College Algebra Exam Review 420

College Algebra Exam Review 420 - H is Fix.H D f a 2 L W...

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430 9. FIELD EXTENSIONS – SECOND LOOK Let us quickly recall our investigation of splitting fields of irreducible cubic polynomials in Chapter 8, where we found the properties of the Ga- lois group corresponded to properties of the splitting field. The only pos- sibilities for the Galois group are A 3 D Z 3 and S 3 . The Galois group is A 3 if, and only if, the field extension K ± L is of dimension 3, and this occurs if, and only if, the element ı defined in Chapter 8 belongs to the ground field K ; in this case there are no intermediate fields between K and L . The Galois group is S 3 if, and only if, the field extension K ± L is of dimension 6. In this case subgroups of the Galois group correspond one to one with fields intermediate between K and L . We are aiming at obtaining similar results in general. Definition 9.4.6. Let H be a subgroup of Aut .L/ . Then the fixed field of
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Unformatted text preview: H is Fix .H/ D f a 2 L W ±.a/ D a for all ± 2 H g . Proposition 9.4.7. Let L be a field, H a subgroup of Aut .L/ and K ² L a subfield. Then (a) Fix .H/ is a subfield of L . (b) Aut Fix .H/ .L/ ³ H . (c) Fix . Aut K .L// ³ K . Proof. Exercise 9.4.1 . n Proposition 9.4.8. Let L be a field, H a subgroup of Aut .L/ , and K ² L a subfield. Introduce the notation H ı D Fix .H/ and K D Aut K .L/ . The previous exercise showed that H ı ³ H and L ı ³ L . (a) If H 1 ± H 2 ± Aut .L/ are subgroups, then H ı 1 ³ H ı 2 . (b) If K 1 ± K 2 ± L are fields, then K 1 ³ K 2 . Proof. Exercise 9.4.2 . n Proposition 9.4.9. Let L be a field, H a subgroup of Aut .L/ , and K ² L a subfield. (a) .H ı / 0ı D H ı . (b) .K / ı0 D K ....
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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.

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