Lemma 15.112.1. Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a finite Galois extension with Galois group $G$. Then $G$ acts on the ring $B$ of Remark 15.111.6 and acts transitively on the set of maximal ideals of $B$.

## 15.112 Galois extensions and ramification

In the case of Galois extensions, we can elaborate on the discussion in Section 15.111.

**Proof.**
Observe that $A = B^ G$ as $A$ is integrally closed in $K$ and $K = L^ G$. Hence this lemma is a special case of Lemma 15.110.8.
$\square$

Lemma 15.112.2. Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a finite Galois extension. Then there are $e \geq 1$ and $f \geq 1$ such that $e_ i = e$ and $f_ i = f$ for all $i$ (notation as in Remark 15.111.6). In particular $[L : K] = n e f$.

**Proof.**
Immediate consequence of Lemma 15.112.1 and the definitions.
$\square$

Definition 15.112.3. 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

*decomposition group of $\mathfrak m$*is the subgroup $D = \{ \sigma \in G \mid \sigma (\mathfrak m) = \mathfrak m\} $.The

*inertia group of $\mathfrak m$*is the kernel $I$ of the map $D \to \text{Aut}(\kappa (\mathfrak m)/\kappa _ A)$.

Note that the field $\kappa (\mathfrak m)$ may be inseparable over $\kappa _ A$. In particular the field extension $\kappa (\mathfrak m)/\kappa _ A$ need not be Galois. If $\kappa _ A$ is perfect, then it is.

Lemma 15.112.4. Let $A$ be a discrete valuation ring with fraction field $K$ and residue field $\kappa $. Let $L/K$ be a finite Galois extension with Galois group $G$. Let $B$ be the integral closure of $A$ in $L$. Let $\mathfrak m$ be a maximal ideal of $B$. Then

the field extension $\kappa (\mathfrak m)/\kappa $ is normal, and

$D \to \text{Aut}(\kappa (\mathfrak m)/\kappa )$ is surjective.

If for some (equivalently all) maximal ideal(s) $\mathfrak m \subset B$ the field extension $\kappa (\mathfrak m)/\kappa $ is separable, then

$\kappa (\mathfrak m)/\kappa $ is Galois, and

$D \to \text{Gal}(\kappa (\mathfrak m)/\kappa )$ is surjective.

Here $D \subset G$ is the decomposition group of $\mathfrak m$.

**Proof.**
Observe that $A = B^ G$ as $A$ is integrally closed in $K$ and $K = L^ G$. Thus parts (1) and (2) follow from Lemma 15.110.9. The “equivalently all” part of the lemma follows from Lemma 15.112.1. Assume $\kappa (\mathfrak m)/\kappa $ is separable. Then parts (3) and (4) follow immediately from (1) and (2).
$\square$

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

such that

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

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

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

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

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

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. $\square$

Definition 15.112.6. With assumptions and notation as in Lemma 15.112.5.

The

*wild inertia group of $\mathfrak m$*is the subgroup $P$.The

*tame inertia group of $\mathfrak m$*is the quotient $I \to I_ t$.

We denote $\theta : I \to \mu _ e(\kappa (\mathfrak m))$ the surjective map (15.112.5.1) whose kernel is $P$ and which induces the isomorphism $I_ t \to \mu _ e(\kappa (\mathfrak m))$.

Lemma 15.112.7. With assumptions and notation as in Lemma 15.112.5. The inertia character $\theta : I \to \mu _ e(\kappa (\mathfrak m))$ satisfies the following property

for $\tau \in D$ and $\sigma \in I$.

**Proof.**
The formula makes sense as $I$ is a normal subgroup of $D$ and as $\tau $ acts on $\kappa (\mathfrak m)$ via the map $D \to \text{Aut}(\kappa (\mathfrak m))$ discussed in Lemma 15.112.4 for example. Recall the construction of $\theta $. Choose a uniformizer $\pi $ of $B_\mathfrak m$ and for $\sigma \in I$ write $\sigma (\pi ) = \theta _\sigma \pi $. Then $\theta (\sigma )$ is the image $\overline{\theta }_\sigma $ of $\theta _\sigma $ in the residue field. For any $\tau \in D$ we can write $\tau (\pi ) = \theta _\tau \pi $ for some unit $\theta _\tau $. Then $\theta _{\tau ^{-1}} = \tau ^{-1}(\theta _\tau ^{-1})$. We compute

However, since $\sigma $ acts trivially modulo $\pi $ we see that the product $\tau (\sigma (\tau ^{-1}(\theta _\tau ^{-1}))) \theta _\tau $ maps to $1$ in the residue field. This proves the lemma. $\square$

We will generalize the following lemma in Fundamental Groups, Lemma 58.12.5.

Lemma 15.112.8. Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a finite Galois extension. Let $\mathfrak m \subset B$ be a maximal ideal of the integral closure of $A$ in $L$. Let $I \subset G$ be the inertia group of $\mathfrak m$. Then $B^ I$ is the integral closure of $A$ in $L^ I$ and $A \to (B^ I)_{B^ I \cap \mathfrak m}$ is étale.

**Proof.**
Write $B' = B^ I$. It follows from the definitions that $B' = B^ I$ is the integral closure of $A$ in $L^ I$. Write $\mathfrak m' = B^ I \cap \mathfrak m = B' \cap \mathfrak m \subset B'$. By Lemma 15.110.8 the maximal ideal $\mathfrak m$ is the unique prime ideal of $B$ lying over $\mathfrak m'$. As $I$ acts trivially on $\kappa (\mathfrak m)$ we see from Lemma 15.110.2 that the extension $\kappa (\mathfrak m)/\kappa (\mathfrak m')$ is purely inseparable (perhaps an easier alternative is to apply the result of Lemma 15.110.9). Since $D/I$ acts faithfully on $\kappa (\mathfrak m')$, we conclude that $D/I$ acts faithfully on $\kappa (\mathfrak m)$. Of course the elements of the residue field $\kappa $ of $A$ are fixed by this action. By Galois theory we see that $[\kappa (\mathfrak m') : \kappa ] \geq |D/I|$, see Fields, Lemma 9.21.6.

Let $\pi $ be the uniformizer of $A$. Since $\text{Norm}_{L/K}(\pi ) = \pi ^{[L : K]}$ we see from Algebra, Lemma 10.121.8 that

as there are $n = |G/D|$ maximal ideals of $B$ which are all conjugate under $G$, see Remark 15.111.6 and Lemma 15.112.1. Applying the same reasoning to the finite extension the finite extension $L/L^ I$ of degree $|I|$ we find

We conclude that

Since the left hand side is a positive integer and since the right hand side is $\leq 1$ by the above, we conclude that we have equality, $\text{ord}_{B'_{\mathfrak m'}}(\pi ) = 1$ and $\kappa (\mathfrak m')/\kappa $ has degree $|D/I|$. Thus $\pi B'_{\mathfrak m'} = \mathfrak m' B_\mathfrak m'$ and $\kappa (\mathfrak m')$ is Galois over $\kappa $ with Galois group $D/I$, in particular separable, see Fields, Lemma 9.21.2. By Algebra, Lemma 10.143.7 we find that $A \to B'_{\mathfrak m'}$ is étale as desired. $\square$

Remark 15.112.9. Let $A$ be a discrete valuation ring with fraction field $K$. Let $L/K$ be a finite Galois extension. Let $\mathfrak m \subset B$ be a maximal ideal of the integral closure of $A$ in $L$. Let

be the wild inertia, inertia, decomposition group of $\mathfrak m$. Consider the diagram

Observe that $B^ P, B^ I, B^ D$ are the integral closures of $A$ in the fields $L^ P$, $L^ I$, $L^ D$. Thus we also see that $B^ P$ is the integral closure of $B^ I$ in $L^ P$ and so on. Observe that $\mathfrak m^ P = \mathfrak m \cap B^ P$, $\mathfrak m^ I = \mathfrak m \cap B^ I$, and $\mathfrak m^ D = \mathfrak m \cap B^ D$. Hence the top line of the diagram corresponds to the images of $\mathfrak m \in \mathop{\mathrm{Spec}}(B)$ under the induced maps of spectra. Having said all of this we have the following

the extension $L^ I/L^ D$ is Galois with group $D/I$,

the extension $L^ P/L^ I$ is Galois with group $I_ t = I/P$,

the extension $L^ P/L^ D$ is Galois with group $D/P$,

$\mathfrak m^ I$ is the unique prime of $B^ I$ lying over $\mathfrak m^ D$,

$\mathfrak m^ P$ is the unique prime of $B^ P$ lying over $\mathfrak m^ I$,

$\mathfrak m$ is the unique prime of $B$ lying over $\mathfrak m^ P$,

$\mathfrak m^ P$ is the unique prime of $B^ P$ lying over $\mathfrak m^ D$,

$\mathfrak m$ is the unique prime of $B$ lying over $\mathfrak m^ I$,

$\mathfrak m$ is the unique prime of $B$ lying over $\mathfrak m^ D$,

$A \to B^ D_{\mathfrak m^ D}$ is étale and induces a trivial residue field extension,

$B^ D_{\mathfrak m^ D} \to B^ I_{\mathfrak m^ I}$ is étale and induces a Galois extension of residue fields with Galois group $D/I$,

$A \to B^ I_{\mathfrak m^ I}$ is étale,

$B^ I_{\mathfrak m^ I} \to B^ P_{\mathfrak m^ P}$ has ramification index $|I/P|$ prime to $p$ and induces a trivial residue field extension,

$B^ D_{\mathfrak m^ D} \to B^ P_{\mathfrak m^ P}$ has ramification index $|I/P|$ prime to $p$ and induces a separable residue field extension,

$A \to B^ P_{\mathfrak m^ P}$ has ramification index $|I/P|$ prime to $p$ and induces a separable residue field extension.

Statements (1), (2), and (3) are immediate from Galois theory (Fields, Section 9.21) and Lemma 15.112.5. Statements (4) – (9) are clear from Lemma 15.112.1. Part (12) is Lemma 15.112.8. Since we have the factorization $A \to B^ D_{\mathfrak m^ D} \to B^ I_{\mathfrak m^ I}$ we obtain the étaleness in (10) and (11) as a consequence. The residue field extension in (10) must be trivial because it is separable and $D/I$ maps onto $\text{Aut}(\kappa (\mathfrak m)/\kappa _ A)$ as shown in Lemma 15.112.4. The same argument provides the proof of the statement on residue fields in (11). To see (13), (14), and (15) it suffices to prove (13). By the above, the extension $L^ P/L^ I$ is Galois with a cyclic Galois group of order prime to $p$, the prime $\mathfrak m^ P$ is the unique prime lying over $\mathfrak m^ I$ and the action of $I/P$ on the residue field is trivial. Thus we can apply Lemma 15.112.5 to this extension and the discrete valuation ring $B^ I_{\mathfrak m^ I}$ to see that (13) holds.

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

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

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 15.110.11. 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

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. $\square$

Remark 15.112.11. In order to use the inertia character $\theta : I \to \mu _ e(\kappa (\mathfrak m))$ for infinite Galois extensions, it is convenient to scale it. Let $A, K, L, B, \mathfrak m, G, P, I, D, e, \theta $ be as in Lemma 15.112.5 and Definition 15.112.6. Then $e = q |I_ t|$ with $q$ is a power of the characteristic $p$ of $\kappa (\mathfrak m)$ if positive or $1$ if zero. Note that $\mu _ e(\kappa (\mathfrak m)) = \mu _{|I_ t|}(\kappa (\mathfrak m))$ because the characteristic of $\kappa (\mathfrak m)$ is $p$. Consider the map

This map induces an isomorphism $\theta _{can} : I_ t \to \mu _{|I_ t|}(\kappa (\mathfrak m))$. We have $\theta _{can}(\tau \sigma \tau ^{-1}) = \tau (\theta _{can}(\sigma ))$ for $\tau \in D$ and $\sigma \in I$ by Lemma 15.112.7. Finally, if $M/L$ is an extension such that $M/K$ is Galois and $\mathfrak m'$ is a prime of the integral closure of $A$ in $M$ lying over $\mathfrak m$, then we get the commutative diagram

by Lemma 15.112.10.

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