Let $B/A$ be a extension of Dedekind rings, where $B$ is the integral closure of $A$ in a finite and separable algebraic extension $L$ of the fraction field $K$ of $A$. Let $\theta$ be a primitive element of $L/K$, and let $\mathscr{F}$ be the conductor of $A[\theta]$ in $B$. Let $\mathfrak{p}$ be a prime ideal of $A$ such that $\mathfrak{p}B+\mathscr{F}=B$.
Neukirch states that $\mathfrak{p}B\cap A[\gamma]=\mathfrak{p}A[\gamma]$. The inclusion $(\supseteq)$ is clear. For the other inclusion, Neukirch argues as follows:
"Since $(\mathfrak{p},\mathscr{F}\cap A)=1$, it follows that $\mathfrak{p}B\cap A[\gamma]=(\mathfrak{p}+\mathscr{F})(\mathfrak{p}B\cap A[\gamma])\subseteq \mathfrak{p}A[\gamma]$".
It may be something easy, but I can't understand the last inclusion. I'm having trouble understanding why $\mathscr{F}(\mathfrak{p}B\cap A[\gamma])\subseteq \mathfrak{p}A[\gamma]$.