I found this problem from Aluffi's Algebra Chapter 0 and only managed to find one reference (without proof) to this statement online, so I'm posting this here for reference/verification.
Problem
Let $\mathcal{A} = \{(a_i)\}$ be a family of principal ideals in a UFD $R$.
Show that the ideal $I = \bigcap_i (a_i)$ is principal.
Solution (?)
$I$ is an ideal since it's an intersection of ideals.
Pick any $r_0 \in I$. If $(r_0) = I$ we are done, so suppose $(r_0) \subsetneq I$.
Then pick any $s \in I – (r_0)$.
Since $R$ is a UFD, greatest common divisors always exist so we can set $r_1 = \mathrm{gcd}(r_0, s)$.
By definition of gcd, the ideal $(r_1)$ is the smallest principal ideal containing the ideal $(r_0, s)$.
Since $(r_0, s) \subseteq (a_i)$ for each $i$,
we therefore have $(r_1) \subseteq (a_i)$ for each $i$.
Thus $(r_1) \subseteq I$ and we get
$$
(r_0) \subsetneq(r_0, s) \subseteq (r_1) \subseteq I.
$$
As long as $(r_n) \subsetneq I$ this process can be repeated fo find $r_{n+1}$ such that
$(r_n) \subsetneq (r_{n+1}) \subseteq I$, yielding an ascending sequence of principal ideals
$$
(r_0) \subsetneq \cdots \subsetneq (r_{n}) \subsetneq (r_{n+1}) \subseteq I.
$$
But since $R$ is a UFD this sequence must stabilize, which can only happen if we eventually have $(r_n) = I$.
Question
Does my solution look ok?
Isn't the generator of the ideal $I$ the least common multiple of the $a_i$'s? If this is true, does this statement imply that lcms exist for arbitrary families in UFDs? Can we say that gcds of arbitrary families exist in UFDs by the same logic?
Best Answer
Your solution looks fine. And you are correct that any generator of the intersection will be an lcm. You can indeed also find arbitrary gcds in a UFD $R$. Take some subset $S\subseteq R$. Let $D$ be the set of principal ideals containing $S$. Of course, the unit ideal $(1)\in D$ so it's nonempty. As you've shown in your proof, the intersection $\bigcap_{(d) \in D} (d)$ has a principal generator, say $e$. Then $S\subseteq (e)$ so for all $s \in S$, $e \mid S$, i.e. $e$ is a common divisor of the elements of $S$. It is the greatest common divisor because if $d$ is another common divisor of the elements of $S$ then $S \subseteq (d)$, i.e. $(d) \in D$. Hence, $(e)\subseteq (d)$ and $d\mid e$.