ON FUNCTIONAL EQUATIONS RELATED TO ITERATION OF YI’S POLYNOMIALS

In this paper, we show that the equation P(f1, ..., fs+1) = Q(g1, ..., gs+1), where P, Q are polynomials in a class of homogeneous polynomials of FermatWaring type, has entire solutions f1, · · · , fs+1; g1, · · · , gs+1. Some classes of unique range sets for linearly nondegenerate holomorphic curves are also obtained

ON FUNCTIONAL EQUATIONS RELATED TO ITERATION OF

YI’S POLYNOMIALS

HA HUY KHOAI, VU HOAI AN, AND PHAM NGOC HOA

Abstract In this paper, we show that the equation P (f 1 , , f s+1 ) = Q(g 1 , , g s+1 ), where P, Q are polynomials in a class of homogeneous polynomials of

Fermat-Waring type, has entire solutions f 1 , · · · , f s+1 ; g 1 , · · · , g s+1 Some classes of unique

range sets for linearly non-degenerate holomorphic curves are also obtained.

1 Introduction

We consider the function equation

P (f1, f2, · · · , fs+1) = Q(g1, g2, · · · , gs+1), (1) where P, Q are polynomials, and fi, gi are entire functions

Since the paper of Ritt ([11]), the functional equation P (f ) = Q(g) has been investigated by many authors (see [1], [2], [3], [4], [6], [8], [9], [10]) This equation related to some other questions, as to find the conditions under which f−1(X) =

g−1(Y ), where f, g are polynomials, and X, Y are compact sets; or the question of finding unique range sets for meromorphic functions

It is shown in [4] that for a generic pair of polynomials P, Q the equation has no non-constant solutions in the set of meromorphic functions Therefore one should interested to find the solutions for some classes of polynomials For example, in [2] and [6] the authors show some classes of pairs P, Q so that the equation has solutions

In this paper we consider the equation P (f ) = Q(g), where P, Q are polynomials

of several variables, and of the Fermat-Waring type, obtained by iteration of Yi’s polynomials Recall that Yi’s polynomials are defined by

P (z) = azn+ bzn−m+ c, where a, b, c are complex numbers It is well-known that the zeros sets of Yi’s polynomials, with some additional conditions on m, n, are unique range sets for meromorphic functions

We first show a class of such polynomials for which the equation (1) has solutions

As a consequence we establish some conditions on holomorphic curves f and g, such that the pull-back divisors νf(X) and νg(Y ) coincide, where X and Y are hypersurfaces defined by polynomials of the mentioned above class Some classes of unique range sets for meromorphic functions are also obtained

2010 Mathematics Subject Classification 30D35.

Key words and phrases Diophantine Equations, unique range sets, holomorphic curves.

1 The work was supported by the National Foundation for Science and Technology Development (NAFOSTED) and the Vietnam Institute for Advanced study in Mathematics (VIASM).

Now let us describe the class of polynomials of Fermat-Waring type considered in this paper

Set:

P1(z1, z2) = a11z1n+ b11z1n−mz2m+ c11z2n,

P2(z1, z2, z3) = a12P1n(z1, z2) + b12P1n−m(z1, z2)z3nm+ c12z3n2,

Pi(z1, z2, , zi+1) = a1iPi−1n (z1, z2, , zi) + b1iPi−1n−m(z1, z2, , zi)zi+1ni−1m+ c1izi+1ni ,

Ps(z1, z2, , zs+1) = a1sPs−1n (z1, z2, , zs) + b1sPs−1n−m(z1, z2, , zs)zs+1ns−1m+ c1szs+1ns ,

Q1(z1, z2) = a21z1n+ b21z1n−mz2m+ c21z2n,

Q2(z1, z2, z3) = a22Qn1(z1, z2) + b22Qn−m1 (z1, z2)z3nm+ c22zn32,

Qi(z1, z2, , zi+1) = a2iQni−1(z1, z2, , zi) + b2iQn−mi−1 (z1, z2, , zi)zi+1ni−1m+ c2izi+1ni ,

Qs(z1, z2, , zs+1) = a2sQns−1(z1, z2, , zs) + b2sQn−ms−1 (z1, z2, , zs)zns+1s−1m+ c2szns+1s , where a1i6= 0, b1i 6= 0, c1i6= 0, a2i 6= 0, b2i 6= 0, c2i 6= 0, i = 1, , s; m, n ∈ N∗, m < n Note that Pi(z1, z2, , zi+1), Qi(z1, z2, , zi+1) are polynomials of degree ni

Set P (z1, z2, , zs+1) = Ps(z1, z2, , zs+1), Q(z1, z2, , zs+1) = Qs(z1, z2, , zs+1) For entire functions f1, · · · , fs+1; g1, · · · , gs+1 over C we consider the following equa-tions:

P (f1, , fs+1) = Q(g1, , gs+1), (1.1)

P (f1, , fs+1) = P (g1, , gs+1) (1.2) Denote by X ( resp Y ) the hypersurface of Fermat-Waring type in PN(C), which

is defined by the equation

PN(z1, , zN +1) = 0(resp, QN(z1, , zN +1) = 0) (1.3) For a holomorphic map f from C to PN(C) we denote by νf(X) the pull-back of the divisor X in PN(C) by f We shall prove the following theorems

Theorem 1.1 Let P, Q be polynomials defined as above, n ≥ 2m + 9, and either

m ≥ 2, (n, m) = 1, or m ≥ 4, and let f1, · · · , fs+1; g1, · · · , gs+1 be two families

of linearly independent entire functions over C, satisfying the equation (1.1) Then there exist constants li such that gi = lifi, i = 1, , s + 1

Theorem 1.2 Let f and g be two linearly non-degenerate holomorphic map-pings from C to PN(C) with reduced representations ˜f = (f1, , fN +1) and ˜g = (g1, , gN +1), respectively Let X, Y be the Fermat-Waring hypersurfaces defined

as (1.3), and let n ≥ 2m + 9, and either m ≥ 2, (n, m) = 1 or m ≥ 4 If

νf(X) = νg(Y ), then there exist a non-zero entire function h(z) and constants li, such that gi = lihfi, i = 1, , N + 1

Corollary 1.3 In the above notations, assume additionally that n ≥ 2m + 9, and either m ≥ 2, (n, m) = 1 or m ≥ 4 Let f1, · · · , fs+1; g1, · · · , gs+1 be two families

of linearly independent entire functions over C, satisfying the equation (1.2) Then there exist constants li such that gi = lifi, i = 1, , s + 1

Next we give an application of Corollary 1.3 Recall that q polynomials of N + 1 variables are said to be in general position if no set of N + 1 polynomials in this family has common zeros in CN +1− {0}

Now let given q linear forms of N + 1 variables (q > N ) in general position:

Li = Li(z1, , zN +1) = αi,1z1+ αi,2z2+ + αi,N +1zN +1, i = 1, 2, · · · , q Consider q homogeneous polynomials:

Y (z1, z2) = azn1 + bz1n−mzm2 + cz2n,

T1(z1, , zN +1) = Y (L1, L2), T2(z1, , zN +1) = Y (T1(z1, , zN +1), Ln3),

, Ti(z1, , zN +1) = Y (Ti−1(z1, , zN +1), Lni+1i−1),

Tq(z1, , zN +1) = Y (Tq−1(z1, , zN +1), Lnq+1q−1), where a 6= 0, b 6= 0, c 6= 0, q = 2, 3, Then Tq(z1, , zN +1) is a homogeneous polynomial of degree nq, and we set T (z1, z1, · · · , zN +1) = Tq(z1, , zN +1)

Denote by Z the hypersurface of Femat-Waring type in PN(C), defined by the equation

T (z1, z1, · · · , zN +1) = 0

Then we have the following

Corollary 1.4 Let f and g be two non-degenerate holomorphic mappings from

C to PN(C) Assume that:

1/ νf(Z) = νg(Z),

2/ n ≥ 2m + 9, and either m ≥ 2, (n, m) = 1 or m ≥ 4

Then f ≡ g

Notices that the constants li in the above Theorems and Corollaries can be defined exactly in terms of the coefficients of polynomials P and Q Namely, we have Proposition 1.5 In the notations of Theorem 1.1, the constants li are defined by

a11c21l2n= a21c11ln1, b11c21ln−m2 = b21c11l1n−m; a1ic2icn1i−1li+1ni = a2ic1icn2i−1lini,

b1ic2icn−m1i−1li+1ni−1(n−m) = b2ic1icn−m2i−1lini−1(n−m), i = 2, , s − 1;

a1scn1s−1 = a2scn2s−1lsns, b1scn−m1s−1 = b2scn−m2s−1lnss−1(n−m)lns+1s−1m, c1s = c2slns+1s

Proposition 1.6 In the notations of Theorem 1.2, the constants li are defined by:

a11c21l1n= a21c11ln2, b11c21ln−m1 = b21c11l2n−m, a1ic2icn1i−1lini = a2ic1icn2i−1lni+1i ,

b1ic2icn−m1i−1lini−1(n−m) = b2ic1icn−m2i−1lni+1i−1(n−m), i = 2, , N − 1;

a1Ncn1N −1lNnN = a2Ncn2N −1, b1Ncn−m1N −1lN +1nN −1m = b2Ncn−m2N −1lnNN −1(n−m), c1N = c2NlnN +1N Proposition 1.7 In the notations of Corollary 1.3, the constants li satisfy the following conditions:

ln1 = ln2, ln−m1 = ln−m2 ; lnii = lni+1i , lini−1(n−m) = lni+1i−1(n−m), i = 2, , s − 1;

lnss = 1, lnss−1(n−m)lns+1s−1m = 1, lns+1s = 1

Remark Propositions 1.5, 1.6 and 1.7 show that the equations (1.1), (1.2) have non-constant solutions only in case where the coefficients of P, Q satisfy some restricted relations As in [4] this means that the equations have no solutions in generic case

2 Preliminaries.

Let f be a non-constant holomorphic function on C For every a ∈ C, expanding

f around a as f =P

ai(z − a)i ,we define νf(a) = min{i : ai 6≡ 0} For a point ξ ∈ C

we define the function νfξ : C → N by νfξ(a) = νf −ξ(a) Let f be a holomorphic curve from C to PN(C) For an arbitrary fixed homogeneous coordinate system (z1 :

· · · : zn+1) in PN(C) we take a reduced representation of f : ˜f = (f1 : · · · : fN +1) Let H be a hypersurface of PN(C) such that the image of f is not contained in H, and H is defined by the equation F = 0 For every a ∈ C set

νf(H, a) = νF ◦ ˜f(a), νf(H) = νf(H, 0)

We assume that the reader is familiar with the notations in the Nevanlinna theory (see [5 ])

The following lemmas were proved in [5]

Lemma 2.1 Let f be a non-constant meromorphic function on C and let a1, a2, , aq

be distinct points of C ∪ {∞} Then

(q − 2)T (r, f ) ≤

q

X

i=1

N1(r, 1

f − ai) + S(r, f ), where S(r, f ) = o(T (r, f )) for all r, except for a set of finite Lebesgue measure Lemma 2.2 Let f be a non-constant meromorphic function on C and let a1, a2, , aq

be distinct points of C ∪ {∞} Suppose either f − ai has no zeros, or f − ai has zeros, in which case all the zeros of the functions f − ai have multiplicity at least

mi, i = 1, , q Then

q

X

i=1

(1 − 1

mi) ≤ 2.

3 Functional Equations and unique range sets

We first need the following Lemmas:

Lemma 3.1 [7 ] Let xd−qi

i Di(x1, x2, , xN +1),1 ≤ i ≤ N + 1, be homogeneous poly-nomials of degree d which determine hypersurfaces in general position of PN(C) Suppose there exists a holomorphic curve f from C to PN(C) with the reduced rep-resentation ˜f = (f1 : · · · : fN +1), such that its image lies in the curve defined by

N +1

X

i=1

xd−qi

i Di(x1, x2, , xN +1) = 0, d ≥ N2+

N +1

X

i=1

qi Then the polynomials xd−q1

1 D1(x1, x2, , xN +1), , xd−qN

N DN(x1, x2, , xN +1) are lin-early dependent on the image of f

Lemma 3.2 Let n, n1, n2, , nq, q ∈ N∗, a1, , aq, c ∈ K, c 6= 0, and q > 2+P q

i=1nni Then the functional equation

(f − a1)n1(f − a2)n2 (f − aq)nq = cgn has no non-constant meromorphic solutions (f, g)

Proof Suppose that (f, g) is a non-constant meromorphic solution of the equation:

(f − a1)n1(f − a2)n2 (f − aq)nq = cgn From this we see that if z0 ∈ C is a zero of f − ai for some 1 ≤ i ≤ q, then z0 is a zero of g and niνai

f (z0) = nνg(z0) So

N1(r, 1

f − ai) ≤

ni

nN (r,

1

f − ai) ≤

ni

nT (r, f ) + O(1).

From this and by Lemma 2.1,

(q − 2)T (r, f ) ≤

q

X

i=1

N1(r, 1

f − ai) + S(r, f ) ≤

q

X

i=1

ni

nN (r,

1

f − ai) + S(r, f ) + O(1) ≤

q

X

i=1

ni

nT (r, f ) + S(r, f ); (q − 2 −

q

X

i=1

ni

n)T (r, f ) ≤ S(r, f ).

Since q > 2 +P q

i=1nni, we obtain a contradiction

Lemma 3.3 Let n, m ∈ N∗, n ≥ 2m+9, a1, b1, c1, a2, b2, c2 ∈ C, a1, b1, c1, a2, b2, c2 6= 0

1 Suppose that f1

f 2 is a non-constant meromorphic function, and

f1n+ a1f1n−mf2m+ b1f2n = b2g2n (3.1) Then there exists a constant h such that g2 = hf2

2 Suppose that f1

f 2 and g1

g 2 are non-constant meromorphic functions, and either

m ≥ 2, (m, n) = 1, or m ≥ 4 and

a1f1n+ b1f1n−mf2m+ c1f2n= a2gn1 + b2g1n−mg2m+ c2g2n (3.2) Then there exist constants h and l, satisfying a1 = a2hn, b1 = b2hn−mlm, c1 = c2ln, such that g1 = hf1, g2 = lf2

Proof

1 From (3.1) we have

f1n−m(f1m+ a1f2m) + b1f2n− b2g2n= 0 (3.3) Note that xn−m1 (xm

1 + a1xm

2 ), b1xn

2, −b2xn

3 are the homogeneous polynomials of degree

n in general position Since n ≥ 2m + 9 and by Lemma 3.1, there exists c1 6= 0 such that c1b2gn

2 = b1fn

2 Therefore g2 = hf2 with b1 = c1b2hn

2 From (3.2 ) we have

a1f1n+ b1f1n−mf2m+ c1f2n− a2g1n− b2g1n−mg2m− c2gn2 = 0, (3.4) and then

c1f2n+ f1n−m(a1f1m+ b1f2m) − c2g2n− gn−m

1 (a2g1m+ b2g2m) = 0 (3.5) Note that c1xn

1, xn−m2 (a1xm

2 + b1xm

1 ), −c2xn

3, −xn−m4 (a2xm

4 + b2xm

3 ) are the homoge-neous polynomials of degree n in general position Since n ≥ 2m + 9 and by Lemma 3.1, there exist constants C1, C2, C3, (C1, C2, C3) 6= (0, 0, 0), such that

C1c1f2n+ C2f1n−m(a1f1m+ b1f2m) + C3c2gn2 = 0 (3.6)

We consider the following possible subcases:

Subcase 1: C3 = 0 Then from (3.6) we have

C1c1f2n+ C2f1n−m(a1f1m+ b1f2m) = 0

Since f2 is a non-zero entire function, we have C2 6= 0 If C1 = 0, then f1

f 2 is a constant, a contradiction So C1, C2 6= 0 Then f1

f 2 is a constant, a contradiction So

C3 6= 0

Subcase 2: C2 = 0 Then from (3.6) we have C1c1fn

2 + C3c2gn

2 = 0 Because f2, g2 are non-zero entire functions, we have C1 6= 0, C3 6= 0 From this and (3.5) it follows that gn

2 = −C1 c 1

C 3 c 2fn

2,g2

f2 = l, l 6= 0, and

c1



1 + C1

C3



f2n+ f1n−m(a1f1m+ b1f2m) − g1n−m(a2g1m+ b2g2m) = 0

−a2gn1 + f1n−m(a1f1m+ b1f2m) +



c1(1 + C1

C3)f

n−m

2 − b2lmg1n−m



f2m = 0 (3.7)

Suppose that 1 +C1

C3 6= 0 Then, from the similarity of (3.7) and (3.5), by a similar argument as in (3.9), there exist constants C10, C20, (C10, C20) 6= (0, 0), such that

C20a2g1n+ C10f1n−m(a1f1m+ b1f2m) = 0 (3.8) Since g1 is a non-zero entire function and f1

f 2 is not a constant, by (3.8) we obtain

C10 6= 0, C0

2 6= 0 We have

C10f1n−m(a1f1m+ b1f2m) = −C20a2g1n, C10a1

f1

f2

 n

+ C10b1

f1

f2

 n−m

= −C20a2

g1

f2

 n

C10a1

f1

f2

 n−m 

(f1

f2)

m+ b1

a1



= −C20a2

g1

f2

 n

Note that the equation zm + b1

a 1 = 0 has m distinct roots d1, d2, , dm Set f =

f 1

f 2, g = g1

f 2 Consequently, by (3.9) we have

fn−m(f − d1) (f − dm) = Cgn, C 6= 0 (3.10) Since f1

f 2 is not a constant, neither is g1

f 2 If (m, n) = 1, then from (3.10) we see that the multiplicity of each zero of f and f − di is a multiple of n By n ≥ 2m +

9, m ≥ 2 and Lemma 2.2 we conclude that the equation (3.10) has no non-constant meromorphic solutions If m ≥ 4, then from (3.10) and we have m + 1 > 2 +n−mn +

P m

i=1

1

n Then applying Lemma 3.2 to (3.10) with q = m + 1, n = n, n1 = n − m, n2 =

1 = n3 = = nm, we have a contradiction So 1 + C1

C3

= 0 Therefore c2gn

2 = c1fn

2, and g2 = lf2 with c1 = c2ln

Subcase 3 C1 = 0 From (3.6) we have C2f1n−m(a1fm

1 + b1fm

2 ) + C3c2gn

2 = 0 Then, from the similarity of this equation and (3.8), by a similar argument as in (3.8) we have a contradiction

Subcase 4 C1 6= 0, C2 6= 0, C3 6= 0

By a similar argument as in (3.7) we obtain a contradiction So c2gn

2 = c1fn

2, and

g2 = lf2 with c1 = c2ln

Moreover from (3.2) we have

c1f2n(a1

c1f

n+ b1

c1f

n−m+ 1) = c2g2n(a2

c2g

n+b2

c2g

n−m+ 1),

where f = f1

f2, g =

g1

g2 Set

a1

b1 = a3,

b1

c1 = b3,

a2

c2 = a4,

b2

c2 = b4 Since c2g

n

2 = c1fn

2,

a3fn+ b3fn−m = a4gn+ b4gn−m Set h1 = g

f From this we obtain

a3fm+ b3 = a4(g

f)

n

fm+ b4(g

f)

n−m

, a3fm+ b3 = a4hn1fm+ b4hn−m1 ,

fm(a3− a4hn1) = b4hn−m1 − b3, −

a4(hn

1 −a3

a4)

b4(hn−m1 − b3

b4)

= (1

f)

Assume that h1 is not a constant Consider the following possible cases:

Case 1 m ≥ 2, (m, n) = 1 If hn1 −a3

a4 and h

n−m

b4 have no common zeros, then all zeros of hn

1 −a3

a4 have multipcities ≥ m Then

N1(r, 1

hn

1 − a3

a4

) ≤ 1

mN (r,

1

hn

1 −a3

a4

)

By Lemma 2.1 we obtain

T (r, hn1) ≤ N1(r, hn1) + N1(r, 1

hn 1

) + N1(r, 1

hn

1 − a3

a4 ) + S(r, f ),

nT (r, h1) ≤ 2T (r, h1) + n

mN (r,

1

h1− a3

a4

) + S(r, f ) ≤ (2 + n

m)T (r, h1) + S(r, f )

(n − 2 − n

m)T (r, h1) ≤ S(r, f ), which leads to n(m − 1) ≤ 2m, a contradiction to the condition n ≥ 2m + 9 If

hn

1−a3

a4 and h

n−m

1 −b3

b4 have common zeros, then there exists z such that h

n

1(z) = a3

a4,

hn−m1 (z) = b3

b4 Set a = h1(z) From (3.11) we get

− a4(h

n

1 − an)

b4(hn−m1 − an−m) = (

1

f)

m, −

a4an((h1

a )

n− 1)

b4an−m((h1

a )

n−m− 1)

= (1

f)

m, −a4a

m(hn2 − 1)

b4(hn−m2 − 1) = (

1

f)

m,

where h2 = h1

a Since (m, n) = 1, the equations z

n− 1 = 0 and zn−m− 1 = 0 have different roots except for z = 1 Let ri, i = 1, , 2n − m − 2 be all the roots of them Then all zeros of h2− ri have multipcities ≥ m Therefore, by Lemma 2.2 we obtain

(1 − 1

m)(2n − m − 2) ≤ 2, 2n(m − 1) ≤ m

2+ 3m − 2, which contradicts n ≥ 2m + 9 Thus h1 is a constant

Case 2 m ≥ 4 Note that equation zn − a3

a4 = 0 has n simple zeros, equation

zn−m− b3

b4 = 0 has n − m simple zeros Then z

n− a3

a4 = 0, z

n−m− b3

b4 = 0 have

at most n − m common simple zeros Therefore the equation zn− a3

a4 = 0 has at least m distinct roots, which are not roots of zn−m − b3

b4 = 0 Let r1, r2, , rm be all these roots Then all the simple zeros of h1− rj, j = 1, , m, have multiplicities

≥ m By Lemma 2.2 we have m(1 − 1

m) ≤ 2 Therefore m ≤ 3 From m ≥ 4,

we obtain a contradiction Thus h1 is a constant By g = h1f, g2 = lf2 we have

g1 = hf1 From (3.2) and since f1

f 2 and g1

g 2 are non-constant functions we obtain

a1 = a2hn, b1 = b2hn−mlm, c1 = c2ln Lemma 3.3 is proved

Now we use the above Lemmas to prove the main result of the paper

Proof of Theorem 1 and Proposition 1.5

Set Pi( ˜f ) = Pi(f1, , fi+1), Pi(˜g) = Pi(g1, , gi+1), i = 1, , s We first prove

Pi( ˜f ) 6≡ 0, i = 1, 2, , s, by induction on i With i = 1 assume that

P1( ˜f ) = a11f1n+ b11f1n−mf2m+ c11f2n≡ 0

Therefore, f1

f2

is a constant, and we have a contradiction to the linearly independence

of f1, , fs+1 With i = 2, assume that

P2( ˜f ) = a12P1n( ˜f ) + b12P1n−m( ˜f )f3nm+ c12f3n2 ≡ 0

Since P1( ˜f ) 6≡ 0, f3n 6≡ 0 we see that P1( ˜f )

fn 3

is a constant Hence

a11f1n+ b11f1n−mf2m+ c11f2n− Afn

3 ≡ 0

Applying Lemma 3.3 we obtain that f2

f3 is a constant, and we have a contradiction

to the linearly independence of f1, , fs+1

Now we consider Pi( ˜f ) ≡ 0 Then

a1iPi−1n ( ˜f ) + b1iPi−1n−m( ˜f )fi+1ni−1m+ c1ifi+1ni ≡ 0 (3.12) Applying the induction hypothesis and by a similar argument as above we have a contradiction

So Pi( ˜f ) 6≡ 0, i = 1, 2, , s Similarly, Qi( ˜f ) 6≡ 0, i = 1, 2, , s

Next we consider

We will show that gi = lifi, i = 1, , s + 1, where li satisfy following conditions:

a11c21l2n= a21c11ln1, b11c21ln−m2 = b21c11l1n−m; a1ic2icn1i−1li+1ni = a2ic1icn2i−1lini,

b1ic2icn−m1i−1li+1ni−1(n−m) = b2ic1icn−m2i−1lini−1(n−m), i = 2, , s − 1;

a1scn1s−1 = a2scn2s−1lsns, b1scn−m1s−1 = b2scn−m2s−1lnss−1(n−m)lns+1s−1m, c1s = c2slns+1s

From (3.13) we have

a1sPs−1n ( ˜f ) + b1sPs−1n−m( ˜f )fs+1ns−1m+ c1sfs+1ns = a2sQns−1(˜g) + b2sQn−ms−1 (˜g)gs+1ns−1m+ c1sgns+1s

(3.14) Applying Lemma 3.3 for (3.14) we have

Qs−1(˜g) = hs−1Ps−1( ˜f ), gs+1 = ls+1fs+1; a1s = a2shns−1, b1s = b2shn−ms−1 lns+1s−1m,

From (3.15) we obtain

hs−1a1s−1Ps−2n ( ˜f ) + hs−1b1s−1Ps−2n−m( ˜f )fsns−2m+ hs−1c1s−1fsns−1 =

a2s−1Qns−2(˜g) + b2s−1Qn−ms−2 (˜g)gsns−2m+ c2s−1gnss−1 (3.16) Applying Lemma 3.3 for (3.16) we have

Qs−2(˜g) = hs−2Ps−2( ˜f ), gs= lsfs; hs−1a1s−1 = a2s−1hns−2,

hs−1b1s−1 = b2s−1hn−ms−2 lsns−2m, hs−1c1s−1 = c2s−1lnss−1 Since and (3.15) we give

hs−1 = c2s−1

c1s−1l

n s−1

s , a1s = a2s(c2s−1

c1s−1l

n s−1

s )n, b1s = b2s(c2s−1

c1s−1l

n s−1

s )n−mlns+1s−1m,

a1scn1s−1 = a2scn2s−1lnss, b1scn−m1s−1 = b2scn−m2s−1lnss−1(n−m)lns+1s−1m, c1s = c2sls+1ns (3.17) Now we consider following

Qi(˜g) = hiPi( ˜f ), i = 2, , s − 1;

hia1iPi−1n ( ˜f ) + hib1iPi−1n−m( ˜f )fi+1ni−1m+ hic1ifi+1ni =

a2iQni−1(˜g) + b2iQn−mi−1 (˜g)gi+1ni−1m+ c2igni+1i (3.18) Applying Lemma 3.3 for (3.18) we give

Qi−1(˜g) = hi−1Pi−1( ˜f ), gi+1= li+1fi+1; hia1i= a2ihni−1,

hib1i= b2ihn−mi−1 lni+1i−1m, hic1i= c2ili+1ni ; hi = c2i

c1i

lni+1i (3.19) From Qi−1(˜g) = hi−1Pi−1( ˜f ) we give

hi−1a1i−1Pi−2n ( ˜f ) + hi−1b1i−1Pi−2n−m( ˜f )fini−2m+ hi−1c1i−1fini−1 =

a2i−1Qni−2(˜g) + b2i−1Qn−mi−2 (˜g)gin m+ c2i−1gni (3.20) Applying Lemma 3.3 for (3.20) we obtain

Qi−2(˜g) = hi−2Pi−2( ˜f ), gi = lifi; hi−1a1i−1 = a2i−1hni−2,

hi−1b1i−1= b2i−1hn−mi−2 lnii−2m, hi−1c1i−1= c2i−1lini−1; hi−1= c2i−1

c1i−1l

n i−1

From this and (3.19) we obtain

hi−1 = c2i−1

c1i−1l

n i−1

i , hi = c2i

c1il

n i

i+1, c2i

c1il

n i

i+1a1i= a2i(c2i−1

c1i−1l

n i−1

i )n,

c2i

c1il

n i

i+1b1i= b2s(c2i−1

c1i−1l

n i−1

i )n−mlni+1i−1m, a1ic2icn1i−1li+1ni = a2ic1icn2i−1lini;

b1ic2icn−m1i−1lni+1i−1(n−m)= b2ic1icn−m2i−1lini−1(n−m) (3.21) Now we consider the following equations:

Q1(˜g) = h1P1( ˜f ),

h1a11f1n+ h1b11f1n−mf2m+ h1c11f2n = a21gn1 + b22gn−m1 g2m+ c22gn2 (3.22) Applying Lemma 3.3 for (3.22) we have

g1 = l1f1, g2 = l2f2, h1a11= a21ln1, h1b11= b21l1n−ml2m, h1c11 = c21ln2

Form this it follows

h1 = c21

c11l

n

2, c21

c11a11l

n

2 = a21ln1, c21

c11b11l

n

2 = b21ln−m1 lm2 Therefore

a11c21l2n= a21c11ln1, b11c21l2n−m= b21c11l1n−m (3.23)

By (3.17), (3.21), (3.23), Theorem 1.1 is proved

Now we are going to complete the proof of Theorem 2

Proof of Theorem 2 and Propositions 1.6

Let ˜f = (f1, , fN +1) and ˜g = (g1, , gN +1) be reduced representations of f and

g, respectively

Since νf(X) = νg(Y ), it is easy to see that there exists a non-zero entire function h(z) such that such that Pq( ˜f ) = hQq(˜g) Take b such that bn N

= h and set

ϕ = (bg1 : · · · : bgN +1) Then ϕ is a reduced representation of g and PN( ˜f ) = QN(ϕ)

By Theorem 2 then follows from Theorem 1

Corollary 1.3 and Proposition 1.7 are consequences of Theorem 1.1 and Proposi-tion 1.5

Now we are going to complete the proof of Corollary 1.4

Proof of Corolary 1.4

Let ˜f = (f1, , fN +1) and ˜g = (g1, , gN +1) be reduced representations of f and

g, respectively

Since νf(X) = νg(X), it is easy to see that there exists an non-zero entire function

h such that Pq( ˜f ) = hPq(˜g) Take b such that bnN = h and set Gi = bgi, i =

1, , N + 1, ˜G = (G1 : · · · : GN +1) Then ˜G is a reduced representation of g and

Pq( ˜f ) = Pq( ˜G)