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Let $A\subseteq B$ be integral domains, where $B$ is contained in the quotient field $K$ of $A$, and $B$ is finitely generated as an $A$-module. Let $\mathfrak{f}=\{a\in A\mid aB\subseteq A\}$ denote the conductor. (As is well known, it is an ideal in both $A$ and $B$.)

If $\mathfrak{p}$ is a prime ideal of $A$ such that $\mathfrak{f}\not\subseteq\mathfrak{p}$, can its extension $\mathfrak{p}B$ fail to be prime in $B$?

Note: if $\mathfrak{p}$ is a maximal ideal with $\mathfrak{f}\not\subseteq\mathfrak{p}$, then $\mathfrak{p}B$ is a maximal ideal of $B$. Indeed, there is a one-to-one correspondence between the ideals $I$ of $A$ such that $I+\mathfrak{f}=A$ and the ideals $J$ of $B$ with $J+\mathfrak{f}=B$, given by $I\mapsto IB$ and $J\mapsto J\cap A.$ This "extension/contraction" correspondence preserves products, sums and intersections of ideals, and one has $A/I=B/J$ for corresponding ideals. It holds for every extension $A\subseteq B$ of commutative rings.

It is not difficult to find counterexamples when $B$ is not assumed to be finitely generated over $A$.

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And Noetherian too... Nice example, thanks! –  Matthé van der Lee Nov 20 '13 at 23:40
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In these notes you can find a counterexample where $B$ is not only a finite extension of $A$, but it is even the integral closure of $A$. (The example was considered on M.SE in this topic and comes from Matsumura, Commutative Algebra, where it is given as a counterexample to the Going-Down Theorem in the absence of integrally closed condition.)

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