Proposition 5.15 of Atiyah-Macdonald (Commutative Algebra) says that if $A\subset B$ are integral domains, $A$ integrally closed, $I$ an ideal of $A$, and $x\in B$ is integral over $I$, then the coefficients of the minimal polynomial $f$ of $x$ over the field of fractions of $A$ are in $\sqrt{I}$.
Is there any kind of converse to this result? That is, is there a condition on $f$ that holds if and only if $x$ is integral over $I$?
By 5.14 in A-M, if $A \subseteq B$, $I$ is an ideal and $x \in B$, then $x$ is integral over $I$ iff $x$ is integral over $\sqrt{I}$.
Indeed, letting $C$ be the integral closure of $A$ in $B$, 5.14 tells you that the integral closure of $I$ in $B$ is $\sqrt{IC}$, and the integral closure of $\sqrt{I}$ in $B$ is $\sqrt{(\sqrt{I}C)}$. It is trivial to check that these are the same.
Therefore the converse of 5.15 is also true.