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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 φ : F ( s ) → F ( s ) , with φ ( s ) = s and φ ( b ) = b for all b ∈ F . Notice f ( x ) ∈ F [ x ] ⊂ F ( s )[ x ], φ ( f ( x )) = f ( x ) ∈ F [ x ] ⊂ F ( s )[ x ], and since K is a splitting field for f ( x ) = φ ( f ( x )), by theorem 1.10 φ can be extended to an isomorphism θ : K → K . In particular, θ ( b ) = b for all b ∈ F , so θ ∈ G . But θ ( 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 φ ∈ G to K [ x ] by setting φ ( x ) = x (this perserves isomorphism by theorem II.3.4). Then for any φ ∈ G , φ ( h ( x )) = Q s ∈ S ( x φ ( s )). If φ ( S ) = S for all φ ∈ G or φ ( S ) = S \ S for all φ ∈ G , G would not be transitive, so there exists a φ such that φ ( S ) 6 = S ,S \ S . Thus since h ( x ) is a product of distinct linear factors, φ ( 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 φ is an Fautomorphism, so φ ( h ( x )) = h ( x ), a contradiction. 4.2. If η ∈ K is an n th root of unity, η 6 = 1 , then 1 + η + η 2 + ··· + η n 1 = 0 . This follows since η n = 1, η 6 = 1, and 1 + x + x 2 + ··· + x n 1 = x n 1 x 1 ....
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 Spring '08
 MORRISON
 Algebra, Galois group, Field extension, splitting field

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