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Additional info for Algebra II [Lecture notes]

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Nach Schritt 2 ist G(E/E H ) abgeschlossen, woraus die Inklusion H ≤ G(E/E H ) folgt. Wir behaupten, dass f¨ ur eine beliebige Untergruppe H = G(E/E H ) (12) gilt, woraus insbesondere f¨ ur abgeschlossene Untergruppen die Behauptung folgt. Dazu sei σ ∈ G(E/E H ). Um σ ∈ H zu zeigen, m¨ ussen wir f¨ ur jede Basisumgebung von σ der Form σG(E/L) zeigen, dass σG(E/L) ∩ H = ∅ , wobei L die endlichen galoischen Zwischenk¨orper von E/K durchl¨auft. Betrachte dazu die Surjektion G(E/E H ) ∪| H G(LE H /E H ) ∪| H0 Dann gilt E H ⊆ (LE H )H0 ⊆ E H , also ist E H = (LE H )H0 .

Iii) Ist PE/K (α) = X n + an−1 X n−1 + · · · + a0 ∈ K[X] das charakteristische Polynom, so gilt SpE/K (α) = −an−1 NE/K (α) = (−1)n a0 . 3. 61 (i) Sei F ein Zwischenk¨orper von E/K und m := [E : F ]. Dann gilt f¨ ur α ∈ F: PE/K (α) = PF/K (α)m SpE/K (α) = m SpF/K (α) NE/K (α) = NF/K (α)m (ii) Sei E/K eine endliche Erweiterung, α ∈ E und f = minK (α) = X n + . . + 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.

Nach diesem Satz gibt es ein β ∈ E, so dass K(β) = E. Sei f (X) = minK (β) das Minimalpolynom von β, das separabel ist und u ¨ber E in paarweise verschiedene Linearfaktoren zerf¨allt: f (X) = (X − σβ) ∈ E[X] . σ∈G 48 F¨ ur jedes Element σ ∈ G der Galoisgruppe betrachte das Polynom g σ (X) := f (X) X − σβ ∈ E[X] . Es gilt g σ (β) = 0 f¨ ur σ = e, aber g e (β) = 0. Betrachte die quadratische Matrix von Polynomen in E[X]: gτ −1 σ (X) f (X) X − τ −1 σβ = τ,σ∈G . τ,σ∈G Ihre Determinante, die nat¨ urlich auch im Polynomring E[X] liegt, d(X) := det(g τ −1 σ (X))τ,σ ∈ E[X] verschwindet nicht: durch Einsetzen des primitiven Elements β ∈ E erh¨alt man n¨amlich eine Diagonalmatrix, also gilt f¨ ur die Determinante d(β) = det(g τ −1 σ (β)) = g(β)n = 0 .

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