Embedding noetherian domains in a PID with finite index

Question

The starting point of this post is the following question:

Embedding number fields in fields with class number 1

It is shown that in the answers that , given an number field $K$, we cannot necessarily find a finite extension $L/K$ such that $O_L$ is a PID.

So my questions are:

1. can we find a PID $A$ containing $O_K$ as a subring such that $[A: O_K]$ is finite ( the condition is here to avoid obvious answers such that $A=K$ !)

2. More generally, if $B$ is a Noetherian domain, can we find a PID $A$ containing $B$ as a subring such that $[A:B]$ is finite ? I am happy to assume that $A$ is a free abelian group of finite rank if necessary, but I would prefer not to if possible.

Apologies if this is well-known...

Side remark: I think I have encountered years ago a similar question on MO or MS, but I cannot find the link...

Answer 1

The answer to the first question (and therefore also to the second question) is negative, for an elementary reason: If the ring of integers $O_K$ of a number field $K$ has finite index (as an additive group) in an integral domain $A$, then by tensoring with the rationals, the field of fractions of $A$ is a field containing $K$ of the same degree as $K$ - i.e., it is $K$ itself. But any subring of $K$ which properly contains $O_K$ will contain elements which aren't algebraic integers, and thus $O_K$ can't have finite index in it.

(On the other hand, if you drop the finite index condition you can always find finitely generated subrings of $K$ which are PIDs, by inverting sufficiently many elements to kill the finite class group of $O_K$: Pick representative ideals for a set of generators of the class group and adjoin the inverses of their norms.)