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Unformatted text preview: Ma 5 c HOMEWORK 5 SOLUTION SPRING 09 The exercises are taken from the text, Abstract Algebra (third edi tion) by Dummit and Foote, unless stated otherwise. Page 595, 4 . Suppose f ( x ) = Q ( x α i ) ∈ L [ x ], and f ( x ) = Q m j =1 f j ( x ) with f j ( x ) irreducible over K . Now each the coefficients of each f j is a polynomial in { α i } . Therefore f j ( x ) ∈ ( L ∩ K )[ x ], and the factorization of f ( x ) over K is the same as over L ∩ K . Suppose α 1 is a root for f 1 , then deg f 1 = [ K ( α 1 ) : K ]. Since f is irreducible over F , there is an isomorphism from K ( α 1 ) → K ( α i ) for all roots α i . Therefore all [ K ( α i ) : K ] are equal. Now we need to show that m = [ F ( α ) ∩ K : F ]. Consider the fields F ( α ), where α is any root of f , and K ∩ L . Then F ( α ) ∩ ( K ∩ L ) = F ( α ) ∩ K , and ( F ( α ))( K ∩ L ) = ( K ∩ L )( α )....
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 Spring '09
 SusamaAgarwala
 Algebra, Group Theory, li, Cyclic group, pai th root

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