College Algebra Exam Review 288

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

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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  a2 R  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  an R  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  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 ...
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