By H. A. Nielsen

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The k-linear map d : k(X) → Ω(X), f → df is the universal k-derivation on k(X), that is any derivation is the composite of d and a k(X)-linear map. 4. 2. The dimension of differentials dimk(X) Ω(X) = 1 Any function t ∈ k(X) such that k(t) ⊂ k(X) is finite separable gives a basis dt for Ω(X). If t is a local parameter at x, then dt is a basis. Proof. 2 we may assume t transcendental over k and k(t) ⊂ k(X) finite separable generated by u. ∂g So k(X) = k(t)[u]/(g) with ∂u = 0. By implicit differentiation ∂g ∂t du = − ∂g dt ∂u so dt is a generator and moreover f→ ∂f − ∂t ∂g ∂t ∂g ∂u ∂f ∂u is a nonzero derivation, so dt is a basis.

Lines in P3 , n = 3, m = 1 give one nontrivial relation y01 y23 − y02 y13 + y03 y12 = 0 3,1 defining G as a hypersurface in P5 . 5. By Gauss elimination the Grassmann variety is covered by open affine spaces A(m+1)(n−m) . It follows that Gn,m is an irreducible nonsingular projective variety of dimension dim Gn,m = (m + 1)(n − m) 19. 1. Let X ⊂ PN be a nonsingular projective variety of dimension n. Let W ⊂ PN × PN × PN be points on lines W = {(x, y, z)|z on the line through x, y} Let ∆X ⊂ X × X be the diagonal.

1 (Riemann-Roch). Let K be a canonical divisor on a curve X of genus g. Then for any divisor D l(D) − l(K − D) = deg(D) + 1 − g Proof. 4. give the equality. 2. Let X be a curve of genus g, K a canonical divisor and D any divisor. (1) deg(K) = 2g − 2. (2) l(K) = g. (3) If deg(D) ≥ 2g − 1 then l(D) = deg(D) + 1 − g. (4) If deg(D) = 2g − 2 and l(D) = g then D = K + div(g). Proof. 1. 44 III. 3 (Clifford). If D is a divisor with l(D) > 0 and l(K − D) > 0 then 1 l(D) ≤ 1 + deg(D) 2 Proof. By the hypothesis assume D ≥ 0 and K − D ≥ 0 and such that l(D − x) < l(D) for all x.