Let $K$ be a complete discrete valuation field, let $L$ be an algebraic closure of $K$. Let $a\in L\setminus K$ be any element. Why the Galois conjugates of $a$ all have the same valuation?
The valuation ring $\mathcal{O}_L$ is the integral closure of $\mathcal{O}_K$ in $L$, so since Galois automorphism maps integral elements to integral elements, it preserves $\mathcal{O}_L$. But I don't know how to extend this argument further. Any hint or help is much appreciated.
It has to do with the Extension Theorem, according to which (using the additive valuation $v$), we must have, for $z\in L$, $v_L(z)=\frac1{[L:K]}v_K\bigl(\mathcal N^L_K(z)\bigr)$, as long as you want the restriction of $v_L$ to $K$ to be equal to $v_K$. Here, $\mathcal N$ is the Galois-theoretic norm. Since $z$ and any of its conjugates have the same norm down to $K$, their $L$-valuations must be the same.
The above description of $v_L$ doesn’t depend on normality or separability of the extension.