Algebraic Geometry

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By J. Sander et al.

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Ist B2 = {β1 , β2 , . . , βd } ⊆ F mit d βk = qi,k αi i=1 (k = 1, . . , d; qi,k ∈ Q) , so gilt: B2 ist Q-Basis von F ⇐⇒ det(qi,k ) = 0. Beweis: =⇒“ ” Sei B2 Q-Basis von F . 35 ist discr (B2 ) = D2 · discr (B1 ) mit D = det(qi,k ). 36 ist discr (B2 ) = 0, also auch D = 0. ⇐=“ ” Sei det(qi,k ) = 0. Wir haben zu zeigen, dass β1 , β2 , . . , βd linear unabh¨angig u ¨ber Q sind. Die Annahme γ1 β1 + · · · + γd βd = 0 f¨ ur gewisse γi ∈ Q impliziert d 0= d d γk γk βk = k=1 k=1 d αi qi,k αi = i=1 d i=1 k=1 Wegen der linearen Unabh¨angigkeit der αi folgt d k=1 γk · qi,k = 0 (1 ≤ i ≤ d) .

Nach dem Gauß-Lemma muß dann auch Φn (x) ∈ Z[x] gelten. ur jedes Behauptung: Ist ζ eine primitive n-te Einheitswurzel, so gilt Mζx ,Q (ζ p ) = 0 f¨ p ∈ P, p n. Wir k¨ urzen ab m1 (x) := mζn ,Q (x) und m2 (x) := mζ p ,Q (x). Da ζ n-te Einheitswurzel ist, gilt m1 (x) | (xn − 1). Da auch ζ p n-te Einheitswurzel ist, folgt m2 (x) | (xn − 1). Mit m1 (ζ p ) = 0 w¨are unsere Behauptung bewiesen. Sei also m1 (ζ p ) = 0. h. (∗) xn − 1 = m1 (x) · m2 (x) · g(x) ur ein g(x) ∈ Z[x]. h. es gibt h(x) ∈ Z[x] mit m2 (xp ) = m1 (x) · h(x) .

F¨ F¨ ur 2 r2 kommt a = 0, also ∆F = (bi)2 = −b2 < 0. 5. h. die multiplikativ invertierbaren Elemente) eine multiplikative (Unter-)Gruppe UR bilden. Wir bezeichnen mit g die von einem Element g einer Gruppe erzeugte zyklische (Unter-)Gruppe. 46 √ ur Ist ∆F < 0 f¨ ur den komplexen quadratischen Zahlk¨orper F = Q( ∆F ), so gilt f¨ die Einheitengruppe UF := UOF  √   ζ6 := 21 (1 − i 3)   = ζ4 := i     ζ := −1 2 Beweis: f¨ ur ∆F = −3 , ur ∆F = −4 , f¨ sonst . h. 43. 26 folgt NF (u) = ±1. Mit NF (u) = a2 − b2 D > 0 ergibt sich NF (u) = 1.

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