College Algebra Exam Review 290

College Algebra Exam Review 290 - 300 6. RINGS Theorem...

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300 6. RINGS Theorem 6.5.19. Every principal ideal domain is a unique factorization domain. Proof. Let R be a principal ideal domain, and let a 2 R be a nonzero, nonunit element. According to Lemma 6.5.17 , a has at least one factoriza- tion by irreducibles. We have to show that in any two such factorizations, the factors are the same, up to order and multiplication by units. So suppose that 0 ± r ± s , and that p 1 ;p 2 ;:::;p r ;q 1 ;q 2 ;:::;q s are irreducibles, and p 1 p 2 ²²² p r D q 1 q 2 ²²² q s : (6.5.1) I claim that r D s , and (possibly after permuting the q i ) there exist units c 1 ;:::;c r such that q i D c i p i for all i . If r D 0 , Equation ( 6.5.1 ) reads 1 D q 1 ²²² q s . It follows that s D 0 , since irreducibles are not invertible. If r D 1 , Equation ( 6.5.1 ) reads p 1 D q 1 ²²² q s ; it follows that s D 1 and p 1 D q 1 , since otherwise p 1 has a proper factorization. Suppose r ³ 2 . We can assume inductively that the assertion holds when r and s are replaced with r 0 and s 0 with r 0 < r and s
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

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