Homework 1 Solutions

Problem 4.1.2. Let p be a prime , and let n be a positive . How many are there of degree n over Zp? Solution. A 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 n are (p − 1)p 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 3 (d) f(x) = x + 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, multi- plication, and division. √ √ √ √ √ Proof. Given α1 = x1 + y√1 3, α2 = √x2 + y2 3 ∈ Q( 3) then α1 ± α2 = (√x1 ± x2)√ + (y1 ± y2√) 3 ∈ Q( 3). Also, α1α2 = (x1 + y1 3)(x2 + y2 3) = (x1x2 + 3y1y2) + (x1y2 + x2y1) 3 ∈ Q( 3), so Q( 3) is closed under multiplication. √ √ 2 2 To show that Q( 3) is closed√ under√ division note that if α1 6= 0 then x1 −√3y1 6= 0 since√ 3 is not rational. x1 y1 −1 Hence β = ( 2 2 ) − ( 2 2 ) 3 ∈ Q( 3) and α1β = 1, hence α1 = β ∈ Q( 3). Since Q( 3) is closed under x1−3y1 x1−3y1 −1 √ products we have α1/α2 = α1 · α2 ∈ 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) = (q0(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).

n Problem 4.2.10. Let p(x) = anx +···+a1x+a0 be a polynomial with rational coefficients such that an and a0 are n nonzero. Show that p(x) is irreducible over the field of rational if and only if q(x) = a0x +···+an−1x+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. −1 −n n −1 n −1 −1 n −1 i −1 j Observe that q(x) = (a0 + a1x + ··· + anx )x = p(x )x = a(x )b(x )x = (a(x )x )(b(x )x ) 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.

2 Problem 4.2.14. Show that there are exactly (p − 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, 2 there are p 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 p(p−1) are p such polynomials with a repeated factor and 2 such polynomials with distinct factors. Thus there are p(p−1) p(p+1) p + 2 = 2 reducible monic polynomials of degree 2 over Zp. 2 p(p+1) p(p−1) A polynomial which is not reducible is irreducible. Thus there are p − 2 = 2 monic irreducible polynomials of degree 2 over Zp.

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