Lecture 16: Bezout’s Theorem

Lecture  18.725 Algebraic Geometry I Lecture 16: Bezout’s Theorem Definition 1. Two (Cartier) divisors are linearly equivalent if D1 - D2 are principal. Given an effective divisor D, we have an associated line bundle L = O(D) given (on each open set U ) by the sections of K whose locus of poles (i.e. locus of zeroes in the dual sheaf) is contained in D. Now suppose X is complete, then given an invertible sheaf L on X, a section σ is uniquely (up to multiplication (O (D),1) by a constant) determined by its corresponding divisor Z(σ), so we have a correspondence D ←→ (L, σ). Z(σ) Now if σ1 , σ2 are nonzero sections, then f = σ1 /σ2 is an rational function on X, and if Z(σ1 ) and Z(σ2 ) are linearly equivalent, then f has no pole and no zero; in other words, linearly equivalent divisors correspond to isomorphic line bundles. So the set of all effective divisors linearly equivalent to a fixed effective divisor D form a projective space PΓ(O(D)), and is called a complete linear system of divisors. Proposition 1. X irreducible curve, deg(D) = 0 if D is a principal divisor. Proof. D is principal, so let D = (f ) = D0 − D∞ where f : X → P1 , X = U1 ∪ U2 , f ∈ k[U1 ], 1/f ∈ k[U2 ], (This is clear for X normal: all local rings are DVR, so either f or 1/f is in OX,x .) where D0 ⊆ f (P1 − {∞}) is the divisor of zeroes of f , and similarly D∞ ⊆ 1/f (P1 − {0}) is the divisor of zeroes of 1/f . We need to same as that of D∞ , and that the degree of both slices are that of deg(f ). check that degree of D0 is theX Recall that D0 = mx x, where mx = length(O/f O)x = dim(Γ((O/f O)x )).1 Clearly x∈f −1 (P1 −{∞}),f (x)=0 f : U = f −1 (A1 ) → A1 is finite, and that f∗ (OX |U ) is a locally free sheaf of rank equal to the degree of f . From classification of finitely generated modules over k[t], we know that every module is the sum of its torsion and a free module; but this one cannot have torsion because there can be no function of X that vanishes away from finitely many points, so it’s free. f∗ O is coherent follows from f being finite, which follows from that f is complete and has finite fibers. Now suppose k[f −1 (A1 )] is a free module of rank d over k[t] = k[A1 ]. Then [K(X) : K(A1 )] = d, which is the degree of the map. Thus d = dim(k[f −1 (A1 )]/t) (dimension of fiber of f∗ O at 0) = dim(Γ(OU1 /f OU1 )) = X dim(Γ((OU1 /f OU1 )x )) = deg(D0 ), where U1 = f −1 (A1 ). The other half is dealt with similarly. Remark 1. k = C, X normal, X(C) (the set X equipped with the complex topology) is a smooth compact Riemann surface (1-dimensional C-manifold). f ∈ K(X) defines a meromorphic function on X(C), (f ) = X X df df nx x, n being the order of zero/pole, or just Resx , which tells us that Resx = 0. f f x∈X(C) Proof of Bezout’s Theorem The multiplicity of intersection of two curves X, Y in P2 at x (X, Y have no common components) is defined as multx (X, Y ) = length(i∗ OX ⊗O(P2 ) j∗ OY )x = dim Γ((i∗ OX ⊗O(P2 ) M j∗ OY )x ). Note that (i∗ OX ⊗O(P2 ) j∗ OY ) = (i∗ OX ⊗O(P2 ) j∗ OY )x . This agrees with earlier definition. x∈X∩Y Theorem 1.1 (Bezout’s Theorem). X multx (X, Y ) = deg(X) deg(Y ). x∈X∩Y Proof. Both sides are additive under X = X1 ∪ X2 where the two curves have no common components. (Clear for RHS, LHS as exercise.) Now we can assume X is irreducible, and we’ll show LHS = deg(O(Y )|X ). O(Y ) is a line bundle with a section σ such that (σ) = Y . We know that OY = O/O(−Y ) from which σ it follows that OX ⊗ OY = OX /im σ|X (where σ denotes O(−Y ) − → O). Compare with X the definition of multiplicity above, it follows that the divisor of zeroes of σ|X , i.e. the pullback of σ, is multx (X, Y )x. ∼ O(d) where d = deg(Y ), so the isomorphism class and hence the degree of Now we know that O(Y ) = O(Y )|X depends only on the degree of Y . Now we can take Y to be the union of d lines; by additivity, we reduce to the case where Y is a line. Since Y and X are symmetric, also reduce to X is a line, from which the result follows. 1 The subscript here refers to the canonical split of sheaves supported at finitely many points, NOT stalks; the same for below. 1 Lecture  18.725 Algebraic Geometry I The analytic story Let X be an irreducible normal curve over C, then X(C) is a compact 1-dimensional C-manifold homeomorphic to a sphere with g handles, g being the genus of the curve. One can look at the topological homology H 1 (X, Z) = Z2g . The important variant here is the space of differential forms. Define Ω1 to be the sheaf of holomorphic 1-forms, e.g. f (z)dz. The global section Γ(Ω1 ) ∼ = Cg . Now, since we have Poincare duality, we can define a map from de Rham classes to singular cohomology as follows: Z given an 1-form ω, we map it to Hom(H1 (X, C), C) = H 1 (X, C) = C2g as [c] 7→ ω. Thus we have c H 1 (X, C) = Im(Γ(Ω1 )) ⊕ Im(Γ(Ω1 )) = H 1,0 ⊕ H 0,1 , usually called the Hodge decomposition. Recall the GAGA theorem, which states that holomorphic line bundles are the same as algebraic line bundles, which are parametrized by the Picard group. Now Picard group is (Divisors) / (Principle Divisors), and there is a degree homomorphism Pic → Z, with the kernel denoted Pic◦ . It turns out that Pic◦ ∼ = Γ(Ω1 )∗ /H1 (X, Z) (image of H1 (X, Z) ⊆ H1 (X, C) under the integral map) ∼ = Cg /Z2g . The structure 1 ∗ Γ(Ω ) /H1 (X, Z) is usually called the Jacobian of the curve, and the isomorphism the Abel-Jacobi map. If D = (f ) is a principal divisor, D gets mapped into 0 by the Abel-Jacobi map above. Sketch of proof: given f from X → P1 , consider a family of divisors D0 − Dz , z ∈ P1 . If z = 0, then this is the 0 divisor; when z = ∞, we get our divisor D = (f ). Easy to see that z 7→ AJ(D0 − Dz ) is a holomorphic function CP1 → Cg /Z2g . Since CP1 is simply connected, it lifts to CP1 → Cg , which is constant by maximal principle. Our next topic is smoothness, which is a local property. Let X be an algebraic variety, and x be a point. Define dimx (X) to be the maximum of dimensions of components passing through x. Definition 2. x is a smooth point on X if dimx (X) = dim(mx /m2x ), where mx is the maximal ideal in OX,x . Example 1. Suppose X in An is a hypersurface (so codimension 1), IX = (f ). Then x is a smooth point 6 0 at x for some i. iff ∂f /∂zi = Corollary 1. For X, Y curves in P2 , the intersection multiplicity is greater than 1 if either X or Y is not smooth at x. To see this, suppose x = (0, 0) ∈ A2 , then OX  k[x, y]/(x, y)2 , then OX ⊗ OY  OY /m2OY . 2