I have a problem in understanding the *inertia group* of an infinite extension. I am studying it in this context.

Let $K$ be a field, $v$ a discrete valuation on $K$, and $\mathcal{O}_v$ the discrete valuation ring of $v$. Let $k$ be the residue field of $v$, we assume $k$ to be perfect. Let $K_s$ be a separable closure of $K$ and $\bar{v}$ an extension of $v$ to $K_s$. The discrete valuation ring of $\bar{v}$ will be $\mathcal{O}_\bar{v}$ and the residue field $\bar{k}$.

Then $K_s/K$ is a galois extension with galois group $G=Gal(K_s/K)$

Now I see two ways to define the **inertia group**.

- The subgroup $I \subset G$ given by the elements $\sigma\in G$ such that $$\bar{v}(\sigma(a)-a)\geq0$$ for every $a\in \mathcal{O}_\bar{v}$.
- The inverse limit of the inertia groups $$I=\varprojlim I_{F/K}(v_F)$$ where the inverse limit is taken over the finite galois extensions $F/K$, and $v_F$ is the restriction of $\bar{v}$ to $F$.

I am convinced that these two definitions are equivalent.

Now comes my actual question: is it true that the subfield of $(K_s)^I\subset K_s$ given by the elements fixed by the action of $I$ is the maximal unramified extension of $K$ contained in $K_s$? In case, how is it proved?