Abstract Algebra Theory and Applications - Computer Science ...

16.2 FACTORIZATION IN INTEGRAL DOMAINS 285By Lemma 16.10, there exists a positive integer N such that 〈a n 〉 = 〈a N 〉for all n ≥ N. Consequently, a N must be irreducible. We have now shownthat a is the product of two elements, one of which must be irreducible.Now suppose that a = c 1 p 1 , where p 1 is irreducible. If c 1 is not a unit,we can repeat the preceding argument to conclude that 〈a〉 ⊂ 〈c 1 〉. Eitherc 1 is irreducible or c 1 = c 2 p 2 , where p 2 is irreducible and c 2 is not a unit.Continuing in this manner, we obtain another chain of ideals〈a〉 ⊂ 〈c 1 〉 ⊂ 〈c 2 〉 ⊂ · · · .This chain must satisfy the ascending chain condition; therefore,a = p 1 p 2 · · · p rfor irreducible elements p 1 , . . . , p r .Uniqueness of the factorization. To show uniqueness, leta = p 1 p 2 · · · p r = q 1 q 2 · · · q s ,where each p i and each q i is irreducible. Without loss of generality, we canassume that r < s. Since p 1 divides q 1 q 2 · · · q s , by Corollary 16.9 it mustdivide some q i . By rearranging the q i ’s, we can assume that p 1 | q 1 ; hence,q 1 = u 1 p 1 for some unit u 1 in D. Therefore,ora = p 1 p 2 · · · p r = u 1 p 1 q 2 · · · q sp 2 · · · p r = u 1 q 2 · · · q s .Continuing in this manner, we can arrange the q i ’s such that p 2 = q 2 , p 3 =q 3 , . . . , p r = q r , to obtainu 1 u 2 · · · u r q r+1 · · · q s = 1.In this case q r+1 · · · q s is a unit, which contradicts the fact that q r+1 , . . . , q sare irreducibles. Therefore, r = s and the factorization of a is unique. □Corollary 16.12 Let F be a field. Then F [x] is a UFD.Example 5. Every PID is a UFD, but it is not the case that every UFDis a PID. In Corollary 16.22, we will prove that Z[x] is a UFD. However,Z[x] is not a PID. Let I = {5f(x) + xg(x) : f(x), g(x) ∈ Z[x]}. We caneasily show that I is an ideal of Z[x]. Suppose that I = 〈p(x)〉. Since 5 ∈ I,5 = f(x)p(x). In this case p(x) = p must be a constant. Since x ∈ I,x = pg(x); consequently, p = ±1. However, it follows from this fact that〈p(x)〉 = Z[x]. But this would mean that 3 is in I. Therefore, we can write3 = 5f(x) + xg(x) for some f(x) and g(x) in Z[x]. Examining the constantterm of this polynomial, we see that 3 = 5f(x), which is impossible.