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Algebra [Lecture notes] by Eva Zerz

By Eva Zerz

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Zλ = G endlich ist (und L¨ange λ hat). 5. 16 Es ist G genau dann nilpotent, wenn es ein λ ≥ 0 gibt mit Zλ = G. Ist G nilpotent vom Grad l, so ist l die L¨ange der aufsteigenden Zentralreihe. Beweis: Sei G nilpotent vom Grad l. Wir zeigen durch Induktion, dass G(l−i) ⊆ Zi . F¨ ur i = 0 sind beide Gruppen gleich {e}. Sei die Aussage f¨ ur i − 1 gezeigt, also G(l−i+1) ⊆ Zi−1 . Dann ist G(l−i+1) = [G(l−i) , G] ⊆ Zi−1 . Daraus folgt G(l−i) Zi−1 /Zi−1 ⊆ Z(G/Zi−1 ), denn f¨ ur x ∈ G(l−i) , y ∈ G gilt [x, y] ∈ Zi−1 und daher xyZi−1 = yxZi−1 , also kommutiert xZi−1 mit jedem yZi−1 .

Dann hat f eine Nullstelle im K¨orper K[x]/ f ∼ = Fpd ⊆ Fpd! = Kd . Da Kd |K normal ist, zerf¨allt f u ¨ber Kd und dann erst recht u ¨ber L. 28 KAPITEL 1. GALOIS-THEORIE Schließlich bleibt zu zeigen, dass L algebraisch abgeschlossen ist. 10: Sei M |L eine endliche KE und α ∈ M . Zu zeigen ist α ∈ L. Da L|K algebraisch ist, ist α algebraisch u ¨ber K. Das Minimalpolynom µα ∈ K[x] zerf¨allt laut Vor¨ uberlegung bereits u ¨ber L ⊆ M , also sind seine Nullstellen in L. Insbesondere ist α ∈ L. 28 Seien M |L|K algebraische KE, K ein algebraischer Abschluss von K und φ : L → K ein K-Algebren-Homomorphismus.

Da f u ¨ber K(ζ) zerf¨allt, gilt dies auch f¨ ur fi . Also enth¨alt K(ζ) f¨ ur alle pi eine primitive pi -te Einheitswurzel. Nach dem Hauptsatz der Galois-Theorie entspricht die Kompositionsreihe von H einer K¨orperkette von Galois-Erweiterungen M = Ml ... M0 = K(ζ) mit Aut(Mi+1 |Mi ) ∼ = Hi /Hi+1 ∼ = Cpi . Mit dem n¨achsten Hilfssatz folgt, dass Mi+1 = Mi (βi ) gelten muss mit βipi ∈ Mi . Daraus folgt, dass M |K(ζ) radikal ist. 12 Sei f ∈ K[x] \ K. Dann ist f genau dann aufl¨osbar, wenn Gal(Lf |K) es ist, wobei Lf der Zerf¨allungsk¨orper von f ist.

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