Lemma 58.13.6. Let $A$ be a discrete valuation ring with fraction field $K$. Let $K^{sep}$ be a separable closure of $K$. Let $A^{sep}$ be the integral closure of $A$ in $K^{sep}$. Let $\mathfrak m^{sep}$ be a maximal ideal of $A^{sep}$. Let $\mathfrak m = \mathfrak m^{sep} \cap A$, let $\kappa = A/\mathfrak m$, and let $\overline{\kappa } = A^{sep}/\mathfrak m^{sep}$. Then $\overline{\kappa }$ is an algebraic closure of $\kappa$. Let $G = \text{Gal}(K^{sep}/K)$, $D = \{ \sigma \in G \mid \sigma (\mathfrak m^{sep}) = \mathfrak m^{sep}\}$, and $I = \{ \sigma \in D \mid \sigma \bmod \mathfrak m^{sep} = \text{id}_{\kappa (\mathfrak m^{sep})}\}$. The decomposition group $D$ fits into a canonical exact sequence

$1 \to I \to D \to \text{Gal}(\kappa ^{sep}/\kappa ) \to 1$

where $\kappa ^{sep} \subset \overline{\kappa }$ is the separable closure of $\kappa$. The inertia group $I$ fits into a canonical exact sequence

$1 \to P \to I \to I_ t \to 1$

such that

1. $P$ is a normal subgroup of $D$,

2. $P$ is a pro-$p$-group if the characteristic of $\kappa _ A$ is $p > 1$ and $P = \{ 1\}$ if the characteristic of $\kappa _ A$ is zero,

3. there exists a canonical surjective map

$\theta _{can} : I \to \mathop{\mathrm{lim}}\nolimits _{n\text{ prime to }p} \mu _ n(\kappa ^{sep})$

whose kernel is $P$, which satisfies $\theta _{can}(\tau \sigma \tau ^{-1}) = \tau (\theta _{can}(\sigma ))$ for $\tau \in D$, $\sigma \in I$, and which induces an isomorphism $I_ t \to \mathop{\mathrm{lim}}\nolimits _{n\text{ prime to }p} \mu _ n(\kappa ^{sep})$.

Proof. The field $\overline{\kappa }$ is the algebraic closure of $\kappa$ by Lemma 58.12.1. Most of the statements immediately follow from the corresponding parts of Lemma 58.13.5. For example because $\text{Aut}(\overline{\kappa }/\kappa ) = \text{Gal}(\kappa ^{sep}/\kappa )$ we obtain the first sequence. Then the only other assertion that needs a proof is the fact that with $S$ as in Lemma 58.13.5 the limit $\mathop{\mathrm{lim}}\nolimits _{n \in S} \mu _ n(\overline{\kappa })$ is equal to $\mathop{\mathrm{lim}}\nolimits _{n\text{ prime to }p} \mu _ n(\kappa ^{sep})$. To see this it suffices to show that every integer $n$ prime to $p$ divides an element of $S$. Let $\pi \in A$ be a uniformizer and consider the splitting field $L$ of the polynomial $X^ n - \pi$. Since the polynomial is separable we see that $L$ is a finite Galois extension of $K$. Choose an embedding $L \to K^{sep}$. Observe that if $B$ is the integral closure of $A$ in $L$, then the ramification index of $A \to B_{\mathfrak m^{sep} \cap B}$ is divisible by $n$ (because $\pi$ has an $n$th root in $B$; in fact the ramification index equals $n$ but we do not need this). Then it follows from the construction of the $S$ in the proof of Lemma 58.13.5 that $n$ divides an element of $S$. $\square$

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