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START: D1 D2 D3 C−1 = 1 , D−1 = 0 , C0 = 0 , D0 = 1 100101 : 11001 = 11 (= F1 ) 110010 10111 11001 1110 (= R2 ) , C1 = 1 , D1 = 11 11001 : 1110 = 10 (= F2 ) 11100 101 (= R3 ) , C2 = 10 , D2 = (10)(11) + 1 = 110 + 1 = 111 1110 : 101 = 11 (= F3 ) 1010 100 101 1 (= R4 ) C3 = (11)(10) + 1 = 110 + 1 = 111 D3 = (111)(11) + 11 = (1110 + 0111) + 11 = 1001 + 11 = 1010 Also ist ggT(N, A) = 1 und C3 · N + D 3 · A = 1 . b ist D3 = X 3 + X das zu A inverse Element im Ring FN . b auf Seite 58), also ist FN K¨orper (=GF (25 )).

Sei F = Z2 . Dann ist 12 + 1 = 0, also A(1) = 0. Sei F = Z3 . Dann ist 12 + 1 = 2 = 22 + 1 und A besitzt keine Nullstelle in F . Deshalb ist A irreduzibel. Sei F = R . Da −1 kein Quadrat in R ist, besitzt A keine Nullstelle in F, A ist also irreduzibel. Sei F = C und i die imagin¨are Einheit, also i2 = −1. Dann ist A = (X − i) · (X + i) . 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 , .

H. die Nullstellen von N in F. b) Sei A = 3X 5 + X 2 + 4X + 5 ∈ F[X]. Bestimme P, R ∈ F[X] mit A = P · N + R , grad R < 2. 2) Sei F = Z11 und N = X 11 + X 3 + 8 ∈ F[X]. Wieviel Elemente besitzt der Ring FN ? 3) Bestimme die Additions– und Multiplikationstafel des Rings F N in folgenden F¨allen a) F = Z3 und N = X 2 + 1 ∈ F[X] b) F = Z2 und N = X 3 + X 2 + 1 ∈ F[X] 34 4) Sei FN wie in Beispiel 7). Bestimme ein primitives Element des K o¨rpers FN . ¨ 5) Sei F = Z5 . Uber F l¨ose das lineare Gleichungssystem a0 + a 1 + a 2 = 2 a0 + 2a1 + 4a2 = 1 a0 + 3a1 + 4a2 = 0 in den Variablen a0 , a1 , a2 .

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Algebra I.K. by Hans Kurzweil


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