Let $A$ an integrally closed domain and $B$ a commutative ring extension of $A$ that is finitely generated as an $A$-module.For $f\in B$ is it true that there exists $a_f\in A$ s.t. $\sqrt{(f)\cap A}=\sqrt{a_f}$?
My idea is to localize at $A$ so the rings become $B'/A'$ and $B'$ is a finite dimensional vector space over $A'$. For every $f\in B$ we associate the determinant of the linear map on B' induced by left multiplication by $f$. If I can prove that in fact $\det(f)\in A$, then it's not hard to see that $a_f=\det(f)$ does the job.
This seems to be true if $B$ is a free $A$-module, but I don't know how to prove it in general. (or even if it's true)
PS: I actually just need the case of $B$ being a $k$-algebra and $A=k[x_1,..,x_n$]