Algebra I by Walter Gubler

By Walter Gubler

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Wir haben also in G ein Element der Ordnung 12 gefunden. Da in Z/2Z × Z/2Z × Z/3Z alle Elemente der Ordnung kleiner gleich 6 haben (weil wir komponentenweise rechnen) gilt: G∼ = Z/3Z × Z/4Z Pr¨azise gilt f¨ ur g = (g1 , g2 , g3 ) ∈ Z/2Z × Z/2Z × Z/3Z: g = (g1 , g2 , g3 ) 2g = (2g1 , 2g2 , 2g3 ) .. ng = (ng1 , ng2 , ng3 ) Die erste Komponente ist 0, wenn n ∈ 2Z ist und analog die zweite Komponente. Die dritte Komponente ist 0, wenn n ∈ 3Z. Insgesamt sehen wir also, dass ng = 0 f¨ ur n ∈ 6Z und dies zeigt, dass die Ordnung jedes Element in Z/2Z × Z/2Z × Z/3Z kleiner oder gleich 6 ist.

M eine L-Basis von M bilden, m existieren µ1 , . . , µm ∈ L mit γ = µj γj . Weil β1 , . . , βl eine K-Basis in j=1 l L ist, gibt es λ1j , . . , λlj ∈ K mit µj = λij βi . Zusammen ergibt sich i=1 m γ= m µj γ j = j=1 l ( j=1 i=1 m l λij βi )γj = λij (βi γj ) j=1 i=1 Damit folgt, dass (βi γj )1≤i≤l,1≤j≤m ein K-Erzeugendensystem in M bilden und es folgt der zweite Schritt. Im dritten Schritt zeigen wir die Behauptung, f¨ ur den fall, dass [L : K] oder [M : L] = ∞ gilt. Wir nehmen dir K-linear unabh¨angige Elemente β1 , .

Ur ein g(x) ∈ K[x]. Beweis. Sei γ ∈ L. q(x), r(x) ∈ K[x] mit g(x) = q(x)f (x) + r(x) und grad(r) < grad(f ) =⇒ γ = g(x) = g(x)f (x) + r(x) f (x)=0 = r(x) = r(x) Benutzen wir nun grad(f (x)) =: n und r(x) = an−1 xn−1 + . . + a0 , sowie β := x, dann folgt aus obigem, dass γ = an−1 β n−1 + an−2 β n−2 + . . + a0 gilt. Also ist 1, β, . . , β n−1 ein K-Erzeugendensystem. Die Eindeutigkeit der Koordinaten a0 , . . , an−1 ∈ K folgt leicht aus der Eindeutigkeit von r(x) und der Konstruktion. Damit ist 1, β, .

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