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College Algebra Exam Review 427

# College Algebra Exam Review 427 - 9.5 THE GALOIS...

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9.5. THE GALOIS CORRESPONDENCE 437 determined by .˛/ , which is necessarily a root of f .x/ . But by Propo- sition 9.4.1 , there is a 0 2 Aut K .L/ such that 0 .˛/ D .˛/ ; therefore, D 0 j M . Now, the homomorphism theorem for groups gives Aut K .M/ Š Aut K .L/= Aut M .L/: This completes the proof of part (b). n We shall require the following variant of Proposition 9.5.3 . Proposition 9.5.5. Let L be a field, H a finite subgroup of Aut .L/ , and F D Fix .H/ . Then (a) L is a finite–dimensional Galois field extension of F . (b) H D Aut F .L/ and dim F .L/ D j H j . Proof. We cannot apply Proposition 9.5.3 because it is not given that L is finite–dimensional over F . Let ˇ be any element of L . We can adapt the argument of Proposition 9.4.13 to show that ˇ is algebraic and separable over F , and that the minimal polynomial for ˇ over F splits in L . Namely, let ˇ D ˇ 1 ; : : : ; ˇ r be the distinct elements of f .ˇ/ W 2 H g . Define g.x/ D .x ˇ 1 /.: : : /.x ˇ r / 2 LOExŁ . Every 2 H leaves g.x/ invariant, so the coefficients of g.x/ lie in Fix .H/ D F . Let p.x/ denote the minimal polynomial of ˇ over F . Since ˇ is a root of g.x/ , it follows that p.x/ divides g.x/ . On the other hand, every root of g.x/ is of the form
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