On factorization of p-adic meromorphic functions

TL;DR

This paper investigates the properties of p-adic meromorphic functions, focusing on primeness, pseudo primeness, and permutability, providing conditions under which these functions exhibit such properties.

Contribution

It introduces new criteria for primeness and pseudo primeness of p-adic meromorphic functions and explores their permutability, advancing understanding of their algebraic structure.

Findings

Sufficient conditions for primeness of p-adic meromorphic functions

Criteria for pseudo primeness in p-adic context

Results on permutability of entire functions

Abstract

In this paper, we study primeness and pseudo primeness of p-adic meromorphic functions. We also consider left (resp. right ) primeness of these functions. We give, in particular, sufficient conditions for a meromorphic function to satisfy such properties. Finally, we consider the problem of permutability of entire functions.

Equations

$$\biggl{(}\dfrac{1}{|\alpha|}\biggr{)}^{E((\frac{n+1}{N})^{N}-(\frac{n}{N})^{N})}\leq\biggl{|}\dfrac{a_{n}}{a_{n+1}}\biggr{|}\leq\biggl{(}\dfrac{1}{|\alpha|}\biggr{)}^{E((\frac{n+1}{N})^{N}-(\frac{n}{N})^{N})+1}$$

$$\biggl{(}\dfrac{1}{|\alpha|}\biggr{)}^{E((\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N})}\leq\biggl{|}\dfrac{a_{n}+1}{a_{n+2}}\biggr{|}\leq\biggl{(}\dfrac{1}{|\alpha|}\biggr{)}^{E((\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N})+1}$$

$$\biggl{|}\dfrac{a_{n+1}}{a_{n+2}}\biggr{|}-\biggl{|}\dfrac{a_{n}}{a_{n+1}}\biggr{|}\geq\biggl{(}\dfrac{1}{|\alpha|}\biggr{)}^{E((\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N})}-\biggl{(}\dfrac{1}{|\alpha|}\biggr{)}^{E((\frac{n+1}{N})^{N}-(\frac{n}{N})^{N})+1}$$

Full text

††Mathematics Subject Classification (2010): Primary 30D05, Secondary 30D35.††keywords: p-adic meromorphic function, primeness , right-primeness, left-primeness

1 Introduction

For every prime number $p$, we denote by $\mathbb{Q}_{p}$ the field of $p$-adic numbers, and we denote by $\mathbb{C}_{p}$ the completion of an algebraic closure of $\mathbb{Q}_{p}$, which is endowed with the usual $p$-adic absolute value.  Given $a\in\mathbb{C}_{p}$ and $r>0$,  $d(a,r)$ and $d(a,r^{-})$ are respectively the disks $\{x\in\mathbb{C}_{p}\ /\ |x-a|\leq r\}$ and $\{x\in\mathbb{C}_{p}\ /\ |x-a|<r\}$; and $C(a,r)$ is the circle $\{x\in\mathbb{C}_{p}\ /\ |x-a|=r\}=d(a,r)\setminus d(a,r^{-})$. It is easily seen that, two disks have a non-empty intersection if and only if they are nested (i.e, one of them is contained in the other).

We denote by $\mathcal{A}(\mathbb{C}_{p})$ the $\mathbb{C}_{p}$-algebra of entire functions in $\mathbb{C}_{p}$ and $\mathcal{M}(\mathbb{C}_{p})$ the field of meromorphic functions in $\mathbb{C}_{p}$, i.e, the field of fractions of $\mathcal{A}(\mathbb{C}_{p})$.  Let $f(x)=\sum_{n\geq 0}a_{n}x^{n}$ be a $p$-adic entire function. For all $r>0$, we denote by $|f|(r)=\max_{n\geq 0}|a_{n}|r^{n}$, the maximum modulus of $f$. This is extended to meromorphic functions $h=f/g$ by $|h|(r)=|f|(r)/|g|(r)$.

An element $f\in\mathcal{A}(\mathbb{C}_{p})$ (resp. $f\in\mathcal{M}(\mathbb{C}_{p})$) is said to be transcendental if it is not a polynomial (resp.  rational function).  Thus, a $p$-adic transcendental meromorphic function admits infinitely many zeros or infinitely many poles or both. It should also be noted that a transcendental entire function $f$ has no exceptional value or Picard value in $\mathbb{C}_{p}$, so for every $\beta\in\mathbb{C}_{p}$,  the function $f-\beta$ has infinitely many zeros.

Recall also that, if $D$ is a disk and $f$ an analytic function in $D$, then $f(D)$ is a disk. Moreover if $f^{{}^{\prime}}$ has no zero in $D$, there exist a disk $d\subset D$ such that the restiction $f_{\mid d}:d\longrightarrow f(d)$ is bi-analytic. This means that $f_{\mid d}$ is an analytic bijection and that its reciprocal function $f_{\mid d}^{-1}$ is also analytic $[4]$.

Let $F,f,g$ be $p$-adic meromorphic functions such that $F=f\circ g$.  We say that $f$ and $g$ are respectively left and right factors of $F$. The function $F$ is said to be prime (resp. pseudo-prime, resp. left-prime, resp. right-prime) if every factorization of $F$ of the above form implies that either $f$ or $g$ is a linear rational function (resp. $f$ or $g$ is a rational function, resp. $f$ is a linear rational function whenever $g$ is transcendental, $g$ is linear whenever $f$ is transcendental). If the factors are restricted to entire functions, the factorization is said to be in the entire sense.  The article by Bézivin and Boutabaa [2] is to our knowledge the only work dedicated to the study of factorization of $p$-adic meromorphic functions. They show, in particular, that if an entire function $F$ is prime (resp pseudo-prime) in $\mathcal{A}(\mathbb{C}_{p})$, then $F$ is prime (resp pseudo-prime) in $\mathcal{M}(\mathbb{C}_{p})$. This is false in the field $\mathbb{C}$ of complex numbers.  Indeed, Ozawa in [6] gives examples of complex entire functions that are prime in $\mathcal{A}(\mathbb{\mathbb{C}})$ without being prime in $\mathcal{M}(\mathbb{\mathbb{C}})$.

In this article, we provide sufficient conditions for a $p$-adic meromorphic function to be prime or pseudo-prime. We give examples of meromorphic functions satisfying these conditions.

We also show that almost all $p$-adic transcendental entire functions are prime, in the sense that it is most often enough to add or multiply these functions by an affinity to obtain a prime entire function.

Finally, we briefly discuss the question of permutability of entire functions. Or, in other words, one wonders: when do we have $f\circ g=g\circ f$ for $p$-adic entire functions $f$ and $g$?

Our method is based on the distribution of zeros and poles of the considered functions as well as the properties of their maximum modulus.

2 PRIMALITY AND PSEUDO PRIMALITY

Let us first prove the following result which gives sufficient conditions for a $p$-adic meromorphic function to be pseudo-prime.

Theorem 2.1**.**

Let $F$ be a transcendental meromorphic function in $\mathbb{C}_{p}$ whose all poles are simple except a finite number of them. Suppose moreover that, for every $\beta\in\mathbb{C}_{p}$, all the zeros of the function $F-\beta$ are simple except a finite number of them. Then $F$ is pseudo-prime.

To prove Theorem $2.1$ we need the following lemma, whose proof is given in $[5]$

Lemma 2.2**.**

Let $f_{1},f_{2}\in\mathcal{A}(\mathbb{C}_{p})$ such that $f_{1}^{{}^{\prime}}f_{2}-f_{1}f_{2}^{{}^{\prime}}\equiv c;\,\,\,\,c\in\mathbb{C}_{p}$. So, if one of the functions $f_{1},f_{2}$ is not affine then $c=0$  and  $f_{1}/f_{2}$ is a constant.

We also need the following lemma, whose proof is given in $[2]$

Lemma 2.3**.**

Let $F,f,g$ be three meromorphic functions in $\mathbb{C}_{p}$. Suppose that $f$ is not a rational function and that $F=f\circ g$. Then g entire.

Proof of Theorem 2.1.

Suppose that $F$ is not pseudo-prime. Hence there exist two transcendental meromorphic functions $f$ and $g$ such that $F=f\circ g$. Then, by Lemma $2.3$, the function $g$ is entire. Let us write $f$ in the form $f=f_{1}/f_{2}$ where $f_{1}$, $f_{2}$ are entire functions with no common zeros. As $f$ is transcendental, we see that at least one of the functions $f_{1}$, $f_{2}$ is transcendental. So by Lemma $2.2$,  $f_{1}^{{}^{\prime}}f_{2}-f_{1}f_{2}^{{}^{\prime}}$ is a non-constant entire function, hence admits at least a zero $\alpha$. Let us distinguish the two following cases:

1. If $f_{2}(\alpha)=0$, then $f_{1}(\alpha)\not=0$ and $f_{2}^{{}^{\prime}}(\alpha)=0$, so $\alpha$ is a multiple zero of $f_{2}$ and all element of the set $g^{-1}(\{\alpha\})$ are multiple zeros of $f_{2}\circ g$ and are multiple poles of $F$. Since $g\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$, then the set $g^{-1}(\{\alpha\})$ is infinite and $F$ has infinitely many multiple poles, which is a contradiction.

2. If $f_{2}(\alpha)\not=0$,  then $f^{{}^{\prime}}(\alpha)=(f_{1}^{{}^{\prime}}f_{2}-f_{1}f_{2}^{{}^{\prime}})(\alpha)/f_{2}(\alpha)=0$.  Let $\beta=f(\alpha)$, then $\alpha$ is a multiple zero of $f-\beta$. But the equation $g(x)=\alpha$ admits infinitely many solutions and for every such solution $\omega$ we have:

\left\{\begin{array}[]{ll}(F-\beta)(\omega)=F(\omega)-\beta=f\circ g(\omega)-\beta=f(\alpha)-\beta=0,\\ (F-\beta)^{{}^{\prime}}(\omega)=F^{{}^{\prime}}(\omega)=f^{{}^{\prime}}(g(\omega))\times g^{{}^{\prime}}(\omega)=f^{{}^{\prime}}(\alpha)\times g^{{}^{\prime}}(\omega)=0.\\ \end{array}\right.

Then $F-\beta$ has a infinitely many zeros, which is a contradiction again.

Hence $F$ is pseudo-prime.                       \sqcap$$\sqcup

Theorem 2.1. provides, in particular, necessary conditions for the pseudo-primeness of $p$-adic entire functions, which are summarized in the following corollary:

Corollary 2.4**.**

Let $F\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$ be such that, for every $\beta\in\mathbb{C}_{p}$,  all the zeros of the function $F-\beta$ are simple except a finite number of them.  Then $F$ is pseudo-prime.

The following results give more information about the left factor in any factorization of a $p$-adic entire function that satisfies the above conditions.

Theorem 2.5**.**

Let $F\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$ be such that for every $\beta\in\mathbb{C}_{p}$, only finitely many zeros of $F-\beta$ are multiple. Then $F$ is left-prime.

Proof.

Suppose that $F=f\circ g$ where $g\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$ and $f\in\mathcal{A}(\mathbb{C}_{p})$. By Corollary $2.4$, we know that $F$ is pseudo-prime. So $f$ is a polynomial. Suppose that $\deg f\geq 2$, then $f^{{}^{\prime}}$ has at least one zero $\alpha$. Since $g\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$, the set $W=\{x\in\mathbb{C}_{p}\ /\ g(x)=\alpha\}$ is infinite.

Let $\beta=f(\alpha)$.  Then, for every $\omega\in W$, we have:

\left\{\begin{array}[]{ll}((F-\beta)(\omega)=F(\omega)-\beta=f\circ g(\omega)-\beta=f(\alpha)-\beta=0,\\ (F-\beta)^{{}^{\prime}}(\omega)=F^{{}^{\prime}}(\omega)=f^{{}^{\prime}}(g(\omega))\times g^{{}^{\prime}}(\omega)=f^{{}^{\prime}}(\alpha)\times g^{{}^{\prime}}(\omega)=0.\\ \end{array}\right.

This means that  $F-\beta$ has infinitely many multiple zeros, a contradiction. Hence the left-primeness of $f$ is proven. ∎

Theorem 2.6**.**

Let $F\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$ be such that for every $\beta\in\mathbb{C}_{p}$, the function $F-\beta$ has at most one multiple zero. Then $F$ is prime.

Proof.

By Theorem $2.5$, we already see that $F$ is left-prime. So, it remains to show the right-primeness of $F$. For that, suppose that $F(z)=f\circ g$, where $f\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$ and $g\in\mathcal{A}(\mathbb{C}_{p})$. From Corollary $2.4$, we know that $F$ is pseudo-prime. So $g$ is a polynomial. Suppose that $\deg g=d\geq 2$. We have  $F^{\prime}(z)=f^{{}^{\prime}}(g(z))g^{{}^{\prime}}(z).$  Since $f\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$, the function $f^{{}^{\prime}}$  has infinitely many zeros. So we may choose an element $w\in\mathbb{C}_{p}$ such that $f^{{}^{\prime}}(w)=0$ and $g-w$ has only simple zeros $\gamma_{1},...,\gamma_{d}$.  Then for $i=1,...,d$, we have \left\{\begin{array}[]{ll}F(\gamma_{i})=f(w)=\beta\\ (F-\beta)^{{}^{\prime}}(\gamma_{i})=0\\ \end{array}\right.,   which means $\gamma_{1},...,\gamma_{d}$ are multiple zeros of $F-\beta$, a contradiction. Hence $F(z)$ is right-prime. ∎

Theorem 2.7**.**

Let $F$ be a $p$-adic transcendental meromorphic function that admits at most finitely many poles. Suppose that for every $\beta\in\mathbb{C}_{p}$, only finitely many zeros of $F-\beta$ are multiple. Then $F$ is right-prime.

To prove this theorem, we need the following lemma whose proof is given in Bézivin $[3]$. It is more general than Lemma $2.2.$

Lemma 2.8**.**

Let $n\geq 1$ and let $f_{1},\dots,f_{n}$ be $p$-adic entire functions such that the wronskian $W(f_{1},\dots,f_{n})$ is a non-zero polynomial. Then $f_{1},\dots,f_{n}$ are polynomials.

Proof of Theorem 2.7.

Suppose that $F(z)=f\circ g$, where $f\in\mathcal{M}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}(X)$ and $g\in\mathcal{M}(\mathbb{C}_{p})$. By Theorem $2.1$ the function $F$ is pseudo-prime. Then $g$ is a polynomial function. Suppose that $deg\,g\geq 2$. Let us write $f$ in the form $f={f_{1}}/{f_{2}}$  where $f_{1}$, $f_{2}$ are entire functions with no common zeros.  We have  $\displaystyle f^{\prime}(z)=W(f_{2},f_{1})/{f_{2}^{2}})(g(z))g^{{}^{\prime}}(z).$   As $f$ is transcendental and has a finite number of poles , we see that $f_{1}$ is transcendental and $f_{2}$ is polynomial.  As $f\in\mathcal{M}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}(X)$,  it follows by Lemma 2.8 that $W(f_{2},f_{1})\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}(X)$  and admits then infinitely many zeros $\{w_{n}\}$ ( which are not zeros of $f_{2}$ . For every integer $n$, big enough, the equation $g(z)=w_{n}$ admits at least two distinct roots , which are also common roots of \displaystyle\left\{\begin{array}[]{ll}F(z)=f(w_{n})\\ F^{{}^{\prime}}(z)=0\\ \end{array}\right..  Then we have a contradiction and $F(z)$ is right-prime.            \sqcap$$\sqcup

Theorem 2.9**.**

Let $f(x)=\sum_{n\geq N}a_{n}x^{n}$ be a $p$-adic entire function such that $a_{n}\neq 0$, for every $n\geq N$. Suppose that there exists an integer $n_{0}\geq N$ such that $(|a_{n}/a_{n+1}|)_{n\geq n_{0}}$ is a strictly increasing unbounded sequence. Then the function $f$ is pseudo-prime.

To prove Theorem 2.9 we need the following lemma whose proof is given in [4].

Lemma 2.10**.**

Let $f$ be a $p$-adic entire function defined by $f(x)=\sum_{n\geq N}a_{n}x^{n}$.  Let $\mu$ and $\nu$ be, respectively, the smallest integer and the largest one such that $\displaystyle|f|(r)=|a_{\mu}|r^{\mu}=|a_{\nu}|r^{\nu}$.  Then:

i)   $\mu$  is the number of zeros of $f$ in the disk $d(0,r^{-})$;

ii)    $\nu$  is the number of zeros of $f$ in the disk $d(0,r)$;

iii)   $\nu-\mu$  is the number of zeros of $f$ in the circle ${\mathcal{C}}(0,r)$.

Proof of Theorem 2.9.

Let us, first set $r_{n}=|a_{n}/a_{n+1}|,\ \forall n\geq 0$.  Quit to replace $n_{0}$ by a greater integer, we may assume that $n_{0}$ is such that:  $r_{n_{0}}>\max\{r_{0},\dots,r_{n_{0}-1}\}$.  In the sequel, we will show that $f$ has only simple zeros in $\mathbb{C}_{p}\backslash d(0,r_{n_{0}})$.  Indeed, let $r>r_{n_{0}}$.   We distinguish two cases:

i)  $r=r_{n}$ for some $n\geq n_{0}+1$. As the sequence $(r_{k})_{k\geq n_{0}}$ is strictly increasing, we have $r_{n-1}<r_{n}<r_{n+1}$. We then have:

$|a_{n+1}|r_{n}^{n+1}/|a_{n}|r_{n}^{n}=|a_{n+1}/a_{n}|r_{n}=1$, which means that $|a_{n}|r_{n}^{n}=|a_{n+1}|r_{n}^{n+1}$.

Moreover for every integer $\ell$, $0\leq\ell\leq n-1$, we have:

$|a_{\ell}|r_{n}^{\ell}/|a_{n}|r_{n}^{n}=|a_{\ell}/a_{n}|{1}/{r_{n}^{n-\ell}}=|a_{\ell}/a_{\ell+1}|\dots|a_{n-1}/a_{n}|{1}/{r_{n}^{n-k}}<1$.

Finally, for every integer $\ell$,   $\ell>n+1$, we have:

$|a_{\ell}|r_{n}^{\ell}/|a_{n+1}|r_{n}^{n+1}=|a_{\ell}/a_{n+1}|r_{n}^{\ell-n-1}$

$<|a_{\ell}/a_{n+1}||a_{n+1}/a_{n+2}|\dots|a_{\ell-1}a_{\ell}|=1.$

Hence $|f|(r_{n})=\max_{\ell\geq 0}|a_{\ell}|r_{n}^{\ell}$ is reached for the two values $\ell=n$ and $\ell=n+1$. This implies, by Lemm 2.10., that $f$ has only one zero in the circle $C(0,r_{n}).$

ii)  Suppose now that $r$ is different from $r_{n}$ for all $n>n_{0}$. Let $n\geq n_{0}$ be the sole integer such that $r_{n}<r<r_{n+1}$. We then have, for every integer $\ell$, $0\leq\ell\leq n$:

$\displaystyle|a_{\ell}|r^{\ell}=|\dfrac{a_{\ell}}{a_{\ell+1}}||\dfrac{a_{\ell+1}}{a_{\ell+2}}|\dots|\dfrac{a_{n}}{a_{n+1}}||a_{n+1}|r^{\ell}\leq|a_{n+1}|r_{n}^{n+1-\ell}r^{\ell}<|a_{n+1}|r^{n+1}$.

Moreover for every integer $\ell>n+1$, we have:

$\displaystyle\dfrac{|a_{\ell}|r^{\ell}}{|a_{n+1}|r^{n+1}}=\dfrac{|a_{\ell}|}{|a_{n+1}|}r^{\ell-n-1}<\dfrac{|a_{\ell}|}{|a_{n+1}|}r_{n+1}^{\ell-n-1}\leq|\dfrac{a_{\ell}}{a_{n+1}}||\dfrac{a_{n+1}}{a_{n+2}}|\dots|\dfrac{a_{\ell-1}}{a_{\ell}}|=1.$

Then $|f|(r)=\max_{\ell\geq 0}|a_{\ell}|r^{\ell}$ is reached only for $\ell=n+1$. This implies, by Lemma 2.10., that $f$ has no zero in the circle $C(0,r)$.   It follow that all the zeros of $f$ in $\mathbb{C}_{p}\backslash d(0,r_{n_{0}})$ are simple.  Now, for every $\beta\in\mathbb{C}_{p},$ there exists $r_{\beta}>0$ such that:   $\displaystyle|f-\beta|(r)=|f|(r),\,\,\forall r>\max(r_{n_{0}},r_{\beta}).$  It follows that $f-\beta$ has only simple zeros in $\mathbb{C}_{p}\backslash d(0,\max(r_{n_{0}},r_{\beta}))$.  This means that all the possible multiple zeros of $f-\beta$ lie in $d(0,\max(r_{n_{0}},r_{\beta}))$ and are therefore finitely many.  Using Corollary 2.4,  we complete the proof of Theorem 2.9.      \sqcap$$\sqcup

Corollary 2.11**.**

Let $f$ be a $p$-adic entire function satisfying the conditions of Theorem $2.9.$, then for any non-zero polynomial $P$, the $p$-adic meromorphic function $g=f/P$ is pseudo-prime.

Proof.

Note first that we may suppose that $f$ and $P$ have no common zeros. It is clear that $g$ has finitely many poles. Now let $\beta\in\mathbb{C}_{p}$. We see that the zeros of $g-\beta$ are the same as those of$f-\beta P$. Moreover, there exists $r_{\beta}>0$ such that for every $r>r_{\beta}$ we have: $\displaystyle|f-\beta P|(r)=|f|(r).$ It follows, by Theorem 2.9, that the function $f-\beta P$ (and therefore also $g-\beta$) has at most a finite number of multiple zeros. Thus by Theorem $2.1.$ the meromorphic function $g$ is pseudo-prime.

∎

Corollary 2.12**.**

Let $f(x)=\sum_{n\geq N}a_{n}x^{n}$ and $g(x)=\sum_{n\geq N}b_{n}x^{n}$ be two $p$-adic entire functions such that $a_{n}b_{n}\neq 0,\forall n\geq N$. Let $n_{0}\geq N$ be an integer such that the sequences $(|a_{n}/a_{n+1}|)_{n\geq n_{0}}$ and $(|b_{n}/b_{n+1}|)_{n\geq n_{0}}$ are strictly increasing and unbounded. Suppose moreover that $\lim_{r\to+\infty}(|f|(r)/|g|(r))=+\infty$.  Then the meromorphic function  $h=f/g$  is pseudo-prime.

Proof.

Note first that we may suppose that the entire functions $f$ and $g$ have no common zeros. By Theorem 2.9, we see that the entire functions $f$ and $g$ have at most finitely many multiple zeros. Hence the meromorphic function $h=f/g$ has at most finitely many multiple zeros and poles.

Now let $\beta\in\mathbb{C}_{p}$. We see that the zeros of $h-\beta$ are the same as those of$f-\beta g$. Moreover, as $\lim_{r\to+\infty}(|f|(r)/|g|(r))=+\infty$, there exists $r_{\beta}>0$ such that for every $r>r_{\beta}$ we have: $\displaystyle|f-\beta g|(r)=|f|(r).$ It follows, by Theorem 2.9, that the function $f-\beta g$ (and therefore also $h-\beta$) has at most a finite number of multiple zeros. Thus, by Theorem $2.1$, the meromorphic function $h$ is pseudo-prime.

∎

In what follows, we will provide examples of meromorphic p-adic functions satisfying the conditions of Corollary 2.12.  Let’s first recall that, given a real number $x$, we call integer part of $x$ and we denote by $E(x)$ the unique integer such that $E(x)\leq x<E(x)+1$.   It is easily shown that:

Lemma 2.13**.**

For all real numbers $x$ and $y$, we have:

$E(x-y)\leq E(x)-E(y)\leq E(x-y)+1.$

Proposition 2.14**.**

Let $N$ be an integer$\geq 3$ and let $\alpha,\beta\in\mathbb{C}_{p}$ be such that $|\beta|<|\alpha|<1$. Let $f$ and $g$ the functions defined by $f(x)=\sum_{n\geq 0}\alpha^{E((n/N)^{N})}x^{n}$ and $g(x)=\sum_{n\geq 0}\beta^{E((n/N)^{N})}x^{n}$. The meromorphic function $h={f}/{g}$ is a pseudo-prime.

Proof.

We have  $f(x)=\sum_{n\geq 0}a_{n}x^{n}$, where  $\displaystyle a_{n}=\alpha^{E((n/N)^{N})}$ for every $n\geq 0$.

We easily check that, $\lim_{n\to+\infty}|a_{n}|r^{n}=\lim_{n\to+\infty}|\alpha^{E((n/N)^{N})}|r^{n}=0$, for every $r>0$; which means that $f$ is an entire function in $\mathbb{C}_{p}$.

Let us now show that, if $n_{0}$ is an integer $\geq({2N^{N-1}}/({N-1}))^{{1}/({N-2})}$,  the sequence $(|{a_{n}}/{a_{n+1}}|)_{n\geq n_{0}}$ is strictly increasing.

For every $n\geq 0$, we have:    $|{a_{n}}/{a_{n+1}}|=({1}/{|\alpha|})^{E(({n+1}/{N})^{N})-E(({n}/{N})^{N})}$.

As the real function $x\mapsto({1}/{|\alpha|})^{x}$  is strictly increasing, it follows by Lemma 2.10. that:

[TABLE]

In the same way, we have:

[TABLE]

It follows from these last two inequalities that:

[TABLE]

But, by Lemma 2.10., We have:

$\displaystyle E((\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N})-E((\frac{n+1}{N})^{N}-(\frac{n}{N})^{N})-1\geq$

$\displaystyle\qquad\qquad\qquad\qquad\qquad E([(\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N}]-[(\frac{n+1}{N})^{N}-(\frac{n}{N})^{N}])-1$.

As :  $\displaystyle(\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N}=\dfrac{1}{N^{N}}\sum_{i=0}^{N-1}(n+2)^{i}(n+1)^{N-i-1}$ and

$\displaystyle(\frac{n+1}{N})^{N}-(\frac{n}{N})^{N}=\dfrac{1}{N^{N}}\sum_{i=0}^{N-1}n^{i}(n+1)^{N-i-1}$,  we see that:

$[(\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N}]-[(\frac{n+1}{N})^{N}-(\frac{n}{N})^{N}]=\dfrac{1}{N^{N}}\sum_{i=0}^{N-1}(n+1)^{N-i-1}[(n+2)^{i}-n^{i}]$. Hence:

$[(\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N}]-[(\frac{n+1}{N})^{N}-(\frac{n}{N})^{N}]=$

$\dfrac{2}{N^{N}}\sum_{i=0}^{N-1}(n+1)^{N-i-1}\sum_{j=0}^{i-1}(n+2)^{j}n^{i-j-1}$.

It follows that:

$[(\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N}]-[(\frac{n+1}{N})^{N}-(\frac{n}{N})^{N}]\geq$

$\dfrac{2}{N^{N}}\sum_{i=0}^{N-1}n^{N-i-1}\sum_{j=0}^{i-1}n^{j}n^{i-j-1}$,

then that:

$[(\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N}]-[(\frac{n+1}{N})^{N}-(\frac{n}{N})^{N}]\geq\dfrac{2n^{N-2}}{N^{N}}\sum_{i=0}^{N-1}i$,

and finally:

$[(\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N}]-[(\frac{n+1}{N})^{N}-(\frac{n}{N})^{N}]\geq\dfrac{n^{N-2}(N-1)}{N^{N-1}}$.

It follows that for every $n\geq n_{0}$, we have:

$[(\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N}]-[(\frac{n+1}{N})^{N}-(\frac{n}{N})^{N}]\geq 2$.

Consequently for every $n\geq n_{0}$, we have:

$E((\frac{n+2}{N})^{N}-(\frac{n+1}{N})^{N})-E((\frac{n+1}{N})^{N}-(\frac{n}{N})^{N})-1>0$.

It follows that for every $n\geq n_{0}$, we have: \biggl{|}\dfrac{a_{n+1}}{a_{n+2}}\biggr{|}>\biggl{|}\dfrac{a_{n}}{a_{n+1}}\biggr{|}.

We complete the proof by applying Theorem 2.9. ∎

In what follows, one aims to show that, given a $p$-adic transcendental entire function $f$, it is always easy to transform it into a prime entire function. For this, most often, we have just to add, or multiply $f$ by an affinity.

Theorem 2.15**.**

Let $f$ be a $p$-adic transcendental entire function. Then the set $\displaystyle\{a\in\mathbb{C}_{p};\,f(x)-ax\,\,\,is\,not\,prime\}$ is at most a countable set.

To prove Theorem 2.15 we need the following lemma.

Lemma 2.16**.**

Let $f\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$. Then there exists a countable set $E\subset\mathbb{C}_{p}$ such that for every $a\in\mathbb{C}_{p}\backslash E$ and every $b\in\mathbb{C}_{p}$, the function $f(x)-(ax+b)$ has at most one multiple zero.

Proof.

Let $Z(f^{{}^{\prime\prime}})$ be the set of zeros of $f^{{}^{\prime\prime}}$. Since $\mathbb{C}_{p}$ is a separable space, there exists a countable family of disks $(d_{i})_{i\geq 1}$ such that $\mathbb{C}_{p}\setminus Z(f^{{}^{\prime\prime}})=\cup_{i\geq 1}d_{i}$ and, for every $i\geq 1$, the restriction $f^{\prime}_{i}$ of $f^{\prime}$ to $d_{i}$ is a bi-analytic function on $d_{i}$. Then we have  $\displaystyle\mathbb{C}_{p}=(\cup_{i\geq 1}D_{i})\cup f^{{}^{\prime}}(Z(f^{{}^{\prime\prime}})).$

Let $g$ be the function defined on $\mathbb{C}_{p}$ by $g(x)=f(x)-xf^{\prime}(x)$. It should be noted that even if the family $(d_{i})_{i\geq 1}$ is chosen so that the disks $d_{i}$ are pairwise disjoint, there is no guarantee that the family $(D_{i})_{i\geq 1}$ retains this property. In other words some of the disks $D_{i}$ could be nested. To take account of this fact, let us set:

$\Gamma=\{(i,j)\in(\mathbb{N}^{\ast})^{2}\ /\ D_{i}\subsetneqq D_{j}\ \text{and}\ g\circ(f^{\prime}_{i})^{-1}\neq g\circ(f^{\prime}_{j})^{-1}\ \text{in}\ D_{i}\}$;

$\Delta_{ij}=\{x\in D_{i}\ /\ g\circ(f^{\prime}_{i})^{-1}(x)=g\circ(f^{\prime}_{j})^{-1}(x)\},\forall(i,j)\in\Gamma$  and $\Delta=\cup_{(i,j)\in\Gamma}\Delta_{ij}$.

Then $E=\mathbb{C}_{p}\setminus[(\cup_{i\geq 1}D_{i})\setminus[(f^{\prime}(Z(f^{\prime\prime})))\cup\Delta]]$ is a countable subset of $\mathbb{C}_{p}$.  Indeed it is easily seen that $E\subset(f^{\prime}(Z(f^{\prime\prime})))\cup\Delta$.

Let $a\in\mathbb{C}_{p}\backslash E$. Let $I_{a}=\{i\in\mathbb{N}^{\ast}\ /\ a\in D_{i}\}$.  Hence the disks $D_{i}$, for i\in I_{a}\, are nested. We easily show that the set of multiple zeros of the function  $f(x)-(ax+b)$  is equal to $A=\{(f^{\prime}_{i})^{-1}(a)\ /i\in I_{a}\ \text{and}\ f\circ(f^{\prime}_{i})^{-1}(a)=a(f^{\prime}_{i})^{-1}(a)+b\}$.

Suppose that $(f^{\prime}_{i})^{-1}(a)$ and $(f^{\prime}_{j})^{-1}(a)$ are two distinct elements of $A$. We may assume that $D_{i}\subsetneqq D_{j}$. The fact that each of these elements is a solution of the equation $f(x)-ax=b$ implies that: $f((f^{\prime}_{i})^{-1}(a))-(f^{\prime}_{i})^{-1}(a)f^{\prime}((f^{\prime}_{i})^{-1}(a))=f((f^{\prime}_{j})^{-1}(a))-(f^{\prime}_{j})^{-1}(a)f^{\prime}((f^{\prime}_{j})^{-1}(a))$  or,   in other words,  $g\circ(f^{\prime}_{i})^{-1}(a)=g\circ(f^{\prime}_{j})^{-1}(a)$.  As $a\notin\Delta$, we deduce that $g\circ(f^{\prime}_{i})^{-1}(x)=g\circ(f^{\prime}_{j})^{-1}(x),\forall x\in D_{i}$.

By derivation of this last equation, we have:  $(f^{\prime}_{i})^{-1}(x)=(f^{\prime}_{j})^{-1}(x),\forall x\in D_{i}$  and particularly $(f^{\prime}_{i})^{-1}(a)=(f^{\prime}_{j})^{-1}(a)$, which is a contradiction. Hence the set $A$ admits at most one element. This completes the proof Lemma 2.16. ∎

*Proof of Theorem 2.15 *

Let $E$ be the countable set of Lemme 2.16. and let $a\in\mathbb{C}_{p}\backslash E$.  Since $f$ is a transcendental entire function, we see that the function $h(x)=f(x)-ax$ is also a transcendental entire function. Lemma 2.16 then ensures that, for every $b\in\mathbb{C}_{p}$,   the function $h(x)-b=f(x)-ax-b$ admits at most one multiple zero. Theorem 2.6 finally enables us to conclude that the function $f(x)-ax=h(x)$ is prime.

Theorem 2.17**.**

Let $f$ be a $p$-adic transcendental entire function. Then the set $\displaystyle\{a\in\mathbb{C}_{p};\,f(x)(x-a)\,\,\,is\,not\,prime\}$ is at most a countable set.

To prove Theorem 2.17 we need the following lemma.

Lemma 2.18**.**

Let $f\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$. Then there exists a countable set $E^{\prime}\subset\mathbb{C}_{p}$ such that for every $a\in\mathbb{C}_{p}\backslash E^{\prime}$ and  every  $b\in\mathbb{C}_{p}$, the   function $(x-a)f(x)-b$ has at most one multiple zero.

Proof.

The procedure is very similar to that used in the   proof  of Lemma 2.16. We easily check that an element    $\zeta$ of $\mathbb{C}_{p}$    is a multiple zero of the function $(x-a)f(x)-b$  if and only if  \left\{\begin{array}[]{ll}a=g(\zeta)\\ b=h(\zeta)\\ \end{array}\right.,   where $g$ and $h$ are meromorphic functions,  defined in $\mathbb{C}_{p}$,  by:

\left\{\begin{array}[]{ll}g(x)=x+f(x)/f^{\prime}(x)\\ h(x)=(x-g(x))f(x)\\ \end{array}\right.                     (1)

Let  $S(g^{\prime})$  be the set of zeros and poles of $g^{\prime}$  and let $(d_{i})_{i\geq 1}$ be a countable family of disks  such that  $\mathbb{C}_{p}\setminus S(g^{\prime})=\cup_{i\geq 1}d_{i}$  and that,  for every $i\geq 1$,   the restriction   $g_{i}$  of   $g$ to $d_{i}$ is a bi-analytic function on $d_{i}$.  Then we have:  $\mathbb{C}_{p}=(\cup_{i\geq 1}D_{i})\cup g(S(g^{\prime})),\ \text{where}\ D_{i}=g(d_{i}).$

As noted before, the disks $D_{i}$ are not necessarily pairwise disjoint. In other words, some of them could be nested. To take account of this fact, let us set:

$\Gamma=\{(i,j)\in(\mathbb{N}^{\ast})^{2}/D_{i}\subsetneqq D_{j}\ \text{and}\ h\circ g_{i}^{-1}\neq h\circ g_{j}^{-1}\ \text{in}\ D_{i}\}$;

$\Delta_{ij}=\{x\in D_{i}/h\circ g_{i}^{-1}(x)=h\circ g_{j}^{-1}(x)\},\forall(i,j)\in\Gamma$  and $\Delta=\cup_{(i,j)\in\Gamma}\Delta_{ij}$.

Then  $E^{\prime}=\mathbb{C}_{p}\setminus[(\cup_{i\geq 1}D_{i})\setminus[(g(S(g^{\prime})))\cup\Delta\cup(\cup_{i\geq 1}Z(f\circ g_{i}^{-1}))]]$  is a  countable subset of $\mathbb{C}_{p}$.  Indeed, it is clear that  $E^{\prime}\subset(g(S(g^{\prime})))\cup\Delta\cup(\cup_{i\geq 1}Z(f\circ g_{i}^{-1}))]$.

Let $a\in\mathbb{C}_{p}\backslash E^{\prime}$. Let $I_{a}=\{i\in\mathbb{N}^{\ast}\ /\ a\in D_{i}\}$.  Hence the disks $D_{i}$, for i\in I_{a}\, are nested. We easily show, by relation (1), that the set of multiple zeros of the function $(x-a)f(x)-b$  is equal to $A=\{g_{i}^{-1}(a)\ /\ b=h\circ g_{i}^{-1}(a)\ \text{for}\ i\in I_{a}\}$.

Suppose that $g_{i}^{-1}(a)$ and $g_{j}^{-1}(a)$ are two distinct elements of $A$. Hence, we have: $h(g_{i}^{-1}(a))=h(g_{j}^{-1}(a))=b$.  Assuming that $D_{i}\subsetneqq D_{j}$ and using the fact that $a\notin\Delta$, we have:

$h\circ g_{i}^{-1}(x)=h\circ g_{j}^{-1}(x),\forall x\in D_{i}$                  (2)

From relation (1), we have:

$h^{\prime}=-fg^{\prime}$                           (3)

By derivation of relation (2) and using relation (3), we obtain:

$f\circ g_{i}^{-1}(x)=f\circ g_{j}^{-1}(x),\forall x\in D_{i}$                   (4)

Particularly, we have  $f\circ g_{i}^{-1}(a)=f\circ g_{j}^{-1}(a).$   But, since $a\in E^{\prime}$, we have  $f\circ g_{i}^{-1}(a)=f\circ g_{j}^{-1}(a)\neq 0$.  Using this, we deduce from relation  (1)  that: $g_{i}^{-1}(a)=g_{j}^{-1}(a)$, a contradiction.  Hence the set $A$ admits at most one element. This completes the proof Lemma 2.18. ∎

*Proof of Theorem 2.17 *

Let $E$ be the countable set of Lemme 2.18 and let $a\in\mathbb{C}_{p}\backslash E^{\prime}$.  Since $f$ is a transcendental entire function, we see that the function $h(x)=f(x)(x-a)$ is also a transcendental entire function. Lemma 2.18 then ensures that, for every $b\in\mathbb{C}_{p}$,   the function $h(x)-b=f(x)(x-a)-b$ admits at most one multiple zero. Theorem 2.6 finally enables us to conclude that the function $f(x)(x-a)=h(x)$ is prime.

3 PERMUTABILITY OF ENTIRE FUNCTIONS

In this part, we consider the question of permutability of $p$-adic meromorphic functions. In other words, given two $p$-adic meromorphic functions $f$ and $g$, under what conditions can we have $f\circ g=g\circ f$? The very particular situation that we are going to study gives an idea of the extreme difficulty of exploring this problem in a general way.  More precisely, we will show the following result:

Theorem 3.1**.**

Let $P$ and $f$ be respectively a non-constant polynomial and a transcendental entire function on $\mathbb{C}_{p}$ such that $P\circ f=f\circ P$. Then the polynomial $P$ is of one of the following forms:

i)  $P(x)=x$,

ii)  $P(x)=ax+b$, where $a,b\in\mathbb{C}_{p}$ such that $a$ is an $n$-th root of unity for some non-zero integer $n$.

To prove this theorem, we neen the following lemmas:

Lemma 3.2**.**

Let $f\in\mathcal{A}(\mathbb{C}_{p})$. Then the two following assertions are equivalent:

i)  $f$ is a polynomial,

ii)  there exist $c>0$ and $d\geq 0$ such that $|f|(r)\leq cr^{d}$, $r\to+\infty$.

Proof.

Suppose that $f$ is a polynomial of degree $n$: $f(x)=a_{0}+...+a_{n}x^{n},a_{n}\not=0$.

Since, for $r$ large enough, we have $|f|(r)=|a_{n}|r^{n}$, just take $c=|a_{n}|$ and $d=n$ to have the desired inequality.

Conversely, if there are $c>0$ and $d\geq 0$ such that $|f|(r)\leq cr^{d}$ as $r\to+\infty$, we have for every integer $n>d$:  $|f^{(n)}|(r)\leq{|f|(r)}/{r^{n}}\leq cr^{d-n}\longrightarrow 0.$  So $f^{(n)}\equiv 0$ and $f$ is a polynomial. ∎

Lemma 3.3**.**

Let $f\in\mathcal{A}(\mathbb{C}_{p})$ and suppose that there exist real numbers $\alpha,\beta>0$ and an integer $n\geq 2$ such that, $|f|(\alpha r^{n})<\beta(|f|(r))^{n},$ when $r\to+\infty$.  Then $f$ is a polynomial.

Proof.

We will construct real numbers $c,d>0$ such that $|f|(r)\leq cr^{d}$, $r\to+\infty$. Let $r_{0}$ be a real number such that $r_{0}>\max\{1,\alpha^{{1}/({1-n})}\}.$  Let us choose $c,\ d>0$ such that \left\{\begin{array}[]{ll}|f|(r_{0})<cr_{0}^{d}\\ \beta<{\alpha^{d}}/{c^{n-1}}\\ \end{array}\right.,    and let us set $r_{k+1}=\alpha r_{k}^{n}$ for $k\geq 0$.

The sequence $(r_{k})$ est strictly increasing and tends to $+\infty$.  We prove by induction that: $|f|(r_{k})<cr_{k}^{d},\,\,\forall k\geq 0$.  Indeed, the property holds for $k=0$.  Suppose that it holds for some integer $k\geq 0$. Then we have:

$|f|(r_{k+1})=|f|(\alpha r_{k}^{n})<\beta(|f|(r_{k}))^{n}<\beta c^{n}r_{k}^{dn}=c(\beta c^{n-1})r_{k}^{dn}$

$<c\alpha^{d}r_{k}^{dn}=c(\alpha r_{k}^{n})^{d}=cr_{k+1}^{d}$,

which means that this inequality is true for $k+1$.

For every $r\in[r_{k},r_{k+1}]$, there exists $t\in[0;\ 1]$ such that $r=r_{k}^{t}r_{k+1}^{1-t}$.   As $\log r\mapsto\log|f|(r)$ is a convex function, we deduce that:

\begin{array}[]{cl}|f|(r)&\leq(|f|(r_{k}))^{t}(|f|(r_{k+1}))^{1-t}=(|f|(r_{k}))^{t}(|f|(\alpha r_{k}^{n}))^{1-t}\\ \\ &<(|f|(r_{k}))^{t}(\beta(|f|(r_{k}))^{n})^{1-t}<(cr_{k}^{d})^{t}(\beta(cr_{k}^{d})^{n})^{1-t}\\ \\ &=c^{t}r_{k}^{dt}c^{1-t}(\beta c^{n-1})^{1-t}(r_{k}^{n(1-t)})^{d}<cr_{k}^{dt}\alpha^{d(1-t)}r_{k}^{n(1-t)d}\\ \\ &=c(r_{k}^{t}(\alpha r_{k}^{n})^{1-t})^{d}=c(r_{k}^{t}r_{k+1}^{1-t})^{d}=cr^{d}.\end{array}

Hence $|f|(r)<cr^{d},\,\,\forall r\geq r_{0}$.  By Lemma $3.2$, $f$ is then a polynomial. ∎

Proof of Theorem 3.1.

Let us set $P(x)=a_{0}+\dots+a_{k}x^{k}$, where $a_{0},\dots,a_{k}$ are elements of $\mathbb{C}_{p}$ and $a_{k}\neq 0$.  Then: $(f\circ P)(x)=(P\circ f)(x)=a_{0}+\dots+a_{k}(f(x))^{k}$.  Suppose that $k\geq 2$. Then, for $r>0$, we have $|f|(|P|(r))=|P|(|f|(r)).$  It follows that, for sufficiently large $r>0$, we have $|f|(|a_{k}|r^{k})=|a_{k}|(|f|(r))^{k}<2|a_{k}|(|f|(r))^{k}$.  Lemma 3.3 then implies that $f$ is a polynomial, , which is a contradiction. Hence, we have $k=1$ and $P$ is of the form $P(x)=ax+b$, where $a,b\in\mathbb{C}_{p}$ and $a\neq 0$.  Two cases can then arise:

i) $a=1$. Then we have $b=0$. Indeed suppose that $b\neq 0$. Then we have, for every $x\in\mathbb{C}_{p}$, $f(x+b)=f(x)+b$. It follows that $f^{\prime}(x+b)=f^{\prime}(x)$, for every $x\in\mathbb{C}_{p}$. Let $\zeta$ be a zero of $f^{\prime}$ such that $|\zeta|>|b|$. Then $\zeta,\zeta+b,\zeta+2b,\dots$ are infinitely many zeros of $f^{\prime}$ included in the disk $d(0;|\zeta|)$, a contradiction. Hence in this case we have $P(x)=x$.

ii) $a\neq 1$.  Let $\sigma$ be the affine application $\sigma(t)=t+{b}({a-1})^{-1}$, hence its inverse$\sigma^{-1}$ is given by $\sigma^{-1}(t)=t+{b}({1-a})^{-1}$.  Let $F$ be the function given by $F=\sigma\circ f\circ\sigma^{-1}$.  It is easily seen that:  $F(ax)=aF(x),\ \forall x\in\mathbb{C}_{p}$.

If $F(x)=\sum_{n\geq 0}b_{n}x^{n}$, we have $F(ax)=\sum_{n\geq 0}b_{n}a^{n}x^{n}$. It follows that:

$\sum_{n\geq 0}(a^{n}-a)b_{n}x^{n}=0$, and hence $b_{0}=0$ and $(a^{n}-a)b_{n+1}=0,\,\,\forall n\geq 1$.

Suppose that there exist two relatively prime integers $m,\ n\geq 2$ such that $b_{n+1}\not=0$, and $b_{m+1}\not=0$ we would have $a^{n}=a^{m}=1$, and therefore $a=1$, which excluded. Hence $F$ has the form $F(X)=\sum_{k\geq 1}b_{nk+1}x^{nk+1}$,  where $n$ is the smallest positive integer such that $a^{n}=1$.  It follows that:

$f(x)=\sum_{k\geq 1}b_{nk+1}(x+{b}/({a-1}))^{nk+1}+{b}/({1-a}).$           \sqcap$$\sqcup

Corollary 3.4**.**

Let $f\in\mathcal{A}(\mathbb{C}_{p})\backslash\mathbb{C}_{p}[X]$, $f(x)=\sum_{n\geq 0}a_{n}x^{n}$. Suppose that there exist two relatively prime integers $m,l$ such that $a_{m}a_{l}\not=0$. Then the only non-constant polynomial $P$ such that $P\circ f=f\circ P$ is $P(x)=x$.