Decomposition groups - Local Fields

Definition 11.1 (Ramification). Let $0 \neq 𝔭$ be a prime ideal of $O_{K}$ , and

𝔭 O_{L} = P_{1}^{e_{1}} \dots P_{r}^{e_{r}}

with $P_{i}$ distinct prime ideals in $O_{L}$ , and $e_{i} > 0$ .

(i) $e_{i}$ is the ramification index of $P_{i}$ over $𝔭$ .
(ii) We say $𝔭$ ramifies in $L$ if some $e_{i} > 1$ .

Proof. Let $S = O_{K} ∖ (𝔭)$ . Exercise (properties of localisation):

(1) $S^{- 1} O_{L}$ is the integral closure of $S^{- 1} O_{K}$ in $L$ .
(2) $msup 𝔭 S^{- 1} O_{L} ≅ S^{- 1} P_{1}^{e_{1}} \dots S^{- 1} P_{r}^{e_{r}}$ .
(3) $S^{- 1} O_{L} ∕ S^{- 1} P_{i} ≅ O_{L} ∕ P_{i}$ and $S^{- 1} O_{K} ∕ S^{- 1} 𝔭 ≅ O_{K} ∕ 𝔭$ .

In particular, (2) and (3) imply $e_{i}$ and $f_{i}$ don’t change when we replace $O_{K}$ and $O_{L}$ by $S^{- 1} O_{K}$ and $S^{- 1} O_{L}$ .

Thus we may assume that $O_{K}$ is a discrete valuation ring (hence a PID). By Chinese remainder theorem, we have

O_{L} ∕ 𝔭 O_{L} ≅ \prod_{i = 1}^{r} O_{L} ∕ P_{i}^{e_{i}} .

We count dimension as $k : = O_{K} ∕ 𝔭$ vector spaces.

RHS: for each $i$ , there exists a decreasing sequence of $k$ -suibspaces

0 \subseteq P_{i}^{e_{i} - 1} ∕ P_{i}^{e_{i}} \subseteq \dots \subseteq P_{i} ∕ P_{i}^{e_{i}} \subseteq O_{L} ∕ P_{i}^{e_{i}} .

Thus $\dim_{k} O_{L} ∕ P_{i}^{e_{i}} = \sum_{j = 0}^{e_{i} - 1} \dim_{k} (P_{i}^{j} ∕ P_{i}^{j + 1})$ . Note that $P_{i}^{j} ∕ P_{i}^{j + 1}$ is an $O_{L} ∕ P_{i}$ -module and $x \in P_{i}^{j} ∖ P_{i}^{j + 1}$ is a generator (for example can prove this after localisation at $P_{i}$ ).

Then $\dim_{k} P_{i}^{j} ∕ P_{i}^{j + 1} = f_{i}$ and we have

\dim_{k} O_{L} ∕ P_{i}^{e_{i}} = e_{i} f_{i},

and hence

\dim_{k} \prod_{i = 1}^{r} O_{L} ∕ P_{i}^{e_{i}} = \sum_{i = 1}^{r} e_{i} f_{i} .

LHS: Structure theorem for finitely generated modules over PIDs tells us that $O_{L}$ is a free module over $O_{K}$ of rank $n$ .

Thus $O_{L} ∕ 𝔭 O_{L} ≅ {(O_{K} ∕ 𝔭)}^{n}$ as $k$ -vector spaces, hence $\dim_{k} O_{L} ∕ 𝔭 O_{L} = n$ . □

where

e_{x}

is the ramification index of

x

. Now assume

L ∕ K

is Galois. Then for any

σ \in Gal (L ∕ K)

σ (P_{i}) \cap O_{K} = 𝔭

and hence

σ (P_{i}) \in {P_{1}, \dots, P_{r}}

Proof. Suppose not, so that there exists $i \neq j$ such that $σ (P_{i}) \neq P_{j}$ for all $σ \in Gal (L ∕ K)$ .

By Chinese remainder theorem, we may choose $x \in O_{L}$ such that $x \equiv 0 m o d P_{i}$ , $x \equiv 1 m o d σ (P_{i})$ for all $σ \in Gal (L ∕ K)$ . Then

N_{L ∕ K} (x) = \prod_{σ \in Gal (L ∕ K)} σ (x) \in O_{K} \cap P_{i} = 𝔭 \subseteq P_{j} .

Since $P_{j}$ prime, there exists $τ \in Gal (L ∕ K)$ such that $τ (x) \in P_{j}$ . Hence $x \in τ^{- 1} (P_{j})$ , i.e. $x \equiv 0 m o d τ^{- 1} (P_{j})$ , contradiction. □

Proof. For any $σ \in Gal (L ∕ K)$ we have

(i) $𝔭 O_{L} = σ (𝔭) O_{L} = σ {(P_{1})}^{e_{1}} \dots σ {(P_{r})}^{e_{r}}$ , hence $e_{1} = \dots = e_{r}$ .
(ii) $O_{L} ∕ P_{i} ≅ O_{L} ∕ σ (P_{i})$ via $σ$ . Hence $f_{1} = \dots = f_{r}$ . □

L ∕ K

is an extension of complete discretely valued fields with normalised valuations

v_{L}

v_{K}

and uniformisers

π_{L}, π_{K}

, then the ramification index is

e = e_{L ∕ K} = v_{L} (π_{K})

. The residue class degree is

f : = f_{L ∕ K} = [k_{L} : k]

Proposition 11.8. Assuming that:

$O_{K}$ a Dedekind domain
$L ∕ K$ a finite Galois extension
$0 \neq P \subseteq O_{L}$ a prime ideal
$P | 𝔭 \subseteq O_{K}$

Then

(i) $L_{P} ∕ K_{𝔭}$ is Galois.
(ii) There is a natural map
$res : Gal (L_{P} ∕ K_{𝔭}) \to Gal (L ∕ K)$

which is injective and has image $G_{P}$ .

Proof.

(i) $L ∕ K$ Galois implies that $L$ is a splitting field of a separable polynomial $f (X) \in K [X]$ . Hence $L_{P}$ is the splitting field of $f (X) \in K_{[} X]$ , hence $L_{P} ∕ K_{𝔭}$ is Galois.
(ii) Let $σ \in Gal (L_{P} ∕ K_{𝔭})$ , then $σ (L) = L$ since $L ∕ K$ is normal, hence we have a map $res : Gal (L_{P} ∕ K_{𝔭}) \to Gal (L ∕ K)$ , $σ \mapsto σ |_{L}$ . Since $L$ is dense in $L_{P}$ , $res$ is injective. By Lemma 8.2, we have
$| σ (x) |_{P} = | x |_{P}$

for all $σ \in Gal (L_{P} ∕ K_{𝔭})$ and $x \in L_{P}$ . Hence $σ (P) = P$ for all $σ \in Gal (L_{P} ∕ K_{𝔭})$ and hence $res (σ) \in G_{P}$ for all $σ \in Gal (L_{P} ∕ K_{𝔭})$ .

To show surjectivity, it suffices to show that
$| G_{P} | = e f = [L_{P} : K_{𝔭}] .$

Write $𝔭 O_{L} = P_{1}^{e_{1}} \dots P_{r}^{e_{r}}$ , $f = [O_{L} ∕ P : O_{K} ∕ 𝔭]$ . Then
- $| G_{P} | = \frac{| Gal (L ∕ K) |}{r} = \frac{e f r}{r} = e f$ (using Corollary 11.5).
- $[L_{P} : K_{𝔭}] = e f$ . Apply Corollary 11.6 to $L_{P} ∕ K_{𝔭}$ , noting that $e, f$ don’t change when we take completions. □

11 Decomposition groups