Unformatted text preview: 332 7. FIELD EXTENSIONS – FIRST LOOK as well as with the quartic polynomial before coming to terms with the general case. Let us assume that E K is the splitting field of an irreducible cubic polynomial in KOExŁ . For each subgroup H of Aut K .E/ , consider Fix .H/ D f a 2 E W .a/ D a for all 2 H g . Proposition 7.4.6. Let E K be the splitting field of an irreducible cubic polynomial in KOExŁ . (a) For each subgroup H of Aut K .E/ , K Fix .H/ E is a field. (b) Fix . Aut K .E// D K . Proof. We leave part (a) as an exercise. For part (b), let K D Fix . Aut K .E// . We have K ¤ E , since E admits nontrivial K –automorphisms. In case dim K .E/ D 3 , it follows that K D K , as there are no fields strictly intermediate between K and E . Suppose now that dim K .E/ D 6 . Let f.x/ D x 3 C px C q 2 KOExŁ be an irreducible cubic polynomial with splitting field E . Let ı D .˛ 1 ˛ 2 /.˛ 1 ˛ 3 /.˛ 2 ˛ 3 / , where ˛ i are the roots of f.x/ . By Propo sition 7.4.5 , Aut K .E/ Š S 3 , acting as permutations of the three roots of...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
 Fall '08
 EVERAGE
 Algebra

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