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.