Algebraic dependences and uniqueness problem of meromorphic mappings sharing moving hyperplanes without counting multiplicities

0.18 MB·

by  Le Ngoc Quynh

ALGEBRAIC DEPENDENCES AND UNIQUENESS PROBLEM OF
MEROMORPHIC MAPPINGS SHARING MOVING HYPERPLANES
WITHOUT COUNTING MULTIPLICITIES
6
LE NGOC QUYNH
1
0
2

n
a
J Abstract. This articledealswith the multiple valuesandalgebraicdependences prob-
lem of meromorphic mappings sharing moving hyperplanes in projective space. We give
1
3 some algebraic dependences theorems for meromorphic mappings sharing moving hy-
perplanes without counting multiplicity, where all zeros with multiplicities more than a

] certain number are omitted. Basing on these results, some unicity theorems regardless
V
of multiplicity for meromorphic mappings in severalcomplex variables are given. These
C
results are extensions and strong improvements of some recent results.
.
h
t
a
m 1. Introduction
[
The theory on algebraic dependences of meromorphic mappings in several complex

1
variables into the complex projective spaces for fixed targets was first studied by W. Stoll
v
1 [7]. Later, M. Ru [6] generalized W. Stoll’s result to the case of holomorphic curves into
9 the complex projective spaces sharing moving hyperplanes. Recently, by using the new
1
second main theorem given by Thai-Quang [11], P. D. Thoan, P. V. Duc and S. D. Quang
0
0 [3, 10, 8] gave some improvements of the results of W. Stoll and M. Ru. In order to state
.
2 some of their result, we first recall the following.
0
We call a meromorphic mapping of Cn into PN(C)∗ a moving hyperplane in PN(C).
6
1 Let a ,...,a (q ≥ N + 1) be q moving hyperplanes with reduced representations a =
1 q j
v: (aj0 : ··· : ajN) (1 6 j 6 q). We say that a1,...,aq are in general position if det(ajkl) 6≡ 0
i for any 1 6 j < j < ... < j 6 q.
X 0 1 N
r Let f : Cn → PN(C) (1 6 i 6 λ) be meromorphic mappings with reduced represen-
i
a
tations f := (f : ··· : f ). Let g : Cn → PN(C)∗ (0 6 j 6 q − 1) be meromor-
i i0 iN j
phic mappings in general position with reduced representations g := (g : ··· : g ).
j j0 jN
Put (f ,g ) := N f g 6= 0 for each 1 6 i 6 λ, 0 6 j 6 q − 1 and assume
i j s=0 is js
that min{1,ν0 } = ··· = min{1,ν0 }. Put A = Supp(ν0 ). As-
(f1,gPj),6kj (fλ,gj),6kj j (f1,gj),6kj
sume that each A has an irreducible decomposition as follows A = tj A . Set
j j s=1 js
A = Ajs6≡Aj′s′{Ajs ∩Aj′s′} with 1 6 s 6 tj,1 6 s′ 6 tj′,0 6 j,j′ 6 q −1. S
Denote by T[N+1,q] the set of all injective maps from {1,··· ,N+1} to {0,··· ,q−1}.
S
For each z ∈ Cn \ { {z|g (z) ∧ ···∧ g (z) = 0} ∪ A ∪ λ I(f )}, we
β∈T[N+1,q] β(1) β(N+1) i=1 i
define ρ(z) = ♯{j|z ∈ A }. Then ρ(z) 6 N. For any positive number r > 0, define ρ(r) =
j
S S
2010 Mathematics Subject Classification: Primary 32H30, 32A22; Secondary 30D35.
Key words and phrases: algebraic dependence, unicity problem, meromorphic mapping, truncated
multiplicity.
1
2 LE NGOC QUYNH
sup{ρ(z)||z| 6 r}, where the supremum is taken over all z ∈ Cn\{ {z|g (z)∧
β∈T[N+1,q] β(1)
···∧g (z) = 0}∪A∪ λ I(f )}. Then ρ(r) is a decreasing function. Let
β(N+1) i=1 i S
S d := lim ρ(r).
r→+∞
Then d 6 N. If for each i 6= j, dim{A ∩A } 6 n−2, then d = 1.
i j
When all k = +∞, in [3], P. V. Duc and P. D. Thoan proved the following.
j
Theorem A (see [3, Theorem 1]). Let f ,··· ,f : Cn → PN(C) be non-constant mero-
1 λ
morphic mappings. Let g : Cn → PN(C) (0 6 i 6 q − 1) be moving targets located
i
in general position and T(r,g ) = o(max T(r,f )) (0 6 i 6 q − 1). Assume that
i 16j6λ j
(f ,g ) 6≡ 0 for 1 6 i 6 λ, 0 6 j 6 q−1 and A := (f ,g )−1{0} = ··· = (f ,g )−1{0} for
i j j 1 j λ j
each 0 6 j 6 q−1. Denote A = ∪q−1A . Let l, 2 6 l 6 λ, be an integer such that for any
j=0 j
increasing sequence 1 6 j < ··· < j 6 λ, f (z)∧···∧f (z) = 0 for every point z ∈ A.
1 l j1 jl
dN(2N +1)λ
If q > , then f ,··· ,f are algebraically over C, i.e. f ∧···∧f ≡ 0.
1 λ 1 λ
λ−l+1
Furthermore, in the case of d = 1, P. D. Thoan, P. V. Duc and S. D. Quang [10] proved
an better algebraic dependences theorem as follows.
Theorem B (see [10, Theorem 1]). Let f ,··· ,f : Cn → PN(C) be non-constant
1 λ
meromorphic mappings. Let g : Cn → PN(C)∗ (0 6 i 6 q−1) be moving hyperplanes in
i
general position such that T(r,g ) = o(max T(r,f )) (0 6 i 6 q −1) and (f ,g ) 6≡ 0
i 16j6λ j i j
for 1 6 i 6 λ, 0 6 j 6 q −1. Assume that the following conditions are satisfied.
(a) min{1,ν } = ··· = min{1,ν } for each 0 6 j 6 q −1,
(f1,gj) (fλ,gj)
(b) dim{z|(f ,g )(z) = (f ,g )(z) = 0} 6 n−2 for each 0 6 i < j 6 q −1,
1 i 1 j
(c) there exists an integer number l, 2 6 l 6 λ, such that for any increasing se-
quence 1 6 j < ··· < j 6 λ, f (z) ∧ ··· ∧ f (z) = 0 for every point
1 l j1 jl
z ∈ q−1(f ,g )−1{0}.
i=0 1 i
If q > N(2N+S1)λ−(N−1)(λ−1), then f ∧···∧f ≡ 0.
λ−l+1 1 λ
The above results are the best results on the algebraic dependences of meromorphic
mappings sharing moving hyperplanes available at the present. However, in our opinion,
they are still weak. Also in the above results, all intersecting points of the mappings and
the moving hyperplanes are considered. Actually, there are many authors consider the
multiple values for meromorphic mappings sharing hyperplanes, i.e., consider only the
intersecting points of the mappings f and the hyperplanes g with the multiplicity not
i j
exceed a certain number k < +∞. For example, in 2010, T. B. Cao and H. X. Yi gave
j
some uniqueness theorems for meromorphic mappings sharing fixed hyperplanes where
all intersecting points more than a certain number are omitted (see [1, Theorems 1.4 and
1.5]). Recently, H. H. Giang consider the multiple values and uniqueness problems for the
mappings sharing moving hyperplanes (see [4, Theorems 1.1 and 1.3]).
Our first purpose in this paper is to generalize and improve Theorems A and B by
considering themultiplevaluesproblemandreducing thenumber ofhyperplanes. Namely,
we will prove the following.
ALGEBRAIC DEPENDENCES OF MEROMORPHIC MAPPINGS 3
Theorem 1.1. Let f ,··· ,f : Cn → PN(C) be non-constant meromorphic mappings.
1 λ
Let {g }q−1 be moving hyperplanes of PN(C) in general position satisfying T(r,g ) =
j j=0 j
o(max T(r,f )) (0 6 j 6 q −1). Let k (0 6 j 6 q −1) be positive integers or +∞.
16i6λ i j
Assume that (f ,g ) 6≡ 0 for 1 6 i 6 λ, 0 6 j 6 q −1 and A := Suppν0 = ··· =
i j j (f1,gj),6kj
Suppν0 for each 0 6 j 6 q − 1. Denote A = q−1A . Let l, 2 6 l 6 λ, be an
(fλ,gj),6kj j=0 j
integer such that for any increasing sequence 1 6 i < ··· < i 6 λ, f (z)∧···∧f (z) = 0
1 S l i1 il
for every point z ∈ A. If
q−1
1 q dλ
< − ,
k N(N +2) λ−l+1
j
j=0
X
then f ∧···∧f ≡ 0.
1 λ
In the above result, letting k = +∞ (0 6 j 6 q−1), we get the conclusion of Theorem
j
A with q > dλN(N+2). Also, letting d = 1, we obtain the conclusion of Theorem B with
(λ−l+1)
q > λN(N+2). Hence our result is improvement of Theorems A and B.
λ−l+1
Let λ = l = 2 and k = +∞ (0 6 j 6 q −1), the above theorem implies the following
j
unicity theorem.
Corollary 1.2. Let f ,f : Cn → PN(C) be two non-constant meromorphic mappings.
1 2
Let {g }q−1 be moving hyperplanes of PN(C) in general position satisfying T(r,g ) =
j j=0 j
o(max T(r,f )) (0 6 j 6 q −1). Assume that the following conditions are satisfied.
16i62 i
(a) min{1,ν0 } = min{1,ν0 } for each 0 6 j 6 q −1,
(f1,gj) (f2,gj)
(b) f (z) = f (z) for each z ∈ q−1Suppν0 ,
1 2 j=0 (f1,gj)
(c) q > 2dN(N +2).
S
Then f ≡ f .
1 2
If d = 1, then from the above corollary, we get a uniqueness theorem for meromorphic
mappings, which are not assumed to be nondegenerate, sharing q > 2N2 + 4N moving
hyperplanes in general position.
Now, if we assume further that the linearly closures of the images of the mappings f
i
in Theorem 1.1 have the same dimension then we will get a better result as follows.
Theorem 1.3. Let f ,··· ,f : Cn → PN(C) be non-constant meromorphic mappings.
1 λ
Let {g }q−1 be q meromorphic mappings g : Cn → PN(C)∗ (0 6 j 6 q − 1) in general
j j=0 j
position satisfying T(r,g ) = o(max T(r,f )) (0 6 j 6 q −1). Let k (0 6 j 6 q −1)
j 16i6λ i j
be positive integers or +∞. Assume that (f ,g ) 6≡ 0 for 1 6 i 6 λ, 0 6 j 6 q −1, and
i j
the following conditions are satisfied.
(a) min{1,ν0 } = ··· = min{1,ν0 } for each 0 6 j 6 q −1,
(f1,gj)6kj (fλ,gj)6kj
(b) dimf −1(g )∩f −1(g ) 6 n−2 for each 0 6 j < j 6 q−1,
1 j1 1 j2 1 2
(c) there exists an integer number l, 2 6 l 6 λ, such that for any increasing sequence
q−1
1 6 i < ··· < i 6 λ, f (z)∧···∧f (z) = 0 for every point z ∈ f −1(g ).
1 l i1 il 1 j
j=0
S
4 LE NGOC QUYNH
We assume further that rank f = ··· = rank f = m+1, where m is a positive
R{gj} 1 R{gj} λ
integer. If
q−1
1 q λq
(*) < −
k +1−m m(2N −m+2) q(λ−l +1)+λ(m−1)
j
j=0
X
then f ∧···∧f ≡ 0.
1 λ
Remark 1: i) When k = ··· = k = +∞, the condition (*) becomes
0 q−1
λ(2Nm−m2 +m+1)
q > .
(λ−l +1)
We see that this inequality is satisfied with q > λ(N2+N+1), since its right hand side
(λ−l+1)
attains maximum at m = N. Hence in this case (d = 1 and k = ··· = k = +∞), the
0 q−1
conclusion of Theorem 1.3 is better than that of Theorem 1.1.
ii) Let λ = l = 2 and k = ··· = k = +∞. We may show that if f and f satisfy
0 q−1 1 2
the conditions (i)-(iii) of Theorem 1.3 and q > N(N +2) then rank f = rank f .
R{gj} 1 R{gj} 2
Indeed, supposethattherearea ,...,a ∈ R{g },notallzeros, satisfying a f ≡
0 N j 06i6N i 1i
0. Set P = a f . Then P = 0 on (f ,g )−1(0). If P 6≡ 0, then by using
06i6N i 2i 16j6q−1 1 j P
Remark 2 below, we have
P S
q−1
T (r) ≥ N(r,ν0)+o(T (r)) ≥ N[1](r,ν0 )+o(T (r))
f2 P f2 (f1,gj) f2
j=0
X
q−1
q
= N[1](r,ν0 )+o(T (r)) ≥ T (r)+o(T (r)).
(f2,gj) f2 N(N +2) f2 f2
j=0
X
Letting r −→ +∞, we get q 6 N(N + 2). This is a contradiction. Then we must have
P ≡ 0. This implies that rank f ≥ rank f .
R{gj} 1 R{gj} 2
Similarly, we have rank f 6 rank f . Hence rank f = rank f
R{gj} 1 R{gj} 2 R{gj} 1 R{gj} 2
Then from Theorem 1.3 and the above remarks, we get a uniqueness theorem as follows.
Corollary 1.4. Let f ,f : Cn → PN(C) be non - constant meromorphic mappings.
1 2
Let {g }q−1 be moving hyperplanes of PN(C) in general position satisfying T(r,g ) =
j j=0 j
o(max T(r,f )) (0 6 j 6 q−1). Assume that (f ,g ) 6≡ 0 for 1 6 i 6 2, 0 6 j 6 q−1,
16i62 i i j
and the following conditions are satisfied.
(a) min{1,ν0 } = min{1,ν0 } for each 0 6 j 6 q −1,
(f1,gj) (f2,gj)
(b) dimf −1(g )∩f −1(g ) 6 n−2 for each 0 6 j < j 6 q−1,
1 j1 1 j2 1 2
q−1
(c) f = f on Suppν0 .
1 2 (f1,gj)
j=0
S
If q > 2N2 +2N +2, then f ≡ f .
1 2
We would like to note that there are several results on the uniqueness problem of mero-
morphic mappings sharing moving hyperplanes regardless of multiplicity. For example,
in 2007, with the same assumption of Corollary 1.4 (and plus N ≥ 2), Z. Chen, Y. Li and
ALGEBRAIC DEPENDENCES OF MEROMORPHIC MAPPINGS 5
Q. Yan [2] get the uniqueness theorem with q ≥ 4N2 +2N. In 2013, P. D. Thoan, P. V.
Duc and S. D. Quang improved the result of these authors to the case of q ≥ 4N2 + 2
(for any N). Therefore, our above uniqueness theorem is much stronger improvements of
many previous results.
Acknowledgements: This paper was supported in part by a NAFOSTED grant of
Vietnam.
2. Basic notions and auxiliary results from Nevanlinna theory
2.1. We set ||z|| = |z |2 +···+|z |2 1/2 for z = (z ,...,z ) ∈ Cn and define
1 n 1 n
B(r) := {z ∈ Cn : ||z|| < r}, S(r) := {z ∈ Cn : ||z|| = r} (0 < r < ∞).
(cid:0) (cid:1)
Define
v (z) := ddc||z||2 n−1 and
n−1
σn(z) := dclog||z||2 ∧(cid:0) ddclog||(cid:1)z||2 n−1on Cn \{0}.
2.2. For a divisor ν on Cn, we denote by N(r,ν) the counting function of the divisor ν
(cid:0) (cid:1)
as usual in Nevanlinna theory (see [5]).
For a positive integer M or M = ∞, we define the truncated divisors of ν by
ν[M](z) if ν[M](z) 6 k,
ν[M](z) = min {M,ν(z)}, ν[M](z) :=
6k 0 if ν[M](z) > k.
(
Similarly, we define ν[M]. We will write N[M](r,ν), N[M](r,ν), N[M](r,ν) for N(r,ν[M]),
>k 6k >k
N(r,ν[M]), N(r,ν[M]) as respectively. We will omit the character [M] if M = ∞.
6k >k
2.3. Let f : Cn −→ PN(C) be a meromorphic mapping. For arbitrarily fixed ho-
mogeneous coordinates (w : ··· : w ) on PN(C), we take a reduced representation
0 N
f = (f : ··· : f ), which means that each f is a holomorphic function on Cn and
0 N i
f(z) = f (z) : ··· : f (z) outside the analytic set {f = ··· = f = 0} of codimension
0 N 0 N
≥ 2. Let a be a meromorphic mapping of Cn into PN(C)∗ with reduced representation
(cid:0) (cid:1)
a = (a : ··· : a ). We denote by T(r,f) the characteristic function of f and by m (r)
0 N f,a
the proximity function of f with respect to a (see [9]).
If (f,a) 6≡ 0, then the first main theorem for moving targets in value distribution theory
states
T(r,f)+T(r,a) = m (r)+N (r).
f,a (f,a)
Theorem 2.1 (The First Main Theorem for general position [7, p. 326]). Let f : Cn →
j
PN(C), 1 6 j 6 k be meromorphic mappings located in general position. Assume that
1 6 k 6 N. Then
N(r,µ )+m(r,f ∧···∧f ) 6 T(r,f )+O(1).
f1∧···∧fλ 1 λ i
16i6λ
X
Here, by µ we denote the divisor associated with f ∧···∧f . We also denote
f1∧···∧fλ 1 λ
by N (r) the counting function associated with the divisor µ .
f1∧···∧fλ f1∧···∧fλ
Let V be a complex vector space of dimension N ≥ 1. The vectors {v ,··· ,v } are said
1 k
to be in general position if for each selection of integers 1 6 i < ··· < i 6 k with p 6 N,
1 p
6 LE NGOC QUYNH
then v ∧···∧v 6= 0. The vectors {v ,··· ,v } are said to be in special position if they
i1 ip 1 k
are not in general position. Take 1 6 p 6 k. Then {v ,··· ,v } are said to be in p-special
1 k
position if for each selection of integers 1 6 i < ··· < i 6 k, the vectors v ,··· ,v are
1 p i1 ip
in special position.
Theorem 2.2 (The Second Main Theorem for general position [7, p. 320]). Let M be a
connected complex manifold of dimension m. Let A be a pure (m−1)-dimensional analytic
subset of M. Let V be a complex vector space of dimension n + 1 > 1. Let p and k be
integers with 1 6 p 6 k 6 n + 1. Let f : M → P(V),1 6 j 6 k, be meromorphic
j
mappings. Assume that f ,...,f are in general position. Also assume that f ,...,f are
1 k 1 k
in p-special position on A. Then we have
µ ≥ (k −p+1)ν .
f1∧···∧fk A
Here by ν we denote the reduced divisor whose support is the set A.
A
The following is a new second main theorem given by S. D. Quang [9], which is an
improvement the second main theorem of Thai-Quang in [11].
Theorem 2.3 (The Second Main Theorem for moving target [9]). Let f : Cn → PN(C)
be a meromorphic mapping. Let {a }q (q ≥ 2N −m+2) be meromorphic mappings of
i i=1
Cn into PN(C)∗ in general position such that (f,a ) 6≡ 0 (1 6 i 6 q), where m + 1 =
i
rank (f). Then we have
R{ai}
q
q
|| T (r) 6 N[m] (r)+o(T (r))+O(maxT (r)).
2N −m+2 f (f,ai) f 16i6q ai
i=1
X
As usual, by the notation “|| P” we mean the assertion P holds for all r ∈ [0,∞)
excluding a Borel subset E of the interval [0,∞) with dr < ∞.
E
Remark 2: With the assumption of Theorem 2.3, weRsee that
q
m(2N −m+2)
|| T (r) 6 N[1] (r)+o(T (r))+O(maxT (r))
f q (f,ai) f 16i6q ai
i=1
X
q
N(N +2)
(2.4) 6 N[1] (r)+o(T (r))+O(maxT (r)).
q (f,ai) f 16i6q ai
i=1
X
3. Proofs of Main Theorems
3.1. Proof of Theorem 1.1.
It suffices to prove Theorem 1.1 in the case of λ 6 N +1.
Suppose that f ∧···∧f 6≡ 0. We now prove the following.
1 λ
Claim 3.1. For every 1 6 i 6 λ, 0 6 j 6 q −1 and 1 6 m 6 N, we have
k +1 m
N[m](r,ν0 ) ≥ j N[m](r,ν0 )− T(r,f ).
6kj (fi,gj) k +1−m (fi,gj) k +1−m i
j j
ALGEBRAIC DEPENDENCES OF MEROMORPHIC MAPPINGS 7
Indeed, we have
N[m](r,ν0 ) = N[m](r,ν0 )−N[m](r,ν0 )
6kj (fi,gj) (fi,gj) >kj (fi,gj)
m
≥ N[m](r,ν0 )− N (r,ν0 )
(fi,gj) k +1 >kj (fi,gj)
j
m m
= N[m](r,ν0 )− N(r,ν0 )+ N (r,ν0 )
(fi,gj) k +1 (fi,gj) k +1 6kj (fi,gj)
j j
m m
≥ N[m](r,ν0 )− T(r,f )+ N[N](r,ν0 ).
(fi,gj) k +1 i k +1 6kj (fi,gj)
j j
Thus
k +1 m
N[m](r,ν0 ) ≥ j N[m](r,ν0 )− T(r,f ).
6kj (fi,gj) k +1−m (fi,gj) k +1−m i
j j
Then the claim is proved.
Claim 3.2. For every 1 6 i 6 λ, we have
q−1
d
min{1,ν0 } 6 µ (z)+q. µ (z)
(fi,gj),6kj λ−l+1 f1∧···∧fλ gβ(1)∧···∧gβ(N+1)
j=0 β
X X
for each z 6∈ A λ I(f ), where the sum is over all injective maps β : {1,2,··· ,N+1} →
i=1 i
{1,2,··· ,q}
S
Indeed, for each regular point z ∈ A\(A∪ λ I(f )∪ {z|g (z)∧···∧
0 i=1 i β∈T[N+1,q] β(1)
g (z) = 0}) and for each increasing sequence 1 6 i < ··· < i 6 λ, we have
β(N+1) 1 l
S S
f (z )∧···∧f (z ) = 0.
i1 0 il 0
By the Second Main Theorem for general position [7, p. 320], we have
µ (z ) ≥ λ−(l−1).
f1∧···∧fλ 0
Hence
q−1 q−1
d
min{1,ν0 (z )} 6 min{1,ν0 (z )} 6 d 6 µ (z ).
(fi,gj),6kj 0 (fi,gj) 0 λ−l +1 f1∧···∧fλ 0
j=0 j=0
X X
If z ∈ {z|g (z)∧···∧g (z) = 0}, then we have
0 β∈T[N+1,q] β(1) β(N+1)
q−1 S q−1
min{1,ν0 (z )} 6 min{1,ν0 (z )} 6 q µ (z ).
(fi,gj),6kj 0 (fi,gj) 0 gβ(1)∧···∧gβ(N+1) 0
j=0 j=0 β∈T[N+1,q]
X X X
Thus, for each z 6∈ A∪ λ I(f ), we have
i=1 i
q−1 S d
min{1,ν0 (z)} 6 µ (z)+q µ (z).
(fi,gj),6kj λ−l+1 f1∧···∧fλ gβ(1)∧···∧gβ(N+1)
j=0 β∈T[N+1,q]
X X
The Claim 3.2 is proved.
8 LE NGOC QUYNH
The above Claim yields that
q−1
d
|| N[1] (r,ν0 ) 6 N (r)+q N (r)
6kj (fi,gj) λ−l+1 f1∧···∧fλ gβ(1)∧···∧gβ(N+1)
j=0 β∈T[N+1,q]
X X
λ N+1
d
6 T(r,f )+q T(r,g )
λ−l+1 i β(i)
i=1 β∈T[N+1,q] i=1
X X X
d
= T(r)+o(max T(r,f )).
λ−l +1 16i6λ i
where T(r) = λ T(r,f ). Then, by Claim 3.1 and the inequality (2.4) we have
i=1 i
λ q−1
Pdλ
|| T(r) ≥ N[1] (r,ν0 )+o(T(r))
λ−l +1 6kj (fi,gj)
i=1 j=0
XX
λ q−1
k +1 1
≥ j N[1](r,ν0 )− T(r,f ) +o(T(r))
k (fi,gj) k i
i=1 j=0 (cid:18) j j (cid:19)
XX
λ q−1 q−1
1
≥ N[1](r,ν0 )− T(r)+o(T(r))
(fi,gj) k
j
i=1 j=0 j=0
XX X
λ q−1
q 1
≥ T(r,f )− T(r)+o(T(r))
i
N(N +2) k
j
i=1 j=0
X X
q−1
q 1
= − T(r)+o(T(r)).
N(N +2) k
(cid:18) j=0 j(cid:19)
X
Letting r → +∞, we get
q−1
1 q dλ
≥ − .
k N(N +2) λ−l +1
j
j=0
X
This is a contradiction. Thus, f ∧···∧f = 0. The theorem is proved. (cid:3)
1 λ
3.2. Proof of Theorem 1.3.
It suffices to prove Theorem 1.3 in the case of λ 6 N+1. We set m = rank (f )−1.
R{gj} i
Similar as [3, Claim 3.1], we have the following claim.
Claim 3.3. Let h : Cn → PN(C) (1 6 i 6 p 6 N + 1) be meromorphic mappings
i
with reduced representations h := (h : ··· : h ). Let a : Cn → PN(C)∗ (1 6 i 6
i i0 iN i
N + 1) be moving hyperplanes with reduced representations a := (a : ··· : a ). Put
i i0 iN
h˜ := ((h ,a ) : ··· : (h ,a )) Assume that a ,··· ,a are located in general position
i i 1 i N+1 1 N+1
such that (h ,a ) 6≡ 0 (1 6 i 6 p, 1 6 j 6 N +1). Let S be a pure (n−1)-dimensional
i j
analytic subset of Cn such that S 6⊂ (a ∧···∧a )−1{0}. Then h ∧···∧h = 0 on S
1 N+1 1 p
if and only if h˜ ∧···∧h˜ = 0 on S.
1 p
We now prove Claim 3.3 .
ALGEBRAIC DEPENDENCES OF MEROMORPHIC MAPPINGS 9
(⇒) Suppose that {h˜ ∧···∧h˜ } 6≡ 0 on S. Then there exists z ∈ S such that h˜ (z )∧
1 p 0 1 0
···∧h˜ (z ) 6= 0. This means that the family {h˜ (z ),··· ,˜h (z )} is linearly independent
p 0 1 0 p 0
on C, i.e., the following matrix is of rank p
(h ,a )(z ) ··· (h ,a )(z )
1 1 0 p 1 0
. . .
A = .. .. ..
 
(h ,a )(z ) ··· (h ,a )(z )
1 N+1 0 p N+1 0
 
a (z ) ··· a (z ) h (z ) ··· h (z )
10 0 1N 0 10 0 p0 0
. . . . . .
= .. .. .. · .. .. .. .
   
a (z ) ··· a (z ) h (z ) ··· h (z )
N+10 0 N+1N 0 1N 0 pN 0
   
Hence the matrix
h (z ) ··· h (z )
10 0 p0 0
. . .
. . .
. . .
 
h (z ) ··· h (z )
1N 0 pN 0
 
is of rank p, i.e., h (z )∧···∧h (z ) 6= 0. This yields that h ∧···∧h 6≡ 0 on S. This is
1 0 p 0 1 p
a contradiction.
(⇐) We see that the following matrix is of rank 6 p−1 for each z ∈ S
(h ,a )(z) ··· (h ,a )(z)
1 1 p 1
. . .
A = .. .. .. .
 
(h ,a )(z) ··· (h ,a )(z)
1 N+1 p N+1
 
On the other hand, we have
a (z) ··· a (z) h (z) ··· h (z)
10 1N 10 p0
. . . . . .
A = .. .. .. · .. .. .. .
   
a (z) ··· a (z) h (z) ··· h (z)
N+10 N+1N 1N pN
   
Since the family {a } is located in general position and S 6⊂ (a ∧···∧a )−1{0}, this
i 1 N+1
implies that the matrix
h (z) ··· h (z)
10 p0
. . .
. . .
. . .
 
h (z) ··· h (z)
1N pN
 
is of rank 6 p−1 for each z ∈ S, i.e., h ∧···∧h ≡ 0 on S.
1 p
The Claim 3.3 is proved.
We now continue to prove the theorem. Suppose that f ∧···∧f 6≡ 0.
1 λ
(f ,g ) ··· (f ,g )
1 j0 λ j0
(f ,g ) ··· (f ,g )
For λ indices 0 = j < j < ··· < j 6 N such that  1 . j1 . λ. j1 
0 1 λ−1 . . .
. . .
 (f ,g ) ··· (f ,g ) 
 1 jλ−1 λ jλ−1 
is nondegenerate.  
10 LE NGOC QUYNH
(f ,g ) ··· (f ,g )
1 j0 λ j0
(f ,g ) ··· (f ,g )
PutJ = {j ,··· ,j },Jc = {0,...,q−1}\J andB =  1 . j1 . λ. j1 .
0 λ−1 J . . .
. . .
 (f ,g ) ··· (f ,g ) 
 1 jλ−1 λ jλ−1 
We now prove the following claim.  
Claim 3.4. If B is nondegenerate, i.e., detB 6≡ 0 then
J J
(min{ν0 }−min{1,ν0 })
16i6λ (fi,gj),6kj (f1,gj),6kj
j∈J
X
q−1
+ (λ−l+1)min{1,ν0 } 6 µ
(f1,gj),6kj f˜1∧···∧f˜λ
j=0
X
on the setCn\(A∪ λ I(f )∪(g ∧···∧g )−1(0)),where f˜ := ((f ,g ) : ··· : (f ,g ))
i=1 i j0 jλ−1 i i j0 i jλ−1
and A = Z ∩Z .
06i<j6q−S1 (f1,gi) (f1,gj)
Indeed,Sput A := Z ,Ac := Z . We consider the following two cases.
j∈J (f1,gj) j∈Jc (f1,gj)
Case 1. Let z ∈ A\(A∪ λ I(f )∪(g ∧···∧g )−1(0)) be a regular point of
0 S i=1 Si j0 jλ−1
A. Then z is a zero of one of the meromorphic functions {(f ,g )} . Without loss of
0 1 j j∈J
S
generality we may assume that z is a zero of (f ,g ). Let S be an irreducible component
0 1 j0
of A containing z . Suppose that U is an open neighborhood of z in Cn such that
0 0
U ∩ {A \ S} = ∅. Choose a holomorphic function h on an open neighborhood U′ ⊂ U
of z such that ν (z) = min {ν0 (z)} if z ∈ S and ν (z) = 0 if z 6∈ S. Then
0 h 16i6λ (fi,gj0),6kj0 h
(f ,g ) = a h (1 6 i 6 λ), where a are holomorphic functions. Therefore, the matrix
i j0 i i
(f ,g ) ··· (f ,g )
1 j1 λ j1
. . .
.. .. .. is of rank 6 λ − 1. Hence, there exist λ holomorphic
 
(f ,g ) ··· (f ,g )
1 jλ−1 λ jλ−1
functions b ,··· ,b , not all zeros, such that
1 λ
λ
b ·(f ,g ) = 0 (1 6 k 6 λ−1).
i i jk
i=1
X
Without loss of generality, we may assume that the set of common zeros of {b }λ is an
i i=1
analytic subset of codimension ≥ 2. Then there exists an index i ,1 6 i 6 λ, such that
1 1
S 6⊂ b−1{0}. We may assume that i = λ. Then for each z ∈ (U′ ∪S)\b−1{0}, we have
i1 1 λ
λ−1
b
f˜(z)∧···∧f˜(z) = f˜(z)∧···∧f˜ (z)∧ f˜(z)+ if˜(z)
1 λ 1 λ−1 λ i
b
(cid:18) i=1 λ (cid:19)
X
= f˜(z)∧···∧f˜ (z)∧(V(z)h(z))
1 λ−1
= h(z)·(f˜(z)∧···∧f˜ (z)∧V(z)),
1 λ−1
b
where V(z) := (a + λ−1 ia ,0,··· ,0).
λ i=1 b i
λ
By the assumptionPand by Claim 3.3, for any increasing sequence 1 6 i < ··· < i 6
1 l
λ−1, we have f˜ ∧···∧f˜ = 0 on S. This implies that the family {f˜,···f˜ ,V} is in
i1 il 1 λ−1