Why are the extensions of valuations in a Galois extension conjugate under the action of the Galois group?

Question

I'm reading the following paper by Hansen about Galois Coverings of Curves: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.31.9776&rep=rep1&type=pdf (see page 37) Unfortunately, I'm having problems with the following argument:

Let $K/L$ be a finite Galois extension and let $\nu$ be a valuation of $L$. $G=\operatorname{Gal}(K/L)$ acts on the valuations $\nu^\prime$ of K by $\sigma(\nu^\prime)=\nu^\prime\circ \sigma$ for all $\sigma\in G$. We see that $\sigma(\nu^\prime)|_L=\nu^\prime\circ\sigma_L=\nu^\prime|_L=\nu$ for a valuation $v^\prime$ of $K$ with $\nu^\prime|_L=\nu$

Hansen argues that, since $G$ consists of all automorphisms of $K$ fixing $L$, the valuations $\{\nu^\prime\text{ of }K: \nu^\prime|_L=\nu\}$ are conjugate under $G$'s action. I don't understand why this follows from the above. I would be very thankful if you could give me a hint.

Answer 1

For $K/L$ a (finite) Galois extension of any kind of fields, $w$ a valuation on $L$ and $v_j$ (the equivalence classes) of valuations on $K$ extending $w$ . There are finitely many such valuations.

$G=Gal(K/L)$ permutes the $v_j$ as $v_j \circ \sigma$ is another valuation above $w$.

The main claim is that there is $\pi_1\in K$ such that $v_1(\pi_1)>0,v_j(\pi_1)=0$.

Since $K/L$ is Galois then $N_{K/L}(\pi_1)=\prod_{\sigma\in G} \sigma(\pi_1)\in L$.

$$w(N_{K/L}(\pi_1))=v_1(N_{K/L}(\pi_1))=\sum_{\sigma\in G}v_1 \circ \sigma(\pi_1)\ge v_1(\pi_1) > 0$$

Whence for all $j$, $v_j(N_{K/L}(\pi_1))> 0$ so that $v_j \circ \sigma(\pi_1)>0$ for some $\sigma$ which implies that $v_1 = v_j\circ \sigma$.

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If $w$ is a discrete valuation then once we know that there are finitely $v_j$ the main claim follows easily. The $v_j$ are discrete valuations because $[L:K]$ is finite. Look at $S_1 = \{ a\in L, v_1(a)=0,\forall j, v_j(a)\ge 0\}$. For each $j$, if it exists take some $a_{j,1}\in S_1,v_j(a_{j,1})>0$ and let $A_1 = \prod_j a_{j,1}$. If some $a_{i,1}$ doesn't exist then take some element $b\in L,v_1(b)>v_i(b)$. Letting $c=bA_1^n$, for $n$ large enough then the minimum valuation of $c$ is at some $v_m$ such that $a_{m,1}$ doesn't exist. Taking $k\in K,w(k)\ne 0$ then for some $r> 0,s$, $r v_m(c) =s v_m(k)$, $d=c^rk^{-s} \in S_m$ and $a_{m,1} = 1-d$ exists. Thus in fact all the $a_{m,1}$ exist and $\pi_1=1-A_1$ works in the main claim.