Chapter6 - Chapter 6 Galois Theory 6.1 Fixed Fields and...

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Chapter 6 Galois Theory 6.1 Fixed Fields and Galois Groups Galois theory is based on a remarkable correspondence between subgroups of the Galois group of an extension E/F and intermediate Felds between E and F . In this section we will set up the machinery for the fundamental theorem. [A remark on notation: Throughout the chapter, the composition τ σ of two automorphisms will be written as a product τσ .] 6.1.1 Defnitions and Comments Let G = Gal( ) be the Galois group of the extension .I f H is a subgroup of G , the fxed feld of H is the set of elements Fxed by every automorphism in H , that is, F ( H )= { x E : σ ( x x for every σ H } . If K is an intermediate Feld, that is, F K E , deFne G ( K ) = Gal( E/K { σ G : σ ( x x for every x K } . I like the term “ fxing group oF K ” for G ( K ), since G ( K ) is the group of automorphisms of E that leave K Fxed. Galois theory is about the relation between Fxed Felds and Fxing groups. In particular, the next result suggests that the smallest subFeld F corresponds to the largest subgroup G . 6.1.2 Proposition Let be a Fnite Galois extension with Galois group G = Gal( ). Then (i) The Fxed Feld of G is F ; (ii) If H is a proper subgroup of G , then the Fxed Feld of H properly contains F . 1
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2 CHAPTER 6. GALOIS THEORY Proof. (i) Let F 0 be the fxed feld oF G .I F σ is an F -automorphism oF E , then by defnition oF F 0 , σ fxes everything in F 0 . Thus the F -automorphisms oF G coincide with the F 0 -automorphisms oF G . Now by (3.4.7) and (3.5.8), E/F 0 is Galois. By (3.5.9), the size oF the Galois group oF a fnite Galois extension is the degree oF the extension. Thus [ E : F ]=[ E : F 0 ], so by (3.1.9), F = F 0 . (ii) Suppose that F = F ( H ). By the theorem oF the primitive element (3.5.12), we have E = F ( α ) For some α E . Defne a polynomial f ( X ) E [ X ]by f ( X )= Y σ H ( X σ ( α )) . IF τ is any automorphism in H , then we may apply τ to f (that is, to the coefficients oF f ; we discussed this idea in the prooF oF (3.5.2)). The result is ( τf )( X Y σ H ( X ( τσ )( α )) . But as σ ranges over all oF H , so does , and consequently = f . Thus each coefficient oF f is fxed by H , so f F [ X ]. Now α isarooto F f , since X σ ( α ) is 0 when X = α and σ is the identity. We can say two things about the degree oF f : (1) By defnition oF f , deg f = | H | < | G | =[ E : F ], and, since f is a multiple oF the minimal polynomial oF α over F , (2) deg f [ F ( α ): F E : F ], and we have a contradiction. There is a converse to the frst part oF (6.1.2). 6.1.3 Proposition Let be a fnite extension with Galois group G . IF the fxed feld oF G is F , then is Galois. Proof. Let G = { σ 1 ,...,σ n } , where σ 1 is the identity. To show that is normal, we consider an irreducible polynomial f F [ X ] with a root α E . Apply each au- tomorphism in G to α , and suppose that there are r distinct images α = α 1 = σ 1 ( α ), α 2 = σ 2 ( α ) ,...,α r = σ r ( α ). IF σ is any member oF G , then σ will map each α i to some α j , and since σ is an injective map oF the fnite set { α 1 r } to itselF, it is surjective as well. To put it simply, σ permutes the α i . Now we examine what σ does to the elementary symmetric functions oF the α i , which are given by e 1 = r
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This note was uploaded on 11/14/2011 for the course MATH 367 taught by Professor Sdd during the Spring '11 term at Middle East Technical University.

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Chapter6 - Chapter 6 Galois Theory 6.1 Fixed Fields and...

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