# Heights of Function Field Points on Curves Given by Equations with Separated
Variables

Ta Thi Hoai An  Institute of Mathematics, 18 Hoang Quoc Viet, Cau Giay
10307 Hanoi
Viet Nam<EMAIL_ADDRESS>and  Nguyen Thi Ngoc Diep  Department of
Mathematics
Vinh University
Vietnam<EMAIL_ADDRESS>
###### Abstract.

Let $P$ and $Q$ be polynomials in one variable over an algebraically closed
field $k$ of characteristic zero. Let $f$ and $g$ be elements of a function
field ${\bf K}$ over $k$ such that $P(f)=Q(g).$ We give conditions on $P$ and
$Q$ such that the height of $f$ and $g$ can be effectively bounded, and
moreover, we give sufficient conditions on $P$ and $Q$ under which $f$ and $g$
must be constant.

###### Key words and phrases:

Heights of Functions, Uniqueness polynomial, Functional equation

Financial support provided to the first named author by Alexander von Humboldt
Foundation and ICTP and to the both authors by Vietnam’s National Foundation
for Science and Technology Development (NAFOSTED)

2000 Mathematics Subject Classification. Primary 14H05 Secondary 30D35 14H55.

##  1\. Introduction

Let $k$ be an algebraically closed field of characteristic zero, and let $P$
and $Q$ be polynomials in $k[X].$ Determining when algebraic curves defined by
equations of the form $P(X)-Q(Y)=0$ have irreducible components of geometric
genus zero or one is or interest both in arithmetic and complex function
theory because it is related to when such curves may have infinitely many
rational solutions in a number field or non-constant meromorphic function
solutions. Genus zero components are also related to the study of what are
known as “uniqueness polynomials.” Special cases of this problem have been
considered, but to date there has been no complete characterization of when an
algebraic curve of the form $P(x)-Q(y)=0$ has no irreducible components of
genus at most one. For example, in [1], [5] and [7], authors considered the
problem for cases of genus zero over non-archimedean field. Fujimoto in [15]
gave some sufficient conditions for the problem over complex number field
under assumption $Q=cP$ for some constant $c$ and the polynomial $P$ satisfies
Hypothesis I, that is $P$ is injective on the set of distinct zeros of
$P^{\prime}.$

Let $C$ be a smooth curve of genus $\mathfrak{g}$ over $k,$ and let ${\bf K}$
be its function field. Let $F_{1}(X,Y,Z)$ be the homogenization of
$[P(X)-P(Y)]/[X-Y]$ and $F_{c}(X,Y,Z)$, $c\neq 0,1$, be the homogenization of
$P(X)-cP(Y)$. If $f$ and $g$ are algebraic functions in ${\bf K}$ such that
$P(f)=cP(g)$ for some nonzero constant $c$, then the morphism $\Phi:=(f,g,1)$
is a morphism from $C$ into the curve $[F_{c}(X,Y,Z)=0]$. By the Hurwitz
theorem this cannot happen if the curves $[F_{c}(X,Y,Z)=0]$ have no components
of genus $\leq\mathfrak{g}$. In [3], An and Wang gave sufficient conditions on
a polynomial $P$ satisfying Hypothesis I such that the height of any solution
$(f,g)$ with $f,g\in{\bf K}$ such that $P(f)=cP(g)$ could be effectively
bounded above. The purpose of this paper is to consider more general separated
variable equations of the form $P(x)-Q(y)$ and to give some conditions on the
polynomials $P$ and $Q$ such that if $f$ and $g$ are elements of ${\bf K}$
satisfying the equation $P(f)=Q(g),$ then the heights of $f$ and $g$ can be
effectively bounded.

For each point $\mathbf{p}\in C$, we may choose a uniformizer $t_{\mathbf{p}}$
to define a normalized order function

$v_{\mathbf{p}}:=\text{ord}_{\mathbf{p}}:{\bf K}\to\mathbb{R}\cup\\{\infty\\}$

at $\mathbf{p}$. For a non-zero element $f\in{\bf K}$, the height $h(f)$
counts the number of poles of $f$ with multiplicities, i.e.

$h(f):=\displaystyle\sum_{\mathbf{p}\in
C}-\mathop{\mathrm{min}}\\{0,\,v_{\mathbf{p}}(f)\\}.$

For $[f,g]\in\mathbb{P}^{1}({\bf K})$, its height is defined by

$h(f,g):=\displaystyle\sum_{\mathbf{p}\in
C}-\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(f),v_{\mathbf{p}}(g)\\}.$

Clearly, $h(f)=h(f,1)$.

From now we will let $P(X)$ and $Q(X)$ be nonconstant polynomials of degree
$n$ and $m$, respectively, in $k[X]$. Without loss of generality, throughout
the paper we will assume that $n\geq m$. We will denote by
$\alpha_{1},\alpha_{2},...,\alpha_{l}$ and $\beta_{1},\beta_{2},...,\beta_{h}$
the distinct roots of $P^{\prime}(X)$ and $Q^{\prime}(X)$, respectively. We
will use $p_{1},p_{2},...,p_{l}$ and $q_{1},q_{2},...,q_{h}$ to denote the
multiplicities of the roots in $P^{\prime}(X)$ and $Q^{\prime}(X)$,
respectively. Thus, for some $a$, $b$ in $k$,

$P^{\prime}(X)=a{(X-\alpha_{1})}^{p_{1}}{(X-\alpha_{2})}^{p_{2}}...{(X-\alpha_{l})}^{p_{l}}$
$Q^{\prime}(X)=b{(X-\beta_{1})}^{q_{1}}{(X-\beta_{2})}^{q_{2}}...{(X-\beta_{h})}^{q_{h}}.$

Recall, the polynomial $P(X)$ satisfies Hypothesis I if

$P(\alpha_{i})\neq P(\alpha_{j})\ {\rm whenever}\ i\neq j,$

or in other words $P$ is injective on the roots of $P^{\prime}$.

If one of the polynomials $P$ or $Q$ is linear, say $P(X)=aX+b,$ then
$(\frac{1}{a}Q(f)-b,f)$ is a solution of the equation $P(X)=Q(Y),$ where $f$
is some nonconstant element of ${\bf K}$. Hence, from now we always assume
that both $P$ and $Q$ are not linear polynomials. The main results are as
follows.

######  Theorem 1.

Suppose that $f$ and $g$ are two distinct non-constant rational functions in
${\bf K}$ such that $P(f)=Q(g)$. Let

$\displaystyle B_{0}=\\{i\mid 1\leq i\leq l,\ P(\alpha_{i})\neq
Q(\beta_{j})\text{ for all $j=1,...,h\\}$\quad and }$ $\displaystyle
B_{1}=\\{i\mid 1\leq i\leq h,\ Q(\beta_{i})\neq P(\alpha_{j})\text{ for all
$j=1,...,l\\}$}.$

Then

  1. (a)

$nh(f)={m}h(g);$

  2. (b)

$\Big{(}{\displaystyle\sum_{i\in
B_{0}}p_{i}}-\frac{m+n}{m}\Big{)}h(f)\leq{2\mathfrak{g}-2};$

  3. (c)

$\Big{(}\displaystyle\sum_{i\in B_{1}}q_{i}-\frac{2m}{n}\Big{)}\ h(g)\leq
2\mathfrak{g}-2.$

As a consequence of Theorem 1 we have the following corollary.

######  Corollary 2.

With the same hypothesis as in Theorem 1.

  1. (a)

If either ${\displaystyle\sum_{i\in B_{0}}p_{i}}-\frac{m+n}{m}>0$ or
$\displaystyle\sum_{i\in B_{1}}q_{i}-\frac{2m}{n}>0$, then the heights of $f$
and $g$ are effectively bounded above;

  2. (b)

If either ${\displaystyle\sum_{i\in
B_{0}}p_{i}}-\frac{m+n}{m}>\max\\{0,2\mathfrak{g}-2\\}$ or
$\displaystyle\sum_{i\in
B_{1}}q_{i}-\frac{2m}{n}>\max\\{0,2\mathfrak{g}-2\\}$, then $f$ and $g$ are
constant.

In order to state Theorem 3 clearly, we need to introduce the following
notation:

Notation. We put:

$\displaystyle A_{0}:=\\{(i,j)\mid 1\leq i\leq l,1\leq j\leq h,\
P(\alpha_{i})=Q(\beta_{j})\\},$ $\displaystyle A_{1}:=\\{(i,j)\mid(i,j)\in
A_{0},\ p_{i}>q_{j}\\},$ $\displaystyle A_{2}:=\\{(i,j)\mid(i,j)\in A_{0},\
p_{i}<q_{j}\\}.$

######  Theorem 3.

Let $P(X)$ and $Q(X)$ satisfy Hypothesis I. Suppose that $f$ and $g$ are two
distinct non-constant rational functions in ${\bf K}$ such that $P(f)=Q(g)$.
Then $\Big{(}\Big{(}\displaystyle\sum_{(i,j)\in
A_{1}}p_{i}-\frac{n}{m}q_{j}\Big{)}+\displaystyle\Big{(}\sum_{1\leq i\leq
l,(i,j)\notin A_{0}}p_{i}\Big{)}-\frac{m+n}{m}\Big{)}\ h(f)\leq
2\mathfrak{g}-2.$

######  Corollary 4.

Let $P(X)$ and $Q(X)$ satisfy Hypothesis I. Suppose that $f$ and $g$ are two
rational functions in ${\bf K}$ such that $P(f)=Q(g)$. If

$\displaystyle\sum_{(i,j)\in
A_{1}}(p_{i}-\frac{n}{m}q_{j})+\displaystyle\sum_{1\leq i\leq l,(i,j)\notin
A_{0}}p_{i}-\frac{m+n}{m}>\max\\{0,2\mathfrak{g}-2\\}.$

Then $f$ and $g$ are constants:

We will see that Hypothesis I tells us that there are not too many $(i,j)\in
A_{0}$ (see Lemma 9).

Especially, when the genus $\mathfrak{g}=0$ and the degrees of $P(X)$ and
$Q(X)$ are the same, the following theorem gives a necessary and sufficient
condition such that the equation $P(X)=Q(Y)$ has no non-constant rational
function solution.

######  Theorem 5.

Let $\mathfrak{g}=0$ and $P(X),\ Q(X)$ satisfy Hypothesis I and suppose $n=m$.
Suppose that $f$ and $g$ are two rational functions in ${\bf K}$ such that
$P(f)=Q(g)$. Then $f$ and $g$ are constant if and only if $P(X)$ and $Q(X)$
satisfy none of the following conditions

  1. (A)

$P(X)-Q(Y)$ has a linear factor.

  2. (B)

$l=1$, $h=2$, $p_{1}=q_{1}+1$, $q_{2}=1$ and $P(\alpha_{1})=Q(\beta_{1})$; or
$h=1$, $l=2$, $q_{1}=p_{1}+1$, $p_{2}=1$ and $P(\alpha_{1})=Q(\beta_{1})$.

  3. (C)

$l=h=2$, $p_{2}=q_{2}=1$, $p_{1}=q_{1}$ and $P(\alpha_{1})=Q(\beta_{1})$.

  4. (D)

$l=h=3$, $p_{i}=q_{i}=1$ for any $i=1,2,3$ and $P(\alpha_{i})=Q(\beta_{i})$
for any $i=1,2,3$ (after changing the indices).

  5. (E)

$l=h=p_{1}=q_{1}=1.$

Acknowledgments. We would like to thank the referee for her/his very careful
reading and helpful suggestions.

##  2\. Some lemmas

For simplicity of notation, for $i\geq 1$, $t\in{\bf K}\setminus k$ and
$\eta\in{\bf K}$, we denote by

$d_{t}^{i}\eta:=\frac{d^{i}\eta}{dt^{i}},\qquad
d_{\mathbf{p}}^{i}\eta:=\frac{d^{i}\eta}{dt_{\mathbf{p}}^{i}}.$

We first recall the following well-known properties, which follow from the
Riemann-Roch theorem and the sum formula.

######  Proposition 6.

Let $\eta\neq 0\in{\bf K}$ and $[f,g]\in\mathbb{P}^{1}({\bf K})$. We have

  1. (i)

$\displaystyle\sum_{\mathbf{p}\in
C}v_{\mathbf{p}}(d_{\mathbf{p}}\eta)=2\mathfrak{g}-2$ if $\eta$ is not
constant.

  2. (ii)

$\displaystyle\sum_{\mathbf{p}\in C}v_{\mathbf{p}}(\eta)=0$.

  3. (iii)

$h(\eta f,\eta g)=h(f,g).$

In order to study some sufficient conditions ensuring that the equation
$P(X)=Q(Y)$ has no non-constant rational function solutions, the basic idea is
as follows. Suppose there are two distinct non-constant rational functions $f$
and $g$ in ${\bf K}$ such that $P(f)=Q(g)$. We will study the height of $f$
and $g$ and give upper bounds for $h(f)$ and $h(g)$. We first give an upper
bound for $h(P^{\prime}(f),Q^{\prime}(g))$ thanks to the following lemma.

######  Lemma 7.

Suppose that $f$ and $g$ are distinct non-constant rational functions in ${\bf
K}$ such that $P(f)=Q(g)$. Then

  1. (i)

$nh(f)=mh(g);$

  2. (ii)

$h({P^{\prime}(f)},{Q^{\prime}(g)})+\displaystyle\sum_{\mathbf{p}\in
C}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\}\leq\frac{m+n}{m}h(f)+2\mathfrak{g}-2;$

where $v_{\mathbf{p}}^{0}(\eta):=\max\\{0,v_{\mathbf{p}}(\eta)\\}$ for
$\eta\in{\bf K}^{*}.$

###### Proof.

Since $P(f)=Q(g)$, for any $\mathbf{p}\in C$ such that $v_{\mathbf{p}}(f)<0$,
we have

$nv_{\mathbf{p}}(f)=v_{\mathbf{p}}(P(f))=v_{\mathbf{p}}(Q(g))=mv_{\mathbf{p}}(g).$

Hence, $nh(f)=mh(g)$. This also yields that

$d_{t}fP^{\prime}(f)=d_{t}gQ^{\prime}(g),$

for $t$ in ${\bf K}\setminus{\bf k}$, and hence

$h(P^{\prime}(f),Q^{\prime}(g))=h(\frac{P^{\prime}(f)}{Q^{\prime}(g)},1)=h(\frac{d_{t}g}{d_{t}f},1)=h(d_{t}f,d_{t}g).$

Since $d_{\mathbf{p}}f=d_{t}fd_{\mathbf{p}}t,$ it follows that
$v_{\mathbf{p}}(d_{\mathbf{p}}f)=v_{\mathbf{p}}(d_{t}f)+v_{\mathbf{p}}(d_{\mathbf{p}}t),$
and hence

$\displaystyle
v_{\mathbf{p}}(d_{t}f)=v_{\mathbf{p}}(d_{\mathbf{p}}f)-v_{\mathbf{p}}(d_{\mathbf{p}}t).$

We have

$\displaystyle
h({P^{\prime}(f)},{Q^{\prime}(g)})=h(d_{t}f,d_{t}g)=\displaystyle\sum_{\mathbf{p}\in
C}-\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(d_{t}f),v_{\mathbf{p}}(d_{t}g)\\}$
$\displaystyle=\displaystyle\sum_{\mathbf{p}\in
C}-\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(d_{\mathbf{p}}f)-v_{\mathbf{p}}(d_{\mathbf{p}}t),v_{\mathbf{p}}(d_{\mathbf{p}}g)-v_{\mathbf{p}}(d_{\mathbf{p}}t)\\}$
$\displaystyle=\displaystyle\sum_{\mathbf{p}\in
C}v_{\mathbf{p}}(d_{\mathbf{p}}t)+\displaystyle\sum_{\mathbf{p}\in
C}-\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(d_{\mathbf{p}}f),v_{\mathbf{p}}(d_{\mathbf{p}}g)\\}$
$\displaystyle=\displaystyle\sum_{\mathbf{p}\in
C}v_{\mathbf{p}}(d_{\mathbf{p}}t)+\displaystyle\sum_{v_{\mathbf{p}}(f)<0}-\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(d_{\mathbf{p}}f),v_{\mathbf{p}}(d_{\mathbf{p}}g)\\}-\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(d_{\mathbf{p}}f),v_{\mathbf{p}}(d_{\mathbf{p}}g)\\}.$

If $v_{\mathbf{p}}(f)<0$, then
$v_{\mathbf{p}}(d_{\mathbf{p}}f)=v_{\mathbf{p}}(f)-1$ and
$v_{\mathbf{p}}(d_{\mathbf{p}}g)=v_{\mathbf{p}}(g)-1=\frac{n}{m}v_{\mathbf{p}}(f)-1.$
If $v_{\mathbf{p}}(f)\geq 0$, then $v_{\mathbf{p}}(d_{\mathbf{p}}f)\geq 0$
hence
$v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f)=\max\\{0,v_{\mathbf{p}}(d_{\mathbf{p}}f)\\}=v_{\mathbf{p}}(d_{\mathbf{p}}f)$
and $v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)=v_{\mathbf{p}}(d_{\mathbf{p}}g).$

All together, we have

$\displaystyle h({P^{\prime}(f)},{Q^{\prime}(g)})$
$\displaystyle=2\mathfrak{g}-2+\displaystyle\sum_{v_{\mathbf{p}}(f)<0}-\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(f)-1,\frac{n}{m}v_{\mathbf{p}}(f)-1\\}-\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\}$
$\displaystyle=2\mathfrak{g}-2+\displaystyle\sum_{v_{\mathbf{p}}(f)<0}(-\frac{n}{m}v_{\mathbf{p}}f+1)-\displaystyle\sum_{\mathbf{p}\in
C}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\}$
$\displaystyle=2\mathfrak{g}-2+\frac{n}{m}h(f)+\\#\\{\mathbf{p}\in C\mid
v_{\mathbf{p}}(f)<0\\}-\displaystyle\sum_{\mathbf{p}\in
C}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\}.$

Clearly, $\\#\\{\mathbf{p}\in C\mid v_{\mathbf{p}}(f)<0\\}\leq h(f)$.
Therefore,

$\displaystyle h({P^{\prime}(f)},{Q^{\prime}(g)})\leq
2\mathfrak{g}-2+\frac{m+n}{m}h(f)-\displaystyle\sum_{\mathbf{p}\in
C}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\}.$

Hence

$\displaystyle
h({P^{\prime}(f)},{Q^{\prime}(g)})+\displaystyle\sum_{\mathbf{p}\in
C}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\}\leq\frac{m+n}{m}h(f)+2\mathfrak{g}-2.$

∎

Remark. We note that Lemma 7 gives an upper bound for
$h(P^{\prime}(f),Q^{\prime}(g)):$

(1) $h({P^{\prime}(f)},{Q^{\prime}(g)})\leq\frac{m+n}{m}h(f)+2\mathfrak{g}-2$

since $v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f)\geq 0$ and
$v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\geq 0$.

######  Lemma 8.

Suppose that $f$ and $g$ are distinct non-constant rational functions in ${\bf
K}$ such that $P(f)=Q(g)$. Then

$-\displaystyle\sum_{\mathbf{p}\in
C,v_{\mathbf{p}}(f)<0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}({P^{\prime}(f)}),v_{\mathbf{p}}({Q^{\prime}(g)})\\}=(n-1)h(f)=(n-1)\frac{m}{n}h(g).$

###### Proof.

Since $P(f)=Q(g)$ it follows that at a point $\mathbf{p}\in C$ such that
$v_{\mathbf{p}}(f)<0$, $v_{\mathbf{p}}(P^{\prime}(f))=(n-1)v_{\mathbf{p}}(f),$
$v_{\mathbf{p}}(g)<0$ and $nv_{\mathbf{p}}(f)=mv_{\mathbf{p}}(g)$. We also
have $v_{\mathbf{p}}(Q^{\prime}(g))=(m-1)v_{\mathbf{p}}(g)$. Therefore,

$\displaystyle\mathop{\mathrm{min}}\\{v_{\mathbf{p}}({P^{\prime}(f)}),v_{\mathbf{p}}({Q^{\prime}(g)})\\}$
$\displaystyle=\mathop{\mathrm{min}}\\{(n-1)v_{\mathbf{p}}(f),(m-1)\frac{n}{m}v_{\mathbf{p}}(f)\\}$
$\displaystyle=(n-1)v_{\mathbf{p}}(f)=(n-1)\frac{m}{n}v_{\mathbf{p}}(g),$

which yields

$\displaystyle-\displaystyle\sum_{\mathbf{p}\in
C,v_{\mathbf{p}}(f)<0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}({P^{\prime}(f)}),v_{\mathbf{p}}({Q^{\prime}(g)})\\}$
$\displaystyle=(n-1)h(f)=(n-1)\frac{m}{n}h(g).$

∎

We now turn to finding a lower bound for $h(P^{\prime}(f),Q^{\prime}(g))$ in
terms of $h(f)$. To find a lower bound for $h(P^{\prime}(f),Q^{\prime}(g))$,
we will need to find an element $G$ in ${\bf K}$ such that: firstly the height
of $G$ is not too big and, secondly the vanishing order of $G$ at each point
of the curve is at least the minimum of the vanishing order of $P^{\prime}(f)$
and $Q^{\prime}(g)$.

##  3\. Proof of Theorem 1

Recall that we have set:

$\displaystyle B_{0}=\\{i\mid 1\leq i\leq l,\ P(\alpha_{i})\neq
Q(\beta_{j}),\text{ for all $j=1,...,h\\},$}$ $\displaystyle B_{1}=\\{i\mid
1\leq i\leq h,\ Q(\beta_{i})\neq P(\alpha_{j}),\text{ for all
$j=1,...,l\\}$.}$

By rearranging the indices if necessary, we may assume

$B_{0}=\\{1,2,\ldots,l_{1}\\}\,\,{\rm and}\,\,B_{1}=\\{1,2,\ldots,h_{1}\\}.$

###### Proof of Theorem 1..

The assertion (a) is given in Lemma 7 (i).

We thus prove (b) and (c), beginning with (b). We take

$G(f):=\displaystyle\prod_{i\notin
B_{0}}{(f-\alpha_{i})}^{p_{i}}=\displaystyle\prod_{i=l_{1}+1}^{l}{(f-\alpha_{i})}^{p_{i}},$

and $d:=\displaystyle\sum_{i=l_{1}+1}^{l}p_{i}$. We have

(2) $\displaystyle h(P^{\prime}(f),Q^{\prime}(g))$
$\displaystyle=h\left(\frac{P^{\prime}(f)}{G(f)},\frac{Q^{\prime}(g)}{G(f)}\right)$
(3) $\displaystyle=-\displaystyle\sum_{\mathbf{p}\in
C}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G(f)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G(f)})\\}$
(4)
$\displaystyle=-\displaystyle\sum_{v_{\mathbf{p}}(f)<0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G(f)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G(f)})\\}$
(5) $\displaystyle\quad\quad-\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G(f)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G(f)})\\}.$

If $v_{\mathbf{p}}(f)<0$ then $v_{\mathbf{p}}(G(f))=dv_{\mathbf{p}}(f)$.
Therefore, by Lemma 8,

(6)
$\displaystyle-\displaystyle\sum_{v_{\mathbf{p}}(f)<0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G(f)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G(f)})\\}$
$\displaystyle=-\displaystyle\sum_{v_{\mathbf{p}}(f)<0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}({P^{\prime}(f)}),v_{\mathbf{p}}({Q^{\prime}(g)})\\}$
(7)
$\displaystyle\quad\quad+\displaystyle\sum_{v_{\mathbf{p}}(f)<0}v_{\mathbf{p}}(G)$
(8) $\displaystyle=(n-1-d)h(f).$

Putting (2) and (6) together, we have

(9) $\displaystyle h(P^{\prime}(f),Q^{\prime}(g))$
$\displaystyle=(n-1-d)h(f)-\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G(f)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G(f)})\\}.$

Therefore, to obtain a lower bound on $h(P^{\prime}(f),Q^{\prime}(g)),$ our
goal is to prove

(10) $\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G(f)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G(f)})\\}\leq
0.$

We have

$\frac{P^{\prime}(f)}{G(f)}=\frac{\displaystyle\prod_{i=1}^{l}{(f-\alpha_{i})}^{p_{i}}}{\displaystyle\prod_{i=l_{1}+1}^{l}{(f-\alpha_{i})}^{p_{i}}}=\displaystyle\prod_{i=1}^{l_{l}}{(f-\alpha_{i})}^{p_{i}}.$

For our purpose, taking into account displayed formula (10), we only have to
consider points $\mathbf{p}\in C$ such that $v_{\mathbf{p}}(f)\geq 0$. We
first consider those $\mathbf{p}$ satisfying
$\frac{P^{\prime}(f)}{G(f)}(\mathbf{p})\neq 0$, (i.e
$v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G(f)})=0)$; hence

(11)
$\displaystyle\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G(f)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G(f)})\\}\leq
0.$

At such point satisfying $\frac{P^{\prime}(f)}{G(f)}(\mathbf{p})=0$, there
exists $1\leq i\leq l_{1}$ such that $f(\mathbf{p})-\alpha_{i}=0$. By
definition of the set $B_{0}$, in this case, $g(\mathbf{p})-\beta_{j}\neq 0$
for all $j\in\\{1,2,\ldots,h\\}$. This means
$v_{\mathbf{p}}({Q^{\prime}(g)})=v_{\mathbf{p}}({\displaystyle\prod_{j=1}^{h}(g-\beta_{j})^{q_{j}}})=0,$
from which it follows that $v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G(f)})\leq 0.$
Thus

(12)
$\displaystyle\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G(f)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G(f)})\\}\leq
0.$

Combining (11) and (12) gives

$\displaystyle\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G(f)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G(f)})\\}\leq
0$

for all points $\mathbf{p}\in C$ such that $v_{\mathbf{p}}(f)\geq 0$. Together
with (9), and the facts $n-1=\displaystyle\sum_{i=1}^{l}p_{i},$ and
$d=\displaystyle\sum_{i=l_{1}+1}^{l}p_{i},$ we have

$\displaystyle\sum_{i\in B_{0}}p_{i}h(f)\leq
h({P^{\prime}(f)},{Q^{\prime}(g)}).$

The above inequality and inequality (1) in Lemma 7 (ii) about the upper bound
for $h(P^{\prime}(f),Q^{\prime}(g))$ give

$(\displaystyle\sum_{i\in B_{0}}p_{i}-\frac{m+n}{m})h(f)\leq 2\mathfrak{g}-2,$

which is the assertion (b).

For (c), let

$G_{1}(g):=\displaystyle\prod_{i\notin
B_{1}}{(g-\beta_{i})}^{q_{i}}=\displaystyle\prod_{i=h_{1}+1}^{h}{(g-\beta_{i})}^{q_{i}},$

and $d:=\deg G_{1}=\displaystyle\sum_{i=h_{1}+1}^{h}q_{i}.$ Similar to (b), we
have

(13) $\displaystyle h(P^{\prime}(f),Q^{\prime}(g))$
$\displaystyle=h\left(\frac{P^{\prime}(f)}{G_{1}(g)},\frac{Q^{\prime}(g)}{G_{1}(g)}\right)$
(14)
$\displaystyle=-\displaystyle\sum_{v_{\mathbf{p}}(g)<0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G_{1}(g)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G_{1}(g)})\\}$
(15) $\displaystyle\quad\quad-\displaystyle\sum_{v_{\mathbf{p}}(g)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G_{1}(g)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G_{1}(g)})\\}$
(16)
$\displaystyle=(m-d-\frac{m}{n})h(g)-\displaystyle\sum_{v_{\mathbf{p}}(g)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G_{1}(g)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G_{1}(g)})\\}.$

We still have to prove at a point $\mathbf{p}\in C$ satisfying
$v_{\mathbf{p}}(g)\geq 0$ that

(17)
$\displaystyle\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G_{1}(g)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G_{1}(g)})\\}\leq
0.$

Indeed, we have

$\frac{Q^{\prime}(g)}{G_{1}(g)}=\frac{\displaystyle\prod_{i=1}^{h}{(g-\beta_{i})}^{q_{i}}}{\displaystyle\prod_{i=h_{1}+1}^{h}{(g-\beta_{i})}^{q_{i}}}=\displaystyle\prod_{i=1}^{h_{l}}{(g-\beta_{i})}^{q_{i}}.$

Hence, if $\frac{Q^{\prime}(g)}{G_{1}(g)}(\mathbf{p})\neq 0$ then
$v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G_{1}(g)})=0$, and we are done. If
$\frac{Q^{\prime}(g)}{G_{1}(g)}(\mathbf{p})=0$, then there exists a $1\leq
j\leq h_{1}$ such that $g(\mathbf{p})-\beta_{j}=0$. By definition of the set
$B_{1}$, in this case, $f(\mathbf{p})-\alpha_{i}\neq 0$ for all
$i\in\\{1,2,\ldots,l\\}$. This means
$v_{\mathbf{p}}({P^{\prime}(f)})=v_{\mathbf{p}}({\displaystyle\prod_{i=1}^{l}(f-\alpha_{i})^{p_{i}}})=0,$
which implies $v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G_{1}(g)})\leq 0.$
Therefore

$\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G_{1}(g)}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G_{1}(g)})\\}\leq
0$

for all points $\mathbf{p}\in C$ satisfying $v_{\mathbf{p}}(g)\geq 0$. The
equalities (13) and (17) imply

$(\displaystyle\sum_{i=1}^{h_{1}}q_{i}+1-\frac{m}{n})h(g)\leq
h({P^{\prime}(f)},{Q^{\prime}(g)}),$

which combines with (1) to give

$(\displaystyle\sum_{i\in B_{1}}q_{i}-\frac{2m}{n})\ h(g)\leq
2\mathfrak{g}-2.$

The assertion (c) is therefore proved. ∎

##  4\. Proof of Theorem 3

Notation. Recall that we have set:

$\displaystyle A_{0}:=\\{(i,j)\mid 1\leq i\leq l,1\leq j\leq h,\
P(\alpha_{i})=Q(\beta_{j})\\},$ $\displaystyle A_{1}:=\\{(i,j)\mid(i,j)\in
A_{0},\ p_{i}>q_{j}\\},$ $\displaystyle A_{2}:=\\{(i,j)\mid(i,j)\in A_{0},\
p_{i}<q_{j}\\},$

and we put $l_{0}:=\\#A_{0}$.

When the polynomial $P$ and $Q$ satisfy the hypothesis I, the following lemma
will bound the cardinality of $A_{0}$.

######  Lemma 9 (see [1]).

Let $P(X)$ and $Q(X)$ satisfy Hypothesis I. Then for each $i$, $1\leq i\leq
l$, there exists at most one $j$, $1\leq j\leq h$, such that
$P(\alpha_{i})=Q(\beta_{j})$. Moreover,
$l_{0}\leq\mathop{\mathrm{min}}\\{l,h\\}.$

###### Proof.

For each $i,(1\leq i\leq l)$, assume that there exist $j_{1},j_{2},\ 1\leq
j_{1},j_{2}\leq h$, such that $P(\alpha_{i})=Q(\beta_{j_{1}})$ and
$P(\alpha_{i})=Q(\beta_{j_{2}})$. This implies that
$Q(\beta_{j_{1}})=Q(\beta_{j_{2}})$ and hence $j_{1}=j_{2}$ because $Q$
satisfies Hypothesis I. Similarly, there exists at most one $i,(1\leq i\leq
l)$ such that $P(\alpha_{i})=Q(\beta_{j})$ for each $j,(1\leq i\leq h)$. This
ends the proof of Lemma 9. ∎

By Lemma 9, without loss of generality we may assume from now that

$\displaystyle A_{0}=\\{(1,j(1)),\ldots,(l_{0},j(l_{0}))\\};$ $\displaystyle
A_{1}=\\{(1,j(1)),\ldots,(l_{1},j(l_{1}))\\}.$

###### Proof of Theorem 3..

The idea to prove this theorem is similar to Theorem 1 in that we have to find
a polynomial of low degree which can cancel all the common zeros of
$P^{\prime}(f)$ and $Q^{\prime}(g)$. We take

$G_{2}:=\displaystyle\prod_{i=1}^{l_{1}}{(g-\beta_{j(i)})}^{q_{j(i)}}\displaystyle\prod_{i=l_{1}+1}^{l_{0}}{(f-\alpha_{i})}^{p_{i}}.$

We have

(18) $\displaystyle h(P^{\prime}(f),Q^{\prime}(g))$
$\displaystyle=h\Big{(}\frac{P^{\prime}(f)}{G_{2}},\frac{Q^{\prime}(g)}{G_{2}}\Big{)}$
(19)
$\displaystyle=-\displaystyle\sum_{v_{\mathbf{p}}(f)<0}\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G_{2}}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G_{2}}\Big{)}\Big{\\}}$
(20) $\displaystyle\quad\quad-\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G_{2}}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G_{2}}\Big{)}\Big{\\}}.$

We first consider a point $\mathbf{p}\in C$ such that $v_{\mathbf{p}}(f)<0.$
We have $v_{\mathbf{p}}(P^{\prime}(f))=(n-1)v_{\mathbf{p}}(f).$ Since
$P(f)=Q(g),$ it follows that $v_{\mathbf{p}}(g)<0$ and
$nv_{\mathbf{p}}(f)=mv_{\mathbf{p}}(g)$. We also have
$v_{\mathbf{p}}(Q^{\prime}(g))=(m-1)v_{\mathbf{p}}(g)$. Hence,

$v_{\mathbf{p}}(P^{\prime}(f))<v_{\mathbf{p}}(Q^{\prime}(g)),$

and

$\displaystyle v_{\mathbf{p}}(G_{2})$
$\displaystyle=\Big{(}\displaystyle\sum_{i=1}^{l_{1}}q_{j(i)}\Big{)}\
v_{\mathbf{p}}(g)+\Big{(}\displaystyle\sum_{i=l_{1}+1}^{l_{0}}p_{i}\Big{)}\
v_{\mathbf{p}}(f)$
$\displaystyle=\Big{(}\displaystyle\sum_{i=1}^{l_{1}}\frac{n}{m}q_{j(i)}+\displaystyle\sum_{i=l_{1}+1}^{l_{0}}p_{i}\Big{)}\
v_{\mathbf{p}}(f).$

We remark that

$n-1=\displaystyle\sum_{i=1}^{l}p_{i}=\displaystyle\sum_{i=1}^{l_{1}}p_{i}+\displaystyle\sum_{i=l_{1}+1}^{l_{0}}p_{i}+\displaystyle\sum_{i=l_{0}+1}^{l}p_{i}.$

Therefore

(21) $\displaystyle-$
$\displaystyle\displaystyle\sum_{v_{\mathbf{p}}(f)<0}\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G_{2}}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G_{2}}\Big{)}\Big{\\}}$
(22)
$\displaystyle=-\displaystyle\sum_{v_{\mathbf{p}}(f)<0}(\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(P^{\prime}(f)),v_{\mathbf{p}}(Q^{\prime}(g))\\}-v_{\mathbf{p}}(G_{2}))$
(23)
$\displaystyle=-\displaystyle\sum_{v_{\mathbf{p}}(f)<0}\Big{(}(n-1)-(\displaystyle\sum_{i=1}^{l_{1}}\frac{n}{m}q_{j(i)}+\displaystyle\sum_{i=l_{1}+1}^{l_{0}}p_{i})\Big{)}v_{\mathbf{p}}(f)$
(24)
$\displaystyle=\Big{(}(\displaystyle\sum_{i=1}^{l_{1}}p_{i}-\frac{n}{m}q_{j(i)})+(\displaystyle\sum_{i=l_{0}+1}^{l}p_{i})\Big{)}\
h(f).$

Now we consider $\mathbf{p}\in C$ such that $v_{\mathbf{p}}(f)\geq 0,$ and we
will prove

$\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G_{2}}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G_{2}}\Big{)}\Big{\\}}\leq
0.$

Indeed, we have

$\displaystyle\frac{P^{\prime}(f)}{G_{2}}$
$\displaystyle=\frac{\displaystyle\prod_{i=1}^{l}{(f-\alpha_{i})}^{p_{i}}}{\displaystyle\prod_{i=1}^{l_{1}}{(g-\beta_{j(i)})}^{q_{j(i)}}\prod_{i=l_{1}+1}^{l_{0}}{(f-\alpha_{i})}^{p_{i}}}=\frac{\displaystyle\prod_{i=1}^{l_{1}}{(f-\alpha_{i})}^{p_{i}}{\displaystyle\prod_{i=l_{0}+1}^{l}{(f-\alpha_{i})}^{p_{i}}}}{\displaystyle\prod_{i=1}^{l_{1}}{(g-\beta_{j(i)})}^{q_{j(i)}}},$

and

$\displaystyle\frac{Q^{\prime}(g)}{G_{2}}$
$\displaystyle=\frac{\displaystyle\prod_{j=1}^{h}{(g-\beta_{j})}^{q_{j}}}{\displaystyle\prod_{i=1}^{l_{1}}{(g-\beta_{j(i)})}^{q_{j(i)}}\displaystyle\prod_{i=l_{1}+1}^{l_{0}}{(f-\alpha_{i})}^{p_{i}}}$
$\displaystyle=\frac{\displaystyle\prod_{j\notin\\{j(1),\dots,j(l_{1})\\}}{(g-\beta_{j})}^{q_{j}}\displaystyle\prod_{i=l_{1}+1}^{l_{0}}{(g-\beta_{j(i)})^{q_{j(i)}}}}{\displaystyle\prod_{i=l_{1}+1}^{l_{0}}{(f-\alpha_{i})}^{p_{i}}}$

If $\frac{P^{\prime}(f)}{G_{2}}(\mathbf{p})\neq 0$ then
$v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G_{2}}\Big{)}=0.$ Hence

$\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G_{2}}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G_{2}}\Big{)}\Big{\\}}\leq
0.$

If $\frac{P^{\prime}(f)}{G_{2}}(\mathbf{p})=0$ then, since we are considering
a point $\mathbf{p}\in C$ satisfying $v_{\mathbf{p}}(g)\geq 0,$ i.e.
$\mathbf{p}$ is not a pole of $g$, $\frac{P^{\prime}(f)}{G_{2}}(\mathbf{p})=0$
only when the numerator is equal to zero at $\mathbf{p}$, in which case there
exists an $i$ with either $i\in\\{1,\dots,l_{1}\\}$ or
$i\in\\{l_{0}+1,\dots,l\\}$ such that $f(\mathbf{p})-\alpha_{i}=0.$ Suppose
that $f(\mathbf{p})-\alpha_{i}=0$ for some $i$ such that
$i\in\\{1,\dots,l_{1}\\}$. By definition of $A_{1}$, we have
$g(\mathbf{p})-\beta_{j(i)}=0,$ and by Lemma 9, for each $i$ there exits at
most one $j(i)$ such that $g(\mathbf{p})-\beta_{j(i)}=0$. Looking at the ratio
$\frac{Q^{\prime}(g)}{G_{2}}$, we see that the factor of the form
$g-\beta_{j(i)}$ with $(i,j(i))\in A_{1}$ is canceled, which means
$\frac{Q^{\prime}(g)}{G_{2}}(\mathbf{p})\neq 0.$ So, we have
$v_{\mathbf{p}}\big{(}\frac{Q^{\prime}(g)}{G_{2}}\big{)}=0$. Suppose that
$f(\mathbf{p})-\alpha_{i}=0$ for some $i$ such that
$i\in\\{l_{0}+1,\dots,l\\}$. By definition of the set $A_{0}$,
$P(\alpha_{i})\neq Q(\beta_{j})$ for all $1\leq j\leq h.$ Therefore,
$g(\mathbf{p})-\beta_{j}\neq 0$ for any $j$, which means
$\frac{Q^{\prime}(g)}{G_{2}}(\mathbf{p})\neq 0,$ i.e
$v_{\mathbf{p}}\big{(}\frac{Q^{\prime}(g)}{G_{2}}\big{)}=0$. Hence, in either
case, we always have
$\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\big{(}\frac{P^{\prime}(f)}{G_{2}}\big{)},v_{\mathbf{p}}\big{(}\frac{Q^{\prime}(g)}{G_{2}}\big{)}\Big{\\}}\leq
0.$

Therefore, for all of $\mathbf{p}\in C$ satisfying $v_{\mathbf{p}}(g)\geq 0,$

(25)
$\displaystyle\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\big{(}\frac{P^{\prime}(f)}{G_{2}}\big{)},v_{\mathbf{p}}\big{(}\frac{Q^{\prime}(g)}{G_{2}}\big{)}\Big{\\}}\leq
0$

holds.

Combining (18), (21) and (25) gives

$\Big{(}\displaystyle\sum_{i=1}^{l_{1}}\big{(}p_{i}-\frac{n}{m}q_{j(i)}\big{)}+\displaystyle\sum_{i=l_{0}+1}^{l}p_{i}\Big{)}\
h(f)\leq h({P^{\prime}(f)},{Q^{\prime}(g)}).$

Together with (1), we have

$\Big{(}(\displaystyle\sum_{i=1}^{l_{1}}p_{i}-\frac{n}{m}q_{j})+(\displaystyle\sum_{i=l_{0}+1}^{l}p_{i})-\frac{m+n}{m}\Big{)}\
h(f)\leq 2\mathfrak{g}-2.$

The theorem is proved. ∎

##  5\. Proof of Theorem 5

In the proof of Theorem 5, we will need the following lemmas.

######  Lemma 10.

Suppose there are non-constant functions $f$ and $g$ in ${\bf K}$ such that
$P(f)=Q(g)$. If $v_{\mathbf{p}}(f-\alpha_{i})>0$ and
$v_{\mathbf{p}}(g-\beta_{j})>0$ at a point $\mathbf{p}\in C$, then

$(p_{i}+1)v_{\mathbf{p}}(f-\alpha_{i})=(q_{j}+1)v_{\mathbf{p}}(g-\beta_{j}).$

###### Proof.

If $v_{\mathbf{p}}(f-\alpha_{i})>0$ and $v_{\mathbf{p}}(g-\beta_{j})>0$, then
$P(\alpha_{i})=Q(\beta_{j})$ since $P(f)=Q(g)$. Since $\alpha_{i},\beta_{j}$
are zeros of $P^{\prime},Q^{\prime}$, with the multiplicities $p_{i},q_{j}$
respectively, we have the following expansions of $P$ at $\alpha_{i}$ and $Q$
at $\beta_{j}$:

$\displaystyle
P(X)=P(\alpha_{i})+b_{i,p_{i}+1}(X-\alpha_{i})^{p_{i}+1}+\ldots+b_{i,n}(X-\alpha_{i})^{n},$
$\displaystyle
Q(X)=Q(\beta_{j})+c_{j,q_{j}+1}(X-\beta_{j})^{q_{j}+1}+\ldots+c_{j,m}(X-\beta_{j})^{m}.$

We have

$\displaystyle 0$ $\displaystyle=P(f)-Q(g)$
$\displaystyle=[b_{i,p_{i}+1}(f-\alpha_{i})^{p_{i}+1}+\\{\text{Higher order
terms in}\ (f-\alpha_{i})\\}]$ $\displaystyle\qquad\
-[c_{j,q_{j}+1}(g-\beta_{j})^{q_{j}+1}+\\{\text{Higher order terms in}\
(g-\beta_{j})\\}].$

Therefore

$(p_{i}+1)v_{\mathbf{p}}(f-\alpha_{i})=(q_{j}+1)v_{\mathbf{p}}(g-\beta_{j}).$

∎

###### Proof of Theorem 5..

By [1, Lemma 6], if $P(X)$ and $Q(X)$ satisfy one of the conditions (A), (B),
(C) or (D) then the curve $P(X)-Q(Y)$ either has a linear factor or it is
irreducible of genus 0. If the case (E) holds, then $n=m=2$ and
$P(X)-Q(Y)=(X-\alpha_{1})^{2}-(Y-\beta_{1})^{2}+c$ for some constant $c$.
Hence either the curve $P(X)-Q(Y)$ has a linear factor or it is irreducible of
genus 0. Therefore, for all of these cases, there exist two non-constant
rational functions $f$ and $g$ in ${\bf K}$ such that $P(f)=Q(g)$. The
necessary condition is proved.

We now turn to the proof of the sufficient condition of the theorem. Suppose
that $P(X)$ and $Q(X)$ satisfy none of the conditions (A), (B), (C), (D) or
(E). Assume that $f$ and $g$ are two non-constant rational functions in ${\bf
K}$ such that $P(f)=Q(g)$. When the polynomials $P$ and $Q$ satisfy Hypothesis
I, by Lemma 9, without loss of generality we may assume that $A_{0}$ is of the
form $\\{(1,j(1)),\ldots,(l_{0},j(l_{0}))\\}$ such that the $p_{i}$ are non-
increasing, it means $p_{1}\geq p_{2}\geq\ldots\geq p_{l_{0}}.$

By Theorem 3, together with the hypothesis $\mathfrak{g}=0,n=m$, the right-
hand sides of the inequalities in Theorem 3 are negative, therefore

(26) $\displaystyle(\sum_{(i,j(i))\in
A_{1}}p_{i}-q_{j(i)})+(\sum_{i=l_{0}+1}^{l}p_{i})-2<0,$

which also implies

(27) $\displaystyle(\sum_{(i,j)\in
A_{2}}q_{j(i)}-p_{i})+(\sum_{j\notin\\{{j(1)},\dots,{j(l_{0})}\\}}q_{j})-2<0.$

From the inequality (26) and (27), we have

(${\mathcal{H}}$) $\displaystyle p_{i}=1\text{ for all }i\geq l_{0}+1;\
q_{j}=1\text{ for all }j\notin\\{{j(1)},\dots,{j(l_{0})},\
|p_{i}-q_{j(i)}|\leq 1\text{ for }i\leq l_{0};$ $\displaystyle|l-h|\leq
1\text{ and }l_{0}\geq\max\\{l,h\\}-1.$

We will consider the following cases.

Case 1. $l_{0}=0.$

By the statement ($\mathcal{H}$), we have $\max\\{l,h\\}\leq l_{0}+1=1.$

If $l=0$, then $P(X)$ is of the form $uX+v$ with $uv\neq 0$, i.e $n=1.$ Since
$n=m$, it follows that $m=1$ and $Q(Y)$ is of the form $sY+t$ with $st\neq 0.$
Therefore, $P(X)-Q(Y)$ has a linear factor. This is the exceptional case
corresponding to condition (A), which is excluded.

If $l=1$, then, by the hypothesis $n=m$ and the condition ($\mathcal{H}$),
$h=l=1$ and $q_{1}=p_{1}=1$. This is the exceptional case (E).

Case 2. $l_{0}=1.$

Then, by the statement ($\mathcal{H}$), we have $\max\\{l,h\\}\leq l_{0}+1=2.$

Suppose first that $l=1$. By the conditions $n=m$ and ($\mathcal{H}$), the
case $p_{1}<q_{j(1)}$ cannot happen. We only have to consider the following
possibilities. If $p_{1}=q_{j(1)}$ then, since $n=m$ and the statement
($\mathcal{H}$), we have $h=1$. Since $l_{0}=1$ we have
$P(\alpha_{1})=Q(\beta_{j(1)})$, therefore $P(X)-Q(Y)$ is of the form
$u(X-\alpha_{1})^{p_{1}+1}-v(Y-\beta_{j(1)})^{p_{1}+1}$ with $uv\neq 0$.
Therefore, $P(X)-Q(Y)$ has a linear factor. This is the exceptional case (A).
If $p_{1}>q_{j(1)}$ then $p_{1}=q_{j(1)}+1,h=2$ and $q_{2}=1$. This is the
exceptional case (B).

Suppose that $l=2$. Then $p_{2}=1$ because of the condition $p_{i}\leq 1$ for
$i\geq l_{0}+1$ in the statement ($\mathcal{H}$). On the other hand, the case
$p_{1}>q_{j(1)}$ cannot hold. Therefore, we consider the following
possibilities. If $p_{1}=q_{j(1)}$ then $h=2$ and $q_{2}=1.$ This is the
exceptional case (C). If $p_{1}<q_{j(1)}$ then $q_{j(1)}=p_{1}+1$ and $h=1$.
This is the exceptional case (B).

Case 3. $l_{0}\geq 2.$

For each $(i_{1},j(i_{1}))$ and $(i_{2},j(i_{2}))$ in $A_{0}$, we define
$L_{i_{1},i_{2}}(f,g)=L_{i_{1},i_{2}}$ as follows

(28) $\displaystyle
L_{i_{1},i_{2}}:=(g-\beta_{j(i_{1})})-\frac{\beta_{j(i_{1})}-\beta_{j(i_{2})}}{\alpha_{i_{1}}-\alpha_{i_{2}}}(f-\alpha_{i_{1}}),$

which can also be expressed as

(29) $\displaystyle
L_{i_{1},i_{2}}:=(g-\beta_{j(i_{2})})-\frac{\beta_{j(i_{1})}-\beta_{j(i_{2})}}{\alpha_{i_{1}}-\alpha_{i_{2}}}(f-\alpha_{i_{2}}).$

Now we take

$\displaystyle
G:=L_{1,2}^{p_{1}+p_{2}-2+{\sum_{i=l_{0}+1}^{l}}p_{i}}\prod_{i=3}^{l_{0}}(f-\alpha_{i})^{p_{i}}.$

We have

$\displaystyle h(P^{\prime}(f),Q^{\prime}(g))$
$\displaystyle=h\Big{(}\frac{P^{\prime}(f)}{G},\frac{Q^{\prime}(g)}{G}\Big{)}$
$\displaystyle=-\displaystyle\sum_{v_{\mathbf{p}}(f)<0}\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G}\Big{)}\Big{\\}}$
$\displaystyle\quad\quad-\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G}\Big{)}\Big{\\}}.$

We first consider a point $\mathbf{p}\in C$ such that $v_{\mathbf{p}}(f)<0.$
We have $v_{\mathbf{p}}(P^{\prime}(f))=(n-1)v_{\mathbf{p}}(f).$ Since
$P(f)=Q(g)$ and $n=m$ by hypothesis, we have
$v_{\mathbf{p}}(f)=v_{\mathbf{p}}(g)=v_{\mathbf{p}}(f-\alpha_{1})=v_{\mathbf{p}}(g-\beta_{j(1)})$
and $v_{\mathbf{p}}(g)<0$. Thus $v_{\mathbf{p}}(L_{1,2})\geq
v_{\mathbf{p}}(f)$, therefore

$\displaystyle v_{\mathbf{p}}(G)$ $\displaystyle\geq\Big{(}\displaystyle
p_{1}+p_{2}-2+\sum_{i=l_{0}+1}^{l}p_{i}+\sum_{i=3}^{l_{0}}p_{i}\Big{)}\
v_{\mathbf{p}}(f)$ $\displaystyle\geq(n-1-2)\ v_{\mathbf{p}}(f)=(n-3)\
v_{\mathbf{p}}(f).$

Therefore,

(30)
$\displaystyle-\displaystyle\sum_{v_{\mathbf{p}}(f)<0}\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G}\Big{)}\Big{\\}}$
(31)
$\displaystyle=-\displaystyle\sum_{v_{\mathbf{p}}(f)<0}\Big{(}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(P^{\prime}(f)),v_{\mathbf{p}}(Q^{\prime}(g))\\}-v_{\mathbf{p}}(G)\Big{)}\geq
2h(f).$

Together with Lemma 7(ii) and $\mathfrak{g}=0$, we have

$\displaystyle 2h(f)-\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G}\Big{)}\Big{\\}}$
$\displaystyle\leq h({P^{\prime}(f)},{Q^{\prime}(g)})$ $\displaystyle\leq
2h(f)-2-\displaystyle\sum_{\mathbf{p}\in
C}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\}$
$\displaystyle\leq 2h(f)-2-\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\},$

since if $v_{\mathbf{p}}(f)<0$ then
$v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f)=v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)=0$.
Therefore, if we can prove

(32) $\displaystyle\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G}\Big{)}\Big{\\}}$
$\displaystyle<2+\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\}$

then we can get a contradiction.
Let $\mathbf{p}\in C$ such that $v_{\mathbf{p}}(f)\geq 0.$ For our purpose, we
only have to consider those $\mathbf{p}\in C$ such that
$v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)}>0$ and also
$v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G}\Big{)}>0.$ Since
$v_{\mathbf{p}}(f)\geq 0$ and

$\displaystyle\frac{P^{\prime}(f)}{G}$
$\displaystyle=\frac{\displaystyle(f-\alpha_{1})^{p_{1}}(f-\alpha_{2})^{p_{2}}\prod_{i=l_{0}+1}^{l}{(f-\alpha_{i})}^{p_{i}}}{\displaystyle
L_{1,2}^{p_{1}+p_{2}-2+{\sum_{i=l_{0}+1}^{l}}p_{i}}},$

we have $v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)}>0$ when
$(f-\alpha_{i})(\mathbf{p})=0$ for $i$ is one index in the set
$\\{1,2,l_{0}+1,\dots,l\\}$. However, if $i\in\\{l_{0}+1,\dots,l\\}$ then, by
the definition of the set $A_{0}$, $(g-\beta_{j})(\mathbf{p})\neq 0$ for any
$j=1,...,h$, which means

$v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G}\Big{)}=v_{\mathbf{p}}\Big{(}\frac{\displaystyle\prod_{j=1}^{h}{(g-\beta_{j})}^{q_{j}}}{\displaystyle
L_{1,2}^{p_{1}+p_{2}-2+{\sum_{i=l_{0}+1}^{l}}p_{i}}\prod_{i=3}^{l_{0}}(f-\alpha_{i})^{p_{i}}}\Big{)}\leq
0.$

Therefore, we only have to check at points $\mathbf{p}\in C$ such that
$v_{\mathbf{p}}(f-\alpha_{i})>0$ and $v_{\mathbf{p}}(g-\beta_{j(i)})>0$ for
$i=1$ or 2.

We first consider $i=1$. By Lemma 10,

(33)
$\displaystyle(p_{1}+1)v_{\mathbf{p}}(f-\alpha_{1})=(q_{j(1)}+1)v_{\mathbf{p}}(g-\beta_{j(1)}).$

If $p_{1}\geq q_{j(1)}$ then $v_{\mathbf{p}}(f-\alpha_{1})\leq
v_{\mathbf{p}}(g-\beta_{j(1)})$ and hence
$v_{\mathbf{p}}(L_{1,2})\geq\mathop{\mathrm{min}}\\{v_{\mathbf{p}}(f-\alpha_{1}),v_{\mathbf{p}}(g-\beta_{j(1)}\\}=v_{\mathbf{p}}(f-\alpha_{1})$.
Therefore

$v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)}\leq-\Big{(}p_{2}-2+\sum_{i=l_{0}+1}^{l}p_{i}\Big{)}v_{\mathbf{p}}(f-\alpha_{1}),$

which is not positive if $p_{2}\geq 2$ or if we have both $p_{2}=1$ and
$l_{0}+1\leq l.$ Hence

$\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}(\frac{P^{\prime}(f)}{G}),v_{\mathbf{p}}(\frac{Q^{\prime}(g)}{G})\Big{\\}}\leq
0<2+\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\},$

which means the inequality (32) holds, so we can get a contradiction for this
case. If $p_{1}<q_{j(1)}$ then, by the assumption ($\mathcal{H}$),
$q_{j(1)}=p_{1}+1.$ Therefore, from the equality (33) we have

$v_{\mathbf{p}}(f-\alpha_{1})=\frac{p_{1}+2}{p_{1}+1}\
v_{\mathbf{p}}(g-\beta_{j(1)}).$

Hence $v_{\mathbf{p}}(f-\alpha_{1})>v_{\mathbf{p}}(g-\beta_{j(1)})\geq
p_{1}+1,$ the last inequality follows from the fact that $p_{1}+1$ and
$p_{1}+2$ are coprime. So
$v_{\mathbf{p}}(L_{1,2})=v_{\mathbf{p}}(g-\beta_{j(1)}).$ Therefore

$\displaystyle v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)}$
$\displaystyle\leq\Big{(}\frac{p_{1}(p_{1}+2)}{p_{1}+1}-p_{1}-p_{2}+2-\sum_{i=l_{0}+1}^{l}p_{i}\Big{)}v_{\mathbf{p}}(g-\beta_{j(1)})$
$\displaystyle\leq
v_{\mathbf{p}}(g-\beta_{j(1)})-\frac{1}{p_{1}+1}v_{\mathbf{p}}(g-\beta_{j(1)}),$

if either $p_{2}\geq 2$ or $p_{2}=1$ and $l_{0}+1\leq l.$ Since
$v_{\mathbf{p}}(g-\beta_{j(1)})\geq p_{1}+1$ , it follows that
$v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)}\leq
v_{\mathbf{p}}(g-\beta_{j(1)})-1=v_{\mathbf{p}}(d_{\mathbf{p}}g)$. Therefore,

$\displaystyle\mathop{\mathrm{min}}\Big{\\{}v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)},v_{\mathbf{p}}\Big{(}\frac{Q^{\prime}(g)}{G}\Big{)}\Big{\\}}$
$\displaystyle\leq v_{\mathbf{p}}\Big{(}\frac{P^{\prime}(f)}{G}\Big{)}\leq
v_{\mathbf{p}}(d_{\mathbf{p}}g)=\displaystyle\sum_{v_{\mathbf{p}}(f)\geq
0}\mathop{\mathrm{min}}\\{v_{\mathbf{p}}^{0}(d_{\mathbf{p}}f),v_{\mathbf{p}}^{0}(d_{\mathbf{p}}g)\\},$

which shows that (32) holds.

Using the same arguments for the case $i=2$, we also have a contradiction if
either $p_{2}\geq 2$ or $p_{2}=1$ and $l_{0}+1\leq l.$ Thus, $f$ and $g$ must
be constants.

The remaining case is when $p_{2}=1$ and $l_{0}=l.$

Subcase 1. $p_{2}=1$ and $l_{0}=l=2$.

Then $P^{\prime}(X)$ is of the form $(X-\alpha_{1})^{p_{1}}(X-\alpha_{2})$ and
$n=p_{1}+2$. We have $h\geq l_{0}=2.$ Since $n=m$ and by the assumption
($\mathcal{H}$), it follows that $Q^{\prime}(Y)$ is only one of the following
forms

$\displaystyle(Y-\beta_{j(1)})^{p_{1}}(Y-\beta_{j(2)}),\ {\rm i.e}\quad
h=2,q_{j(1)}=p_{1},q_{j(2)}=1;$
$\displaystyle(Y-\beta_{j(1)})^{p_{1}-1}(Y-\beta_{j(2)})^{2},\ {\rm i.e}\quad
h=2,q_{j(1)}=p_{1}-1,q_{j(2)}=2;$
$\displaystyle(Y-\beta_{j(1)})^{p_{1}-1}(Y-\beta_{j(2)})(Y-\beta_{3}),\ {\rm
i.e}\quad h=3,q_{j(1)}=p_{1}-1,q_{j(2)}=1,q_{3}=1.$

The first form corresponds to condition (C), which is excluded. For two
remaining cases, we take

$G:=L_{1,2}(g-\beta_{j(1)})^{p_{1}-2}.$

By an argument analogous to the previous, we get $f$ and $g$ are constants.

Subcase 2. $p_{2}=1$ and $l_{0}=l\geq 3$.

If $p_{1}=1$ and $l_{0}=l=3$ then, since $p_{1}\geq p_{2}\geq p_{3}$, it
follows that $P^{\prime}(X)$ is of the form
$(X-\alpha_{1})(X-\alpha_{2})(X-\alpha_{3})$ and $n=4.$ Since $n=m$ and
$l_{0}=l=3$, hence $h=3$ and $q_{1}=q_{2}=q_{3}=1$. Therefore, $Q^{\prime}(Y)$
is of the form $(Y-\beta_{j(1)})(Y-\beta_{j(2)})(Y-\beta_{j(3)})$ and
$P(\alpha_{i})=Q(\beta_{j(i)})\text{\;for any\;}i=1,2,3.$ This is the
exceptional case (D).

If $p_{1}=1$ and $l_{0}=l\geq 4$, then we take

$G:=L_{1,2}L_{3,4}\prod_{i=4}^{l_{0}}(f-\alpha_{i})^{p_{i}}.$

If $p_{1}\geq 2$, then we take

$G:=L_{1,2}^{p_{1}-1}L_{1,3}\prod_{i=4}^{l_{0}}(f-\alpha_{i})^{p_{i}}.$

For the above two cases, by repeating arguments similarly to the Case 1, we
also get $f$ and $g$ are constants. Theorem 5 is therefore proved. ∎

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