What is an example of UFD such that a gcd of an infinite set does not exist?

Question

Let $R$ be a unique factorization domain and $S$ be an infinite subset of $R$. What would be an example of $R$, $S$ such that a gcd of $S$ does not exist in $R$?

That is, is there an infinite set $S$ such that there is no least principal ideal containing it? I think it is really hard to find this example..

Answer 1

Let $S$ be an infinite subset of $R$. We prove that $\gcd S$ does exist.

If there exists a finite subset $S'\subset S$ with $\gcd S'=1$ then $\gcd S=1$.

Suppose $\gcd S'\ne1$ for any finite subset $S'\subset S$.

Fix an element $x\in S$. It has finitely many prime factors $p_1,\dots,p_r$, and one of these occurs in every other element of $S$. Otherwise, let $s_i\in S$ be such that $p_i\nmid s_i$. Then $s_1,\dots,s_r,x$ is a finite subset of $S$ with $\gcd=1$. Say $p_1\mid s$ for any $s\in S$. Now replace $S$ by $(1/p_1)S$ and continue the reasoning.

Answer 2

Let $L$ be a bounded lattice where every element has only a finite number of predecessors. Then $L$ is $\land$-complete, in the sense that every subset has an infimum.

Indeed, if $S$ is a nonempty subset of $L$, fix $x\in S$. Then the set $$ S'=\{x\land s:s\in S\} $$ is finite, so its infimum exists and $\inf S'=\inf S$.

Now, let $L=R/{\sim}$, where $\sim$ is the “being associate” relation. Then $L$ is a lattice with respect to "is a divisor of” relation and every element has only a finite number of divisors.

Answer 3

Hint $\ \ldots (s_3,s_2,s_1) \mid (s_2,s_1)\mid s_1\, $ must stabilize (at $\,\gcd \{s_i\})$