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 10 LECTURES ABHINAV KUMAR Recall that we had left to show that there are no surfaces in characteristic p 0 satisfying 1 Pic X is generated by X OX K and the anticanonical bundle is ample In particular X doesn t have any nonsingular rational curves 2 Every divisor of K is an integral curve of arithmetic genus 1 3 K 2 5 b2 5 Lemma 1 Let X be as above Then a nonsingular curve D K Proof Suppose not Since pa D 1 we may assume that every D in K has exactly one singular point a node or a cusp Then dim K K 2 1 Let L K be a one dimensional linear subsystem The ber s of L X P1 are exactly the curves in L because L has no xed components all the elements of L are integral Let Y be the set of all singular points on curves in L and Y This is because if let x be a base point of L if any We claim that x x Y then for D L a curve singular at x X X the blowup of X at x P1 After then D is a nonsingular rational curve and a ber of X further at most K 2 blowups we get an X s t X P1 is a morphism and one ber of is a smooth rational curve So X is geometrically ruled over P1 implying that X is rational and so X is rational This however is impossible by the classi cation of rational surfaces Pic X is never Z X giving the desired contradiction P1 is a quasi elliptic bration So blowup the base points of so that X by Tate all singular points are cusps in characteristic 2 or 3 e g y 2 x3 t All the bers are integral rational curves with one singular point and the set of singularities Y of is a nonsingular irreducible curve with Y Y P1 a bijective purely inseparable morphism Thus Y P1 Y P1 contradicting the fact that there are no smooth rational curves on X Lemma 2 Let X be as above Then H 2 X TX 0 for TX 1X k the tangent sheaf 1 2 LECTURES ABHINAV KUMAR Proof a nonsingular elliptic curve D K by the above lemma The short exact sequence 1 0 OX n 1 D TX OX nD TX OX nD TX OD 0 gives the long exact sequence in cohomology 2 H 1 X OX nD TX OD H 2 X OX n 1 D TX H 2 D OX nD TX 0 for large n by Serre vanishing since D is ample By reverse induction it is enough to show that H 1 X OX nD TX OD 0 for n 1 This will show 3 H 2 OX n 1 D TX 0 H 2 OX n 2 D TX 0 H 2 OX TX 0 Dualizing the conormal exact sequence 0 I I 2 X OD D 0 we get 0 OD TD TX OD N OX D OD 0 4 where we used that TD D OD since D is an elliptic curve Taking coho mology gives 5 H 1 OX nD OD H 1 OX nD TX OD H 1 OX nD N D2 0 so we get H 1 OX nD OD 0 H 1 OX nD N as desired since OX nD OD and OX nD N are sheaves of large positive degree on D for n 0 We now return to the proof of the proposition for p char k 0 Let A W k be the Witt vectors of k A is a complete DVR of characteristic 0 with maximal ideal m and residue eld A m k The idea is to lift X to characteristic 0 and use the result already proved Note that H 2 X TX 0 in addition X is projective and H 2 X OX 0 so by SGA1 Theorem III 7 3 there is a smooth projective morphism f U V Spec A which closed ber isomorphic to X Let X be the general ber Then X is a nonsingular projective surface over the fraction eld K of A which is unfortunately not algebraically closed The bers of f are 2 dimensional so Ri f OU 0 for i 3 The base change theorem gives 6 R2 f OU A A m H 2 f 1 m OU mOU H 2 X OX 0 is an isomorphism By Nakayama s lemma we get that R2 f OU 0 and sim ilarly for R1 Thus H 1 X OX H 2 X OX 0 See Mumford s Abelian Varieties or Chapters on Algebraic Surfaces for more details Now let K be an algebraic closure of K and Ki the family of nite extensions of K inside K Let X X K k Xi X K Ki Let A be the integral ALGEBRAIC SURFACES LECTURE 10 3 closure of A inside K and Ai A Ki Let m be a maximal ideal of A lying over m and B A m its localization Similarly let mi m Ai Bi Ai mi and set n m B ni mi Bi Since K A m is algebraically closed we see B n Bi ni K Now let Vi Spec Bi Ui U A Bi fi f A Bi Ui Vi Xi 7 Spec Ki Ui fi Vi U f Spec A X Spec K Since Bi ni k the closed ber of fi is canonically isomorphic to X The generic ber of fi is isomorphic to Xi and since Ki K is nite Bi is a DVR and Vi is an inductive system By EGA and general nonsense lim Pic Xi Pic X is an isomorphism Lemma 3 There is a group isomorphism b Pic Xi Pic X de ned by the following for L an invertible OXi module an invertible OUi module Li s t Li Xi L and we set b L Li X Proof Omitted Proof of theorem So we get a canonical isomorphism between Pic X and Pic X which takes X to X Since Pic X Z X we see that Pic X Z X and 1 X is ample giving us 1 for X Also X X X X X X using atness and the de nition of by so X X 5 and b2 X 5 by Noether s formula Since q X 0 q X 0 But X is over an algebraically closed eld of characteristic 0 which as shown last time is impossible De nition 1 A surface X is unirational if there is a dominant morphism Y X from a rational surface Corollary 1 In characteristic zero a unirational surface is rational Note that this is not true in characteristic 0 e g the Zariski surfaces z p f x y Proof Given f Y …