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Homework 1 Solutions Problem 4.1.2. Let p be a prime number, and let n be a positive integer. How many polynomials are there of degree n over Zp ? Solution. A polynomial of degree n is determined by it’s n + 1 coefficients where the coefficient of xn is not zero. Thus there are (p − 1) choices for the leading coefficient and p choices for the remaining coefficients. Hence there are (p − 1)pn polynomials of degree n over Zp . Problem 4.1.5. Over the given field F , write f (x) = q(x)(x − c) + f (c) for (b) f (x) = x3 − 5x2 + 6x + 5; c = 2; F = Q (d) f (x) = x3 + 2x + 3; c = 2; F = Z5 Solution. (b): Following the method in Theorem 4.1.9 we have f (x) − f (1) = 2x3 − 2 + x2 − 1 − 4x + 4 = 2(x − 1)(x2 + x + 1) + (x − 1)(x + 1) − 4(x − 1) = (x − 1)(2x2 + 3x − 1). Hence f (x) = (x − 1)(2x2 + 3x − 1) + 2. (d) Similarly one obtains f (x) = (x − 2)(x2 + 2x + 1). Problem 4.1.7. Show that if c is any element of the field F and k > 2 is an odd integer, then x + c is a factor of xk + ck . Proof. By theorem 4.1.9 it suffices to check that f (−c) = 0 for f (x) = xk +ck . Since k is odd f (−c) = (−c)k +ck = −ck + ck = 0. Problem 4.1.9. Let a be a nonzero element of a field F . Show that (a−1 )−1 = a and (−a)−1 = −a−1 . Proof. Using the field axioms we have a = a · 1 = a(a−1 · (a−1 )−1 ) = (a · a−1 )(a−1 )−1 = 1 · (a−1 )−1 = (a−1 )−1 . By proposition 4.1.3(e) we have (−a)−1 = (−a)−1 · 1 = (−a)−1 (a · a−1 ) = (−a)−1 ((−a)(−a−1 )) = ((−a)−1 (−a))(−a−1 ) = 1 · (−a−1 ) = −a−1 . √ √ Problem 4.1.11. Show that the set Q( 3) = {a + b 3| a, b ∈ Q} is closed under addition, subtraction, multiplication, and division. √ √ √ √ √ x2 + y2 3 ∈ Q( 3) then α1 ± α2 = (x ) 3 ∈ Q( 3). Proof. Given α1 = x1 + y√ 1 3, α2 = √ √ 1 ± x2 )√+ (y1 ± y2√ Also, α1 α2 = (x1 + y1 3)(x2 + y2 3) = (x1 x2 + 3y1 y2 ) + (x1 y2 + x2 y1 ) 3 ∈ Q( 3), so Q( 3) is closed under multiplication. √ √ under√division note that if α1 6= 0 then x21 −√3y12 6= 0 since√ 3 is not rational. To show that Q( 3) is closed √ y1 x1 Hence β = ( x2 −3y 3 ∈ Q( 3) and α1 β = 1, hence α1−1 = β ∈ Q( 3). Since Q( 3) is closed under 2 ) − ( x2 −3y 2 ) 1 1 1 1 √ products we have α1 /α2 = α1 · α2−1 ∈ Q( 3) for α2 6= 0. Problem 4.2.9. Let a ∈ R, and let f (x) ∈ R[x], with derivative f 0 (x). Show that the remainder when f (x) is divided by (x − a)2 is f 0 (a)(x − a) + f (a). Proof. By the division algorithm there exist unique polynomials q(x), r(x) ∈ F [x] such that f (x) = q(x)(x−a)2 + r(x) where deg(x) < 2. Write r(x) = bx + c then observe that f (a) = 0 + r(a) = ba + c. Taking the derivative we have f 0 (x) = (q 0 (x)(x − a) + 2q(x))(x − a) + b so f 0 (a) = b. Consequently, c = f (a) − ba = f (a) − f 0 (a)a and hence r(x) = f 0 (a)x + f (a) − f 0 (a)a = f 0 (a)(x − a) + f (a). Problem 4.2.10. Let p(x) = an xn +· · ·+a1 x+a0 be a polynomial with rational coefficients such that an and a0 are nonzero. Show that p(x) is irreducible over the field of rational numbers if and only if q(x) = a0 xn +· · ·+an−1 x+an is irreducible over the field of rational numbers. 1 Proof. We prove the contrapositive of one direction, the other direction is similar. Suppose that p(x) is reducible then p(x) = a(x)b(x) for some a(x), b(x) ∈ Q[x] with deg(a(x)) = i < n, deg(b(x)) = j < n and i + j = n. Observe that q(x) = (a0 + a1 x−1 + · · · + an x−n )xn = p(x−1 )xn = a(x−1 )b(x−1 )xn = (a(x−1 )xi )(b(x−1 )xj ) where a(x−1 )xi ∈ Q[x] of degree i < n and b(x−1 )xj ∈ Q[x] of degree j < n. Thus q(x) is also reducible. Problem 4.2.11. Find the irreducible factors of x6 − 1 over R. Proof. Observe that x6 − 1 = (x − 1)(x2 + x + 1)(x + 1)(x2 − x + 1). The factors x − 1 and x + 1 are clearly irreducible. The factors x2 + x + 1 and x2 − x + 1 have no root in R since they both have discriminant −3 and hence they are irreducible by proposition 4.2.7. Thus x − 1, x + 1, x2 + x + 1 and x2 − x + a are the irreducible factors of x6 − 1 over R[x]. Problem 4.2.13. Find all the monic irreducible polynomials of degree ≤ 3 over Z3 . Using your list, write each of the following polynomials as a product of irreducible polynomials. (b) x4 + 2x2 + 2x + 2 Proof. One can verify that the following polynomials of degree 2 and 3 are irreducible by checking that they have no roots in Z3 . Moreover, one can check that the list is exhaustive by checking the finite list (albeit lenghty) of monic polynomials of degree less than or equal to 3 or by counting the number of monic irreducibles of a given degree (see problem 4.2.14 for a method of counting). Monic irreducible polynomials of degree 1: x, x − 1, x − 2. Monic irreducible polynomials of degree 2: x2 + x + 2, x2 + 2x + 2, x2 + 1. Monic irreducible polynomials of degree 3: x3 + x2 + 2x + 1, x3 + 2x + 1, x3 + 2x2 + x + 1, x3 + 2x2 + 1, x3 + x + 2, x3 + x2 + 2, x3 + x2 + x + 2, x3 + 2x2 + 2x + 2. (b): Note that x4 + 2x2 + 2x + 2 = (x + 1)2 (x2 + x + 2) where each of the factors are irreducible. Problem 4.2.14. Show that there are exactly (p2 − p)/2 monic irreducible polynomials of degree 2 over Zp . Proof. A monic polynomial of degree 2 over Zp is determined by the two non-leading coefficients. Since |Zp | = p, there are p2 monic polynomials of degree 2 over Zp . A monic polynomial of degree 2 which is reducible must be of the form (x − a)(x − b) for a, b ∈ Zp . There are p such polynomials with a repeated factor and p(p−1) such polynomials with distinct factors. Thus there are 2 p(p+1) = reducible monic polynomials of degree 2 over Zp . p + p(p−1) 2 2 A polynomial which is not reducible is irreducible. Thus there are p2 − p(p+1) = p(p−1) monic irreducible 2 2 polynomials of degree 2 over Zp . 2