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By Heinz Spindler

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Example text

Mit O(U) werde die Menge aller in U regularen Funktionen bezeichnet. O(U) = ff : U ! 7 O(U) ist Unterring von C(U; K). Beweis: Ist f : U ! K regular in p, so ist f auch stetig in p. Sei dazu W go ene Umgebung von f(p) in K = A 1 ; W = A 1 n V(q); q 2 K z] mit q(f(p)) 6= 0. Sei f = h in einer o enen Umgebung V von p in U. Dann ist g(x) = 6 0 : f 1 (W) \ V = x 2 V j q h(x) q= d X =0 az X ) q hg = h1d a hd g : | {z } q~2K z1 ;:::;zn ] Also gilt: U~ = f 1 (W) \ V = V n V(~q) ist o ene Umgebung von p in U mit f(U~ ) V .

8: Unter der kanonischen Abbildung A(X) ! O(Xf ) wird f auf die nirgends verschwindende Funktion f jXf abgebildet; f jXf ist Einheit in O(Xf ). Die universelle Eigenschaft fur A(X) f1 ] gibt uns den Homomorphismus : A(X) 1f ] ! O(Xf ): Wir zeigen zunachst: a) ist injektiv. Es sei g = fhn 2 A(X) f1 ] und g(x) = 0 fur alle x 2 Xf . h. fh = 0 in A(X), also g = 0 in A(X) 1f ]: Das war trivial. Tiefliegender ist die Surjektivitat: b) Es sei r 2 O(Xf ) eine regulare Funktion auf Xf .

Dann gilt: Der kanonische Ringhomomorphismus A(X) ! O(Xf ); g + I(X) 7 ! gjXf induziert einen Ringisomorphismus A(X) f1 ] ! O(Xf ): Wir schreiben O(Xf ) = A(X) f1 ] = fgn j g 2 A(X); n 2 N : Insbesondere (wahle f 1) gilt O(X) = A(X): Weiter gilt: Fur jede ausgezeichnete o ene Menge U = Xf ist O(U) noetherscher Ring. Sind f1 ; : : :; fm 2 K z1 ; : : :; zn] Erzeuger des Ideals I(X) und ist fm+1 2 K z1 ; : : :; zn] Reprasentant von f 2 A(X), so ist O(Xf ) = K z1 ; : : :; zn+1] hf1 ; : : :; fm ; 1 zn+1 fm+1 i: Xf kann also als abgeschlossene Untervarietat von A n+1 aufgefa t werden!

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Algebraische Geometrie I by Heinz Spindler

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