If any "dividing" chain "terminates" at some point, does that imply an integral domain being P.I.D.? I'm working on Dummit & Foote's Abstract Algebra. In Section 8.2 of P.I.D.s I found the following question:

If $R$ is an integral domain, prove that $R$ is a P.I.D. if the following two conditions hold:
(a) Any two nonzero elements $a,b$ have a g.c.d in $(a,b)$;
(b) For a chain of nonzero elements $a_1,a_2,a_3,\dots$, if $a_{i+1}|a_i$, then there exists a positive integer $N$ such that for all $n>N$, $a_n$ is a unit times $a_N$.

A simple proof goes by taking elements that are g.c.d. with all previous elements successively, which forms a chain, and by (b) it must terminate, hence giving a generating element.
My question would be, is condition (b) sufficient on itself?
Condition (b) reminds me of a lot of proofs using Zorn's Lemma, so I tried to outline a proof below:
Define an equivalence relation on $R$ by $a\sim b\iff a=ub$ where $u$ is a unit. Define a partial order on equivalence classes so that $\bar a\leq \bar b\iff a|b$. Then condition (b) states that given an ideal $I$, all decreasing chains in $I$ adopt a minimal element, thus by Zorn's Lemma, $I$ has a minimal element, which would be its generator.
Are there any mistakes in this argument? If it is wrong, are there any counterexamples?
Edit: Yes, the proof is wrong because I misunderstood Zorn's Lemma. A minimal element $m$ in Zorn's Lemma means that for any $x\neq m$ there can't be $x\leq m$, but it doesn't necessarily mean that $m\leq y$ for all $y$.
 A: 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.
