Journal of Number Theory 87, 211221 (2001)

doi:10.1006jnth.2000.2591, available online at http:www.idealibrary.com on

On Uniqueness Polynomials and Bi-URS for p-adic

Meromorphic Functions

Ha Huy Khoai

Institute of Mathematics, P.O. BOX 631 BO HO, 10000 Hanoi, Vietnam

E-mail: hhkhoaihanimath.ac.vn

and

Ta Thi Hoai An

Vinh Pedagogical Institute, Vinh, Nghe An, Vietnam

E-mail: tthanhanimath.ac.vn

Communicated by M. Waldschmidt

Received September 13, 1999

Let W be an algebraically closed field of characteristic zero, and let K be an

algebraically closed field of characteristic zero, complete for an ultrametric absolute
value. A(K) will denote the ring of entire functions in K and M(K) will denote the
field of meromorphic functions in K. In this paper we give some classes of unique-
ness polynomials for M(K) and show the existence of a bi-URS for M(K) of the
form ([a 1 , a 2 , a 3 , a 4 ], [|]). Also a sufficient condition of uniqueness range sets for
M(K) in terms of uniqueness polynomials is given.  2001 Academic Press

1. INTRODUCTION

Definitions and Notations. We first recall some notations and definitions

(see [9]). Let W be an algebraically closed field of characteristic zero,
W*=W "[0] and W be the projective space of dimension 1 over W. Let
K be an algebraically closed field of characteristic zero, complete for an
ultrametric absolute value. Let L be a field which is either C or K. We
denote by A(L) the ring of entire functions in L and by M(L) the field of
meromorphic functions, i.e., the field of fractions of A(L).
211
0022-314X01 35.00
Copyright  2001 by Academic Press
All rights of reproduction in any form reserved.
212 KHOAI AND AN

For f # M(K) and S a subset of K, we denote

E( f, S)= . [(q, z) # N*_K | z a zero of order q of f (z)&a].

a#S

Besides, given a subset of K containing [], we denote by E( f, S) the

set of N*_K : E( f, S & K) _ [(q, z) | z a pole of order q of f ]. Usually, the
notation E f (S) express the set which contains the same points as E f (S) but
without counting multiplicities.
Let F be a nonempty subset of M(K). A subset S of K is called a unique
range set (a URS in short) for F if for any f, g # F such that E( f, S)=
E( g, S), one has f =g.
In the same way, a couple of sets S, T in K such that S & T=< will be
called a bi-URS for F if for any f, g # F such that E( f, S)=E( g, S) and
E( f, T)=E(g, T ), one has f =g.
The problem of determining a meromorphic (or entire) function on W
by its single pre-images, counting multiplicities, of finite sets is an impor-
tant one and it has been studied by many mathematicians. For instance, in
1921, G. Polya [22] showed that an entire function on C is determined by
the inverse images, counting multiplicities, of 3 distinct non-omitted values.
In 1926, R. Nevanlinna [20] showed that a meromorphic function on the
complex plane C is uniquely determined by the inverse images, ignoring
multiplicities, of 5 distinct values. A few years later, by the argument of
G. Polya [22], he showed that when multiplicities are considered, 4 points
are sufficient (with one exceptional situation).
In recent years, much attention has been given to URS for M(C) or
A(C) and various results have been obtained by Yi [23], Li and Yang
[17, 18], Mues and Reinders [19] and recently by Shiffman [21], Frank
and Reinders [11], and Fujimoto [10].
For the p-adic case, several interesting results about URS for p-adic
entire and meromorphic functions also have been obtained (see [2, 46,
13], etc.). In [2], A. Boutabaa and A. Escassut proved that, for every
n5, there exist bi-URS for M(K) of the form ([a 1 , a 2 , ..., a n ], []). In
[9], A. Escassut et al. proved that, there exist no bi-URS for M(K) of the
form ([a 1 , a 2 , a 3 ], [|]), and posed the question whether there exist bi-
URS for M(K) of the form ([a 1 , a 2 , a 3 , a 4 ], [|]). In this paper, we give
a positive answer to the above question of Escassut, Haddad and Vidal
[9]. Namely, we prove that for a generic set [a 1 , a 2 , a 3 , a 4 ], ([a 1 , a 2 , a 3 ,
a 4 ], [|]) is a bi-URS for M(K). Then we will give a sufficient condition
of URS in terms of uniqueness polynomials. The main tool is the p-adic
Nevanlinna theory, developed in [1, 14, 16].
Throughout this paper, we shall make use of standard notations of
Nevanlinna Theory such as m(r, f ), N(r, f ), T(r, f ), N 0(r, f ), N(r, f ), etc.
 ON UNIQUENESS POLYNOMIALS 213

(see [3, 12]). For a certain condition C, we denote by N(r, 1f ) | C the

valence function of the zeros of f, where the zero z is taken if z satisfies
condition C.

2. UNIQUENESS POLYNOMIALS AND bi-URS FOR

MEROMORPHIC FUNCTIONS

Definition. We say that a non-constant polynomial P(z) is a strong

uniqueness polynomial for F if the identity P( f )=cP(g) implies f =g for
any pair of non-constant functions f, g # F and for any non-zero constant
c # K.
Similarly, we say that a non-constant polynomial P(z) is a weak unique-
ness polynomial for F if for any pair of non-constant functions f, g # F the
identity P( f )=P(g) implies f =g.
Throughout this paper, we assume that P(z) is a non-zero polynomial of
degree q without multiple zeros and its derivative is given by

P$(z)=q(z&d 1 ) q1 } } } (z&d k ) qk,

where q 1 + } } } +q k =q&1 and d 1 , ..., d k are distinct zeros of P$. The

number k is called the derivative index of P.

Definition. A non-zero polynomial P(z) is said to satisfy the condition

(H) if P(d l ){P(d m ) for 1l<mk. (see [10]).
We can assume that d 1 , ..., d k # K"[0].
Suppose k=1. Then we can write P(z) in the form P(z)=(z&a) q &c for
some points a and c with c{0. For an arbitrary non-constant function
f # M(K) and a constant ! ( {1) with ! q =1, the function g :=! f +
(1&!) a satisfies the condition P( f )=P(g), but g{ f. Thus, P(z) cannot
be a uniqueness polynomial. Therefore, we shall assume that k2.
Let f be a non-constant meromorphic function. For a point a # K we
define the function / af : K Ä Z by

0 if f (z){a
/ af(z) :=
{ 1 if f (z)=a.

If a=, define / f (z) := &1 if z is a pole of f.

For a condition C, we define

/ 0f $(z) if z satisfies the condition C and f (z){d j for any j

/*
f $ | C (z) :=
{ 0 otherwise.
214 KHOAI AND AN

It is shown in [9] that a URS for A(K) (resp., M(K)) is also a URS for
W[x] (resp., W(x)). Therefore, in this paper, the results which are valid
for A(K) (resp., M(K)) are also valid for W[x] (resp., W(x)).

Theorem 1. Let P(z) # K[z] have no multiple zeros, have derivative

index k3, and satisfy the condition (H). Then P(z) is a weak uniqueness
polynomial.
Proof. Suppose that there are two distinct non-constant meromorphic
functions f and g such that P( f )=P( g).
Define

1 1
.= & .
f g

Then, .0 and

T(r, .)T(r, f )+T(r, g)+O(1).

From P( f )=P( g) we conclude that f and g have poles of the same

order. Therefore /  
f (z)=/ g (z). On the other hand, we have f $(z) P$( f (z))
d d
= g$(z) P$( g(z)). As P satisfies the condition (H), we have / f j(z)/ gj(z)+
/*
g$ | f =dj (z). It implies that

k k
d  d 
: / f j(z)&/ f (z) : (/ gj(z)+/*g$ | f =dj (z))&/ g (z)
j=1 j=1

k
/ 0.(z)+ : /*
g$ | f =dj (z).
j=1

Therefore, applying the Second Main Theorem (see [3]) to the function f
and the values d 1 , ..., d k , we have
k
1 1
(k&1) T(r, f ) : N r,
j=1
\ f &d j + \ +
+N(r, f )&N 0 r,
f$
&log r+O(1)

k
1 1 1
\ .+ + : N \r, g$+} &N \r, f $+ &log r+O(1).
N r,
j=1
0
f =dj
0

We also have similar inequalities for the function g,

k
1 1 1
(k&1) T(r, g)N r,
\ +.
+ : N 0 r,
j=1
f$ \ +} g=dj
&N 0 r,
\ +g$
&log r+O(1).
 ON UNIQUENESS POLYNOMIALS 215

Summing up these inequalities and using the First Main Theorem (see
[1]), we get

1 1
\ f $+ &N \r g$+
(k&1)(T(r, f )+T(r, g))2(T(r, f )+T(r, g))&N 0 r, 0

k
1 1
\ \ g$+}
+ : N r,
j=1
0
\ g$+} +
+N r,
f =dj
0
g=dj

&2 log r+O(1).

Since

k
1 1
j=1
\ +}
: N 0 r,
g$ f =dj \ +
N 0 r,
g$
,

and

k
1 1
j=1
\ f $+}
: N 0 r,
g=dj \ f $+ ,
N 0 r,

we have

(k&3)(T(r, f )+T(r, g)) &log r+O(1).

It implies that k&3<0, a contradiction. K

Remark. Without the condition (H) in Theorem 1, we can construct a

polynomial which is not a weak uniqueness polynomial (and hence, it is
not a strong uniqueness polynomial).

In fact, we can easily choose points a j (1 jn), where n1, such that
the polynomial

P(z)=z 2n +a 1 z 2n&2 + } } } +a n&1 z 2 +a n ,

has no multiple zeros and its derivative

P$(z)=2nz 2n&1 +a 1(2n&2) z 2n&3 + } } } +2a n&1 z,

has 2n&1 distinct zeros 0, \d 1 , ..., \d n&1 .

We have P(d l )=P(&d l ) and P( f )=P(&f ) for any non-constant
meromorphic function f. Thus P is not a weak uniqueness polynomial.
216 KHOAI AND AN

Definition. A non-zero polynomial P(z) is said to satisfy the condition

(G) if  ki=1 P(d i ){0.

Theorem 2. Let P(z) # K[z] be a polynomial having no multiple zeros.

Let further the derivative index k3 and P(z) satisfy the conditions (H) and
(G). Then P(z) is a strong uniqueness polynomial.

Proof. By Theorem 1, P(z) is a weak uniqueness polynomial.

Assume that P(z) is not a strong uniqueness polynomial. Then there
exist two distinct non-constant meromorphic functions f and g such that
P( f )=cP( g) for some non-zero constant c{1.
We consider the set

4 :=[(l, m) : P(d l )=cP(d m )],

and denote the number of elements of 4 by k 0 . We set k 0 =0 if 4=<.

In order to complete the proof we need the following lemmas.

Lemma 1. In the above hypothesis, assume additionally that f{(c 0 g+c 1 )

(c 2 g+c 3 ) for any c 0 , c 1 , c 2 , c 3 # K, c 0 {0. Then k 0 =k.

Proof. It is easy to see that if (m 1 , l 1 ), (m 2 , l 2 ) are arbitrary elements of

4 such that m 1 =m 2 or l 1 =l 2 , then (m 1 , l 1 )=(m 2 , l 2 ). This implies k 0 k.
We consider the possible cases:

Case 1. k 0 =0. Define

1 1
.= & .
f g

Then . is a non-zero meromorphic function. As in the proof of Theorem 1,

we have k<3, a contradiction.

Case 2. k 0 =1. Then there exists a unique element (l, m) such that
P(d l )=cP(d m ).
Define

1 dm
.= & .
f dl g

Since f{(c 0 g+c 1 )(c 2 g+c 3 ) for any c 0 , c 1 , c 2 , c 3 # K, c 0 {0, . is a

non-zero meromorphic function.
 ON UNIQUENESS POLYNOMIALS 217

It is easy to see that g(z)= for any z # K with f(z)=, and g(z)=d j $
or g$(z)=0 for any z # K with f (z)=d j . If j=l, by the hypothesis 4 consists
of only one element, then j $=m. It implies that .(z)=0. On the other
hand, if j{l, then g$(z)=0. We conclude that

k k
: / df j(z)&/ 
j=1
dm
f (z)/ g (z)+ : /*
g$
j=1
} f =dj
(z)&/ 
g (z)

k
/ 0.(z)+ : /*
g$ | f =dj (z).
j=1

Using the Second Main Theorem and repeating the argument in the proof
of Theorem 1, we get a contradiction.

Case 3. k 0 2. After a suitable change of indices, we may assume that

P(d 1 )=cP(d t(1) ), ..., P(d k0 )=cP(d t(k0 ) ). (1)

Define

1 d t(2) &d t(1)

.= & .
f (d 2 &d 1 )( g&d t(1) )+d 1(d t(2) &d t(1) )

Then . is a non-zero meromorphic function.

If f (z)=, z # K, then g(z)=. If f (z)=d j , 1 jk 0 , z # K, then, by
(1), g(z)=d t( j) or g$(z)=0, because P satisfies the condition (H).
Moreover, if f (z)=d 1 or f (z)=d 2 , then g$(z)=0 or .(z)=0. If f (z)=d j ,
k 0 +1 jk, z # K, then, by the definition of 4, P(d j ){cP(d j $ ) for every
1 j $k. Hence g(z){d j $ for every 1 j $k. It implies g$(z)=0. We
conclude that

k k0
: / df j(z)&/  dt ( j )
f (z) : (/ g (z)+/*
g$ | f =dj (z))
j=1 j=1

k
+ :
j=ko +1
/*
g$
} f =dj
(z)&/ 
g (z)

k0 k
/ 0.(z)+ : / dgt ( j )(z)+ : /*
j=3
g$
j=1
} f =dj
(z).
218 KHOAI AND AN

Therefore, applying the Second Main Theorem to the function f and the
values d 1 , ..., d k , we have
k
1 1
(k&1) T(r, f ) : N r,
j=1
\ f &d + +N(r, f )&N \r, f $+ &log r+O(1)
j
0

k0
1 1
\ .+ + : N \r, g&d +
N r,
j=3 t( j)

k
1 1
+ : N r,
\ g$+}
j=1
0 &N r,
\ f $+ f =dj
0

&log r+O(1).

We have similar inequalities for the function g,

k0
1 1
(k&1) T(r, g)N r,
\ + .
+ : N r,
j=3
f &d \
t( j) +
k
1 1
+ : N 0 r,
j=1
\ f $+} g=dj \ g$+
&N 0 r,

&log r+O(1).

Summing up these inequalities and using the First Main Theorem, we get

(k&1)(T(r, f )+T(r, g))2(T(r, f )+T(r, g))

+(k 0 &3+1)(T(r, f )+T(r, g))
&2 log r+O(1).

Thus,

(k&k 0 &1)(T(r, f )+T(r, g)) &2 log r+O(1).

It implies that k 0 >k&1 and hence, k 0 =k.

Lemma 2. Assume that k3 and P(z) satisfies the condition (H). If
there are two distinct non-constant meromorphic functions f and g such that
P( f )=cP( g) for some non-zero constant c, then we can find a permutation
(t(1), ..., t(k)) of (1, ..., k) such that

P(d 1 ) P(d k )
c= = }}} = .
P(d t(1) ) P(d t(k) )
 ON UNIQUENESS POLYNOMIALS 219

Proof. It suffices to show that k=k 0 . We consider the following cases:

Case 1. f{(c 0 g+c 1 )(c 2 g+c 3 ) for any c 0 , c 1 , c 2 , c 3 # K, c 0 {0. From
Lemma 1, it implies immediately that k=k 0 .
Case 2. f =(c 0 g+c 1 )(c 2 g+c 3 ) for some c 0 , c 1 , c 2 , c 3 # K.
For the proof of k=k 0 , see [10]. K
We now continue to prove Theorem 2.
By Lemma 2, there is a permutation (t(1), t(2), ..., t(k)) of (1, 2, ..., k)
such that

P(d 1 ) P(d k )
c= = }}} = {1.
P(d t(1) ) P(d t(k) )

Since P satisfies the condition (G), we have an absurd identity

P(d 1 )+P(d 2 )+ } } } +P(d k )

c= =1.
P(d t(1) )+P(d t(2) )+ } } } +P(d t(k) )

The proof of Theorem 2 is completed. K

Corollary 1. For every q4, there exist strong uniqueness polyno-

mials of degree q.

Example. The polynomial P(z)=3z 4 &8z 3 &6z 2 +24z is a strong

uniqueness polynomial (and hence, it is also a weak polynomial).
Indeed, it is easy to check that P satisfies the hypothesis of Theorem 2.

Corollary 2. Let P(z) be a polynomial satisfying the hypothesis of

Theorem 2 and | # K. If S is the set of roots of P(z)=0 and | Â S, then
(S, [|]) is a bi-URS for M(K).
Proof. Without loss of generality, we can assume that |=.
Assume that f, g are two non-constant meromorphic functions satisfying
E( f, S)=E(g, S) and E( f, )=E(g, ). This implies immediately that
P( f )P( g)=c for some non-zero constant c. On the other hand, it follows
from Theorem 2 that P(z) is a uniqueness polynomial. Therefore, f =g.
Thus (S, []) is a bi-URS for M(K). K
Remark. The number 4 in Corollary 1 is the best-possible. Indeed,
assume that there exists a uniqueness polynomial P of degree 3. Let S :=
[a 1 , a 2 , a 3 ] be the set of roots of P. Suppose E( f, S)=E( g, S) and E( f, )
=E(g, ) for two non-constant meromorphic functions f, g. Then, we
have P( f )=cP(g) for some non-zero constant c. By the assumption, it
220 KHOAI AND AN

implies f =g. Therefore, there exist bi-URS for M(K) of the form ([a 1 , a 2 ,
a 3 ], [|]). It is a contradiction (see [9]).

3. SUFFICIENT CONDITION FOR UNIQUE RANGE SETS

Consider the polynomial

P(z)=(z&a 1 )(z&a 2 ) } } } (z&a q ).

Assume that the derivative of P is given by

P$(z)=q(z&d 1 ) q1 (z&d 2 ) q2 } } } (z&d k ) qk,

where k2.
It is known that if S :=[a 1 , a 2 , ..., a q ] is URS for M(K) (A(K) resp.),
then P is a strong uniqueness polynomial for meromorphic (entire resp.)
functions. It would be nice to give suitable conditions such that the
converse assertion holds. In this section, we give partial answers to this
problem.

Theorem 3. Let P(z) be a strong uniqueness polynomial, satisfy the

condition (H), have derivative index k3, or k=2 and inf(q 1 , q 2 )2. Let
S be the set of the roots of P. If q>2k+5, (q>2k+11 resp.), then for any
pair of non-constant meromorphic functions f and g the equality E( f, S)=
E( g, S), (E( f, S)=E( g, S) resp.), implies f =g.
Theorem 3 is a p-adic version of Theorem 7.2 in [10], and the proof is
omitted (we refer the reader to [15]). Note that from Theorem 3, it follows
that there exist unique range sets for meromorphic functions if q10 (see
[13]) and unique range sets for meromorphic functions ignoring multi-
plicities if q16 (see [4]).

ACKNOWLEDGMENTS

The paper was completed during a stay of the second-named author at the Laboratoire de
Mathematiques Emile Picard (CNRS UMR 5580, Universite Paul Sabatier, Toulouse,
France). This author thanks Professor Pascal Thomas and Professor Nguyen Thanh Van for
the invitation and helpful discussions. Some techniques in this paper are inspired from the
preprint [10] and the authors would like to thank Professor Fujimoto. Sincere thanks are due
also to the refree for hisher careful reading of the first and second versions of this paper and
for useful suggestions.
 ON UNIQUENESS POLYNOMIALS 221

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