Conductor of a ring

Question

An easy (possibly trivial) question from Neukirch's Algebraic Number Theory, p.47.

Let $A$ be a Dedekind domain, $K$ its fraction field, $L$ a finite separable extension of $K$ and $B$ the integral closure of $A$ in $L$. If $L=K(\theta)$ where $\theta\in B$ then the conductor $\mathfrak F$ of the ring $A[\theta]$ is defined to be the largest $B$-ideal which is contained in $A[\theta]$, so $\mathfrak F=\{x\in B\mid xB\subset A[\theta]\}$.

The book says that as $B$ is a finitely generated $A$-module, $\mathfrak F\neq 0$. I know that as $A$ is noetherian, $\mathfrak F$ is also f.g. over $A$, but I'm not sure why it must be nonzero.

Many thanks for your help.

Answer 1

Let $(b_1, \dots ,b_k)$ be a generating set of $B$ as an $A$ module.

As $L= K(\theta)$ and the extension is finite, $L=K[\theta]$, and recall $K$ is the quotient field of $A$.

So $b_i= \sum_{j=1}^{l_i} d_{ij} \theta^l $ where $d_{ij} \in K$ yet then $d_{ij}= \frac{a_{ij}}{c_{ij}}$ with ${a_{ij}}, {c_{ij}} \in A$. So with $c_i = \prod_{j=1}^{l_i} c_{ij}$ you have $c_ib_i$ in $A[\theta]$. Setting $c= \prod_{i= 1}^k c_i$ you have $cb_i \in A[\theta]$ for all $i$. And thus $c b \in A[\theta]$ for every $b$ in the $A$-module generated by the $b_i$ that is $B$. So $c$ is in your ideal.

Since $c$ is nonzero the claim is proved.

Answer 2

See Lemma $1$ and Lemma $2$ here, in Osserman's lecture notes.