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Unformatted text preview: Problem Set #3 Solutions Chapter III. 4.1 If f ( x ) F [ x ] and K is a splitting field for f ( x ) over F , denote by S the set of distinct roots of f ( x ) in K and let G = G ( K : F ) . i. If f ( x ) is irreducible over F , then G is transitive on S . Since K is a splitting field for f ( x ), f ( x ) = c Q s S ( x- s ) n s in K [ x ] for some n s Z 1 , c F . Thus 1 c f ( x ) is a minimal polynomial over F since it is irreducible and monic with roots S , so all the s S are conjugate over F . If S consists of a single element G is trivially transitive, so let s,s S . By the corollary to proposition 1.9, since s,s K are conjugate there is an isomorphism : F ( s ) F ( s ) , with ( s ) = s and ( b ) = b for all b F . Notice f ( x ) F [ x ] F ( s )[ x ], ( f ( x )) = f ( x ) F [ x ] F ( s )[ x ], and since K is a splitting field for f ( x ) = ( f ( x )), by theorem 1.10 can be extended to an isomorphism : K K . In particular, ( b ) = b for all b F , so G . But ( s ) = s , so G is transitive. ii. If f ( x ) has no repeated roots and G is transitive on S , then f ( x ) is irre- ducible over F . If | S | = 1 then f ( x ) is linear and hence irreducible. Otherwise, if f ( x ) = Q s S ( x- s ) were reducible over F , then for some subset S S we would have h ( x ) = Q s S ( x- s ) F [ x ]. Extend each G to K [ x ] by setting ( x ) = x (this perserves isomorphism by theorem II.3.4). Then for any G , ( h ( x )) = Q s S ( x- ( s )). If ( S ) = S for all G or ( S ) = S \ S for all G , G would not be transitive, so there exists a such that ( S ) 6 = S ,S \ S . Thus since h ( x ) is a product of distinct linear factors, ( h ( x )) 6 = h ( x ). (Note that if h ( x ) were not a product of distinct linear factors, it is easy to construct an example where this would fail by using exercise 3.1.) But h ( x ) F [ x ] and is an F-automorphism, so ( h ( x )) = h ( x ), a contradiction. 4.2. If K is an n th root of unity, 6 = 1 , then 1 + + 2 + + n- 1 = 0 . This follows since n = 1, 6 = 1, and 1 + x + x 2 + + x n- 1 = x n- 1 x- 1 ....
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This note was uploaded on 08/06/2008 for the course MATH 220 taught by Professor Morrison during the Spring '08 term at UCSB.
- Spring '08