The Stacks project

Proof. Recall that $|G| = [L : K] = nef$, see Lemma 15.114.2. Since $G$ acts transitively on the set $\{ \mathfrak m_1, \ldots , \mathfrak m_ n\} $ of maximal ideals of $B$ (Lemma 15.114.1) and since $D$ is the stabilizer of an element we see that $|D| = ef$. By Lemma 15.114.4 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 $P = \mathop{\mathrm{Ker}}(\theta )$. By construction the elements of $P$ are exactly the elements of $I$ which act trivially on $\mathfrak m/\mathfrak m^2 = \text{Gr}_\mathfrak m^1(B)$, i.e., $P = \{ \sigma \in I \mid \sigma |_{\mathfrak m/\mathfrak m^2} = \text{id}_{\mathfrak m/\mathfrak m^2}\} $. Also $I$ consists of the elements of $D$ which act trivially on $\kappa (\mathfrak m) = B/\mathfrak m$. Since the graded ring $\text{Gr}_\mathfrak m(B)$ is generated by $\text{Gr}^1_\mathfrak m(B) = \mathfrak m/\mathfrak m^2$ over $\text{Gr}^0_\mathfrak m(B) = B/\mathfrak m$, we conclude that $P = \{ \sigma \in D \mid \sigma \text{ acts trivially on } \text{Gr}_\mathfrak m(B)\} $. Thus (1) is true. This implies (2) as $P$ is the kernel of the homomorphism $D \to \text{Aut}(\text{Gr}_\mathfrak m(B))$. If we can prove (3), then parts (4) and (5) 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.

Proof of (6). Assume $\kappa (\mathfrak m)/\kappa $ is separable. In this case $f = |D/I|$ by Lemma 15.114.4 and $I$ has order $e$. If $e = 1$, then $P = I = \{ \text{id}\} $ and the result is true. If $e > 1$, then $B/\mathfrak m^2 = C/\pi _ C^2C$ is a $\kappa $-algebra and a small extension of $\kappa (\mathfrak m)$. Because $\kappa (\mathfrak m)$ is formally étale over $\kappa $ (Algebra, Lemma 10.158.1) there is a unique $\kappa $-algebra map $\kappa (\mathfrak m) \to B/\mathfrak m^2$ right inverse to $B/\mathfrak m^2 \to \kappa (\mathfrak m)$. In other words, there is a $D$-equivariant isomorphism $B/\mathfrak m^2 = \mathfrak m/\mathfrak m^2 \oplus \kappa (\mathfrak m)$ of $\kappa $-algebras. This immediately shows that $P = \{ \sigma \in D \mid \sigma |_{B/\mathfrak m^2} = \text{id}_{B/\mathfrak m^2}\} $ in this case. $\square$