ALGEBRAIC GEOMETRY I, FALL 2016.

ALGEBRAIC GEOMETRY I, FALL 2016. DIVISORS. 1. Weil and Cartier divisors Let X be an algebraic variety. Define a Weil divisor on X as a formal (finite) linear combination of irreducible subvarieties of codimension 1 with integral coefficients: D = m a i D i, where a i Z. Define a Cartier divisor on X as a collection of rational i=1 functions f i on U i, where X = U i is an open cover, such that f i /f j has no zeroes or poles on U i U j, up to the following equivalence relation: we say that {(U i, F i )} is equivalent to {(W j, g j )} if the covers {U i } and {W j } have a common refinemenent {Z k }, and if Z k U i, Z k W j, then (f i Zk )/(g j Zk ) has neither zeroes nor poles on Z k. 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 order of a rational function f along a subvariety Y X of codimension 1. If X is the affine line k 1, 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 divisor by taking the sum a i X i over all subvarieties X i X of codimension 1 where a i is the order of the function f along X i. This sum will be finite. Then we can define a Weil divisor for any Cartier divisor {(U i, f i )} by taking the above sum on every U i and noticing that these Weil divisors agree on the intersections U i U j. 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 group (we just add the linear combinations formally). Cartier divisors also form an abelian group: 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 Picard group 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 line bundle with a rational section: having {(U i, f i )} take the bundle that is trivial over each U i, and define the transition functions as g ij = f j /f i (regular function on U i U j ). Then {f i } define a rational section. A sum of two divisors corresponds to the tensor product 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 {(U i, s i )} and {(W j, t j )} of the same line bundle. By choosing a common refinement, we can assume that the coverings {U i } and {W i } are the same. Then on each open set s i /t i defines a rational function, and these functions coincide on the intersections U i U j, 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: {(U i, f i )} and {(U i, g i )} with f i = fg i for some rational function f on X. Then the corresponding line bundles have transition functions g i /g j and f i /f j = fg i /fg j = g i /g j, 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. Example 1. We have established that every line bundle on P n is of the form O(k) for some k Z. Since O(k) O(m) O(k + m), we have an isomorphism of abelian groups Pic(P n ) Z. 3. Linear systems Definition 3. A divisor D = a i D i on an algebraic variety X is effective if a i > 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 D that is linearly equivalent to D is the divisor of another section s 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 vector space, 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 projective space 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. Examples of complete linear systems are the space of all hyperplanes in P n, the space of all quadric curves on P 2, etc. An example of a linear system that is not a complete linear system would be all quadric curves on P 2 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) 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 s 0,..., s N of a line bundle L on X such that for every x X there is i such that s i (x) 0 determines a map X P N. Indeed, let {U i } be a covering of X such that L is trivial on every U i. Each s k is given by a collection of functions f ik on U i. Then to each x U i we can associate the point (f i0 (x) :... : f in (x)) in P N. 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 f ik (x) and f jk (x) differ by the same scalar factor g ij (x), where g ij are transition functions for L. The space P N here is naturally the dual space to the projectivization of the space spanned by s i 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))). Example 2. If D is a divisor of degree d on P 1, then the map defined by D is the Veronese embedding P 1 v d P d. Example 3. If X P n and D is the intersection of X and a hyperplane (equivalently, O(D) is the restriction of O(1) to X), then the map defined by D is the embedding X P n that we started with. Example 4. If X π P n is the blow-up of a point p in P n and D is the preimage of a hyperplane in P n 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 the restriction of the line bundle O(1) from P N to X for some embedding X P N. A divisor D is ample, if md is very ample for some m > 0. Example 5. For X = P n 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 ample line bundle on a smooth projective variety X. Then for any line bundle F, there is n 0 > 0 such that for every n > n 0 the bundle F L n is generated by a finite number of global sections.

4 DIVISORS. Note: this theorem holds for any coherent sheaf 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. (4) If E is ample, then there is n 0 such that F E n is very ample for n > n 0. 5. Intersection of curves on a smooth surface 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 T p C T p X and T p D T p X are distinct. Then the intersection number 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) (C 1 + C 2 ) D = C 1 D + C 2 D; (4) If C 1 C 2, then C 1 D = C 2 D. The strategy to prove this theorem is to show that C and D are always linearly equivalent to some divisors C and D (not necessarily effective) with smooth components such that the components of C and the components of D intersect transversally. To do that, we will need Proposition 1 and the following theorem: Theorem 3. (Bertini) Let X P n be a smooth projective variety over an algebraically closed field. Then there is a hyperplane H P n such that H does not contain X and H X is smooth. Moreover, such hyperplanes H form a dense subset in the projective space (P n ) of all hyperplanes in P n. Lemma 2. Let C 1,..., C r 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 C i transversally away from the points C i C j. Proof. Use the linear system L to embed X into P N. 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 intersect C i P N transversally are all dense in the set of all hyperplanes in P N, 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 C 1 C + nh and C 2 nh. Then C C 1 C 2. By Lemma 2 we can choose smooth curves D 1 D + mh and D 2 mh that intersect C 1 and C 2 transversally. Then D D 1 D 2, and the divisors C 1 C 2 and D 1 D 2 intersect transversally, so we can define C D = (C 1 C 2 ) (D 1 D 2 ) = C 1 D 1 + C 2 D 2 C 1 D 2 C 2 D 1. This definition is consistent, because for a different representation C C 1 C 2 and D D 1 D 2 we ll have C 1 (D 1 D 2 ) = C 1 (D 1 D 2 ) as long as D i, D i all intersect C 1 transversally, because D 1 D 2 D 1 D 2, so their difference will be a principal divisor, which on a smooth projective curve C 1 has equal number of zeroes and poles. Similarly, C 2 (D 1 D 2 ) = C 2 (D 1 D 2 ), so (C 1 C 2 ) (D 1 D 2 ) = (C 1 C 2 ) (D 1 D 2 ). Likewise, (C 1 C 2 ) (D 1 D 2 ) = (C 1 C 2 ) (D 1 D 2 ). If some of the intersections between C i and D j are not transversal, we can add an intermediate step between C 1 C 2 and C 1 C 2 that will be transversal both to D j and D j. 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 O X (C) to C. Let X be a blow-up of the point (0 : 0 : 1) on P 2, and let C be the exceptional divisor. Introduce the coordinates (x 0 : x 1 : x 2 ) on P 2 and (u : v) on C. Then X can be covered by the following open sets: U 0 = {x 0 0}, U 1 = {x 1 0}, U 2u = {x 2 0, u 0}, and U 2v = {x 2 0, v 0}. The equation for X in P 2 C will be x 0 v = x 1 u. The Cartier divisor defining C can be taken to be 1 on U 0 and U 1 (since these sets do not intersect C); on U 2u the coordinates are (x 0 /x 2, x 1 /x 2, v/u), the equation for X is (x 0 /x 2 )(v/u) = x 1 /x 2, and the equations for C are x 0 /x 2 = x 1 /x 2 = 0, so we can take the function x 0 /x 2. Likewise, on U 2v we take the function x 1 /x 2. Then on U 2u U 2v the transition function for O(C) will be x 1/x 2 x 0 /x 2 = x 1 x 0 = v u since x 0v = x 1 u on X. Then the restriction of O(C) to C has transition function v/u (from the chart u 0 to the chart v 0), and is isomorphic to O( 1). Hence C C = 1.