Let $E/\mathbb{Q}$ be a finite-dimensional Galois Extension and $B\subset E$ the integral closure of $\mathbb{Z}$ in $E$; i.e. the ring of algebraic integers in $E$. Then $B$ is a Dedekind domain; take $P$ any nonzero prime ideal such that $P\cap \mathbb{Z}=(p)$ for a prime number $p$. Then $F:=B/P$ is an algebraic extension of $\mathbb{F}_p$. If $\xi$ is a primitive generator for this extension so that $F=\mathbb{F}_p(\xi)$, one must prove there exists $x\in B$ such that $x\equiv \xi$ (mod $P$) and such that for every $\sigma \in Gal(E/\mathbb{Q})$ satisfying $\sigma(P)\neq P$, we have $x\in \sigma(P)$.
Furthermore, we wish to prove the following map is surjective: \begin{equation} \{ \sigma \in Gal(E/\mathbb{Q}) \mid \sigma(P)=P \} \to Gal(F/\mathbb{F}_p). \end{equation} (It might help to know the right-hand side is a cyclic group? EDIT: It would suffice to find $\sigma\in Gal(E\mathbb{Q})$ such that $\sigma(x)=x^p$ mod $P$, for $x\in B$ as above).
My approach is as follows... We can decompose the ideal $pB$ as prime ideals $\prod_{i=1}^m P_i$ where WLOG $P_1=P$. Furthermore the primes appearing in this decomposition should be precisely those of the form $\sigma(P)$. For the surjectivity part I'm very lost.
Any help would be greatly appreciated!
EDIT: Thanks to Jyrki Lahtonen for the following remark: any two distinct ideals of the form $\tau_1(P)$ and $\tau_2(P)$ must be coprime since they are distinct maximal ideals of $B$. Thus we can use the Chinese Remainder Theorem to find our desired $x$ which is $\xi$ mod $P$ and $0$ modulo all the other ideals of that form.