# HW9_soln - Homework #9 Solutions Handout, #1 As suggested,...

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Homework #9 Solutions Handout, #1 As suggested, we induct on m . When m = 1 we must prove the following statement: if p 1 ,q 1 ,...,q n D ( n Z + ) are primes and p 1 = q 1 q 2 ··· q n then n = 1. So, suppose we have the stated hypotheses and assume that n 2. Since p 1 is prime and divides q 1 ··· q n it divides q 1 (without loss of generality). So q 1 = ap 1 for some a D . But then the irreducibility of q 1 implies that a is a unit (since p 1 is not). Therefore we have p 1 = ( ap 1 ) q 2 ··· q n = p 1 aq 2 ··· q n . As we are working in a domain we can cancel p 1 from both sides to obtain 1 = aq 2 ··· q n , implying that q 2 is a unit. As q 2 is prime this is a contradiction and we conclude therefore that our assumption that n 2 is false. Thus, n = 1 and p 1 = q 1 . We now prove the induction step. Let m Z + be at least 2 and assume that the statement of the problem is true for m - 1 and any n Z + . Let p 1 ,...,p m ,q 1 ,...,q n D ( n Z + ) be primes with p 1 p 2 ··· p m = q 1 q 2 ··· q n . Since p m is prime and divides q 1 ··· q n it divides q n (without loss of generality). Since p m and q n are both primes (and hence irreducible) we may argue as above and conclude that p m and q n are associate. Writing q n = ap m for some unit a D we have

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## This note was uploaded on 02/29/2012 for the course MATH 4363 taught by Professor Ryandaileda during the Spring '07 term at Trinity University.

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HW9_soln - Homework #9 Solutions Handout, #1 As suggested,...

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