Download Algebraische Zahlentheorie [Lecture notes] by Rainer Vogt PDF

By Rainer Vogt

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Sommersemester 2010

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Das Minimalpolynom f von α ist separabel und hat daher nur verschiedene Nullstellen. Es folgt direkt aus der Definition, dass D(f ) = 0. 7 folgt DF/K (1, α, α2 , . . 4 auch DF/K (β1 , . . , βn ) = 0. Wir wollen nun eine Methode zur Berechnung von DF/K (1, α, . . , αn−1 ) angeben, f¨ ur die man α nicht explizit zu kennen braucht. 10 Satz: Sei F = K(α), α separabel u ¨ber K n n−1 X + an−1 X + . . + a1 X + a0 . Dann gilt  p0 p1  p1 p2  DF/K (1, α, . . , αn−1 ) = det  ..  . mit Minimalpolynom f = p2 p3 ..

N ) = det(σi (αj ))2 . Beweis: DF/K (α1 , . . , αn ) = det(SF/K (αi · αj )) = det( k σk (αi · αj )) = det( k σk (αi ) · σk (αj )) = det((σk (αi ))i,k · (σk (αj ))k,j ) = det(σk (αj ))2 . 6 Definition: Sei f ∈ K[X] ein Polynom vom Grad n und Leitkoeffizin ent c. Sei f = c · dann nennt man i=1 (X − αi ) seine Faktorisierung in Linearfaktoren in K[X], D(f ) = 1≤i

10 sind damit alle ci ganz u ¨ber R. Da ci ∈ K und R ganz abgeschlossen in K ist, sind alle ci ∈ R, also f ∈ R[X]. 18 Aufgabe: (1) Zeigen Sie: Sei R ganz abgeschlossener Integrit¨atsring und K sein Quotientenk¨orper. Sei f ∈ R[X] normiert und seien g, h ∈ K[X] normierte Polynome, so dass f =g·h Dann gilt g, h ∈ R[X]. 30 in K[X]. 17 mit Hilfe von (1). 19 Satz: Sei R ⊂ F ganze Ringerweiterung und F ein K¨orper. Dann ist auch R ein K¨orper. Beweis: Sei a = 0 aus R. Da a1 ∈ F ganz u ullt es eine Ganz¨ber R ist, erf¨ heitsgleichung mit Koeffizienten ci ∈ R 1 a n + cn−1 n−1 1 a + .

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