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]. Prov Prove that that α is an integer. Let α = b/c, b/c, wh where ere b and c are are coprim coprimee integ integer ers. s. Let α be a root of p(x) = −1 −2 x + a −1 x + a −2 x + · · · + 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 power term. Thus b | a0 and c | 1. It follows follows that c = ±1 ± 1 and α = ±b ± b, an integer. n
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[6] Show that if α is a root of a x + a −1x −1 + · · · + a1x + a + a0 then a α is a root of −1 −2 the monic polynomial x + a −1x + a a −2 x + · · · + a −2 a1 x + a + a −1 a0. n
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Let α be a root of p(x) = a x + a −1 x a −1 x −1 + a a −2 x −2 + · · · + a −2 a1 x + a + a n
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+ · · · + a + a 1 x + a + a 0. Let q (x) = x + −1 a0 . Calculating,
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q (a α) = (a α) + a −1(a α) −1 + a a −2 (a α) −2 + · · · + a −2 a1 (a α) + a + a −1a0 = a −1 a α + a −1 a −1 α −1 + a −1 a −2 α −2 + · · · + a −1 a1 α + a + a −1 a0 = a −1 (a α + a −1 α −1 + a −2 α −2 + · · · + a1 α + a + a0 ) = a −1 ( p( p(α)) =0 n
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Thus a Thus a α is a root of q of q (x).
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[7] Prove that x3 − nx + nx + 2 is irreducible for n = −1, − 1, 3, 5. Let p(x) = x3 − nx + nx + 2. We show show p(x) is irreducible in Q[x], assuming n ∈ Z for n = −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 of n make p(x) have no roots in Q. Sup Suppos posee x were a root of p(x). By the Rational Rational Root Theorem, Theorem, x = b/c with b/c with b = ±2, ±1 and c = ±1. ± 1. Thus x = ±2, ±1. Note Note p(1) = 3 − n, which is equal to 0 only if n = 3. Similarly Similarly,, p(−1) = 1 + n + n = = 0 only if n = −1, − 1, p(2) = 10 − 2n = 0 only if n = 5, and p(−2) = −6 − 6 + 2n 2 n = 0 only if n = 3. Thus, if n = −1, − 1, 3, 5, then 3 x − nx + 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 = 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 = 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 + a2x2 + 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 = a 0 b0 + (a1 b0 + a0 b1 )x + (a2 b0 + a1 b1 + a0 b2 )x2 + (a3 b0 + a2 b1 + a1 b2 )x3 + (a3b1 + a2 b2 )x4 + (a3 b2 )x5 By comparing coefficients, we have the following equations: (x0 ) → a0 b0 = −1, (x1) → a 1 b0 +a0 b1 = −a, (x2 ) → a 2 b0 +a1 b1 +a0 b2 = 0, (x3 ) → a 3 b0 +a2 b1 +a1 b2 = 0, (x4) → a3b1 + a2 b2 = 0, and (x5 ) → a3 b2 = 1. Note a3b2 = 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 ), a 1 = 1+b21 , and by (x2 ), b 1 (a1 +1)+1 = 0. After substituting, we have b31 + 2b1 + 1 = 0. By the Rational Root Theorem, the numerator p/q = b 1 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 = b 0 = 1. By (x3), a1 = b 21 − 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 a 2 = 1 and a 1 = 0. This leads to p(x) = (−1 + x2 + x3 )(1 − x + x2 ) = x 5 + x − 1. Hence a = −1, a contradiction, since we assumed a = 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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