# 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 $$S$$.

• $\operatorname{Spec}\mathscr{O}_K=(\operatorname{Spec}\mathscr{O}_F)/G$.
– abx
Jan 25 at 14:34
• $K \otimes_F F_P$ 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). Jan 25 at 15:05
• Is this a homework forum now too?
– user498412
Jan 26 at 1:15

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 $$R$$.