Algebra by Markus Junker

By Markus Junker

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B)⇒(c) ist klar. (c)⇒(a): Sei L = K(a1 , . . , an ) mit u ¨ber K algebraischen ai . Dann ist [L : K] = [K(a1 ) : K] · [K(a1 , a2 ) : K(a1 )] · . . · [L : K(a1 , . . , an−1 ] [K(a1 ) : K] · [K(a2 ) : K] · . . 3 (a) M/L und L/K algebraisch ⇐⇒ M/K algebraisch. (b) Sind a1 , . . ,an ) P Q ∈ K(X1 , . . , Xn ) mit Q(a1 , . . , an ) = 0, so algebraisch u ¨ber K. Insbesondere sind dies a1 ± a2 , a1 a2 und a1 a2 f¨ ur a2 = 0. Beweis: (a) ⇐“ ist klar. ” ⇒“: Sei a ∈ M und c0 , . . , ck die Koeffizienten von mina/L .

Beweis: Falls f (X) = c(X − a1 ) . . (X − an ), so f (X) = c · (X − aj ). Man sieht: Ist i=1 j=i a1 = a2 doppelte Nullstelle, so auch Nullstelle von f und X − a1 ist gemeinsamer Teiler. E. ein Linearfaktor X − ai , dann teilt X − ai auch (X − aj ), also folgt ai = aj f¨ ur ein j = i. 5 Sei K ein K¨ orper und αi ∈ Aut(K) f¨ ur i ∈ I. Dann ist Fix({αi | i ∈ I}) := {x ∈ K | αi (x) = x f¨ ur alle i ∈ I} ein K¨ orper. Beweis: Ist αi (x) = x und αi (y) = y, so auch αi (x + y) = αi (x) + αi (y) = x + y, ebenso f¨ ur die anderen K¨ orperoperationen.

Es gibt also soviele Bilder von K(a) unter Aut(K/K) wie Konjugierte von a u ¨ber K. h. in L[X] in Linearfaktoren zerf¨ allt. L heißt Aut(K/K)-invariant, falls α(L) = L f¨ ur alle α ∈ Aut(K/K). 3 F¨ ur eine endliche Erweiterung L/K sind gleichwertig: (a) L/K ist normal; (b) L ist Aut(K/K)-invariant; (c) L ist Zerf¨ allungsk¨ orper u ¨ber K eines Polynoms f ∈ K[X]. Beweis: (a)⇒(c): Sei L = K(b1 , . . , bn ), so ist L Zerf¨ allungsk¨orper von minb1 /K (X) · · · minbn /K (X) u ¨ber K. (c)⇒(b): Sei α ∈ Aut(K/K).

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