{1}$ is a subgroup of the archimedean ordered group
${\mathbb{R}}$ implies that $\Gamma_{1}$ is cofinal in ${\mathbb{R}}$. It
follows that $\\{\delta_{1,\nu}\\}_{\nu<\lambda}$ is cofinal in $\Gamma_{1}$
if and only if $\\{\delta_{1,\nu}\\}_{\nu<\lambda}$ is cofinal in
${\mathbb{R}}$, that is, $\sup\\{\delta_{1,\nu}\\}_{\nu<\lambda}=\infty$.
Consider an embedding of $\overline{v}\overline{K}$ into the ordered abelian
group $({\mathbb{Z}}\oplus\overline{v}\overline{K})_{\mathrm{lex}\,}$ given by
$\beta\mapsto(0,\beta)$ for all $\beta\in\overline{v}\overline{K}$. Set
$\alpha:=(1,0)$ and take the extension $\overline{v}_{a,\alpha}$ of
$\overline{v}$ to $\overline{K}(X)$. In light of Lemma 3.8 we observe that $X$
is a limit of $E$. It then follows from the corollary to Theorem 7.2 that
$\overline{v}_{a,\alpha}=\overline{v}_{E}$. Observe further that
$(\overline{K}(X)|\overline{K},\overline{v}_{E})$ is a pure value
transcendental extension and hence
$\overline{v}_{E}\overline{K}(X)=\overline{v}\overline{K}\oplus{\mathbb{Z}}\alpha$.
It follows that
$\overline{v}_{E}\overline{K}(X)=({\mathbb{Z}}\oplus\overline{v}\overline{K})_{\mathrm{lex}\,}$.
Hence,

$\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1\text{ when $E$ is a Cauchy
sequence}.$

We now assume that $E$ is not a Cauchy sequence. Hence there exists some
$r_{1}\in{\mathbb{R}}$ such that

$r_{1}=\sup\\{\delta_{1,\nu}\\}_{\nu<\lambda}.$

We first assume that $r_{1}\notin\Gamma_{1}$. Hence $\delta_{1,\nu}<r_{1}$ for
all $\nu<\lambda$ and $\\{\delta_{1,\nu}\\}_{\nu<\lambda}$ is not ultimately
constant. Consider the ordered abelian group
$(\Gamma_{1}^{\prime}\oplus\Gamma_{2}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$
where $\Gamma_{1}^{\prime}:=\Gamma_{1}\oplus{\mathbb{Z}}r_{1}$. Take
$\alpha:=(r_{1},0,\dotsc,0)\notin\overline{v}\overline{K}$ and the extension
$\overline{v}_{a,\alpha}$ of $\overline{v}$ to $\overline{K}(X)$. Observe that
$\alpha>\delta_{\nu}$ for all $\nu<\lambda$. It thus follows from Lemma 3.8
that $X$ is a limit of $E$. Take any
$(\beta_{1},\dotsc,\beta_{n})\in\overline{v}\overline{K}$. If
$\beta_{1}=\delta_{1,\nu}$ for some $\nu$, then the fact that
$\\{\delta_{1,\nu}\\}_{\nu<\lambda}$ is not ultimately constant implies that
$\beta_{1}<\delta_{1,\mu}$ for all $\mu$ sufficiently large. It follows that
$(\beta_{1},\dotsc,\beta_{n})>\delta_{\nu}$ for all $\nu<\lambda$ if and only
if $\beta_{1}>\delta_{1,\nu}$ for all $\nu<\lambda$. The facts that
$r_{1}=\sup\\{\delta_{1,\nu}\\}_{\nu<\lambda}$, $r_{1}\notin\Gamma_{1}$ and
$\beta_{1}\in\Gamma_{1}$ imply that $\beta_{1}>\delta_{1,\nu}$ for all
$\nu<\lambda$ if and only if $\beta_{1}>r_{1}$. It follows that

$(\beta_{1},\dotsc,\beta_{n})>\delta_{\nu}\text{ for all }\nu<\lambda\text{ if
and only if }(\beta_{1},\dotsc,\beta_{n})>\alpha.$

We can then conclude from Theorem 7.2 that
$\overline{v}_{a,\alpha}=\overline{v}_{E}$. The observation
$\overline{v}_{E}\overline{K}(X)=\overline{v}\overline{K}\oplus{\mathbb{Z}}\alpha$
implies that
$\overline{v}_{E}\overline{K}(X)=(\Gamma_{1}^{\prime}\oplus\Gamma_{2}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$.
The fact that $\Gamma_{1}^{\prime}$ is a subgroup of ${\mathbb{R}}$ implies
that $\mbox{\rm rk}\,\Gamma_{1}^{\prime}=1$. It then follows from (1) that
$\mbox{\rm rk}\,\overline{v}_{E}\overline{K}(X)=n$. As a consequence,

$\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK\text{ when
}r_{1}\in{\mathbb{R}}\setminus\Gamma_{1}.$

We now assume that $r_{1}\in\Gamma_{1}$. There are two cases:
$r_{1}>\delta_{1,\nu}$ for all $\nu<\lambda$, and, $r_{1}=\delta_{1,\nu}$ for
all $\nu\geq\nu_{1}$. We first assume that

$r_{1}>\delta_{1,\nu}\text{ for all }\nu<\lambda.$

As a consequence, the sequence $\\{\delta_{1,\nu}\\}_{\nu<\lambda}$ is not
ultimately constant. Take any
$(\beta_{1},\dotsc,\beta_{n})\in\overline{v}\overline{K}$. Our prior
observations show that

$(\beta_{1},\dotsc,\beta_{n})>\delta_{\nu}\text{ for all }\nu<\lambda\text{ if
and only if }\beta_{1}\geq r_{1}.$

Consider an embedding of $\overline{v}\overline{K}$ into the ordered abelian
group
$(\Gamma_{1}\oplus{\mathbb{Z}}\oplus\Gamma_{2}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$
given by
$(\beta_{1},\dotsc,\beta_{n})\mapsto(\beta_{1},0,\beta_{2},\dotsc,\beta_{n})$.
Take $\alpha:=(r_{1},-1,0,\dotsc,0)$. By definition,
$(\beta_{1},0,\beta_{2},\dotsc,\beta_{n})>\alpha$ if and only if
$\beta_{1}\geq r_{1}$. We have thus obtained that

$(\beta_{1},\dotsc,\beta_{n})>\delta_{\nu}\text{ for all }\nu<\lambda\text{ if
and only if }(\beta_{1},0,\beta_{2},\dotsc,\beta_{n})>\alpha.$

Consider the valuation $\overline{v}_{a,\alpha}$ on $\overline{K}(X)$. The
fact that $r_{1}>\delta_{1,\nu}$ for all $\nu<\lambda$ implies that
$\alpha>\delta_{\nu}$ for all $\nu<\lambda$. Consequently, $X$ is a limit of
$E$. It then follows from Theorem 7.2 that
$\overline{v}_{a,\alpha}=\overline{v}_{E}$. The observation
$\overline{v}_{E}\overline{K}(X)=\overline{v}\overline{K}\oplus{\mathbb{Z}}\alpha$
implies that
$\overline{v}_{E}\overline{K}(X)=(\Gamma_{1}\oplus{\mathbb{Z}}\oplus\Gamma_{2}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$.
Hence,

$\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1\text{ when
}r_{1}\in\Gamma_{1}\text{ and }r_{1}>\delta_{1,\nu}\text{ for all
}\nu<\lambda.$

We now consider the final case, that is, $r_{1}=\delta_{1,\nu}$ for all
$\nu\geq\nu_{1}$. The fact that $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is strictly
increasing implies that $\\{\delta_{2,\nu}\\}_{\nu\geq\nu_{1}}$ is an
increasing sequence. We can now use the preceding arguments to draw similar
conclusions. We briefly sketch the arguments in each case without going into
the detailed checks.

We first assume that $\sup\\{\delta_{2,\nu}\\}_{\nu\geq\nu_{1}}=\infty$.
Consider the ordered abelian group
$(\Gamma_{1}\oplus{\mathbb{Z}}\oplus\Gamma_{2}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$
and take $\alpha:=(r_{1},1,0,\dotsc,0)$. Embed $\overline{v}\overline{K}$ in
$(\Gamma_{1}\oplus{\mathbb{Z}}\oplus\Gamma_{2}\oplus\dotsc\Gamma_{n})_{\mathrm{lex}\,}$
by setting
$(\beta_{1},\dotsc,\beta_{n})\mapsto(\beta_{1},0,\beta_{2},\dotsc,\beta_{n})$.
Observe that $\alpha>\delta_{\nu}$ for all $\nu<\lambda$. In this case we
obtain that
$\overline{v}_{E}\overline{K}(X)=(\Gamma_{1}\oplus{\mathbb{Z}}\oplus\Gamma_{2}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$.
Consequently, $\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1$.

We now assume that there exists $r_{2}\in{\mathbb{R}}$ such that

$r_{2}=\sup\\{\delta_{2,\nu}\\}_{\nu\geq\nu_{1}}.$

Further, assume that $r_{2}\notin\Gamma_{2}$. Take
$\alpha:=(r_{1},r_{2},0,\dotsc,0)$ in the ordered abelian group
$(\Gamma_{1}\oplus\Gamma_{2}^{\prime}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$,
where $\Gamma_{2}^{\prime}:=\Gamma_{2}\oplus{\mathbb{Z}}r_{2}$. The fact that
$r_{2}\notin\Gamma_{2}$ implies that $r_{2}>\delta_{2,\nu}$ for all
$\nu\geq\nu_{1}$ and consequently, $\alpha>\delta_{\nu}$ for all
$\nu<\lambda$. In this case we obtain that
$\overline{v}_{E}\overline{K}(X)=(\Gamma_{1}\oplus\Gamma_{2}^{\prime}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$
and hence $\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK$.

We now assume that $r_{2}\in\Gamma_{2}$ and $r_{2}>\delta_{2,\nu}$ for all
$\nu\geq\nu_{1}$. Hence $\\{\delta_{2,\nu}\\}_{\nu\geq\nu_{1}}$ is not
ultimately constant. Embed $\overline{v}\overline{K}$ in the ordered abelian
group
$(\Gamma_{1}\oplus\Gamma_{2}\oplus{\mathbb{Z}}\oplus\Gamma_{3}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$
by setting
$(\beta_{1},\dotsc,\beta_{n})\mapsto(\beta_{1},\beta_{2},0,\beta_{3},\dotsc,\beta_{n})$.
Take $\alpha:=(r_{1},r_{2},-1,0,\dotsc,0)$. Observe that $\alpha>\delta_{\nu}$
for all $\nu<\lambda$. Take any
$(\beta_{1},\dotsc,\beta_{n})\in\overline{v}\overline{K}$. The fact that
$\\{\delta_{2,\nu}\\}_{\nu\geq\nu_{1}}$ is not a ultimately constant sequence
implies that $(\beta_{1},\dotsc,\beta_{n})>\delta_{\nu}$ for all $\nu<\lambda$
if and only if $\beta_{1}>r_{1}$, or, $\beta_{1}=r_{1}$ and
$\beta_{2}>\delta_{2,\nu}$ for all $\nu\geq\nu_{1}$. Observe that
$\beta_{2}>\delta_{2,\nu}$ for all $\nu\geq\nu_{1}$ if and only if
$\beta_{2}\geq r_{2}$. It follows that
$(\beta_{1},\dotsc,\beta_{n})>\delta_{\nu}$ for all $\nu<\lambda$ if and only
if $(\beta_{1},\beta_{2})>(r_{1},r_{2})$. On the other hand,
$(\beta_{1},\beta_{2},0,\beta_{3},\dotsc,\beta_{n})>\alpha=(r_{1},r_{2},-1,0,\dotsc,0)$
if and only if $(\beta_{1},\beta_{2})>(r_{1},r_{2})$. We have thus obtained
that

$(\beta_{1},\dotsc,\beta_{n})>\delta_{\nu}\text{ for all }\nu<\lambda\text{ if
and only if }(\beta_{1},\beta_{2},0,\beta_{3},\dotsc,\beta_{n})>\alpha.$

Similar arguments as above now imply that
$\overline{v}_{E}\overline{K}(X)=(\Gamma_{1}\oplus\Gamma_{2}\oplus{\mathbb{Z}}\oplus\Gamma_{3}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$
and as a consequence, $\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1$.

The remaining case is when $r_{2}=\delta_{2,\nu}$ for all $\nu\geq\nu_{2}$. It
follows that $\\{\delta_{3,\nu}\\}_{\nu\geq\nu_{2}}$ is an increasing
sequence. We can again study the separate cases in a similar fashion. We now
assume that

$\displaystyle\sup\\{\delta_{1,\nu}\\}_{\nu<\lambda}$
$\displaystyle=r_{1}=\delta_{1,\nu}\text{ for all }\nu\geq\nu_{1},$
$\displaystyle\sup\\{\delta_{i,\nu}\\}_{\nu\geq\nu_{i-1}}$
$\displaystyle=r_{i}=\delta_{i,\nu}\text{ for all }\nu\geq\nu_{i},$

for all $i=2,\dotsc,n-1$. Then the sequence
$\\{\delta_{n,\nu}\\}_{\nu\geq\nu_{n-1}}$ is an increasing sequence. The cases
$\sup\\{\delta_{n,\nu}\\}_{\nu\geq\nu_{n-1}}=\infty$ and
$r_{n}:=\sup\\{\delta_{n,\nu}\\}_{\nu\geq\nu_{n-1}}\notin\Gamma_{n}$ are
treated similarly as before. We now assume that $r_{n}\in\Gamma_{n}$. If
$r_{n}=\delta_{n,\nu}$ for all $\nu\geq\nu_{n}$, then the sequnece
$\\{(\delta_{1,\nu},\dotsc,\delta_{n,\nu})\\}_{\nu<\lambda}$ is ultimately
constant, contradicting the fact that $\\{\delta_{\nu}\\}_{\nu<\lambda}$ is
strictly increasing. Thus $r_{n}>\delta_{n,\nu}$ for all $\nu\geq\nu_{n-1}$,
in which case we obtain $\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1$ as
above.
We compile our observations in the tree diagram Figure 1.

$\mbox{\rm
rk}\,vK<\infty$$\sup\\{\delta_{1,\nu}\\}_{\nu<\lambda}=\infty$($*$)$r_{1}:=\sup\\{\delta_{1,\nu}\\}_{\nu<\lambda}\in{\mathbb{R}}$$r_{1}\notin\Gamma_{1}$($\\#$)$r_{1}\in\Gamma_{1}$$r_{1}>\delta_{1,\nu}$
for all $\nu<\lambda$($*$)$r_{1}=\delta_{1,\nu}$ for all $\nu\geq\nu_{1}$
Figure 1. $E$ pcs of alg type: base level

If we arrive at a hollow node in Figure 1 with a branch marked $(*)$ we have
$\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1$. If we arrive at a hollow node
with a branch marked $(\\#)$ we have $\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm
rk}\,vK$. We keep going when we arrive at the square node. Assume that we have
arrived at the $i$-th square node for some $i<n$. Note that this implies we
have arrived at the square node in each of the previous steps. We then
continue as in Figure 2.

$\sup\\{\delta_{i,\nu}\\}_{\nu\geq\nu_{i-1}}=r_{i}=\delta_{i,\nu}$ for all
$\nu\geq\nu_{i}$$\sup\\{\delta_{i+1,\nu}\\}_{\nu\geq\nu_{i}}=\infty$($*$)$r_{i+1}:=\sup\\{\delta_{i+1,\nu}\\}_{\nu\geq\nu_{i}}\in{\mathbb{R}}$$r_{i+1}\notin\Gamma_{i+1}$($\\#$)$r_{i+1}\in\Gamma_{i+1}$$r_{i+1}>\delta_{i+1,\nu}$
for all $\nu\geq\nu_{i}$($*$)$r_{i+1}=\delta_{i+1,\nu}$ for all
$\nu\geq\nu_{i+1}$ Figure 2. $E$ pcs of alg type: $(i+1)$-th level

The dual result holds when $E$ is a pds in $(K,v)$. This is precisely due to
the dual nature of Theorem 7.2$(iii)$ and Theorem 8.1$(iii)$. Since the
arguments are completely similar to the case when $E$ is a pcs of algebraic
type, we only state the final results here. Recall that we identify
$\overline{v}\overline{K}$ with
$(\Gamma_{1}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$ where $\Gamma_{i}$
are subgroups of ${\mathbb{R}}$. Further, for each $\nu<\lambda$,
$\delta_{\nu}$ corresponds to $(\delta_{1,\nu},\dotsc,\delta_{n,\nu})$ under
this identification. We then obtain the tree diagram Figure 3.

$\mbox{\rm
rk}\,vK<\infty$$\inf\\{\delta_{1,\nu}\\}_{\nu<\lambda}=-\infty$($*$)$r_{1}^{\prime}:=\inf\\{\delta_{1,\nu}\\}_{\nu<\lambda}\in{\mathbb{R}}$$r_{1}^{\prime}\notin\Gamma_{1}$($\\#$)$r_{1}^{\prime}\in\Gamma_{1}$$r_{1}^{\prime}<\delta_{1,\nu}$
for all $\nu<\lambda$($*$)$r_{1}^{\prime}=\delta_{1,\nu}$ for all
$\nu\geq\nu_{1}$ Figure 3. $E$ pds: base level

Whenever we arrive at a hollow node in Figure 3 with a branch marked $(*)$ we
have $\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1$. We have $\mbox{\rm
rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK$ whenever we arrive at a hollow node with a
brach marked $(\\#)$. We arrive at the square node if we have arrived at the
square node in each of the preceding steps. In this case we continue as in
Figure 4.

$\inf\\{\delta_{i,\nu}\\}_{\nu\geq\nu_{i-1}}=r_{i}^{\prime}=\delta_{i,\nu}$
for all
$\nu\geq\nu_{i}$$\inf\\{\delta_{i+1,\nu}\\}_{\nu\geq\nu_{i}}=-\infty$($*$)$r_{i+1}^{\prime}:=\inf\\{\delta_{i+1,\nu}\\}_{\nu\geq\nu_{i}}\in{\mathbb{R}}$$r_{i+1}^{\prime}\notin\Gamma_{i+1}$($\\#$)$r_{i+1}^{\prime}\in\Gamma_{i+1}$$r_{i+1}^{\prime}<\delta_{i+1,\nu}$
for all $\nu\geq\nu_{i}$($*$)$r_{i+1}^{\prime}=\delta_{i+1,\nu}$ for all
$\nu\geq\nu_{i+1}$ Figure 4. $E$ pds: $(i+1)$-th level

Observe that
$(\tilde{{\mathbb{R}}}\oplus\dotsc\oplus\tilde{{\mathbb{R}}})_{\mathrm{lex}\,}$
is a complete totally ordered set [References, Proposition 9.16], where
$\tilde{{\mathbb{R}}}:=\\{-\infty\\}\cup{\mathbb{R}}\cup\\{\infty\\}$. Hence
the set
$\mathcal{D}:=\\{(\delta_{1,\nu},\dotsc,\delta_{n,\nu})\\}_{\nu<\lambda}$ has
a unique supremum and infimum in
$(\tilde{{\mathbb{R}}}\oplus\dotsc\oplus\tilde{{\mathbb{R}}})_{\mathrm{lex}\,}$.
We define,

$\sup(E,\overline{v}):=\sup\mathcal{D}\text{ and
}\inf(E,\overline{v}):=\inf\mathcal{D}.$

We first assume that $E$ is a pcs of algebraic type. Hence $\mathcal{D}$ is
strictly increasing. Let $r_{1}:=\sup\\{\delta_{1,\nu}\\}_{\nu<\lambda}$ where
$r_{1}\in\tilde{{\mathbb{R}}}$ (we allow $r_{1}=\infty$ in this case). If
$r_{1}>\delta_{1,\nu}$ for all $\nu<\lambda$, then it is a straightforward
check that $\sup(E,\overline{v})=(r_{1},-\infty,\dotsc,-\infty)$. If
$r_{1}=\delta_{1,\nu}$ for all $\nu\geq\nu_{1}$, then we observe that
$\\{\delta_{2,\nu}\\}_{\nu\geq\nu_{1}}$ is an increasing sequence. Let
$r_{2}:=\sup\\{\delta_{2,\nu}\\}_{\nu\geq\nu_{1}}\in\tilde{{\mathbb{R}}}$. If
$r_{2}>\delta_{2,\nu}$ for all $\nu\geq\nu_{1}$, then
$\sup(E,\overline{v})=(r_{1},r_{2},-\infty,\dotsc,-\infty)$. Otherwise,
$r_{2}=\delta_{2,\nu}$ for all $\nu\geq\nu_{2}$ and consequently
$\\{\delta_{3,\nu}\\}_{\nu\geq\nu_{2}}$ is an increasing sequence. Repeated
iterations of this argument yield the following result:

######  Proposition 9.1.

Assume that $E$ is a pcs of algebraic type in $(K,v)$ with $\mbox{\rm
rk}\,vK=n$ and fix an extension $\overline{v}$ of $v$ to $\overline{K}$.
Identify $\overline{v}\overline{K}$ with
$(\Gamma_{1}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$ where $\Gamma_{i}$
are subgroups of ${\mathbb{R}}$. Consider the identification (2). Then the
following statements hold true:

(1) $\sup(E,\overline{v})=(r_{1},-\infty,\dotsc,-\infty)$ if and only if
$\sup\\{\delta_{1,\nu}\\}_{\nu<\lambda}=r_{1}>\delta_{1,\nu}$ for all
$\nu<\lambda$.

(2) $\sup(E,\overline{v})=(r_{1},\dotsc,r_{i},-\infty,\dotsc,-\infty)$ for
some $i\geq 2$ if and only if the following hold true:

$\displaystyle\sup\\{\delta_{1,\nu}\\}_{\nu<\lambda}=r_{1}$
$\displaystyle=\delta_{1,\nu}\text{ for all }\nu\geq\nu_{1},$
$\displaystyle\sup\\{\delta_{j,\nu}\\}_{\nu\geq\nu_{j-1}}=r_{j}$
$\displaystyle=\delta_{j,\nu}\text{ for all }\nu\geq\nu_{j},\text{ for all
}j=2,\dotsc,i-1,$
$\displaystyle\sup\\{\delta_{i,\nu}\\}_{\nu\geq\nu_{i-1}}=r_{i}$
$\displaystyle>\delta_{i,\nu}\text{ for all }\nu\geq\nu_{i-1}.$

Observe that $r_{j}\in\Gamma_{j}$ for all $j<i$.

(3) $\sup(E,\overline{v})\in\overline{v}\overline{K}$ if and only if
$\sup(E,\overline{v})=(r_{1},\dotsc,r_{n})$ where $r_{n}\in\Gamma_{n}$ and
$r_{n}>\delta_{n,\nu}$ for all $\nu\geq\nu_{n-1}$.

The dual result when $E$ is a pds in $(K,v)$ is presented below:

######  Proposition 9.2.

Assume that $E$ is a pds in $(K,v)$ with $\mbox{\rm rk}\,vK=n$ and fix an
extension $\overline{v}$ of $v$ to $\overline{K}$. Identify
$\overline{v}\overline{K}$ with
$(\Gamma_{1}\oplus\dotsc\oplus\Gamma_{n})_{\mathrm{lex}\,}$ where $\Gamma_{i}$
are subgroups of ${\mathbb{R}}$. Consider the identification (2). Then the
following statements hold true:

(1) $\inf(E,\overline{v})=(r_{1}^{\prime},\infty,\dotsc,\infty)$ if and only
if $\inf\\{\delta_{1,\nu}\\}_{\nu<\lambda}=r_{1}^{\prime}<\delta_{1,\nu}$ for
all $\nu<\lambda$.

(2)
$\inf(E,\overline{v})=(r_{1}^{\prime},\dotsc,r_{i}^{\prime},\infty,\dotsc,\infty)$
for some $i\geq 2$ if and only if the following hold true:

$\displaystyle\inf\\{\delta_{1,\nu}\\}_{\nu<\lambda}=r_{1}^{\prime}$
$\displaystyle=\delta_{1,\nu}\text{ for all }\nu\geq\nu_{1},$
$\displaystyle\inf\\{\delta_{j,\nu}\\}_{\nu\geq\nu_{j-1}}=r_{j}^{\prime}$
$\displaystyle=\delta_{j,\nu}\text{ for all }\nu\geq\nu_{j},\text{ for all
}j=2,\dotsc,n-1,$
$\displaystyle\inf\\{\delta_{i,\nu}\\}_{\nu\geq\nu_{i-1}}=r_{i}^{\prime}$
$\displaystyle<\delta_{i,\nu}\text{ for all }\nu\geq\nu_{i-1}.$

Observe that $r_{j}^{\prime}\in\Gamma_{j}$ for all $j<i$.

(3) $\inf(E,\overline{v})\in\overline{v}\overline{K}$ if and only if
$\inf(E,\overline{v})=(r_{1}^{\prime},\dotsc,r_{n}^{\prime})$ where
$r_{n}^{\prime}\in\Gamma_{n}$ and $r_{n}^{\prime}<\delta_{n,\nu}$ for all
$\nu\geq\nu_{n-1}$.

Observe that if $E$ is a pcs of algebraic type with
$\sup(E,\overline{v})\in\overline{v}\overline{K}$, then we arrive at a hollow
node in Figure 2 at the $n$-th level and at the square node in each preceding
level. Analogously, if $E$ is a pds with
$\inf(E,\overline{v})\in\overline{v}\overline{K}$, then we arrive at a hollow
node in Figure 4 at the $n$-th level and at the square node in each preceding
level. We thus have the following result:

######  Theorem 9.3.

Assume that $(K,v)$ is a valued field of finite rank and fix an extension
$\overline{v}$ of $v$ to $\overline{K}$. Further, assume that $E$ is a pms in
$(K,v)$ of one of the following types:

(i) $E$ is a Cauchy sequnce,

(ii) $E$ is a pcs of algebraic type with
$\sup(E,\overline{v})\in\overline{v}\overline{K}$,

(iii) $E$ diverges to infinity,

(iv) $E$ is a pds with $\inf(E,\overline{v})\in\overline{v}\overline{K}$.
Then,

$\mbox{\rm rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1.$

If $\mbox{\rm rk}\,vK=1$, then the above conditions are also necessary.

It is clear from our preceding discussions that the above conditions are
sufficient but not necessary for rank incrementation for valued fields of
higher rank. An example illustrating this fact is furnished below.

######  Example 9.4.

Let $(K,v)$ be a valued field with
$vK=(\Gamma\oplus{\mathbb{Z}})_{\mathrm{lex}\,}$ where $\Gamma$ is a
nontrivial subgroup of ${\mathbb{Q}}$. Fix an extension $\overline{v}$ of $v$
to $\overline{K}$. Then
$\overline{v}\overline{K}=({\mathbb{Q}}\oplus{\mathbb{Q}})_{\mathrm{lex}\,}$.
Take an element $g\in\Gamma$ and set $\delta_{i}:=(g,i)$ for all
$i\in{\mathbb{N}}$. Take $z_{i}\in K$ such that $vz_{i}=\delta_{i}$. By
definition, $E:=\\{z_{i}\\}_{i\in{\mathbb{N}}}$ is a pcs in $(K,v)$ with $0$
as a limit, hence $E$ is a pcs of algebraic type. Observe that
$\sup(E,\overline{v})=(g,\infty)\notin\overline{v}\overline{K}$. Further, we
observe that we arrive at the square node in Figure 1 and we arrive at the top
left hollow node in Figure 2 at the second level. As a consequence, $\mbox{\rm
rk}\,v_{E}K(X)=\mbox{\rm rk}\,vK+1=3$.

## References

  * [1] Shreeram S. Abhyankar, Ramification theoretic methods in algebraic geometry, Princeton Univ. Press, 1959.
  * [2] V. Alexandru and A. Zaharescu, Sur une classe de prolongements à $K(x)$ d’une valuation sur une corp $K$, Rev. Roumaine Math. Pures Appl., 5 (1988), 393–400.
  * [3] V. Alexandru, N. Popescu and A. Zaharescu, A theorem of characterization of residual transcendental extensions of a valuation, J. Math. Kyoto University, 28 (1988), 579–592.
  * [4] N. Bourbaki, Commutative Algebra, Hermann, Paris (1972).
  * [5] A. Dutta, Minimal pairs, minimal fields and implicit constant fields, https://arxiv.org/abs/2107.04828.
  * [6] I. Kaplansky, Maximal fields with valuations I, Duke Math. Journal 9 (1942), 303–321,
  * [7] F.-V. Kuhlmann, The defect, in Commutative Algebra - Noetherian and non-Noetherian perspectives. Marco Fontana, Salah-Eddine Kabbaj, Bruce Olberding and Irena Swansoneds. Springer, 2011.
  * [8] F.-V. Kuhlmann, Value groups, residue fields and bad places of rational function fields, Trans. Amer. Math. Soc., 356 (2004), 4559–4600.
  * [9] S. MacLane, A construction for absolute values in polynomial rings, Trans. Amer. Math. Soc., 40 (1936), 363–395.
  * [10] J. Novacoski, Key polynomials and minimal pairs, Journal of Algebra, 523 (2019), 1–14.
  * [11] J. Novacoski and M. Spivakovsky, Key polynomials and pseudo-convergent sequences, Journal of Algebra, 495 (2018), 199–219.
  * [12] A. Ostrowski, Unterschungen zur arithmetischen Theorie der Körper, Math. Z. 39 (1935), 269–404.
  * [13] G. Peruginelli and D. Spirito, Extending valuations to the field of rational functions using pseudo-monotone sequences, Journal of Algebra, 586 (2021), 756–786.
  * [14] Bernd S. W. Schröder, Ordered sets: an introduction, Birkhäuser Boston, 2003.
  * [15] M. Vaquié, Extension d’une valuation, Trans. Amer. Math. Soc., 359 no. 7 (2007), 3439–3481.
  * [16] O. Zariski and P. Samuel, Commutative Algebra, Volume II, Van Nostrand, 1958.