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1) any 2 elements [tex] a, b \in R [/tex] have a greatest common divisor which can be written as ra + sb for some [tex] r, s \in R [/tex].

2) If [tex] a_1, a_2, ... [/tex] are nonzero elements of R such that [tex] a_{i+1}|a_i [/tex] for all i, then there is a positive integer N such that [tex] a_n [/tex] is a unit times [tex] a_N [/tex] for all [tex] n \geq N [/tex].

Letting I be any ideal in R with the ordering [tex] x \leq y [/tex] iff y|x, condition #2 implies that each chain of elements in I has an upper bound and thus by Zorn's Lemma, there is a maximal element in I with respect to this ordering, namely, an element that divides every other element in I.

what i am having trouble understanding is why condition #1 is necessary. it seems that if we say a is the element in I that divides every other element, then can't we conclude that I = (a), since (a) is clearly in I and I is in (a) since every element x of I must be contained in (a)?

if there was another such element b in I not equal to a, then since b and a are both maximal elements with respect to this ordering, a|b and b|a which means b = ac for some c and a = bd for some d, which implies that cd = 1 since R is an integral domain. Thus c and d are units so a and b are associates so (a) = (b). Why is the condition that the gcd of a and b exist necessary for R to be a PID?