Field of fractions of etale stalk of Dedekind domain (Example from Milne's LEC)

Question

Let $X=\operatorname{Spec}(A)$ be an affine Dedekind domain 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 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.

Answer 1

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 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.