College Algebra Exam Review 289

College Algebra Exam Review 289 - pR prime ” p prime”...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
6.5. FACTORIZATION 299 irreducibles; suppose without loss of generality that b has this property. Now we have aR ± ¤ bR , where a and b are nonunits that do not admit a factorization by irreducibles. An induction based on the previous paragraph gives a sequence a 1 ;a 2 ;::: in R such that for all n , a n is a nonunit that does not admit a factorization by irreducibles, and a 1 R ± ¤ a 2 R ± ¤ ²²² ± ¤ a n R ± ¤ a n C 1 R ± ¤ ²²² : But the existence of such a sequence contradicts Lemma 6.5.15 . n In the next lemma, we show that principal ideal domains also have the property that every irreducible element is prime. Lemma 6.5.18. Let R be an integral domain. Consider the following prop- erties of an nonzero nonunit element p of R : ³ pR is a maximal ideal. ³ pR is a prime ideal. ³ p is prime. ³ p is irreducible. (a) The following implications hold: pR maximal H) pR prime p prime H) p irreducible (b) If R is a principal ideal domain, then the four conditions are equivalent. Proof. Let R be an integral domain and p 2 R a nonzero nonunit element. The implication “ pR maximal H) pR prime” follows from Corollary 6.4.10 . The equivalence “
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: pR prime ” p prime” is tautological. Finally, we prove the implication “ p prime H) p irreducible.” Sup-pose p is prime and p has a factorization p D ab . We have to show that either a or b is a unit. Because p is prime and divides ab , p divides a or b . Say p divides a , namely a D pr . Then p D ab D prb . Since R is an integral domain, we can cancel p , getting 1 D rb . Therefore, b is a unit. Now suppose that R is a principal ideal domain. To show the equiva-lence of the four conditions, we have only to establish the implication “ p irreducible H) pR maximal.” Suppose that p is irreducible and that J is an ideal with pR ´ J ´ R . Because R is a principal ideal domain, J D aR for some a 2 R . It follows that p D ar for some element r . Since p is irreducible, one of a and r is a unit. If a is a unit, then J D aR D R . If r is a unit, then a D r ± 1 p , so J D aR D pR . n...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online