Adv Alegbra HW Solutions 117

Adv Alegbra HW Solutions 117 - 117 Solution. Since a = 0...

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Unformatted text preview: 117 Solution. Since a = 0 and a = ±1, there is a prime p dividing a ; since a is squarefree, p2 a . As all coefficients of x i for 0 < i < n are 0, Eisenstein’s criterion applies to show that x n − a is irreducible in Q [x ]. √ (ii) If a = ±1 is a squarefree integer, prove that n a is irrational. Solution. Since x n − a is irreducible in Q [x ], it has no rational roots. 3.91 Let k be a field, and let f (x ) = a0 + a1 x + · · · + an x n ∈ k [x ] have degree n . If f (x ) is irreducible, then so is an + an −1 x + · · · + a0 x n . Solution. If f ∗ (x ) denotes the polynomial f (x ) with coefficients reversed, then a factorization f ∗ (x ) = g (x )h (x ) gives a factorization f (x ). One sees this just by using the definition of multiplication of polynomials. Let p q g (x ) = i =0 bi x i and h (x ) = j =0 c j x j , where p + q = n . Thus, an −m = bi c j . i+ j = m It follows that b p−i cq − j = an −(n −m ) = ak . i + j = n −m Therefore, if we define g ∗ (x ) = i =0 b p−i x i and h ∗ (x ) = j =0 cq − j x j , then f (x ) = g ∗ (x )h ∗ (x ), contradicting the irreducibility of f (x ). Note that f (x ) → f ∗ (x ), which reverses coefficients, is not a welldefined function k [x ] → k [x ], because it is not clear how to define f ∗ (x ) if the constant term of f (x ) is zero. And even if one makes a bona fide definition, the function is not a homomorphism. For example, let f (x ) = x 5 + 3x 4 ; that is, in sequence notation, p q f (x ) = (0, 0, 0, 0, 3, 1, 0, . . .). Let g (x ) = x 3 + x ; in sequence notation, g (x ) = (0, 1, 0, 1, 0, . . .). Now f (x )g (x ) = [x 8 + 3x 7 + x 6 + 4x 5 + 3x 4 ; in sequence notation, f (x )g (x ) = (0, 0, 0, 0, 3, 4, 1, 3, 1, 0, . . .). Therefore, [ f (x )g (x )]∗ = 3x 4 + 4x 3 + x 2 + 3x + 1, which is a quartic. But f ∗ (x ) = 3x + 1 and g ∗ (x ) = x 2 + 1, so that f ∗ (x )g ∗ (x ) is a cubic. Therefore, [ f g ]∗ = f ∗ g ∗ . ...
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