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By Norbert Klingen

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Bj ∗ bj und folglich R[a] = ⊕j R · f (a) . f (a) n n−1 i j alt man rekursiv (mit bn = 0) i=0 ci X := f (X) = (X − a) j=0 bj X erh¨ Die Dualbasis zu ar ist also a∗j = Aus ci = bi−1 − abi bzw. bi−1 = abi + ci (i = 0, . . , n) und daraus explizit bn−1 = 1 , bn−2 = a + cn−1 , bn−3 = abn−2 + cn−2 = a2 + acn−1 + cn−2 , .. i−2 bn−i = ai−1 + ai−2 cn−1 + . . + acn−i + cn−i+1 = ai−1 + aj cn−j+1 . ) und auf der Hauptdiagonalen lauter Einsen, also Determinante 1 und somit invertierbar u ¨ber R. Also gilt R[a] ∗ = n−1 n−1 1 1 · ⊕ Rbj = · ⊕ Raj .

Um also die Idealgruppe zu studieren, muss man sich eine ¨ Ubersicht u ¨ber die Primideale des Ringes verschaffen. Dazu betrachten wir (einfachere) Unterringe und setzen die Idealgruppen, speziell die Primideale dieser Ringe miteinander in Beziehung. Dabei gehen wir grunds¨ atzlich von folgender Situation aus: R ist ein Dedekindring mit Quotientenk¨ orper k. Wir betrachten R als (den fixierten) ausgezeichneten Dedekindring mit Quotientenk¨ orper k. In endlich separablen K¨ orpererweiterungen K, K , N, .

Es sei f das Minimalpolynom von a u ¨ber k und f die (formale) Ableitung. a) Dann gilt: n−1 aj 1 ∗ R[a] = ⊕ R · = · R[a] . f (a) f (a) j=0 b) Die Differente DK|k enth¨ alt alle Elementdifferenten ganzer primitiver Elemente: δK|k (a) | a ∈ S 48 S ⊂ DK|k c) Die Diskriminante dK|k enth¨ alt alle Elementdiskriminanten ganzer primitiver Elemente: dK|k (a) | a ∈ S R ⊂ dK|k . Zusatz (ohne Beweis): In b) gilt Gleicheit, in c) jedoch i. a. nicht. Beweis: a) R[a] ist freier R-Modul mit der Basis aj (j = 0, .

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