College Algebra Exam Review 323

# College Algebra Exam Review 323 - D Because g.x splits into...

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7.4. SPLITTING FIELD OF A CUBIC POLYNOMIAL 333 ± the cyclic group A 3 , and ± the trivial subgroup f e g . Now, suppose K ² M ² E is some intermediate ﬁeld. Let H D Aut M .E/ ² Aut K .E/ Š S 3 : Then H must be one of the subgroups just listed. Put M D Fix .H/ ; then M ² M . I want to show that M D M (and that M must be one of the “known” intermediate ﬁelds.) The ﬁrst step is the following proposition: Proposition 7.4.7. If K ² M ³ ¤ E , then Aut M .E/ ¤ f e g . Proof. It is no loss of generality to assume K ¤ M . Because dim M .E/ divides dim K .E/ D 6 , it follows that dim M .E/ is either 2 or 3. Let a 2 E n M . Then E D M.a/ ; there are no more intermediate ﬁelds because of the multiplicativity of dimension. Consider the polynomial g.x/ D Y ± 2 Aut K .E/ .x ´ ±.a//: This polynomial has coefﬁcients in E that are invariant under Aut K .E/ ; because Fix . Aut K .E// D K , the coefﬁcients are in K . Now we can re- gard g.x/ as an element of MŒxŁ ; since g.a/ D 0 , g.x/ has an irreducible factor h.x/ 2 MŒxŁ such that h.a/
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Unformatted text preview: D . Because g.x/ splits into lin-ear factors in EŒxŁ , so does h.x/ . Thus E is a splitting ﬁeld for h.x/ . Since deg .h/ D dim M .E/ µ 2 , and since the roots of h.x/ are distinct by Exercise 7.4.1 , h.x/ has at least one root b 2 E other than a and, by Proposition 7.4.2 , there is an M –automorphism of E that takes a to b . Therefore, Aut M .E/ ¤ f e g : n Proposition 7.4.8. Let K ² M ² E be an intermediate ﬁeld. Let H D Aut M .E/ and let M D Fix .H/ . (a) If H is one of the H i , then M D M D K.˛ i / . (b) If H D A 3 , then M D M D K.ı/ . (c) If H D Aut K .E/ , then M D M D K . (d) If H D f e g , then M D M D E . Proof. If H D Aut K .E/ Š S 3 , then M D M D K , by Proposition 7.4.6 . If H D f e g , then M D M D E , by Proposition 7.4.7 . To complete...
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