Adv Alegbra HW Solutions 151

Adv Alegbra HW Solutions 151 - if the only nonzero coef f...

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151 (ii) Find the Galois group of f ( x ) = x 3 2 Q [ x ] . Solution. f ( x ) is irreducible because it has no rational root. Its discriminant is D =− 108, and so its Galois group is S 3 . (iii) Find a cubic polynomial g ( x ) Q [ x ] whose Galois group has order 3. Solution. Try g ( x ) = 3 x 3 3 x + 1. 5.30 (i) If k is a f eld and f ( x ) k [ x ] has derivative f 0 ( x ) , prove that either f 0 ( x ) = 0ordeg ( f 0 )< deg ( f ) . Solution. If f ( x ) = a n x n + a n 1 x n 1 + ··· , then f ( x ) = na n x n 1 + ( n 1 ) a n 1 x n 2 +···+ a 1 . Hence, if f 0 ( x ) ±= 0, then deg ( f 0 )< n . (ii) If k is a f eld of characteristic 0, prove that an irreducible polyno- mial p ( x ) k [ x ] has no repeated roots; that is, if E is the splitting f eld of p ( x ) , then there is no a E with ( x a ) 2 | p ( x ) in E [ x ] . Solution. We may assume that p ( x ) is monic. By Exercise 3.67, p ( x ) has no repeated roots if and only if gcd ( p , p 0 ) = 1, where p 0 ( x ) is the derivative of p ( x ) .Bu t n ±= 0in k because k has char- acteristic 0, so that p 0 ( x ) ±= 0 and deg ( p 0 )< deg ( p ) . Since p ( x ) is irreducible, its only monic divisors are 1 and itself; therefore, gcd ( p , p 0 ) = 1 and p ( x ) has no repeated roots. 5.31 Let k be a f eld of characteristic p . (i) Prove that if f ( x ) = i a i x i k [ x ] , then f 0 ( x ) =
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Unformatted text preview: if the only nonzero coef f cients are those a i with p | i . Solution. If f ( x ) = 0, then any nonzero term a i x i of f ( x ) gives the term ia i x i 1 ; if p-a i , however, then ia i = 0, and so f ( x ) = 0. Conversely, if f ( x ) = j b j x pj , then f ( x ) = j pjb j x pj 1 = 0. (ii) If k is f nite and f ( x ) = i a i x i k [ x ] , prove that f ( x ) = 0 if and only if there is g ( x ) k [ x ] with f ( x ) = g ( x ) p . Solution. By part (i), if f ( x ) = 0, then f ( x ) = j a jp x jp . For each j , let b p j = a jp . If g ( x ) = j b j x j , then g ( x ) p = ( X j b j x j ) p = X j b p j x pj = f ( x ). (iii) Prove that if k is a f nite f eld, then every irreducible polynomial p ( x ) k [ x ] has no repeated roots....
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