Part III — Positivity in Algebraic Geometry
Based on lectures by S. Svaldi
Notes taken by Dexter Chua
Lent 2018
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
This class aims at giving an introduction to the theory of divisors, linear systems and
their p ositivity properties on projective algebraic varieties.
The first part of the class will be dedicated to introducing the basic notions and results
regarding these objects and special attention will be devoted to discussing examples in
the case of curves and surfaces.
In the second part, the course will cover classical results from the theory of divisors
and linear systems and their applications to the study of the geometry of algebraic
varieties.
If time allows and based on the interests of the participants, there are a number of
more advanced topics that could possibly be covered: Reider’s Theorem for surfaces,
geometry of linear systems on higher dimensional varieties, multiplier ideal sheaves and
invariance of plurigenera, higher dimensional birational geometry.
Pre-requisites
The minimum requirement for those students wishing to enroll in this class is their
knowledge of basic concepts from the Algebraic Geometry Part 3 course, i.e. roughly
Chapters 2 and 3 of Hartshorne’s Algebraic Geometry.
Familiarity with the basic concepts of the geometry of algebraic varieties of dimension 1
and 2 — e.g. as covered in the preliminary sections of Chapters 4 and 5 of Hartshorne’s
Algebraic Geometry — would be useful but will not be assumed — besides what was
already covered in the Michaelmas lectures.
Students should have also some familiarity with concepts covered in the Algebraic
Topology Part 3 course such as cohomology, duality and characteristic classes.
1 Divisors
1.1 Projective embeddings
Our results here will be stated for rather general schemes, since we can, but at
the end, we are only interested in concrete algebraic varieties.
We are interested in the following problem — given a scheme
X
, can we
embed it in projective space? The first observation that
P
n
comes with a very
special line bundle
O
(1), and every embedding
X → P
n
gives pulls back to a
corresponding sheaf L on X.
The key property of
O
(1) is that it has global sections
x
0
, . . . , x
n
which
generate O(1) as an O
X
-module.
Definition
(Generating section)
.
Let
X
be a scheme, and
F
a sheaf of
O
X
-
modules. Let
s
0
, . . . , s
n
∈ H
0
(
X, F
) be sections. We say the sections generate
F if the natural map
n+1
M
i=0
O
X
→ F
induced by the s
i
is a surjective map of O
X
-modules.
The generating sections are preserved under pullbacks. So if
X
is embedded
into
P
n
, then it should have a corresponding line bundle generated by
n
+ 1
global sections. More generally, if there is any map to
P
n
at all, we can pull
back a corresponding bundle. Indeed, we have the following theorem:
Theorem. Let A be any ring, and X a scheme over A.
(i)
If
ϕ
:
X → P
n
is a morphism over
A
, then
ϕ
∗
O
P
n
(1) is an invertible sheaf
on X, generated by the sections ϕ
∗
x
0
, . . . , ϕ
∗
x
n
∈ H
0
(X, ϕ
∗
O
P
n
(1)).
(ii)
If
L
is an invertible sheaf on
X
, and if
s
0
, . . . , s
n
∈ H
0
(
X, L
) which generate
L
, then there exists a unique morphism
ϕ
:
X → P
n
such that
ϕ
∗
O
(1)
∼
=
L
and ϕ
∗
x
i
= s
i
.
This theorem highlights the importance of studying line bundles and their
sections, and in some sense, understanding these is the whole focus of the course.
Proof.
(i)
The pullback of an invertible sheaf is an invertible, and the pullbacks of
x
0
, . . . , x
n
generate ϕ
∗
O
P
n
(1).
(ii) In short, we map x ∈ X to [s
0
(x) : · · · : s
n
(x)] ∈ P
n
.
In more detail, define
X
s
i
= {p ∈ X : s
i
6∈ m
p
L
p
}.
This is a Zariski open set, and
s
i
is invertible on
X
s
i
. Thus there is a dual
section
s
∨
i
∈ L
∨
such that
s
i
⊗ s
∨
i
∈ L ⊗ L
∨
∼
=
O
X
is equal to 1. Define
the map X
s
i
→ A
n
by the map
K[A
n
] → H
0
(X
s
i
, O
s
i
)
y
i
7→ s
j
⊗ s
∨
i
.
Since the
s
i
generate, they cannot simultaneously vanish on a point. So
X
=
S
X
s
i
. Identifying
A
n
as the chart of
P
n
where
x
i
6
= 0, this defines
the desired map X → P
n
.
The theorem tells us we get a map from X to P
n
. However, it says nothing
about how nice these maps are. In particular, it says nothing about whether or
not we get an embedding.
Definition
(Very ample sheaf)
.
Let
X
be an algebraic variety over
K
, and
L
be an invertible sheaf. We say that
L
is very ample if there is a closed immersion
ϕ : X → P
n
such that ϕ
∗
O
P
n
(1)
∼
=
L.
It would be convenient if we had a good way of identifying when
L
is very
ample. In this section, we will prove some formal criteria for very ampleness,
which are convenient for proving things but not very convenient for actually
checking if a line bundle is very ample. In specific cases, such as for curves, we
can obtain some rather more concrete and usable criterion.
So how can we understand when a map
X → P
n
is an embedding? If it were
an embedding, then it in particular is injective. So given any two points in
X
,
there is a hyperplane in
P
n
that passes through one but not the other. But being
an embedding means something more. We want the differential of the map to
be injective as well. This boils down to the following conditions:
Proposition.
Let
K
=
¯
K
, and
X
a projective variety over
K
. Let
L
be an
invertible sheaf on
X
, and
s
0
, . . . , s
n
∈ H
0
(
X, L
) generating sections. Write
V
=
hs
0
, . . . , s
n
i
for the linear span. Then the associated map
ϕ
:
X → P
n
is a
closed embedding iff
(i)
For every distinct closed points
p 6
=
q ∈ X
, there exists
s
p,q
∈ V
such that
s
p,q
∈ m
p
L
p
but s
p,q
6∈ m
q
L
q
.
(ii)
For every closed point
p ∈ X
, the set
{s ∈ V | s ∈ m
p
L
p
}
spans the vector
space m
p
L
p
/m
2
p
L
p
.
Definition
(Separate points and tangent vectors)
.
With the hypothesis of the
proposition, we say that
– elements of V separate points if V satisfies (i).
– elements of V separate tangent vectors if V satisfies (ii).
Proof.
(⇒)
Suppose
φ
is a closed immersion. Then it is injective on points. So suppose
p 6
=
q
are (closed) points. Then there is some hyperplane
H
p,q
in
P
n
passing through
p
but not
q
. The hyperplane
H
p,q
is the vanishing locus
of a global section of
O
(1). Let
s
p,q
∈ V ⊆ H
0
(
X, L
) be the pullback of
this section. Then s
p,q
∈ m
p
L
p
and s
p,q
6∈ m
q
L
q
. So (i) is satisfied.
To see (ii), we restrict to the affine patch containing
p
, and
X
is a closed
subvariety of
A
n
. The result is then clear since
m
p
L
p
/m
2
p
L
p
is exactly the
span of
s
0
, . . . , s
n
. We used
K
=
¯
K
to know what the closed points of
P
n
look like.
(⇐)
We first show that
ϕ
is injective on closed points. For any
p 6
=
q ∈ X
,
write the given s
p,q
as
s
p,q
=
X
λ
i
s
i
=
X
λ
i
ϕ
∗
x
i
= ϕ
∗
X
λ
i
x
i
for some
λ
i
∈ K
. So we can take
H
p,q
to be given by the vanishing set
of
P
λ
i
x
i
, and so it is injective on closed point. It follows that it is also
injective on schematic points. Since
X
is proper, so is
ϕ
, and in particular
ϕ is a homeomorphism onto the image.
To show that
ϕ
is in fact a closed immersion, we need to show that
O
P
n
→ ϕ
∗
O
X
is surjective. As before, it is enough to prove that it holds
at the level of stalks over closed points. To show this, we observe that
L
p
is trivial, so
m
p
L
p
/m
2
p
L
p
∼
=
m
p
/m
2
p
(unnaturally). We then apply the
following lemma:
Lemma. Let f : A → B be a local morphism of local rings such that
◦ A/m
A
→ B/m
B
is an isomorphism;
◦ m
A
→ m
B
/m
2
B
is surjective; and
◦ B is a finitely-generated A-module.
Then f is surjective.
To check the first condition, note that we have
O
p,P
n
m
p,P
n
∼
=
O
p,X
m
p,X
∼
=
K.
Now since
m
p,P
n
is generated by
x
0
, . . . , x
n
, the second condition is the
same as saying
m
p,P
n
→
m
p,X
m
2
p,X
is surjective. The last part is immediate.
Unsurprisingly, this is not a very pleasant hypothesis to check, since it requires
us to really understand the structure of
V
. In general, given an invertible sheaf
L
, it is unlikely that we can concretely understand the space of sections. It
would be nice if there is some simpler criterion to check if sheaves are ample.
One convenient place to start is the following theorem of Serre:
Theorem
(Serre)
.
Let
X
be a projective scheme over a Noetherian ring
A
,
L
be
a very ample invertible sheaf, and
F
a coherent
O
X
-module. Then there exists
a positive integer
n
0
=
n
0
(
F, L
) such that for all
n ≥ n
0
, the twist
F ⊗ L
n
is
generated by global sections.
The proof of this is straightforward and will be omitted. The idea is that
tensoring with
L
n
lets us clear denominators, and once we have cleared all the
denominators of the (finitely many) generators of
F
, the resulting sheaf
F ⊗ L
n
will be generated by global sections.
We can weaken the condition of very ampleness to only require the condition
of this theorem to hold.
Definition
(Ample sheaf)
.
Let
X
be a Noetherian scheme over
A
, and
L
an
invertible sheaf over
X
. We say
L
is ample iff for any coherent
O
X
-module
F
,
there is an
n
0
such that for all
n ≥ n
0
, the sheaf
F ⊗ L
n
is generated by global
sections.
While this seems like a rather weak condition, it is actually not too bad.
First of all, by taking
F
to be
O
X
, we can find some
L
n
that is generated by
global sections. So at least it gives some map to
P
n
. In fact, another theorem of
Serre tells us this gives us an embedding.
Theorem
(Serre)
.
Let
X
be a scheme of finite type over a Noetherian ring
A
,
and
L
an invertible sheaf on
X
. Then
L
is ample iff there exists
m >
0 such
that L
m
is very ample.
Proof.
(⇐)
Let
L
m
be very ample, and
F
a coherent sheaf. By Serre’s theorem, there
exists n
0
such that for all j ≥ j
0
, the sheafs
F ⊗ L
mj
, (F ⊗ L) ⊗ L
mj
, . . . , (F ⊗ L
m−1
) ⊗ L
mj
are all globally generated. So F ⊗ L
n
is globally generated for n ≥ mj
0
.
(⇒)
Suppose
L
is ample. Then
L
m
is globally generated for
m
sufficiently large.
We claim that there exists t
1
, . . . , t
n
∈ H
0
(X, L
N
) such that L|
X
t
i
are all
trivial (i.e. isomorphic to O
X
t
i
), and X =
S
X
t
i
.
By compactness, it suffices to show that for each
p ∈ X
, there is some
t ∈ H
0
(
X, L
n
) (for some
n
) such that
p ∈ X
t
i
and
L
is trivial on
X
t
i
. First
of all, since
L
is locally free by definition, we can find an open affine
U
containing p such that L|
U
is trivial.
Thus, it suffices to produce a section
t
that vanishes on
Y
=
X − U
but
not at
p
. Then
p ∈ X
t
⊆ U
and hence
L
is trivial on
X
t
. Vanishing on
Y
is the same as belonging to the ideal sheaf
I
Y
. Since
I
Y
is coherent,
ampleness implies there is some large
n
such that
I
Y
⊗ L
n
is generated by
global sections. In particular, since
I
Y
⊗ L
n
doesn’t vanish at
p
, we can
find some
t ∈
Γ(
X, I
Y
⊗ L
N
) such that
t 6∈ m
p
(
I
Y
⊗ L
n
)
p
. Since
I
Y
is a
subsheaf of O
X
, we can view t as a section of L
n
, and this t works.
Now given the
X
t
i
, for each fixed
i
, we let
{b
ij
}
generate
O
X
t
i
as an
A
-algebra. Then for large
n
,
c
ij
=
t
n
i
b
ij
extends to a global section
c
ij
∈
Γ(
X, L
n
) (by clearing denominators). We can pick an
n
large enough
to work for all
b
ij
. Then we use
{t
n
i
, c
ij
}
as our generating sections
to construct a morphism to
P
N
, and let
{x
i
, x
ij
}
be the corresponding
coordinates. Observe that
S
X
t
i
=
X
implies the
t
n
i
already generate
L
n
.
Now each
x
t
i
gets mapped to
U
i
⊆ P
N
, the vanishing set of
x
i
. The map
O
U
i
→ ϕ
∗
O
X
t
i
corresponds to the map
A[y
i
, y
ij
] → O
X
t
i
,
where
y
ij
is mapped to
c
ij
/t
n
i
=
b
ij
. So by assumption, this is surjective,
and so we have a closed embedding.
From this, we also see that
Proposition.
Let
L
be a sheaf over
X
(which is itself a projective variety over
K). Then the following are equivalent:
(i) L is ample.
(ii) L
m
is ample for all m > 0.
(iii) L
m
is ample for some m > 0.
We will also frequently make use of the following theorem:
Theorem
(Serre)
.
Let
X
be a projective scheme over a Noetherian ring
A
, and
L is very ample on X. Let F be a coherent sheaf. Then
(i) For all i ≥ 0 and n ∈ N, H
i
(F ⊗ L
n
) is a finitely-generated A-module.
(ii)
There exists
n
0
∈ N
such that for all
n ≥ n
0
,
H
i
(
F ⊗ L
n
) = 0 for all
i > 0.
The proof is exactly the same as the case of O(1) on P
n
.
As before, this theorem still holds for ample sheaves, and in fact characterizes
them.
Theorem.
Let
X
be a proper scheme over a Noetherian ring
A
, and
L
an
invertible sheaf. Then the following are equivalent:
(i) L is ample.
(ii) For all coherent F on X, there exists n
0
∈ N such that for all n ≥ n
0
, we
have H
i
(F ⊗ L
n
) = 0.
Proof. Proving (i) ⇒ (ii) is the same as the first part of the theorem last time.
To prove (ii) ⇒ (i), fix a point x ∈ X, and consider the sequence
0 → m
x
F → F → F
x
→ 0.
We twist by
L
n
, where
n
is sufficiently big, and take cohomology. Then we have
a long exact sequence
0 → H
0
(m
x
F(n)) → H
0
(F(n)) → H
0
(F
x
(n)) → H
1
(m
x
F(n)) = 0.
In particular, the map
H
0
(
F
(
n
))
→ H
0
(
F
x
(
n
)) is surjective. This mean at
x
,
F
(
n
) is globally generated. Then by compactness, there is a single
n
large
enough such that F(n) is globally generated everywhere.
1.2 Weil divisors
If
X
is a sufficiently nice Noetherian scheme,
L
is a line bundle over
X
, and
s ∈ H
0
(
X, L
), then the vanishing locus of
s
is a codimension-1 subscheme. More
generally, if
s
is a rational section, then the zeroes and poles form codimension 1
subschemes. Thus, to understand line bundles, a good place to start would be
to understand these subschemes. We will see that for suitably nice schemes, we
can recover L completely from this information.
For the theory to work out nicely, we need to assume the following technical
condition:
Definition
(Regular in codimension 1)
.
Let
X
be a Noetherian scheme. We say
X is regular in codimension 1 if every local ring O
x
of dimension 1 is regular.
The key property of such schemes we will make use of is that if
Y ⊆ X
is
a codimension 1 integral subscheme, then the local ring
O
X,Y
is a DVR. Then
there is a valuation
val
Y
: O
X,Y
→ Z,
which tells us the order of vanishing/poles of our function along Y .
Definition
(Weil divisor)
.
Let
X
be a Noetherian scheme, regular in codimension
1. A prime divisor is a codimension 1 integral subscheme of
X
. A Weil divisor
is a formal sum
D =
X
a
i
Y
i
,
where the
a
i
∈ Z
and
Y
i
are prime divisors. We write
WDiv
(
X
) for the group of
Weil divisors of X.
For K a field, a Weil K-divisor is the same where we allow a
i
∈ K.
Definition
(Effective divisor)
.
We say a Weil divisor is effective if
a
i
≥
0 for
all i. We write D ≥ 0.
Definition
(Principal divisor)
.
If
f ∈ K
(
X
), then we define the principal
divisor
div(f) =
X
Y
val
Y
(f) · Y.
One can show that
val
Y
(
f
) is non-zero for only finitely many
Y
’s, so this is
a genuine divisor. To see this, there is always a Zariski open
U
such that
f|
U
is invertible, So
val
Y
(
f
)
6
= 0 implies
Y ⊆ X \ U
. Since
X
is Noetherian,
X \ U
can only contain finitely many codimension 1 subscheme.
Definition (Support). The support of D =
P
a
i
Y
i
is
supp(D) =
[
Y
i
.
Observe that
div
(
fg
) =
div
(
f
) +
div
(
g
). So the principal divisors form a
subgroup of the Weil divisors.
Definition
(Class group)
.
The class group of
X
,
Cl
(
X
), is the group of Weil
divisors quotiented out by the principal divisors.
We say Weil divisors
D, D
0
are linearly equivalent if
D − D
0
is principal, and
we write D ∼ D
0
.
Example. Take X = A
1
K
, and f =
x
3
x+1
∈ K(X). Then
div(f) = 3[0] − [−1].
A useful result for the future is the following:
Theorem
(Hartog’s lemma)
.
Let
X
be normal, and
f ∈ O
(
X \ V
) for some
V ≥ 2. Then f ∈ O
X
. Thus, div(f ) = 0 implies f ∈ O
×
X
.
1.3 Cartier divisors
Weil divisors do not behave too well on weirder schemes. A better alternative is
Cartier divisors. Let
L
be a line bundle, and
s
a rational section of
L
, i.e.
s
is a
section of L|
U
for some open U ⊆ X. Given such an s, we can define
div(s) =
X
Y
val
Y
(s) · Y.
To make sense of
val
Y
(
s
), for a fixed codimension 1 subscheme
Y
, pick
W
such
that
W ∩ U 6
=
∅
, and
L|
W
is trivial. Then we can make sense of
val
Y
(
s
) using
the trivialization. It is clear from Hartog’s lemma that
Proposition. If X is normal, then
div : {rational sections of L} → WDiv(X).
is well-defined, and two sections have the same image iff they differ by an element
of O
∗
X
.
Corollary. If X is normal and proper, then there is a map
div{rational sections of L}/K
∗
→ WDiv(X).
Proof. Properness implies O
∗
X
= K
∗
.
Example.
Take
X
=
P
1
K
, and
s
=
X
2
X+Y
∈ H
0
(
O
(1)), where
X, Y
are our
homogeneous coordinates. Then
div(s) = 2[0 : 1] − [1 : −1].
This lets us go from line bundles to divisors. To go the other direction, let
X
be a normal Noetherian scheme. Fix a Weil divisor
X
. We define the sheaf
O
X
(D) by setting, for all U ⊆ X open,
O
U
(D) = {f ∈ K(X) : div(f) + D|
U
≥ 0}.
Proposition. O
X
(D) is a rank 1 quasicoherent O
X
-module. .
In general,
O
X
(
D
) need not be a line bundle. If we want to prove that it is,
then we very quickly see that we need the following condition:
Definition
(Locally principal)
.
Let
D
be a Weil divisor on
X
. Fix
x ∈ X
. Then
D
is locally principal at
x
if there exists an open set
U ⊆ X
containing
x
such
that D|
U
= div(f)|
U
for some f ∈ K(X).
Proposition.
If
D
is locally principal at every point
x
, then
O
X
(
D
) is an
invertible sheaf.
Proof. If U ⊆ X is such that D|
U
= div(f)|
U
, then there is an isomorphism
O
X
|
U
→ O
X
(D)|
U
g 7→ g/f.
Definition
(Cartier divisor)
.
A Cartier divisor is a locally principal Weil divisor.
By checking locally, we see that
Proposition. If D
1
, D
2
are Cartier divisors, then
(i) O
X
(D
1
+ D
2
) = O
X
(D
1
) ⊗ O
X
(D
2
).
(ii) O
X
(−D)
∼
=
O
X
(D)
∨
.
(iii) If f ∈ K(X), then O
X
(div(f))
∼
=
O
X
.
Proposition.
Let
X
be a Noetherian, normal, integral scheme. Assume that
X
is factorial, i.e. every local ring
O
X,x
is a UFD. Then any Weil divisor is Cartier.
Note that smooth schemes are always factorial.
Proof.
It is enough to prove the proposition when
D
is prime and effective. So
D ⊆ X is a codimension 1 irreducible subvariety. For x ∈ D
– If x 6∈ D, then 1 is a divisor equivalent to D near x.
–
If
x ∈ D
, then
I
D,x
⊆ O
X,x
is a height 1 prime ideal. So
I
D,x
= (
f
) for
f ∈ m
X,x
. Then f is the local equation for D.
Recall the Class group
Cl
(
X
) was the group of Weil divisors modulo principal
equivalence.
Definition
(Picard group)
.
We define the Picard group of
X
to be the group
of Cartier divisors modulo principal equivalence.
Then if X is factorial, then Cl(X) = Pic(X).
Recall that if
L
is an invertible sheaf, and
s
is a rational section of
L
, then
we can define div(s).
Theorem.
Let
X
be normal and
L
an invertible sheaf,
s
a rational section of
L. Then O
X
(div(s)) is invertible, and there is an isomorphism
O
X
(div(s)) → L.
Moreover, sending L to divs gives a map
Pic(X) → Cl(X),
which is an isomorphism if X is factorial (and Noetherian and integral).
So we can think of
Pic
(
X
) as the group of invertible sheaves modulo isomor-
phism, namely H
1
(X, O
∗
X
).
Proof.
Given
f ∈ H
0
(
U, O
X
(
div
(
s
))), map it to
f · s ∈ H
0
(
U, L
). This gives the
desired isomorphism.
If we have to sections
s
0
6
=
s
, then
f
=
s
0
/s ∈ K
(
X
). So
div
(
s
) =
div
(
s
0
) +
div
(
f
), and
div
(
f
) is principal. So this gives a well-defined map
Pic
(
X
)
→
Cl(X).
1.4 Computations of class groups
To explicitly compute the class group, the following proposition is useful:
Proposition.
Let
X
be an integral scheme, regular in codimension 1. If
Z ⊆ X
is an integral closed subscheme of codimension 1, then we have an exact sequence
Z → Cl(X) → Cl(X \ Z) → 0,
where n ∈ Z is mapped to [nZ].
Proof.
The map
Cl
(
X
)
→ Cl
(
X \ Z
) is given by restriction. If
S
is a Weil divisor
on
X \ Z
, then
¯
S ⊆ X
maps to
S
under the restriction map. So this map is
surjective.
Also, that [
nZ
]
|
X\Z
is trivial. So the composition of the first two maps
vanishes. To check exactness, suppose
D
is a Weil divisor on
X
, principal on
X \ Z
. Then
D|
X\Z
=
dim
(
f
)
|
X\Z
for some
f ∈ K
(
X
). Then
D − div
(
f
) is just
supported along Z. So it must be of the form nZ.
If we remove something of codimension at least two, then something even
simpler happens.
Proposition.
If
Z ⊆ X
has codimension
≥
2, then
Cl
(
X
)
→ Cl
(
X \ Z
) is an
isomorphism.
The proof is the same as above, except no divisor can be supported on Z.
Example. Cl(A
2
\ {0}) = Cl(A
2
).
In general, to use the above propositions to compute the class group, a good
strategy is to remove closed subschemes as above until we reach something we
understand well. One thing we understand well is affine schemes:
Proposition.
If
A
is a Noetherian ring, regular in codimension 1, then
A
is a
UFD iff A is normal and Cl(Spec A) = 0
Proof.
If
A
is a UFD, then it is normal, and every prime ideal of height 1 is
principally generated. So if
D ∈ Spec A
is Weil and prime, then
D
=
V
(
f
) for
some f , and hence (f ) = I
D
.
Conversely, if
A
is normal and
Cl
(
Spec A
) = 0, then every Weil divisor is
principal. So if
I
is a height 1 prime ideal, then
V
(
I
) =
D
for some Weil divisor
D
. Then
D
is principal. So
I
= (
f
) for some
f
. So
A
is a Krull Noetherian
integral domain with principally generated height 1 prime ideals. So it is a
UFD.
Example. Cl(A
2
) = 0.
Example. Consider P
n
. We have an exact sequence
Z → Cl(P
n
) = Cl(P
n
\ P
n−1
) = Cl(A
n
) → 0.
So
Z → Cl
(
P
n
) is surjective. So
Cl
(
P
n
) is generated by a hyperplane. Moreover,
since any principal divisor has degree 0, it follows that
nH 6
= 0 for all
n >
0.
This H corresponds to O(1).
Example.
Let
X
=
V
(
xy − z
2
)
⊆ A
3
. Let
Z
=
V
(
x, z
). We claim that
Cl(X)
∼
=
Z/2Z, and is generated by [Z].
We compute
K[X \ Z] =
K[x, x
−1
, y, z]
(xy − z)
∼
=
K[x, x
−1
, t, z]
(t − z)
= K[x, x
−1
, z],
where t = xy, and this is a UFD. So Cl(X \ Z) = 0.
We now want to compute the kernel of the map Z → Cl(X). We have
O
X,Z
=
K[x, y, z]
xy − z
2
(x,z)
=
K[x, y, y
−1
, z]
x − z
2
(x,y)
= K[y, y
−1
, z]
(z)
.
Unsurprisingly, this is a DVR, and crucially, the uniformizer is
z
. So we know
that
div
(
x
) = 2
Z
. So we know that 2
Z ⊆ ker
(
Z → Cl
(
X
)). There is only one
thing to check, which is that (x, y) is not principal. To see this, consider
T
(0)
X = A
3
, T
(0)
(Z ⊆ X) = A
1
.
But if
Z
were principal, then
I
Z,0
= (
f
) for some
f ∈ O
X,0
. But then
T
0
(
Z
)
⊆
T
0
X
will be
ker
d
f
. But then
ker
d
f
should have dimension at least 2, which is
a contradiction.
Proposition. Let X be Noetherian and regular in codimension one. Then
Cl(X) = Cl(X × A
1
).
Proof. We have a projection map
pr
∗
1
: Cl(X) → Cl(X × A
1
)
[D
i
] 7→ [D
i
× A
1
]
It is an exercise to show that is injective. ‘
To show surjectivity, first note that we can the previous exact sequence and
the 4-lemma to assume X is affine.
We consider what happens when we localize a prime divisor
D
at the generic
point of
X
. Explicitly, suppose
I
D
is the ideal of
D
in
K
[
X × A
1
], and let
I
0
D
be the ideal of K(X)[t] generated b I
D
under the inclusion
K[X × A
1
] = K[X][t] ⊆ K(X)[t].
If
I
0
D
= 1, then
I
D
contains some function
f ∈ K
(
X
). Then
D ⊆ V
(
f
)
as a subvariety of
X × A
1
. So
D
is an irreducible component of
V
(
f
), and in
particular is of the form D
0
× A
1
.
If not, then
I
0
D
= (
f
) for some
f ∈ K
(
X
)[
t
], since
K
(
X
)[
t
] is a PID. Then
divf
is a principal divisor of
X × A
1
whose localization at the generic point is
D
.
Thus,
divf
is
D
plus some other divisors of the form
D
0
× A
1
. So
D
is linearly
equivalent to a sum of divisors of the form D
0
× A
1
.
Exercise. We have Cl(X × P
n
) = Z ⊕ Cl(X).
1.5 Linear systems
We end by collecting some useful results about divisors and linear systems.
Proposition.
Let
X
be a smooth projective variety over an algebraically closed
field. Let D
0
be a divisor on X.
(i)
For all
s ∈ H
0
(
X, O
X
(
D
)),
div
(
s
) is an effective divisor linearly equivalent
to D.
(ii)
If
D ∼ D
0
and
D ≥
0, then there is
s ∈ H
0
(
O
X
(
D
0
)) such that
div
(
s
) =
D
(iii)
If
s, s
0
∈ H
0
(
O
X
(
D
0
)) and
div
(
s
) =
div
(
s
0
), then
s
0
=
λs
for some
λ ∈ K
∗
.
Proof.
(i) Done last time.
(ii)
If
D ∼ D
0
, then
D − D
0
=
div
(
f
) for some
f ∈ K
(
X
). Then (
f
) +
D
0
≥
0.
So f induces a section s ∈ H
0
(O
X
(D
0
)). Then div(s) = D.
(iii) We have
s
0
s
∈ K(X)
∗
. So div
s
0
s
. So
s
0
s
∈ H
0
(O
∗
) = K
∗
.
Definition
(Complete linear system)
.
A complete linear system is the set of all
effective divisors linearly equivalent to a given divisor D
0
, written |D
0
|.
Thus, |D
0
| is the projectivization of the vector space H
0
(X, O(D
0
)).
We also define
Definition
(Linear system)
.
A linear system is a linear subspace of the projective
space structure on |D
0
|.
Given a linear system
|V |
of
L
on
X
, we have previously seen that we get a
rational map
X 99K P
(
V
). For this to be a morphism, we need
V
to generate
L
.
In general, the O
X
action on L induces a map
V ⊗ O
X
→ L.
Tensoring with L
−1
gives a map
V ⊗ L
−1
→ O
X
.
The image is an
O
X
-submodule, hence an ideal sheaf. This ideal
b
|V |
is called
the base ideal of
|V |
on
X
, and the corresponding closed subscheme is the base
locus. Then the map
X 99K P
(
V
) is a morphism iff
b
|V |
=
O
X
. In this case, we
say |V | is free.
Accordingly, we can define
Definition
((Very) ample divisor)
.
We say a Cartier divisor
D
is (very) ample
when O
X
(D) is.
By Serre’s theorem, if
A
is ample and
X
is projective, and
L
is any line
bundle, then for n sufficiently large, we have
h
0
(X, L ⊗ A
n
) =
X
(−1)
i
h
i
(X, L ⊗ A
n
) = χ(X, L ⊗ A
n
).
In the case of curves, Riemann–Roch lets us understand this number well:
Theorem (Riemann–Roch theorem). If C is a smooth projective curve, then
χ(L) = deg(L) + 1 − g(C).
It is often easier to reason about very ample divisors than general Cartier
divisors. First observe that by definition, every projective normal scheme has a
very ample divisor, say
H
. If
D
is any divisor, then by Serre’s theorem,
D
+
nH
is globally generated for large
n
. Moreover, one can check that the sum of a
globally generated divisor and a very ample divisor is still very ample, e.g. by
checking it separates points and tangent vectors. Hence
D
+
nH
is very ample
for large n. Thus, we deduce that
Proposition.
Let
D
be a Cartier divisor on a projective normal scheme .Then
D ∼ H
1
− H
2
for some very ample divisors
H
i
. We can in fact take
H
i
to be
effective, and if
X
is smooth, then we can take
H
i
to be smooth and intersecting
transversely.
The last part is a consequence of Bertini’s theorem.
Theorem
(Bertini)
.
Let
X
be a smooth projective variety over an algebraically
closed field
K
, and
D
a very ample divisor. Then there exists a Zariski open set
U ⊆ |D|
such that for all
H ∈ U
,
H
is smooth on
X
and if
H
1
6
=
H
2
, then
H
1
and H
2
intersect transversely.
2 Surfaces
2.1 The intersection product
We now focus on the case of smooth projective surfaces over an algebraically
closed field. In this case, there is no need to distinguish between Weil and Cartier
divisors.
Let
X
be such a surface. Let
C, D ⊆ X
be smooth curves intersecting
transversely. We want to count the number of points of intersection. This is
given by
|C ∩ D| = deg
C
(O
C
(D)) = h
0
(O
C∩D
) = χ(O
C∩D
),
using that
O
C∩D
is a skyscraper sheaf supported at
C ∩ D
. To understand this
number, we use the short exact sequences
0 O
C
(−D|
C
) O
C
O
C∩D
0
0 O
X
(−C − D) O
X
(−D) O
C
(−D|
C
) 0
0 O
X
(−C) O
X
O
C
0
This allows us to write
χ(O
C∩D
) = −χ(C, O
C
(−D|
C
)) + χ(C, O
C
)
= χ(O
X
(−C − D)) − χ(O
X
(−D)) + χ(C, O
C
)
= χ(O
X
(−C − D)) − χ(O
X
(−D)) − χ(O
X
(−C)) + χ(O
X
).
Thus suggests for for general divisors D
1
, D
2
, we can define
Definition
(Intersection product)
.
For divisors
D
1
, D
2
, we define the intersec-
tion product to be
D
1
· D
2
= χ(O
X
) + χ(−D
1
− D
2
) − χ(−D
1
) − χ(−D
2
).
Proposition.
(i) The product D
1
· D
2
depends only on the classes of D
1
, D
2
in Pic(X).
(ii) D
1
· D
2
= D
2
· D
1
.
(iii) D
1
· D
2
= |D
1
∩ D
2
| if D
1
and D
2
are curves intersecting transversely.
(iv) The intersection product is bilinear.
Proof. Only (iv) requires proof. First observe that if H is a very ample divisor
represented by a smooth curve, then we have
H · D = deg
H
(O
H
(D)),
and this is linear in D.
Next, check that
D
1
·
(
D
2
+
D
3
)
−D
1
·D
2
−D
1
·D
3
is symmetric in
D
1
, D
2
, D
3
.
So
(a)
Since this vanishes when
D
1
is very ample, it also vanishes if
D
2
or
D
3
is
very ample.
(b) Thus, if H is very ample, then D · (−H) = −(D · H).
(c) Thus, if H is very ample, then (−H) · D is linear in D.
(d)
If
D
is any divisor, write
D
=
H
1
− H
2
for
H
1
, H
2
very ample and smooth.
Then D · D
0
= H
1
· D
0
− H
2
· D
0
by (a), and thus is linear in D
0
.
In fact, bilinearity and (iii) shows that these properties uniquely characterize
the intersection product, since every divisor is a difference of smooth very ample
divisors.
Example.
Take
X
=
P
2
. We saw that
Pic
(
P
2
)
∼
=
Z
, generated by a hyperplane
. So all we need to understand is what
2
is. But two transverse lines intersect
at a point. So
2
= 1. Thus, the intersection product on
Pic
(
P
2
) is just ordinary
multiplication. In particular, if
D
1
, D
2
are curves that intersect transversely, we
find that
D
1
· D
2
= deg(D
1
) deg(D
2
).
This is B´ezout’s theorem!
Exercise. Let C
1
, C
2
⊆ X be curves without common components. Then
C
1
· C
2
=
X
p∈C
1
∩C
2
(O
C
1
∩C
2
),
where for p ∈ C
1
∩ C
2
, if C
1
= V (x) and C
2
= V (y), then
(O
C
1
∩C
2
) = dim
k
(O
X,p
/(x, y))
counts multiplicity.
Example. Take X = P
1
× P
1
. Then
Pic(X) = Z
2
= Z[p
∗
1
O(1)] ⊕ Z[p
∗
2
O(1)].
Let A = p
∗
1
O(1), B = p
∗
2
O(1). Then we see that
A
2
= B
2
= 0, A · B = 1
by high school geometry.
The Riemann–Roch theorem for curves lets us compute
χ
(
D
) for a divisor
D. We have an analogous theorem for surfaces:
Theorem (Riemann–Roch for surfaces). Let D ∈ Div(X). Then
χ(X, O
X
(D)) =
D · (D − K
X
)
2
+ χ(O
X
),
where K
X
is the canonical divisor.
To prove this, we need the adjunction formula:
Theorem
(Adjunction formula)
.
Let
X
be a smooth surface, and
C ⊆ X
a
smooth curve. Then
(O
X
(K
X
) ⊗ O
X
(C))|
C
∼
=
O
C
(K
C
).
Proof.
Let
I
C
=
O
X
(
−C
) be the ideal sheaf of
C
. We then have a short exact
sequence on C:
0 → O
X
(−C)|
C
∼
=
I
C
/I
2
C
→ Ω
1
X
|
C
→ Ω
1
C
→ 0,
where the left-hand map is given by d. To check this, note that locally on affine
charts, if
C
is cut out by the function
f
, then smoothness of
C
implies the kernel
of the second map is the span of df.
By definition of the canonical divisor, we have
O
X
(K
X
) = det(Ω
1
X
).
Restricting to C, we have
O
X
(K
X
)|
C
= det(Ω
1
X
|
C
) = det(O
X
(−C)|
C
) ⊗ det(Ω
1
C
) = O
X
(C)|
∨
C
⊗ O
C
(K
C
).
Proof of Riemann–Roch.
We can assume
D
=
H
1
− H
2
for very ample line
bundles
H
1
, H
2
, which are smoothly irreducible curves that intersect transversely.
We have short exact sequences
0 O
X
(H
1
− H
2
) O
X
(H
1
) O
H
2
(H
1
|
H
2
) 0
0 O
X
O
X
(H
1
) O
H
1
(H
1
) 0
where
O
H
1
(
H
1
) means the restriction of the line bundle
O
X
(
H
1
) to
H
1
. We can
then compute
χ(H
1
− H
2
) = χ(O
X
(H
1
)) − χ(H
2
, O
H
2
(H
1
|
H
2
))
= χ(O
X
) + χ(H
1
, O
H
1
(H
1
)) − χ(H
2
, O
H
2
(H
1
|
H
2
)).
The first term appears in our Riemann–Roch theorem, so we leave it alone. We
then use Riemann–Roch for curves to understand the remaining. It tells us
χ(H
i
, O
H
i
(H
1
)) = deg(O
H
i
(H
1
)) + 1 − g(H
i
) = (H
i
· H
1
) + 1 − g(H
i
).
By definition of genus, we have
2g(H
i
) − 2 = deg(K
H
i
).
and the adjunction formula lets us compute deg(K
H
i
) by
deg(K
H
i
) = H
i
· (K
X
+ H
i
).
Plugging these into the formula and rearranging gives the desired result.
Previously, we had the notion of linear equivalence of divisors. A coarser
notion of equivalence is that of numerical equivalence, which is about what the
intersection product can detect:
Definition
(Numerical equivalence)
.
We say divisors
D, D
0
are numerically
equivalent, written D ≡ D
0
, if
D · E = D
0
· E
for all divisors E.
We write
Num
0
= {D ∈ Div(X) : D ≡ 0}.
Definition (N´eron–Severi group). The N´eron–Severi group is
NS(X) = Div(X)/Num
0
(X).
We will need the following important fact:
Fact. NS(X) is a finitely-generated free module.
Note that
Pic
(
X
) is not be finitely-generated in general. For example, for
an elliptic curve,
Pic
(
X
) bijects with the closed points of
X
, which is in general
very large!
Definition (ρ(X)). ρ(X) = dim NS
R
(X) = rk NS(X).
Tensoring with R, the intersection product gives a map
( · , · ) : NS
R
(X) = NS(X) ⊗ R → R.
By definition of
NS
(
X
), this is non-degenerate. A natural question to ask is then,
what is the signature of this form? Since
X
is projective, there is a very ample
divisor
H
, and then
H
2
>
0. So this is definitely not negative definite. It turns
out there is only one positive component, and the signature is (1, ρ(X) − 1):
Theorem
(Hodge index theorem)
.
Let
X
be a projective surface, and
H
be a
(very) ample divisor on
X
. Let
D
be a divisor on
X
such that
D · H
= 0 but
D 6≡ 0. Then D
2
< 0.
Proof.
Before we begin the proof, observe that if
H
0
is very ample and
D
0
is (strictly) effective, then
H
0
· D
0
>
0, since this is given by the number of
intersections between
D
0
and any hyperplane in the projective embedding given
by H
0
.
Now assume for contradiction that D
2
≥ 0.
–
If
D
2
>
0, fix an
n
such that
H
n
=
D
+
nH
is very ample. Then
H
n
·D >
0
by assumption.
We use Riemann–Roch to learn that
χ(X, O
X
(mD)) =
m
2
D
2
− mK
X
· D
2
+ χ(O
X
).
We first consider the
H
2
(
O
X
(
mD
)) term in the left-hand side. By Serre
duality, we have
H
2
(O
X
(mD)) = H
0
(K
X
− mD).
Now observe that for large m, we have
H
n
· (K
X
− mD) < 0.
Since
H
n
is very ample, it cannot be the case that
K
X
− mD
is effective.
So H
0
(K
X
− D) = 0.
Thus, for m sufficiently large, we have
h
0
(mD) − h
1
(mD) > 0.
In particular,
mD
is (strictly) effective for
m
0. But then
H · mD >
0
since H is very ample and mD is effective. This is a contradiction.
–
If
D
2
= 0. Since
D
is not numerically trivial, there exists a divisor
E
on
X such that D · E 6= 0. We define
E
0
= (H
2
)E − (E · H)H.
It is then immediate that
E
0
·H
= 0. So
D
0
n
=
nD
+
E
0
satisfies
D
0
n
·H
= 0.
On the other hand,
(D
0
n
)
2
= (E
0
)
2
+ 2nD · E
0
> 0
for n large enough. This contradicts the previous part.
2.2 Blow ups
Let
X
be a smooth surface, and
p ∈ X
. There is then a standard way to “replace”
p
with a copy of
P
1
. The result is a surface
¯
X
=
Bl
p
X
together with a map
ε
:
¯
X → X
such that
ε
−1
(
p
) is a smoothly embedded copy of
P
1
, and
ε
is an
isomorphism outside of this P
1
.
To construct
¯
X
, pick local coordinates
x, y
in a neighbourhood
U
of
¯
X
such
that
V
(
x, y
) =
p
. Work on
U × P
1
with coordinates [
X
:
Y
] on
P
1
. We then set
¯
U = V (xY − yX).
There is a canonical projection of
¯
U
onto
U
, and this works, since above the
point (0
,
0), any choice of
X, Y
satisfies the equation, and everywhere else, we
must have [X : Y ] = [x : y].
Definition
(Blow up)
.
The blow up of
X
at
p
is the variety
¯
X
=
Bl
p
X
that
we just defined. The preimage of
p
is known as the exceptional curve, and is
denoted E.
We would like to understand
Pic
(
¯
X
) in terms of
Pic
(
X
), and in particular
the product structure on
Pic
(
¯
X
). This will allow us to detect whether a surface
is a blow up.
The first step to understanding Pic(
¯
X) is the localization sequence
Z → Pic(
¯
X) → Pic(
¯
X \ P
1
) → 0.
We also have the isomorphism
Pic(
¯
X \ P
1
) = Pic(X \ {p})
∼
=
Pic(X).
Finally, pulling back along
ε
gives a map
Pic
(
X
)
→ Pic
(
¯
X
), and the push-pull
formula says
π
∗
(π
∗
L) = L ⊗ π
∗
(O
¯
X
).
Hartog’s lemma says
π
∗
O
¯
X
= O
X
.
So in fact
ε
∗
is a splitting. So
Pic
(
¯
X
) is the direct sum of
Pic
(
X
) with a quotient
of
Z
, generated by [
E
]. We will show that in fact
Pic
(
¯
X
)
∼
=
ε
∗
Pic
(
X
)
⊕ Z
[
E
].
This will be shown by calculating E
2
= −1.
In general, if C is a curve in X, then this lifts to a curve
˜
C ⊆
¯
X, called the
strict transform of C. This is given by
˜
C = ε
−1
(C \ {p}).
There is also another operation, we can do, which is the pullback of divisors
π
∗
C.
Lemma.
π
∗
C =
˜
C + mE,
where m is the multiplicity of C at p.
Proof.
Choose local coordinates
x, y
at
p
and suppose the curve
y
= 0 is not
tangent to any branch of
C
at
p
. Then in the local ring
ˆ
O
X,p
, the equation of
C
is given as
f = f
m
(x, y) + higher order terms,
where
f
m
is a non-zero degree
m
polynomial. Then by definition, the multiplicity
is
m
. Then on
¯
U ⊆
(
U × P
1
), we have the chart
U × A
1
where
X 6
= 0, with
coordinates
(x, y, Y/X = t).
Taking (
x, t
) as local coordinates, the map to
U
is given by (
x, t
)
7→
(
x, xt
).
Then
ε
∗
f = f (x, tx) = x
m
f
m
(1, t) + higher order terms = x
m
· h(x, t),
with
h
(0
,
0)
6
= 0. But then on
¯
U
x6=0
, the curve
x
= 0 is just
E
. So this equation
has multiplicity m along E.
Proposition.
Let
X
be a smooth projective surface, and
x ∈ X
. Let
¯
X
=
Bl
x
X
π
→ X. Then
(i) π
∗
Pic(X) ⊕ Z[E] = Pic(
¯
X)
(ii) π
∗
D · π
∗
F = D · F , π
∗
D · E = 0, E
2
= −1.
(iii) K
¯
X
= π
∗
(K
X
) + E.
(iv) π
∗
is defined on NS(X). Thus,
NS(
¯
X) = NS(X) ⊕ Z[E].
Proof.
(i) Recall we had the localization sequence
Z → Pic(
¯
X) → Pic(X) → 1
and there is a right splitting. To show the result, we need to show the
left-hand map is injective, or equivalently,
mE 6∼
0. This will follow from
the fact that E
2
= −1.
(ii)
For the first part, it suffices to show that
π
∗
D · π
∗
F
=
D · F
for
D, F
very ample, and so we may assume
D, F
are curves not passing through
x
.
Then their pullbacks is just the pullback of the curve, and so the result is
clear. The second part is also clear.
Finally, pick a smooth curve
C
passing through
E
with multiplicity 1.
Then
π
∗
C =
˜
C + E.
Then we have
0 = π
∗
C · E =
˜
C · E + E
2
,
But
˜
C and E intersect at exactly one point. So
˜
C · E = 1.
(iii) By the adjunction formula, we have
(K
¯
X
+ E) · E = deg(K
P
1
) = −2.
So we know that
K
¯
X
· E
=
−
1. Also, outside of
E
,
K
¯
X
and
K
X
agree. So
we have
K
¯
X
= π
∗
(K
X
) + mE
for some m ∈ Z. Then computing
−1 = K
¯
X
· E = π
∗
(K
X
) · E + mE
2
gives m = 1.
(iv)
We need to show that if
D ∈ Num
0
(
X
), then
π
∗
(
D
)
∈ Num
0
(
¯
X
). But this
is clear for both things that are pulled back and things that are
E
. So we
are done.
One thing blow ups are good for is resolving singularities.
Theorem
(Elimination of indeterminacy)
.
Let
X
be a sooth projective surface,
Y
a projective variety, and
ϕ
:
X → Y
a rational map. Then there exists a
smooth projective surface X
0
and a commutative diagram
X
0
X Y
p
q
ϕ
where p : X
0
→ X is a composition of blow ups, and in particular is birational.
Proof.
We may assume
Y
=
P
n
, and
X 99K P
n
is non-degenerate. Then
ϕ
is
induced by a linear system
|V | ⊆ H
0
(
X, L
). We first show that we may assume
the base locus has codimension > 1.
If not, every element of
|V |
is of the form
C
+
D
0
for some fixed
C
. Let
|V
0
|
be the set of all such
D
0
. Then
|V |
and
|V
0
|
give the same rational map. Indeed,
if
V
has basis
f
0
, . . . , f
n
, and
g
is the function defining
C
, then the two maps
are, respectively,
[f
0
: · · · : f
n
] and [f
0
/g : · · · : f
n
/g],
which define the same rational maps. By repeating this process, replacing
V
with V
0
, we may assume the base locus has codimension > 1.
We now want to show that by blowing up, we may remove all points from
the base locus. We pick
x ∈ X
in the base locus of
|D|
. Blow it up to get
X
1
= Bl
x
X → X. Then
π
∗
|D| = |D
1
| + mE,
where
m >
0, and
|D
1
|
is a linear system which induces the map
ϕ ◦ π
1
. If
|D
1
|
has no basepoints, then we are done. If not, we iterate this procedure. The
procedure stops, because
0 ≤ D
2
1
= (π
∗
1
D
2
+ m
2
E
2
) < D
2
.
Exercise.
We have the following universal property of blow ups: If
Z → X
is
a birational map of surfaces, and suppose
f
−1
is not defined at a point
x ∈ X
.
Then f factors uniquely through the blow up Bl
x
(X).
Theorem.
Let
g
:
Z → X
be a birational morphism of surfaces. Then
g
factors
as
Z
g
0
→ X
0
p
→ X
, where
p
:
X
0
→ X
is a composition of blow ups, and
g
0
is an
isomorphism.
Proof.
Apply elimination of indeterminacy to the rational inverse to
g
to obtain
p
:
X
0
→ X
. There is then a lift of
g
0
to
X
0
by the universal property, and these
are inverses to each otehr.
Birational isomorphism is an equivalence relation. So we can partition the
set of smooth projective surfaces into birational isomorphism classes. If
X, X
0
are in the same birational isomorphism class, we can say
X ≤ X
0
if
X
0
is a blow
up of
X
. What we have seen is that any two elements have an upper bound.
There is no global maximum element, since we can keep blowing up. However,
we can look for minima elements.
Definition
(Relatively minimal)
.
We say
X
is relatively minimal if there does
not exist a smooth
X
0
and a birational morphism
X → X
0
that is not an
isomorphism. We say
X
is minimal if it is the unique relative minimal surface
in the birational equivalence class.
If
X
is not relatively minimal, then it is the blow up of another smooth
projective surface, and in this case the exceptional curve
C
satisfies
C
2
=
−
1. It
turns out this criterion itself is enough to determine minimality.
Theorem
(Castelnuovo’s contractibility criterion)
.
Let
X
be a smooth projective
surface over
K
=
¯
K
. If there is a curve
C ⊆ X
such that
C
∼
=
P
1
and
C
2
=
−
1,
then there exists
f
:
X → Y
that exhibits
X
as a blow up of
Y
at a point with
exceptional curve C.
Exercise.
Let
X
be a smooth projective surface. Then any two of the following
three conditions imply the third:
(i) C
∼
=
P
1
(ii) C
2
= −1
(iii) K
X
· C = −1
When these are satisfied, we say C is a (−1) curve.
For example, the last follows from the first two by the adjunction formula.
Corollary.
A smooth projective surface is relatively minimal if and only if it
does not contain a (−1) curve.
Proof.
The idea is to produce a suitable linear system
|L|
, giving a map
f
:
X → P
n
, and then take
Y
to be the image. Note that
f
(
C
) =
∗
is the same as
requiring
O
X
(L)|
C
= O
C
.
After finding a linear system that satisfies this, we will do some work to show
that f is an isomorphism outside of C and has smooth image.
Let
H
be a very ample divisor on
X
. By Serre’s theorem, we may assume
H
1
(X, O
X
(H)) = 0. Let
H · C = K > 0.
Consider divisors of the form H + iC. We can compute
deg
C
(H + iC)|
C
= (H + iC) · C = K − i.
Thus, we know deg
C
(H + KC)|
C
= 0, and hence
O
X
(H + KC)|
C
∼
=
O
C
.
We claim that
O
X
(
H
+
KC
) works. To check this, we pick a fairly explicit
basis of H
0
(H + KC). Consider the short exact sequence
0 → O
X
(H + (i − 1)C) → O
X
(H + iC) → O
C
(H + iC) → 0,
inducing a long exact sequence
0 → H
0
(H + (i − 1)C) → H
0
(H + iC) → H
0
(C, (H + iC)|
C
)
→ H
1
(H + (i − 1)C) → H
1
(H + iC) → H
1
(C, (H + iC)|
C
) → · · ·
We know
O
X
(
H
+
iC
)
|
C
=
O
P
1
(
K − i
). So in the range
i
= 0
, . . . , K
, we have
H
1
(C, (H + iC)|
C
) = 0. Thus, by induction, we find that
H
1
(H + iC) = 0 for i = 0, . . . , K.
As a consequence of this, we know that for
i
= 1
, . . . , K
, we have a short
exact sequence
H
0
(H + (i − 1)C) → H
0
(H + iC) H
0
(C, (H + iC)|
C
) = H
0
(O
P
1
(K − i)).
Thus,
H
0
(
H
+
iC
) is spanned by the image of
H
0
(
H
+ (
i −
1)
C
) plus a lift of a
basis of H
0
(O
P
1
(K − i)).
For each
i >
0, pick a lift
y
(i)
0
, . . . , y
(i)
K−i
of a basis of
H
0
(
O
P
1
(
K − i
)) to
H
0
(
H
+
iC
), and take the image in
H
0
(
H
+
KC
). Note that in local coordinates,
if C is cut out by a function g, then the image in H
0
(H + KC) is given by
g
K−i
y
(i)
0
, . . . , g
K−i
y
(i)
K−i
.
For
i
= 0, we pick a basis of
H
0
(
H
), and then map it down to
H
0
(
H
+
KC
).
Taking the union over all
i
of these elements, we obtain a basis of
H
0
(
H
+
KC
).
Let f be the induced map to P
r
.
For concreteness, list the basis vectors as
x
1
, . . . , x
r
, where
x
r
is a lift of
1 ∈ O
P
1
to H
0
(H + KC), and x
r−1
, x
r−2
are gy
(K−1)
0
, gy
(K−1)
1
.
First observe that x
1
, . . . , x
r−1
vanish at C. So
f(C) = [0 : · · · : 0 : 1] ≡ p,
and C is indeed contracted to a point.
Outside of
C
, the function
g
is invertible, so just the image of
H
0
(
H
) is
enough to separate points and tangent vectors. So
f
is an isomorphism outside
of C.
All the other basis elements of
H
+
KC
are needed to make sure the image is
smooth, and we only have to check this at the point
p
. This is done by showing
that
dim m
Y,p
/m
2
Y,p
≤
2, which requires some “infinitesimal analysis” we will not
perform.
Example.
Consider
X
=
P
2
, and
x, y, z
three non-colinear points in
P
2
given
by [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1].
There are three conics passing through
x, y, z
, given by
x
1
x
2
,
x
0
, x
2
and
x
0
x
1
, which is the union of lines
0
,
1
,
2
, and let
|C
x,y,z
|
be the linear system
generated by these three conics. Then we obtain a birational map
P
2
99K P
2
[x
0
: x
1
: x
2
] 7→ [x
1
x
2
: x
0
x
2
: x
0
x
1
] = [x
−1
0
: x
−1
1
: x
−1
2
]
with base locus {x, y, z}.
Blowing up at the three points
x, y, z
resolves our indeterminacies, and we
obtain a diagram
Bl
x,y,z
P
2
P
2
P
2
C
Then in the blowup, we have
π
∗
i
=
˜
i
+ E
j
+ E
k
for
i, j, k
distinct. We then check that
˜
2
i
=
−
1, and each
˜
i
is isomorphic
to
P
1
. Thus, Castelnuovo tells us we can blow these down. This blow down
map is exactly the right-hand map in the diagram above, which one can check
compresses each line
˜
i
to a point.
3 Projective varieties
We now turn to understand general projective schemes. Our goal will be to
understand the “geometry” of the set of ample divisors as a subspace of
NS
(
X
).
3.1 The intersection product
First, we have to define
NS
(
X
), and to do so, we need to introduce the intersection
product. Thus, given a projective variety
X
over an algebraically closed field
K
with
dim X
=
n
, we seek a multilinear symmetric function
CaDiv
(
X
)
n
→ Z
,
denoted
(D
1
, . . . , D
n
) 7→ (D
1
· . . . · D
n
).
This should satisfy the following properties:
(i)
The intersection depends only on the linear equivalence class, i.e. it factors
through Pic(X)
n
.
(ii)
If
D
1
, . . . , D
n
are smooth hypersurfaces that all intersect transversely, then
D
1
· · · · · D
n
= |D
1
∩ · · · ∩ D
n
|.
(iii)
If
V ⊆ X
is an integral subvariety of codimension 1, and
D
1
, . . . , D
n−1
are
Cartier divisors, then
D
1
· . . . · D
n−1
· [V ] = (D
1
|
V
· . . . · D
n−1
|
V
).
In general, if V ⊆ X is an integral subvariety, we write
D
1
· . . . · D
dim V
· [V ] = (D
1
|
V
· . . . · D
dim V
|
V
).
There are two ways we can define the intersection product:
(i)
For each
D
i
, write it as
D
i
∼ H
i,1
− H
i,2
with
H
i,j
very ample. Then
multilinearity tells us how we should compute the intersection product,
namely as
X
I⊆{1,...,n}
(−1)
|I|
Y
i∈I
H
i,2
·
Y
j6∈I
H
j,1
.
By Bertini, we can assume the
H
i,j
intersect transversely, and then we just
compute these by counting the number of points of intersection. We then
have to show that this is well-defined.
(ii)
We can write down the definition of the intersection product in a more
cohomological way. Recall that for n = 2, we had
D · C = χ(O
X
) + χ(O
X
(−C − D)) − χ(O
X
(−C)) − χ(O
X
(−D)).
For general
n
, let
H
1
, . . . , H
n
be very ample divisors. Then the last property
implies we have
H
1
· · · · · H
n
= (H
2
|
H
1
· . . . · H
n
|
H
1
).
So inductively, we have
H
1
· · · · · H
n
=
X
I⊆{2,...,n}
(−1)
|I|
χ
H
1
, O
H
1
−
X
i∈I
H
i
!!
But we also have the sequence
0 → O
X
(−H
1
) → O
X
→ O
H
1
→ 0.
Twisting this sequence shows that
H
1
· · · · · H
n
=
X
I⊆{2,...,n}
(−1)
|I|
χ
O
X
−
X
i∈I
H
i
− χ
O
X
−H
1
−
X
i∈I
H
i
=
X
J⊆{1,...,n}
(−1)
|J|
χ
O
X
−
X
i∈J
H
i
We can then adopt this as the definition of the intersection product
D
1
· · · · · D
n
=
X
J⊆{1,...,n}
(−1)
|J|
χ
O
X
−
X
i∈J
D
i
!!
.
Reversing the above procedure shows that property (iii) is satisfied.
As before we can define
Definition
(Numerical equivalence)
.
Let
D, D
0
∈ CaDiv
(
X
). We say
D
and
D
0
are numerically equivalent, written
D ≡ D
0
, if
D ·
[
C
] =
D
0
·
[
C
] for all integral
curves C ⊆ X.
In the case of surfaces, we showed that being numerically equivalent to zero
is the same as being in the kernel of the intersection product.
Lemma. Assume D ≡ 0. Then for all D
2
, . . . , D
dim X
, we have
D · D
2
· . . . · D
dim X
= 0.
Proof.
We induct on
n
. As usual, we can assume that the
D
i
are very ample.
Then
D · D
2
· · · · D
dim X
= D|
D
2
· D
3
|
D
2
· · · · · D
dim X
|
D
2
.
Now if D ≡ 0 on X, then D|
D
2
≡ 0 on D
2
. So we are done.
Definition
(Neron–Severi group)
.
The Neron–Severi group of
X
, written
NS(X) = N
1
(X), is
N
1
(X) =
CaDiv(X)
Num
0
(X)
=
CaDiv(X)
{D | D ≡ 0}
.
As in the case of surfaces, we have
Theorem
(Severi)
.
Let
X
be a projective variety over an algebraically closed
field. Then
N
1
(
X
) is a finitely-generated torsion free abelian group, hence of
the form Z
n
.
Definition (Picard number). The Picard number of X is the rank of N
1
(X).
We will not prove Severi’s theorem, but when working over
C
, we can indicate
a proof strategy. Over
C
, in the analytic topology, we have the exponential
sequence
0 → Z → O
an
X
→ O
∗,an
X
→ 0.
This gives a long exact sequence
H
1
(X, Z) → H
1
(X, O
an
X
) → H
1
(X, O
∗,an
X
) → H
2
(X, Z).
It is not hard to see that
H
i
(
X, Z
) is in fact the same as singular cohomology,
and
H
1
(
O
∗,an
X
) is the complex analytic Picard group. By Serre’s GAGA theorem,
this is the same as the algebraic
Pic
(
X
). Moreover, the map
Pic
(
X
)
→ H
2
(
X, Z
)
is just the first Chern class.
Next check that the intersection product [
C
]
·D
is the same as the topological
intersection product [
C
]
·c
1
(
D
). So at least modulo torsion, we have an embedding
of the Neron–Severi group into H
2
(X, Z), which we know is finite dimensional.
In the case of surfaces, we had the Riemann–Roch theorem. We do not have
a similarly precise statement for arbitrary varieties, but an “asymptotic” version
is good enough for what we want.
Theorem
(Asymptotic Riemann–Roch)
.
Let
X
be a projective normal variety
over
K
=
¯
K
. Let
D
be a Cartier divisor, and
E
a Weil divisor on
X
. Then
χ
(
X, O
X
(
mD
+
E
)) is a numerical polynomial in
m
(i.e. a polynomial with
rational coefficients that only takes integral values) of degree at most
n
=
dim X
,
and
χ(X, O
X
(mD + E)) =
D
n
n!
m
n
+ lower order terms.
Proof.
By induction on
dim X
, we can assume the theorem holds for normal
projective varieties of dimension
< n
. We fix
H
on
X
very ample such that
H
+
D
is very ample. Let
H
0
∈ |H|
and
G ∈ |H
+
D|
be sufficiently general. We
then have short exact sequences
0 → O
X
(mD + E) → O
X
(mD + E + H) → O
H
0
((mD + E + H)|
H
0
) → 0
0 → O
X
((m − 1)D + E) → O
X
(mD + E + H) → O
G
((mD + E + H)|
G
) → 0.
Note that the middle term appears in both equations. So we find that
χ(X, O
X
(mD + E)) + χ(H
0
, O
H
0
((mD + E + H)|
H
))
= χ(X, O
X
((m − 1)D + E)) + χ(H
0
, O
G
((mD + E + H)|
G
))
Rearranging, we get
χ(O
X
(mD − E)) − χ(O
X
((m − 1)D − E))
= χ(G, O
G
(mD + E + H)) − χ(H
0
, O
H
0
(mD + E + H)).
By induction (see proposition below) the right-hand side is a numerical polyno-
mial of degree at most n − 1 with leading term
D
n−1
· G − D
n−1
· H
(n − 1)!
m
n−1
+ lower order terms,
since
D
n−1
· G
is just the (
n −
1) self-intersection of
D
along
G
. But
G − H
=
D
.
So the LHS is
D
n
(n − 1)!
m
n−1
+ lower order terms,
so we are done.
Proposition.
Let
X
be a normal projective variety, and
|H|
a very ample
linear system. Then for a general element
G ∈ |H|
,
G
is a normal projective
variety.
Some immediate consequences include
Proposition. Let X be a normal projective variety.
(i) If H is a very ample Cartier divisor, then
h
0
(X, mH) =
H
n
n!
m
n
+ lower order terms for m 0.
(ii) If D is any Cartier divisor, then there is some C ∈ R
>0
such that
h
0
(mD) ≤ C · m
n
for m 0.
Proof.
(i)
By Serre’s theorem,
H
i
(
O
x
(
mH
)) = 0 for
i >
0 and
m
0. So we apply
asymptotic Riemann Roch.
(ii)
There exists a very ample Cartier divisor
H
0
on
X
such that
H
0
+
D
is
also very ample. Then
h
0
(mD) ≤ h
0
(m(H
0
+ D)).
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
n−1
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.
3.3 Nef divisors
We can weaken the definition of an ample divisor to get an ample divisor.
Definition
(Numerically effective divisor)
.
Let
X
be a proper scheme, and
D
a
Cartier divisor. Then
D
is numerically effective (nef) if
D · C ≥
0 for all integral
curves C ⊆ X.
Similar to the case of ample divisors, we have
Proposition.
(i) D is nef iff D|
X
red
is nef.
(ii) D is nef iff D|
X
i
is nef for all irreducible components X
i
.
(iii) If V ⊆ X is a proper subscheme, and D is nef, then D|
V
is nef.
(iv)
If
f
:
X → Y
is a finite morphism of proper schemes, and
D
is nef on
Y
,
then f
∗
D is nef on X. The converse is true if f is surjective.
The last one follows from the analogue of Nakai’s criterion, called Kleinmann’s
criterion.
Theorem
(Kleinmann’s criterion)
.
Let
X
be a proper scheme, and
D
an
R
-
Cartier divisor. Then
D
is nef iff
D
dim V
[
V
]
≥
0 for all proper irreducible
subvarieties.
Corollary.
Let
X
be a projective scheme, and
D
be a nef
R
-divisor on
X
, and
H be a Cartier divisor on X.
(i) If H is ample, then D + εH is also ample for all ε > 0.
(ii) If D + εH is ample for all 0 < ε 1, then D is nef.
Proof.
(i)
We may assume
H
is very ample. By Nakai’s criterion this is equivalent
to requiring
(D + εH)
dim V
· [V ] =
X
dim V
p
ε
p
D
dim V −p
H
p
[V ] > 0.
Since any restriction of a nef divisor to any integral subscheme is also nef,
and multiplying with
H
is the same as restricting to a hyperplane cuts, we
know the terms that involve
D
are non-negative. The
H
p
term is positive.
So we are done.
(ii)
We know (
D
+
εH
)
· C >
0 for all positive
ε
sufficiently small. Taking
ε → 0, we know D · C ≥ 0.
Corollary. Nef(X) = Amp(X) and int(Nef(X)) = Amp(X).
Proof.
We know
Amp
(
X
)
⊆ Nef
(
X
) and
Amp
(
X
) is open. So this implies
Amp(X) ⊆ int(Nef(X)), and thus Amp(X) ⊆ Nef(X).
Conversely, if
D ∈ int
(
Nef
(
X
)), we fix
H
ample. Then
D − tH ∈ Nef
(
X
)
for small
t
, by definition of interior. Then
D
= (
D − tH
) +
tH
is ample. So
Amp(X) ⊇ int(Nef(X)).
Proof of Kleinmann’s criterion.
We may assume that
X
is an integral projective
scheme. The
⇐
direction is immediate. To prove the other direction, since the
criterion is a closed condition, we may assume
D ∈ Div
Q
(
X
). Moreover, by
induction, we may assume that
D
dim V
[
V
]
≥
0 for all
V
strictly contained in
X
,
and we have to show that D
dim X
≥ 0. Suppose not, and D
dim X
< 0.
Fix a very ample Cartier divisor H, and consider the polynomial
P (t) = (D + tH)
dim X
= D
dim X
+
dim X−1
X
i=1
t
i
dim X
i
H
i
D
dim X−i
+ t
dim X
H
dim X
.
The first term is negative; the last term is positive; and the other terms are
non-negative by induction since H is very ample.
Then on R
>0
, this polynomial is increasing. So there exists a unique t such
that P (t) = 0. Let
¯
t be the root. Then P (
¯
t) = 0. We can also write
P (t) = (D + tH) · (D + tH)
dim X−1
= R(t) + tQ(t),
where
R(t) = D · (D + tH)
dim X−1
, Q(t) = H · (D + tH)
dim X−1
.
We shall prove that R(
¯
t) ≥ 0 and Q(
¯
t) > 0, which is a contradiction.
We first look at Q(t), which is
Q(t) =
dim X−1
X
i=0
t
i
dim X − 1
i
H
i+1
D
dim X−i
,
which, as we argued, is a sum of non-negative terms and a positive term.
To understand R(t), we look at
R(t) = D · (D + tH)
dim X−1
.
Note that so far, we haven’t used the assumption that
D
is nef. If
t >
¯
t
, then
(
D
+
tH
)
dim X
>
0, and (
D
+
tH
)
dim V
[
V
]
>
0 for a proper integral subvariety,
by induction (and binomial expansion). Then by Nakai’s criterion,
D
+
tH
is
ample. So the intersection (
D
+
tH
)
dim X−1
is essentially a curve. So we are
done by definition of nef. Then take the limit t →
¯
t.
It is convenient to make the definition
Definition
(1-cycles)
.
For
K
=
Z, Q
or
R
, we define the space of 1-cycles to be
Z
1
(X)
K
=
n
X
a
i
C
i
: a
i
∈ K, C
i
⊆ X integral proper curves
o
.
We then have a pairing
N
1
K
(X) × Z
1
(X)
K
→ K
(D, E) 7→ D · E.
We know that
N
1
K
(
X
) is finite-dimensional, but
Z
1
(
X
)
K
is certainly uncountable.
So there is no hope that this intersection pairing is perfect
Definition
(Numerical equivalence)
.
Let
X
be a proper scheme, and
C
1
, C
2
∈
Z
1
(X)
K
. Then C
1
≡ C
2
iff
D · C
1
= D · C
2
for all D ∈ N
1
K
(X).
Definition
(
N
1
(
X
)
K
)
.
We define
N
1
(
X
)
K
to be the
K
-module of
K
1-cycles
modulo numerical equivalence.
Thus we get a pairing N
1
(X)
K
× N
1
(X)
K
→ K.
Definition (Effective curves). We define the cone of effective curves to be
NE(X) = {γ ∈ N
1
(X)
R
: γ ≡
X
[a
i
C
i
] : a
i
> 0, C
i
⊆ X integral curves}.
This cone detects nef divisors, since by definition, an
R
-divisor
D
is nef iff
D · γ ≥ 0 for all γ ∈ NE(X). In general, we can define
Definition
(Dual cone)
.
Let
V
be a finite-dimensional vector space over
R
and
V
∗
the dual of
V
. If
C ⊆ V
is a cone, then we define the dual cone
C
∨
⊆ V
∗
by
C
∨
= {f ∈ V
∗
: f(c) ≥ 0 for all c ∈ C}.
Then we see that
Proposition. Nef(X) = NE(X)
∨
.
It is also easy to see that C
∨∨
=
¯
C. So
Proposition. NE(X) = Nef(X)
∨
.
Theorem
(Kleinmann’s criterion)
.
If
X
is a projective scheme and
D ⊆
CaDiv
R
(X). Then the following are equivalent:
(i) D is ample
(ii) D|
NE(X)
> 0, i.e. D · γ > 0 for all γ ∈ NE(X).
(iii) S
1
∩ NE(X) ⊆ S
1
∩ D
>0
, where
S
1
⊆ N
1
(
X
)
R
is the unit sphere under
some choice of norm.
Proof.
– (1) ⇒ (2): Trivial.
– (2) ⇒ (1): If D|
NE(X)
> 0, then D ∈ int(Nef(X)).
– (2) ⇔ (3): Similar.
Proposition.
Let
X
be a projective scheme, and
D, H ∈ N
1
R
(
X
). Assume that
H is ample. Then D is ample iff there exists ε > 0 such that
D · C
H · C
≥ ε.
Proof. The statement in the lemma is the same as (D − εH) · C ≥ 0.
Example. Let X be a smooth projective surface. Then
N
1
(X)
R
= N
1
(X)
R
,
since
dim X
= 2 and
X
is smooth (so that we can identify Weil and Cartier
divisors). We certainly have
Nef(X) ⊆ NE(X).
This can be proper. For example, if
C ⊆ X
is such that
C
is irreducible with
C
2
< 0, then C is not nef.
What happens when we have such a curve? Suppose
γ
=
P
a
i
C
i
∈ NE
(
X
)
and
γ · C <
0. Then
C
must be one of the
C
i
, since the intersection with any
other curve is non-negative. So we can write
γ = a
C
C +
k
X
i=2
a
i
C
i
,
where a
C
, a
i
> 0 and C
i
6= C for all i ≥ 2.
So for any element in γ ∈ NE(X), we can write
γ = λC + γ
0
,
where λ > 0 and γ
0
= NE(X)
C≥0
, i.e. γ
0
· C ≥ 0. So we can write
NE(C) = R
+
[C] + NE(X)
C≥0
.
Pictorially, if we draw a cross section of
NE(C)
, then we get something like this:
γ · C = 0
C
To the left of the dahsed line is
NE(X)
C≥0
, which we can say nothing about,
and to the right of it is a “cone” to
C
, generated by interpolating
C
with the
elements of NE(X)
C≥0
.
Thus, [C] ∈ N
1
(X)
R
is an extremal ray of NE(X). In other words, if
λ[C] = µ
1
γ
1
+ µ
2
γ
2
where µ
i
> 0 and γ
i
∈ NE(X), then γ
i
is a multiple of [C].
In fact, in general, we have
Theorem
(Cone theorem)
.
Let
X
be a smooth projective variety over
C
. Then
there exists rational curves {C
i
}
i∈I
such that
NE(X) = NE
K
X
≥0
+
X
i∈I
R
+
[C
i
]
where
NE
K
X
≥0
=
{γ ∈ NE(X)
:
K
X
· γ ≥
0
}
. Further, we need at most
countably many C
i
’s, and the accumulation points are all at K
⊥
X
.
3.4 Kodaira dimension
Let
X
be a normal projective variety, and
L
a line bundle with
|L| 6
= 0. We
then get a rational map
ϕ
|L|
: X 99K P(H
0
(L)).
More generally, for each
m >
0, we can form
L
⊗m
, and get a rational map
ϕ
L
⊗m
. What can we say about the limiting behaviour as m → ∞?
Theorem
(Iitaka)
.
Let
X
be a normal projective variety and
L
a line bundle
on
X
. Suppose there is an
m
such that
|L
⊗m
| 6
= 0. Then there exists
X
∞
, Y
∞
,
a map
ψ
∞
:
X
∞
→ Y
∞
and a birational map
U
∞
:
X
∞
99K X
such that for
K 0 such that |L
⊗K
| 6= 0, we have a commutative diagram
X Im(ϕ
|L
⊗K
)
X
∞
Y
∞
ϕ
|L
⊗k
|
U
∞
ψ
∞
where the right-hand map is also birational.
So birationally, the maps ψ
K
stabilize.
Definition (Kodaira dimension). The Kodaira dimension of L is
K(X, L) =
(
−∞ h
0
(X, L
⊗m
) = 0 for all m > 0
dim(Y
∞
) otherwise
.
This is a very fundamental invariant for understanding
L
. For example, if
K(X, L) = K ≥ 0, then there exists C
1
, C
2
∈ R
>0
such that
C
2
m
K
≤ h
0
(X, L
⊗m
) ≤ C
1
m
K
for all m. In other words, h
0
(X, L
⊗m
) ∼ m
K
.