The following question seems natural to ask in view of this question and its comments/answers:
Let $R \subseteq S$ be commutative Noetherian rings, let $q$ be a maximal ideal of $S$, $p$ a maximal of $R$ and $p=R \cap q$.
The localization of $R$ at $p$, $R_p$, is a local ring with unique maximal ideal $pR_p$ and the localization of $S$ at $q$, $S_q$, is a local ring with unique maximal ideal $qS_q$.
Since $p=R \cap q$, we can view $R_p$ as a subring of $S_q$.
Let $pS \subseteq I \subseteq q$ be an ideal of $S$ (not necessarily prime), and then $I_q \subseteq S_q$ is an ideal of $S_q$ (where $I_q:=IS_q$).
Assume that $(pR_p)S_q = I_q = qS_q$, where $(pR_p)S_q$ is the ideal of $S_q$ generated by $pR_p$ (this makes sense, since $pR_p \subset R_p \subseteq S_q$).
Question: Is it true that $I=q$?
If not, would it help to further assume one or more of the following conditions:
(i) $R \subseteq S$ is flat. (ii) $R \subseteq S$ is integral. (iii) $\dim(R)=\dim(S) < \infty$ ($\dim$ is Krull dimension). (iv) $R$ and $S$ are regular rings.
Remark: The name 'sandwich' comes from the assumption that $(pR_p)S_q = I_q = qS_q$; call it 'the sandwich equation'.
A non-counterexample: Without condition (iii), the following is not a counterexample: $R=k[x]$, $S=k[x,y]$, $p=(x)$, $q=(x,y)$, $I=(x(x-1),y)$. Indeed, the sandwich equation is not satisfied: only the right equality is satisfied, but not the left equality.
Any hints and comments are welcome! Thank you.