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By Hans Kurzweil

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Man sagt, ein Polynom A ∈ F[X] zerf¨allt in Linearfaktoren, wenn es Produkt von linearen Polynomen ist; bis auf eine Normierung hat dann A in F[X] die Form A = (X − λ1 ) · (X − λ2 ) · · · (X − λn ) , n = grad A ; dabei sind λ1 , . . , λn die (nicht notwendig verschiedenen) Nullstellen von A. Existiert ein solche Zerlegung, so ist sie offenbar die Primfaktorzerlegung von A im Ring F[X]. 6 Sei n ∈ N und seien λ1 , λ2 , . . , λn verschiedene Zahlen aus F. Es gelte λni = 1 f¨ur i = 1, 2, . . , n .

Im vorigen Kapitel haben wir den Ring F N gebildet, es ist FN = Fn [X] und die Multiplikation ist modulo N . h. A ·N B = N (A · B) = 0 . Der Ring FN besitzt also Nullteiler. 7 analoge Aussage – mit demselben Beweis. 6 Genau dann ist der Ring FN nullteilerfrei, wenn N ein irreduzibles Polynom ist. 7 SATZ: Genau dann ist der Ring FN K¨orper, wenn N ein irreduzibles Polynom ist. In diesem Fall ist FN = GF (q n ) , wenn F = GF (q) . Wir werden sehen, daß dieser Satz alle endlichen K¨orper beschreibt.

Ein Polynom A ∈ F[X] ist dann entweder ein Vielfaches von N oder es ist teilerfremd zu N . Sei A ∈ F[X], grad A ≥ 1 . Mit P(A) bezeichnen wir die Menge der normierten, irreduziblen Teiler von A. 5 SATZ: Im Ring F[X] gilt PRIM. Beweis: Die Inklusion P(A) ∪ P(B) ⊆ P(A · B) ist klar. Fu¨r den Beweis der umgekehrten Inklusion sei P ∈ P(A · B), und sei P ∈ P(A) angenommen, wir zeigen P ∈ P(B). Wegen P ∈ P(A) sind P und A teilerfremd . 2 kgV(A, P ) = A · P . h. P ∈ P(B). Sei N ∈ F[X] vom Grad n ≥ 1. Im vorigen Kapitel haben wir den Ring F N gebildet, es ist FN = Fn [X] und die Multiplikation ist modulo N .

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Algebra I.K. by Hans Kurzweil

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