College Algebra Exam Review 429

College Algebra Exam Review 429 - L D [f M W K M L and M is...

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9.5. THE GALOIS CORRESPONDENCE 439 inductive assumption applied to the original Equation ( 9.5.1 ) gives that all the coefficients ± i are zero. n Proposition 9.5.8. Let K ± L be a field extension with dim K .L/ finite. Then j Aut K .L/ j ² dim K .L/ . Proof. Suppose that dim K .L/ D n and f ± 1 ;:::;± n g is a basis of L over K . Suppose also that f ² 1 ;:::;² n C 1 g is a subset of Aut K .L/ . (We do not assume that the ² i are all distinct!) The n -by- n C 1 matrix ± ² j i / ² 1 ± i ± n; 1 ± j ± n C 1 has a nontrivial kernel by basic linear algebra. Thus, there exist b 1 ;:::;b n C 1 in L , not all zero, such that X j ² j i /b j D 0 for all i . Now, if k 1 ;:::;k n are any elements of K , 0 D X i k i . X j ² j i / b j / D X j b j ² j . X i k i ± i /: But P i k i ± i represents an arbitrary element of L , so the last equation gives P j b j ² j D 0 . Thus, the collection of ² j is linearly dependent. By the previous proposition, the ² j cannot be all distinct. That is, the cardinality of Aut K .L/ is no more than n . n Exercises 9.5 9.5.1. Suppose that K ± L is an algebraic field extension. Show that
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Unformatted text preview: L D [f M W K M L and M is nitedimensional g . If there is an N 2 N such that dim K .M/ N whenever K M L and M is nitedimensional, then also dim K .L/ N . 9.5.2. This exercise gives the proof of Proposition 9.5.6 . We suppose that K L is a nitedimensional eld extension, that A;B are elds interme-diate between K and L , and that B is Galois over K . Let be an element of B such that B D K./ (Proposition 9.5.1 ). Let p.x/ 2 Kx be the minimal polynomial for . Then B is a splitting eld for p.x/ over K , and the roots of p.x/ are distinct, by Theorem 9.4.14 . (a) Show that A B is Galois over A . Hint: A B D A./ ; show that A B is a splitting eld for p.x/ 2 Ax ....
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