Why is a Galois group of a finite Galois extension of global fields K/F transitive on the set of primes of K lying above P, a prime of F? Let $K/F$ be a finite Galois extension of global extensions, and let $P$ be a prime of $F$. Letting $S$ be
$$
\{Q : Q \ \text{is a prime of} \ K \ \text{that lies above} \  P\}
$$
show that $G={\rm Gal}(K/F)$ is transitive on $S$.
 A: This is due to the Chinese remainder theorem in a suitable ring. Do you already know the argument in number fields (see an algebraic number theory book)? I ask because it is ultimately the same argument.
Let $G = {\rm Gal}(K/F)$ and $O$ be the valuation ring in $F$ at $P$, with integral closure $R$ in $K$. Assume two primes $Q$ and $Q’$ lying over $P$ in $R$ are not in the same $G$-orbit.
By the Chinese remainder theorem, there is an $\alpha \in R$ such that
$$
\alpha \equiv 1 \bmod g^{-1}(Q), \ \ \alpha \equiv 0 \bmod Q’
$$
where $g$ runs over $G$. (These congruences are compatible because $Q’$ is not in the $G$-orbit of $Q$.) Set $a= {\rm N}_{K/F}(\alpha)$, so $a\in O$. We have $a \equiv 1 \bmod P$ because $a$ is the product of all $g(\alpha)$ and each of them is $1\bmod Q$, so the norm is in $(1+Q)\cap O = 1+P$. However, since $\alpha \equiv 0 \bmod Q’$ and $a$ is a multiple of $\alpha$ in $R$, we also have $a\equiv 0 \bmod Q’$, so $a$ is in $Q’\cap O = P$. That contradicts $a$ being in $1+P$. So $Q’$ must be in the $G$-orbit of $Q$.
This result is not a special feature of global fields. For any field $F$ with a discrete valuation $v$ on it and a finite Galois extension $K/F$ with Galois group $G$, the action of $G$ on the valuations of $K$  extending $v$ is transitive: letting $O$ be the valuation ring of $v$ in $K$ and $P$ be the maximal ideal of $O$, the valuations on $K$ extending $v$ are in bijection with the prime ideals $Q$ lying over $P$ in the integral closure $R$ of $O$ in $K$. Now just run through the above proof with these rings $O$ and $R$.
Remark. I don’t think questions should be posed in the form “Show that…”, as it comes across on a Q&A site like a demand.
