Consider the following setup: Let $R$ be a ring (for simplicity, let us assume at least a UFD), $I<R$ a (proper) ideal. Suppose that every two elements of $I$ have a non-trivial $\operatorname{gcd}$. I am interested in the criterions under which the following holds:
Claim. Then there is some $d\in R\setminus R^\times$ such that $I\subseteq\langle d\rangle<R$.
I.e., if any two elements have a non-trivial $\operatorname{gcd}$ then the ideal has a non-trivial common divisor. This conditions is in general satisfied for all ideals contained in principal ideals (trivially), hence the statement is of no interest in PIDs. I am interested in reasonable conditions to put on $R$ making this work.
Actually, I know a ring for which this works, namely the Iwasawa algebra $\Lambda=\mathbb Z_p[[T]]$ (this is also where I stumbled upon this statement originally). However, the proof I found depends on several properties of this ring and is, as far as I can tell, not noteably generalizable. See below.
I made some observations from some attempts at proving/disproving the claim and talking to some people about this. The main obstruction for almost everything seems to be that we have a priori no control over the $\operatorname{gcd}$s of elements in $I$; we do not even know if they are still in $I$ (it is not in general). However, requiring this property trivializes the proof (if we assume Noetherianess too) and so I would like to avoid this. As non-trivial (=not contained in a principal ideal by construction) examples of ideals satisfying this property are hard to come by, it is difficult to look for explicit counterexamples. E.g., it also does not suffice to just consider the condition on generators as $\langle 2x,xy,2y\rangle<\mathbb Z[x,y]$ shows (this ideal is also of height $>1$ while height $<1$ is necessary for the claim to work).
I am looking for a proof and/or a counterexample for the claim under reasonable assumptions (e.g. Noetherianess but not principality of ideals; take rings to similar $\Lambda$ as something I am willing to accept). I would also appreciate any kind of input on this as was not able to find anything related.
Thanks in advance!
Proof for $\mathit\Lambda$. We argue over the height $\operatorname{ht}$ of $I$. Recall that the height of a non-prime ideal is the minimum of the height of primes containing $I$. If $\operatorname{ht}(I)\le1$ the result follows from the classification of height $1$ prime ideals of $\Lambda$ (these are all principal). Now recall that $\Lambda$ is local and $2$-dimensional with maximal ideal $\langle p,T\rangle$ of height $2$. The case $\operatorname{ht}(I)=2$ occurs iff the only prime above $I$ is $\langle p,T\rangle$, whence $\sqrt I=\langle p,T\rangle$. Then there are $n,m>0$ such that $p^n,T^m\in I$. But $p^n$ and $T^m$ are coprime contradicting our assumption on $I$. $\square$
Best Answer
The titular question was answered in the linked MathOverflow post. It boils down to a clever application of the prime avoidance lemma or similar arguments (e.g. as found in common proofs of the Chinese Remainder Theorem).
The main takeaway: a UFD certainly suffices for the proof but the assumption on the existence of general unique factorizations seems to be crucial!