Homework 7 Solutions to Selected Prolems May 9, 01 1 Chapter 16, Prolem 17 Let D e an integral domain and f(x) = a n x n +... + a 0 and g(x) = m x m +... + 0 e polynomials with coecients in D, where a n and m are nonzero. Then f(x) has degree n and g(x) has degree m. Their product is f(x)g(x) = a n m x n+m, and since a n and m are nonzero, a n m is nonzero since D is an integral domain (in D, the product of two nonzero elements is nonzero). Hence f(x)g(x) has degree n + m. What if we had a commutative ring that had zero-divisors? For example consider these two polynomials with degree 1 in Z 4 [x]: f(x) = x, g(x) = x + 1. Their product is ( )x + x = 0x + x = x, which has degree 1, less than deg(f(x)) + deg(g(x)) = 1 + 1 =. In general, the degree of a product f(x)g(x) can e less than the sum of the degrees of f(x) and g(x) if the leading coecients of f(x) and g(x) are zero-divisors whose product is zero. Chapter 16, Prolem 3 Let us nd one nonzero polynomial rst. If p(x) has a as a zero for a = 0, 1, in Z 3, then p(x) has degree at least 3, and x, x 1, and x divide p(x). Thus, we can start with p(x) = x(x 1)(x ) = x 3 + x. Let g(x) e any other polynomial in Z 3 [x]. Then for a = 0, 1,, p(a) = 0, so g(a)p(a) = g(a) 0 = 0. Therefore, there are innitely many polynomials f(x) in Z 3 [x] such that f(a) = 0 for a = 0, 1,. They are polynomial multiples of any polynomial p(x) I[x]. We know there is an innite amount, since x n Z 3 [x] for any positive integer n. 1
3 Chapter 16, Prolem 38 Let I e an ideal of a commutative ring R (see the note to this prolem). Then I is closed under sutraction and multiplication y anything in R: a, I imply a I, a I, r R imply ra I. We need to prove that I[x] is an ideal of R[x]. We note that the constant polynomial that equals zero is in I[x], so we need to show: f(x), g(x) I[x] imply f(x) g(x) I[x], f(x) I[x], r(x) R[x] imply r(x)f(x) I[x]. Let f(x) = a n x n +... + a 0, g(x) = m x m +... + 0 e polynomials with coecients in I (so they are in I[x]). Let s e the maximum of m and n, a i = 0 for i > n, and i = 0 for i > m. Since I is an ideal, 0 I, and a i i I for 0 i s. Therefore f(x) g(x) = (a s s )x s +... + (a 0 0 ) I[x], so I[x] is closed under sutraction. Let r(x) = r l x l +... + r 0 e any polynomial in R[x] (the coecients might not e in I). Then for any 0 i l, 0 j n, Therefore, for k = 0,..., n + l, so Thus, I[x] is an ideal of R[x]. Note: r i a j I. c k = r k a 0 + r k 1 a 1 +... + r 1 a k 1 + r 0 a k I, r(x)f(x) = c n+l x n+l +... + c 0 I[x]. Although the prolem does not specify that R is a commutative ring, e aware that the ook has only dened polynomial rings over a commutative ring. What happens if we try to make polynomials over a ring that is not commutative? You have to e more careful when multiplying polynomials. In Z[x], we can write (x)(3x ) = ( 3)x x = 6x 3, allowing the 3 and an x to switch places, ut if the coecients are matrices, then for example, y denition, [ ] [ ] ([ ] [ ]) [ ] 1 0 1 1 x x 1 0 1 1 = x 1+ 1 1 = x 3, 1 1 1 1 1 1 3 3 [ ] 1 1 ut we are not making x and switch places. 1 1
4 Chapter 16, Prolem 40 In prolem 38, we have shown that I[x] is an ideal. Now suppose that I is a prime ideal of R as well. To show that I[x] is a prime ideal of R[x], we need to show that given f(x), g(x) R[x], if f(x)g(x) I[x], then f(x) or g(x) is in I[x]. We will proceed y induction on the sum of the degrees of f(x) and g(x). 4.1 Base Case: f(x) and g(x) oth have degree zero We have f(x) = a 0 and g(x) = 0, and suppose f(x)g(x) = a 0 0 I[x]. Then the constant term a 0 0 must e in I. Since I is a prime ideal, a 0 or 0 is in I, so either f(x) or g(x) has its constant term in I. Hence either f(x) or g(x) is in I[x]. 4. Induction Step Suppose that if the sum of the degrees of f(x) and g(x) is at most n, then f(x)g(x) I[x] implies f(x) or g(x) is in I[x]. That is, if the coecients of f(x)g(x) are in I, then all the coecients of either f(x) or g(x) are elements of I. Now let f(x) and g(x) e polynomials with coecients known to e in R, and say f(x) has degree k and g(x) has degree n + 1 k, so their degrees sum to n + 1. f(x) = a k x k + a k 1 x k 1 +... + a 0, and a k, n+1 k are nonzero. Then g(x) = n+1 k x n+1 k +... + 0, f(x)g(x) = a k n+1 k x n+1 +... + a 0 0. Suppose f(x)g(x) I[x]. Then all the coecients of f(x)g(x) are in I. In particular, the coecient of x n+1, namely a k n+1 k, must e in I (the other coecients of f(x)g(x) can involve sums of products, and are more complicated). Hence either a k or n+1 k is in I, and without loss of generality, assume a k is the coecient in I. Since a k I, this implies that the polynomial a k x k I[x]. Then since I[x] is an ideal of R[x] (y prolem 38), it is closed under multiplication y any other polynomial in R[x]. Hence ( a k x k) g(x) I[x]. In addition, I[x] is closed under sutraction, so f(x)g(x) I[x] and ( a k x k) g(x) I[x] imply f(x)g(x) ( a k x k) g(x) I[x], so y distriuting, ( f(x) a k x k) g(x) = ( a k 1 x k 1 +... + a 0 ) g(x) I[x]. Let f 1 (x) = f(x) a k x k = a k 1 x k 1 +... + a 0. Its degree is at most k 1 (it could e smaller since some of the coecients like a k 1 could equal zero). Now we can use our inductive hypothesis on the two polynomials f 1 (x) and g(x), since the sum of their degrees are at most k 1 + n + 1 k = n: either f 1 (x) or g(x) is in I[x]. If f 1 (x) I[x], then since a k x k I[x], f(x) = f 1 (x) + a k x k I[x]. Otherwise, g(x) is in I[x], so either way, f(x) or g(x) is in I[x]. 3
4.3 Conclusion By induction, given f(x), g(x) R[x], if f(x)g(x) I[x], then f(x) or g(x) is in I[x]. Therefore, I[x] is a prime ideal. 4.4 Note If R is an integral domain, we can actually perform induction on the degree of f(x)g(x) y Prolem 17. 5 Chapter 16, Prolem 46 5.1 Method 1: Mimic the proof that is irrational Before we egin, let us look at the rings involved. Let F (x) denote the eld of fractions of F [x], where F is a eld. Then elements of F (x) are of the form f(x)/g(x) where f(x), g(x) F [x] and g(x) is not the constant polynomial that always equals zero. Furthermore, f 1 (x)/g 1 (x) = f (x)/g (x) in F (x)if and only if f 1 (x)g (x) = g 1 (x)f (x) in F [x]. Now we will solve the prolem. We will use a proof y contradiction and assume there is an element in F (x) whose square is x/1 (where 1 denotes a constant polynomial that equals the unity of F, although y prolem 8 from the previous homework, we do not have to worry aout the distinction). That is, there exist polynomials f(x), g(x) in F [x] with g(x) 0 and (f(x)/g(x)) = f(x) /g(x) = x/1. If f(x) and g(x) had any zeros in common, y Corollary on page 98, if a is a common zero of f(x) and g(x), then f(x) = (x a)f 1 (x) and g(x) = (x a)g 1 (x). Hence f(x)/g(x) = f 1 (x)/g 1 (x), ecause f(x)g 1 (x) = (x a)f 1 (x)g 1 (x) = (x a)g 1 (x)f 1 (x) = g(x)f 1 (x). Therefore, we can cancel any common zeros (common factors of the form (x a)), and we can assume f(x) and g(x) have no zeros in common. Essentially, we are assuming that f(x)/g(x) is in lowest terms. Now, f(x) /g(x) = x/1 implies 1f(x) = f(x) = xg(x). Hence x divides f(x), so there exists a polynomial a n x n +... + a 0 such that f(x) = (a n x n +... + a 0 ) x = a n x n+1 +... + a 0 x + 0. Most importantly, the constant term of f(x) is zero. Now suppose f(x) = m x m +... + 0. Could 0 e nonzero? By denition, f(x) = f(x)f(x) = m m x m +... + 0 0 = a n x n+1 +... + a 0 x + 0, 4
so 0 and 0 = 0. Since F is a eld, it has no zero-divisors, so 0 = 0. Hence f(x) = m x m +... + 1 x + 0 = ( m x m 1 +... + 1 ) x, so x divides f(x). Hence x 0 is a factor of f(x), so 0 F is a zero of f(x). ( However, since f(x) ) = m x m +... + 1 x, f(x) = mx m +... + 1x = m x m +... + 1 x, so f(x) = xg(x) implies ( m x m +... + ) 1 x = xg(x). Since F [x] is an integral domain and x is not the additive identity of F [x], we can cancel an x on oth sides: ( m x m +... + ) 1 x = g(x). Hence x divides g(x), and y a similar argument for f(x), we have x = x 0 is a factor of g(x). But that means g(x) also has 0 as a zero, contradicting the assumption that f(x) and g(x) had no zeros in common. Therefore, there is no element in F (x) which squares to x/1. 5. Method : Use Prolem 17 and compare degrees We will use a proof y contradiction and assume there is an element in F (x) whose square is x/1. That is, there exist polynomials f(x), g(x) in F [x] with g(x) 0 and (f(x)/g(x)) = f(x) /g(x) = x/1. Now, f(x) /g(x) = x/1 implies 1f(x) = f(x) = xg(x). Suppose f(x) has degree n and g(x) has degree m. By Prolem 17, f(x) has degree n, while g(x) has degree m. Hence xg(x) has degree 1 + m = m + 1. Then f(x) = xg(x) implies their degrees are equal, so n = m+1 for some integers n and m. However, there is no integer which is oth even (n) and odd (m+1). Thus, f(x) and g(x) do not exist, and there is no element in F (x) which squares to x/1. 5