The following proposition guarantees that any

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The following proposition guarantees that any factorization of an element of a principal ideal domain as the product of one or more prime elements of that domain is unique up to the order of the factors and the replacement of any factor by an associate of that factor. Proposition 2.21 Let R be an integral domain, and let x be a non-zero element of R that is not a unit of R . Suppose that x = p 1 p 2 · · · p k = q 1 q 2 , · · · , q l , where p 1 , p 2 , . . . , p k are prime elements of R and q 1 , q 2 , . . . , q l are irreducible elements of R . Then l = k , and there exists some permutation σ of the set { 1 , 2 , . . . , k } such that q i and p σ ( i ) are associates for i = 1 , 2 , . . . , k . 23
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Proof The result holds when k = 1, because every prime element of R is irreducible, and therefore cannot be factored as a product of two or more irreducible elements. Let k be an integer greater than 1, and suppose that the stated result holds for all non-zero elements of R that are not units of R and that can be factored as a product of fewer than k prime elements of R . We shall prove that the result then holds for any non-zero element x of R that is not a unit of R and that can be factored as a product p 1 p 2 · · · p k of k prime elements p 1 , p 2 , . . . , p k of R . The required result will then follow by induction on k . So, suppose that x is an non-zero element of R that is not a unit of R , and that x = p 1 p 2 · · · p k = q 1 q 2 , · · · , q l , where p 1 , p 2 , . . . , p k are prime elements of R and q 1 , q 2 , . . . , q l are irreducible elements of R . Now p 1 divides the product q 1 q 2 , · · · , q l , and therefore p 1 divides at least one of the factors q i of this product. We may reorder and relabel the irreducible elements q 1 , q 2 , . . . q l to ensure that p 1 divides q 1 . The irreducibility of q 1 then ensures that p 1 is an associate of q 1 , and therefore there exists some unit u in R such that q 1 = p 1 u . But then p 1 ( p 2 p 3 · · · p k ) = p 1 ( uq 2 q 3 · · · q l ) and p 1 6 = 0 R , and therefore p 2 p 3 · · · p k = ( uq 2 ) q 3 · · · q l . (see Lemma 2.1). Moreover uq 2 is an irreducible element of R that is an as- sociate of q 2 . Now it follows from the induction hypothesis that the de- sired result holds for the product p 2 p 3 · · · p k . Therefore l = k and moreover q 2 , q 3 , . . . , q k can be reordered and relabeled so that p i and q i are associates for i = 2 , 3 , . . . , k . The stated result therefore follows by induction on the number of prime factors occuring in the product p 1 p 2 · · · p k . Definition An integral domain R is said to be a unique factorization domain if every non-zero element of R that is not a unit of R can be factored as the product of one or more prime elements of R . Lemma 2.22 An integral domain R is a unique factorization domain if and only if it has the following two properties: (i) any non-zero element of R that is not a unit of R can be factored as the product of one or more irreducible elements of R ; (ii) every irreducible element of R is a prime element of R .
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  • Fall '16
  • Jhon Smith
  • Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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