Adv Alegbra HW Solutions 148

Adv Alegbra HW Solutions 148 - 148 (ix) The alternating...

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Unformatted text preview: 148 (ix) The alternating group A 5 is a solvable group. Solution. False. 5.16 Let : A H be a group homomorphism. If B C A and B ker , prove that the induced map : A / B H , given by aB 7 ( a ) , is a well-de fi ned homomorphism with im = im . Solution. If aB = cB , then c 1 a B ker , and so ( c 1 a ) = 1. Therefore, ( c ) = ( a ) , and is well-de fi ned. 5.17 If z C is a constructible number, prove that Q ( i , z )/ Q is a radical exten- sion. Solution. Absent. 5.18 Let k be a fi eld and let f ( x ) k [ x ] . Prove that if E and E are splitting fi elds of f ( x ) over k , then Gal ( E / k ) = Gal ( E / k ) . Solution. Absent. 5.19 Prove that F 3 [ x ] /( x 3 x 2 1 ) = F 3 [ x ] /( x 3 x 2 + x 1 ) . Solution. We saw, in Chapter 3, that both cubics are irreducible in F 3 [ x ] , and so the quotient rings are fi elds of order 3 3 = 27. But any two fi nite fi elds of the same order are isomorphic.elds of the same order are isomorphic....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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