It follows that if p and q are non zero prime ideals

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It follows that if P and Q are non-zero prime ideals of R , and if Q P , then Q = P . We can now prove the uniqueness of the factorization of the principal ideal (6) as a product P 2 1 P 2 P 3 of prime ideals, where P 1 , P 2 and P 3 are defined as described above. Suppose that (6) = P 2 1 P 2 P 3 = Q 1 Q 2 · · · Q l . Then Q 1 Q 2 · · · Q l P 1 . It follows from repeated applications of Lemma 2.20 that at least one of the prime ideals Q 1 , Q 2 , . . . , Q k is contained in the prime ideal P 1 . We reorder Q 1 , Q 2 , . . . , Q l , if necessary, so that Q 1 P 1 . Then Q 1 = P 1 , and thus P 2 1 P 2 P 3 = P 1 Q 2 Q 3 · · · Q l . If we then multiply both sides of this identity by the ideal P 1 , we find that (2) P 1 P 2 P 3 = P 3 1 P 2 P 3 = P 2 1 Q 2 Q 3 · · · Q l = (2) Q 2 Q 3 · · · Q l . But the ideals (2) P 1 P 2 P 3 and (2) Q 2 Q 3 · · · Q l are the images of the ideals P 1 P 2 P 3 and Q 2 Q 3 · · · Q l under the injective function from R to itself that mul- tiplies all elements of R by 2. It therefore follows that P 1 P 2 P 3 = Q 2 Q 3 · · · Q l . We now note that at least one of the ideals Q 2 , Q 3 , . . . , Q l must be contained in the ideal P 1 . We can therefore reorder these ideals, if necessary, to ensure that Q 2 P 1 . Then Q 2 = P 1 . If we then multiply both sides of the identiy by P 1 , we find that (2) P 2 P 3 = P 2 1 P 2 P 3 = P 2 1 Q 3 Q 4 · · · Q l = (2) Q 3 Q 4 · · · Q l . It follows that P 2 P 3 = Q 3 Q 4 · · · Q l . Repetition of the argument shows that at least one of the ideals Q 3 Q 4 · · · Q l is in P 2 . We may suppose that Q 3 P 2 . Then Q 3 = P 2 , and thus P 2 P 3 = P 2 Q 4 · · · Q l . If we then multiply both sides of this identity by the ideal P 3 , we find that (3) P 3 = P 3 P 2 P 3 = P 3 P 2 Q 3 · · · Q l = (3) Q 4 · · · Q l . It follows that P 3 = Q 4 · · · Q l . Then at least one of the ideals Q 4 , . . . , Q l must be contained in P 3 . We may suppose that Q 4 P 3 , and therefore Q 4 = P 3 . It cannot be the case that l > 4, because multiplying both sides 29
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of the identity P 3 = P 3 Q 5 · · · Q l by the ideal P 2 would lead to the identity (3) = (3) Q 5 · · · Q l , from which it would follow that the product of Q 5 · · · Q l would be the whole of the ring R , which is impossible. Therefore l = 4, and we have shown that Q 1 , Q 2 , Q 3 , Q 4 can be ordered so that Q 1 = Q 2 = P 1 , Q 3 = P 2 and Q 4 = P 3 . This proves that the factorization of the principal ideal (6) as (6) = P 2 1 P 2 P 3 is unique, subject only to reordering of the factors. The integral domain Z [ - 5] is not a unique factorization domain. We have shown that the element 6 of the domain cannot be factorized as a product of prime elements of the domain. Nevertheless the corresponding principal ideal (6) can be factorized uniquely as a product of prime ideals. In fact, every non-zero ideal of Z [ - 5] can be factorized uniquely as a product of prime ideals. The same is true of many analogous integral domains that arise in algebraic number theory.
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  • Fall '16
  • Jhon Smith
  • Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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