I'm aware there are many posts asking this question, but I was wondering if it was possible to prove this fact using ascending chains?
The problem (as in the title) is the following: Let $D$ be a UFD. Prove that for all $a,b \in D \setminus \{0\}$, the least common multiple $\text{lcm}(a,b)$ exists.
My idea was to go by contradiction and try to avoid the messy expansions I've seen in various proofs of this fact (which leads me to believe that my idea is obviously incorrect at some point).
Suppose that $a,b \in D$ are nonzero and that $\text{lcm}(a,b)$ does not exist. Note since $a \mid ab$ and $b \mid ab$, there exist elements of $D$ that $a,b$ both divide.
Thus, given any $\beta_1 \in D$ such that $a \mid \beta_1$ and $b \mid \beta_1$, there must exist some $\beta_2 \in D$ such that $\beta_1 \neq \beta_2$ and $\beta_2 \mid \beta_1$. Repeating inductively gives a chain of elements $\beta_k \in D$ such that $\beta_{k+1} \mid \beta_k$ and $\beta_{k+1} \neq \beta_k$. Thus, we have that $\langle \beta_{k} \rangle \subsetneq \langle \beta_{k+1} \rangle$ for all $k$ which is a chain of principle ideals in $D$ that never stabilizes, a contradiction to the fact that all UFD's satisfy the ascending chain condition on principle ideals (ACCPI). Thus, $a,b$ must have a lcm.
Is this correct, or did I make some obvious mistake somewhere? I imagine someone more experienced than myself can point out any blunder I may have in this argument.
Thank you.
