College Algebra Exam Review 426

College Algebra Exam Review 426 - 9.5.1 Let f.x denote the...

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436 9. FIELD EXTENSIONS – SECOND LOOK Then, by the argument of Proposition 9.4.13 , the minimal polynomial for ˇ over F is g.x/ D .x ˇ 1 /.: : : /.x ˇ n / . Therefore, dim F L D deg .g/ D j H j j Aut F .L/ j D dim F .L/; using Proposition 9.4.16 . n We are now ready for the fundamental theorem of Galois theory: Theorem 9.5.4. Let K L be a Galois field extension. (a) There is an order-reversing bijection between subgroups of Aut K .L/ and intermediate fields K M L , given by H 7! Fix .H/ . (b) The following conditions are equivalent for an intermediate field M : (i) M is Galois over K . (ii) M is invariant under Aut K .L/ (iii) Aut M .L/ is a normal subgroup of Aut K .L/ . In this case, Aut K .M/ Š Aut K .L/= Aut M .L/: Proof. If K M L is an intermediate field, then L is Galois over M (i.e., M D Fix . Aut M .L// ). On the other hand, if H is a subgroup of Aut K .L/ , then, according to Proposition 9.5.3 , H D Aut Fix .H/ .L/ . Thus, the two maps H 7! Fix .H/ and M 7! Aut M .L/ are inverses, which gives part (a). Let M be an intermediate field. There is an ˛ such that M D K.˛/
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Unformatted text preview: 9.5.1 . Let f.x/ denote the minimal polynomial of ˛ over K . M is Galois over K if, and only if, M is the splitting field for f.x/ , by Theorem 9.4.14 . But the roots of f.x/ in L are the images of ˛ under Aut K .L/ by Proposition 9.4.4 . Therefore, M is Galois over K if, and only if, M is invariant under Aut K .L/ . If ± 2 Aut K .L/ , then ±.M/ is an intermediate field with group Aut ±.M/ .L/ D ± Aut M .L/± ± 1 : By part (a), M D ±.M/ if, and only if, Aut M .L/ D Aut ±.M/ .L/ D ± Aut M .L/± ± 1 : Therefore, M is invariant under Aut K .L/ if, and only if, Aut M .L/ is nor-mal. If M is invariant under Aut K .L/ , then ² W ± 7! ± j M is a homo-morphism of Aut K .L/ into Aut K .M/ , with kernel Aut M .L/ . I claim that this homomorphism is surjective. In fact, an element of ± 2 Aut K .M/ is...
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