Ak
n
·
·
Стр.1
n
Стр.2
℄
Cn
n = 2
℄
f
Ak
Ak
f : (Cn, a) → (C, 0)
{f = 0} ⊂ Cn
Cn
C2
f(z1, . . . , zn) = zk+1
k ∈ N,
℄
Ak
a
1 +z2
n = 3
kn(d) dn.
lim
d→∞
℄
2 +. . .+z2
lim
d→∞
kn(d)
dn 112
kn(d)
dn 1
d→∞
lim
lim
d→∞
k2(d)
d2
n = 3
209 · 2n−2 .
n 2
1
C3
28.
k2(d) ∼ d2
n 2
2n−1 .
n = 3
Ak
Ak
C3).
C2
z1, . . . , zn,
n.
Ak
k2(d)
d2
3
15
4.
d → ∞.
k ∈ N.
Ak
C3
d
C2
Cn
n 2
d ∈ N
Ak
a
kn(d)
Стр.3
λ ∈ U(0) \ {0}
0} ⊂ CP3.
B,
A = B ∩ C3,
m = α+β + γ
m+3
d
m− d − k + 1
d−2
m 2d−4
d − 2
0 m d−2
0 m d−2,
[m+3
{xαyβzγ | 0 α,β, γ d−2, l(xαyβzγ) /
℄
℄
d − 1 m 2d − 3,
m = α + β + γ
2
3d3 +O(d2)
d(d −1)
2 +
C2
m > 2d−3
d ]
{f(x1,x2,x3) = ε}
{Fλ = 0}
U(0) ⊂ C
F(x0 : x1 : x2 : x3) = 0,
C3 ⊂ CP3
deg f = d,
F
α,β, γ
CP3;
F
C = {f(x1,x2,x3) = 0} ⊂ C3
Ak
d→∞
lim
x0 = 0
{Fλ(x0 : x1 : x2 : x3) = 0}
ε ∈ C \ {0}
xd
d→∞.
1 +xd
2 +xd
xαyβzγ
0 α,β, γ d− 2.
d−2+2 = d(d−1)
C2
m = α+β +γ
d−1 m 2d−3
m+2 −
3(m−d+2)(m−d+1)
2
m > 3d − 6
m+3
d
2d−4
m=d−1
2d−3 m 3d−6
m 2d − 3
m = α+β +γ
k=−1
m−d
(m−d−k +1) = (m−d+2)(m−d+1)
d−1 β m
C2
d− 2
m+2
m− (d + k)
= (m+2)(m+1)
2
3(m−d+2)(m−d+1)
2
m+3
d
−
2
2 .
m
d−1 α m,
.
∈ Z, [l(xαyβzγ)]
m
3 = 1
C {Fλ |λ ∈ U(0)}
d.
k3(d)
d3
Bδ(0)
d
d
CP3
CP3
}.
l(xαyβzγ) = α+β+γ+3
d−2
m ∈ N,
m
3(m−d+2)(m−d+1)
2
m = α + β + γ
d−1 γ m
d+k, k −1
{Fλ = 0}
2
3 .
d
B = {xd
δ > 0,
Bδ(0)
Ak
0 + xd
B
d
(x0 : x1 : x2 : x3)
λ ∈ U(0) \ {0}
δ
λ
1 + xd
m = d − 2.
[·]
H2(A;R)
C2
m+2.
2 + xd
(m+2)(m+1)
2
−
3 =
d − 1
3(m−d+2)(m−d+1)
2
= 2
3d3 +O(d2), d→∞.
Стр.4