3Projective varieties

III Positivity in Algebraic Geometry



3.2 Ample divisors
We begin with the following useful lemma:
Lemma.
Let
X, Y
be projective schemes. If
f
:
X Y
is a finite morphism of
schemes, and D is an ample Cartier divisor on Y , then so is f
D.
Proof.
Let
F
be a coherent sheaf on
X
. Since
f
is finite, we have
R
i
f
F
= 0
for all i > 0. Then
H
i
(F f
O
X
(mD)) = H
i
(f
F O
Y
(mD)) = 0
for all i > 0 and m 0. So by Serre’s theorem, we know f
D is ample.
Another useful result is
Proposition.
Let
X
be a proper scheme, and
L
an invertible sheaf. Then
L
is
ample iff L|
X
red
is ample.
Proof.
()
If
L
induces a closed embedding of
X
, then map given by
L|
X
red
is given
by the composition with the closed embedding X
red
X.
()
Let
J O
X
be the nilradical. By Noetherianness, there exists
n
such that
J
n
= 0.
Fix F a coherent sheaf. We can filter F using
F J F J
2
F · · · J
n1
F J
n
F = 0.
For each j, we have a short exact sequence
0 J
j+1
F J
j
F G
j
0.
This
G
i
is really a sheaf on the reduced structure, since
J
acts trivially.
Thus
H
i
(
G
j
L
m
) for
j >
0 and large
m
. Thus inducting on
j
0, we
find that for i > 0 and m 0, we have
H
i
(J
j
F L
m
) = 0.
The following criterion for ampleness will give us a very good handle on how
ample divisors behave:
Theorem
(Nakai’s criterion)
.
Let
X
be a projective variety. Let
D
be a Cartier
divisor on
X
. Then
D
is ample iff for all
V X
integral proper subvariety
(proper means proper scheme, not proper subset), we have
(D|
V
)
dim V
= D
dim V
[V ] > 0.
Before we prove this, we observe some immediate corollaries:
Corollary.
Let
X
be a projective variety. Then ampleness is a numerical
condition, i.e. for any Cartier divisors
D
1
, D
2
, if
D
1
D
2
, then
D
1
is ample iff
D
2
is ample.
Corollary.
Let
X, Y
be projective variety. If
f
:
X Y
is a surjective finite
morphism of schemes, and
D
is a Cartier divisor on
Y
. Then
D
is ample iff
f
D
is ample.
Proof.
It remains to prove
. If
f
is finite and surjective, then for all
V Y
,
there exists
V
0
f
1
(
V
)
X
such that
f|
V
0
:
V
0
V
is a finite surjective
morphism. Then we have
(f
D)
dim V
0
[V
0
] = deg f |
V
0
D
dim V
[V ],
which is clear since we only have to prove this for very ample D.
Corollary.
If
X
is a projective variety,
D
a Cartier divisor and
O
X
(
D
) globally
generated, and
Φ : X P(H
0
(X, O
X
(D))
)
the induced map. Then D is ample iff Φ is finite.
Proof.
()
If
X
Φ(
X
) is finite, then
D
= Φ
O
(1). So this follows from the previous
corollary.
()
If Φ is not finite, then there exists
C X
such that Φ(
C
) is a point.
Then
D ·
[
C
] = Φ
O
(1)
·
[
C
] = 0, by the push-pull formula. So by Nakai’s
criterion, D is not ample.
Proof of Nakai’s criterion.
()
If
D
is ample, then
mD
is very ample for some
m
. Then by multilinearity,
we may assume D is very ample. So we have a closed embedding
Φ : X P(H
0
(D)
).
If
V X
is a closed integral subvariety, then
D
dim V
·
[
V
] = (
D|
V
)
dim V
.
But this is just deg
Φ(V )
O(1) > 0.
()
We proceed by induction on
dim X
, noting that
dim X
= 1 is trivial. By
induction, for any proper subvariety V , we know that D|
V
is ample.
The key of the proof is to show that
O
X
(
mD
) is globally generated for
large
m
. If so, the induced map
X P
(
|mD|
) cannot contract any curve
C
, or else
mD · C
= 0. So this is a finite map, and
mD
is the pullback of
the ample divisor O
P(|mD|)
(1) along a finite map, hence is ample.
We first reduce to the case where
D
is effective. As usual, write
D H
1
H
2
with H
i
very ample effective divisors. We have exact sequences
0 O
X
(mD H
1
) O
X
(mD) O
H
1
(mD) 0
0 O
X
(mD H
1
) O
X
((m 1)D) O
H
2
((m 1)D) 0.
We know
D|
H
i
is ample by induction. So the long exact sequences implies
that for all m 0 and j 2, we have
H
j
(mD)
=
H
j
(mD H
1
) = H
j
((m 1)D).
So we know that
χ(mD) = h
0
(mD) h
1
(mD) + constant
for all
m
0. On the other hand, since
X
is an integral subvariety of
itself,
D
n
>
0, and so asymptotic Riemann–Roch tells us
h
0
(
mD
)
>
0 for
all
m
0. Since
D
is ample iff
mD
is ample, we can assume
D
is effective.
To show that
mD
is globally generated, we observe that it suffices to show
that this is true in a neighbourhood of
D
, since outside of
D
, the sheaf
is automatically globally generated by using the tautological section that
vanishes at D with multiplicity m.
Moreover, we know
mD|
D
is very ample, and in particular globally gener-
ated for large
m
by induction (the previous proposition allows us to pass
to D|
red
if necessary). Thus, it suffices to show that
H
0
(O
X
(mD)) H
0
(O
D
(mD))
is surjective.
To show this, we use the short exact sequence
0 O
X
((m 1)D) O
X
(mD) O
D
(mD) 0.
For
i >
0 and large
m
, we know
H
i
(
O
D
(
mD
)) = 0. So we have surjections
H
1
((m 1)D) H
1
(mD)
for
m
large enough. But these are finite-dimensional vector spaces. So for
m
sufficiently large, this map is an isomorphism. Then
H
0
(
O
X
(
mD
))
H
0
(O
D
(mD)) is a surjection by exactness.
Recall that we defined
Pic
K
(
X
) for
K
=
Q
or
R
. We can extend the
intersection product linearly, and all the properties of the intersection product
are preserved. We write
N
1
K
(X) = Pic
K
(X)/Num
0,K
(X) = NS(X) K
where, unsurprisingly,
Num
0,K
(X) = {D Pic
K
(X) : D · C = 0 for all C X} = Num
0
K.
We now want to define ampleness for such divisors.
Proposition. Let D CaDiv
Q
(X) Then the following are equivalent:
(i) cD is an ample integral divisor for some c N
>0
.
(ii) D =
P
c
i
D
i
, where c
i
Q
>0
and D
i
are ample Cartier divisors.
(iii) D
satisfies Nakai’s criterion. That is,
D
dim V
[
V
]
>
0 for all integral
subvarieties V X.
Proof.
It is easy to see that (i) and (ii) are equivalent. It is also easy to see that
(i) and (iii) are equivalent.
We write Amp(X) N
1
Q
(X) for the cone given by ample divisors.
Lemma. A positive linear combination of ample divisors is ample.
Proof. Let H
1
, H
2
be ample. Then for λ
1
, λ
2
> 0, we have
(λ
1
H
1
+ λ
2
H
2
)
dim V
[V ] =
X
dim V
p
λ
p
1
λ
dim V p
2
H
p
1
· H
dim V p
2
[V ]
Since any restriction of an ample divisor to an integral subscheme is ample, and
multiplying with
H
is the same as restricting to a hyperplane cuts, we know all
the terms are positive.
Proposition.
Ampleness is an open condition. That is, if
D
is ample and
E
1
, . . . E
r
are Cartier divisors, then for all
|ε
i
|
1, the divisor
D
+
ε
i
E
i
is still
ample.
Proof.
By induction, it suffices to check it in the case
n
= 1. Take
m N
such
that
mD ± E
1
is still ample. This is the same as saying
D ±
1
m
E
1
is still ample.
Then for |ε
1
| <
1
m
, we can write
D + εE
1
= (1 q)D + q
D +
1
m
E
1
for some q < 1.
Similarly, if D CaDiv
R
(X), then we say D is ample iff
D =
X
a
i
D
i
,
where D
i
are ample divisors and a
i
> 0.
Proposition.
Being ample is a numerical property over
R
, i.e. if
D
1
, D
2
CaDiv
R
(X) are such that D
1
D
2
, then D
1
is ample iff D
2
is ample.
Proof.
We already know that this is true over
Q
by Nakai’s criterion. Then for
real coefficients, we want to show that if
D
is ample,
E
is numerically trivial
and
t R
, then
D
+
tE
is ample. To do so, pick
t
1
< t < t
2
with
t
i
Q
, and
then find λ, µ > 0 such that
λ(D
1
+ t
1
E) + µ(D
1
+ t
2
E) = D
1
+ tE.
Then we are done by checking Nakai’s criterion.
Similarly, we have
Proposition.
Let
H
be an ample
R
-divisor. Then for all
R
-divisors
E
1
, . . . , E
r
,
for all kε
i
k 1, the divisor H +
P
ε
i
E
i
is still ample.