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Unformatted text preview: Ma 5 c HOMEWORK 6 SOLUTION SPRING 09 The exercises are taken from the text, Abstract Algebra (third edi tion) by Dummit and Foote, unless stated otherwise. Page 618, 16 . The polynomial is irreducible by brute force. The resol vent cubic is x 3 16 x 2 +48 x +64, which is also irreducible by rational root theorem. Furthermore, D = 200704 = 448 2 . Therefore G = A 4 . The only normal subgroup of A 4 is V , with  A 4 /V  = 3. Therefore, the only Galois subfiled is the splitting field of the resolvent cubic. Page 618, 19 . a)Since √ D = Q i<j ( α i α j ), where α i ,α j are roots of f ( x ). Since α i ,α j ∈ K , so are ( α i α j ). Therefore, √ D ∈ K , and Q ⊂ K . b) Since the complex conjugate of a root of f is also a root, τ K ∈ Gal ( K/ Q ). Obviously, the order is 1 if K has all real elements, and order 2 otherwise. c) Assume K is cyclic, and G = Gal ( K/ Q ) = { 1 , (1234) , (13)(24) , (1432) } ....
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 Spring '09
 SusamaAgarwala
 Algebra, Group Theory, α, )bn .

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