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Extra info for Algebra II [Lecture notes]
A0 das Minimalpolynom. Setze m := [E : K(α)]. Dann gilt PK(α)/K (α) PE/K (α) SpE/K (α) NE/K (α) = = = = f fm −man−1 (−1)mn am 0 Dies erlaubt es, Norm und Spur eines Elements α ∈ E aus seinem Minimalpolynom abzulesen. Beweis. (i) folgt daraus, dass E ∼ = F m als K-Vektorraum. (ii) Da das Minimalpolynom f das charakteristische Polynom PK(α)/K (α) teilt, und grad f = [K(α) : K] = grad PK(α)/K (α) folgt f = PK(α)/K und aus (i) die weiteren Behauptungen. 4. Sei E/K eine endliche und separable K¨orpererweiterung, C ein algebraischer Abschluss von E und setze G = HomK (E, C).
Von Q(ζ)/Q zeigt, dass qZ[ζ] ∩ Z = qZ gilt. 7 (i) verwendet wurde. Wir finden q p∗ = p q −1 = q p−1 2 p−1 q−1 p p = (−1) 2 2 , q q wobei in der letzten Gleichheit noch einmal q = 2 benutzt wurde. (ii) Um die Gleichung 2 p 2 = (−1) p 8−1 zu zeigen, betrachten wir den K¨orper Q(ζ) mit ζ = ζ8 = e2πi/8 und rechnen im Ring Q(ζ8 ). Wegen (ζ + ζ −1 )2 = ζ 2 + 2 + ζ −2 = 2 + i − i = 2 ist y = ζ + ζ −1 = 43 √ 2 Ferner gilt y p = ζ p + ζ −p mod p . h. 2 2 = 1 mod p. h. es ist 2 −1. 5 p−1 2 = −1 mod p.
7 ist die Abbildung injektiv, denn dort wurde im Beweis Endlichkeit nicht benutzt. 4. Schritt: Sei H ⊆ G(E/K) eine offene Untergruppe und F = E H der Fixk¨orper. Dann ist die K¨orpererweiterung F/K endlich. 8 (ii) gibt es eine endliche galoische Erweiterung L/K mit 58 L ⊆ E, so dass G(E/L) ⊆ H gilt. Damit gilt aber auch die Inklusion F = E H ⊆ E G(E/L) = L, also ist auch die K¨orpererweiterung F/K endlich. 5. Schritt: Sei H ⊆ G(E/K) eine abgeschlossene Untergruppe. Dann ist H = G(E/E H ). Beweis des fu ¨ nften Schritts: Sei H eine nicht notwendigerweise abgeschlossene Untergruppe von G(E/K).