Lemma 15.112.10. Let $A$ be a discrete valuation ring with fraction field $K$. Let $M/L/K$ be a tower with $M/K$ and $L/K$ finite Galois. Let $C$, $B$ be the integral closure of $A$ in $M$, $L$. Let $\mathfrak m' \subset C$ be a maximal ideal and set $\mathfrak m = \mathfrak m' \cap B$. Let

$P \subset I \subset D \subset \text{Gal}(L/K) \quad \text{and}\quad P' \subset I' \subset D' \subset \text{Gal}(M/K)$

be the wild inertia, inertia, decomposition group of $\mathfrak m$ and $\mathfrak m'$. Then the canonical surjection $\text{Gal}(M/K) \to \text{Gal}(L/K)$ induces surjections $P' \to P$, $I' \to I$, and $D' \to D$. Moreover these fit into commutative diagrams

$\vcenter { \xymatrix{ D' \ar[r] \ar[d] & \text{Aut}(\kappa (\mathfrak m')/\kappa _ A) \ar[d] \\ D \ar[r] & \text{Aut}(\kappa (\mathfrak m)/\kappa _ A) } } \quad \text{and}\quad \vcenter { \xymatrix{ I' \ar[r]_-{\theta '} \ar[d] & \mu _{e'}(\kappa (\mathfrak m')) \ar[d]^{(-)^{e'/e}} \\ I \ar[r]^-\theta & \mu _ e(\kappa (\mathfrak m)) } }$

where $e'$ and $e$ are the ramification indices of $A \to C_{\mathfrak m'}$ and $A \to B_\mathfrak m$.

Proof. The fact that under the map $\text{Gal}(M/K) \to \text{Gal}(L/K)$ the groups $P', I', D'$ map into $P, I, D$ is immediate from the definitions of these groups. The commutativity of the first diagram is clear (observe that since $\kappa (\mathfrak m)/\kappa _ A$ is normal every automorphism of $\kappa (\mathfrak m')$ over $\kappa _ A$ indeed induces an automorphism of $\kappa (\mathfrak m)$ over $\kappa _ A$ and hence we obtain the right vertical arrow in the first diagram, see Lemma 15.112.4 and Fields, Lemma 9.15.7).

The maps $I' \to I$ and $D' \to D$ are surjective by Lemma 15.110.8. The surjectivity of $P' \to P$ follows as $P'$ and $P$ are p-Sylow subgroups of $I'$ and $I$.

To see the commutativity of the second diagram we choose a uniformizer $\pi '$ of $C_{\mathfrak m'}$ and a uniformizer $\pi$ of $B_\mathfrak m$. Then $\pi = c' (\pi ')^{e'/e}$ for some unit $c'$ of $C_{\mathfrak m'}$. For $\sigma ' \in I'$ the image $\sigma \in I$ is simply the restriction of $\sigma '$ to $L$. Write $\sigma '(\pi ') = c \pi '$ for a unit $c \in C_{\mathfrak m'}$ and write $\sigma (\pi ) = b \pi$ for a unit $b$ of $B_\mathfrak m$. Then $\sigma '(\pi ) = b \pi$ and we obtain

$b \pi = \sigma '(\pi ) = \sigma '(c' (\pi ')^{e'/e}) = \sigma '(c') c^{e'/e} (\pi ')^{e'/e} = \frac{\sigma '(c')}{c'} c^{e'/e} \pi$

As $\sigma ' \in I'$ we see that $b$ and $c^{e'/e}$ have the same image in the residue field which proves what we want. $\square$

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