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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 $S$.

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    $\begingroup$ $\operatorname{Spec}\mathscr{O}_K=(\operatorname{Spec}\mathscr{O}_F)/G $. $\endgroup$
    – abx
    Jan 25 at 14:34
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    $\begingroup$ $K \otimes_F F_P$ is the product of the completions of $K$ at the places of $S$. So you need to show that the Galois group of $K/F$ acts transitively on the spectrum of $K \otimes_F F_P$. That’s classical (the norm of any non-trivial idempotent is an idempotent in $F_P$ which isn’t invertible, so it must be zero). $\endgroup$
    – Aphelli
    Jan 25 at 15:05
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    $\begingroup$ Is this a homework forum now too? $\endgroup$
    – user498412
    Jan 26 at 1:15

1 Answer 1

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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.

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