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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 positivity 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.
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Contents
III Positivity in Algebraic Geometry
Contents 1 Divisors 1.1 Projective embeddings . . . . 1.2 Weil divisors . . . . . . . . . 1.3 Cartier divisors . . . . . . . . 1.4 Computations of class groups 1.5 Linear systems . . . . . . . .
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3 3 7 9 11 13
2 Surfaces 15 2.1 The intersection product . . . . . . . . . . . . . . . . . . . . . . . 15 2.2 Blow ups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3 Projective varieties 3.1 The intersection product 3.2 Ample divisors . . . . . 3.3 Nef divisors . . . . . . . 3.4 Kodaira dimension . . .
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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 Pn comes with a very special line bundle O(1), and every embedding X ,→ Pn gives pulls back to a corresponding sheaf L on X. The key property of O(1) is that it has global sections x0 , . . . , xn which generate O(1) as an OX -module. Definition (Generating section). Let X be a scheme, and F a sheaf of OX modules. Let s0 , . . . , sn ∈ H 0 (X, F) be sections. We say the sections generate F if the natural map n+1 M OX → F i=0
induced by the si is a surjective map of OX -modules. The generating sections are preserved under pullbacks. So if X is embedded into Pn , then it should have a corresponding line bundle generated by n + 1 global sections. More generally, if there is any map to Pn 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 → Pn is a morphism over A, then ϕ∗ OPn (1) is an invertible sheaf on X, generated by the sections ϕ∗ x0 , . . . , ϕ∗ xn ∈ H 0 (X, ϕ∗ OPn (1)). (ii) If L is an invertible sheaf on X, and if s0 , . . . , sn ∈ H 0 (X, L) which generate L, then there exists a unique morphism ϕ : X → Pn such that ϕ∗ O(1) ∼ =L and ϕ∗ xi = si . 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 x0 , . . . , 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∈ mp Lp }. This is a Zariski open set, and si is invertible on Xsi . Thus there is a dual ∨ ∨ ∨ ∼ section s∨ i ∈ L such that si ⊗ si ∈ L ⊗ L = OX is equal to 1. Define n the map Xsi → A by the map K[An ] → H 0 (Xsi , Osi ) yi 7→ sj ⊗ s∨ i . 3
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SinceSthe si generate, they cannot simultaneously vanish on a point. So X = 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 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 → Pn such that ϕ∗ OPn (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 ,→ Pn 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 Pn 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: ¯ and X a projective variety over K. Let L be an Proposition. Let K = K, invertible sheaf on X, and s0 , . . . , sn ∈ H 0 (X, L) generating sections. Write V = hs0 , . . . , sn i for the linear span. Then the associated map ϕ : X → Pn is a closed embedding iff 6 q ∈ X, there exists sp,q ∈ V such that (i) For every distinct closed points p = sp,q ∈ mp Lp but sp,q 6∈ mq Lq . (ii) For every closed point p ∈ X, the set {s ∈ V | s ∈ mp Lp } spans the vector space mp Lp /m2p Lp . 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 Hp,q in Pn passing through p but not q. The hyperplane Hp,q is the vanishing locus of a global section of O(1). Let sp,q ∈ V ⊆ H 0 (X, L) be the pullback of this section. Then sp,q ∈ mp Lp and sp,q 6∈ mq Lq . So (i) is satisfied. To see (ii), we restrict to the affine patch containing p, and X is a closed subvariety of An . The result is then clear since mp Lp /m2p Lp is exactly the ¯ to know what the closed points of Pn span of s0 , . . . , sn . We used K = K look 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 X X X sp,q = λi si = λi ϕ∗ xi = ϕ∗ λi xi forP some λi ∈ K. So we can take Hp,q to be given by the vanishing set λi xi , and so it is injective on closed point. It follows that it is also of 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 OPn → ϕ∗ OX 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 Lp is trivial, so mp Lp /m2p Lp ∼ = 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 ∼ Op,X ∼ = = K. mp,Pn mp,X Now since mp,Pn is generated by x0 , . . . , xn , the second condition is the same as saying mp,X mp,Pn → 2 mp,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 OX -module. Then there exists a positive integer n0 = n0 (F, L) such that for all n ≥ n0 , the twist F ⊗ Ln is generated by global sections. The proof of this is straightforward and will be omitted. The idea is that tensoring with Ln lets us clear denominators, and once we have cleared all the denominators of the (finitely many) generators of F, the resulting sheaf F ⊗ Ln will be generated by global sections. We can weaken the condition of very ampleness to only require the condition of this theorem to hold. 5
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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 OX -module F, there is an n0 such that for all n ≥ n0 , the sheaf F ⊗ Ln 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 OX , we can find some Ln that is generated by global sections. So at least it gives some map to Pn . 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 Lm is very ample. Proof. (⇐) Let Lm be very ample, and F a coherent sheaf. By Serre’s theorem, there exists 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. N We claim that there exists t1 , . . . , tn ∈ H 0 (X, S L ) such that L|Xti are all trivial (i.e. isomorphic to OXti ), and X = Xti . By compactness, it suffices to show that for each p ∈ X, there is some t ∈ H 0 (X, Ln ) (for some n) such that p ∈ Xti and L is trivial on Xti . 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 ∈ Xt ⊆ U and hence L is trivial on Xt . Vanishing on Y 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 by global sections. In particular, since IY ⊗ Ln doesn’t vanish at p, we can find some t ∈ Γ(X, IY ⊗ LN ) such that t 6∈ mp (IY ⊗ Ln )p . Since IY is a subsheaf 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 OXti as an A-algebra. Then for large n, cij = tni bij extends to a global section cij ∈ Γ(X, Ln ) (by clearing denominators). We can pick an n large enough to work for all bij . Then we use {tni , cij } as our generating sections N to construct a morphism to S P , and let {xi , xij } nbe the corresponding coordinates. Observe that Xti = X implies the ti already generate Ln . Now each xti gets mapped to Ui ⊆ PN , the vanishing set of xi . The map OUi → 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 6
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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) 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, 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 ⊗ Ln ) is a finitely-generated A-module. (ii) There exists n0 ∈ N such that for all n ≥ n0 , H i (F ⊗ Ln ) = 0 for all i > 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 an invertible 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 , we have H i (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 → mx F → F → Fx → 0. We twist by Ln , where n is sufficiently big, and take cohomology. Then we have a long exact sequence 0 → H 0 (mx F(n)) → H 0 (F(n)) → H 0 (Fx (n)) → H 1 (mx F(n)) = 0. In particular, the map H 0 (F(n)) → H 0 (Fx (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: 7
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Definition (Regular in codimension 1). Let X be a Noetherian scheme. We say X 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 is a codimension 1 integral subscheme, then the local ring OX,Y is a DVR. Then there is a valuation valY : OX,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 X D= ai Yi , where the ai ∈ Z and Yi 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 ai ∈ K. Definition (Effective divisor). We say a Weil divisor is effective if ai ≥ 0 for all i. We write D ≥ 0. Definition (Principal divisor). If f ∈ K(X), then we define the principal divisor X div(f ) = valY (f ) · Y. Y
One can show that valY (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 valY (f ) 6= 0 implies Y ⊆ X \ U . Since X is Noetherian, X \ U can only contain finitely many codimension 1 subscheme. P Definition (Support). The support of D = ai Yi is [ supp(D) = Yi . Observe that div(f g) = 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, D0 are linearly equivalent if D − D0 is principal, and we write D ∼ D0 . 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 some × V ≥ 2. Then f ∈ OX . Thus, div(f ) = 0 implies f ∈ OX .
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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 X div(s) = valY (s) · Y. Y
To make sense of valY (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 valY (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 OX . Corollary. If X is normal and proper, then there is a map div{rational sections of L}/K ∗ → WDiv(X). ∗ Proof. Properness implies OX = K ∗.
Example. Take X = P1K , and s = homogeneous coordinates. Then
X2 X+Y
∈ H 0 (O(1)), where X, Y are our
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 OX (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.
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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. 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 OX (D) is an invertible sheaf. Proof. If U ⊆ X is such that D|U = div(f )|U , then there is an isomorphism OX |U → OX (D)|U g 7→ g/f. Definition (Cartier divisor). A Cartier divisor is a locally principal Weil divisor. 9
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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 X is 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. 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 ID,x ⊆ OX,x is a height 1 prime ideal. So ID,x = (f ) for f ∈ mX,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 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 H 1 (X, OX ). Proof. Given f ∈ H 0 (U, OX (div(s))), map it to f · s ∈ H 0 (U, L). This gives the desired isomorphism. If we have to sections s0 6= s, then f = s0 /s ∈ K(X). So div(s) = div(s0 ) + div(f ), and div(f ) is principal. So this gives a well-defined map Pic(X) → Cl(X).
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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(A2 \ {0}) = Cl(A2 ). 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 ) = ID . 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(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 − z 2 ) ⊆ A3 . 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] ∼ K[x, x−1 , t, z] = K[x, x−1 , z], = (xy − z) (t − 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 K[x, y, y −1 , z] K[x, y, z] = = K[y, y −1 , z](z) . OX,Z = xy − z 2 (x,z) x − z2 (x,y) Unsurprisingly, this is a DVR, and crucially, the uniformizer is z. So we know that div(x) = 2Z. So we know that 2Z ⊆ 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 = A3 ,
T(0) (Z ⊆ X) = A1 .
But if Z were principal, then IZ,0 = (f ) for some f ∈ OX,0 . But then T0 (Z) ⊆ T0 X will be ker df . But then ker df 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 × 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 0 point of X. Explicitly, suppose ID is the ideal of D in K[X × A1 ], and let ID be the ideal of K(X)[t] generated b ID under the inclusion K[X × A1 ] = K[X][t] ⊆ K(X)[t]. 0 If ID = 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 in particular is of the form D0 × A1 . 0 If not, then ID = (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 D0 × A1 . So D is linearly equivalent to a sum of divisors of the form D0 × A1 .
Exercise. We have Cl(X × Pn ) = Z ⊕ Cl(X).
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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 D0 be a divisor on X. (i) For all s ∈ H 0 (X, OX (D)), div(s) is an effective divisor linearly equivalent to D. (ii) If D ∼ D0 and D ≥ 0, then there is s ∈ H 0 (OX (D0 )) such that div(s) = D (iii) If s, s0 ∈ H 0 (OX (D0 )) and div(s) = div(s0 ), then s0 = 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 ∈ H 0 (OX (D0 )). Then div(s) = D. 0 0 0 (iii) We have ss ∈ K(X)∗ . So div ss . So ss ∈ 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 D0 , written |D0 |. Thus, |D0 | is the projectivization of the vector space H 0 (X, O(D0 )). We also define Definition (Linear system). A linear system is a linear subspace of the projective space structure on |D0 |. 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 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 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 | = OX . 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 OX (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 X h0 (X, L ⊗ An ) = (−1)i hi (X, L ⊗ An ) = χ(X, L ⊗ An ). In the case of curves, Riemann–Roch lets us understand this number well: 13
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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 ∼ H1 − H2 for some very ample divisors Hi . We can in fact take Hi to be effective, and if X is smooth, then we can take Hi 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 H1 6= H2 , then H1 and H2 intersect transversely.
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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| = degC (OC (D)) = h0 (OC∩D ) = χ(OC∩D ), using that OC∩D is a skyscraper sheaf supported at C ∩ D. To understand this number, 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 intersection 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 divisor represented 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 is very ample. 15
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(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 · D0 = H1 · D0 − H2 · D0 by (a), and thus is linear in D0 . 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 = 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 intersect at a point. So `2 = 1. Thus, the intersection product on Pic(P2 ) is just ordinary multiplication. In particular, if D1 , D2 are curves that intersect transversely, we find that D1 · D2 = deg(D1 ) deg(D2 ). This is B´ezout’s theorem! Exercise. Let C1 , C2 ⊆ X be curves without common components. Then X C1 · C2 = `(OC1 ∩C2 ), p∈C1 ∩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∗1 O(1)] ⊕ Z[p∗2 O(1)]. Let A = p∗1 O(1), B = p∗2 O(1). Then we see that A2 = 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, OX (D)) =
D · (D − KX ) + χ(OX ), 2
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 a smooth curve. Then (OX (KX ) ⊗ OX (C))|C ∼ = OC (KC ). 16
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Proof. Let IC = OX (−C) be the ideal sheaf of C. We then have a short exact sequence on C: 0 → OX (−C)|C ∼ = IC /IC2 → Ω1X |C → Ω1C → 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 OX (KX ) = det(Ω1X ). Restricting to C, we have OX (KX )|C = det(Ω1X |C ) = det(OX (−C)|C ) ⊗ det(Ω1C ) = OX (C)|∨ C ⊗ OC (KC ). Proof of Riemann–Roch. We can assume D = H1 − H2 for very ample line bundles 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. We then 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 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, D0 are numerically equivalent, written D ≡ D0 , if D · E = D0 · E for all divisors E. We write Num0 = {D ∈ Div(X) : D ≡ 0}. 17
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Definition (N´eron–Severi group). The N´eron–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, 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 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 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 D2 < 0. Proof. Before we begin the proof, observe that if H 0 is very ample and D0 is (strictly) effective, then H 0 · D0 > 0, since this is given by the number of intersections between D0 and any hyperplane in the projective embedding given by H 0 . Now assume for contradiction that D2 ≥ 0. – If D2 > 0, fix an n such that Hn = D + nH is very ample. Then Hn · D > 0 by assumption. We use Riemann–Roch to learn that χ(X, OX (mD)) =
m2 D2 − mKX · D + χ(OX ). 2
We first consider the H 2 (OX (mD)) term in the left-hand side. By Serre duality, we have H 2 (OX (mD)) = H 0 (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 H 0 (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 > 0 since H is very ample and mD is effective. This is a contradiction. 18
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– If D2 = 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 Dn0 = nD + E 0 satisfies Dn0 · H = 0. On the other hand, (Dn0 )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” ¯ = Blp X together with a map p with a copy of P1 . The result is a surface X −1 ¯ ε : X → X such that ε (p) is a smoothly embedded copy of P1 , and ε is an isomorphism outside of this P1 . ¯ pick local coordinates x, y in a neighbourhood U of X ¯ such To construct X, 1 1 that V (x, y) = p. Work on U × P with coordinates [X : Y ] on P . We then set ¯ = V (xY − yX). U ¯ onto U , and this works, since above the There is a canonical projection of U point (0, 0), any choice of X, Y satisfies the equation, and everywhere else, we must have [X : Y ] = [x : y]. ¯ = Blp X that Definition (Blow up). The blow up of X at p is the variety X we just defined. The preimage of p is known as the exceptional curve, and is denoted E. ¯ in terms of Pic(X), and in particular We would like to understand Pic(X) ¯ the product structure on Pic(X). This will allow us to detect whether a surface is a blow up. ¯ is the localization sequence The first step to understanding Pic(X) ¯ → Pic(X ¯ \ P1 ) → 0. Z → Pic(X) We also have the isomorphism ¯ \ P1 ) = Pic(X \ {p}) ∼ Pic(X = Pic(X). ¯ and the push-pull Finally, pulling back along ε gives a map Pic(X) → Pic(X), formula says π∗ (π ∗ L) = L ⊗ π∗ (OX¯ ). Hartog’s lemma says π∗ OX¯ = OX . ¯ is the direct sum of Pic(X) with a quotient So in fact ε∗ is a splitting. So Pic(X) ¯ ∼ of Z, generated by [E]. We will show that in fact Pic(X) = ε∗ Pic(X) ⊕ Z[E]. 2 This will be shown by calculating E = −1.
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¯ called the In general, if C is a curve in X, then this lifts to a curve C˜ ⊆ X, 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 ˆX,p , the equation of C tangent to any branch of C at p. Then in the local ring O is given as f = fm (x, y) + higher order terms, where fm is a non-zero degree m polynomial. Then by definition, the multiplicity ¯ ⊆ (U × P1 ), we have the chart U × A1 where X 6= 0, with is m. Then on U 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) = xm fm (1, t) + higher order terms = xm · h(x, t), ¯x6=0 , the curve x = 0 is just E. So this equation with h(0, 0) 6= 0. But then on U has multiplicity m along E. ¯ = Proposition. Let X be a smooth projective surface, and x ∈ X. Let X π Blx X → 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) KX¯ = π ∗ (KX ) + E. (iv) π ∗ is defined on NS(X). Thus, ¯ = NS(X) ⊕ Z[E]. NS(X) Proof. (i) Recall we had the localization sequence ¯ → Pic(X) → 1 Z → Pic(X) 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.
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(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 (KX¯ + E) · E = deg(KP1 ) = −2. So we know that KX¯ · E = −1. Also, outside of E, KX¯ and KX agree. So we have KX¯ = π ∗ (KX ) + mE for some m ∈ Z. Then computing −1 = KX¯ · E = π ∗ (KX ) · E + mE 2 gives m = 1. ¯ But this (iv) We need to show that if D ∈ Num0 (X), then π ∗ (D) ∈ Num0 (X). 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 X0 q p
X
ϕ
Y
where p : X 0 → 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 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 + D0 for some fixed C. Let |V 0 | be the set of all such D0 . Then |V | and |V 0 | give the same rational map. Indeed, if V has basis f0 , . . . , fn , and g is the function defining C, then the two maps are, respectively, [f0 : · · · : fn ] and [f0 /g : · · · : fn /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. 21
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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 X1 = Blx X → 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. The procedure stops, because 0 ≤ D12 = (π1∗ D2 + m2 E 2 ) < D2 . 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 Blx (X). Theorem. Let g : Z → X be a birational morphism of surfaces. Then g factors g0
p
as Z → X 0 → 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 ¯ If there is a curve C ⊆ X such that C ∼ surface over K = K. = P1 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 ∼ = P1 (ii) C 2 = −1 (iii) KX · C = −1 When these are satisfied, we say C is a (−1) curve. 22
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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 → Pn , and then take Y to be the image. Note that f (C) = ∗ is the same as requiring OX (L)|C = OC . 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, 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 explicit basis of H 0 (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 → 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 OX (H + iC)|C = OP1 (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 (OP1 (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 (OP1 (K − i)). (i) (i) For each i > 0, pick a lift y0 , . . . , yK−i of a basis of H 0 (OP1 (K − i)) to 0 0 H (H + iC), and take the image in H (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 (i)
(i)
g K−i y0 , . . . , g K−i yK−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 Pr . 23
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For concreteness, list the basis vectors as x1 , . . . , xr , where xr is a lift of (K−1) (K−1) 1 ∈ OP1 to H 0 (H + KC), and xr−1 , xr−2 are gy0 , gy1 . 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 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 mY,p /m2Y,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 given by [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1]. There are three conics passing through x, y, z, given by x1 x2 , x0 , x2 and x0 x1 , which is the union of lines `0 , `1 , `2 , and let |Cx,y,z | be the linear system generated by these three conics. Then we obtain a birational map P2 99K P2 −1 −1 [x0 : x1 : x2 ] 7→ [x1 x2 : x0 x2 : x0 x1 ] = [x−1 0 : x1 : x2 ]
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,z P2 P2
C
P2
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 isomorphic to P1 . 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.
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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 (D1 , . . . , Dn ) 7→ (D1 · . . . · Dn ). 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 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 are Cartier divisors, then D1 · . . . · Dn−1 · [V ] = (D1 |V · . . . · Dn−1 |V ). In general, if V ⊆ X is an integral subvariety, we write D1 · . . . · Ddim V · [V ] = (D1 |V · . . . · Ddim V |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. Then multilinearity tells us how we should compute the intersection product, namely as X Y Y (−1)|I| Hi,2 · Hj,1 . i∈I
I⊆{1,...,n}
j6∈I
By Bertini, we can assume the Hi,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(OX ) + χ(OX (−C − D)) − χ(OX (−C)) − χ(OX (−D)). For general n, let H1 , . . . , Hn be very ample divisors. Then the last property implies we have H1 · · · · · Hn = (H2 |H1 · . . . · Hn |H1 ).
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So inductively, we have !! X
H1 · · · · · Hn =
(−1)|I| χ H1 , OH1
−
X
Hi
i∈I
I⊆{2,...,n}
But we also have the sequence 0 → OX (−H1 ) → OX → OH1 → 0. Twisting this sequence shows that X X H1 · · · · · Hn = (−1)|I| χ OX −
i∈I
Hi
I⊆{2,...,n}
X
=
X − χ OX −H1 − Hi X i∈I |J| (−1) χ OX − Hi i∈J
J⊆{1,...,n}
We can then adopt this as the definition of the intersection product !! X X |J| D1 · · · · · Dn = (−1) χ OX − Di . J⊆{1,...,n}
i∈J
Reversing the above procedure shows that property (iii) is satisfied. As before we can define Definition (Numerical equivalence). Let D, D0 ∈ CaDiv(X). We say D and D0 are numerically equivalent, written D ≡ D0 , if D · [C] = D0 · [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 D2 , . . . , Ddim X , we have D · D2 · . . . · Ddim X = 0. Proof. We induct on n. As usual, we can assume that the Di are very ample. Then D · D2 · · · · Ddim X = D|D2 · D3 |D2 · · · · · Ddim X |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, written NS(X) = N 1 (X), is N 1 (X) =
CaDiv(X) CaDiv(X) = . Num0 (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 Zn . 26
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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 ∗,an an 0 → Z → OX → OX → 0. This gives a long exact sequence ∗,an an H 1 (X, Z) → H 1 (X, OX ) → H 1 (X, OX ) → H 2 (X, Z).
It is not hard to see that H i (X, Z) is in fact the same as singular cohomology, ∗,an and H 1 (OX ) 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]·c1 (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 ¯ Let D be a Cartier divisor, and E a Weil divisor on X. Then over K = K. χ(X, OX (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 Dn n χ(X, OX (mD + E)) = m + lower order terms. n! 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 → OX (mD + E) → OX (mD + E + H) → OH 0 ((mD + E + H)|H 0 ) → 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 0 , OH 0 ((mD + E + H)|H )) = χ(X, OX ((m − 1)D + E)) + χ(H 0 , 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 0 , OH 0 (mD + E + H)). By induction (see proposition below) the right-hand side is a numerical polynomial of degree at most n − 1 with leading term Dn−1 · G − Dn−1 · H n−1 m + lower order terms, (n − 1)! 27
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since Dn−1 · G is just the (n − 1) self-intersection of D along G. But G − H = D. So the LHS is Dn mn−1 + lower order terms, (n − 1)! 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 h0 (X, mH) =
Hn n m + lower order terms for m 0. n!
(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, H i (Ox (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 h0 (mD) ≤ h0 (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 Ri f∗ F = 0 for all i > 0. Then H i (F ⊗ f ∗ OX (mD)) = H i (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 Proposition. Let X be a proper scheme, and L an invertible sheaf. Then L is ample iff L|Xred is ample. Proof. (⇒) If L induces a closed embedding of X, then map given by L|Xred is given by the composition with the closed embedding Xred ,→ X. 28
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(⇐) Let J ⊆ OX 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 → Gj → 0. This Gi is really a sheaf on the reduced structure, since J acts trivially. Thus H i (Gj ⊗ Lm ) 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 ⊗ Lm ) = 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 = Ddim 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 D1 , D2 , if D1 ≡ D2 , then D1 is ample iff D2 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 0
(f ∗ D)dim V [V 0 ] = deg f |V 0 Ddim 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 OX (D) globally generated, and Φ : X → P(H 0 (X, OX (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.
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(⇒) 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 Ddim V · [V ] = (D|V )dim V . But this is just degΦ(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 OX (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 OP(|mD|) (1) along a finite map, hence is ample. We first reduce to the case where D is effective. As usual, write D ∼ 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|Hi 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 − H1 ) = H j ((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 of itself, Dn > 0, and so asymptotic Riemann–Roch tells us h0 (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 generated 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 (OX (mD)) → H 0 (OD (mD)) is surjective. To show this, we use the short exact sequence 0 → OX ((m − 1)D) → OX (mD) → OD (mD) → 0. 30
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For i > 0 and large m, we know H i (OD (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 (OX (mD)) → H 0 (OD (mD)) is a surjection by exactness. Recall that we defined PicK (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 1 NK (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 . P (ii) D = ci Di , where ci ∈ Q>0 and Di are ample Cartier divisors. (iii) D satisfies Nakai’s criterion. That is, Ddim 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) ⊆ NQ1 (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 X dim V V −p p dim V −p (λ1 H1 + λ2 H2 )dim V [V ] = λp1 λdim H · H [V ] 2 1 2 p 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 E1 , . . . Er are Cartier divisors, then for all |εi | 1, the divisor D + εi Ei is still ample. Proof. By induction, it suffices to check it in the case n = 1. Take m ∈ N such 1 that mD ± E1 is still ample. This is the same as saying D ± m E1 is still ample. 1 Then for |ε1 | < m , we can write 1 D + εE1 = (1 − q)D + q D + E1 m for some q < 1. 31
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Similarly, if D ∈ CaDivR (X), then we say D is ample iff X D= ai Di , 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 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 t1 < t < t2 with ti ∈ Q, and then find λ, µ > 0 such that λ(D1 + t1 E) + µ(D1 + t2 E) = D1 + tE. Then we are done by checking Nakai’s criterion. Similarly, we have Proposition. Let H be an ample PR-divisor. Then for all R-divisors E1 , . . . , Er , for all kεi k ≤ 1, the divisor H + εi Ei 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|Xred is nef. (ii) D is nef iff D|Xi is 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’s criterion. Theorem (Kleinmann’s criterion). Let X be a proper scheme, and D an RCartier divisor. Then D is nef iff Ddim 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.
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(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 X dim V p dim V −p p dim V (D + εH) · [V ] = ε D H [V ] > 0. p 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 ∈ DivQ (X). Moreover, by induction, we may assume that Ddim V [V ] ≥ 0 for all V strictly contained in X, and we have to show that Ddim X ≥ 0. Suppose not, and Ddim X < 0. Fix a very ample Cartier divisor H, and consider the polynomial dim X
P (t) = (D + tH)
=D
dim X
+
dim X−1 X i=1
i
t
dim X H i Ddim X−i i + tdim 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
dim X − 1 t H i+1 Ddim X−i , i i
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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 nX o Z1 (X)K = ai Ci : ai ∈ K, Ci ⊆ X integral proper curves . We then have a pairing 1 NK (X) × Z1 (X)K → K
(D, E) 7→ D · E. 1 We know that NK (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 1 for all D ∈ NK (X).
Definition (N1 (X)K ). We define N1 (X)K to be the K-module of K 1-cycles modulo numerical equivalence. Thus we get a pairing N 1 (X)K × N1 (X)K → K. Definition (Effective curves). We define the cone of effective curves to be X NE(X) = {γ ∈ N1 (X)R : γ ≡ [ai Ci ] : ai > 0, Ci ⊆ 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)∨ . ¯ So It is also easy to see that C ∨∨ = C. Proposition. N E(X) = Nef(X)∨ . 34
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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|N E(X) > 0, i.e. D · γ > 0 for all γ ∈ N E(X). (iii) S1 ∩ N E(X) ⊆ S1 ∩ D>0 , where S1 ⊆ N1 (X)R is the unit sphere under some choice of norm. Proof. – (1) ⇒ (2): Trivial. – (2) ⇒ (1): If D|N E(X) > 0, then D ∈ int(Nef(X)). – (2) ⇔ (3): Similar. Proposition. Let X be a projective scheme, and D, H ∈ NR1 (X). Assume that H is ample. Then D is ample iff there exists ε > 0 such that D·C ≥ ε. H ·C 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 = N1 (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. P ai Ci ∈ NE(X) What happens when we have such a curve? Suppose γ = and γ · C < 0. Then C must be one of the Ci , since the intersection with any other curve is non-negative. So we can write γ = aC C +
k X
ai Ci ,
i=2
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 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: 35
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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] ∈ N1 (X)R is an extremal ray of NE(X). In other words, if λ[C] = µ1 γ1 + µ2 γ2 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. Then there exists rational curves {Ci }i∈I such that X NE(X) = NEKX ≥0 + R+ [Ci ] i∈I
where NEKX ≥0 = {γ ∈ NE(X) : KX · γ ≥ 0}. Further, we need at most ⊥ countably many Ci ’s, and the accumulation points are all at KX .
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 ϕL⊗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
ϕ|L⊗k |
Im(ϕ|L⊗K )
U∞
X∞
ψ∞
Y∞
where the right-hand map is also birational. So birationally, the maps ψK stabilize.
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Definition (Kodaira dimension). The Kodaira dimension of L is ( −∞ h0 (X, L⊗m ) = 0 for all m > 0 K(X, L) = . dim(Y∞ ) otherwise This is a very fundamental invariant for understanding L. For example, if K(X, L) = K ≥ 0, then there exists C1 , C2 ∈ R>0 such that C2 mK ≤ h0 (X, L⊗m ) ≤ C1 mK for all m. In other words, h0 (X, L⊗m ) ∼ mK .
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Index
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Index (−1) curve, 22 KX , 16 N1 (X)K , 34 Z1 (X)R , 34 Num0 , 17 ≡, 34 |D|, 13 OX (D), 9 NE(X), 34 div(f ), 8 ρ(X), 18 1-cycle, 34
Hodge index theorem, 18 intersection product, 15 Kleinmann’s criterion, 32 Kodaira dimension, 37 linear system, 13 linearly equivalent, 8 locally principal, 9 minimal, 22 N´eron–Severi group, 18 Nakai’s criterion, 29 nef divisor, 32 Neron–Severi group, 26 numerical equivalence, 17, 26, 34 numerically effective divisor, 32
adjunction formula, 16 ample, 13 ample sheaf, 6 Asymptotic Riemann–Roch, 27 base ideal, 13 base locus, 13 Bertini’s theorem, 14 blow up, 19
Picard group, 10 Picard number, 27 prime divisor, 8 principal divisor, 8
canonical divisor, 16 Cartier divisor, 9 Castelnuovo’s contractibility criterion, 22 class group, 8 complete linear system, 13
regular in codimension 1, 8 relatively minimal, 22 Riemann–Roch for surfaces, 16 Riemann–Roch theorem, 14 separate points, 4 separate tangent vectors, 4 sheaf very ample, 4 strict transform, 20 support, 8
dual cone, 34 effective curves, 34 effective divisor, 8 exceptional curve, 19 factorial, 10 free, 13
very ample, 13 very ample sheaf, 4
generating section, 3 Weil K-divisor, 8 Weil divisor, 8
Hartog’s lemma, 8
38