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New PDF release: Algebra II [Lecture notes]

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Additional resources for Algebra II [Lecture notes]

Sample text

N } 62 und es gilt (X − ρi α) . 3 folgt n SpK(α)/K (α) = −an−1 = ρi α i=1 n N NK(α)/K (α) = (−1) a0 = ρi α . 10 m = [E : F ] Fortsetzungen zu einem Morphismus σ ∈ G = HomK (E, C). 3 (i) folgt. Ebenso rechnet man f¨ ur die Norm m n ρi α σα = σ∈G = (NK(α)/K (α))m = NE/K (α) . 5. Es gilt der folgende Schachtelungssatz: sei E/K eine endliche K¨orpererweiterung, F ein Zwischenk¨orper. Dann gilt SpE/K = SpF/K ◦ SpE/F NE/K = NF/K ◦ NE/F Beweis: ausgelassen. 6. Sei E/K eine endliche separable K¨orpererweiterung.

Dann gilt p−1 q−1 q p = (−1) 2 2 p q In Worten: Gilt p = 1 mod 4 oder q = 1 mod 4, so ist q ein quadratischer Rest modulo p genau dann, wenn p ein quadratischer Rest modulo q ist. Gilt p = q = 3 mod 4, so ist q quadratischer Rest modulo p genau dann, wenn p nicht quadratischer Rest q ist. (ii) F¨ ur p = 2 gelten die folgenden Erg¨anzungss¨atze : p−1 −1 = (−1) 2 = p p2 −1 2 = (−1) 8 = p +1 f¨ ur p = 1 mod 4 −1 f¨ ur p = 3 mod 4 1 −1 p = 1, −1 mod 8 p = 3, −3 mod 8 Beispiel: Ist 29 ein Quadrat modulo 43?

Ein allgemeines Element dieses Ring R ist von der Form aσ σ(ζ) mit aσ ∈ Z . α= σ∈G(Q(ζ)/Q) 42 Wir rechnen nun modulo dem Hauptideal qR: aσ σ(ζ)q = σq (α) = aqσ σ(ζ)q mod qR q = aσ σ(ζ) mod qR Also gilt σq (α) = αq mod qR. 9 √ √ p∗ , so √ ∗ q q √ ∗ p = p mod qR . p √ Diese Gleichung multiplizieren wir mit p∗ und erhalten σq p∗ = q+1 q ∗ p = (p∗ ) 2 mod qR . p Nach unseren Voraussetzungen gilt (p∗ , q) = 1, also ist p∗ mod q invertierbar in Z/q ⊆ R/q. Es folgt q−1 q = (p∗ ) 2 mod qR , p also q p − (p∗ ) q−1 2 ∈ qR ∩ Z, da q ungerade sein sollte.

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