Let $R$ be a UFD. Suppose that there exist $a_1, a_2, ... \in R$ such that $(a_1) \subseteq (a_2) \subseteq \cdots $ . Show that there exists an $i$ such that $(a_i ) = (a_{i+1})$.
My attempt: Suppose that $(a_1) \subseteq(a_2) \subseteq \cdots $. Then for all positive integer $i$, $a_{i+1}\mid a_i$. So $a_i=a_{i+1}b_i$ for some nonzero nonunit $b_i \in R$. But then, in particular, for $a_1$ we have that $a_1=a_2b_1=a_3b_2b_1=\cdots $. That is, $a_1$ can be written as a product of infinitely many irreducibles. But by definition of UFD, the factorization of $a_1$ is finite. Thus for some $i$, we must have that $(a_i ) = (a_{i+1})$.
Here is an answer to this very exercise. But I wanted to bring it to a notation a bit closer to the one I use. I must say that the idea behind the solution is clear to me, but at the time of writing the proof it has been quite difficult for me. Do I need to add more details to my proof? When I make sure in the last part that the definition of UFD completes the proof, is this correct? Are there any other arguments? Thanks in advance
Your "proof" is not a proof. YOu didn't prove that $a_1$ can be written as a product of infinitely many irreducibles. What you proved is that for all $n$, $a_1=a_{n+1}b_nb_{n-1}\cdots b_1$ where the $b_i's$ are non zero and non units, which is totally different. For example,all the $b_i$'s could have a power of the same irreducible. And you do not use fully the fact that $R$ is a UFD, just the existence of a decomposition.
One could imagine a ring $R$ which is not a UFD, but for which a decomposition into irreducibles exist for any nonzero element but for which uniquness fails, and such that there is an infinite chain of principal ideals (something like $R=\mathbb{C}[X_n,Y_n, n\geq 1]/(X_n-X_{n+1}Y_n,n \geq 0)$ might work but I didn't check).
The neater argument is given in the link you provided....