By H. A. Nielsen

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Additional info for Algebraic varieties [Lecture notes]

Sample text

Then V (F ) ∩ V (G) = {x1 , . . , xs }. Let L ∈ k[X, Y, Z] be a linear form such that V (L) ∩ {x1 , . . , xs } = ∅ and define a divisor div(G) = vxi ( i G )xi Lm independent of L. This satisfies the product rule div(GH) = div(G) + div(H) and therefore defines a divisor div( also satisfying the product rule. 2 (Bezout). (1) div(G) ≥ 0. (2) deg(div(G)) = mn. G ) = div(G) − div(H) H 46 III. ALGEBRAIC CURVES Proof. (2) Assume by change of coordinates and product rule that G = Z. 3. 3. The genus of X is g= 1 (n − 1)(n − 2) 2 Proof.

9. Let Y an irreducible variety. The set of points where Y is normal is a nonempty open subset. Proof. The normalization f : X → Y gives the open subset y ∈ V ⊂ Y where |f −1 (y)| = 1. 10. Let X be a normal irreducible projective variety. If dim X = n then X is isomorphic to the normalization of a closed hypersurface V (F ) ⊂ Pn+1 . Proof. Assume X ⊂ PN and let L0 , . . , Ln+1 be linear forms such that X∩V (L0 , . . , Ln+1 ) = ∅, then the projection X → Pn+1 is finite with image V (F ). Conclude by X is normal.

Let | Aut(X)| = n and choose by normalization a curve Y together with a morphism f : X → Y of deg(f ) = n identifying f ∗ (k(Y )) = k(X)Aut(X) as the fixed field. By the Zeuthen-Hurwitz formula 1 2g − 2 = 2gY − 2 + (1 − ) n ry y giving 2g − 2 1 1 1 1 ≥ −2 + (1 − ) + (1 − ) + (1 − ) = n 2 3 7 42 Examples and comments k is a fixed algebraically closed ground field of characteristic 0 when necessary. 1. 1. The open subset A1 −{0} of A1 is a variety isomorphic to the affine variety V (Y1 Y2 − 1) ⊂ A2 .