Homework 9 Solutions to Selected Problems June 11, 2012 1 Chapter 17, Problem 12 Since x 2 + x + 4 has degree 2 and Z 11 is a eld, we may use Theorem 17.1 and show that f(x) is irreducible because it has no zeros: x x 2 + x + 4 mod 11 0 4 1 6 2 10 3 16 = 5 4 24 = 2 5 34 = 1 6 46 = 2 7 60 = 5 8 76 = 10 9 94 = 6 10 114 = 4 Since the polynomial has no zeros in Z 11, it is irreducible. 1
2 Chapter 17, Problem 14 Let us start by nding zeros of f(x) in Z 2. It turns out that f(0) = 1, while f(1) = 4 = 0 (mod 2). Thus, x 1 divides f(x). By using long division (remember to reduce mod 2), we see that x 2 +1 x 1 ) x 3 +x 2 +x +1 x 3 x 2 0 0 x +1 x 1 0 f(x) = (x 1)(x 2 + 1). Is x 2 + 1 irreducible? Let us nd zeros again. Since 1 2 + 1 = 2 = 0 (mod 2), x 1 divides x 2 + 1. We can use long division again, but since we are working mod 2, 1 = 1, so we can write and therefore x 2 + 1 = x 2 1 = (x 1)(x + 1) = (x 1)(x 1) f(x) = (x 1)(x 2 + 1) = (x 1)(x 1)(x 1) = (x 1) 3. 3 Chapter 17, Problem 30 Note that f(x) = x p 1 x p 2 +x p 3... x 1 +1 = ( x) p 1 +( x) p 2 +( x) p 3 +...+( x)+1 = Φ p ( x). That is, f(x) is the pth cyclotomic polynomial, but with x replaced by x. Therefore, if f(x) = g(x)h(x), then Φ p (x) = f( x) = g( x)h( x). By the Corollary on page 310, Φ p (x) is irreducible, so either g( x) or h( x) is a unit, so either g( x) = a or h( x) = b where a, b Z p and a, b 0. Thus, g(x) = a, or h(x) = b, so either g(x) is a unit or h(x) is a unit. Therefore, f(x) is irreducible. 2
4 Chapter 17, Problem 32 4.1 x 2 + 1 is a prime ideal Let f(x), g(x) Z[x], and suppose f(x)g(x) x 2 + 1. Then there is a polynomial q(x) Z[x] such that f(x)g(x) = q(x)(x 2 + 1). Since x 2 + 1 is monic (in particular its leading coecient is a unit in Z), we can use the division algorithm from page 296. Although the book proves the division algorithm for polynomials with coecients in a eld, if you look at the proof, the only multiplicative inverse needed is for the leading coecient of the dividing polynomial. Thus, f(x) = q 1 (x)(x 2 + 1) + r 1 (x), g(x) = q 2 (x)(x 2 + 1) + r 2 (x), where r 1 (x) and r 2 (x) have degree less than 2 (the degree of x 2 + 1). Then f(x)g(x) = q 1 (x)q 2 (x)(x 2 + 1) 2 + q 1 (x)(x 2 + 1)r 2 (x) +q 2 (x)(x 2 + 1)r 1 (x) + r 1 (x)r 2 (x) = q(x)(x 2 + 1). We can get rid of the terms involving x 2 + 1 by plugging in x = i. i 2 + 1 = 0, we get Since f(i)g(i) = r 1 (i)r 2 (i) = q(i)(i 2 + 1) = 0. Since r 1 (x) and r 2 (x) have degree less than 2, write r 1 (x) = a 1 x + a 0 and r 2 (x) = b 1 x + b 0. Then r 1 (i)r 2 (i) = (a 1 i + a 0 )(b 1 i + b 0 ) = 0. Since C is a eld, and hence an integral domain, either a 1 i+a 0 = 0 or b 1 i+b 0 = 0. Without loss of generality, say a 1 i + a 0 = 0. Then a 1 = 0 and a 0 = 0, so and therefore r 1 (x) = 0x + 0 = 0, f(x) = q 1 (x)(x 2 + 1) + r 1 (x) = q 1 (x)(x 2 + 1) x 2 + 1, so x 2 + 1 is a prime ideal. 4.2 x 2 + 1 is not a maximal ideal Consider x 2 + 1, 2 = { p 1 (x)(x 2 + 1) + p 2 (x) 2: p 1 (x), p 2 (x) Z[x] }. It is the ideal generated by x 2 + 1 and 2. We need to show that x 2 + 1 x 2 + 1, 2 Z[x]. 3
4.2.1 x 2 + 1 x 2 + 1, 2 Note that every nonzero element of x 2 + 1 has the form q(x)(x 2 + 1) where q(x) 0, so the degree of a nonzero element is the sum of the degree of q(x) and the degree of x 2 + 1. Hence the degree of any nonzero element of x 2 + 1 is greater than or equal to 2, so there are no nonzero constant polynomials in x 2 + 1. Thus 2 / x 2 + 1, but 2 x 2 + 1, 2, so x 2 + 1 x 2 + 1, 2. 4.2.2 x 2 + 1, 2 Z[x] We need to show that 1 / x 2 + 1, 2. We will proceed by contradiction and assume that 1 x 2 + 1, 2, so there exist polynomials p 1 (x) and p 2 (x) in Z[x] such that 1 = p 1 (x)(x 2 + 1) + p 2 (x) 2. Warning: We cannot conclude that f(x) = 0 because the left hand side is a constant polynomial. I apologize for this mistake from Problem 24, Chapter 16 of HW6 (a corrected proof has been posted). We can get rid of p 1 (x)(x 2 + 1) by plugging in x = i: 1 = p 1 (i)(i 2 + 1)p 2 (i) 2 = p 2 (i) 2, so p 2 (i) = 2 1 = 1 2. However, the coecients of p 2 (x) are integers, and since i 2 = 1 Z, p 2 (i) must be a complex number of the form a + bi where a and b are integers (that 1 is, p 2 (i) Z[i]. Thus, p 2 (i) cannot equal 2, a contradiction. Therefore, 1 / x 2 + 1, 2, so x 2 + 1, 2 Z[x]. Alternate Method: Take 1 = p 1 (x)(x 2 + 1) + p 2 (x) 2 and reduce both sides mod 2 (so we will be working in Z 2 [x]). This gets rid of the p 2 (x) 2 term and leaves us with 1 = p 1 (x)(x 2 + 1) where the coecients of p 1 (x) are those of p 1 (x) reduced mod 2. If p 1 (x) = 0, then we get a contradiction (1 = 0). If p 1 (x) 0, then by Problem 17 from Chapter 16, the degree of p 1 (x)(x 2 + 1) equals the sum of the degrees of p 1 (x) and x 2 + 1, so the degree of p 1 (x)(x 2 + 1) is greater than or equal to the degree of x 2 + 1, which is 2. However, the constant polynomial 1 has degree zero, a contradiction. 4
4.2.3 Conclusion Since x 2 + 1 x 2 + 1, 2 Z[x], x 2 + 1 is strictly contained in an ideal which is not equal to Z[x], so x 2 + 1 is not a maximal ideal. 5 Chapter 17, Problem 34 Suppose r is a real number and r + r 1 = 2m + 1 where m is an integer (so 2m + 1 is an odd integer). Let us multiply both sides by r: r 2 + 1 = (2m + 1)r. Move everything to the left: Thus, r is a zero of the polynomial r 2 (2m + 1)r + 1 = 0. f(x) = x 2 (2m + 1)x + 1. Does f(x) have any zeros in Q? Since it has degree two, this is equivalent (by Theorem 17.1) to asking whether f(x) is irreducible in Q[x]. Since the coecients are integers, we can use Theorem 17.3 and reduce the polynomial mod 2 (I chose 2 because the coecient of x is only known to be odd, so it will denitely reduce to 1 mod 2): Note that f(x) has no zeros in Z 2 : f(x) = x 2 (1)x + 1 = x 2 + x + 1 mod 2. f(0) = 0 + 0 + 1 = 1 0, f(1) = 1 2 + 1 + 1 = 3 = 1 0. Since f(x) has degree 2, by Theorem 17.1, f(x) is irreducible over Z2, and since it has the same degree as f(x), by Theorem 17.3, f(x) is irreducible over Q, so by Theorem 17.1, f(x) has no zeros in Q. Since r is a zero of f(x), r cannot be in Q, so it must be irrational. 5