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By this I mean the following

Suppose we have a ring extension $A\rightarrow B$, $B$ integral over $A$, and we have an arbitrary ideal $I\subset A$. For which rings $A$ can we say that for any $B$ and any $I$, $IB \cap A=I$?

I know this holds for PIDs, since if $x \in (a)B\cap A$ where $(a)$ is the ideal of $A$ generated by some $a\in A$, then $x=ab$ for some $b \in B$, and since a PID is integrally closed in its fraction field, we can conclude $b \in A$ and $x \in(a)$.

On the other hand, it's easy to see that for some domains that are not integrally closed like $\mathbb{Z}[\sqrt{-7}]$, this does not hold. If we take our integral extension to be $\mathbb{Z}[\frac{1+\sqrt{-7}}{2}]$, and our ideal to be $(2)$, we clearly see that $1+\sqrt{-7}$ lies in $(2)\mathbb{Z}[\frac{1+\sqrt{-7}}{2}]\cap\mathbb{Z}[\sqrt{-7}]$ but not in $(2)$ as an ideal of $\mathbb{Z}[\sqrt{-7}]$.

Is there any criterion more inclusive than being a PID which allows this condition to hold for $A$? I'm particularly interested in the question of whether, if it holds for $A$, it can be taken to hold for $A[x]$.

user26857
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Without a Compass
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    If $A$ is Dedekind, I think it works when $A \rightarrow B$ is injective. Indeed, $B/A$ is torsion-free (because no element of $Frac(A)$ is integral over $A$), hence flat, thus $0 \rightarrow A \rightarrow B \rightarrow B/A \rightarrow 0$ remains exact when tensored with any $A$-module and in particular $A/I \rightarrow B/IB$ is always injective. – Aphelli Oct 10 '20 at 20:55
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    If $A$ is a normal Noetherian ring such that its injective integral extensions are always flat, then they’re (the finite ones at least) always locally free and thus $B/A$ is always flat and the same thing should happen. I wonder if it works for regular rings... – Aphelli Oct 10 '20 at 21:09
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    It's true when $A \to B$ is faithfully flat: see here. This post is related and David Speyer's comment seems especially relevant. – Viktor Vaughn Oct 11 '20 at 01:24

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