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 $A[\theta]$.
Best Answer
It appears that the claim in the exercise is false. I think this is a counterexample:
Let $A=\mathbb Z_{(2)}$, i.e. $\mathbb Z$ localized at the prime $(2)$. Let $K=\mathbb Q$ and let $L=\mathbb Q(\zeta)$, where $\zeta^3-\zeta^2-2\zeta-8=0$. Let $B$ be the integral closure of $A$ in $L$, so $B=S^{-1}\mathcal O_L$, where $S=\mathbb Z \smallsetminus 2\mathbb Z$. Let $J=2B$.
$A$ is a DVR with $2A$ its only prime, which factors as $2B=P_1^{e_1}\cdots P_r^{e_r}$, and every prime of $B$ is one of the $P_i$. Thus, the conductor $\mathfrak f$ is relatively prime to $J=2B$ iff $\mathfrak f=B.$ So we must find a $\theta \in B$ such that $B=A[\theta]$. Without loss of generality, we can assume that $\theta \in \mathcal O_L$.
But this is impossible. Indeed, it can be shown that $2$ divides $[\mathcal O_L:\mathbb Z[\theta]]$ for every $\theta$ in $\mathcal O_L$ (as noted in Keith Conrad's notes "Factoring after Dedekind", pg. 5-6). Choose $b \in \mathcal O_L$ such that $b\mod \mathbb Z[\theta]$ is an element of even order in $\mathcal O_L/\mathbb Z[\theta]$. Then for any odd integer $k$, $kb\notin \mathbb Z[\theta]$, so $b \notin A[\theta]$. Thus $B \not= A[\theta]$.
If anyone finds a flaw in this counterexample, or thinks the claim in the exercise should be correct, please let me know! Thanks.