Abstract Algebra Theory and Applications - Computer Science ...

284 CHAPTER 16 INTEGRAL DOMAINSConsequently, either a and p are associates or a is a unit. Therefore, p isirreducible.Conversely, let p be irreducible. If 〈a〉 is an ideal in D such that 〈p〉 ⊂〈a〉 ⊂ D, then a | p. Since p is irreducible, either a must be a unit or aand p are associates. Therefore, either D = 〈a〉 or 〈p〉 = 〈a〉. Thus, 〈p〉 is amaximal ideal.□Corollary 16.9 Let D be a PID. If p is irreducible, then p is prime.Proof. Let p be irreducible and suppose that p | ab. Then 〈ab〉 ⊂ 〈p〉. ByCorollary 14.17, since 〈p〉 is a maximal ideal, 〈p〉 must also be a prime ideal.Thus, either a ∈ 〈p〉 or b ∈ 〈p〉. Hence, either p | a or p | b.□Lemma 16.10 Let D be a PID. Let I 1 , I 2 , . . . be a set of ideals such thatI 1 ⊂ I 2 ⊂ · · · . Then there exists an integer N such that I n = I N for alln ≥ N.Proof. We claim that I = ⋃ ∞i=1is an ideal of D. Certainly I is not empty,since I 1 ⊂ I and 0 ∈ I. If a, b ∈ I, then a ∈ I i and b ∈ I j for some i and jin N. Without loss of generality we can assume that i ≤ j. Hence, a and bare both in I j and so a − b is also in I j . Now let r ∈ D and a ∈ I. Again,we note that a ∈ I i for some positive integer i. Since I i is an ideal, ra ∈ I iand hence must be in I. Therefore, we have shown that I is an ideal in D.Since D is a principal ideal domain, there exists an element a ∈ D thatgenerates I. Since a is in I N for some N ∈ N, we know that I N = I = 〈a〉.Consequently, I n = I N for n ≥ N.□Any commutative ring satisfying the condition in Lemma 16.10 is saidto satisfy the ascending chain condition, or ACC. Such rings are calledNoetherian rings, after Emmy Noether.Theorem 16.11 Every PID is a UFD.Proof. Existence of a factorization. Let D be a PID and a be a nonzeroelement in D that is not a unit. If a is irreducible, then we are done. If not,then there exists a factorization a = a 1 b 1 , where neither a 1 nor b 1 is a unit.Hence, 〈a〉 ⊂ 〈a 1 〉. By Lemma 16.7, we know that 〈a〉 ̸= 〈a 1 〉; otherwise, aand a 1 would be associates and b 1 would be a unit, which would contradictour assumption. Now suppose that a 1 = a 2 b 2 , where neither a 2 nor b 2 is aunit. By the same argument as before, 〈a 1 〉 ⊂ 〈a 2 〉. We can continue withthis construction to obtain an ascending chain of ideals〈a〉 ⊂ 〈a 1 〉 ⊂ 〈a 2 〉 ⊂ · · · .