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 pa 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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 Fall '11
 KeithCornell
 Group Theory, Complex number, Integral domain, finite field, Galois group

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