Download e-book for kindle: Algebraische Zahlentheorie [Lecture notes] by Rainer Vogt

By Rainer Vogt

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Sommersemester 2010

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Extra resources for Algebraische Zahlentheorie [Lecture notes]

Example text

En } in V hat das Volumen 1, (ii) das von n Vektoren v1 , . . , vn aufgespannte Parallelepiped Φ := {x1 · v1 + . . + xn · vn ; 0 ≤ xi < 1} hat das Volumen vol(Φ) = | det(ϕ(v1 ), . . , ϕ(vn ))|. Wie man aus der linearen Algebra weiß, h¨angt dieser Volumenbegriff nicht von der Wahl der Orthonormalbasis ab. Ist n¨amlich T die Transformationsmatrix zwischen zwei Orthonormalbasen, dann ist T eine orthogonale Matrix, ihre Determinante also det T = ±1. 9 Definition: Sei Γ ein vollst¨andiges Gitter in V mit Basis {v1 , .

1. Also folgt OK = Z[α] . (2) f = X 3 − X − 1. Wieder ist {1, α, α2 } ganze Basis, da D(f ) = (−1)3·2/2 (33 (−1)2 + (−1)2 · 22 · (−1)3 ) = −23. OK = Z[α]. (3) f = X 5 − X − 1 D(f ) = (−1)5·4/2 (55 · (−1)4 + (−1)4 · 44 · (−1)5 ) = 3125 − 256 = 2869 D(f ) = 19 · 151. Also ist D(f ) wieder quadratfrei, so dass OK = Z[α] mit ganzer Basis {1, α, α2 , α3 , α4 }. ¨ Die Beispiele belegen, dass es ohne den Einsatz theoretischer Uberlegungen oft schwierig sein kann, ganze Basen zu finden. Das folgende Ergebnis soll dabei helfen.

Da R noethersch ist, gibt es ein maximales Gegenbeispiel J. h. es gibt Elemente x, y ∈ R, so dass x · y ∈ J, aber x ∈ / J und y ∈ / J. Dann gilt J J + (x), J J + (y) und (J + (x)) · (J + (y)) ⊂ J + (x · y) = J. Da J ein maximales Gegenbeispiel ist, enthalten J +(x) und J +(y) Produkte von Primidealen = 0 und damit aber auch J, ein Widerspruch. 13 Lemma: Sei p Primideal in einem Dedekindring R. Sei K der Quotientenk¨orper von R und p−1 = {x ∈ K; x · p ⊂ R}. Dann gilt: (1) p−1 ist ein R-Modul. (2) Ist J ⊂ R ein Ideal, dann ist J · p−1 := { ni=1 ai · xi ; n ∈ N, ai ∈ J, xi ∈ p−1 } ebenfalls ein R-Modul, und es gilt J · p−1 = J, falls J = 0.

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