Algebraische Zahlentheorie [Lecture notes] by J. Sander et al.

By J. Sander et al.

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H. α ∈ UF , also gilt die Zwischenbehauptung. 46 gilt f¨ u = ±1. Nach Ausmultiplizieren der rechten Seite ergibt Koeffizientenvergleich y = ±c(c2 − 6d2 ) und 1 = ±d(3c2 − 2d2 ) . Die zweite Gleichung liefert d = ±1, also 1 = ±(3c2 − 2) und somit c = ±1. Einsetzen in die erste Gleichung ergibt y = ±(1 − 6) = ±5, also x = 3. 1. 43). 43). 46). 50). 1 Ist F komplexer quadratischer Zahlk¨orper mit Diskriminante ∆F < −12, so ist OF kein euklidischer Ring. Beweis: Wir nehmen an, dass OF ein euklidischer Ring mit euklidischer Funktion f sei.

F¨ ur α ∈ OF irreduzibel ist zu zeigen: α ist prim. h. es gibt σ ∈ OF mit βγ = ασ. Nach Voraussetzung haben β, γσ eindeutige Zerlegungen r β =u· t s βj j=1 , γ=v· γj , j=1 σ=w· σj j=1 mit u, v, w ∈ UF und βj , γj , σj alle irreduzibel. Also t α·w· r σj = ασ = βγ = uv j=1 j=1 s βj · γj . j=1 Da α irreduzibel ist, folgt aus der eindeutigen Faktorisierung, dass α ∈ {βj : 1 ≤ j ≤ r} ∩ {γj : 1 ≤ j ≤ s}. h. α ist prim. ⇐=“ ” Sei jedes irreduzible Element von OF prim. Sei f¨ ur irreduzible αj , βj und u, v ∈ UF mit 1 ≤ s ≤ r (∗) uα1 · .

Qd,1 · · · qd,d Θ1 (αd ) · · · Θd (αd )   also durch quadrieren der Determinanten     ,  discr (B2 ) = D2 · discr (B1 ) . ✷ Beispiel: √ √ √ Sei F = Q( 13), und seien α = 12 (1 + 13), β = 13. Dann sind B1 = {1, α} und B2 = {1, β} Q-Basen von F (aber B2 ist keine Ganzheitsbasis von F ). Die Einbettungen vo F in C sind Θ1 : Wir haben und √ √ √ √ 13 −→ 13 und Θ2 : 13 −→ − 13 . 2 2  1 1 Θ1 (1) Θ2 (1)  = det  √ discr (B2 ) = det  √ √  = 52 √ Θ1 ( 13) Θ2 ( 13) 13 − 13  2 Θ1 (1) Θ2 (1)  discr (B1 ) = det  √ √ 1 1 Θ1 ( 2 (1 + 13)) Θ2 ( 2 (1 + 13))  2 1 1  = 13.

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