Higher Ramification Groups - Local Fields

14 Higher Ramification Groups

Let $L ∕ K$ be a finite Galois extension of local fields, and $π_{L} \in O_{L}$ a uniformiser.

Definition 14.1 ( $s$ -th ramification group). Let $v_{L}$ be a normalised valuation in $O_{L}$ . For $s \in ℝ_{\geq - 1}$ , the $s$ -th ramification group is

G_{s} (L ∕ K) = {σ \in Gal (K) | v_{L} (σ (x) - x) \geq s + 1 \forall x \in O_{L}} .

Remark. $G_{s}$ only changes at integers.

$G_{s}$ , $s \in ℝ_{\geq - 1}$ used to define upper numbering.

Example.

\begin{array}{l} G_{- 1} (L ∕ K) & = Gal (L ∕ K) \\ G_{0} (L ∕ K) & = {σ \in Gal (L ∕ K) | σ (x) \equiv x mod π_{L} \forall x \in O_{L}} \\ = \ker (Gal (L ∕ K) ↠ Gal (k_{L} ∕ k)) \\ = I_{L ∕ K} \end{array}

Note. For $s \in ℤ_{\geq 0}$ ,

G_{s} (L ∕ K) = \ker (Gal (L ∕ K) ↠ Aut (O_{L} ∕ π_{L}^{s + 1} O_{L}))

hence $G_{s} (L ∕ K)$ is normal in $G_{- 1}$ .

\dots \subseteq G_{s} \subseteq G_{s - 1} \subseteq \dots \subseteq G_{- 1} = Gal (L ∕ K) .

Theorem 14.2.

(i) For $s \geq 1$ ,
$G_{s} = {σ \in G_{0} | v_{L} (σ (π_{L}) - π_{L}) \geq s + 1} .$
(ii) $⋂_{n = 0}^{\infty} G_{n} = {1}$ .
(iii) Let $s \in ℤ_{\geq 0}$ . Then there exists an injective group homomorphism
$G_{s} ∕ G_{s + 1} ↪ U_{L}^{(s)} ∕ U_{L}^{(s + 1)}$

induced by $σ \mapsto \frac{σ (π_{L})}{π_{L}}$ . This map is independent of the choice of $π_{L}$ .

Proof. Let $K_{0} \subseteq L$ be a maximal unramified extension of $K$ in $L$ . Upon replacing $K$ by $K_{0}$ , we may assume that $L ∕ K$ is totally ramified.

(i) Theorem 13.8 implies $O_{L} ∕ O_{K} [π_{L}]$ . Suppose $v_{L} (σ (π_{L}) - π_{L}) \geq s + 1$ . Let $x \in O_{L}$ , then $x = f (π_{L})$ , $f (X) \in O_{K} [X]$ . $\begin{array}{l} σ (x) - x & = σ (f (π_{L})) - f (π_{L}) \\ = f (σ (π_{L})) - f (π_{L}) \\ = (σ (π_{L}) - π_{L}) g (π_{L}) \end{array}$
for some $g (X) \in O_{K} [X]$ , using the fact that $X^{n} - Y^{n} = (X - Y) (X^{n - 1} + \dots + Y^{n - 1})$ . Thus
$v_{L} (σ (x) - x) = v_{L} (σ (π_{L}) - π_{L}) + \underset{\geq 0}{\underset{⏟}{v_{L} (g (π_{L}))}} \geq s + 1 .$
(ii) Suppose $σ \in Gal (L ∕ K)$ , $σ \neq 1$ . Then $σ (π_{L}) \neq π_{L}$ , because $L = K (π_{L})$ and hence $v_{L} (σ (π_{L}) - π_{L}) < \infty$ . Thus $σ \notin G_{s}$ for some $s ≫ 0$ by (i).
(iii) Note: for $σ \in G_{s}$ , $s \in ℤ_{\geq 0}$ ,
$σ (π_{L}) \in π_{L} + π_{L}^{s + 1} O_{L}$

hence
$\frac{σ (π_{L})}{π_{L}} \in 1 + π_{L}^{s} O_{L} = U_{L}^{(s)} .$

We claim
$\begin{array}{l} φ : G_{s} & \to U_{L}^{(s)} ∕ U_{L}^{(s + 1)} \\ σ & \mapsto \frac{σ (π_{L})}{π_{L}} \end{array}$
is a group homomorphism with kernel $G_{s + 1}$ . For $σ, τ \in G_{s}$ , let $τ (π_{L}) = u π_{L}$ , $u \in O_{L}^{\times}$ . Then
$\begin{array}{l} \frac{σ τ (π_{L})}{π_{L}} & = \frac{σ (τ (π_{L}))}{τ (π_{L})} \frac{τ (π_{L})}{π_{L}} \\ = \frac{σ (u)}{u} \frac{σ (π_{L})}{π_{L}} \frac{τ (π_{L})}{π_{L}} \end{array}$
But $σ (u) \in u + π_{L}^{s + 1} O_{L}$ since $σ \in G_{s}$ . Thus $\frac{σ (u)}{u} \in U_{L}^{(s + 1)}$ and hence
$\frac{σ τ (π_{L})}{π_{L}} \equiv \frac{σ (π_{L})}{π_{L}} \cdot \frac{τ (π_{L})}{π_{L}} mod U_{L}^{(s + 1)} .$

Hence $φ$ is a group homomorphism. Moreover,
$\ker (φ) = {σ \in G_{s} | σ (π_{L}) \equiv π_{L} mod π_{L}^{s + 1}} = G_{s + 1} .$

If $π_{L}^{'} = a π_{L}$ is another uniformiser, $a \in O_{L}^{\times}$ . Then
$\frac{σ (π_{L}^{'})}{π_{L}^{'}} = \frac{σ (a)}{a} \cdot \frac{σ (π_{L})}{π_{L}} \equiv \frac{σ (π_{L})}{π_{L}} mod U_{L}^{(s + 1)} . □$

Corollary 14.3. $Gal (L ∕ K)$ is solvable.

Proof. By Proposition 13.11, Theorem 14.2 and Theorem 13.4, for $s \in ℤ_{\geq - 1}$ ,

G_{s} ∕ G_{s + 1} ≅ a subgroup {\begin{matrix} Gal (k_{L} ∕ k) & if s = - 1 \\ (k_{L}^{\times}, \times) & if s = 0 \\ (k_{L}, +) & if s \geq 1 \end{matrix}

Thus $G_{s} ∕ G_{s + 1}$ is solvable for $s \geq - 1$ . Conclude using Theorem 14.2(ii). □

Let $characteristic k = p$ . Then $p ∤ | G_{0} ∕ G_{1} |$ and $| G_{1} | = p^{n}$ . Thus $G_{1}$ is the unique (since normal) Sylow $p$ -subgroup of $G_{0} = I_{L ∕ K}$ .

Definition 14.4. $G_{1}$ is called the wild inertial group, and $G_{0} ∕ G_{1}$ is called the tame quotient.

Suppose $L ∕ K$ is finite separable. Say $L ∕ K$ is tamely ramified if $characteristic k ∤ e_{L ∕ K}$ . Otherwise it is wildly ramified.

Theorem 14.5. Assuming that:

$[K : ℚ_{p}] < \infty$
$L ∕ K$ finite
$D_{L ∕ K} = (π^{δ (L ∕ K)})$

Then

δ (L ∕ K) \geq e_{L ∕ K} - 1

, with equality if and only if tamely ramified. In particular,

L ∕ K

unramified if and only if

D_{L ∕ K} = O_{L}

Proof. Example Sheet 3 shows $D_{L ∕ K} = D_{L ∕ K_{0}} \cdot D_{K_{0} ∕ K}$ . Suffices to check $2$ cases:

(i) $L ∕ K$ unramified. Then ?? gives that $O_{L} = O_{K} [α]$ , for some $α \in O_{L}$ with $k_{L} = k (\bar{a})$ .
Let $g (X) \in O_{K} [X]$ be the minimal polynomial of $α$ . Since $[L : K] = [k_{L} : k]$ , we have that $\bar{g} (X) \in k [X]$ is the minimal polynomial of $\bar{a}$ . $\bar{g} (X)$ separable and hence $g^{'} (α) ⁄ \equiv 0 {(m o d π)}_{L}$ . Theorem 12.8 implies $D_{L ∕ K} = (g (α)) = O_{L}$ .
(ii) $L ∕ K$ totally ramified. Say $[L : K] = e$ , $O_{L} = O_{K} [π_{L}]$ , $π_{L}$ a root of
$g (X) = X^{e} + \sum_{i = 0}^{e - 1} a_{i} X^{i} \in O_{K} [X]$

is Eisenstein. Then
$g^{'} (π_{L}) = \underset{\geq e - 1}{\underset{⏟}{e π_{L}^{e - 1}}} + \underset{v_{L} \geq e}{\underset{⏟}{\sum_{i = 1}^{e - 1} i a_{i} π_{L}^{i - 1}}} .$

Thus $v_{L} (y^{'} (π_{L})) \geq e - 1$ . Equality if and only if $p ∤ e$ . □

Corollary 14.6. Suppose $L ∕ K$ is an extension of number fields. Let $P \subseteq O_{L}$ , $P \cap O_{K} = 𝔭$ . Then $e (P ∕ 𝔭) > 1$ if and only if $P | D_{L ∕ K}$ .

Proof. Theorem 12.9 implies $D_{L ∕ K} = \prod_{P} D_{L_{P} ∕ K_{𝔭}}$ . Then use $e (P ∕ 𝔭) = e_{L_{P} ∕ K_{𝔭}}$ and Theorem 14.5. □

Example.

$K = ℚ_{p}$ , $ζ_{p^{n}}$ a primitive $p^{n}$ -th root of unity. $L = ℚ_{p} (ζ_{p^{n}})$ . The $p^{n}$ -th cyclotomic polynomial is
$Φ_{p^{n}} (X) = X^{p^{n - 1} (p - 1)} + X^{p^{n - 1} (p - 2)} + \dots + 1 \in ℤ_{p} [X] .$

See Example Sheet 3.
$Φ_{p^{n}} (X)$ irreducible (hence $Φ_{p^{n}} (X)$ is the minimal polynomial of $ζ_{p^{n}}$ ).
$L ∕ ℚ_{p}$ is Galois, totally ramified of degree $p^{n + 1} (p - 1)$ .
$π : = ζ_{p^{n}} - 1$ a uniformiser in $O_{L}$ $⇝$ $O_{L} = ℤ_{p} [ζ_{p^{n}} - 1] = ℤ_{p} [ζ_{p^{n}}]$ .
$Gal (L ∕ ℚ_{p}) \tilde{\to} {(ℤ ∕ p^{n} ℤ)}^{\times}$ (abelian). $σ_{m} \leftrightarrow m$ where $σ_{m} (ζ_{p^{n}}) = ζ_{p^{n}}^{m}$ .

$v_{L} (σ_{m} (π) - π) = v_{L} (ζ_{p^{n}}^{m} - ζ_{p^{n}}) = v_{L} (ζ_{p^{n}}^{m - 1} - 1) .$

Let $k$ be maximal such that $p^{k} | m - 1$ . Then $ζ_{p^{n}}^{m - 1}$ is a primitive $p^{n - k}$ -th root of unity, and hence $ζ_{p^{n}}^{m - 1} - 1$ is a uniformiser $π^{'}$ in $L^{'} = ℚ_{p} (ζ_{p^{n}}^{m - 1})$ . Hence
$v_{L} (ζ_{p^{n}}^{m - 1} - 1) = e_{L ∕ L^{'}} = \frac{e_{L ∕ ℚ_{p}}}{e_{L^{'} ∕ ℚ_{p}}} = \frac{[L : ℚ_{p}]}{[L^{'} : ℚ_{p}]} = \frac{p^{n - 1} (p - 1)}{p^{n - k - 1} (p - 1)} = p^{k} .$

Theorem 14.2(i) implies that $σ_{m} \in G_{i}$ if and only if $p^{k} \geq i + 1$ . Thus
$G_{i} ≅ {\begin{matrix} {(ℤ ∕ p^{n} ℤ)}^{\times} & i \leq 0 \\ (1 + p^{k} ℤ) ∕ p^{n} ℤ & p^{k - 1} - 1 < i \leq p^{k} - 1 (1 \leq k \leq i + 1) \\ {1} & p^{n - 1} - 1 < i \end{matrix} .$