Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A$.

share|cite|improve this question
And Noetherian too... Nice example, thanks! – Matthé van der Lee Nov 20 '13 at 23:40

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.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.