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.)

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

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Is isomorphic to the . ,dr) such that n-r>/~ and e > (al,. max(dl,... ,dr). Then there is a Zariski open (resp. an ope n in the complex topology) nei6hbourhood T' of o in T such that for every, k-rational point Iresp. ,dr) (but may be not isomorphic to Y .......... Proof. Let Y~ = Y). O >pn(s) be the closed embedding given by the complete li- near system IOy(s)~ (n(1) - n). Then the cone X = C(Y,O. (s)) lles in pn(s)+l. x n(s)+l Step I. Every small (embedded) deformation of X in P is again a cone of type C(¥t,OYt(s)) , where Yt is a complete intersection in pn of type (dl,.

The fibres of fK have degree 4 and are not multiple. b) X=BLI(Y) where Y is of type 13 in ~n+l , 56 (II6) x is Castelnuovo. Proof. 5) we have HK even ~6 and Pa(X)=HK/2. Suppose HK=0. with M=0 and Pa=0, Suppose HK=2. Since tl2Kl~0 we get 12K~0. So X must be minimal hence X is Enriques. Since Pa(X)=l there exists an effective canonical divisor K. There are two possibilities: l) K is a smooth conic. exceptional. By adjunction we get K2=-I hence K is Let fH+2K:X--Y be the corresponding K~f'Ky+K hence Ky~0.

Surface in pn with 0Ss~n-3 X is a r i t h m e t i c a l l y normal and q(X)=0. Suppose X is not minimal and choose an e x c e p t i o n a l divisor t E. 9). Then Y has degree d'=d+k 2 and n ' = n + k ( k + l ) / 2 hence d'=2n'- -2+s' with st=s-k. If s'<0,Y must be ruled, hence so is X, contradic- 5~ ring pg(X)>l. NOW o b v i o u s l y s'

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