# Download Algebraic Geometry Bucharest 1982: Proceedings of the by Lucian BĂdescu (auth.), Lucian Bădescu, Dorin Popescu (eds.) PDF

By Lucian BĂdescu (auth.), Lucian Bădescu, Dorin Popescu (eds.)

**Read or Download Algebraic Geometry Bucharest 1982: Proceedings of the International Conference held in Bucharest, Romania, August 2–7, 1982 PDF**

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**Additional info for Algebraic Geometry Bucharest 1982: Proceedings of the International Conference held in Bucharest, Romania, August 2–7, 1982**

**Example text**

Now we treat the case HK=3 in general. Since Pa(X)=l there exists an e f f e c t i v e canonical divisor K. There are several possibilities: i) K is integral. We get Pa(K)~l. If Pa(K)=0 we get that K is an e x c e p t i o n a l line and X is of type III3a. If Pa(K)=l we get K2=0 hence X is of type III3e. 2) K=L+Q where L is a straight line and Q is a smooth conic. By adjunction, - 2 = 2 L 2 + L Q and -2=2Q2+LQ. If LQ=0 then both L and Q are e x c e p t i o n a l and c o n t r a c t i n g them we get that X is of type III3b.

IIIg) X is Castelnuovo. Proof. 5) we have HK oddS9 and Pa(X)= (HE-I)/2. Suppose HK=I. We get Pa=0. We cannot have pg~l because we would get there exists an effective canonical divisor which is a straight line. By adjunction formula this line is exceptional and we get 59 X=BLI(Y) where Y is a K3 surface which is absurd since pa(Y)=Pa(X)=0. It follows that pg=q=0. By C a s t e l n u o v o ' s c r i t e r i o n of rationality, second p l u r i g e n u s P2 cannot be 0. Take DeI2KI; we have DH=2. the If D w o u l d be reduced irreducible we w o u l d get by a d j u n c t i o n - 2 = D 2 + D K hence - 4 = 2 D 2 + 2 D K = 3 D 2 w h i c h is impossible.

Let X be a nondegenerate s=d-2n+2. i) LEMM~. section on X. Then H is a 2n-2+s in pn-i hence by Castelnuovo's genus we get: If s<0 then Pa(H)~n+s-l. If 0Ss~n-3 then Pa(H)S ~n~2s. 2) LEMMA. divisor on X. 3) L E M M A . I f d~2n-3 of ruledness we get: then X is ruled. Let us also note that for 0~s~n-3 and only if equality holds in nuovo curve. 4) LEMMA. of X is s+l. surface hence we get: For 0~s~n-3 Harris" bound on the g e o m e t r i c genus If X is linearly normal in en with 0gs~n-3 and H is then H2(Ox(H))=0 and Pa(X)=l+(i/2) (HK-s).