The Stacks project

Lemma 15.112.5. Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a finite Galois extension with Galois group $G$. Let $B$ be the integral closure of $A$ in $L$. Let $\mathfrak m \subset B$ be a maximal ideal. The inertia group $I$ of $\mathfrak m$ sits in a canonical exact sequence

\[ 1 \to P \to I \to I_ t \to 1 \]

such that

  1. $P = \{ \sigma \in D \mid \sigma |_{B/\mathfrak m^2} = \text{id}_{B/\mathfrak m^2}\} $ where $D$ is the decomposition group,

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

  3. $P$ is a $p$-group if the characteristic of $\kappa _ A$ is $p > 0$ and $P = \{ 1\} $ if the characteristic of $\kappa _ A$ is zero,

  4. $I_ t$ is cyclic of order the prime to $p$ part of the integer $e$, and

  5. there is a canonical isomorphism $\theta : I_ t \to \mu _ e(\kappa (\mathfrak m))$.

Here $e$ is the integer of Lemma 15.112.2.

Proof. Recall that $|G| = [L : K] = nef$, see Lemma 15.112.2. Since $G$ acts transitively on the set $\{ \mathfrak m_1, \ldots , \mathfrak m_ n\} $ of maximal ideals of $B$ (Lemma 15.112.1) and since $D$ is the stabilizer of an element we see that $|D| = ef$. By Lemma 15.112.4 we have

\[ ef = |D| = |I| \cdot |\text{Aut}(\kappa (\mathfrak m)/\kappa )| \]

where $\kappa $ is the residue field of $A$. As $\kappa (\mathfrak m)$ is normal over $\kappa $ the order of $\text{Aut}(\kappa (\mathfrak m)/\kappa )$ differs from $f$ by a power of $p$ (see Fields, Lemma 9.15.9 and discussion following Fields, Definition 9.14.7). Hence the prime to $p$ part of $|I|$ is equal to the prime to $p$ part of $e$.

Set $C = B_\mathfrak m$. Then $I$ acts on $C$ over $A$ and trivially on the residue field of $C$. Let $\pi _ A \in A$ and $\pi _ C \in C$ be uniformizers. Write $\pi _ A = u \pi _ C^ e$ for some unit $u$ in $C$. For $\sigma \in I$ write $\sigma (\pi _ C) = \theta _\sigma \pi _ C$ for some unit $\theta _\sigma $ in $C$. Then we have

\[ \pi _ A = \sigma (\pi _ A) = \sigma (u) (\theta _\sigma \pi _ C)^ e = \sigma (u) \theta _\sigma ^ e \pi _ C^ e = \frac{\sigma (u)}{u} \theta _\sigma ^ e \pi _ A \]

Since $\sigma (u) \equiv u \bmod \mathfrak m_ C$ as $\sigma \in I$ we see that the image $\overline{\theta }_\sigma $ of $\theta _\sigma $ in $\kappa _ C = \kappa (\mathfrak m)$ is an $e$th root of unity. We obtain a map
\begin{equation} \label{more-algebra-equation-inertia-character} \theta : I \longrightarrow \mu _ e(\kappa (\mathfrak m)),\quad \sigma \mapsto \overline{\theta }_\sigma \end{equation}

We claim that $\theta $ is a homomorphism of groups and independent of the choice of uniformizer $\pi _ C$. Namely, if $\tau $ is a second element of $I$, then $\tau (\sigma (\pi _ C)) = \tau (\theta _\sigma \pi _ C) = \tau (\theta _\sigma ) \theta _\tau \pi _ C$, hence $\theta _{\tau \sigma } = \tau (\theta _\sigma ) \theta _\tau $ and since $\tau \in I$ we conclude that $\overline{\theta }_{\tau \sigma } = \overline{\theta }_\sigma \overline{\theta }_\tau $. If $\pi '_ C$ is a second uniformizer, then we see that $\pi '_ C = w \pi _ C$ for some unit $w$ of $C$ and $\sigma (\pi '_ C) = w^{-1}\sigma (w)\theta _\sigma \pi '_ C$, hence $\theta '_\sigma = w^{-1}\sigma (w)\theta _\sigma $, hence $\theta '_\sigma $ and $\theta _\sigma $ map to the same element of the residue field as before.

Since $\kappa (\mathfrak m)$ has characteristic $p$, the group $\mu _ e(\kappa (\mathfrak m))$ is cyclic of order at most the prime to $p$ part of $e$ (see Fields, Section 9.17).

Let $P = \mathop{\mathrm{Ker}}(\theta )$. The elements of $P$ are exactly the elements of $D$ acting trivially on $C/\pi _ C^2C \cong B/\mathfrak m^2$. Thus (a) is true. This implies (b) as $P$ is the kernel of the map $D \to \text{Aut}(B/\mathfrak m^2)$. If we can prove (c), then parts (d) and (e) will follow as $I_ t$ will be isomorphic to $\mu _ e(\kappa (\mathfrak m))$ as the arguments above show that $|I_ t| \geq |\mu _ e(\kappa (\mathfrak m))|$.

Thus it suffices to prove that the kernel $P$ of $\theta $ is a $p$-group. Let $\sigma $ be a nontrivial element of the kernel. Then $\sigma - \text{id}$ sends $\mathfrak m_ C^ i$ into $\mathfrak m_ C^{i + 1}$ for all $i$. Let $m$ be the order of $\sigma $. Pick $c \in C$ such that $\sigma (c) \not= c$. Then $\sigma (c) - c \in \mathfrak m_ C^ i$, $\sigma (c) - c \not\in \mathfrak m_ C^{i + 1}$ for some $i$ and we have

\begin{align*} 0 & = \sigma ^ m(c) - c \\ & = \sigma ^ m(c) - \sigma ^{m - 1}(c) + \ldots + \sigma (c) - c \\ & = \sum \nolimits _{j = 0, \ldots , m - 1} \sigma ^ j(\sigma (c) - c) \\ & \equiv m(\sigma (c) - c) \bmod \mathfrak m_ C^{i + 1} \end{align*}

It follows that $p | m$ (or $m = 0$ if $p = 1$). Thus every element of the kernel of $\theta $ has order divisible by $p$, i.e., $\mathop{\mathrm{Ker}}(\theta )$ is a $p$-group. $\square$

Comments (2)

Comment #3267 by on

Typo 1: in the second paragraph, not .

Typo 2: at the end of the second to last paragraph, the arguments above actually show the reverse inequality, , not ≤. (You need ≥.)

Typo 3: in the last paragraph it is stated but besides being false what you need and what you have (since is in the kernel of ) is that the restriction of to coincides with the identity modulo .

Comment #3362 by on

Thanks for your very precise explanation of what was wrong and how to fix it. The fix is here and will be online in a couple of days.

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