The Stacks project

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

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

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

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

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. $\square$