Let $A$ be a Dedekind domain, $K, L, B$ the usual designations of $A$'s quotient field, a finite separable extension of $K$, and integral closure of $A$ in $L$ respectively. If $\alpha \in B$ generates $L/K$ with minimal polynomial $\mu \in A[X]$, the following are equivalent for a prime ideal $P$:
(i) $B_P = A_P[\alpha]$
(ii) $\mathfrak D_P = \mu'(\alpha)B_P$, where $\mathfrak D$ is the different of $B/A$.
(iii) $\nu_P(\mathfrak d(A[\alpha]) \mathfrak d(B)^{-1}) = 0$, where $\mathfrak d(-)$ is the ideal discriminant.
So it's obvious that for any integral generator $\alpha$, you have $B_P = A_P[\alpha]$ for all but finitely many prime ideals $P$. This is because $A[\alpha] \subseteq B$ implies $\mathfrak d(A[\alpha]) \subseteq \mathfrak d(B)$, making $I = \mathfrak d(A[\alpha]) \mathfrak d(B)^{-1}$ an integral ideal, and hence $\nu_P(I) = 0$ for all but possibly finitely many $P$.
My question is about what happens if you vary the integral generators. If you fix a prime ideal $P$, when does it happen that $A_P[\alpha] \subsetneq B_P$ for EVERY integral generator $\alpha$ of $L/K$? In other words, when does $P$ divide $\mathfrak d(A[\alpha]) \mathfrak d (B)^{-1}$ for each integral generator $\alpha$?
A partial answer in terms of (ii): The different $\mathfrak D$ is the greatest common divisor of the ideals $\mu'(\alpha)B$, as $\alpha$ runs through the integral generators of $L/K$. So for any prime ideal $\mathfrak P$ lying over $P$, $\nu_{\mathfrak P}(\mathfrak D_P)$ is the minimum of all the numbers $\nu_{\mathfrak P}(\mu'(\alpha))$.
To say that $A_P[\alpha] \subsetneq B_P$ for all $\alpha$ is to say that $\mu'(\alpha)B_P \subsetneq \mathfrak D_P$ for all $\alpha$, which is to say that for all $\alpha$, there exists a prime ideal $\mathfrak P$ lying over $P$ such that $\nu_{\mathfrak P}(\mu'(\alpha)) > \nu_{\mathfrak P}(\mathfrak D)$.