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 eld, H a subgroup of Aut .L/ and K L a subeld. Then (a) Fix .H/ is a subeld of L . (b) Aut Fix .H/ .L/ H . (c) Fix . Aut K .L// K . Proof. Exercise 9.4.1 . n Proposition 9.4.8. Let L be a eld, H a subgroup of Aut .L/ , and K L a subeld. Introduce the notation H D Fix .H/ and K D Aut K .L/ . The previous exercise showed that H H and L L . (a) If H 1 H 2 Aut .L/ are subgroups, then H 1 H 2 . (b) If K 1 K 2 L are elds, then K 1 K 2 . Proof. Exercise 9.4.2 . n Proposition 9.4.9. Let L be a eld, H a subgroup of Aut .L/ , and K L a subeld. (a) .H / 0 D H . (b) .K / 0 D K ....
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