I am trying to understand the intuition behind this condition for when a UFD is a PID. I understand the steps in the proof, but I don't understand why not having this property of prime ideals should be the obstruction to being a PID. Any help is appreciated.
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This will be clarified greatly if you study the proofs of the equivalent characterizations in the linked dupe. – Bill Dubuque Nov 21 '23 at 11:35
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Suppose your ring $R$ is a PID and let $I = \langle p\rangle$ be a nonzero prime ideal that is not maximal and let $M = \langle m\rangle$ be a maximal ideal containing $I$. Then since $p\in\langle m\rangle$, there exists some element $a\in R$ such that $am=p$. Since $I$ is prime this means that $a\in I$ and in particular $a=pb$ for some $b\in R$. Then we finally get $p = pbm\Leftrightarrow p(bm-1)=0\Leftrightarrow m\text{ is a unit}$ which contradicts the fact that $M$ is a maximal ideal (the last iff is due to the fact that $p$ is not a nonzero divisor, the latter being the case since your ring is integral).
Scott
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It seems that this is a proof that the property holds for PIDs? But I'm asking about the linked proof that says prime ideals being maximal is the condition that makes a UFD a PID. – Ray James Nov 21 '23 at 10:55
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You are right, that is what is shown. Proving this would have been for me the reason why "not having this property of prime ideals should be an obstruction to being a PID". I am not sure I understand your question. Could you maybe rephrase it ? – Scott Nov 21 '23 at 11:02
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Well, I just don't find the linked answer to when UFD=PID to be very enlightening. Why a UFD with this property must be a PID. The use of Zorn's lemma is not intuitive to me, for instance. – Ray James Nov 21 '23 at 11:21