Proof it follows directly from the definition of

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Proof It follows directly from the definition of unique factorization domains given above that every integral domain with these two properties is a unique factorization domain. Also every prime element of an integral domain is irreducible (Lemma 2.4), and therefore every unique factorization domain 24
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satisfies the first of these two properties. Moreover an irreducible element of a unique factorization domain R must factor as a product of one or more prime elements of R . But, being irreducible, it can only have one prime factor, and therefore it must itself be prime. Thus every unique factorization domain satisfies property (ii). Lemma 2.23 Every principal ideal domain is a unique factorization do- main. Proof Every ideal of a principal ideal domain can be generated by a single element of the domain, and is thus finitely generated. A direct application of Proposition 2.5 therefore shows that any non-zero element of a principal ideal domain that is not a unit can be factored as a finite product of irreducible elements of the domain. Moreover Theorem 2.9 guarantees that every irre- ducible element of a principal ideal domain is prime. The result therefore follows from Lemma 2.22. 2.7 Prime Ideals of Principal Ideal Domains Lemma 2.24 Let R be a principal ideal domain. Then every non-zero prime ideal of R is a maximal ideal of R . Moreover every non-zero prime ideal of R is a principal ideal generated by some prime element of R . Proof Let P be a non-zero prime ideal of R . Then there exists some non- zero element x of R such that P = ( x ). The ideal ( x ) is a non-zero proper ideal of R . It follows from Lemma 2.19 that x is a prime element of R . Then x is an irreducible element of R , because all prime elements of a principal ideal domain are irreducible. Let I be an ideal of R satisfying ( x ) I R . Then I = ( y ) for some element y of R . But then ( x ) ( y ), and therefore y | x . It follows from the irreducibility of x that either y is a unit, in which case I = R , or y is an associate of x , in which case I = ( x ). Therefore the ideal ( x ) is a maximal ideal of R , as required. Remark Let R be a principal ideal domain, and let x be a prime element of R . The ideal ( x ) generated by x is then a non-zero prime ideal of R . It follows from Lemma 2.24 that the ideal ( x ) generated by x is a maximal ideal of R . It then follows from Lemma 2.13 that the quotient ring R/ ( x ) is a field. These results therefore combine to provide an alternative proof of Lemma 2.8. 25
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Lemma 2.25 Let R be a principal ideal domain. Then every non-zero proper ideal I of R factors as a product I = P 1 P 2 · · · P k where P 1 , P 2 , . . . , P k are non-zero prime ideals of R . Moreover this factor- ization of I as a product of prime ideals is unique: if P 1 P 2 · · · P k = Q 1 Q 2 · · · Q l , where P 1 , P 2 , . . . , P k and Q 1 , Q 2 , . . . , Q l are prime ideals of R , then k = l , and there exists some permutation σ of the set { 1 , 2 , . . . , k } such that Q i = P σ ( i ) for i = 1 , 2 , . . . , k .
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