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

  • 1
    $\begingroup$ By definition $\mathscr{F}B\subseteq A[\gamma]$. Thus ${\frak p}\mathscr{F}B\subseteq {\frak p}A[\gamma]$ $\endgroup$
    – leoli1
    Aug 27 '21 at 7:23

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