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 21 LECTURES ABHINAV KUMAR From last time f X B is an elliptic quasi elliptic bration Fbi mi Pi multiple bers R1 f OX L T for L invertible on B and T torsion b Supp T h0 OFb 0 h1 OFb 2 Fb is an exceptional wild ber Theorem 1 With the above notation X f L 1 B OX ai Pi where 0 ai m ai mi 1 unless Fbi is exceptional and deg L 1 B 2pa B 2 OX T where T is its length as an OB module 1 Proof We have proved most of this speci cally we have that X f L pq B OX ai Pi for 0 ai m We have a Leray spectral sequence E2 H p B Rq f OX H p q X OX The smaller order terms give us a short exact sequence 1 0 H 0 OB H 1 OX H 0 R1 f OX H 2 OB 0 0 H 2 OX H 1 R1 f OX 0 Using this we see that OX h0 OX h1 OX h2 OX 2 h0 OB h1 OB h0 L T h1 L T OB L h0 T deg L T by Riemann Roch so deg L OX T Since deg B 2pa B 2 we have deg L 1 B 2pa B 2 OX T It remains to show that ai mi 1 if Fbi is not exceptional If fact we can prove something stronger let i be the order of OX Pi OPi in Pic Pi Then we claim that 1 i divides mi and ai 1 2 h0 Pi O i 1 Pi 2 and h0 Pi O i Pi 1 and 3 h0 Pi nPi is a nondecreasing function of n Assuming this if ai mi 1 then i mi so mi Pi is exceptional by b and c since then h0 Omi Pi 2 We now prove the claim If m n 1 then OmP OnP 0 gives H 1 P OmP H 1 P OnP 0 implying that n h1 P OnP is nondecreas ing But by Riemann Roch and the de nition of canonical type OnP 0 so 1 2 LECTURES ABHINAV KUMAR h0 h1 is also nondecreasing Now by the de nition of i OX i Pi OPi OPi implying that OX ni Pi OPi O as well We thus obtain an exact Pi sequence 0 OX i Pi OPi OPi O i 1 Pi O i Pi 0 inducing a long exact sequence 0 k 3 H 0 OP H 0 O 1 P H 0 O P i i i i i and h0 O i 1 Pi 2 But for 1 j i Lj OX jPi OPi is an invertible OPi module whose degree in each component of Pi equals 0 Since OPi H 0 Lj 0 and 0 Lj O j 1 Pi OjPi 0 gives H 0 O j 1 Pi Lj 0 0 0 0 H OjPi Since H OP k for P icoct H O2P H O P k as well Finally OX Pi OP mi OX Fb OP OP 4 i i i i This is proved as follows Since the ber is cut out by a rational function f H 0 OX Fbi OPi 0 Via the exact sequence 5 0 OX OX Fbi 1 f 1 f OX Fbi OFbi 0 we get a global section of OX Fbi OFbi But this also has degree 0 along the components So it must be trivial but what we proved for icoct We also have OX ai 1 Pi OPi X OX Pi OPi 6 Pi O Pi implying that i ai 1 as desired Corollary 1 K 2 0 Corollary 2 If h1 OX 1 then either ai 1 mi or ai i 1 mi Proof Exercise Remark Raynaud showed that mi i is a power of p char k or is 1 if char k 0 Therefore there are no exceptional bers in characteristic 0 1 Classification contd If f X B is an elliptic quasi elliptic bration then 7 X f L 1 B OX ai Pi 0 ai mi If n 1 is a multiple of m1 mr then 8 n H 0 X X H 0 B L n Bn OB ai n mi bi Now we recall the 4 classes of surfaces a an integral curve C on X s t K C 0 ALGEBRAIC SURFACES LECTURE 21 3 b K 0 c K 2 0 K C 0 for all integral curves C and C s t K C 0 d K 2 0 and K C 0 for all integral curves C Lemma 1 If X is in a then X i e pn 0 for all n 1 If X is in b then X 0 If X has an elliptic or quasielliptic ai bration f X B and if we let f 2pa B 2 OX T then X is not in class mi d and X is in a i f 0 in which case X X is in b i f 0 in which case X 0 X is in c i f 0 in which case X 1 Proof If K C 0 then X is ruled and X We did this before and there is an easy way to see that pn 0 for all n 1 For every divisor D Div X nD s t D nK for n nD Since D nK C D C n K C becomes negative eventually Now C is e ective We claim that C 2 0 so by our useful lemma 6 D nK can t have an e ective divisor If C 2 0 then C K 0 would imply that C was an exceptional curve of the rst kind contradicting the minimality of X Thus C 2 0 In particular D K gives nK for large enough n implying that nK for all n since pn pmn Next assume K 0 case b If pn 2 then dim nK 1 a strictly positive divisor 0 in nK Then H 0 for a hypersurface section contradicting nK H 0 since K 0 So pn 1 for all n implying that X 0 Now assume X has an elliptic quasielliptic bration and let M f X L 1 B from last time Then M has degree f Let H be a very ample divisor on X Then f H H B is some nite map of degree H F 0 n H deg H M deg deg B M H F f So Now n K H deg X if f 0 then K H 0 and X is in a Similarly f 0 K H 0 for every irreducible hyperplane section H and any curve C can be written up to as the di erence of 2 such This implies that K C …