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By J. Sander et al.

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Example text

F¨ B = {1, α, α2 , . . , αd−1 } heißt det(Θj (αj−1 )) 1≤i≤d Vandermonde-Determinante. 34 Die Vandermondsche-Determinante hat den Wert det(Θj (αj−1 )) 1≤i≤d = 1≤j≤d (αi − αj ) , 1≤i

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) . Wir betrachten die Polynomidentit¨aten u ¨ber Z nun modulo p. Man zeigt leicht mit dem Binomischen Lehrsatz, dass f¨ ur beliebige Polynome f (x) gilt f (xp ) ≡ f (x)p mod p.

Die gegebene Bachet-Gleichung liefert in Z[ −2] die Faktorisierung (y + Behauptung: ggT(y + Sei dazu f¨ ur a, b ∈ Z √ √ −2) · (y − −2, y − √ −2) = x3 . √ −2) = 1. √ √ √ α := (a + b −2) | ggT(y + −2, y − −2) . Es folgt NF (α) | NF ((y + √ √ √ −2) − (y − −2)) = NF (2 −2) = 8 und NF (α) | NF (x3 ) = x6 . h. α ∈ UF , also gilt die Zwischenbehauptung. 46 gilt f¨ u = ±1. Nach Ausmultiplizieren der rechten Seite ergibt Koeffizientenvergleich y = ±c(c2 − 6d2 ) und 1 = ±d(3c2 − 2d2 ) . Die zweite Gleichung liefert d = ±1, also 1 = ±(3c2 − 2) und somit c = ±1.

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Algebraische Zahlentheorie [Lecture notes] by J. Sander et al.


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