If any "dividing" chain "terminates" at some point, does that imply an integral domain being P.I.D.?

Solution 1:

Your idea is good: given an ideal $I$ of $R$, you can consider $$ \tilde{I}=\{\tilde{a}:a\in I\} $$ and this is a partially ordered set that satisfies the condition that every chain has a lower bound. Thus, by Zorn's lemma, there is a minimal element $\tilde{c}$.

You'd like to show that $c$ is a generator for $I$, but this cannot be proved in general. Let's see why.

You can avoid considering divisibility by realizing that $a\mid b$ is the same as saying that $bR\subseteq aR$, so you can instead consider the set of principal ideals contained in $I$ and condition (b) tells you that this set of ideals has a maximal element, say $aR$.

Now this is where condition (a) enters the scene. Take any $b\in I$; then a (greatest) common divisor $d$ of $a$ and $b$ belongs to $aR+bR\subseteq I$. Then $a,b\in dR$ and $aR\subseteq dR$ forces $dR=aR$ by maximality; hence $b\in aR$.

You see that condition (a) can be relaxed to “for any $a,b\in R$, there exists a common divisor $d$ of $a$ and $b$ such that $d\in aR+bR$”.

Note that condition (b) surely holds in any Noetherian domain, but such domains need not be UFD.