Let $R$ be a ring and $I$ an ideal. I am interested under which conditions the following holds:
Claim. Suppose that any two elements in $I$ have a non-trivial $\operatorname{gcd}$. Then $I$ is contained in some non-trivial principal ideal.
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) 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.
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).
Thanks in advance!
Best Answer
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.