TITLE: Find primitive element such that conductor is relatively prime to an ideal (exercise from Neukirch)
QUESTION [7 upvotes]: This is an exercise from Neukirch, "Algebraic Number Theory", Ch I, Sec 8, Exercise 2, pg 52.  It really has me stumped.
Suppose $A$ is a Dedekind domain, $K$ its field of fractions, $L$ a finite, separable extension of $K$, and $B$ the integral closure of $A$ in $L$ (so $B$ is also Dedekind).
Now suppose $J$ is any integral ideal of $B$.  The exercise asks to show that there exists a $\theta \in B$ such that $\theta$ is a primitive element (i.e. $L=K(\theta)$) and such that $J$ is relatively prime to the conductor $\mathfrak f = \{\alpha \in B|\alpha B \subseteq A[\theta] \}$ of $A[\theta]$.

REPLY [3 votes]: The claim is indeed false, as John's counterexample shows.
It is, however, true if $J$ is assumed to be a prime ideal of $B$ - at least in the number field case. A proof can be found in Hecke's "Vorlesungen über die Theorie der Algebraischen Zahlen" (1923), Hilfssatz c on page 135 in the second edition (Chelsea Publishing Co, 1970, available from the AMS).
The proof uses the finiteness of the residue class field, so it does not generalize directly to the case of general Dedekind domains.
Oh, just found it in Serre's "Local Fields", Ch. III, exercise 2 on p. 59. It holds in the general Dedekind domain case if the residue extension $A/J\cap A \subset B/J$ is separable.