Algebraic Geometry

Download e-book for iPad: Algebraic geometry I-V by Shafarevich I.R. (ed.)

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By Shafarevich I.R. (ed.)

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Extra resources for Algebraic geometry I-V

Example text

C >, 0. C + nC2 0. 4 Step 1: H2(S,OS) = 0. Put F = p-1(x). 8(i) and the genus formula). 11). F 3 0 by the useful remark, a contradiction. F = 1. Let f be the class of F in H2(S,Z). 10). Thus it suffices to show that there exists a class h E H2(S,7L) with h. f = 1. As a runs through H2(S, 7L), the set of integers (a. f) is an ideal in Z, of the form dZ (d > 1). The map a '-1 f) is a linear form on H2(S,7L); now Poincare duality d(a. says that the cup product H2(S,7L) 0 H2(S,7L) -. H4(S,Z) Z Z is a duality, in other words that the associated map H2(S,Z) --r Hom(H2(S,7L),Z) is surjective, with kernel equal to the torsion subgroup.

8(i) and the genus formula). 11). F 3 0 by the useful remark, a contradiction. F = 1. Let f be the class of F in H2(S,Z). 10). Thus it suffices to show that there exists a class h E H2(S,7L) with h. f = 1. As a runs through H2(S, 7L), the set of integers (a. f) is an ideal in Z, of the form dZ (d > 1). The map a '-1 f) is a linear form on H2(S,7L); now Poincare duality d(a. says that the cup product H2(S,7L) 0 H2(S,7L) -. H4(S,Z) Z Z is a duality, in other words that the associated map H2(S,Z) --r Hom(H2(S,7L),Z) is surjective, with kernel equal to the torsion subgroup.

F = 0, every component of E is a fibre of p, so E is the inverse image by p of a divisor on C. Part (ii) can be proved either directly from the topology (using exact sequences of S2-bundles), or from part (i): since H2(S, Z) is a quotient of Pic S (cf. h = 1. M = L-1 M-1 = X(Os) - X(L) - X(M) + X(L 0 M) X(Os) - X(E') + X(A2E') . M) depends only on E'; we will denote it by c2(E'). We now apply this to the bundle p*E on S. p*M)=0. [N] = 0. From this exact sequence we obtain an isomorphism N ® Os(1) = p* A2 E, whence [N] = -h + p* e, writing e for the class of A2E in Pic C.

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