Part Solution Of Dummit And Foote 2lg61

James Pringle June 20, 2011

Abstract Algebra Dummit and Foote Third Edition

Section 13.1 Problems: 5, 6, 7, 8 [5] Suppose α is a rational root of a monic polynomial in Z[x]. Prove that α is an integer. Let α = b/c, where b and c are coprime integers. Let α be a root of p(x) = xn + an−1 xn−1 + an−2 xn−2 + · · · + a0 , a monic polynomial in Z[x]. By the Rational Root Theorem, b divides the constant term, and c divides the coefficient of the highest power term. Thus b | a0 and c | 1. It follows that c = ±1 and α = ±b, an integer. [6] Show that if α is a root of an xn + an−1 xn−1 + · · · + a1 x + a0 then an α is a root of the monic polynomial xn + an−1 xn−1 + an an−2 xn−2 + · · · + ann−2 a1 x + an−1 n a0 . Let α be a root of p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 . Let q(x) = xn + an−1 xn−1 + an an−2 xn−2 + · · · + ann−2 a1 x + ann−1 a0 . Calculating, q(an α) = (an α)n + an−1 (an α)n−1 + an an−2 (an α)n−2 + · · · + ann−2 a1 (an α) + an−1 n a0 n−1 n n−1 n−1 n−1 n−2 n−1 n−1 = an an α + an an−1 α + an an−2 α + · · · + an a1 α + an a0 n−1 n n−1 n−2 = an (an α + an−1 α + an−2 α + · · · + a1 α + a0 ) n−1 = an (p(α)) =0 Thus an α is a root of q(x).



[7] Prove that x3 − nx + 2 is irreducible for n 6= −1, 3, 5. Let p(x) = x3 − nx + 2. We show p(x) is irreducible in Q[x], assuming n ∈ Z for n 6= −1, 3, 5. Since it is of degree 3, if p(x) were reducible, it would have a linear factor. Hence, it is sufficient to show what values of n make p(x) have no roots in Q. Suppose x were a root of p(x). By the Rational Root Theorem, x = b/c with b = ±2, ±1 and c = ±1. Thus x = ±2, ±1. Note p(1) = 3 − n, which is equal to 0 only if n = 3. Similarly, p(−1) = 1 + n = 0 only if n = −1, p(2) = 10 − 2n = 0 only if n = 5, and p(−2) = −6 + 2n = 0 only if n = 3. Thus, if n 6= −1, 3, 5, then x3 − nx + 2 is irreducible.

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James Pringle June 20, 2011 [8] Prove that x5 − ax − 1 ∈ Z[x] is irreducible unless a = 0, 2, or −1. The first two correspond to linear factors, the third corresponds to the factorization (x2 − x + 1)(x3 + x2 − 1). Let p(x) = x5 − ax − 1. First note that if a = 0, then p(x)l can be written as (x − 1)(x4 + x3 + x2 + x + 1), and if a = 2, then p(x) can be written as (x + 1)(x4 − x3 +x2 −x−1). Now, suppose to the contrary, a 6= 0, 2, or −1 and p(x) is reducible. Then either it has a linear factor or it is the product of a degree 2 polynomial and a degree 3 polynomial. Case 1: p(x) has a linear factor. By the Rational Root Theorem, the only possible roots—which correspond to linear factors—are ±1. But, as confirmed by the above, p(1) = −a = 0 only if a = 0 and p(−1) = a − 2 = 0 only if a = 2. Thus we have a contradiction, since we assumed a 6= 0 or 2. Case 2: p(x) can be written as a product of two polynomials of degree greater than 1 (the complement of “p(x) has a linear factor”). Hence p(x) = (a0 + a1 x + a2 x2 + a3 x3 )(b0 + b1 x + b2 x2 ) with coefficients in Z. Without loss of generality, we assume that a3 is positive. Expanding, we have x5 − ax − 1 = a0 b0 + (a1 b0 + a0 b1 )x + (a2 b0 + a1 b1 + a0 b2 )x2 + (a3 b0 + a2 b1 + a1 b2 )x3 + (a3 b1 + a2 b2 )x4 + (a3 b2 )x5 By comparing coefficients, we have the following equations: (x0 ) → a0 b0 = −1, (x1 ) → a1 b0 +a0 b1 = −a, (x2 ) → a2 b0 +a1 b1 +a0 b2 = 0, (x3 ) → a3 b0 +a2 b1 +a1 b2 = 0, (x4 ) → a3 b1 + a2 b2 = 0, and (x5 ) → a3 b2 = 1. Note a3 b2 = 1. Since we assumed a3 = 1, it must be that b2 = 1 as well. Therefore (x4 ) implies b1 = −a2 . Now, our subcases are based on a0 b0 = −1. Subcase 2.1: a0 = −b0 = 1. Hence, by (x3 ), a1 = 1+b21 , and by (x2 ), b1 (a1 +1)+1 = 0. After substituting, we have b31 + 2b1 + 1 = 0. By the Rational Root Theorem, the numerator p/q = b1 with p and q coprime has to be ±1. Since b1 ∈ Z, q = ±1. However, b1 = ±1 leads to b1 = 1 ⇒ 4 = 0 and b1 = −1 ⇒ −2 = 0. This is a contradiction, since we assumed that p(x) was reducible. Subcase 2.2: −a0 = b0 = 1. By (x3 ), a1 = b21 − 1, and by (x2 ), −b1 + a1 b1 − 1 = 0. After substituting, we have b31 −2b1 −1 = 0. By the Rational Root Theorem and the reasoning from subcase 2.1, the only possible solutions are b1 = ±1. Only b1 = −1 is valid, so by the preceding equations, we now know a2 = 1 and a1 = 0. This leads to p(x) = (−1 + x2 + x3 )(1 − x + x2 ) = x5 + x − 1. Hence a = −1, a contradiction, since we assumed a 6= 0, 2, or −1. We therefore reject the hypothesis of case 2. In total, we reject the overall hypothesis and conclude x5 − ax − 1 is irreducible unless a = 0, 2, or −1.

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