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Algebra I.K. by Hans Kurzweil

By Hans Kurzweil

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Mit dem Kalk¨ ul der Funktionentheorie bewiesen werden 53 Teiler P von N mit grad P = 2, der keine Nullstelle in F besitzt. 5 ist P irreduzibel und nach dem schon bewiesenem ist P = X 2 + X + 1. Wir haben aber in Beispiel II auf Seite 22, gezeigt, dass X 2 + X + 1 kein Teiler von N ist. Also ist N irreduzibel. V¨ollig analog beweist man dies in den F¨allen N = 1 + X + X 3 + X 4 und N = 1 + X + X 2 + X 3 + X 4 . Nun gibt es noch weitere fu¨nf Polynome vom Grad 4 . Diese werden, wenn sie keine Nullstele in F besitzen, von dem irreduziblen Polynom 1 + X + X 2 geteilt, sind also nicht irreduzibel; wir unterdru¨cken die dazugeh¨orenden Rechnungen.

A(λ) = 0. Dann existiert ein P ∈ F[X] , so dass A = (X − λ) · P . Beweis: Im Fall grad A ≤ 0 ist A = 0, da A(λ) = 0; wir k¨onnen also P = 0 setzen. Sei grad A ≥ 1. Wir teilen A durch X − λ A = P · (X − λ) + R , grad R < 1 , und zeigen R = 0. Dazu setzen wir λ in A ein und erhalten 0 = A(λ) = P (λ)(λ − λ) + R(λ) = R(λ) . Wegen grad R < 1 ist R = 0 . Sei λ Nullstelle von A, also A = (X − λ) · P . Ist µ = λ eine weitere Nullstelle von A, so ist 0 = A(µ) = (µ − λ)P (µ), also P (µ) = 0. 1 hat P die Form P = (X − µ) · Q, Q ∈ F[X].

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 . 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.

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