Unramified and totally ramified extensions of local fields - Local Fields

13 Unramified and totally ramified extensions of local fields

Let $L ∕ K$ be a finite separable extension of non-archimedean local fields. Corollary 11.6 implies

[L : K] = e_{L ∕ K} f_{L ∕ K} . (∗)

Lemma 13.1. Assuming that:

$M ∕ L ∕ K$ finite separable extensions of local fields

Then

(i) $f_{M ∕ K} = f_{L ∕ K} f_{M ∕ L}$
(ii) $e_{M ∕ K} = e_{L ∕ K} f_{M ∕ L}$

Proof.

(i) $f_{M ∕ K} = [k_{M} : k] = [k_{M} : k_{L}] [k_{L} : k] = f_{M ∕ L} f_{L ∕ K}$ .
(ii) (i) and ( $*$ ). □

Definition 13.2 (Unramified / ramified / totally ramified). The extension $L ∕ K$ is said to be:

unramified if $e_{L ∕ K} = 1$ (equivalently $f_{L ∕ K} = [L : K]$ ).
ramified if $e_{L ∕ K} > 1$ (equivalently $f_{L ∕ K} < [L : K]$ ).
totally ramified if $e_{L ∕ K} = [L : K]$ (equivalently $f_{L ∕ K} = 1$ ).

From now on in this course: if unspecified $L ∕ K$ is a finite separable extension of (non-archimedean) local fields. Also, all local fields that we consider from now on will be non-archimedean.

Theorem 13.3. Assuming that:

$L ∕ K$ a finite separable extension of non-archimedean local fields

Then there exists a field

K_{0}

K \subseteq K_{0} \subseteq L

and such that

(i) $K_{0}$ is unramified
(ii) $L ∕ K_{0}$ is totally ramified

Moreover $[L : K_{0}] = e_{L ∕ K}$ , $[K_{0} : K] = f_{L ∕ K}$ and $K_{0} ∕ K$ is Galois.

Proof. Let $k = 𝔽_{q}$ , so that $k_{L} = 𝔽_{q^{f}}$ , $f_{L ∕ K} = f$ . Set $m = q^{f} - 1$ , $[∙] : 𝔽_{q^{f}} \to L$ the Teichmüller map for $L$ .

Let $ζ_{m} : = [α]$ for $α$ a generator of $𝔽_{q^{f}}^{\times}$ . $ζ_{m}$ a primitive $m$ -th root of unity. Set $K_{0} = K (ζ_{m}) \subseteq L$ , then $K_{0} ∕ K$ is Galois and has residue field $k_{0} = 𝔽_{q} (α) = k_{L}$ . Hence $f_{L ∕ K_{0}} = 1$ , i.e. $L ∕ K_{0}$ is totally ramified.

Let $res : Gal (K_{0} ∕ K) \to Gal (k_{0} ∕ k)$ be the natural map. For $σ \in Gal (K_{0} ∕ K)$ . We have $σ (ζ_{m}) = ζ_{m}$ if $σ (ζ_{m}) \equiv ζ_{m} mod m$ (since $μ_{m} (K_{0}) ≅ μ_{m} (k_{0})$ by Hensel’s Lemma version 1). Hence $res$ is injective. Thus $| Gal (K_{0} ∕ K) | \leq | Gal (k_{0} ∕ k) | = f_{K_{0} ∕ K}$ , so $[K_{0} : K] = f_{K_{0} ∕ K}$ .

Hence $res$ is an isomorphism, and $K_{0} ∕ K$ is unramified. □

Theorem 13.4. Assuming that:

$k = 𝔽_{q}$
$n \geq 1$

Then there exists a unique unramified

L ∕ K

of degree

n

. Moreover,

L ∕ K

is Galois and the natural

Gal (L ∕ K) \to Gal (k_{L} ∕ k)

is an isomorphism. In particular,

Gal (L ∕ K) ≅ ⟨ {Frob}_{L ∕ K} ⟩

is cyclic, where

{Frob}_{L ∕ K} (x) = x^{q} mod m_{L}

for all

x \in O_{L}

Proof. For $n \geq 1$ , take $L = K (ζ_{m})$ where $m = q^{n} - 1$ .

As in Theorem 13.3:

Gal (L ∕ K) \tilde{\to} Gal (k_{L} ∕ K) ≅ Gal (𝔽_{q^{n}} ∕ 𝔽_{q}) .

Hence $Gal (L ∕ K)$ is cyclic, generated by a lift of $x \mapsto x^{q}$ .

Uniqueness: $L ∕ K$ of degree $n$ unramified. Then Teichmüller gives $ζ_{m} \in L$ , so $L = K (ζ_{m})$ . □

Corollary 13.5. $L ∕ K$ a finite Galois extension. Then $res : Gal (L ∕ K) \to Gal (k_{L} ∕ k)$ is surjective.

Proof. $res$ factorises as

Gal (L ∕ K) ↠ Gal (K_{0} ∕ K) \tilde{\to} Gal (k_{L} ∕ k) . □

Definition 13.6 (Inertial subgroup). The inertial subgroup is

I_{L ∕ K} = \ker (Gal (L ∕ K) ↠ Gal (k_{L} ∕ k)) .

Since $e_{L ∕ K} f_{L ∕ K} = [L : K]$ , we have $| I_{L ∕ K} | = e_{L ∕ K}$ .
$I_{L ∕ K} = Gal (L ∕ K_{0})$ – $K_{0}$ as in Theorem 13.3.

Definition 13.7 (Eisenstein polynomial). $f (x) = x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0} \in O_{K} [x]$ is Eisenstein if $v_{K} (a_{i}) \geq 1$ for all $i$ , and $v_{K} (a_{0}) = 1$ .

Fact: $f (x)$ Eisenstein implies $f (x)$ irreducible.

Theorem 13.8.

(i) Let $L ∕ K$ finite totally ramified, $π_{L} \in O_{L}$ a uniformiser. Then the minimal polynomial of $π_{L}$ is Eisenstein and $O_{L} = O_{K} [π_{L}]$ (hence $L = K (π_{L})$ )
(ii) Conversely, if $f (x) \in O_{K} [x]$ is Eisenstein and a root of if f, then $L : = K (α) ∕ K$ is totally ramified and $α$ is a uniformiser of $L$ .

Proof.

(i) $[L : K] = e = e_{L ∕ K}$ . Let
$f (x) = x^{m} + a_{m - 1} x^{m - 1} + \dots + a_{0} \in O_{K} [x]$

the minimal polynomial for $π_{L}$ . Then $m \leq e$ . Since $v_{L} (K^{\times}) = e ℤ$ , we have $v_{L} (a_{i} π^{i}) \equiv e mod e$ , for $i < m$ . Hence these terms have distinct valuations. As
$π_{L}^{m} = - \sum_{i = 0}^{m - 1} a_{i} π_{L}^{i} .$

we have
$m = v_{L} (π_{L}^{m}) = \min_{0 \leq i \leq m - 1} (i + e v_{K} (a_{i}))$

hence $v_{K} (a_{i}) \geq 1$ for all $i$ .

Hence $v_{K} (a_{0}) = 1$ and $m = e$ . Thus $f (x)$ is Eisenstein and $L = K (π_{L})$ . For $y \in L$ , we write $y = \sum_{i = 0}^{e - 1} π_{L}^{i} b_{i}$ , $b_{i} \in K$ . Then
$v_{L} (y) = \min_{0 \leq i \leq e - 1} (i + e v_{K} (b_{i})) .$

Thus
$\begin{array}{l} y \in O_{L} & ⟺ v_{L} (y) \geq 0 \\ ⟺ v_{K} (b_{i}) \geq 0 \forall i \\ ⟺ y \in O_{K} [π_{L}] \end{array}$
(ii) Let $f (x) = x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0}$ is Eisenstein and $e = e_{L ∕ K}$ . Thus $v_{L} (a_{i}) \geq e$ and $v_{L} (a_{0}) = e$ . If $v_{L} (α) \leq 0$ , we have
$v_{L} (α^{n}) < v_{L} (\sum_{i = 0}^{n - 1} a_{i} α^{i})$

hence $v_{L} (α) > 0$ . For $i \neq 0$ , $v_{L} (a_{i} α^{i}) > e = v_{L} (a_{0})$ . Therefore
$v_{L} (α^{n}) = v_{L} (- \sum_{i = 0}^{n - 1} a_{i} α^{i}) = v_{L} (a_{0}) = e .$

Hence $n v_{L} (α) = e$ . But $n = [L : K] \geq e$ , so $n = e$ and $v_{L} (α) = 1$ .

□

13.1 Structure of Units

Let $[K : ℚ] < \infty$ , $e : = e_{K ∕ ℚ_{p}}$ , $π$ a uniformiser in $K$ .

Proposition 13.9. Assuming that:

$r > \frac{e}{p - 1}$

Then

\exp (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}

converges on

π^{r} O_{K}

and induces an isomorphism

(π^{r} O_{K}, +) \tilde{\to} (1 + π^{r} O_{K}, \times) .

Proof.

\begin{array}{l} v_{K} (n!) & = e v_{p} (n!) \\ = \frac{e (n - s_{p} (n))}{p - 1} & Example Sheet 1 \\ \leq e (\frac{n - 1}{p - 1}) \end{array}

For $x \in π^{r} O_{K}$ and $n \geq 1$ ,

\begin{array}{l} v_{K} (\frac{x^{n}}{n!}) & \geq n r - \frac{e (n - 1)}{p - 1} \\ = r - (n - 1) \underset{> 0}{\underset{⏟}{(r - \frac{e}{p - 1})}} \end{array}

Hence $v_{K} (\frac{x^{n}}{n!}) \to \infty$ as $n \to \infty$ . Thus $\exp (x)$ converges.

Since $v_{K} (\frac{x^{n}}{n!}) \geq r$ for all $n \geq 1$ , $\exp (x) \in 1 + π^{r} O_{K}$ .

Consider $\log$ : $1 + π^{r} O_{K} \to π^{r} O_{K}$ .

\log (1 + x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{n}

which converges as before.

Recall identities in $ℚ [[X, Y]]$ :

\begin{array}{l} \exp (X + Y) & = \exp (X) \exp (Y) \\ \exp (\log (1 + X)) & = 1 + X \\ \log (\exp (X)) & = X \end{array}

Thus $\exp; (π^{r} O_{K}, +) \tilde{\to} (1 + π^{r} O_{K}, \times)$ is an isomorphism. □

$K$ any local field: $U_{K} : = O_{K}^{\times}$ , $π \in O_{K}$ uniformiser.

Definition 13.10 ( $s$ -th unit group). For $s \in ℤ$ , the $s$ -th unit group $U_{K}^{(s)}$ is defined by

U_{K}^{(s)} = (1 + π^{s} O_{K}, \times) .

Set $U_{K}^{(0)} = U_{K}$ . Then we have

\dots \subseteq U_{K}^{(s)} \subseteq U_{K}^{(s - 1)} \subseteq \dots \subseteq U_{K}^{(0)} = U_{K} .

Proposition 13.11.

(i) $U_{K}^{(0)} ∕ U_{K}^{(i)} ≅ (k^{\times}, \times)$ ( $k ≅ O_{K} ∕ π$ )
(ii) $U_{K}^{(s)} ∕ U_{K}^{(s + 1)} ≅ (k, +)$ for $s \geq 1$

Proof.

(i) Reduction modulo $π$ . $O_{K}^{\times} \to k^{\times}$ is surfective with kernel $1 + π O_{K} = U_{K}^{(1)}$ .
(ii) $f; U_{K}^{(s)} \to k$ , $1 + π^{s} x \mapsto x m o d π$ .
$(1 + π^{s} x) (1 + π^{s} y) = 1 + π^{s} (x + y + π^{s} x y) .$

$x + y + π^{s} x y \equiv x + y mod π$ , hence $f$ is a group homomorphism, surjective with kernel $U_{K}^{(s + 1)}$ .

□

Remark. Let $[K : ℚ_{p}] < \infty$ . Proposition 13.9, ?? implies that there exists finite index subgroup of $O_{K}^{\times}$ isomorphism to $(O_{K}, +)$ .

Example. $ℤ_{p}$ , $p > 2$ , $e = 1$ , take $r = 1$ . Then

\begin{array}{l} {ℤ_{p}}^{\times} & \tilde{\to} {(ℤ ∕ p ℤ)}^{\times} \times (1 + p ℤ_{p}) ≅ ℤ ∕ (p - 1) ℤ \times ℤ_{p} \\ x & \mapsto (x mod p, \frac{x}{[x mod p]}) \end{array}

$p = 2$ , take $r = 2$ .

\begin{array}{l} ℤ_{2}^{\times} & \tilde{\to} {(ℤ ∕ 4 ℤ)}^{\times} \times (1 + p^{2} ℤ_{p}) ≅ ℤ ∕ 2 ℤ \times ℤ_{2} \\ x & \mapsto (x mod 4, \frac{x}{𝜀 (x)}) \end{array}

where

𝜀 (x) = {\begin{matrix} + 1 & x \equiv 1 (m o d 4) \\ - 1 & x \equiv - 1 (m o d 4) \end{matrix}

So:

{ℤ_{p}}^{\times} ∕ {({ℤ_{p}}^{\times})}^{2} ≅ {\begin{matrix} ℤ ∕ 2 ℤ & if p > 2 \\ {(ℤ ∕ 2 ℤ)}^{2} & if p = 2 \end{matrix}