Let $X=\operatorname{Spec}(A)$ be an affine [Dedekind domain](https://en.m.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions) with field of fractions $K$. Let $\widetilde{A}$ be the integral closure of $A$ in separable closure $ K^{\text{sep}}$. A closed point $x$ of $X$ is a nonzero prime ideal $\mathfrak{p}$ of $A$, and the choice of a prime ideal $\widetilde{p}$ of $\widetilde{A}$ lying over $\mathfrak{p}$ determines a geometric point $\overline{x}= \operatorname{Spec}k(x)^{\text{sep}} \to X$ a geometric point of $X$. The etale stalk with respect to $\overline{x}$ is given by


$$ \mathcal{O}\_{X,\overline{x}}^{\text{et}}= \varinjlim\_{(A\_U,u) \text{ etale nbhds of } A \text{ over }\overline{x}} (A\_U,u)$$


**Question:** How to see that the field of fractions of the ring $\mathcal{O}\_{X,\overline{x}}^{\text{et}}$ equals to the fixed field $(K^{\text{sep}})^{I(\widetilde{\mathfrak{p}})}$ where the inertia group $I(\widetilde{\mathfrak{p}})$ of $\widetilde{\mathfrak{p}}$ is defined as


$$I(\widetilde{\mathfrak{p}})=:\{ g \in G \ | \ g(\widetilde{\mathfrak{p}})=\widetilde{\mathfrak{p}} \text{ and } g | \_{\kappa(\widetilde{\mathfrak{p}})} = \text{id}\_{\kappa(\widetilde{\mathfrak{p}})} \} \subset \text{Gal}(K^{\text{sep}}/K)$$


Source: Milne's LEC script on Etale Cohomology (Example 12.4, page 82 last paragraph)


Note that previous considerations in the linked script show that the field of fractions of the ring $\mathcal{O}\_{X,\overline{x}}^{\text{et}}$ is given by inductive limit of those finite separable field extensions $L/K$ ( contained in $ K^{\text{sep}}$) such that the normalization $X\_L$ of $ X$ in $L$ is unramified at some point lying over $x$, but I not see why this inductive limit equals in this case to $(K^{\text{sep}})^{I(\widetilde{\mathfrak{p}})}$ as claimed. Especially, how the requirement that normalization $X\_L$ of $ X$ in $L$ is unramified at some point lying over $x$ is related to the property of $L$ being fixed by the inertia group $I(\widetilde{\mathfrak{p}})$.


Thematically the problem is closely related to [this](https://mathoverflow.net/questions/452621/calculate-stalk-of-etale-derived-pushforward-sheaf-milnes-lec) question I asked before.


Ideas: Here a way to reduce the problem at least to finite extension case. There exist a finite extension $M/K$, such that the associated map $\overline{x}= \operatorname{Spec}k(x)^{\text{sep}} \to X$ to the considered geometric point factors scheme theoretically through $X\_M$, the Spec of integral closure of $A$ in $M$. So we can assume that all field extensions discussed about about to be finite and we can replace avove everywhere $ K^{\text{sep}}$ by finite extension $M$, letting the inertia group living inside a finite Galois group.


But not see how to proceed from here.



Field of fractions of etale stalk of Dedekind domain (Example from Milne's LEC)
As Milne (sort of) explains, the following two data are exactly the same:


1. A connected étale neighbourhood $U \to X$ with a lift $\bar x \to U$ of $\bar x \to X$, up to shrinking Zariski-locally around the image of $\bar x$ in $U$ (I suppose you could call these 'Zariski germs of connected étale neighbourhoods');
2. A finite separable field extension $K \to L$ together with a prime $\mathfrak q$ above $\mathfrak p$ in the integral closure $B$ of $A$ in $L$, together with a map $\kappa(\mathfrak q) \to \kappa(\bar x)$, such that the map $\operatorname{Spec} B \to \operatorname{Spec} A$ is étale at $\mathfrak q$.


Indeed, to pass from (1) to (2), we saw in [my answer](https://mathoverflow.net/q/452621/82179) to your previous question that the generic fibre of $U \to X$ is a field if $U$ is connected, and $\operatorname{Spec} B \to \operatorname{Spec} A = X$ is étale at $\mathfrak q$ by assumption. Conversely, to go from (2) to (1), we may for instance take $U$ to be the (open) locus of $\operatorname{Spec} B$ where $\operatorname{Spec} B \to \operatorname{Spec} A$ is étale, which is nonempty since it contains the generic point (as $K \to L$ is separable).


The choice of $\mathfrak q$ above $\mathfrak p$ together with the map $\kappa(\mathfrak q) \to \kappa(\bar x)$ can be taken care of by choosing a prime ideal $\widetilde{\mathfrak p}$ in $\widetilde A$ together with an isomorphism $\kappa\big(\widetilde{\mathfrak p}\big) \stackrel\sim\to \kappa(\bar x)$. So the only thing to think about is when $\operatorname{Spec} B \to \operatorname{Spec} A$ is étale at $\mathfrak q := \widetilde{\mathfrak p} \cap B$. Standard Galois ramification theory tells us that $(\bar K)^{I(\widetilde{\mathfrak p})}$ is the unique largest subfield of $\bar K$ such that for any finite subextension $K \to L \to (\bar K)^{I(\widetilde{\mathfrak p})}$, the map $\operatorname{Spec} B \to \operatorname{Spec} A$ is étale at $\mathfrak q := \widetilde{\mathfrak p} \cap B$ (where as usual $B$ is the integral closure of $A$ in $L$). A sample reference is Neukirch's *Algebraic Number Theory*, Ch. I, Prop. 9.6.


