HW6_soln - Homework #6 Solutions p 315, #4 Let f ( x ) = x...

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Unformatted text preview: Homework #6 Solutions p 315, #4 Let f ( x ) = x n + a n- 1 x n- 1 + a Z [ x ] and suppose that x- r divides f ( x ) for some r Q . Then we must have f ( r ) = 0. We will use this fact to prove that in fact r Z . If r = 0 then there is nothing to prove. So assume r 6 = 0 and write r = a/b with a, b Z , b 6 = 0 and ( a, b ) = 1. Then 0 = f ( r ) = f ( a/b ) = a b n + a n- 1 a b n- 1 + + a 1 a b + a . Multiplying both sides by b n then yields 0 = a n + a n- 1 a n- 1 b + a n- 2 a n- 2 b 2 + + a 1 ab n- 1 + a b n which is equivalent to a n =- ( a n- 1 a n- 1 b + a n- 2 a n- 2 b 2 + + a 1 ab n- 1 + a b n ) =- b ( a n- 1 a n- 1 + a n- 2 a n- 2 b + + a 1 ab n- 2 + a b n- 1 ) . Since a n- 1 a n- 1 + a n- 2 a n- 2 b + + a 1 ab n- 2 + a b n- 1 Z , this implies that b divides a n . Since ( a, b ) = 1, this can only occur if b = 1. But then r = a/b = a Z , as claimed. p 315, #6 If p is prime and f ( x ) Z p [ x ] is irreducible then Z p [ x ] / h f ( x ) i is a field by Corollary 1 to Theorem 17.5, since Z p is a field. Moreover, we proved in class that since deg f ( x ) = n the each element of Z p [ x ] / h f ( x ) i can be expressed uniquely as a n- 1 x n- 1 + a n- 2 x n- 2 + + a + h f ( x ) i for some a i F . Since there are p choices for each coefficient a i and n coefficients, there are exactly p n such cosets. That is, Z p [ x ] / h f ( x ) i is a field with p n elements. p 316, #8 Let f ( x ) = x 3 + 2 x + 1 Z 3 [ x ]. It is easy to show that f ( x ) has no roots in Z 3 and as deg f ( x ) = 3, this implies f ( x ) is irreducible in Z 3 [ x ]. So according to Exercise #6, Z 3 [ x ] / h x 3 + 2 x + 1 i is a field with 3 3 = 27 elements. p 316, #10 a. x 5 + 9 x 4 + 12 x 2 + 6 is irreducible according Eisensteins criterion with p = 3....
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HW6_soln - Homework #6 Solutions p 315, #4 Let f ( x ) = x...

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