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ahw5 - Math 5126 Monday February 18 Fifth Homework...

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Math 5126 Monday, February 18 Fifth Homework Solutions 1. Section 13.1, Exercise 2 on page 519. Show that x 3 - 2 x - 2 is irreducible over Q and let θ be a root. Compute ( 1 + θ )( 1 + θ + θ 2 ) and ( 1 + θ ) / ( 1 + θ + θ 2 ) in Q ( θ ) . If x 3 - 2 x - 2 was not irreducible, then it would have a degree one factor, and then by Gauss’s lemma we could assume that this factor was of the form x - a where a Z and a 2. Thus ± 1 or ± 2 would be a root, which is clearly not the case. From x 3 - 2 x - 2 = 0, we see that θ 3 = 2 θ + 2. Therefore ( 1 + θ )( 1 + θ + θ 2 ) = θ 3 + 2 θ 2 + 2 θ + 1 = 2 θ + 2 + 2 θ 2 + 2 θ + 1 = 2 θ 2 + 4 θ + 3 . Let us write ( 1 + θ ) / ( 1 + θ + θ 2 ) = a + b θ + c θ 2 , where a , b , c Q . Then we have 1 + θ = c θ 4 +( b + c ) θ 3 +( a + b + c ) θ 2 +( a + b ) θ + a . Using x 3 - 2 x - 2 = 0, we see that θ 3 = 2 θ + 2 and θ 4 = 2 θ 2 + 2 θ . Substituting back in the above, we obtain 1 + θ = ( a + 2 b + 2 c )+( a + 3 b + 4 c ) θ +( a + b + 3 c ) θ 2 . Equating the coefficients of θ i for i = 0 , 1 , 2, we find that 1 = a + 2 b + 2 c , 1 = a + 3 b + 4 c , 0 = a + b + 3 c . Therefore a = 1 / 3, b = 2 / 3, c = - 1 / 3 and we conclude that ( 1 + θ ) / ( 1 + θ + θ 2 ) = ( 1 + 2 θ - θ 2 ) / 3 .

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