Effective base point freeness on a normal surface

16 Pages·0.14 MB·English

by  Takeshi Kawachi

EFFECTIVE BASE POINT FREENESS ON A NORMAL SURFACE
Takeshi Kawachi
7
9
9 Department of Mathematics, Tokyo Institute of Technology
1

n
a Abstract. We prove effective base point freeness of the adjoint linear system on normal
J surfaces with a boundary.

0
2

2
v
0. Introduction
8
1
Let Y be a compact normal two dimensional projective variety over C (we call it “normal
0
2 surface” for short). Let y ∈ Y be a given point, and D be a nef and big Q-divisor on Y
1
such that K +⌈D⌉ is a Cartier divisor, There is various numerical criteria on D when the
6 Y
9 adjoint linear system |KY +⌈D⌉| is free at y. The main theorem includes all earlier result
m/ and gives the first effective version if y is log-terminal singularity.
To describe the main theorem we prepare notation.
o
e
g Definition. Let Y be a normal surface, and B be an effective Q-divisor on Y. Let y be a
-
g point on Y. Then the triple (Y,B,y) is said to be a germ of quasi-log-terminal singularity
l if the following conditions are satisfied:
a
:
v (1) [B] = 0, where [B] is the integral part of B.
i
X (2) Let f: X → Y be the minimal resolution of y (if y is a smooth point then let f be
r the blowing-up at y). If we write
a
K = f∗(K +B)+ a F (for a ∈ Q)
X Y j j j
X
then a > −1 whenever F is f-exceptional.
j j
Remark. The differences with “log-terminal” are:
(1) We don’t mention where K +B is Q-Cartier or not. The pull-back and the inter-
Y
section are used by Mumford’s Q-valued pull-back and intersection theory.
(2) f is only “minimal resolution”. Thus the Q-divisor a F may not be normal
j j
crossings. Hence “quasi-log-terminal” may not be “log-terminal” (cf. [KMM]).
P
(3) If (Y,B,y) is quasi-log-terminal then (Y,0,y) which is ignored the boundary, is log-
terminal.
Typeset by AMS-TEX
1
Let (Y,B,y) be a germ of normal surface singularity with a boundary B, and let f: X →
Y be the minimal resolution of y (if y is a smooth point then the blowing-up at y). Let Z
is the fundamental cycle of y (if y is a smooth point then the exceptional divisor of f). Let
∆ = f∗K −K . ∆ is an effective Q-divisor if y is singular, and ∆ = −Z if y is smooth.
y Y X y y
Definition.
−(Z −∆ )2, if y is quasi-log-terminal
y
δ =
y
0, if y is not quasi-log-teminal
(cid:26)
Definition. We define
µ(B,y) = max{µ | µ(Z −∆ ) ≤ f∗B}
y
Remark.
(1) This δ satisfies 0 ≤ δ ≤ 4 ([KM]).
y y
(2) If y is a smooth point then 2µ = mult B. If y is quasi-log-terminal then µ < 1.
y
The main result is the following.
Theorem. Let (Y,B) be a normal surface with a boundary B and assume that [B] = 0.
Let M be a Cartier divisor on Y such that M −(K +B) is nef big. Let D be an effective
Y
Q-divisor on Y which is Q-linearly equivalent to M − (K + B). Let y be a point on Y.
Y
Let µ = µ(B,y). Assume that D2 > (1−µ)2δ . If DC ≥ (1−µ)δ /2 for all curves C in
y y
Y such that y ∈ C then the linear system |M| is free at y. Furthermore if y is a singular
point which is not of type A then |M| is free at y.
n
We can get the following immediately.
Corollary. Let Y be a normal surface. Let D be a nef Q-divisor on Y such that K +⌈D⌉
Y
is Cartier. Let y be a point on Y. Let B = ⌈D⌉ − D and µ = µ(B,y) Assume that
D2 > (1−µ)2δ . If DC ≥ (1−µ)δ /2 for all curves C in Y such that y ∈ C then the linear
y y
system |K +⌈D⌉ is free at y.
Y
Remark. If y is a smooth point or a rational double point then this theorem is essentially
the theorem of Ein-Lazarsfeld ([EL]). If y is not a quasi-log-teminal point then the result
are proved in [ELM]. If B = 0 then the result is described in [KM].
In section 1, we remark the facts about quasi-log-terminal singularity. These are used by
the proof of the main theorem. In section 2, we prove the theorem.
1. Facts about quasi-log-terminal singularity
Throughout this section, let (Y,B,y) be a germ of quasi-log-terminal surface singularity.
Let ∆ = f∗K − K = a ∆ where ∆ is an irreducible exceptional divisor lying
y Y X j j i
over y. Let B = B + B′ and f∗B = B + B exc = B + b ∆ where B consists
y P y y y y j j y
of the components which passing through y, B′ is disjoint from y and B is the proper
P y
transformation of B . Let Z be the fundamfental cycle of yf. Hence we have
y
f
f∗(K +B) = K +B +∆ +B exc +f∗B′.
Y X y y y
2
f
First we bound the value of µ(B,y). we recall the definition of µ(B,y).
µ(B,y) = max{µ | µ(Z −∆ ) ≤ f∗B}
y
Lemma 1. If (Y,B,y) is quasi-log-terminal then µ(B,y) < 1.
Proof. If µ(B,y) ≥ 1 then B exc ≥ Z −∆ . Hence we have ∆ +B exc ≥ Z. On the other
y y y y
hand, since (Y,B,y) is quasi-log-terminal, all coefficients of ∆ +B exc must be less than 1.
y y
That is contradiction. (cid:3)
Since B is an effective divisor, B exc is also effective. Therefore all coefficients of ∆ are
y y
less than 1. This is the case of classical “log-terminal” surface singularity without boundary.
We defined δ = −(Z −∆ )2. This is bounded by the following lemma (cf. [KM]).
y y
Lemma 2. 0 < δ < 2 if y is log-terminal but not a smooth or a rational double point.
y
Proof. Since the intersection matrix k∆ ∆ k is negative definite and Z−∆ is effective and
i j y
non-zero, we have δ > 0. On the other hand, we have
y
(Z −∆ )2 = Z(Z −∆ )−∆ (Z −∆ )
y y y y
= Z(Z +K )+K (Z −∆ )
X X y
> Z(Z +K ) = 2Pa(Z)−2.
X
Since the log-terminal singularity is the rational singularity, we have Pa(Z) = 0 by [A].
Therefore we have δ < 2. (cid:3)
y
It is easy to prove that δ = 4 if y is a smooth point and that δ = 2 if y is a rational
y y
double point.
Let C be an irreducible divisor on Y and y ∈ C. Let f∗C = C + c ∆ , and C∆ = 1
i,j j i
for an i and C∆ = 0 for j 6= i. That is, we assume that C meets only ∆ . Next, we study
j i
P
the value a +c . Since all log-terminal singularities are classifieed in [B], we exameine it in
i i,i
each cases. wee set indexes as in Appendix. e
1.1 The case of type A . Let A = A(w ,...,w ) be the intersection matrix of type A
n 1 n n
where w = −∆2.
i i
−w 1 0 0
1
.
.
1 −w . 0
A(w ,...,w ) =  2 .
1 n . .
. .
0 . . 1
 
 0 0 1 −w 
 n
 
Let a(w ,...,w ) = detA(w ,...,w ), and we define a() = 1 for convention.
1 n 1 n
3
Let A be the (i,j)-component of A−1. Suppose j < i then we have
ij
−w 1 0 0
e 1
.
.
1 . 1 0
(cid:12) (cid:12)
(cid:12) (cid:12)
0 1 −w 1 0 0
(cid:12) j−1 (cid:12)
(cid:12) (cid:12)
0 0 1 −w 1 0
(cid:12) j+1 (cid:12)
1 (cid:12) . (cid:12)
Aij = detA(−1)i+j (cid:12)(cid:12) 0 .. −wi−1 0 (cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) 0 1 1 0 (cid:12)
e (cid:12)(cid:12) 0 −w ... 0 (cid:12)(cid:12)
(cid:12) i (cid:12)
(cid:12) . (cid:12)
(cid:12) 0 1 .. 1 (cid:12)
(cid:12) (cid:12)
(cid:12) 0 0 1 −w (cid:12)
(cid:12) n(cid:12)
1 (cid:12) (cid:12)
= (−1)i+ja(cid:12)(w ,...,w )a(w ,...,w ). (cid:12)
(cid:12) 1 j−1 i+1 n (cid:12)
detA
If j > i then we have the following as same as above.
1
A = (−1)i+ja(w ,...,w )a(w ,...,w ).
ij 1 i−1 j+1 n
detA
Also we have e
1
A = a(w ,...,w )a(w ,...,w ).
ii 1 i−1 i+1 n
detA
Since
e
a 2−w
1 1
. .
A(w ,...,w ) . = . ,
1 n . .
   
a 2−w
n n
   
we have the following.
i−1 n
a = A (2−w )+A (2−w )+ A (2−w )
i ij j ii i ij j
j=1 j=i+1
X X
1 e e e
= a(w ,...,w ) (−1)i+1(2−w )+···+(−1)2i−1a(w ,...,w )(2−w ) +
i+1 n 1 1 i−2 i−1
detA
a(w(cid:8) ,...,w )a(w(cid:0) ,...,w )(2−w )+ (cid:1)
1 i−1 i+1 n i
a(w ,...,w ) (−1)2i+1a(w ,...,w )(2−w )+···+(−1)i+n(2−w ) .
1 i−1 i+2 n i+1 n
Since a(w ,...,w ) = −w(cid:0) a(w ,...,w )−a(w ,...,w ), we have (cid:1)(cid:9)
1 j j 1 j−1 1 j−2
(−1)i+1(2−w )+(−1)i+2a(w )(2−w )+···+(−1)2i−1a(w ,...,w )(2−w )
1 1 2 1 i−2 i−1
= (−1)i+1 2−w +(−1)(2a(w )+a()+a(w ,w ))+···+
1 1 1 2
(−1)(cid:8)i−2(2a(w1,...,wi−2)+a(w1,...,wi−3)+a(w1,...,wi−1))
= (−1)i+1 1+(−1)i−2(a(w1,...,wi−2)+a(w1,...,wi−1)) (cid:9)
= (−1)i+1(cid:0)−a(w1,...,wi−2)−a(w1,...,wi−1). (cid:1)
4
We also have the following by same calculation.
(−1)2i+1a(w ,...,w )(2−w )+···+(−1)i+n(2−w )
i+2 n i+1 n
= −a(w ,...,w )−a(w ,...,w )+(−1)i+n.
i+1 n i+2 n
Hence we have
1
a = a(w ,...,w )((−1)i+1 −a(w ,...,w )−a(w ,...,w ))+
i i+1 n 1 i−2 1 i−1
detA
a(w(cid:8) ,...,w )a(w ,...,w )(2−w )+
1 i−1 i+1 n i
a(w ,...,w )((−1)i+n −a(w ,...,w )−a(w ,...,w ))
1 i−1 i+1 n i+2 n
1
= (−1)i+1a(w ,...,w )+(−1)i+na(w ,...,w ) (cid:9)
i+1 n 1 i−1
detA
−(cid:8)a(w ,...,w )a(w ,...,w )−a(w ,...,w )a(w ,...,w )w
1 i−2 i+1 n 1 i−1 i+1 n i
−a(w ,...,w )a(w ,...,w )
1 i−1 i+2 n
1
= 1+ {(−1)i+1a(w ,...,w )+((cid:9)−1)i+na(w ,...,w )}.
i+1 n 1 i−1
detA
Since (A(w ,...,w )(c )) = −1 if j = i and = 0 if j 6= i, we have
1 n i,j j
c = −A
i,j ji
1 a(w ,...,w )a(w ,...,w ), if j ≤ i
= e (−1)i+j+1 1 j−1 i+1 n
detA a(w ,...,w )a(w ,...,w ), if j > i.
(cid:26) 1 i−1 j+1 n
Therefore we have the following.
1
a +c = 1+ (−|a(w ,...,w )|−|a(w ,...,w )|+
i i,i i+1 n 1 i−1
|detA|
|a(w ,...,w )||a(w ,...,w )|)
1 i−1 i+1 n
1
= 1+ (|a(w ,...,w )|−1)(|a(w ,...,w )|−1)−1
1 i−1 i+1 n
|detA|
(cid:0) (cid:1)
Hence we get the following proposition.
Proposition 1. Let C as above. Assume y is of type A . If n ≥ 3 and i 6= 1,n then
n
a +c ≥ 1.
i i,i
If i = 1 or n then a +c = 1−1/|detA| (for n ≥ 1).
i i,i
In the case of type A we have δ = 2 − a − a . Indeed, since (∆ − Z)∆ = 2 −
n y 1 n y i
w − (−w + 2) = 0 if i 6= 1,n and (∆ − Z)∆ = 1 if i = 1 or n, we get the equation
i i y i
δ = 2−a −a .
y 1 n
5
1.2 The case of type D . Let D = D(w ,...,w ) be the intersection matrix of type
n 1 n−2
D where w = −∆2.
n i i
−w 1 0 0
1
.
.
1 −w . 0
 2 
. .
. .
D(w ,...,w ) = 0 . . 1 0 0 .
1 n  
 0 1 −w 1 1 
 n−2 
 0 1 −2 0 
 
 0 0 1 0 −2
 
 
Let d(w ,...,w ) = detD(w ,...,w ), and we define d() = 4 for convention.
1 n−2 1 n−2
Let D be the (i,j)-component of D−1. As same as of type A , we have
ij n
e 1 (−1)i+ja(w1,...,wj−1)d(wi+1,...,wn−2), i ≤ n−2,j ≤ i
D =
ij detD (−1)i+ja(w ,...,w )d(w ,...,w ), i ≤ n−2,i < j ≤ n−2.
(cid:26) 1 i−1 j+1 n−2
e
Let i ≤ n−2,j = n−1. In this case we have
−w 1 0 0
1
.
.
(cid:12) 1 . 1 0 (cid:12)
(cid:12) . (cid:12)
1 (cid:12)(cid:12) 0 .. −wi−1 0 (cid:12)(cid:12)
Di,n−1 = (−1)i+n−1(cid:12) 0 1 1 0 (cid:12)
detD (cid:12) (cid:12)
(cid:12) .. (cid:12)
(cid:12) 0 −w . 0 0 (cid:12)
e (cid:12) i+1 (cid:12)
(cid:12) 0 1 −w 1 1 (cid:12)
(cid:12) n−2 (cid:12)
(cid:12) 0 0 1 0 −2(cid:12)
(cid:12) (cid:12)
1 (cid:12) (cid:12)
= (−1)i+n−1((cid:12)−2)a(w ,··· ,w ). (cid:12)
(cid:12) 1 i−1 (cid:12)
detD
Also we have
D = D = D
in n−1,i ni
1
= (−1)i+n−1(−2)a(w ,··· ,w ).
e e e 1 i−1
detD
Now we suppose i,j ≥ n−1. In this case, we have
D = D
n−1,n−1 n,n
−w 1 0 0
1
e e .
1 . 1
1 . 1 0
= (cid:12) (cid:12) = a(w ,...,w ,2).
1 n−2
detD (cid:12) 0 1 −w 1 (cid:12) detD
(cid:12) n−2 (cid:12)
(cid:12) 0 0 1 −2(cid:12)
(cid:12) (cid:12)
(cid:12) 6 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
D = D
n−1,n n,n−1
−w 1 0 0
1
e e .
1 1 .. 1 0 1
= (−1)(cid:12) (cid:12) = a(w ,...,w ).
1 n−3
detD (cid:12) 0 1 −w 1(cid:12) detD
(cid:12) n−2 (cid:12)
(cid:12) 0 0 1 0(cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Note that there is an equation (cid:12) (cid:12)
(cid:12) (cid:12)
d(w ,...,w ) = −4a(w ,...,w ,1).
1 n−2 1 n−2
Now we calculate the value a . The calculation will proceed as same as of type A . But
i n
one thing differs, that is d(w ) = −4w +4. For i ≤ n−2, we have the following.
n−2 2
1
a = d(w ,...,w )((−1)i+1 −a(w ,...,w )−a(w ,...,w ))+
i i+1 n−2 1 i−2 1 i−1
detD
a(w(cid:8) ,...,w )d(w ,...,w )(2−w )
1 i−1 i+1 n−2 i
−a(w ,...,w )(d(w ,...,w )+d(w ,...,w ))
1 i−1 i+1 n−2 i+2 n−2
1
= (−1)i+1d(w ,...,w )−a(w ,...,w )d(w ,...,(cid:9)w )
i+1 n−2 1 i−2 i+1 n−2
detD
−(cid:8)a(w ,...,w )d(w ,...,w )w −a(w ,...,w )d(w ,...,w )
1 i−1 i+1 n−2 i 1 i−1 i+2 n−2
1 1
= 1+ (−1)i+1d(w ,...,w ) = 1− |d(w ,...,w )|. (cid:9)
i+1 n−2 i+1 n−2
detD |detD|
Since c = −D , we have the following.
i,i ii
1
a +c = 1+ |d(w ,...,w )| |a(w ,...,w )|−1 .
ie i,i i+1 n−2 1 i−1
|detD|
(cid:0) (cid:1)
For i = n−1,n, we have a = a = a /2. Hence we have
n−1 n n−2
1 2 1 2
a = a = +(−1)n−1 = − .
n−1 n
2 detD 2 |detD|
Therefore we have the following.
a +c = a +c
n−1 n−1,n−1 n n,n
1 1
= 1+ {2(−1)n−1 − d(w ,...,w )−a(w ,...,w ,2)}
1 n−2 1 n−2
detD 2
1
= 1+ {2(−1)n−1 +2a(w ,...,w ,1)−a(w ,...,w ,2)}
1 n−2 1 n−2
detD
1
= 1+ {2(−1)n−1 −2a(w ,...,w )−2a(w ,...,w )+
1 n−2 1 n−3
detD
2a(w ,...,w )+a(w ,...,w )}
1 n−2 1 n−3
1
= 1+ {2(−1)n−1 −a(w ,...,w )}
1 n−3
detD
1
= 1+ (|a(w ,...,w )|−2) ≥ 1.
1 n−3
|detD|
Hence we get the following proposition.
7
Proposition 2. Let C as above. If y is of type D then a +c ≥ 1 for all i.
n i i,i
1.3 The case of type E . If the dual graph is in the Appendix then we write
n
(m;a,b,c;d,e)or may write (m;a,b,c;d,e;−2). We set indexes i as in the Appendix. There
are only 15 types in all by the classification ([B]). Hence we can calculate a +c directly.
i i,i
The result is in Appendix.
Hence we get the following proposition.
Proposition 3. Let C as above. If y is of type E then a +c > 1 for all i.
n i i
2. Proof of the Theorem
We recall notation. Let (Y,B) be a normal surface with a boundary B such that [B] = 0.
Let M be a Cartier divisor on Y. Let D be an effective Q-divisor on Y, Q-linearly equivalent
to the nef big divisor M − (K + B). Let y be a point of Y. Let f: X → Y be the
Y
global minimal resolution, if y is a smooth point, f factors the blowing-up at y. Let ∆ =
f∗K −K . Let ∆ be the components of ∆ supported on f−1(y), and ∆′ = ∆−∆ . Let Z
Y X y y
bethefundamentalcycleofy. Letµ = µ(B,y),whereµ(B,y) = max{µ | µ(Z−∆ ) ≤ f∗B}.
y
Let δ = −(Z−∆ )2 if y is quasi-log-terminal, and δ = 0 if y is not quasi-log-terminal. Let
y y y
B = B +B′ where B consists of the component which contain y, and B′ is disjoint from
y y
y. Let f∗B = B +B exc where B is the proper transformation of B , and f(B exc) = {y}.
y y y y y
Since M ∼ K +D +B is Cartier, f∗(K +D +B) = K +f∗(D +B)+∆ is integral.
Y Y X
Hence D +B isfalso integral coeffiecients, thus ⌈D⌉ = D +B.
First we treat the case which y is not quasi-log-terminal. In this case, we assume that
D2 > 0 and DC ≥ 0 for all curve C ∋ y, because we defined δ = 0. Since D is nef and big,
y
this always hold. Hence it is sufficient to prove that y is not a base point of |M|.
Since f∗D is nef and big, we have H1(X,K +⌈f∗D⌉) = 0 by Sakai’s lemma [S1] where
X
⌈⌉ means the “round-up”. On the other hand,
K +⌈f∗D⌉ = ⌈K +f∗D⌉ = ⌈f∗(K +B)−∆−f∗B +f∗D⌉
X X Y
= ⌈f∗(K +⌈D⌉)−(∆ +B exc)−∆′ −f∗B′⌉
Y y y
= f∗(K +⌈D⌉)−⌊∆ +B exc⌋−⌊∆′ +f∗B′⌋.
Y y y
Since y is not quasi-log-terminal, the round-down ⌊∆ +B exc⌋ 6= 0. Let S = ⌊∆ +B exc⌋
y y y y
and T = ⌊∆′ +f∗B′⌋. Hence we have a surjection
H0(X,f∗(K +⌈D⌉)−T) → H0(S,(f∗(K +⌈D⌉)−T)| ) → 0.
Y Y S
Since f(S) = {y}, O ((f∗(K + ⌈D⌉) − T)| ) is trivial. Hence we get a section of
S Y S
O (f∗(K + ⌈D⌉) − T) which does not vanish at any point of SuppS. Therefore we
X Y
get a global section of O (f∗(K +⌈D⌉)) which does not vanish on f−1(y) by multiplying
X Y
the global section of O (T).
X
From now we assume that y is quasi-log-terminal. Let D = D +D′ where all irreducible
y
divisors in D contain y and D′ is the others. Let f∗D = D +D exc = d D + d′∆ .
y y y y i i j j
8
P P
f
Let f∗B = b D + b′∆ . Let ∆ = a ∆ . Since D2 > −((1 − µ)(Z − ∆ ))2,
y i i j j y j j y
f∗D −(1−µ)(Z −∆ ) is big. Hence if necessary, we change D by Q-linearly equivalence
y
P P P
and we assume D 6= 0. We define the number
y
1−bi 1−aj −b′j
c = min ,
d d′
( i j )
Since d ,d′ > 0, 0 ≤ b < 1 and a +b′ < 1 we have c > 0.
i j i j j
Lemma 3. c ≤ 1/2.
Proof. Since D2 > −((1 + ε)(1 − µ)(Z − ∆ ))2 for sufficiently small ε, we have D − (1 +
y
ε)(1−µ)(Z −∆ ) > 0. Hence we have d′ −(1−µ)(z −a ) > 0 where Z = z ∆ . Thus
y j j j j j
d′ +a −z +µ(z −a ) > 0. By the definition of µ, we have b′ ≥ µ(z −a ). Hence we have
j j j j j j j j
P
d′ +a −z +b′ > 0. Therefore d′ +a +b′ > z ≥ 1 because z is positive integer. On the
j j j j j j j j j
otherhand, SinceK +⌈D⌉ = K +D+B isCartier, f∗D+f∗B+∆ = f∗(K +D+B)−K
Y Y Y X
is integral divisor. Hence d′ + a + b′ is also integral. Since d′ + a + b′ ≥ 2, we have
j j j j j j
2(1−a /2−b′/2) ≤ d′. Thus we have
j j j
c ≤ 1−aj −b′j ≤ 1 − aj +b′j ≤ 1. (cid:3)
d′ 2 2d′ 2
j j
Let
R = f∗D−cf∗D = f∗⌈D⌉+∆−f∗B −cf∗D−∆
= f∗⌈D⌉+∆− (b +cd )D + (b′ +cd′ +a )∆ −(f∗B′ +f∗D′ +∆′).
i i i j j j j
(cid:16)X X (cid:17)
Since f∗⌈D⌉+∆ = f∗(K +⌈D⌉)−K is Cartier,
Y X
⌈R⌉ = f∗⌈D⌉+∆− (b +cd )D + (b′ +cd′ +a )∆ −⌊f∗B′ +f∗D′ +∆′⌋
i i i j j j j
= f∗⌈D⌉+∆−(jDX+···+D )−EX′ −Q k
1 s
where ⌊ (b +cd )D ⌋ = D +···+D , (b′ +cd′ +a )∆ = E′, and
i i i 1 s j j j j
⌊f∗B′ +f∗D′ +∆′⌋ = Q.
P (cid:4)P (cid:5)
Suppose E′ 6= 0. Since R = (1−c)f∗D is nef big and D ,...,D are not appear in the
1 s
fractional part of R, we have H1(X,K +⌈R⌉+D +···+D ) = 0 by Kawamata-Viehweg
X 1 s
vanishing theorem. Hence we have the surjection
H0(X,f∗(KY +⌈D⌉)−Q) → H0(E′,(f∗(KY +⌈D⌉)−Q)|E′) → 0.
Since f(E′) = {y} and Q is disjoint from y, we get a global section of K +⌈D⌉ which is
Y
not vanish at y.
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Now we assume E′ = 0. Note that the minimality of c induce that b +cd ≤ 1. Thus the
i i
minimal value c is obtained by (1−b )/d , hence s ≥ 1. By Kawamata-Viehweg vanishing
i i
theorem, we have
H0(X,K +⌈R⌉+D +···+D ) = H0(K +f∗⌈D⌉+∆−D −Q) = 0.
X 2 s X 1
Thus there is the surjection
H0(X,K +f∗⌈D⌉+∆−Q) → H0(D ,K +(f∗⌈D⌉+∆−D −Q)| ) → 0.
X 1 D1 1 D1
So it is sufficient to prove ⌈R⌉D > 1. Indeed, if ⌈R⌉D > 1 then ⌈R⌉D ≥ 2 because
1 1 1
⌈R⌉ is Cartier. Thus (f∗⌈D⌉+∆−D −Q)D ≥ ⌈R⌉D ≥ 2. Hence we get the nowhere
1 1 1
vanishing global section in H0(D ,K + (f∗⌈D⌉ + ∆ − D − Q)| ). Hence we get the
1 D1 1 D1
global section of O(f∗(K +⌈D⌉)) which is not vanish on D , especially not vanish at y.
Y 1
Hence we prove ⌈R⌉D > 1. Remark that
1
⌈R⌉ = R+ (b +cd )D + (b′ +cd′ +a )∆ +(others)
 i i i j j j j 
i6=X1,...s X 
where {(divisor)} is thefractional part of (divisor). Since E′ = 0, { (b′j +cd′j +aj)∆j} =
(b′ +cd′ +a )∆ . Therefore
j j j j
P
P ⌈R⌉D ≥ (1−c)f∗DD + (b′ +cd′ +a )∆ D .
1 1 j j j j 1
X
Thus it is sufficient to prove
(1) (1−c)(1−µ)δ /2+ (b′ +cd′ +a )∆ D > 1.
y j j j j 1
If D meets more components of ∆ tXhen (b′ + cd′ + a )∆ D is larger. Thus we
1 y j j j j 1
assume ∆ D = 1 for an i and ∆ D = 0 for j 6= i. Let f∗f D = D + c ∆ . Since
i 1 i 1 ∗ 1 1 i,j j
P
f∗B +cf∗D = D +(others), we have c ≤ b′ +cd′. But by Proposition 1 through 3,
y y 1 i,j j j
P
c + a ≥ 1 if y is of type D or of type E , and if y is of type A and i 6= 1,n. They
i,i i n n n
contradict to the assumption E′ = 0. Thus we assume y is of type A and j = 1,n or y is
n
a smooth point.
Thus if y is of type D or E and D2 > (1−µ)2δ then y is not a base point of |M|.
n n y
First we assume that y is a smooth point or of type A . In this case, b′ = µ(1 − a ).
1 1 1
Let p = µ(D,y), hence d′ = p(1−a ). Let w = −∆2. Then we have a = 1−2/w and
1 1 1 1 1 1
δ = 4/w . Hence δ = 2(1−a ). Since f∗D −(1−µ)(Z −∆ ) is big, we have p+µ > 1
y 1 y 1 y
as in the proof of lemma 3. Hence we have
δ
⌈R⌉D ≥ (1−c)(1−µ) y +(b′ +cd′ +a )
1 2 1 1 1
δ δ
= (1−µ) y +b′ +a +(d′ −(1−µ) y)c
2 1 1 1 2
= (1−µ)(1−a )+µ(1−a )+a +(p(1−a )−(1−µ)(1−a ))c
1 1 1 1 1
= 1+(p+µ−1)(1−a )c > 1.
1
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