For any set in a principal ideal ring, can one choose a finite subset so that the generated ideals are the same?

Question

Let $R$ be a principal ideal ring, meaning that every ideal is generated by one element. Given a subset $A\subseteq R$, it is generally not possible to choose one element $x\in A$ so that $I(\{x\})=I(A)$, i.e. the ideal generated by $A$ need not be generated by a single element of $A$.

As an example one can consider $R=\mathbb Z$ and $A=\{2,3\}$, then $I(A)=\mathbb Z$.

My question is whether one can choose a finite subset $X$ of $A$ so that $I(A)=I(X)$.

In $\mathbb Z$ for example this is always possible:

The ideal generated by a set is always the ideal generated by the GCD of that set. If $X\subseteq Y$ then $\mathrm{GCD}(Y)≤\mathrm{GCD}(X)$. If we have a finite sbuset $X$ of $A$ and the GCDs are not equal, there must exist an element of $A$ so that appending it to $X$ will decrease the GCD. This can only happen a finite amount of times since we are decreasing a positive number by at least $1$ each time.

I don't expect this argument to work in general principal ideal rings, since it is using a total ordering structure on the ideal space for which every ideal has only finitely many "lesser" ideals.

Answer 1

"My question is whether one can choose a finite subset $X$ of $A$ so that $I(A)=I(X)$."

Yes. If $I(A) = (p)$, then $p$ is a linear combination of elements of $A$, say $p=r_1a_1+r_2a_2+\cdots +r_ka_k$ with $r_i\in R$ and $a_i\in A$. Now take $X = \{a_1,\ldots,a_k\}$.

Answer 2

PIDs are Noetherian, and Noetherian rings have the acc on ideals. So order some countable subset of A and consider the ideals generated by the first n.