I'm studying number theory at the moment and came across a step in a proof I cannot resolve. The setting is the following:
Let $A \subseteq B$ a finite ring extension of Dedekind domains, i.e. one-dimensional normal noetherian integral domains. Let $(0) \ne \mathfrak{p} \subset A$ be a prime ideal. Consider the localisations $A_\mathfrak{p}$ and $B_\mathfrak{p} := (A \setminus \mathfrak{p})^{-1} B$. Then $A_\mathfrak{p} / \mathfrak{p} A_\mathfrak{p} \cong A / \mathfrak{p}$ and $B_\mathfrak{p} / \mathfrak{p} B_\mathfrak{p} \cong B / \mathfrak{p} B$.
I managed to show the part about $A$: Localisations and quotients commute, namely $A_\mathfrak{p} / \mathfrak{p} A_\mathfrak{p} \cong \mathrm{Quot}(A / \mathfrak{p}) = A / \mathfrak{p}$. The equality holds since $\dim(A) = 1$, so $\mathfrak{p}$ is a maximal ideal and $A / \mathfrak{p}$ is already a field.
Applying the same reasoning to $B$ doesn't seem to be fruitful: We have $B_\mathfrak{p} / \mathfrak{p} B_\mathfrak{p} \cong ((A \setminus \mathfrak{p}) / \mathfrak{p} B)^{-1} (B / \mathfrak{p} B)$. For this to be $B / \mathfrak{p} B$, I'd have to show that $(A \setminus \mathfrak{p}) / \mathfrak{p} B$ are all units.
I think I know the answer. It is important to now what type of isomorphism this is. The above might work for domains, but the conjectured isomorphism is an isomorphism of $A$-modules. Actually, we have $S^{-1} (M/N) = S^{-1}M/S^{-1}N$ for all $A$-modules $M$ and $N$ as well as $S \subseteq A$ any multiplicative subset.