Teichmuller lifts - Local Fields - Cambridge III notes

5 Teichmüller lifts

Definition 5.1 (Perfect). A ring $R$ of characteristic $p > 0$ (prime) is a perfect ring if the Frobenius $x \mapsto x^{p}$ is a bijection. A field of characteristic $p$ is a perfect field if it is perfect as a ring.

Remark. Since $characteristic R = p$ , ${(x + y)}^{p} = x^{p} + y^{p}$ , so Frobenius is a ring homomorphism.

Example.

(i) $𝔽_{p^{n}}$ and $\bar{𝔽_{p}}$ are perfect fields.
(ii) $𝔽_{p} [t]$ is not perfect, because $t \notin Im (Frob)$ .
(iii) $𝔽_{p} (t^{\frac{1}{p^{\infty}}}) : = 𝔽_{p} (t, t^{\frac{1}{p}}, t^{\frac{1}{p^{2}}}, \dots)$ is a perfect field (called the perfection of $𝔽_{p} (t)$ ).

Fact: A field of characteristic $p > 0$ is perfect if and only if any finite extension of $k$ is separable.

Theorem 5.2. Assuming that:

$(K, | ∙ |)$ is a complete discretely valued field
such that $k : = O_{K} ∕ m$ is a perfect field of characterist $p$

Then there exists a unique map

[∙] : k \to O_{K}

such that

(i) $a \equiv [a] mod m$ for all $a \in k$
(ii) $[a b] = [a] [b]$ for all $a, b \in k$

Moreover if $characteristic O_{K} = p$ , then $[∙]$ is a ring homomorphism.

Definition 5.3. The element $[a] \in O_{K}$ constructed in Theorem 5.2 is the Teichmüller lift of $a$ .

Lemma 5.4. Assuming that:

$(K, | ∙ |)$ is a complete discretely valued field
such that $k : = O_{K} ∕ m$ is a perfect field of characterist $p$
$π \in O_{K}$ a fixed uniformiser
$x, y \in O_{K}$ such that $x \equiv y mod π^{k}$ ( $k \geq 1$ )

Then

x^{p} \equiv y^{p} mod π^{k + 1}

Proof. Let $x = y + u π^{k}$ with $u \in O_{K}$ . Then

\begin{array}{l} x^{p} & = \sum_{i = 0}^{p} (\binom{p}{i}) y^{p - i} {(u π^{k})}^{i} \\ = y^{p} + \sum_{i = 1}^{p} (\binom{p}{i}) y^{p - 1} {(u π^{k})}^{i} \end{array}

Since $O_{K} ∕ π O_{K}$ has characteristic $p$ , we have $p \in π O_{K}$ . Thus

(\binom{p}{i}) {(u π^{k})}^{i} y^{p - i} \in π^{k + 1} O_{K} \forall i \geq 1,

hence $x^{p} \equiv y^{p} mod π^{k + 1}$ . □

Proof of Theorem 5.2. Let $a \in k$ . For each $i \geq 0$ we choose a lift $y_{i} \in O_{K}$ of $a^{\frac{1}{p^{i}}}$ , and we define

x_{i} : = y_{i}^{p_{i}} .

We claim that ${(x_{i})}_{i = 1}^{\infty}$ is a Cauchy sequence and its limit is independent of the choice of $y_{i}$ .

By construction, $y_{i} \equiv y_{i + 1}^{p} mod π$ . By Lemma 5.4 and induction on $k$ , we have $y_{i}^{p^{k}} \equiv y_{i + 1}^{p^{k + 1}}$ and hence $x_{i} \equiv x_{i + 1} mod π^{i + 1}$ (take $i = p$ ). Hence ${(x_{i})}_{i = 1}^{\infty}$ is Cauchy, so $x_{i} \to x \in O_{K}$ .

Suppose ${(x_{i}^{'})}_{i = 1}^{\infty}$ arises from another choice of $y_{i}^{'}$ lifting $a_{i}^{\frac{1}{p^{i}}}$ . Then ${(x_{i}^{'})}_{i = 1}^{\infty}$ is Cauchy, and $x_{i}^{'} \to x^{'} \in O_{K}$ . Let

x_{i}^{″} = {\begin{matrix} x_{i} & i even \\ x_{i}^{'} & i odd \end{matrix} .

Then $x_{i}^{″}$ arises from lifting

y_{i}^{″} = {\begin{matrix} y_{i} & i even \\ y_{i} & i odd \end{matrix} .

Then $x_{i}^{″}$ is Cauchy and $x_{i}^{″} \to x$ , $x_{i}^{″} \to x^{'}$ . So $x = x^{'}$ and hence $x$ is independet of the choice of $y_{i}$ . So we may define $[a] = x$ .

Then $x_{i} = y_{i}^{p^{i}} \equiv {(a^{\frac{1}{p^{i}}})}^{p^{i}} \equiv a mod π$ . Hence $x \equiv a mod π$ . So (i) is satisfied.

We let $b \in k$ and we choose $u_{i} \in O_{K}$ a lift of $b^{\frac{1}{p^{i}}}$ , and let $z_{i} : = u_{i}^{p^{i}} t$ . Then $\lim_{i \to \infty} z_{i} = [b]$ .

Now $u_{i} y_{i}$ is a lift of ${(a b)}^{\frac{1}{p^{i}}}$ , hence

[a b] = \lim_{i \to \infty} x_{i} z_{i} = (\lim_{i \to \infty} x_{i}) (\lim_{i \to \infty} z_{i}) = [a] [b] .

So (ii) is satisfied.

If $characteristic K = p$ , $y_{i} + u_{i}$ is a lift of $a^{\frac{1}{p^{i}}} + b^{\frac{1}{p^{i}}} = {(a + b)}^{\frac{1}{p^{i}}}$ . Then

\begin{array}{l} [a + b] & = \lim_{i \to \infty} {(y_{i} + u_{i})}^{p^{i}} \\ = \lim_{i \to \infty} y_{i}^{p^{i}} + u_{i}^{p^{i}} \\ = \lim_{i \to \infty} x_{i} + z_{i} \\ = [a] + [b] \end{array}

Easy to check that $[0] = 0$ , $[1] = 1$ , and hence $[∙]$ is a ring homomorphism.

Uniqueness: let $ϕ : k \to O_{K}$ be another such map. Then for $a \in k$ , $ϕ (a^{\frac{1}{p^{i}}})$ is a lift of $a^{\frac{1}{p^{i}}}$ . It follows that

\begin{array}{l} [a] & = \lim_{i \to \infty} ϕ {(a^{\frac{1}{p^{i}}})}^{p^{i}} \\ = \lim_{i \to \infty} ϕ (a) \\ = ϕ (a) □ \end{array}

Example. $K = ℚ_{p}$ , $[∙] : 𝔽_{p} \to ℤ_{p}$ , $a \in 𝔽_{p}^{\times}$ , ${[a]}^{p - 1} = [a^{p - 1}] = [1] = 1$ . So $[a]$ is a $(p - 1)$ -th root of unity.

Lemma 5.5. Assuming that:

$(K, | ∙ |)$ complete discretely valued field
$k = O_{K} ∕ m \subseteq \bar{𝔽_{p}}$
$a \in k^{\times}$

Then

[a]

is a root of unity.

Proof.

\begin{array}{l} a \in k^{\times} & ⟹ a \in 𝔽_{p^{n}}^{\times} for some n \\ ⟹ {[a]}^{p^{n} - 1} = [a^{p^{n} - 1}] = [1] = 1 □ \end{array}

Theorem 5.6. Assuming that:

$(K, | ∙ |)$ complete discretely valued field
$characteristic (K) = p > 0$
$k$ is perfect

Then

K = k ((t))

(

k = O_{K} ∕ m

Proof. Since $K = Frac (O_{K})$ , it suffices to show $O_{K} ≅ k [[t]$ . Fix $π \in O_{K}$ a uniformiser, and let $[∙] : k \to O_{K}$ be the Teichmüller map and define

\begin{array}{l} φ : k [[t]] & \to O_{K} \\ φ (\sum_{i = 0}^{\infty} a_{i} t^{i}) & = \sum_{i = 0}^{\infty} [a_{i}] π^{i} \end{array}

Then $φ$ is a ring homomorphism since $[∙]$ is, and it is a bijection by Proposition 3.4(ii). □