Adv Alegbra HW Solutions 147

Adv Alegbra HW - n ≥ 1 Solution False(ii There are no 5th roots of unity in a f eld of characteristic 5 Solution False(iii R is a splitting f eld

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147 Thus, m =− 2, ` = 3, and we have the factorization x 4 15 x 2 20 x 6 = ( x 2 + 4 x + 3 )( x 2 4 x 2 ). The quadratic formula applied to each of the two factors gives the desired roots: 3 , 1 , 2 + 6 , 2 6 . In particular, all the roots are real. Remark. Given the discussion of roots in Chapter 3, especially Corol- lary 3.91, it is natural to look for rational, hence integral roots, and to f nd that 1 and 3 are roots (the candidates are ± 1, ± 2, ± 3, and ± 6). There- fore, long division gives x 4 15 x 2 20 x 6 = ( x + 1 )( x + 3 )( x 2 4 x 2 ). The other two roots are thus 2 + 6 and 2 6. We now understands the unpopularity of the quartic formula. J 5.15 True or false with reasons. (i) Every algebraically closed f eld contains n distinct n th roots of unity, where
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Unformatted text preview: n ≥ 1. Solution. False. (ii) There are no 5th roots of unity in a f eld of characteristic 5. Solution. False. (iii) R is a splitting f eld of x 2 − 5 over Q . Solution. False. (iv) Q ( √ 5 ) is a normal extension of Q . Solution. False. (v) No polynomial of degree ≥ 5 in Q [ x ] is solvable by radicals. Solution. False. (vi) F 2 ( x ) = Frac ( F 2 [ x ] ) is an in f nite f eld of characteristic 2. Solution. True. (vii) A polynomial f ( x ) ∈ Q [ x ] can have two splitting f elds inside of C . Solution. False. (viii) The alternating group A 4 is a solvable group. Solution. True....
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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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