Let $A$ be a Dedekind domain, $K$ its fraction field, $L|K$ a finite separable field extension of degree $n$ and $B$ the integral closure of $A$ in $L$. Let $q\subset B$ a nonzero prime ideal such that $\kappa(q)= B/q$ is separable over $\kappa(p)=A/p$ where $p = q \cap A$. Show that there is an element $b\in B$ such that $L = K(b)$ and the conductor $\mathfrak{f}$ of $A[b]$ is coprime with $q$.
Note: here $\mathfrak{f}:=\{a\in B\mid aB\subset A[b]\}$
The only facts I know that seem useful here are that $B$ is a Dedekind domain and can be generated by $b_1,...,b_n\in B$ as a free $A$-module. I imagine one of these $b_1,...,b_n$ (or a combination of them) should be a good candidate for $b$. I don't know how can I use the fact that $\kappa(q)|\kappa(p)$ is separable and have no idea how to show that $\mathfrak{f}+q=1$.
Any tips? Thanks!