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 9 LECTURES ABHINAV KUMAR 1 Castelnuovo s Criterion for Rationality 2 Theorem 1 Any surface with q h1 X OX 0 and p2 h0 X X 0 is rational Note Every rational surface satis es these they are birational invariants which vanish for P2 Reduction 1 Let X be a minimal surface with q p2 0 It is enough to show there is a smooth rational curve C on X with C 2 0 Proof First observe that 2g C 2 2 C C K and OX C OX 12 C C K Since p2 0 p1 h0 X h2 X OX 0 and OX 1 Since h2 C h0 K C h0 K 0 h0 C 1 12 C C K so h0 C 2 C 2 2 Choose a pencil inside this system containing C i e a subspace of dimension 2 The pencil has no xed component the only possibility is C but C moves in the pencil after blowing up nitely many base points we get a morphism X P1 with a ber isomorphic to C P1 Therefore by the 1 is ruled over P and X is rational as is X Noether Enriques theorem X Reduction 2 Let X be a minimal surface with q p2 0 It is enough to show that an e ective divisor D on X s t K D and K D 0 Proof This implies that some irreducible component C of D satis es K C 0 Clearly K C K D Using Riemann Roch for K C gives 0 h0 U C h0 C h0 K C h2 K C 1 1 1 K C C g C 2 We thus obtain a smooth rational curve C on X 2 2g 2 C C K 1 so C 2 0 as and C K 0 C 2 1 Since X is minimal C 2 desired We now prove our second statement There are three cases 1 2 LECTURES ABHINAV KUMAR Case 1 K 2 0 Riemann Roch gives h0 K h0 k h0 2K h0 K h2 K 2 1 1 K 2K 1 K 2 1 2 so K Take a hyperplane section H of X Then there is an n 0 s t H nK but H n 1 K Since K an e ective nonzero divisor H K 0 and H H nK is eventually negative and H nK is not e ective Let D H nK then D K and K D K H nK K H 0 since K is e ective H very ample Case 2 K 2 0 it is enough to nd an e ective divisor E on X s t K E 0 Then some component C of E will have K C 0 The genus formula gives 2 2g 2 C C K C 2 1 C 2 1 is impossible since X is minimal so C 2 0 Now C nK C is negative for n 0 so C nK is not e ective for n 0 by the useful lemma So n s t C nK but C n 1 K Choosing D C nK gives the desired divisor We now nd the claimed E Again let H be a hyperplane section if K H 0 we can take E H if K H 0 we can take K nH for n 0 so assume H K H 0 Let K 0 so that H K K 0 Also K2 3 H K 2 H 2 2 H K 2 K H 2 K H 2 0 K 2 So take rational and slightly larger than to get 4 H K K H K K 0 since K 2 0 and H K 2 0 Therefore H K H 0 Write rs Then 5 rH sK 2 0 rH sK K 0 rH sK H 0 by equivalent facts for Let D rH sK For m 0 by Riemann Roch we get h0 mD h0 K mD 12 mD mD K 1 Moreover K mD is not e ect over for m 0 since K mD H K H m D H Thus mD is e ective for large m and we can take E mD Case 3 K 2 0 Assume that there is no such D as in reduction 2 i e K D 0 for every e ective divisor D s t K D We will obtain a contradiction Lemma 1 If X is a minimal surface with p2 q 0 K 2 0 and K D 0 for every e ective divisor D on X s t K D then 1 Pic X is generated by X OX K and the anticanonical bundle OX K is ample In particular X doesn t have any nonsingular ra tional curves ALGEBRAIC SURFACES LECTURE 9 3 2 Every divisor of K is an integral curve of arithmetic genus 1 3 K 2 5 b2 5 Here b2 h2et X Q in general Proof First let us see that every element D of K is an irreducible curve If not let C be a component of D s t K C 0 which we can nd since K D K 2 0 If D C C K C D C C since C is e ective Also C K 0 contradicting the hypothesis So D is irreducible and similarly D is not a multiple Furthermore pa D 12 D D K 1 1 showing 2 Next we claim that the only e ective divisor s t D K is the zero divisor Assume not i e D 0 s t K D Let x D then since h0 K 1 K 2 2 there is a C K passing through x C is an integral curve and cannot be a component of D since then 6 K D K C 0 So C D 0 since they meet at least in x Then K D C D 0 contradicting the hypothesis As an aside we claim that pn 0 for all n 1 we know that p2 0 p1 0 if 3K were e ective then 2K would be too since K is e ective which contradicts p2 0 p3 0 and by induction pn 0 for all n 1 We claim that adjuction terminates if D is any divisor on X then there is an integer nD s t D nK for n nD To see this note that D nK K will eventually become negative K is represented by an irreducible curve of positive self intersection so by the useful lemma D nK is not e ective for n 0 Now let be an arbitrary e ective divisor Then n 0 s t nK 0 but n 1 K Take D nK e ective D K D 0 from above Since any divisor is a di erence of e ective divisors Pic X is generated by K If …