Algebra (2006) by Claus Scheiderer PDF

By Claus Scheiderer

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N}. In dem K¨ orperturm K ⊂ K(α1 ) ⊂ K(α1 , α2 ) ⊂ · · · ⊂ K(α1 , . . , αn ) = L hat jeder Schritt endlichen Grad, denn wegen αi+1 algebraisch u ¨ber K ist erst recht αi+1 algebraisch u orpergrad multiplikativ ist ¨ber K(α1 , . . , αi ) (i = 1, . . , n − 1). 2), folgt [L : K] < ∞. Also ist L algebraisch u ¨ber K. 14. Korollar. Jede endlich erzeugte algebraische K¨ orpererweiterung ist endlich (und umgekehrt, nat¨ urlich). 15. Korollar (Transitivit¨ at). Seien K ⊂ L ⊂ M K¨ orpererweiterungen.

Dann ist L algebraisch u ¨ber K. Anders gesagt: Summen und Produkte von u ¨ber K algebraischen Elementen sind wieder algebraisch u ¨ber K. Beweis. Sei etwa L = K(αi : i ∈ I), wobei alle αi algebraisch u ¨ber K seien. Sei α ∈ L. Zu α gibt es eine endliche Teilmenge J ⊂ I mit α ∈ K(αj : j ∈ J). Wir k¨ onnen daher annehmen I = {1, . . , n}. In dem K¨ orperturm K ⊂ K(α1 ) ⊂ K(α1 , α2 ) ⊂ · · · ⊂ K(α1 , . . , αn ) = L hat jeder Schritt endlichen Grad, denn wegen αi+1 algebraisch u ¨ber K ist erst recht αi+1 algebraisch u orpergrad multiplikativ ist ¨ber K(α1 , .

Hier bezeichnet f ◦ g das durch Einsetzen von g(t) f¨ ur die Variable t aus ¨ f (t) entstehende Polynom. Den Beweis u orern als leichte Ubung (Auf¨berlassen wir den H¨ gabe 25). 5. Satz. Sei 0 = f ∈ K[t], sei c ∈ K eine Nullstelle von f (t) von der Vielfachheit e ≥ 1. Dann ist c eine Nullstelle von f (t) von einer Vielfachheit ≥ e − 1. Dabei gilt Gleichheit genau dann, wenn char(K) kein Teiler von e ist (also z. B. stets f¨ ur char(K) = 0). Beweis. Es ist f (t) = (t − c)e · g(t) mit g(c) = 0. Ableiten gibt nach der Produktregel f (t) = e (t − c)e−1 · g(t) + (t − c)e · g (t) = (t − c)e−1 h(t) mit h(t) := eg(t) + (t − c)g (t).

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Algebra (2006) by Claus Scheiderer


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