part iii - positivity in algebraic geometry · part iii | positivity in algebraic geometry based on...

Part III — Positivity in Algebraic Geometry

Based on lectures by S. SvaldiNotes taken by Dexter Chua

Lent 2018

These notes are not endorsed by the lecturers, and I have modified them (oftensignificantly) 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 andtheir positivity properties on projective algebraic varieties.

The first part of the class will be dedicated to introducing the basic notions and resultsregarding these objects and special attention will be devoted to discussing examples inthe case of curves and surfaces.

In the second part, the course will cover classical results from the theory of divisorsand linear systems and their applications to the study of the geometry of algebraicvarieties.

If time allows and based on the interests of the participants, there are a number ofmore advanced topics that could possibly be covered: Reider’s Theorem for surfaces,geometry of linear systems on higher dimensional varieties, multiplier ideal sheaves andinvariance of plurigenera, higher dimensional birational geometry.

Pre-requisites

The minimum requirement for those students wishing to enroll in this class is theirknowledge of basic concepts from the Algebraic Geometry Part 3 course, i.e. roughlyChapters 2 and 3 of Hartshorne’s Algebraic Geometry.

Familiarity with the basic concepts of the geometry of algebraic varieties of dimension 1and 2 — e.g. as covered in the preliminary sections of Chapters 4 and 5 of Hartshorne’sAlgebraic Geometry — would be useful but will not be assumed — besides what wasalready 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.

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Contents III Positivity in Algebraic Geometry

Contents

1 Divisors 31.1 Projective embeddings . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Weil divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Cartier divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Computations of class groups . . . . . . . . . . . . . . . . . . . . 111.5 Linear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Surfaces 152.1 The intersection product . . . . . . . . . . . . . . . . . . . . . . . 152.2 Blow ups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3 Projective varieties 253.1 The intersection product . . . . . . . . . . . . . . . . . . . . . . . 253.2 Ample divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3 Nef divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.4 Kodaira dimension . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Index 38

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1 Divisors III Positivity in Algebraic Geometry

1 Divisors

1.1 Projective embeddings

Our results here will be stated for rather general schemes, since we can, but atthe end, we are only interested in concrete algebraic varieties.

We are interested in the following problem — given a scheme X, can weembed it in projective space? The first observation that Pn comes with a veryspecial line bundle O(1), and every embedding X → Pn gives pulls back to acorresponding sheaf L on X.

The key property of O(1) is that it has global sections x0, . . . , xn whichgenerate O(1) as an OX -module.

Definition (Generating section). Let X be a scheme, and F a sheaf of OX -modules. Let s0, . . . , sn ∈ H0(X,F) be sections. We say the sections generateF if the natural map

n+1⊕i=0

OX → F

induced by the si is a surjective map of OX -modules.

The generating sections are preserved under pullbacks. So if X is embeddedinto Pn, then it should have a corresponding line bundle generated by n + 1global sections. More generally, if there is any map to Pn at all, we can pullback 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 → Pn is a morphism over A, then ϕ∗OPn(1) is an invertible sheafon X, generated by the sections ϕ∗x0, . . . , ϕ

∗xn ∈ H0(X,ϕ∗OPn(1)).

(ii) If L is an invertible sheaf on X, and if s0, . . . , sn ∈ H0(X,L) which generateL, then there exists a unique morphism ϕ : X → Pn such that ϕ∗O(1) ∼= Land ϕ∗xi = si.

This theorem highlights the importance of studying line bundles and theirsections, 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 ofx0, . . . , xn generate ϕ∗OPn(1).

(ii) In short, we map x ∈ X to [s0(x) : · · · : sn(x)] ∈ Pn.

In more detail, define

Xsi = p ∈ X : si 6∈ mpLp.

This is a Zariski open set, and si is invertible on Xsi . Thus there is a dualsection s∨i ∈ L∨ such that si ⊗ s∨i ∈ L ⊗ L∨ ∼= OX is equal to 1. Definethe map Xsi → An by the map

K[An]→ H0(Xsi ,Osi)yi 7→ sj ⊗ s∨i .

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Since the si generate, they cannot simultaneously vanish on a point. SoX =

⋃Xsi . Identifying An as the chart of Pn where xi 6= 0, this defines

the desired map X → Pn.

The theorem tells us we get a map from X to Pn. However, it says nothingabout how nice these maps are. In particular, it says nothing about whether ornot we get an embedding.

Definition (Very ample sheaf). Let X be an algebraic variety over K, and Lbe an invertible sheaf. We say that L is very ample if there is a closed immersionϕ : X → Pn such that ϕ∗OPn(1) ∼= L.

It would be convenient if we had a good way of identifying when L is veryample. In this section, we will prove some formal criteria for very ampleness,which are convenient for proving things but not very convenient for actuallychecking if a line bundle is very ample. In specific cases, such as for curves, wecan obtain some rather more concrete and usable criterion.

So how can we understand when a map X → Pn is an embedding? If it werean embedding, then it in particular is injective. So given any two points in X,there is a hyperplane in Pn that passes through one but not the other. But beingan embedding means something more. We want the differential of the map tobe injective as well. This boils down to the following conditions:

Proposition. Let K = K, and X a projective variety over K. Let L be aninvertible sheaf on X, and s0, . . . , sn ∈ H0(X,L) generating sections. WriteV = 〈s0, . . . , sn〉 for the linear span. Then the associated map ϕ : X → Pn is aclosed embedding iff

(i) For every distinct closed points p 6= q ∈ X, there exists sp,q ∈ V such thatsp,q ∈ mpLp but sp,q 6∈ mqLq.

(ii) For every closed point p ∈ X, the set s ∈ V | s ∈ mpLp spans the vectorspace mpLp/m2

pLp.

Definition (Separate points and tangent vectors). With the hypothesis of theproposition, 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 supposep 6= q are (closed) points. Then there is some hyperplane Hp,q in Pnpassing through p but not q. The hyperplane Hp,q is the vanishing locusof a global section of O(1). Let sp,q ∈ V ⊆ H0(X,L) be the pullback ofthis section. Then sp,q ∈ mpLp and sp,q 6∈ mqLq. So (i) is satisfied.

To see (ii), we restrict to the affine patch containing p, and X is a closedsubvariety of An. The result is then clear since mpLp/m2

pLp is exactly the

span of s0, . . . , sn. We used K = K to know what the closed points of Pnlook like.

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(⇐) We first show that ϕ is injective on closed points. For any p 6= q ∈ X,write the given sp,q as

sp,q =∑

λisi =∑

λiϕ∗xi = ϕ∗

∑λixi

for some λi ∈ K. So we can take Hp,q to be given by the vanishing setof∑λixi, 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 particularϕ is a homeomorphism onto the image.

To show that ϕ is in fact a closed immersion, we need to show thatOPn → ϕ∗OX is surjective. As before, it is enough to prove that it holdsat the level of stalks over closed points. To show this, we observe thatLp is trivial, so mpLp/m2

pLp ∼= mp/m2p (unnaturally). We then apply the

following lemma:

Lemma. Let f : A→ B be a local morphism of local rings such that

A/mA → B/mB is an isomorphism;

mA → mB/m2B is surjective; and

B is a finitely-generated A-module.

Then f is surjective.

To check the first condition, note that we have

Op,Pn

mp,Pn

∼=Op,Xmp,X

∼= K.

Now since mp,Pn is generated by x0, . . . , xn, the second condition is thesame as saying

mp,Pn → mp,Xm2p,X

is surjective. The last part is immediate.

Unsurprisingly, this is not a very pleasant hypothesis to check, since it requiresus to really understand the structure of V . In general, given an invertible sheafL, it is unlikely that we can concretely understand the space of sections. Itwould 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 bea very ample invertible sheaf, and F a coherent OX -module. Then there existsa positive integer n0 = n0(F ,L) such that for all n ≥ n0, the twist F ⊗ Ln isgenerated by global sections.

The proof of this is straightforward and will be omitted. The idea is thattensoring with Ln lets us clear denominators, and once we have cleared all thedenominators of the (finitely many) generators of F , the resulting sheaf F ⊗ Lnwill be generated by global sections.

We can weaken the condition of very ampleness to only require the conditionof this theorem to hold.

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Definition (Ample sheaf). Let X be a Noetherian scheme over A, and L aninvertible sheaf over X. We say L is ample iff for any coherent OX -module F ,there is an n0 such that for all n ≥ n0, the sheaf F ⊗ Ln is generated by globalsections.

While this seems like a rather weak condition, it is actually not too bad.First of all, by taking F to be OX , we can find some Ln that is generated byglobal sections. So at least it gives some map to Pn. In fact, another theorem ofSerre 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 suchthat Lm is very ample.

Proof.

(⇐) Let Lm be very ample, and F a coherent sheaf. By Serre’s theorem, thereexists n0 such that for all j ≥ j0, the sheafs

F ⊗ Lmj , (F ⊗ L)⊗ Lmj , . . . , (F ⊗ Lm−1)⊗ Lmj

are all globally generated. So F ⊗ Ln is globally generated for n ≥ mj0.

(⇒) Suppose L is ample. Then Lm is globally generated for m sufficiently large.We claim that there exists t1, . . . , tn ∈ H0(X,LN ) such that L|Xti

are alltrivial (i.e. isomorphic to OXti

), and X =⋃Xti .

By compactness, it suffices to show that for each p ∈ X, there is somet ∈ H0(X,Ln) (for some n) such that p ∈ Xti and L is trivial on Xti . Firstof all, since L is locally free by definition, we can find an open affine Ucontaining p such that L|U is trivial.

Thus, it suffices to produce a section t that vanishes on Y = X − U butnot at p. Then p ∈ Xt ⊆ U and hence L is trivial on Xt. Vanishing onY is the same as belonging to the ideal sheaf IY . Since IY is coherent,ampleness implies there is some large n such that IY ⊗Ln is generated byglobal sections. In particular, since IY ⊗Ln doesn’t vanish at p, we canfind some t ∈ Γ(X, IY ⊗ LN ) such that t 6∈ mp(IY ⊗ Ln)p. Since IY is asubsheaf of OX , we can view t as a section of Ln, and this t works.

Now given the Xti , for each fixed i, we let bij generate OXtias an

A-algebra. Then for large n, cij = tni bij extends to a global sectioncij ∈ Γ(X,Ln) (by clearing denominators). We can pick an n large enoughto work for all bij . Then we use tni , cij as our generating sectionsto construct a morphism to PN , and let xi, xij be the correspondingcoordinates. Observe that

⋃Xti = X implies the tni already generate Ln.

Now each xti gets mapped to Ui ⊆ PN , the vanishing set of xi. The mapOUi → ϕ∗OXti

corresponds to the map

A[yi, yij ]→ OXti,

where yij is mapped to cij/tni = bij . So by assumption, this is surjective,

and so we have a closed embedding.

From this, we also see that

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Proposition. Let L be a sheaf over X (which is itself a projective variety overK). Then the following are equivalent:

(i) L is ample.

(ii) Lm is ample for all m > 0.

(iii) Lm 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, andL is very ample on X. Let F be a coherent sheaf. Then

(i) For all i ≥ 0 and n ∈ N, Hi(F ⊗ Ln) is a finitely-generated A-module.

(ii) There exists n0 ∈ N such that for all n ≥ n0, Hi(F ⊗ Ln) = 0 for alli > 0.

The proof is exactly the same as the case of O(1) on Pn.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 aninvertible sheaf. Then the following are equivalent:

(i) L is ample.

(ii) For all coherent F on X, there exists n0 ∈ N such that for all n ≥ n0, wehave Hi(F ⊗ Ln) = 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→ mxF → F → Fx → 0.

We twist by Ln, where n is sufficiently big, and take cohomology. Then we havea long exact sequence

0→ H0(mxF(n))→ H0(F(n))→ H0(Fx(n))→ H1(mxF(n)) = 0.

In particular, the map H0(F(n)) → H0(Fx(n)) is surjective. This mean atx, F(n) is globally generated. Then by compactness, there is a single n largeenough 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, ands ∈ H0(X,L), then the vanishing locus of s is a codimension-1 subscheme. Moregenerally, if s is a rational section, then the zeroes and poles form codimension 1subschemes. Thus, to understand line bundles, a good place to start would beto understand these subschemes. We will see that for suitably nice schemes, wecan recover L completely from this information.

For the theory to work out nicely, we need to assume the following technicalcondition:

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Definition (Regular in codimension 1). Let X be a Noetherian scheme. We sayX is regular in codimension 1 if every local ring Ox of dimension 1 is regular.

The key property of such schemes we will make use of is that if Y ⊆ X isa codimension 1 integral subscheme, then the local ring OX,Y is a DVR. Thenthere is a valuation

valY : OX,Y → Z,

which tells us the order of vanishing/poles of our function along Y .

Definition (Weil divisor). LetX be a Noetherian scheme, regular in codimension1. A prime divisor is a codimension 1 integral subscheme of X. A Weil divisoris a formal sum

D =∑

aiYi,

where the ai ∈ Z and Yi are prime divisors. We write WDiv(X) for the group ofWeil divisors of X.

For K a field, a Weil K-divisor is the same where we allow ai ∈ K.

Definition (Effective divisor). We say a Weil divisor is effective if ai ≥ 0 forall i. We write D ≥ 0.

Definition (Principal divisor). If f ∈ K(X), then we define the principal divisor

div(f) =∑Y

valY (f) · Y.

One can show that valY (f) is non-zero for only finitely many Y ’s, so this isa genuine divisor. To see this, there is always a Zariski open U such that f |Uis invertible, So valY (f) 6= 0 implies Y ⊆ X \ U . Since X is Noetherian, X \ Ucan only contain finitely many codimension 1 subscheme.

Definition (Support). The support of D =∑aiYi is

supp(D) =⋃Yi.

Observe that div(fg) = div(f) + div(g). So the principal divisors form asubgroup of the Weil divisors.

Definition (Class group). The class group of X, Cl(X), is the group of Weildivisors quotiented out by the principal divisors.

We say Weil divisors D,D′ are linearly equivalent if D−D′ is principal, andwe write D ∼ D′.

Example. Take X = A1K , and f = x3

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 someV ≥ 2. Then f ∈ OX . Thus, div(f) = 0 implies f ∈ O×X .

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1.3 Cartier divisors

Weil divisors do not behave too well on weirder schemes. A better alternative isCartier divisors. Let L be a line bundle, and s a rational section of L, i.e. s is asection of L|U for some open U ⊆ X. Given such an s, we can define

div(s) =∑Y

valY (s) · Y.

To make sense of valY (s), for a fixed codimension 1 subscheme Y , pick W suchthat W ∩ U 6= ∅, and L|W is trivial. Then we can make sense of valY (s) usingthe 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 elementof O∗X .

Corollary. If X is normal and proper, then there is a map

divrational sections of L/K∗ →WDiv(X).

Proof. Properness implies O∗X = K∗.

Example. Take X = P1K , and s = X2

X+Y ∈ H0(O(1)), where X,Y are ourhomogeneous coordinates. Then

div(s) = 2[0 : 1]− [1 : −1].

This lets us go from line bundles to divisors. To go the other direction, letX be a normal Noetherian scheme. Fix a Weil divisor X. We define the sheafOX(D) by setting, for all U ⊆ X open,

OU (D) = f ∈ K(X) : div(f) +D|U ≥ 0.

Proposition. OX(D) is a rank 1 quasicoherent OX -module. .

In general, OX(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. ThenD is locally principal at x if there exists an open set U ⊆ X containing x suchthat D|U = div(f)|U for some f ∈ K(X).

Proposition. If D is locally principal at every point x, then OX(D) is aninvertible sheaf.

Proof. If U ⊆ X is such that D|U = div(f)|U , then there is an isomorphism

OX |U → OX(D)|Ug 7→ g/f.

Definition (Cartier divisor). A Cartier divisor is a locally principal Weil divisor.

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By checking locally, we see that

Proposition. If D1, D2 are Cartier divisors, then

(i) OX(D1 +D2) = OX(D1)⊗OX(D2).

(ii) OX(−D) ∼= OX(D)∨.

(iii) If f ∈ K(X), then OX(div(f)) ∼= OX .

Proposition. Let X be a Noetherian, normal, integral scheme. Assume that Xis factorial , i.e. every local ring OX,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. SoD ⊆ 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 ID,x ⊆ OX,x is a height 1 prime ideal. So ID,x = (f) forf ∈ mX,x. Then f is the local equation for D.

Recall the Class group Cl(X) was the group of Weil divisors modulo principalequivalence.

Definition (Picard group). We define the Picard group of X to be the group ofCartier 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 ofL. Then OX(div(s)) is invertible, and there is an isomorphism

OX(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 H1(X,O∗X).

Proof. Given f ∈ H0(U,OX(div(s))), map it to f · s ∈ H0(U,L). This gives thedesired isomorphism.

If we have to sections s′ 6= s, then f = s′/s ∈ K(X). So div(s) = div(s′) +div(f), and div(f) is principal. So this gives a well-defined map Pic(X) →Cl(X).

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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 ⊆ Xis 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 divisoron X \ Z, then S ⊆ X maps to S under the restriction map. So this map issurjective.

Also, that [nZ]|X\Z is trivial. So the composition of the first two mapsvanishes. To check exactness, suppose D is a Weil divisor on X, principal onX \Z. Then D|X\Z = dim(f)|X\Z for some f ∈ K(X). Then D− div(f) is justsupported along Z. So it must be of the form nZ.

If we remove something of codimension at least two, then something evensimpler happens.

Proposition. If Z ⊆ X has codimension ≥ 2, then Cl(X) → Cl(X \ Z) is anisomorphism.

The proof is the same as above, except no divisor can be supported on Z.

Example. Cl(A2 \ 0) = Cl(A2).

In general, to use the above propositions to compute the class group, a goodstrategy is to remove closed subschemes as above until we reach something weunderstand well. One thing we understand well is affine schemes:

Proposition. If A is a Noetherian ring, regular in codimension 1, then A is aUFD iff A is normal and Cl(SpecA) = 0

Proof. If A is a UFD, then it is normal, and every prime ideal of height 1 isprincipally generated. So if D ∈ SpecA is Weil and prime, then D = V (f) forsome f , and hence (f) = ID.

Conversely, if A is normal and Cl(SpecA) = 0, then every Weil divisor isprincipal. So if I is a height 1 prime ideal, then V (I) = D for some Weil divisorD. Then D is principal. So I = (f) for some f . So A is a Krull Noetherianintegral domain with principally generated height 1 prime ideals. So it is aUFD.

Example. Cl(A2) = 0.

Example. Consider Pn. We have an exact sequence

Z→ Cl(Pn) = Cl(Pn \ Pn−1) = Cl(An)→ 0.

So Z→ Cl(Pn) is surjective. So Cl(Pn) 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).

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Example. Let X = V (xy − z2) ⊆ A3. Let Z = V (x, z). We claim thatCl(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

OX,Z =

(K[x, y, z]

xy − z2

)(x,z)

=

(K[x, y, y−1, z]

x− z2

)(x,y)

= K[y, y−1, z](z).

Unsurprisingly, this is a DVR, and crucially, the uniformizer is z. So we knowthat div(x) = 2Z. So we know that 2Z ⊆ ker(Z → Cl(X)). There is only onething to check, which is that (x, y) is not principal. To see this, consider

T(0)X = A3, T(0)(Z ⊆ X) = A1.

But if Z were principal, then IZ,0 = (f) for some f ∈ OX,0. But then T0(Z) ⊆T0X will be ker df . But then ker df should have dimension at least 2, which isa contradiction.

Proposition. Let X be Noetherian and regular in codimension one. Then

Cl(X) = Cl(X × A1).

Proof. We have a projection map

pr∗1 : Cl(X)→ Cl(X × A1)

[Di] 7→ [Di × A1]

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 ID is the ideal of D in K[X × A1], and let I0D

be the ideal of K(X)[t] generated b ID under the inclusion

K[X × A1] = K[X][t] ⊆ K(X)[t].

If I0D = 1, then ID contains some function f ∈ K(X). Then D ⊆ V (f)

as a subvariety of X × A1. So D is an irreducible component of V (f), and inparticular is of the form D′ × A1.

If not, then I0D = (f) for some f ∈ K(X)[t], since K(X)[t] is a PID. Then

divf is a principal divisor of X ×A1 whose localization at the generic point is D.Thus, divf is D plus some other divisors of the form D′ × A1. So D is linearlyequivalent to a sum of divisors of the form D′ × A1.

Exercise. We have Cl(X × Pn) = Z⊕ Cl(X).

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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 closedfield. Let D0 be a divisor on X.

(i) For all s ∈ H0(X,OX(D)), div(s) is an effective divisor linearly equivalentto D.

(ii) If D ∼ D0 and D ≥ 0, then there is s ∈ H0(OX(D0)) such that div(s) = D

(iii) If s, s′ ∈ H0(OX(D0)) and div(s) = div(s′), then s′ = λs for some λ ∈ K∗.

Proof.

(i) Done last time.

(ii) If D ∼ D0, then D−D0 = div(f) for some f ∈ K(X). Then (f) +D0 ≥ 0.So f induces a section s ∈ H0(OX(D0)). Then div(s) = D.

(iii) We have s′

s ∈ K(X)∗. So div(s′

s

). So s′

s ∈ H0(O∗) = K∗.

Definition (Complete linear system). A complete linear system is the set of alleffective divisors linearly equivalent to a given divisor D0, written |D0|.

Thus, |D0| is the projectivization of the vector space H0(X,O(D0)).We also define

Definition (Linear system). A linear system is a linear subspace of the projectivespace structure on |D0|.

Given a linear system |V | of L on X, we have previously seen that we get arational map X 99K P(V ). For this to be a morphism, we need V to generate L.In general, the OX action on L induces a map

V ⊗OX → L.

Tensoring with L−1 gives a map

V ⊗ L−1 → OX .

The image is an OX -submodule, hence an ideal sheaf. This ideal b|V | is calledthe base ideal of |V | on X, and the corresponding closed subscheme is the baselocus. Then the map X 99K P(V ) is a morphism iff b|V | = OX . In this case, wesay |V | is free.

Accordingly, we can define

Definition ((Very) ample divisor). We say a Cartier divisor D is (very) amplewhen OX(D) is.

By Serre’s theorem, if A is ample and X is projective, and L is any linebundle, then for n sufficiently large, we have

h0(X,L ⊗An) =∑

(−1)ihi(X,L ⊗An) = χ(X,L ⊗An).

In the case of curves, Riemann–Roch lets us understand this number well:

13


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 Cartierdivisors. First observe that by definition, every projective normal scheme has avery ample divisor, say H. If D is any divisor, then by Serre’s theorem, D+ nHis globally generated for large n. Moreover, one can check that the sum of aglobally generated divisor and a very ample divisor is still very ample, e.g. bychecking it separates points and tangent vectors. Hence D + nH is very amplefor large n. Thus, we deduce that

Proposition. Let D be a Cartier divisor on a projective normal scheme .ThenD ∼ H1 −H2 for some very ample divisors Hi. We can in fact take Hi to beeffective, and if X is smooth, then we can take Hi to be smooth and intersectingtransversely.

The last part is a consequence of Bertini’s theorem.

Theorem (Bertini). Let X be a smooth projective variety over an algebraicallyclosed field K, and D a very ample divisor. Then there exists a Zariski open setU ⊆ |D| such that for all H ∈ U , H is smooth on X and if H1 6= H2, then H1

and H2 intersect transversely.

14

2 Surfaces III Positivity in Algebraic Geometry

2 Surfaces

2.1 The intersection product

We now focus on the case of smooth projective surfaces over an algebraicallyclosed field. In this case, there is no need to distinguish between Weil and Cartierdivisors.

Let X be such a surface. Let C,D ⊆ X be smooth curves intersectingtransversely. We want to count the number of points of intersection. This isgiven by

|C ∩D| = degC(OC(D)) = h0(OC∩D) = χ(OC∩D),

using that OC∩D is a skyscraper sheaf supported at C ∩D. To understand thisnumber, we use the short exact sequences

0 OC(−D|C) OC OC∩D 0

0 OX(−C −D) OX(−D) OC(−D|C) 0

0 OX(−C) OX OC 0

This allows us to write

χ(OC∩D) = −χ(C,OC(−D|C)) + χ(C,OC)

= χ(OX(−C −D))− χ(OX(−D)) + χ(C,OC)

= χ(OX(−C −D))− χ(OX(−D))− χ(OX(−C)) + χ(OX).

Thus suggests for for general divisors D1, D2, we can define

Definition (Intersection product). For divisors D1, D2, we define the intersec-tion product to be

D1 ·D2 = χ(OX) + χ(−D1 −D2)− χ(−D1)− χ(−D2).

Proposition.

(i) The product D1 ·D2 depends only on the classes of D1, D2 in Pic(X).

(ii) D1 ·D2 = D2 ·D1.

(iii) D1 ·D2 = |D1 ∩D2| if D1 and D2 are curves intersecting transversely.

(iv) The intersection product is bilinear.

Proof. Only (iv) requires proof. First observe that if H is a very ample divisorrepresented by a smooth curve, then we have

H ·D = degH(OH(D)),

and this is linear in D.Next, check that D1 ·(D2 +D3)−D1 ·D2−D1 ·D3 is symmetric in D1, D2, D3.

So

(a) Since this vanishes when D1 is very ample, it also vanishes if D2 or D3 isvery ample.

15


(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 = H1−H2 for H1, H2 very ample and smooth.Then D ·D′ = H1 ·D′ −H2 ·D′ by (a), and thus is linear in D′.

In fact, bilinearity and (iii) shows that these properties uniquely characterizethe intersection product, since every divisor is a difference of smooth very ampledivisors.

Example. Take X = P2. We saw that Pic(P2) ∼= Z, generated by a hyperplane`. So all we need to understand is what `2 is. But two transverse lines intersectat a point. So `2 = 1. Thus, the intersection product on Pic(P2) is just ordinarymultiplication. In particular, if D1, D2 are curves that intersect transversely, wefind that

D1 ·D2 = deg(D1) deg(D2).

This is Bezout’s theorem!

Exercise. Let C1, C2 ⊆ X be curves without common components. Then

C1 · C2 =∑

p∈C1∩C2

`(OC1∩C2),

where for p ∈ C1 ∩ C2, if C1 = V (x) and C2 = V (y), then

`(OC1∩C2) = dimk(OX,p/(x, y))

counts multiplicity.

Example. Take X = P1 × P1. Then

Pic(X) = Z2 = Z[p∗1O(1)]⊕ Z[p∗2O(1)].

Let A = p∗1O(1), B = p∗2O(1). Then we see that

A2 = B2 = 0, A ·B = 1

by high school geometry.

The Riemann–Roch theorem for curves lets us compute χ(D) for a divisorD. We have an analogous theorem for surfaces:

Theorem (Riemann–Roch for surfaces). Let D ∈ Div(X). Then

χ(X,OX(D)) =D · (D −KX)

2+ χ(OX),

where KX is the canonical divisor .

To prove this, we need the adjunction formula:

Theorem (Adjunction formula). Let X be a smooth surface, and C ⊆ X asmooth curve. Then

(OX(KX)⊗OX(C))|C ∼= OC(KC).

16


Proof. Let IC = OX(−C) be the ideal sheaf of C. We then have a short exactsequence on C:

0→ OX(−C)|C ∼= IC/I2C → Ω1

X |C → Ω1C → 0,

where the left-hand map is given by d. To check this, note that locally on affinecharts, if C is cut out by the function f , then smoothness of C implies the kernelof the second map is the span of df .

By definition of the canonical divisor, we have

OX(KX) = det(Ω1X).

Restricting to C, we have

OX(KX)|C = det(Ω1X |C) = det(OX(−C)|C)⊗ det(Ω1

C) = OX(C)|∨C ⊗OC(KC).

Proof of Riemann–Roch. We can assume D = H1 − H2 for very ample linebundles H1, H2, which are smoothly irreducible curves that intersect transversely.We have short exact sequences

0 OX(H1 −H2) OX(H1) OH2(H1|H2

) 0

0 OX OX(H1) OH1(H1) 0

where OH1(H1) means the restriction of the line bundle OX(H1) to H1. We can

then compute

χ(H1 −H2) = χ(OX(H1))− χ(H2,OH2(H1|H2

))

= χ(OX) + χ(H1,OH1(H1))− χ(H2,OH2

(H1|H2)).

The first term appears in our Riemann–Roch theorem, so we leave it alone. Wethen use Riemann–Roch for curves to understand the remaining. It tells us

χ(Hi,OHi(H1)) = deg(OHi(H1)) + 1− g(Hi) = (Hi ·H1) + 1− g(Hi).

By definition of genus, we have

2g(Hi)− 2 = deg(KHi).

and the adjunction formula lets us compute deg(KHi) by

deg(KHi) = Hi · (KX +Hi).

Plugging these into the formula and rearranging gives the desired result.

Previously, we had the notion of linear equivalence of divisors. A coarsernotion of equivalence is that of numerical equivalence, which is about what theintersection product can detect:

Definition (Numerical equivalence). We say divisors D,D′ are numericallyequivalent, written D ≡ D′, if

D · E = D′ · E

for all divisors E.We write

Num0 = D ∈ Div(X) : D ≡ 0.

17


Definition (Neron–Severi group). The Neron–Severi group is

NS(X) = Div(X)/Num0(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, foran elliptic curve, Pic(X) bijects with the closed points of X, which is in generalvery large!

Definition (ρ(X)). ρ(X) = dim NSR(X) = rk NS(X).

Tensoring with R, the intersection product gives a map

( · , · ) : NSR(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 ampledivisor H, and then H2 > 0. So this is definitely not negative definite. It turnsout 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 butD 6≡ 0. Then D2 < 0.

Proof. Before we begin the proof, observe that if H ′ is very ample and D′

is (strictly) effective, then H ′ · D′ > 0, since this is given by the number ofintersections between D′ and any hyperplane in the projective embedding givenby H ′.

Now assume for contradiction that D2 ≥ 0.

– If D2 > 0, fix an n such that Hn = D+nH is very ample. Then Hn ·D > 0by assumption.

We use Riemann–Roch to learn that

χ(X,OX(mD)) =m2D2 −mKX ·D

2+ χ(OX).

We first consider the H2(OX(mD)) term in the left-hand side. By Serreduality, we have

H2(OX(mD)) = H0(KX −mD).

Now observe that for large m, we have

Hn · (KX −mD) < 0.

Since Hn is very ample, it cannot be the case that KX −mD is effective.So H0(KX −D) = 0.

Thus, for m sufficiently large, we have

h0(mD)− h1(mD) > 0.

In particular, mD is (strictly) effective for m 0. But then H ·mD > 0since H is very ample and mD is effective. This is a contradiction.

18


– If D2 = 0. Since D is not numerically trivial, there exists a divisor E onX such that D · E 6= 0. We define

E′ = (H2)E − (E ·H)H.

It is then immediate that E′ ·H = 0. So D′n = nD+E′ satisfies D′n ·H = 0.On the other hand,

(D′n)2 = (E′)2 + 2nD · E′ > 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 P1. The result is a surface X = BlpX together with a mapε : X → X such that ε−1(p) is a smoothly embedded copy of P1, and ε is anisomorphism outside of this P1.

To construct X, pick local coordinates x, y in a neighbourhood U of X suchthat V (x, y) = p. Work on U × P1 with coordinates [X : Y ] on P1. We then set

U = V (xY − yX).

There is a canonical projection of U onto U , and this works, since above thepoint (0, 0), any choice of X,Y satisfies the equation, and everywhere else, wemust have [X : Y ] = [x : y].

Definition (Blow up). The blow up of X at p is the variety X = BlpX thatwe just defined. The preimage of p is known as the exceptional curve, and isdenoted E.

We would like to understand Pic(X) in terms of Pic(X), and in particularthe product structure on Pic(X). This will allow us to detect whether a surfaceis a blow up.

The first step to understanding Pic(X) is the localization sequence

Z→ Pic(X)→ Pic(X \ P1)→ 0.

We also have the isomorphism

Pic(X \ P1) = Pic(X \ p) ∼= Pic(X).

Finally, pulling back along ε gives a map Pic(X)→ Pic(X), and the push-pullformula says

π∗(π∗L) = L ⊗ π∗(OX).

Hartog’s lemma saysπ∗OX = OX .

So in fact ε∗ is a splitting. So Pic(X) is the direct sum of Pic(X) with a quotientof Z, generated by [E]. We will show that in fact Pic(X) ∼= ε∗Pic(X) ⊕ Z[E].This will be shown by calculating E2 = −1.

19


In general, if C is a curve in X, then this lifts to a curve C ⊆ X, called thestrict 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 divisorsπ∗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 nottangent to any branch of C at p. Then in the local ring OX,p, the equation of Cis given as

f = fm(x, y) + higher order terms,

where fm is a non-zero degree m polynomial. Then by definition, the multiplicityis m. Then on U ⊆ (U × P1), we have the chart U × A1 where X 6= 0, withcoordinates

(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) = xmfm(1, t) + higher order terms = xm · h(x, t),

with h(0, 0) 6= 0. But then on Ux 6=0, the curve x = 0 is just E. So this equationhas multiplicity m along E.

Proposition. Let X be a smooth projective surface, and x ∈ X. Let X =BlxX

π→ X. Then

(i) π∗Pic(X)⊕ Z[E] = Pic(X)

(ii) π∗D · π∗F = D · F , π∗D · E = 0, E2 = −1.

(iii) KX = π∗(KX) + 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 theleft-hand map is injective, or equivalently, mE 6∼ 0. This will follow fromthe fact that E2 = −1.

20


(ii) For the first part, it suffices to show that π∗D · π∗F = D · F for D,Fvery 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 isclear. 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 have0 = π∗C · E = C · E + E2,

But C and E intersect at exactly one point. So C · E = 1.

(iii) By the adjunction formula, we have

(KX + E) · E = deg(KP1) = −2.

So we know that KX ·E = −1. Also, outside of E, KX and KX agree. Sowe have

KX = π∗(KX) +mE

for some m ∈ Z. Then computing

−1 = KX · E = π∗(KX) · E +mE2

gives m = 1.

(iv) We need to show that if D ∈ Num0(X), then π∗(D) ∈ Num0(X). But thisis clear for both things that are pulled back and things that are E. So weare 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 asmooth projective surface X ′ and a commutative diagram

X ′

X Y

p

q

ϕ

where p : X ′ → X is a composition of blow ups, and in particular is birational.

Proof. We may assume Y = Pn, and X 99K Pn is non-degenerate. Then ϕ isinduced by a linear system |V | ⊆ H0(X,L). We first show that we may assumethe base locus has codimension > 1.

If not, every element of |V | is of the form C +D′ for some fixed C. Let |V ′|be the set of all such D′. Then |V | and |V ′| give the same rational map. Indeed,if V has basis f0, . . . , fn, and g is the function defining C, then the two mapsare, respectively,

[f0 : · · · : fn] and [f0/g : · · · : fn/g],

which define the same rational maps. By repeating this process, replacing Vwith V ′, we may assume the base locus has codimension > 1.

21


We now want to show that by blowing up, we may remove all points fromthe base locus. We pick x ∈ X in the base locus of |D|. Blow it up to getX1 = BlxX → X. Then

π∗|D| = |D1|+mE,

where m > 0, and |D1| is a linear system which induces the map ϕ π1. If |D1|has no basepoints, then we are done. If not, we iterate this procedure. Theprocedure stops, because

0 ≤ D21 = (π∗1D

2 +m2E2) < D2.

Exercise. We have the following universal property of blow ups: If Z → X isa birational map of surfaces, and suppose f−1 is not defined at a point x ∈ X.Then f factors uniquely through the blow up Blx(X).

Theorem. Let g : Z → X be a birational morphism of surfaces. Then g factors

as Zg′→ X ′

p→ X, where p : X ′ → X is a composition of blow ups, and g′ is anisomorphism.

Proof. Apply elimination of indeterminacy to the rational inverse to g to obtainp : X ′ → X. There is then a lift of g′ to X ′ by the universal property, and theseare inverses to each otehr.

Birational isomorphism is an equivalence relation. So we can partition theset of smooth projective surfaces into birational isomorphism classes. If X,X ′

are in the same birational isomorphism class, we can say X ≤ X ′ if X ′ is a blowup 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 doesnot exist a smooth X ′ and a birational morphism X → X ′ that is not anisomorphism. We say X is minimal if it is the unique relative minimal surfacein the birational equivalence class.

If X is not relatively minimal, then it is the blow up of another smoothprojective surface, and in this case the exceptional curve C satisfies C2 = −1. Itturns out this criterion itself is enough to determine minimality.

Theorem (Castelnuovo’s contractibility criterion). Let X be a smooth projectivesurface over K = K. If there is a curve C ⊆ X such that C ∼= P1 and C2 = −1,then there exists f : X → Y that exhibits X as a blow up of Y at a point withexceptional curve C.

Exercise. Let X be a smooth projective surface. Then any two of the followingthree conditions imply the third:

(i) C ∼= P1

(ii) C2 = −1

(iii) KX · C = −1

When these are satisfied, we say C is a (−1) curve.

22


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 itdoes not contain a (−1) curve.

Proof. The idea is to produce a suitable linear system |L|, giving a map f :X → Pn, and then take Y to be the image. Note that f(C) = ∗ is the same asrequiring

OX(L)|C = OC .

After finding a linear system that satisfies this, we will do some work to showthat 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 assumeH1(X,OX(H)) = 0. Let

H · C = K > 0.

Consider divisors of the form H + iC. We can compute

degC(H + iC)|C = (H + iC) · C = K − i.

Thus, we know degC(H +KC)|C = 0, and hence

OX(H +KC)|C ∼= OC .

We claim that OX(H +KC) works. To check this, we pick a fairly explicitbasis of H0(H +KC). Consider the short exact sequence

0→ OX(H + (i− 1)C)→ OX(H + iC)→ OC(H + iC)→ 0,

inducing a long exact sequence

0→ H0(H + (i− 1)C)→ H0(H + iC)→ H0(C, (H + iC)|C)

→ H1(H + (i− 1)C)→ H1(H + iC)→ H1(C, (H + iC)|C)→ · · ·

We know OX(H + iC)|C = OP1(K − i). So in the range i = 0, . . . ,K, we haveH1(C, (H + iC)|C) = 0. Thus, by induction, we find that

H1(H + iC) = 0 for i = 0, . . . ,K.

As a consequence of this, we know that for i = 1, . . . ,K, we have a shortexact sequence

H0(H + (i− 1)C) → H0(H + iC) H0(C, (H + iC)|C) = H0(OP1(K − i)).

Thus, H0(H + iC) is spanned by the image of H0(H + (i− 1)C) plus a lift of abasis of H0(OP1(K − i)).

For each i > 0, pick a lift y(i)0 , . . . , y

(i)K−i of a basis of H0(OP1(K − i)) to

H0(H+ iC), and take the image in H0(H+KC). Note that in local coordinates,if C is cut out by a function g, then the image in H0(H +KC) is given by

gK−iy(i)0 , . . . , gK−iy

(i)K−i.

For i = 0, we pick a basis of H0(H), and then map it down to H0(H + KC).Taking the union over all i of these elements, we obtain a basis of H0(H +KC).Let f be the induced map to Pr.

23


For concreteness, list the basis vectors as x1, . . . , xr, where xr is a lift of

1 ∈ OP1 to H0(H +KC), and xr−1, xr−2 are gy(K−1)0 , gy

(K−1)1 .

First observe that x1, . . . , xr−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 H0(H) is

enough to separate points and tangent vectors. So f is an isomorphism outsideof C.

All the other basis elements of H +KC are needed to make sure the image issmooth, and we only have to check this at the point p. This is done by showingthat dimmY,p/m

2Y,p ≤ 2, which requires some “infinitesimal analysis” we will not

perform.

Example. Consider X = P2, and x, y, z three non-colinear points in P2 givenby [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1].

There are three conics passing through x, y, z, given by x1x2, x0, x2 andx0x1, which is the union of lines `0, `1, `2, and let |Cx,y,z| be the linear systemgenerated by these three conics. Then we obtain a birational map

P2 99K P2

[x0 : x1 : x2] 7→ [x1x2 : x0x2 : x0x1] = [x−10 : x−1

1 : x−12 ]

with base locus x, y, z.Blowing up at the three points x, y, z resolves our indeterminacies, and we

obtain a diagram

Blx,y,zP2

P2 P2C

Then in the blowup, we have

π∗`i = ˜i + Ej + Ek

for i, j, k distinct. We then check that ˜2i = −1, and each ˜

i is isomorphicto P1. Thus, Castelnuovo tells us we can blow these down. This blow downmap is exactly the right-hand map in the diagram above, which one can checkcompresses each line ˜

i to a point.

24

3 Projective varieties III Positivity in Algebraic Geometry

3 Projective varieties

We now turn to understand general projective schemes. Our goal will be tounderstand 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 intersectionproduct. Thus, given a projective variety X over an algebraically closed fieldK with dimX = n, we seek a multilinear symmetric function CaDiv(X)n → Z,denoted

(D1, . . . , Dn) 7→ (D1 · . . . ·Dn).

This should satisfy the following properties:

(i) The intersection depends only on the linear equivalence class, i.e. it factorsthrough Pic(X)n.

(ii) If D1, . . . , Dn are smooth hypersurfaces that all intersect transversely, then

D1 · · · · ·Dn = |D1 ∩ · · · ∩Dn|.

(iii) If V ⊆ X is an integral subvariety of codimension 1, and D1, . . . , Dn−1 areCartier divisors, then

D1 · . . . ·Dn−1 · [V ] = (D1|V · . . . ·Dn−1|V ).

In general, if V ⊆ X is an integral subvariety, we write

D1 · . . . ·DdimV · [V ] = (D1|V · . . . ·DdimV |V ).

There are two ways we can define the intersection product:

(i) For each Di, write it as Di ∼ Hi,1 − Hi,2 with Hi,j very ample. Thenmultilinearity tells us how we should compute the intersection product,namely as ∑

I⊆1,...,n

(−1)|I|∏i∈I

Hi,2 ·∏j 6∈I

Hj,1.

By Bertini, we can assume the Hi,j intersect transversely, and then we justcompute these by counting the number of points of intersection. We thenhave to show that this is well-defined.

(ii) We can write down the definition of the intersection product in a morecohomological way. Recall that for n = 2, we had

D · C = χ(OX) + χ(OX(−C −D))− χ(OX(−C))− χ(OX(−D)).

For general n, let H1, . . . ,Hn be very ample divisors. Then the last propertyimplies we have

H1 · · · · ·Hn = (H2|H1· . . . ·Hn|H1

).

25


So inductively, we have

H1 · · · · ·Hn =∑

I⊆2,...,n

(−1)|I|χ

(H1,OH1

(−∑i∈I

Hi

))

But we also have the sequence

0→ OX(−H1)→ OX → OH1 → 0.

Twisting this sequence shows that

H1 · · · · ·Hn =∑

I⊆2,...,n

(−1)|I|(χ(OX

(−∑

i∈IHi

))− χ

(OX

(−H1 −

∑i∈I

Hi

)))=

∑J⊆1,...,n

(−1)|J|χ(OX

(−∑

i∈JHi

))We can then adopt this as the definition of the intersection product

D1 · · · · ·Dn =∑

J⊆1,...,n

(−1)|J|χ

(OX

(−∑i∈J

Di

)).

Reversing the above procedure shows that property (iii) is satisfied.

As before we can define

Definition (Numerical equivalence). Let D,D′ ∈ CaDiv(X). We say D and D′

are numerically equivalent, written D ≡ D′, if D · [C] = D′ · [C] for all integralcurves C ⊆ X.

In the case of surfaces, we showed that being numerically equivalent to zerois the same as being in the kernel of the intersection product.

Lemma. Assume D ≡ 0. Then for all D2, . . . , DdimX , we have

D ·D2 · . . . ·DdimX = 0.

Proof. We induct on n. As usual, we can assume that the Di are very ample.Then

D ·D2 · · · ·DdimX = D|D2·D3|D2

· · · · ·DdimX |D2.

Now if D ≡ 0 on X, then D|D2 ≡ 0 on D2. So we are done.

Definition (Neron–Severi group). The Neron–Severi group of X, writtenNS(X) = N1(X), is

N1(X) =CaDiv(X)

Num0(X)=

CaDiv(X)

D | D ≡ 0.

As in the case of surfaces, we have

26


Theorem (Severi). Let X be a projective variety over an algebraically closedfield. Then N1(X) is a finitely-generated torsion free abelian group, hence ofthe form Zn.

Definition (Picard number). The Picard number of X is the rank of N1(X).

We will not prove Severi’s theorem, but when working over C, we can indicatea proof strategy. Over C, in the analytic topology, we have the exponentialsequence

0→ Z→ OanX → O

∗,anX → 0.

This gives a long exact sequence

H1(X,Z)→ H1(X,OanX )→ H1(X,O∗,an

X )→ H2(X,Z).

It is not hard to see that Hi(X,Z) is in fact the same as singular cohomology,and H1(O∗,anX ) 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)→ H2(X,Z)is just the first Chern class.

Next check that the intersection product [C] ·D is the same as the topologicalintersection product [C]·c1(D). So at least modulo torsion, we have an embeddingof the Neron–Severi group into H2(X,Z), which we know is finite dimensional.

In the case of surfaces, we had the Riemann–Roch theorem. We do not havea similarly precise statement for arbitrary varieties, but an “asymptotic” versionis good enough for what we want.

Theorem (Asymptotic Riemann–Roch). Let X be a projective normal varietyover K = K. Let D be a Cartier divisor, and E a Weil divisor on X. Thenχ(X,OX(mD + E)) is a numerical polynomial in m (i.e. a polynomial withrational coefficients that only takes integral values) of degree at most n = dimX,and

χ(X,OX(mD + E)) =Dn

n!mn + lower order terms.

Proof. By induction on dimX, we can assume the theorem holds for normalprojective varieties of dimension < n. We fix H on X very ample such thatH +D is very ample. Let H ′ ∈ |H| and G ∈ |H +D| be sufficiently general. Wethen have short exact sequences

0→ OX(mD + E)→ OX(mD + E +H)→ OH′((mD + E +H)|H′)→ 0

0→ OX((m− 1)D + E)→ OX(mD + E +H)→ OG((mD + E +H)|G)→ 0.

Note that the middle term appears in both equations. So we find that

χ(X,OX(mD + E)) + χ(H ′,OH′((mD + E +H)|H))

= χ(X,OX((m− 1)D + E)) + χ(H ′,OG((mD + E +H)|G))

Rearranging, we get

χ(OX(mD − E))− χ(OX((m− 1)D − E))

= χ(G,OG(mD + E +H))− χ(H ′,OH′(mD + E +H)).

27


By induction (see proposition below) the right-hand side is a numerical polyno-mial of degree at most n− 1 with leading term

Dn−1 ·G−Dn−1 ·H(n− 1)!

mn−1 + lower order terms,

since Dn−1 ·G is just the (n− 1) self-intersection of D along G. But G−H = D.So the LHS is

Dn

(n− 1)!mn−1 + lower order terms,

so we are done.

Proposition. Let X be a normal projective variety, and |H| a very amplelinear system. Then for a general element G ∈ |H|, G is a normal projectivevariety.

Some immediate consequences include

Proposition. Let X be a normal projective variety.

(i) If H is a very ample Cartier divisor, then

h0(X,mH) =Hn

n!mn + lower order terms for m 0.

(ii) If D is any Cartier divisor, then there is some C ∈ R>0 such that

h0(mD) ≤ C ·mn for m 0.

Proof.

(i) By Serre’s theorem, Hi(Ox(mH)) = 0 for i > 0 and m 0. So we applyasymptotic Riemann Roch.

(ii) There exists a very ample Cartier divisor H ′ on X such that H ′ + D isalso very ample. Then

h0(mD) ≤ h0(m(H ′ +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 ofschemes, 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 Rif∗F = 0for all i > 0. Then

Hi(F ⊗ f∗OX(mD)) = Hi(f∗F ⊗OY (mD)) = 0

for all i > 0 and m 0. So by Serre’s theorem, we know f∗D is ample.

Another useful result is

28


Proposition. Let X be a proper scheme, and L an invertible sheaf. Then L isample iff L|Xred

is ample.

Proof.

(⇒) If L induces a closed embedding of X, then map given by L|Xredis given

by the composition with the closed embedding Xred → X.

(⇐) Let J ⊆ OX be the nilradical. By Noetherianness, there exists n such thatJ n = 0.

Fix F a coherent sheaf. We can filter F using

F ⊇ JF ⊇ J 2F ⊇ · · · ⊇ J n−1F ⊇ J nF = 0.

For each j, we have a short exact sequence

0→ J j+1F → J jF → Gj → 0.

This Gi is really a sheaf on the reduced structure, since J acts trivially.Thus Hi(Gj ⊗ Lm) for j > 0 and large m. Thus inducting on j ≥ 0, wefind that for i > 0 and m 0, we have

Hi(J jF ⊗ Lm) = 0.

The following criterion for ampleness will give us a very good handle on howample divisors behave:

Theorem (Nakai’s criterion). Let X be a projective variety. Let D be a Cartierdivisor 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 )dimV = DdimV [V ] > 0.

Before we prove this, we observe some immediate corollaries:

Corollary. Let X be a projective variety. Then ampleness is a numericalcondition, i.e. for any Cartier divisors D1, D2, if D1 ≡ D2, then D1 is ample iffD2 is ample.

Corollary. Let X,Y be projective variety. If f : X → Y is a surjective finitemorphism of schemes, and D is a Cartier divisor on Y . Then D is ample iff f∗Dis ample.

Proof. It remains to prove ⇐. If f is finite and surjective, then for all V ⊆ Y ,there exists V ′ ⊆ f−1(V ) ⊆ X such that f |V ′ : V ′ → V is a finite surjectivemorphism. Then we have

(f∗D)dimV ′ [V ′] = deg f |V ′DdimV [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 OX(D) globallygenerated, and

Φ : X → P(H0(X,OX(D))∗)

the induced map. Then D is ample iff Φ is finite.

29


Proof.

(⇐) If X → Φ(X) is finite, then D = Φ∗O(1). So this follows from the previouscorollary.

(⇒) 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’scriterion, 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(H0(D)∗).

If V ⊆ X is a closed integral subvariety, then DdimV · [V ] = (D|V )dimV .But this is just degΦ(V )O(1) > 0.

(⇐) We proceed by induction on dimX, noting that dimX = 1 is trivial. Byinduction, for any proper subvariety V , we know that D|V is ample.

The key of the proof is to show that OX(mD) is globally generated forlarge m. If so, the induced map X → P(|mD|) cannot contract any curveC, or else mD · C = 0. So this is a finite map, and mD is the pullback ofthe ample divisor OP(|mD|)(1) along a finite map, hence is ample.

We first reduce to the case whereD is effective. As usual, writeD ∼ H1−H2

with Hi very ample effective divisors. We have exact sequences

0→ OX(mD −H1)→ OX(mD)→ OH1(mD)→ 0

0→ OX(mD −H1)→ OX((m− 1)D)→ OH2((m− 1)D)→ 0.

We know D|Hiis ample by induction. So the long exact sequences implies

that for all m 0 and j ≥ 2, we have

Hj(mD) ∼= Hj(mD −H1) = Hj((m− 1)D).

So we know that

χ(mD) = h0(mD)− h1(mD) + constant

for all m 0. On the other hand, since X is an integral subvariety ofitself, Dn > 0, and so asymptotic Riemann–Roch tells us h0(mD) > 0 forall 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 showthat this is true in a neighbourhood of D, since outside of D, the sheafis automatically globally generated by using the tautological section thatvanishes 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 passto D|red if necessary). Thus, it suffices to show that

H0(OX(mD))→ H0(OD(mD))

30


is surjective.

To show this, we use the short exact sequence

0→ OX((m− 1)D)→ OX(mD)→ OD(mD)→ 0.

For i > 0 and large m, we know Hi(OD(mD)) = 0. So we have surjections

H1((m− 1)D) H1(mD)

for m large enough. But these are finite-dimensional vector spaces. So form sufficiently large, this map is an isomorphism. Then H0(OX(mD))→H0(OD(mD)) is a surjection by exactness.

Recall that we defined PicK(X) for K = Q or R. We can extend theintersection product linearly, and all the properties of the intersection productare preserved. We write

N1K(X) = PicK(X)/Num0,K(X) = NS(X)⊗K

where, unsurprisingly,

Num0,K(X) = D ∈ PicK(X) : D · C = 0 for all C ⊆ X = Num0 ⊗K.

We now want to define ampleness for such divisors.

Proposition. Let D ∈ CaDivQ(X) Then the following are equivalent:

(i) cD is an ample integral divisor for some c ∈ N>0.

(ii) D =∑ciDi, where ci ∈ Q>0 and Di are ample Cartier divisors.

(iii) D satisfies Nakai’s criterion. That is, DdimV [V ] > 0 for all integralsubvarieties 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) ⊆ N1Q(X) for the cone given by ample divisors.

Lemma. A positive linear combination of ample divisors is ample.

Proof. Let H1, H2 be ample. Then for λ1, λ2 > 0, we have

(λ1H1 + λ2H2)dimV [V ] =

(∑(dimV

p

)λp1λ

dimV−p2 Hp

1 ·HdimV−p2

)[V ]

Since any restriction of an ample divisor to an integral subscheme is ample, andmultiplying with H is the same as restricting to a hyperplane cuts, we know allthe terms are positive.

Proposition. Ampleness is an open condition. That is, if D is ample andE1, . . . Er are Cartier divisors, then for all |εi| 1, the divisor D + εiEi is stillample.

31


Proof. By induction, it suffices to check it in the case n = 1. Take m ∈ N suchthat mD ± E1 is still ample. This is the same as saying D ± 1

mE1 is still ample.Then for |ε1| < 1

m , we can write

D + εE1 = (1− q)D + q

(D +

1

mE1

)for some q < 1.

Similarly, if D ∈ CaDivR(X), then we say D is ample iff

D =∑

aiDi,

where Di are ample divisors and ai > 0.

Proposition. Being ample is a numerical property over R, i.e. if D1, D2 ∈CaDivR(X) are such that D1 ≡ D2, then D1 is ample iff D2 is ample.

Proof. We already know that this is true over Q by Nakai’s criterion. Then forreal coefficients, we want to show that if D is ample, E is numerically trivialand t ∈ R, then D + tE is ample. To do so, pick t1 < t < t2 with ti ∈ Q, andthen find λ, µ > 0 such that

λ(D1 + t1E) + µ(D1 + t2E) = D1 + 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 E1, . . . , Er,for all ‖εi‖ ≤ 1, the divisor H +

∑εiEi 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 aCartier divisor. Then D is numerically effective (nef) if D ·C ≥ 0 for all integralcurves C ⊆ X.

Similar to the case of ample divisors, we have

Proposition.

(i) D is nef iff D|Xredis nef.

(ii) D is nef iff D|Xiis nef for all irreducible components Xi.

(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’scriterion.

32


Theorem (Kleinmann’s criterion). Let X be a proper scheme, and D an R-Cartier divisor. Then D is nef iff DdimV [V ] ≥ 0 for all proper irreduciblesubvarieties.

Corollary. Let X be a projective scheme, and D be a nef R-divisor on X, andH 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 equivalentto requiring

(D + εH)dimV · [V ] =

(∑(dimV

p

)εpDdimV−pHp

)[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, weknow the terms that involve D are non-negative. The Hp term is positive.So we are done.

(ii) We know (D + εH) · C > 0 for all positive ε sufficiently small. Takingε→ 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 impliesAmp(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. SoAmp(X) ⊇ int(Nef(X)).

Proof of Kleinmann’s criterion. We may assume that X is an integral projectivescheme. The ⇐ direction is immediate. To prove the other direction, since thecriterion is a closed condition, we may assume D ∈ DivQ(X). Moreover, byinduction, we may assume that DdimV [V ] ≥ 0 for all V strictly contained in X,and we have to show that DdimX ≥ 0. Suppose not, and DdimX < 0.

Fix a very ample Cartier divisor H, and consider the polynomial

P (t) = (D + tH)dimX = DdimX +

dimX−1∑i=1

ti(

dimX

i

)HiDdimX−i

+ tdimXHdimX .

The first term is negative; the last term is positive; and the other terms arenon-negative by induction since H is very ample.

Then on R>0, this polynomial is increasing. So there exists a unique t suchthat P (t) = 0. Let t be the root. Then P (t) = 0. We can also write

P (t) = (D + tH) · (D + tH)dimX−1 = R(t) + tQ(t),

33


where

R(t) = D · (D + tH)dimX−1, Q(t) = H · (D + tH)dimX−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) =

dimX−1∑i=0

ti(

dimX − 1

i

)Hi+1DdimX−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)dimX−1.

Note that so far, we haven’t used the assumption that D is nef. If t > t, then(D + tH)dimX > 0, and (D + tH)dimV [V ] > 0 for a proper integral subvariety,by induction (and binomial expansion). Then by Nakai’s criterion, D + tH isample. So the intersection (D + tH)dimX−1 is essentially a curve. So we aredone 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

Z1(X)K =∑

aiCi : ai ∈ K,Ci ⊆ X integral proper curves.

We then have a pairing

N1K(X)× Z1(X)K → K

(D,E) 7→ D · E.

We know that N1K(X) is finite-dimensional, but Z1(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 C1, C2 ∈Z1(X)K . Then C1 ≡ C2 iff

D · C1 = D · C2

for all D ∈ N1K(X).

Definition (N1(X)K). We define N1(X)K to be the K-module of K 1-cyclesmodulo numerical equivalence.

Thus we get a pairing N1(X)K ×N1(X)K → K.

Definition (Effective curves). We define the cone of effective curves to be

NE(X) = γ ∈ N1(X)R : γ ≡∑

[aiCi] : ai > 0, Ci ⊆ X integral curves.

This cone detects nef divisors, since by definition, an R-divisor D is nef iffD · γ ≥ 0 for all γ ∈ NE(X). In general, we can define

34


Definition (Dual cone). Let V be a finite-dimensional vector space over R andV ∗ 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 ⊆CaDivR(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) S1 ∩ NE(X) ⊆ S1 ∩ D>0, where S1 ⊆ N1(X)R is the unit sphere undersome 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 ∈ N1R(X). Assume that

H is ample. Then D is ample iff there exists ε > 0 such that

D · CH · 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

N1(X)R = N1(X)R,

since dimX = 2 and X is smooth (so that we can identify Weil and Cartierdivisors). We certainly have

Nef(X) ⊆ NE(X).

This can be proper. For example, if C ⊆ X is such that C is irreducible withC2 < 0, then C is not nef.

What happens when we have such a curve? Suppose γ =∑aiCi ∈ NE(X)

and γ · C < 0. Then C must be one of the Ci, since the intersection with anyother curve is non-negative. So we can write

γ = aCC +

k∑i=2

aiCi,

35


where aC , ai > 0 and Ci 6= C for all i ≥ 2.So for any element in γ ∈ NE(X), we can write

γ = λC + γ′,

where λ > 0 and γ′ = NE(X)C≥0, i.e. γ′ · 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 theelements of NE(X)C≥0.

Thus, [C] ∈ N1(X)R is an extremal ray of NE(X). In other words, if

λ[C] = µ1γ1 + µ2γ2

where µi > 0 and γi ∈ NE(X), then γi is a multiple of [C].

In fact, in general, we have

Theorem (Cone theorem). Let X be a smooth projective variety over C. Thenthere exists rational curves Cii∈I such that

NE(X) = NEKX≥0 +∑i∈I

R+[Ci]

where NEKX≥0 = γ ∈ NE(X) : KX · γ ≥ 0. Further, we need at mostcountably many Ci’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. Wethen get a rational map

ϕ|L| : X 99K P(H0(L)).

More generally, for each m > 0, we can form L⊗m, and get a rational mapϕL⊗m . What can we say about the limiting behaviour as m→∞?

Theorem (Iitaka). Let X be a normal projective variety and L a line bundleon X. Suppose there is an m such that |L⊗m| 6= 0. Then there exists X∞, Y∞,

36


a map ψ∞ : X∞ → Y∞ and a birational map U∞ : X∞ 99K X such that forK 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 ψK stabilize.

Definition (Kodaira dimension). The Kodaira dimension of L is

K(X,L) =

−∞ h0(X,L⊗m) = 0 for all m > 0

dim(Y∞) otherwise.

This is a very fundamental invariant for understanding L. For example, ifK(X,L) = K ≥ 0, then there exists C1, C2 ∈ R>0 such that

C2mK ≤ h0(X,L⊗m) ≤ C1m

K

for all m. In other words, h0(X,L⊗m) ∼ mK .

37

Index III Positivity in Algebraic Geometry

Index

(−1) curve, 22KX , 16N1(X)K , 34Z1(X)R, 34Num0, 17≡, 34|D|, 13OX(D), 9NE(X), 34div(f), 8ρ(X), 181-cycle, 34

adjunction formula, 16ample, 13ample sheaf, 6Asymptotic Riemann–Roch, 27

base ideal, 13base locus, 13Bertini’s theorem, 14blow up, 19

canonical divisor, 16Cartier divisor, 9Castelnuovo’s contractibility

criterion, 22class group, 8complete linear system, 13

dual cone, 35

effective curves, 34effective divisor, 8exceptional curve, 19

factorial, 10free, 13

generating section, 3

Hartog’s lemma, 8

Hodge index theorem, 18

intersection product, 15

Kleinmann’s criterion, 33Kodaira dimension, 37

linear system, 13linearly equivalent, 8locally principal, 9

minimal, 22

Neron–Severi group, 18Nakai’s criterion, 29nef divisor, 32Neron–Severi group, 26numerical equivalence, 17, 26, 34numerically effective divisor, 32

Picard group, 10Picard number, 27prime divisor, 8principal divisor, 8

regular in codimension 1, 8relatively minimal, 22Riemann–Roch for surfaces, 16Riemann–Roch theorem, 14

separate points, 4separate tangent vectors, 4sheaf

very ample, 4strict transform, 20support, 8

very ample, 13very ample sheaf, 4

Weil K-divisor, 8Weil divisor, 8

38

part iii - positivity in algebraic geometry · part iii | positivity in algebraic geometry based on...

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