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Home , Ample line bundle, Divisor, Intersection number, Picard group, Projective variety

I, FALL 2016.

DIVISORS.

1. Weil and Cartier divisors Let X be an . Define a Weil divisor on X as a formal (finite) of irreducible subvarieties of 1 with integral coefficients: m P D = aiDi, where ai ∈ Z. Define a Cartier divisor on X as a collection of rational i=1 S functions fi on Ui, where X = Ui is an open cover, such that fi/fj has no zeroes or poles on Ui ∩ Uj, up to the following equivalence : we say that {(Ui,Fi)} is equivalent to {(Wj, gj)} if the covers {Ui} and {Wj} have a common refinemenent

{Zk}, and if Zk ⊂ Ui, Zk ⊂ Wj, then (fi|Zk )/(gj|Zk ) has neither zeroes nor poles on Zk. Definition 1. A principal divisor is a Cartier divisor that is defined by a single rational function f on X.

On a reasonably good variety (such as a smooth variety) we can define the of a rational function f along a subvariety Y ⊂ X of codimension 1. If X is the affine line k1, Y is a point, and f = P/Q is a rational function, the order of f at (along) Y is k if P has a zero at Y with multiplicity k, and −k if Q has a zero at Y with multiplicity k. Once we have this notion, we can define a Weil divisor for any principal Cartier P divisor by taking the sum aiXi over all subvarieties Xi ⊂ X of codimension 1 where ai is the order of the function f along Xi. This sum will be finite. Then we can define a Weil divisor for any Cartier divisor {(Ui, fi)} by taking the above sum on every Ui and noticing that these Weil divisors agree on the intersections Ui ∩ Uj. On smooth varieties, Weil divisors are in bijection with Cartier divisors. On singular varieties, there may be Weil divisors that cannot be given as Cartier divisors, or non-trivial Cartier divisors for which the operation above produces a zero Weil divisor. Weil divisors naturally form an abelian (we just add the linear combinations formally). Cartier divisors also form an : for two Cartier divisors we can take a common refinement of the open covers and take the product of the functions defining the divisors. It is easy to see that the map from Cartier to Weil divisors described above is a group homomorphism.

Definition 2. Two divisors (Weil or Cartier) are called linearly equivalent if their difference is a principal divisor. The group of all Cartier divisors on X modulo linear equivalence is called the of X and denoted by Pic(X). 1 2 DIVISORS.

2. Cartier divisors and line bundles The data of a Cartier divisor in fact consists of the data of a with a rational section: having {(Ui, fi)} take the bundle that is trivial over each Ui, and define the transition functions as gij = fj/fi (regular function on Ui ∩Uj). Then {fi} define a rational section. A sum of two divisors corresponds to the of their respective ine bundles. Lemma 1. Two Cartier divisors are linearly equivalent if and only if the line bundles that correspond to them are isomorphic.

Proof. Suppose we have two sections {(Ui, si)} and {(Wj, tj)} of the same line bundle. By choosing a common refinement, we can assume that the coverings {Ui} and {Wi} are the same. Then on each open set si/ti defines a rational function, and these functions coincide on the intersections Ui∩Uj, so they define a global rational function that gives the linear equivalence between the corresponding Cartier divisors. On the other hand, if two Cartier divisors are linearly equivalent, we can again assume that they are described by functions on the same covering: {(Ui, fi)} and {(Ui, gi)} with fi = fgi for some rational function f on X. Then the corresponding line bundles have transition functions gi/gj and fi/fj = fgi/fgj = gi/gj, which gives the same line bundle.  As a corollary, we get an alternative definition of Pic(X) as the group of line bundles on X with tensor product as the operation. n Example 1. We have established that every line bundle on P is of the form O(k) for some k ∈ Z. Since O(k) ⊗ O(m) 'O(k + m), we have an isomorphism of abelian n groups Pic(P ) ' Z. 3. Linear systems P Definition 3. A divisor D = aiDi on an algebraic variety X is effective if ai > 0 for all i. Let X be a smooth variety, and let D be an effective divisor on it. Then D is the divisor of a regular section s of the line bundle O(D). Any other effective divisor D0 that is linearly equivalent to D is the divisor of another section s0 of O(D). These sections are defined uniquely up to multiplication by a non-zero scalar, and each section of O(D) defines an effective divisor. The space Γ(O(D)) of all global regular sections of O(D) is a , so effective divisors that are linearly equivalent to D are in bijection with its projectivization P(Γ(O(D))). Definition 4. A complete linear system |D| of divisors on X is the set of all effective divisors that are linearly equivalent to a given divisor D, viewed as a P(Γ(O(D))) (the divisor D itself does not have to be effective). A linear system of divisors is a projective subspace of a complete linear system, i.e. a projectivization of a linear subspace in Γ(O(D)) for some D. n Examples of complete linear systems are the space of all hyperplanes in P , the 2 space of all curves on P , etc. An example of a linear system that is not a 2 complete linear system would be all quadric curves on P passing through a given point. ALGEBRAIC GEOMETRY I, FALL 2016. 3

Definition 5. The base locus of a linear system of divisors is the set of points such that all of the divisors in the linear system pass through them. We say that the line bundle O(D) is generated by global sections if for any point x ∈ X there is a section s ∈ Γ(O(D) such that s(x) 6= 0; equivalently, O(D) is generated by global sections if the base locus of the linear system |D| is empty.

A finite collection of sections s0, . . . , sN of a line bundle L on X such that for N every x ∈ X there is i such that si(x) 6= 0 determines a map X → P . Indeed, let {Ui} be a covering of X such that L is trivial on every Ui. Each sk is given by a collection of functions fik on Ui. Then to each x ∈ Ui we can associate the N point (fi0(x): ... : fiN (x)) in P . This point is well-defined because all these values cannot be zero simutaneously, and it does not depend on i because for every k the values fik(x) and fjk(x) differ by the same scalar factor gij(x), where gij are N transition functions for L. The space P here is naturally the to the projectivization of the space spanned by si themselves. In the case where D is a divisor such that O(D) is generated by global sections ∨ and dim Γ(O(D)) < ∞, we get a map X → P(Γ(O(D))) . 1 Example 2. If D is a divisor of degree d on P , then the map defined by |D| is the 1 vd d Veronese embedding P −→ P . n Example 3. If X ⊂ P and D is the intersection of X and a hyperplane (equiv- alently, O(D) is the restriction of O(1) to X), then the map defined by |D| is the n embedding X → P that we started with. π n n Example 4. If X −→ P is the blow-up of a point p in P and D is the preimage of n a hyperplane in P that does not contain p, the map defined by |D| is π.

4. Ample and very ample divisors So far we have established that any linear system with empty base locus defines a map from X to a projective space. This map is not always injective (see example of the blow-up above), but we are especially interested in the divisors whose complete linear system does define an injective map. ∨ Definition 6. A divisor D on X is very ample if the map X → P(Γ(O(D))) is an embedding. Equivalently, D is very ample if the line bundle O(D) is isomorphic to N N the restriction of the line bundle O(1) from P to X for some embedding X ⊂ P . A divisor D is ample, if mD is very ample for some m > 0. n Example 5. For X = P and a line bundle L'O(k) the following are equivalent: L is ample, L is very ample, L is generated by global sections, k > 0. Example 6. On a smooth projective curve X, a divisor D is ample if and only if its degree is positive. This follows from the Riemann-Roch Theorem. We will not give the proof for the following theorem and proposition. Theorem 1. (Serre) Let L be a very on a smooth X. Then for any line bundle F , there is n0 > 0 such that for every n > n0 the bundle F ⊗ L⊗n is generated by a finite of global sections. 4 DIVISORS.

Note: this theorem holds for any coherent F , of which a line bundle F is a special case. We will only need this theorem for line bundles now. Proposition 1. (Hartshorne, Exercise 7.5) Let E and F be line bundles. (1) If E and F are both ample, then E ⊗ F is ample. (2) If E is ample and F is generated by global sections, then E ⊗ F is ample. (3) If E is very ample and F is generated by global sections, then E ⊗ F is very ample. ⊗n (4) If E is ample, then there is n0 such that F ⊗ E is very ample for n > n0.

5. Intersection of curves on a smooth Let C and D be two curves on a smooth projective surface X that intersect transversally, meaning that at every intersection point p ∈ C ∩ D both C and D are smooth at p, and the tangent spaces TpC ⊂ TpX and TpD ⊂ TpX are distinct. Then the of C and D is simply the number of points in C ∩ D. This notion can be generalized to the situations when C and D do not intersect transversally. Let Div(X) denote the abelian group of all divisors on X. Then Pic(X) = Div(X)/∼, where ∼ denotes linear equivalence.

Theorem 2. There exists a unique pairing Div(X)×Div(X) → Z, denoted by C ·D for C,D ∈ Div(X), such that (1) For smooth C and D intersecting transversally, C · D = |C ∩ D|; (2) C · D = D · C; (3) (C1 + C2) · D = C1 · D + C2 · D; (4) If C1 ∼ C2, then C1 · D = C2 · D. The strategy to prove this theorem is to show that C and D are always lin- early equivalent to some divisors C0 and D0 (not necessarily effective) with smooth components such that the components of C0 and the components of D0 intersect transversally. To do that, we will need Proposition 1 and the following theorem: n Theorem 3. (Bertini) Let X ⊂ P be a smooth projective variety over an alge- n braically closed field. Then there is a hyperplane H ⊂ P such that H does not contain X and H ∩ X is smooth. Moreover, such hyperplanes H form a dense subset n ∨ n in the projective space (P ) of all hyperplanes in P .

Lemma 2. Let C1,...,Cr be smooth curves on X, and let L be a very ample line bundle. Then we can choose a smooth curve in |L| that intersects each Ci transversally away from the points Ci ∩ Cj. N Proof. Use the linear system |L| to embed X into P . Then the divisors in |L| are hyperplane sections of X. Then the sets of hyperplanes that intersect X along smooth curves, hyperplanes that do not contain certain points, and hyperplanes that N N intersect Ci ⊂ P transversally are all dense in the set of all hyperplanes in P , so they have a non-empty (and, moreover, dense) intersection.  Now let H be an ample divisor on X. By Proposition 1 there is n such that both nH and C + nH are very ample. Then by Bertini’s Theorem we can choose smooth ALGEBRAIC GEOMETRY I, FALL 2016. 5 curves C1 ∈ |C + nH| and C2 ∈ |nH|. Then C ∼ C1 − C2. By Lemma 2 we can choose smooth curves D1 ∈ |D + mH| and D2 ∈ |mH| that intersect C1 and C2 transversally. Then D ∼ D1 − D2, and the divisors C1 − C2 and D1 − D2 intersect transversally, so we can define

C · D = (C1 − C2) · (D1 − D2) = |C1 ∩ D1| + |C2 ∩ D2| − |C1 ∩ D2| − |C2 ∩ D1|. 0 0 This definition is consistent, because for a different representation C ∼ C1 − C2 and 0 0 0 0 0 D ∼ D1 −D2 we’ll have C1 ·(D1 −D2) = C1 ·(D1 −D2) as long as Di, Di all intersect 0 0 C1 transversally, because D1 − D2 ∼ D1 − D2, so their difference will be a principal divisor, which on a smooth projective curve C1 has equal number of zeroes and poles. 0 0 0 0 Similarly, C2·(D1−D2) = C2·(D1−D2), so (C1−C2)·(D1−D2) = (C1−C2)·(D1−D2). 0 0 0 0 0 0 Likewise, (C1 − C2) · (D1 − D2) = (C1 − C2) · (D1 − D2). If some of the intersections 0 between Ci and Dj are not transversal, we can add an intermediate step between 0 0 0 C1 − C2 and C1 − C2 that will be transversal both to Dj and Dj.  If C is smooth, the intersection number C · D can be defined as the degree of the restriction of the line bundle O(D) to C.

Example 7. We can compute C · C for a smooth curve C by restricting OX (C) to 2 C. Let X be a blow-up of the point (0 : 0 : 1) on P , and let C be the exceptional 2 divisor. Introduce the coordinates (x0 : x1 : x2) on P and (u : v) on C. Then X can be covered by the following open sets: U0 = {x0 6= 0}, U1 = {x1 6= 0}, U2u = {x2 6= 0, u 6= 0}, and U2v = {x2 6= 0, v 6= 0}. The equation for X in 2 P × C will be x0v = x1u. The Cartier divisor defining C can be taken to be 1 on U0 and U1 (since these sets do not intersect C); on U2u the coordinates are (x0/x2, x1/x2, v/u), the equation for X is (x0/x2)(v/u) = x1/x2, and the equations for C are x0/x2 = x1/x2 = 0, so we can take the function x0/x2. Likewise, on U2v we take the function x1/x2. Then on U2u ∩ U2v the transition function for O(C) will x1/x2 x1 v be = = since x0v = x1u on X. Then the restriction of O(C) to C has x0/x2 x0 u transition function v/u (from the chart u 6= 0 to the chart v 6= 0), and is isomorphic to O(−1). Hence C · C = −1.