This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 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 { i } . Therefore f j ( x ) ( L K )[ x ], and the factorization of f ( x ) over K is the same as over L K . Suppose 1 is a root for f 1 , then deg f 1 = [ K ( 1 ) : K ]. Since f is irreducible over F , there is an isomorphism from K ( 1 ) K ( i ) for all roots i . Therefore all [ K ( i ) : K ] are equal. Now we need to show that m = [ F ( ) K : F ]. Consider the fields F ( ), where is any root of f , and K L . Then F ( ) ( K L ) = F ( ) K , and ( F ( ))( K L ) = ( K L )( )....
View
Full
Document
 Spring '09
 SusamaAgarwala
 Algebra

Click to edit the document details