College Algebra Exam Review 288

# College Algebra Exam Review 288 - 298 6 RINGS coefﬁcients...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 298 6. RINGS coefﬁcients whose constant term is an integer. R is an integral domain, as it is a subring of QŒx. The units in R are ˙1. No rational multiple of x is irreducible, because x D 2. 1 x/ is a proper factorization. If x is written 2 as a product of several elements in R, exactly one of these elements must be a rational multiple of x , while the remaining factors must be integers. Therefore, x has no factorization by irreducibles. Factorization in Principal Ideal Domains We are going to show that a principal ideal domain is a unique factorization domain. The proof has two parts: First we show that every nonzero, nonunit element of a PID has at least one factorization by irreducibles. Then we show that an element cannot have two essentially different factorizations by irreducibles. Lemma 6.5.15. Suppose that R is a principal ideal domain. If a1 R Â a2 R Â a3 R is an increasing sequence of ideals in R, then there exists an n 2 N such that [m 1 am R D an R. Proof. Let I D [n an R. Then I is an ideal of R, by Proposition 6.2.21. Since R is a PID, there exists an element b 2 I such that I D bR. Since b 2 I , there exists an n such that b 2 an R. It follows that bR Â an R Â I D bR; so I D an R. I Lemma 6.5.16. Let R be a commutative ring with multiplicative identity element, and let a; b 2 R. Then ajb , bR Â aR. Moreover, a is a proper factor of b , bR aR R. ¤ Proof. Exercise 6.5.13. ¤ I Lemma 6.5.17. Let R be a principal ideal domain. Then every nonzero element of R that is not a unit has at least one factorization by irreducible elements. Proof. Let R be a principal ideal domain. Suppose that R has a nonzero element a that is not a unit and has no factorization by irreducible elements. Then a itself cannot be irreducible, so a admits a proper factorization a D bc . At least one of b and c does not admit a factorization by ...
View Full Document

## This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

Ask a homework question - tutors are online