the complete linear series of D. Notice that D = PH 0 (X; O X (D)). Given any subvectorspace V H 0 (X; O X (D)) there is a rational map given by V : X


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1 2. Preliminaries 2.1. Divisors and line bundles. Let X be an irreducible complex variety of dimension n. The group of kcycles on X is Z k (X) = fz linear combinations of subvarieties of dimension kg: We will often denote Z k (X) by Z n k (X). The group of Weil divisors is just WDiv(X) = Z n 1 (X): The group of Cartier divisors is Div(X) := (X; K X=O X): Here K X denotes the sheaf of nonzero rational functions on X and O X is the sgeaf of nonvanishing regular functions. Therefore, a Cartier divisor D can be given by specifying an open cover U i of X and sections f i 2 (U i ; KX(U i )) such that f i =f j 2 (U i \ U j ; OX(U i \ U j )). Given divisors D; D 0 2 Div(X), then by denition D +D 0 = fu i ; f i fig. 0 There is a natural homomorphism of groups Div(X)! WDiv(X) given by D! [D] = P ord V (D) [V ] where ord V (D) is the order of vanishing of the f i along the codimension 1 subvariety. This homomorphism is injective when X is normal and an isomorphism when X is smooth. Exercise 2.1. Let X = Spec k[x; y; z]=(xy z 2 ) be the quadric cone given by the equation xy z 2. Show that WDiv(X) 6= Div(X). (In fact WDiv(X) = Z=2Z and Div(X) = 0. There is a short exact sequence 0! O X! K X! K X=O X! 0: Therefore, there is a natural homomorphism Div(X) = H 0 (X; K X=O X)! H 1 (X; O X) = Pic(X) which sends a divisor D = fu i ; f i g to the line bundle O X (D) = fu i \ U j ; f i =f j g. The kernel of this map is the group of principal divisors Princ(X) Div(X): Two divisors D; D 0 are linearly equivalent D D 0 if D D 0 2 Princ(X). We denote by jdj := fd 0 jd 0 Dg 4
2 the complete linear series of D. Notice that D = PH 0 (X; O X (D)). Given any subvectorspace V H 0 (X; O X (D)) there is a rational map given by V : X 99K PV = jv j x! [s 0 (x) : s 1 (x) : ::: : s n (x)] where s 0 ; :::; s n is a basis of V. This map is only dened away from the base locus of V and it sends x to the hyperplane of divisors containing x. We say that D is very ample if D is an embedding. D is ample if there is a positive integer m such that md is ample Rules of coherent cohomology. Let X be a variety (over an algebraically closed eld k), and F a quasicoherent sheaf 1, then there exist kvector spaces H i (X; F) such that: (1) For any homomorphism of O X modules a : F! G, there is an induced klinear map a : H i (X; F)! H i (X; G); (2) H 0 (X; F) = (X; F); (3) If 0! F 0! F! F 00! 0 is a short exact sequence, of quasicoherent sheaves on X, then there is a coboundary map d : H i (X; F 00 )! H i+1 (X; F 0 ) shuch that the sequence H i (X; F 0 )! H i (X; F)! H i (X; F 00 )! H i+1 (X; F 0 ) is exact; (4) H i (X; F) = 0 for all i > dim X; (5) If F is coherent 2, X is proper then H i (X; F) is nite dimensional and we denote its dimension by h i (X; F). (6) h i (P r ; O P r(n)) = 0 for all n 2 Z and all 1 i r 1; h 0 (P r ; O P r(n)) = r+n n if n 0 and 0 if n < 0; hr (P n ; O P r( n r 1)) = h 0 (P n ; O P r(n)). (cf. [Hartshorne] III.5.1. Recall that the Euler characteristic X of a cohrent sheaf F is (X; F) = h i (X; F): i0 1 i.e. it is locally isomorphic to the cokernel of a homomorphism of locally free sheaves 2 i.e. it is locally isomorphic to the cokernel of a homomorphism of locally free sheaves of nite rank 5
3 Theorem 2.2 (Asymptotic RiemannRoch). Let X be an irreducible projective variety of dimension n, and D 2 Div(X). Then (X; F O X (md)) is a polynomial of degree at most n in m such that (X; F O X (md)) = rank(f) Dn n mn + O(m n 1 ): Proof. See HirzebruchRiemannRoch: (X; F O X (md)) = ch(f O X (md)) T d(x): 2.3. Serre's Theorems. Recall that a coherent sheaf F is globally generated if the homomorphism is surjective. H 0 (X; F) O X! F Theorem 2.3 (Serre). Let X be a projective scheme and D be a very ample line bundle and F a coherent sheaf. Then there is an integer n 0 > 0 such that for all n n 0, F O X (nd) is globally generated. Proof. [Hartshorne] II x5. The idea is that we may assume that X = P N. Since F is coherent, it is locally generated by nitely many sections of O. Each local section is the restriction of some global section of O P N (n) for n 0. By compactness, we only need nitely many such sections. Theorem 2.4 (Serre Vanishing). Let X be a projective scheme and D be a very ample line bundle and F a coherent sheaf. Then there is an integer n 0 > 0 such that for all n n 0 and all i > 0, H i (X; F O X (nd)) = 0: Proof. (cf. [Hartshorne] III.5.2) We may assume that X = P N (replace F by D F ). The Theorem is clear if F is a nite direct sum of sheaves of the form O P N (q). We can nd a short exact sequence of coherent sheaves 0! K! O P N (q i )! F! 0 (eg. use (2.3)). We now consider the exact sequence H i (O P N (q i + n))! H i (F O P N (n))! H i+1 (K O P N (n)) and proceed by descending induction on i so that we may assume that h i+1 (K O P N (n)) = 0 for n 0 and i 0. Since h i (O P N (q i + n)) = 0 for n 0 and i 0, we have h i (F O P N (n)) = 0 as required for n 0 and i 0. Recall that we also have the following 6
4 Proposition 2.5. The following are equivalent (1) D is ample; (2) md is ample for all m > 0; (3) md is ample for some m > 0; (4) for any coherent sheaf F, there exists an integer m > 0 such that F O X (nd) is globally generated. Proof. Excercise. (See [Hartshorne] II x7). P ni 2.4. The cone of eective 1cycles. For any D 2 Div(X) and C = C i 2 Z 1 (X) we dene D C = X n i (D C i ) by considering the normalization i : ~C i! C i and letting D C i = deg( i O X (D)): Two Cartier divisors D; D 0 are numerically equivalent D D 0 if D C = D 0 C 8 C irreducible curves on X: Two curves C; C 0 are numerically equivalent C C 0 if C D = C 0 D 8 D 2 Div(X): Remark 2.6. Recall that there is a short exact sequence inducing a short exact sequence 0! Z! O X! O X! 0 Pic 0 (X)! Pic(X)! NS(X)! 0: here Pic 0 (X) = H 1 (X; O X )=H 1 (X; Z) is the group of topologically trivial line bundles and the Neron Severi group is NS(X) = Imm(H 1 (X; O X)! H 2 (X; Z)): Since the group Div(X)= is a quotient of NS(X), it is a free abelian group of nite rank which we denote by (X) the Picard number of X. We will adopt the following notation: N 1 (X) = (Z 1 (X)= ) R; N 1 (X) = (Z 1 (X)= ) R; so that (X) = dim R N 1 (X) = dim R N 1 (X): We dene the cone of eective 1cycles to be NE(X) N 1 (X) 7
5 to be the cone generated by f P n i C i s: t: n i 0g. Given a morphism f : X! Y and an irreducible curve C X, we let f (C) = df(c) where d = deg(c! f(c)). If f(c) is a point, then we set f C = 0. One sees that f D C = D f C 8D 2 Div(Y ): Extending by linearity, we get an injective homomorphism and surjective homomorphisms f : N 1 (Y )! N 1 (X) f : N 1 (X)! N 1 (Y ); NE(X)! NE(Y ): Let W be a subvariety of X of codimension i, and D 2 Div(X) then we may also dene the intersection numbrs D i W 2 Z. Suppose that this has been done for all j > i, then pick a very ample divisor H such that D + H is also very ample. Then (D + H) i W is just the degree of the image of W via H 0 (X;O X (D+H)) and D j H i j W have been previously dened (as the codimension of H i j W is j > i). Hence D i W is also dened. We have the following important results: Theorem 2.7. [NakaiMoishezon criterion] Let D be a Cartier divisor on a projective scheme X, then D is ample if and only if for any 0 i n 1 and any subvariety W of codimension i, one has D i W > 0. It is enough to assume that X is proper. Proof. (cf. [KM98] 1.37) We may assume that X is irreducible. Clearly, if D is ample, then D i W > 0. For the converse implication, we proceed by induction on n = dim X. When dim X = 1, the Theorem is obvious. So we may assume that Dj Z is ample for all for all proper closed subschemes Z ( X. Claim 1. h 0 (X; O X (D)) > 0 for some k > 0 (actually (D) = n). We choose a very ample divisor B such that D+B is very ample. Let A 2 jd + Bj be a general member. Consider the short exact sequence 0! O X (kd B)! O X (kd)! O B (kd)! 0: Since O B (kd) is ample, for all k 0 we have that h i (B; O B (kd)) = 0 for all i > 0, and so h i (O X (kd B)) = h i (O X (kd)) for i 2 and k 0: The same argument applied to the short exact sequence 0! O X (kd B)! O X ((k + 1)D)! O A ((k + 1)D)! 0 shows that h i (O X (kd B)) = h i (O X ((k + 1)D)) for i 2 and k 0: 8
6 Putting this toghether, we see that for i 2 and k 0, one has h i (O X (kd)) = h i (O X ((k + 1)D)) and so the number h i (O X (kd)) is constant. But then for k 0 h 0 (X; O X (kd)) h 0 (X; O X (kd)) h 1 (X; O X (kd)) = (X; O X (kd)) + (constant) = D n =n k n + O(k 1): Claim 2. O X (kd) is generated by global sections for some k > 0. Fix a nonzero section s 2 H 0 (X; O X (md)). Let S be the divisor dened by s. Consider the short exact sequence 0! O X ((k 1)mD)! O X (kmd)! O S (kmd)! 0: By induction O S (kmd) is generated by global sections and so it suf ces to show that for k 0, the homomorphism H 0 (X; O X (kmd))! H 0 (S; O S (kmd)) is surjective. Arguing as in Claim 1, one sees that h 1 (X; O X (kmd)) is a decreasing sequence and is hence eventually constant as required. Conclusion of the proof. kd : X! P N is a nite morphism for k 0. If infact C is a curve contracted by kd, then kd C = O P N (1) kd (C) = 0 which contradicts C D > 0. The Theorem now follows as kd = O kd PN (1) is ample (as it is the pullback of an ample line bundle via a nite map). We have the following useful corollary: Theorem 2.8. [Kleiman's Theorem] Let X be a proper variety, D a nef divisor. Then D dim Z Z 0 for all irreducible subvarieties Z X. Proof. (cf. [Lazarsfeld05] 1.4.9) We assume that X is projective (Chow's Lemma) and irreducible. When dim X = 1, the Theorem is clear. By induction on n = dim X, we may assume that D dim Z Z 0 8Z X irreducible of dim Z < n; and we must show that D n 0. Fix H an ample divisor and consider the polynomial P (t) := (D + th) n 2 Q[t]: We must show that P (0) 0. For 1 k n, the coecient of t k is D n k H k 0: Assume that P (0) < 0, then one sees that P (t) has a unique real root t 0 > 0. For any rational number t > t 0, one sees that (D + th) dim Z Z > 0 8Z X irreducible of dim Z n; 9
7 and so by (2.7), D + th is ample. We write P (t) = Q(t) + R(t) = D (D + th) n 1 + th (D + th) n 1 : As D+tH is ample for t > t 0, one has that (D+tH) n 1 is an eective 1 cycle, so Q(t) > 0 for all rational numbers t > t 0 and so Q(t 0 ) 0. One sees that all the coecients of R(t) are nonnegative and the coecient of t n is H n > 0. It follows that R(t 0 ) > 0 and so P (t 0 ) > 0 which is the required contradiction. Exercise 2.9. Let X be a projective variety, H an ample divisor on X. A divisor D on X is nef if and only if D + H is ample for all rational numbers > 0. Exercise Let X be a projective variety, H an ample divisor on X. A divisor D on X is ample if and only if there exists an > 0 such that D C H C for all irreducible curves C X. Theorem [Nakai's Criterion] Let D be a divisor on a projective scheme X. D is ample if and only if D Z > 0 for any Z 2 NE(X) f0g. Proof. Exercise. Theorem [Seshadri's Criterion] Let D be a divisor on a projective scheme X. D is ample if and only if there exists an > 0 such that C D mult x (C) for all x 2 C X. Proof. Assume that D is ample, then there exists an integer n > 0 such that nd is very ample, but then nd C mult x (C) for all x 2 C X. For the reverse implication, we proceed by induction. Therefore, we may assume that for any irreducible subvariety Z ( X, the divisor Dj Z is ample and so D dim Z Z > 0. By (2.7), it is enough to show that D n > 0. Let : X 0! X be the blow up of X at a smooth point. Then D E is nef. In fact, for any curve C 0 X 0 we either have C = (C 0 ) is a curve and then ( D E) C 0 = D C mult x C 0; 10
8 or (C 0 ) = x and then since C 0 E = P n 1 and O E (E) = O P n 1( 1) ( D E) C 0 = deg C 0 > 0: But then, by (2.8) ( D E) n = D n n 0 and this completes the proof. 11
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