Exercise 20, § 2

Let $\mathbf{A}$ be a Dedekind domain, $\mathbf{K}$ its field of fractions, $\mathbf{L}$ a finite algebraic extension of $\mathbf{K}$ and $\mathbf{B}$ the integral closure of $\mathbf{A}$ in $\mathbf{L}$, which is a Dedekind domain. Let $\mathfrak{f}$ be an ideal of $\mathbf{B}$. For there to exist a ring $\mathbf{C}$ such that $\mathbf{A} \subset \mathbf{C} \subset \mathbf{B}$ and such that $\mathfrak{f}$ is the conductor of $\mathbf{B}$ in $\mathbf{A}$ (Chapter V, § 1, no. 5 ), it is necessary and sufficient that, for every prime ideal $\mathfrak{p}'$ of $\mathbf{B}$ containing $\mathfrak{f}$ such that the field $\mathbf{B}/\mathfrak{p}'$ is isomorphic to $\mathbf{A}/(\mathfrak{p}' \cap \mathbf{A})$ (prime ideals of residue degree 1), the intersections off and the transporter $\mathfrak{f}: \mathfrak{p}'$ of $\mathfrak{p}'$ in $\mathfrak{f}$ with $\mathbf{A}$ be equal. (Note that, if there exists such a ring $\mathbf{C}$, the conductor of $\mathbf{B}$ in the ring $\mathbf{C}_0 = \mathbf{A} + \mathfrak{f}$ is also equal to $\mathfrak{f}$; the existence of $\mathbf{C}$ is therefore equivalent to the fact that $\mathbf{A} + \mathfrak{f}$ contains no ideal of $\mathbf{B}$ distinct from $\mathfrak{f}$ and containing $\mathfrak{f}$. To prove that the latter condition is equivalent to that of the statement, reduce it to the case where $\mathbf{A}$ is a discrete valuation ring.)