By Scheithauer

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Xn ]/I(V ). , ym ) ist. , ym } ist also eine Transzendenzbasis von K(V )/K und trgradK K(V ) = m. 3 Algebraische Charakterisierung der Dimension Sei V eine irreduzible affine Variet¨at. Wir zeigen, dass dim(V ) = trgradK K(V ). , Xn ] irreduzibel ist. , Xn ]/(f ). Wir k¨onnen annehmen, dass f die Variable Xn enth¨alt. h. trgradK K(V ) = n − 1 = dim(V ). Sei V ⊂ AnK eine irreduzible affine Variet¨at. , Xn − an ) ein maximales Ideal mit I(V ) ⊂ I(P ). Der Quotient MP = I(P )/I(V ) ist ein maximales Ideal in K[V ].

Beweis. Wir benutzen affine Koordinaten. ⇐“ Sei P ∈ / C ∩ C . Dann ist P ∈ / C oder ” P ∈ / C . Also ist f oder g eine Einheit in OA,P , also ist dim(OA,P /(f, g)) = 0. ⇒“ ” Sei P ∈ C ∩ C . Dann ist f, g ∈ mP , also (f, g) ∈ mP . Also ist dim(OA,P /(f, g)) ≥ dim(OA,P /(f, g)) = 1. Sei L ⊂ P2K eine Gerade, die nicht in C enthalten ist und P ∈ C. Definiere IP (C, L) = multP (fL ). Es folgt IP (C, C ) ≥ 2 ⇐⇒ L ∈ TP C. 2. Sei f ∈ K[X0 , X1 , X2 ] und homogenes Polynom vom Grad d und C = V (f ) ⊂ P2K .

Xn ] mit Xi = Xi + I(X). Dann sind die Xi algebraisch unabh¨angig u ¨ber K (sonst w¨are trgradK K(X) ≥ 1). Somit kann Xi (P ) = Xi (P ) ∈ K nur endlich viele Werte annehmen. 6. Sei Y eine irreduzible affine Variet¨at, Y ⊂ X eine abgeschlossene Teilmenge. Dann ist dim(Y ) ≤ dim(X). Ist dim(Y ) = dim(X), so ist X = Y . 7. Sei C eine glatte projektive Kurve und f ∈ K(C)∗ . Dann gibt es nur endlich viele Punkte mit νP (f ) = 0. Beweis. Sei C ⊂ PnK . , Xn ] homogen und grad(g) = grad(h). Ist νP (f ) = 0, so ist νP (g) = 0 oder νP (h) = 0.

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Algebraische Geometrie [Lecture notes] by Scheithauer

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