Algebraic varieties [Lecture notes] by H. A. Nielsen

By H. A. Nielsen

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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. Assume X ∩ V (Y, Z) = ∅. For a fraction H such that div( K ) ≥ − div(Z m ) H K of homogeneous forms of degree d K HZ m ) ≥ vx ( d ) Y m+d Y for all x. This gives forms G, E such that vx ( HZ m = GK + EF It follows that the linear map k[X, Y, Z]m → L(div(Z m )), G → G Zm is surjective and the linear map k[X, Y, Z]m−n → k[X, Y, Z]m , H → H · F for large m is an isomorphism onto the kernel of the map above.

1) If Y is normal and f (x) ∈ W ⊂ Y is closed irreducible, then there is an irreducible component x ∈ Z ⊂ f −1 (W ) such that f (Z) = W . (2) If Y is normal then for every open subset U ⊂ X, f(U) is open in Y . For each y ∈ Y the number of points in the fibre |f −1 (y)| ≤ dimk(Y ) k(X) (3) If the extension f ∗ (k(Y )) ⊂ k(X) is separable then there is a nonempty open subset V of Y such that for each y ∈ Y the number of points in the fibre |f −1 (y)| = dimk(Y ) k(X) Proof. Assume X, Y affine. (2) Let g ∈ k[X], y ∈ f (Xg ).

6. (1) For any divisor D, l(D) ≥ deg(D) + 1 − g. (2) Let D0 be a divisor giving g = deg(D0 ) + 1 − l(D0 ). If deg(D) ≥ deg(D0 ) + g then l(D) = deg(D) + 1 − g. Proof. 5. 7. For any divisor D on a curve X (1) Λ(X)/Λ(D) + k(X) has finite dimension over k. (2) l(D) − dimk Λ(X)/Λ(D) + k(X) = deg(D) + 1 − g. Proof. 5. 4. 1. Let X be a curve. The space of rational differentials on X is the k(X)vector space Ω(X) = ⊕f ∈k(X) k(X)df /(da, d(g + h) − dg − dh, d(gh) − hdg − gdh) a ∈ k, g, h ∈ k(X) generate the relations.

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