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Let $L/K$ be a finite Galois extension, $\phi$ a valuation on $K$ and $\psi$ a valuation on $L$ extending $\phi$. I'm trying to parse a very short proof of the theorem that all extensions of $\phi$ to $L$ are conjugate under the Galois group of $L/K$. The proof starts as follows:

Let $\psi_1$ and $\psi_2$ be extensions of $\phi$ lying in different orbits of $G=\textrm{Gal}(L/K)$. Then the $G\psi_i$ are disjoint and by the approximation theorem there exists an $x\in L$ s.t. $\psi(x)<1$ for all $\psi\in G\psi_1$ and $\psi(x)>1$ for all $\psi\in G\psi_2$.

I can't see why the approximation theorem would imply this, because in this case both orbits can have more than one element. The way I've been taught the approximation theorem is that we can pick one valuation $\psi$ for which $\psi(x)<1$ and for all the others the inequality would go the other way.

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Weak approximation is a bit stronger than that, no? – Dylan Moreland Aug 12 '11 at 16:19
Yes, that's enough. This version of weak approximation is stronger than the one I've seen which is just a lemma in that proof. Thanks. – dstt Aug 12 '11 at 16:30
up vote 2 down vote accepted

The approximation theorem states that for any finite set of valuations $S$ on a global field $L$ and any sequence $(a_v)_{\mathbb{v\in S}} \in \prod_{v\in S} L_v,$ there exists for any $\epsilon > 0$ an element $x\in L$ such that $|x - a_v|_v \leq \epsilon$ for all $v\in S.$ Equivalently, it states that the diagonal embedding of $L$ in $\prod_{v\in S} L_v$ is dense. The fact that there is an $x \in L$ s.t. $\psi(x)<1$ for all $\psi \in G\psi_1$ and $\psi(x)>1$ for all $\psi \in G\psi_2$ therefore follows immediately. To see this, apply the approximation theorem to the set of valuations $S = G\{\psi_1,\psi_2\},$ and any element $(a_\psi)_{\mathbb{\psi\in S}} \in \prod_{\psi\in S} L_\psi$ such that $\psi(a_\psi)<1$ for all $\psi \in G\psi_1$ and $\psi(a_\psi)>1$ for all $\psi \in G\psi_2.$

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