MIT OpenCourseWare http ocw mit edu 18 727 Topics in Algebraic Geometry Algebraic Surfaces Spring 2008 For information about citing these materials or our Terms of Use visit http ocw mit edu terms ALGEBRAIC SURFACES LECTURE 4 LECTURES ABHINAV KUMAR We recall the theorem we stated and lemma we proved from last time Theorem 1 Let f X S be a birational morphism of surfaces s t f 1 is g not de ned at a point p S Then f factors as f X S S where g is a birational morphism and is the blowup of S at p Lemma 1 Let S be an irreducible surface possibly singular and S a smooth surface with a birational morphism f S S Suppose f 1 is unde ned at p S Then f 1 p is a curve on S Lemma 2 Let S S be a birational map s t 1 is unde ned at a point p S Then there is a curve C on S s t C p Proof corresponds to a morphism f U S where U is some open set in S Let U S be the graph of f and let S1 denote its closure in S S S1 is irreducible but may be singular S1 1 q S S q The projections q q are birational morphisms and the diagonal morphism com mutes Since 1 q is not de ned q 1 p is not de ned either so C1 S an irreducible curve s t q C1 p Moreover q C1 C is a curve in S if not since S1 S S q C1 a point C1 x S for some x S but such a C1 can only intersect the graph of f in x f x so the closure of the graph of f can t contain the curve C1 By construction C contracts to p under Proof of theorem Let g 1 f be the rational map in question We need to show that g is a morphism Let s g 1 and suppose that g is unde ned at a point q X X 2 q s S f S 1 2 LECTURES ABHINAV KUMAR Applying the second lemma we obtain a curve C S s t s C q Then C f q by composing s C q with f So we must have C E the exceptional divisor for and f q p Let Ox q be the local ring of X at q and let mq be its maximal ideal We claim that there is a local coordinate y on S at p s t f y m2q To see this let x t be a local system of coordinates at p If f t m2q then we are done If not i e f t m2q then f t vanishes on f 1 p with multiplicity 1 so it de nes a local equation for f 1 p in OX q So f x u f t for some u Ox q Let y x u q t then 3 f y f x u q f t uf t u q f t u u q f t m2q Next let e be any point on E where s is de ned Then we have s f y f s y y m2e This holds for all e outside a nite set But y is a local coordinate at every point of E except one by construction giving the desired contradiction This proves the universal property of blowing up Here is another Proposition 1 Every morphism from S to a variety X that contracts E to a point must factor through S Proof We can reduce to X a ne then to X An then to X A1 Then f de nes a function on S E S p which extends Theorem 2 Let f S S0 be a birational morphism of surfaces Then a sequence of blowups k Sk Sk 1 k 1 n and an isomorphism S Sn s t f 1 n u i e f factors through blowups and an isomorphism Proof If f 1 is a morphism we re done Otherwise a point p of S0 where f 1 is not de ned Then f 1 f1 where 1 is the blowup of S0 at p If f1 1 is a morphism we are done otherwise we keep going We need to show that this process terminates Note that the rank of the Neron Severi group rk N S Sk 1 rk N S Sk 1 since rk S is nite this sequence must terminate More simply since f contracts only nitely many curves it can only factor through nitely many distinct blowups Corollary 1 Any birational map S S is dominated by a nonsingular surface S with birational morphisms q q S S S which are compositions of blow up maps i e S 4 q S q S Proof First resolve the indeterminacy of using S and then note that q is a birational morphism i e a composition of blowups by the above ALGEBRAIC SURFACES LECTURE 4 3 1 Minimal Surfaces We say that a surface S1 dominates S2 if there is a birational morphism S1 S2 A surface S is minimal if it is minimal up to isomorphism in its birational equivalence class with respect to this ordering Proposition 2 Every surface dominates a minimal surface Proof Let S be a surface If S is not minimal a birational morphism S S1 that is not an isomorphism so rk N S S rk N S S1 If S1 is minimal we are done if not continue in this fashion which must terminate because rk N S S is nite Note We say that E S is exceptional if it is the exceptional curve of a blowup S S Clearly an exceptional curve E is isomorphism to P1 and satis es E 2 1 and E KS 1 since 2 2g 2 E E K Theorem 3 Castelnuovo Let S be a projective surface and E S a curve P1 with E 2 1 Then a morphism S S s t it is a blowup and E is the exceptional curve classically called an exceptional curve of the rst kind Proof We will nd S as the image of a particular morphism from S to a pro jective space informally we need a nearly ample divisor which will contract E and nothing else Let H be very ample on S s t H 1 S OS H 0 take any hyperplane section H then H nH will have zero higher cohomology by Serre s theorem Let k H E 0 and let M H kE Note that M E H kE E k kE 2 0 This M will de ne the morphism S P H 0 S OS M i e some Pn Now OS H E OE k since E P1 1 and deg OS H E H E k and on P line bundles are determined by degree Thus OS M E OE Now consider the exact sequence 5 …