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Let $L/K$ be an algebraic Galois extension and $w/v$ a non-archimedean extension of valuations.

Let $\mathcal{O}$ and $\mathfrak{P}$ denote the valuation ring and valuation ideal of $(L,w)$.

After defining the decomposition group $G_{w}$ and the inertia group $I_{w}$, Neukirch defines the ramification group $R_{w}$ of $w$ as \begin{equation} R_{w}=\{\sigma\in I_{w}\:|\:\frac{\sigma(x)}{x}\equiv 1\pmod{\mathfrak{P}}\:\:\forall x\in L^{\times}\}, \end{equation} and remarks that, since $w\circ\sigma=w$, we have $\sigma\mathcal{O}=\mathcal{O}$ (which I can see is true) and that $\sigma(x)/x\in\mathcal{O}$. This last part I don't understand.

Question:

Why $w\circ\sigma=w$ imples that $\sigma(x)/x\in\mathcal{O}$?

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Just figured that if $w\circ\sigma=w$ then the value of any conjugate $\sigma(x)$ is equal to the value of $x$ (Didn't realize this obvious fact). Then, since $v(1/x)=-v(x)$ we have that \begin{equation} v(\sigma(x)/x))=v(\sigma(x))-v(x)=v(x)-v(x)=0. \end{equation}

I'll leave this post here in case someone has the same question.

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