Luận Văn Tiến Sỹ Về tập xác định duy nhất cho hàm chỉnh hình nhiều biến - Pdf 19

class="bi x0 y0 w1 h1"
class="bi x0 y0 w1 h1"
class="bi x1 y1 w2 h2"
p
p
p
p
p
p
p
class="bi x8 y9 w3 h5"
C
f, g C
f =
ag + b
cg + d
a, b, c, d
ad − bc = 0
f, g C
f ≡ g.
p
p
f, g : C
m
−→ P
n
(C) 3n + 1
P
n
(C) L
P

(C
p
) f
−1
(H
i
)

f
−1
(H
j
) = ∅
i = j f
−1
(H
i
) = g
−1
(H
i
) i = 1, , 3n + 1 f(z) = g(z)
z ∈
3n+1

i=1
f
−1
(H
i

(S)).
P
n
(C) n C
f f = [f
1
: : f
n+1
] : C −→ P
n
(C)
f
1
, , f
n+1

f = (f
1
, , f
n+1
) : C −→ C
n+1
− {0} f

f = (f
1
, , f
n+1
) g = (g
1

(H) = E
F


f
(0),
E
f
(H,  k) =

z ∈ C : F

˜
f(z) = 0 , v
F

˜
f
(z)  k

.
C ∪ {∞}
S C∪ {∞}
f, g E
f
(S) = E
g
(S), f ≡ g
{S
1

}, w) n ≥ 4
H
1
, , H
q
P
n
k
1
, , k
q
A
C P
n
(C), f, g ∈ A,
1)E
f
(H
i
,  k
i
) = E
g
(H
i
,  k
i
),
2)f = g
q

}
q
i=1
C
m
P
n
(C)
a
i
f (f, a
i
) ≡ 0, i = 1, , q
a) {z ∈ C
m
|(f, a
i
)(z) = (f, a
j
)(z) = 0}  m − 2 (1  i < j  q)},
b) (v
(f,a
j
)
, d) = (v
(g,a
j
)
, d), 1  j  q,
c) f(z) = g(z)


f
−1
(H
j
) = ∅
i = j f
−1
(H
i
) = g
−1
(H
i
) i = 1, , q f(z) = g(z)
z ∈
q

i=1
f
−1
(H
i
).
q = 3n + 2 f ≡ g
q = 2n + 3 f ≡ g
1
k
f, g : C −→ P
n

q

i=1
E
g
(H
i
,  k
i
). (1.5)
q > 2n
2
+ n + 1 +
q

i=1
n
k
i
+ 1
f ≡ g.
E
f
(H
i
,  k
i
)

E

,  k
i
).
n ∈ N

n ≥ 2m + 9 (n, m) = 1 m ≥ 2, P
i
(z) = z
n
− a
i
z
n−m
+ b
i
0 = a
i
, b
i
∈ C, i = 1, 2, , s b
2d
i
= b
d
j
b
d
l
i = j, i = l.
Q

1
+ + Q
d
s
, d ≥ (2s − 1)
2
. (1.11)
P
s+1,d
nd C.
f, g : C −→ P
s
(C)
X P
s
(C)
P
s+1,d
= 0 E
f
(X) = E
g
(X). f ≡ g.
nd
d ≥ (2s − 1)
2
p k
p
p
f, g n+1 d

i
)
f ≡ g.
p
q
p
p
p
p
p
P (x) ∈ C
p
[x]
C
p
C
m
p
p
C
m
p
C
m
p
bi p

{a
1
, , a

f
(a) f a,
f(z) = (z − a)
v
f
(a)
g(z),
g(z) a g(a) = 0
k, l r > 1
v
k
f
: C −→ N v
k
f
(z) =

v
f
(z) v
f
(z)  k,
0 v
f
(z) > k.
n
k
f
(r) =


N
k
l,f
(a, r) =
r

1
n
k
l,f
(a, x)dx
x
, n
k
l,f
(a, r) =

|z|r
min

v
k
f−a
(z), l

.
v
>k
f
: C −→ N v

f
(a, r) =
r

1
n
>k
f
(a, x)
x
dx, N
>k
f
(r) = N
>k
f
(0, r),
N
>k
l,f
(a, r) =
r

1
n
>k
l,f
(a, r)
x
dx, n

q
j=1
n
f : C −→ P
n
(C)
P F z
1
, . . . , z
n+1
P (

f) = 0, F (

f) = 0).
f m
m
P
n
(C) f m
m = n f
f, g : C −→ P
n
(C)

f = (f
1
, . . . , f
n+1
) g = (g



0
log ||

f(re
ıθ
)||dθ − log ||

f(0)||,
||

f|| =

|f
1
|
2
+ + |f
n+1
|
2

1/2
H P
n
(C)
F = 0, f H
T
f

l,F ◦

f
(r).
f : C −→ P
n
(C) m
H
1
, . . . , H
q
P
n
(C)
f(C) ⊂ H
j
j = 1, . . . , q
(q − 2n + m − 1)T
f
(r) 
q

j=1
N
m,f
(H
j
, r) + S
f
(r),

ε = {ε
1
, , ε
n+1
} C
n+1

ε
m+2
, , ε
n+1
E[f] = {α ∈ C
n+1

:<

f, α >≡ 0}
e = {e
1
, , e
n+1
} ε V
e = {e
1
, , e
m+1
} P(V)
V g C P(V)
g = (<


, a
j
> ε
i
=
n+1

k=1
a
jk
ε
k
, j = 1, , q,
c
j
=
m+1

i=1
< e
i
, b
j
> ε
i
=
m+1

k=1
a

.
{H
j
}
q
j=1
{b
j
}
n+1
j=1
C
n+1

x =
n+1

j=1
y
j
b
j
=
n+1

j=1
y
j
n+1


x =
m+1

k=1
n+1

j=1
y
j
a
jk
ε
k
=
n+1

j=1
y
j
m+1

k=1
a
jk
ε
k
=
n+1

j=1

m,g
(X
j
, r) + S
g
(r).
T
f
(r) = T
g
(r)+O(1), S
f
(r) = S
g
(r), N
m,f
(H
j
, r) = N
m,g
(X
j
, r)+O(1).
(q − 2n + m − 1)T
f
(r) 
q

j=1
N

− m + 1
k
i
+ 1
− 2n + m − 1

T
f
(r) 
q

i=1
k
i
k
i
+ 1
N
k
i
m,f
(H
i
, r) + S
f
(r).
S
f
(r) = o(T
f

, r)
=
k
i
k
i
+ 1
N
k
i
m,f
(H
i
, r) +
1
k
i
+ 1
N
k
i
m,f
(H
i
, r) + N
>k
i
m,f
(H
i

, r)

k
i
k
i
+ 1
N
k
i
m,f
(H
i
, r) +
m
k
i
+ 1
N
f
(H
i
, r).
N
f
(H
i
, r)  T
f
(r) + O(1).

i
, r) 
q

i=1
k
i
k
i
+ 1
N
k
i
m,f
(H
i
, r) +
q

i=1
m
k
i
+ 1
T
f
(r) + O(1).
f, g : C −→ P
n
(C)


i=1
E
g
(H
i
,  k
i
).
q > 2n
2
+ n + 1 +
q

i=1
n
k
i
+ 1
f ≡ g
f ≡ g h, l ∈ {1, . . . , n + 1}
h = l f
h
g
l
− f
l
g
h
≡ 0. f, g

i
, r) + S
f
(r)

q

i=1
kk
i
k
i
+ 1
N
k
i
1,f
(H
i
, r) + S
f
(r)
 knN
f
h
g
l
−f
l
g

(r)) + S
f
(r)

q − 2n − 1 −
q

i=1
m
k
i
+ 1
+ m

T
g
(r)  mn(T
f
(r) + T
g
(r)) + S
g
(r)
1  m  k  n.

q − 2n − 1 −
q

i=1
k

g
(r)) + S
g
(r).

q −2n−1−
q

i=1
k
k
i
+ 1
+m−kn−mn


T
f
(r)+T
g
(r)

 S
f
(r)+S
g
(r)).
q − 2n − 1 −
q


+ n + 1 +
q

i=1
n
k
i
+ 1
,
ϕ(k, m) ≥ ϕ(n, n) = q − 2n
2
− n − 1 −
q

i=1
n
k
i
+ 1
> 0.
f ≡ g
E
f
(H
i
,  k
i
)

E

k
1
, , k
q
∈ N

H
1
, , H
q
P
n
(C)
f(C) ⊂ H
i
g(C) ⊂ H
i
i = 1, . . . , q
E
f
(H
i
,  k
i
)

E
f
(H
j

i
+ 1
f ≡ g
k
i
→ ∞ n = 1 k
i
→ ∞
P (x) ∈ C[x]
C f, g
C P (f) = P (g) f = g.
P ∈ C[z]
f, g
C c = 0 P (f) = cP (g), f = g
p
P z
1
, . . . , z
n+1
f, g : C −→ P
n
(C)

f g P (

f) = P (g)
f = g
P z
1
, . . . , z

n+1
f
d
1
+ + f
d
n+1
= 0 d ≥ n
2
i ∈ {1, , n + 1}, j = i f
i
= c
ij
f
j
, c
ij
[58]. D
i
(x
1
, , x
s+1
) q
i
1  i  s + 1 f : C → P
s
(C)
f
s+1


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