An *overring* of an integral domain is a domain lying between it and its quotient field. Is it possible to have an overring $S$ of an integral domain $R$ such that $S$ has two maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$ with $R \cap \mathfrak{m} \subsetneq R \cap \mathfrak{n}$? I think this should be possible but I can't come up with an example. Such an overring $S$ of $R$ cannot be flat or integral over $R$.


I feel like it should be straightforward to come up with an example or prove that there can't be such an example, but thus far all of my attempts have failed. I posted the question first on Math StackExchange but didn't get any replies.



An overring of an integral domain having maximal ideals contracting to strictly comparable primes
Let $k$ be a field and start with $S\_0:=k[x,y]$. (Ultimately, $S$ will be a localization of $S\_0$). Let $\mathfrak{m}\_0=(y)$ and $\mathfrak{n}\_0=(x,y-1)$. In the affine plane $P:=\mathrm{Spec}(S\_0)$, these are respectively the generic point of the $x$-axis and the closed point $(0,1)$. 


Put $z=xy\in S\_0$ and $R=k[x,z]\subset S\_0$: the inclusion corresponds to the map $\phi:(a,b)\mapsto(a,ab)$ from $P$ to $P':=\mathrm{Spec}(R)$ (also isomorphic to $\mathbb{A}^2\_k$, with coordinates $x,z$). Clearly, $\phi$ contracts the whole $y$-axis to the point $(0,0)$, and sends the $x$-axis to the $x$-axis. In other words, $\mathfrak{n}\_0\cap R=(x,z)$ and $\mathfrak{m}\_0\cap R=(z)$, so we have $\mathfrak{m}\_0\cap R\subsetneq\mathfrak{n}\_0\cap R$. 


Now of course $\mathfrak{m}\_0$ is not maximal. But let $S:=(S\_0)\_{\mathfrak{m}\_0}\cap(S\_0)\_{\mathfrak{n}\_0}\subset k(x,y)=k(x,z)$: you can check that this is a semilocal ring having $\mathfrak{m}:=\mathfrak{m}\_0S$ and $\mathfrak{n}:=\mathfrak{n}\_0S$ as distinct maximal ideals, sitting respectively above $\mathfrak{m}\_0$ and $\mathfrak{n}\_0$ (in $S\_0$) and above $(z)$ and $(x,z)$ (in $R$). 


