Homework 8 Solutions to Selected Problems


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1 Homework 8 Solutions to Selected Problems June 7, 01 1 Chapter 17, Problem Let f(x D[x] and suppose f(x is reducible in D[x]. That is, there exist polynomials g(x and h(x in D[x] such that g(x and h(x are not units in D[x] (so they do not have multiplicative inverses and f(x = g(xh(x. Now we use the fact that D F to note that f(x, g(x, and h(x are in F [x] as well. However, f(x is irreducible in F [x]. Thus, f(x = g(xh(x implies that g(x or h(x is a unit in F [x]. Without loss of generality, g(x is a unit in F [x]. What can we say about g(x? Since g(x is a unit, there is a polynomial g(x 1 F [x] with g(xg(x 1 = 1. However, by problem 17 of chapter 16, the degree of g(xg(x 1 is the sum of the degree of g(x and the degree of g(x 1. Since g(xg(x 1 = 1, the degrees of g(x and g(x 1 must sum to zero. The only way for that to happen is for g(x and g(x 1 to have degree zero, and hence be nonzero constant polynomials. That is, g(x = a and g(x 1 = a 1 where a F [x] and a 0. In particular, a is a unit in F (although every nonzero element of F is a unit. However, f(x is reducible in D[x], and g(x = a was originally in D[x], so a D. Also, g(x is not a unit in D[x], so a cannot be a unit in D. In other words, a D but a 1 / D. 1
2 Example Let f(x = x Z[x]. f(x is reducible: a factorization is f(x = g(xh(x where g(x = and h(x = x. Although g(x is a nonzero constant polynomial, it is not a unit in Z[x] since does not have a multiplicative inverse in Z. However, this factorization does not prove that f(x is reducible over Q[x], because g(x = is a unit in Q[x]: its inverse is g(x 1 = 1 Q[x]. In fact, f(x = x is irreducible over Q[x]. Since it has degree one, if you try to write f(x as a product of two polynomials in Q[x], one of the polynomials will be a nonzero constant polynomial, and hence it is a unit in Q[x] whose multiplicative inverse is its reciprocal. In general Let D be an integral domain. Then f(x D[x] is a unit if and only if f(x is a nonzero constant polynomial which equals a unit in D. In particular, if D is a eld, then any nonzero constant polynomial is a unit. Chapter 17, Problem 4 Let r = p q, where p and q are integers (q 0 with no common prime factors (no positive prime number divides both p and q. Since x r divides f(x = x n + a n 1 x n a 1 x + a 0, by Corollary on page 98, f(r = r n + a n 1 r n a 1 r + a 0 = 0. Thus, ( p q n + a n 1 ( p q Let us multiply both sides by q n : n a 1 ( p q + a 0 = 0. p n + qa n 1 p n q n 1 a 1 p + q n a 0 = q n 0 = 0. q divides 0, so q must divide the left hand side, which is p n + qa n 1 p n q n 1 a 1 p + q n a 0 = p n + q ( a n 1 p n q n a 1 p + q n 1 a 0. q divides q ( a n 1 p n q n a 1 p + q n 1 a 0, so q must divide p n + q ( a n 1 p n q n a 1 p + q n 1 a 0 q ( a n 1 p n q n a 1 p + q n 1 a 0 = p n. Hence q divides p n. Suppose q > 1 (or q < 1. Let s be a prime which divides q. Then s divides p n, so by Euclid's lemma, s divides p, and hence s is a common prime factor of p and q, a contradiction. Thus, 1 q 1, and since q 0, q is either 1 or 1, and therefore r = p or p. In either case, r is an integer.
3 3 Chapter 17, Problem 13 Let us try Theorem 17.1 and determine the zeros of f(x = x 3 + 6: x f(x mod = 0 14 = = = = 5 6 = 5 Thus, x 1, x, and x 4 divide x These three polynomials are irreducible, since they have degree one (since Z 7 is a eld, trying to write a degree one polynomial in Z 7 [x] as a product of polynomials would require one of the factors to be a nonzero constant polynomial, which is a unit in Z 7 [x]. We can check that (x 1(x (x 4 = (x 3x + (x 4 = x 3 3x + x 4x + 1x 8 = x 3 7x + 14x 8 = x mod 7. 4 Chapter 17, Problem 17 In Example 10 on page 31, the book constructs a eld of order 8 = 3 by taking the quotient of Z [x] by the ideal generated by an irreducible monic polynomial of degree 3. Elements of Z [x]/ < x 3 + x + 1 > have the form ax + bx + c+ < x 3 + x + 1 >, where a, b, c Z, because given any polynomial f(x Z [x], by the division algorithm, f(x = q(x(x 3 + x r(x, where r(x has degree less than 3, so it has the form r(x = ax + bx + c. Hence f(x+ < x 3 + x + 1 >= q(x(x 3 + x r(x+ < x 3 + x + 1 >, and since q(x(x 3 + x + 1 < x 3 + x + 1 >, q(x(x 3 + x r(x+ < x 3 + x + 1 >= r(x+ < x 3 + x + 1 >, so there is one coset for each r(x = ax + bx + c. Hence there are 8 cosets total. 3
4 In general, if p(x is an irreducible polynomial in Z p [x] of degree n, then Z p [x]/ < p(x > is a eld of order p n. Thus, to show that there is a eld of degree p for any prime p, we need to show that there is an irreducible polynomial of degree in Z p [x] for any prime p. First, let us try to work with monic polynomials. Note that there are p monic polynomials of degree (both reducible and irreducible: x + bx + c, where b, c Z p. What does a reducible polynomial in degree look like? If f(x has degree and is reducible, then f(x = g(xh(x where neither g(x not h(x are units in Z p [x]. Then g(x and h(x cannot be nonzero constant polynomials (since those are units in Z p [x], and since the sum of their degrees is two, both g(x and h(x are degree 1 polynomials: Are g(x and h(x monic? so a 1 b 1 = 1. Hence g(x = a 1 x + a 0, h(x = b 1 x + b 0, a 1, b 1 0. g(xh(x = a 1 b 1 x + (a 1 b 0 + a 0 b 1 x + a 0 b 0 = x + bx + c, g(x = a 1 (x + a 1 1 a 0, h(x = b 1 (x + b 1 1 b 0, f(x = a 1 b 1 (x + a 1 1 a 0(x + b 1 1 b 0 = (x + a 1 1 a 0(x + b 1 1 b 0. Thus, without loss of generality, if f(x has degree and is reducible, f(x = (x α(x β where α, β Z p. How many reducible polynomials are there? When choosing α and β, we have two cases: 1. If α β, then we have p choices (in Z p for α and p 1 choices for β. However, we must divide by, since we do not want to count (x α(x β and (x β(x α as two separate polynomials. Thus, we have p(p 1 polynomials of the form (x α(x β where α β.. If α = β, then we have p choices for α (so there are p polynomials of the form (x α. Thus, there are p(p 1 + p = p p + p = p +p reducible monic polynomials of degree, so there are p p +p = p p = p(p 1 > 0 irreducible monic polynomials of degree. Let p(x be one of these monic irreducible polynomials of degree. Then Z p [x]/ < p(x > is a eld of order p. 4
5 5 Chapter 17, Problem 6 Let f(x F [x] be a nonzero polynomial which is not a unit (so it has degree 1 or higher: f(x = a n x n a 0, where a n 0. Since a n 0, it has an inverse in F. Let f 1 (x = a 1 n f(x = x n a 1 n a 0. We need to prove that f(x is irreducible if and only if f 1 (x is irreducible. 5.1 f(x irreducible implies f 1 (x irreducible We start not with a factorization of f(x, but with a factorization of f 1 (x: f 1 (x = g(xh(x. We need to prove that either g(x or h(x is a unit in F [x]. Equivalently, we need to prove that either g(x or h(x is a nonzero constant polynomial. We proceed by multiplying both sides by a n in order to introduce f(x: f(x = a n f 1 (x = a n g(xh(x. Before we can use the hypothesis that f(x is irreducible, we need to write the right hand side as a product of two factors: f(x = [a n g(x]h(x. Since f(x is irreducible, either a n g(x or h(x is a unit in F [x]. That is, either a n g(x = a or h(x = b where a, b F and a, b 0 (so either a n g(x or h(x is a nonzero constant polynomial. If h(x = b, then in the original factorization of f 1 (x, h(x is a unit. However, if a n g(x = a, then we need to show that g(x itself is a nonzero constant polynomial. But we have g(x = a 1 n a, and since a 1 n 0, a 0, and F is a eld (and hence an integral domain, g(x = a 1 n a 0, so g(x is a nonzero constant polynomial and hence a unit. Thus, if f 1 (x = g(xh(x, there either g(x or h(x is a unit in F [x], so f 1 (x is irreducible. 5. f 1 (x irreducible implies f(x irreducible We start with a factorization of f(x: f(x = g(xh(x. We need to prove that either g(x or h(x is a nonzero constant polynomial. We proceed by multiplying both sides by a 1 n in order to introduce f 1 (x: f 1 (x = a 1 n f(x = a 1 g(xh(x = [a 1 g(x]h(x. n Since f 1 (x is irreducible, either a 1 n g(x = a or h(x = b where a, b F and a, b 0. If h(x = b, then in the original factorization of f(x, h(x is a unit. However, if a 1 n g(x = a, then g(x = a n a, and since a n 0, a 0, and F is a eld (and hence an integral domain, g(x = a n a 0, so g(x is a unit. Hence if f(x = g(xh(x, there either g(x or h(x is a unit in F [x], so f(x is irreducible. n 5
6 5.3 What happens if instead of a eld, we had an integral domain where not every nonzero element has an inverse? For example, in Z[x], f 1 (x = x is irreducible, because if f 1 (x = x = g(xh(x, then since the sum of the degrees of g(x and h(x is one, without loss of generality, g(x is a nonzero constant polynomial (g(x = a and h(x = b 1 x+b 0. Is g(x a unit? That is, does a have an inverse in Z? g(xh(x = a(b 1 x + b 0 = (ab 1 x + ab 0 = x, so ab 1 = 1, and therefore a and b 1 are units in Z. Hence a = 1 or a = 1, so g(x = 1 or g(x = 1, and either way, it is a unit in Z[x]. However, f(x = x is reducible, because f(x = x and neither nor x are units in Z[x]. is not a unit in Z[x], because the integer has no multiplicative inverse in Z. 6
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