Let $R$ be a ring and $I$ an ideal. I am interested under which conditions the following holds:



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> **Claim.** Suppose that any two elements in $I$ have a non-trivial $\operatorname{gcd}$. Then $I$ is contained in some non-trivial principal ideal.
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For simplicity you may assume that $R$ is a UFD so that $\operatorname{gcd}$s exist and are well-defined (a GCD domain probably suffices for this too but I have no experience with them). Although this question looks quite simple, I only know of one non-trivial example where this holds. Namely $R=\mathbb Z\_p[[T]]$ and by extension $R=\mathcal O[[T]]$ for $\mathcal O$ the ring of integers of some local field extension.


I asked this question some while ago on Math.SE ([here](https://math.stackexchange.com/questions/4317163/a-criterion-for-whether-an-ideal-is-contained-in-a-principal-ideal)) but have received no comment/answer. There you can also find a proof of the above claim for $R=\mathbb Z\_p[[T]]$. As far as I can tell, this proof does not generalize notably.



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> Are there some reasonable conditions one can put on $R$ to make the above claim work? By reasonable I mean something not making the assertion trivial (e.g. not taking a PID but maybe UFD+Noetherian suffices).
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Thanks in advance!



A criterion for whether an ideal is contained in a principal ideal
This is true in UFDs. Let $f$ be any element of $I$ and let $\pi\_1,\dots, \pi\_n$ be its irreducible factors (choosing one from each class of irreducible factors that differ by a unit). If, for some $i$, $I \subseteq (\pi\_i)$ then we are done. Otherwise, for each $i$ let $g\_i$ be an element of $I$ that is nonzero mod $\pi\_i$.


Then $$ h= \sum\_{i=1}^n g\_i \prod\_{ 1\leq j \leq n, j \neq i} \pi\_j$$ is an element of $I$. Because $R$ is a UFD, $\pi\_i$ is a prime ideal, so $R / \pi\_i$ is an integral domain. For $j\neq i$, since $\pi\_j$ is irreducible and not a unit times $\pi\_i$, $\pi\_j$ is not a multiple of $\pi\_i$, so $\pi\_j$ is not a zero-divisor mod $\pi\_i$. Thus


$$ h \equiv g\_i \prod\_{ 1\leq j \leq n, j \neq i} \pi\_j \not\equiv 0 \mod \pi\_i$$


Since $h$ is not a multiple of $\pi\_i$ for any irreducible factor $\pi\_i$ of $f$, $f$ and $h$ do not have a gcd, contradicting the assumption.


