Proposition 8.3, Neukirch- Algebraic Number Theory

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 $$\mathscr{F}(\mathfrak{p}B\cap A[\gamma])\subseteq \mathfrak{p}A[\gamma]$$.

• By definition $\mathscr{F}B\subseteq A[\gamma]$. Thus ${\frak p}\mathscr{F}B\subseteq {\frak p}A[\gamma]$ Aug 27 '21 at 7:23