TITLE: When does $\mathfrak{a}B\cap A = \mathfrak{a}$?
QUESTION [6 upvotes]: Let $A\subset B$ be rings, and let $\mathfrak{a}$ be an ideal of $A$. Under what circumstances does $\mathfrak{a}B\cap A = \mathfrak{a}$?  More precisely, are there conditions on $A,B$ that guarantee this for all ideals $\mathfrak{a}\triangleleft A$? If so, what are the most general such conditions?
(I had a hunch that if $A$ is integrally closed and $B$ is integral over $A$, this would be guaranteed, but have thus far been unable either to prove it or form a counterexample.)

REPLY [3 votes]: Your hunch is in the right direction, but further conditions are required. Below is an excerpt from the introduction of  R. Gilmer's Contracted ideals with respect to integral extensions.
Theorem $\ $Let $D$ be an integrally closed domain with quotient field $K$; let $L$ be an extension field of $K$ such hat $L/K$ is separable algebraic, and let $D’$ be the integral closure of $D$ in $L.$ Then each ideal of $D$ is the contraction of an ideal of $D’$ -- that is, if $A$ is an ideal of $D,$ there is an ideal $B$ of $D’$ such that $A = B \cap D.$
We later show (Example 2) that this conjecture is false in general. But we
show that in several important cases, each ideal of $ D$ is the contraction of an
ideal of $D’$. Notably, this is true in case $D$ is a Prüfer domain (Corollary 2) or
if $D’$ has an integral basis over $D$ (Theorem 6).