The p-adic numbers - Local Fields

3 The $p$ -adic numbers

Recall that $ℚ_{p}$ is the completion of $ℚ$ with respect to ${| ∙ |}_{p}$ . On Example Sheet 1, we will show that $ℚ_{p}$ is a field. We also show that ${| ∙ |}_{p}$ extends to $ℚ_{p}$ and the associated valuation is discrete.

Definition 3.1. The ring of $p$ -adic integers $ℤ_{p}$ is the valuation ring

ℤ_{p} = {x \in ℚ_{p} | {| x |}_{p} \leq 1} .

Facts: $ℤ_{p}$ is a discrete valuation ring, with maximal ideal $p ℤ_{p}$ , and non-zero ideals are given by $p^{n} ℤ_{p}$ .

Proposition. $ℤ_{p}$ is the closure of $ℤ$ inside $ℚ_{p}$ . In particular, $ℤ_{p}$ is the completion of $ℤ$ with respect to ${| ∙ |}_{p}$ .

Proof. Need to show $ℤ$ is dense in $ℤ_{p}$ . Note $ℚ$ is dense in $ℚ_{p}$ . Since $ℤ_{p} \subseteq ℚ_{p}$ is open, we have that $ℤ_{p} \cap ℚ$ is dense in $ℤ_{p}$ . Now:

ℤ_{p} \cap ℚ = {x \in ℚ | {| x |}_{p} \leq 1} = {\frac{a}{b} \in ℚ | p ∤ b} = ℤ_{(p)} .

Thus it suffices to show $ℤ$ is dense in $ℤ_{(p)}$ .

Let $\frac{a}{b} \in ℤ_{(p)}$ , $a, b \in ℤ$ , $p ∤ b$ . For $n \in ℕ$ , choose $y_{n} \in ℤ$ such that $b y_{n} \equiv a (m o d p^{n})$ . THen $y_{n} \to \frac{a}{b}$ as $n \to \infty$ .

In particular, $ℤ_{p}$ is complete and $ℤ \subseteq ℤ_{p}$ is dense. □

Definition (Inverse limit). Let ${(A_{n})}_{n = 1}^{\infty}$ be a sequence of sets / groups / rings together with homomorphisms $φ_{n} : A_{n + 1} \to A_{n}$ (transition maps). Then the inverse limit of ${(A_{n})}_{n = 1}^{\infty}$ is the set / group / ring defined by

\lim_{\overset{[}{n}] \leftarrow} A_{n} = {{(a_{n})}_{n = 1}^{\infty} \in \prod_{n = 1}^{\infty} A_{n} | φ (a_{n + 1}) = a_{n} \forall n} .

Define the group / ring operation componentwise.

Notation. Let $𝜃_{m} : \lim_{\overset{[}{n}] \leftarrow} A_{n} \to A_{m}$ denote the natural projection.

The inverse limit satisfies the following universal property:

Proposition 3.2 (Universal property of inverse limits). Assuming that:

$B$ is a set / group / ring
$ψ_{n}$ are homomorphisms $ψ_{n} : B \to A_{n}$ such that
commutes for all $n$

Then there exists a unique homomorphism

ψ : B \to \lim_{\overset{[}{n}] \leftarrow} A_{n}

such that

𝜃_{n} \circ ψ = ψ_{n}

Proof. Define

\begin{array}{l} ψ : B & \to \prod_{n = 1}^{\infty} A_{n} \\ b & \mapsto \prod_{n = 1}^{\infty} ψ_{n} (b) \end{array}

Then $ψ_{n} = φ_{n} \circ ψ_{n + 1}$ implies that $ψ (b) \in \lim_{\overset{[}{n}] \leftarrow} A_{n}$ . The map is clearly unique (determined by $ψ_{n} = 𝜃_{n} \circ ψ$ ) and is a homomorphism of sets / groups / rings. □

Definition 3.3 ( $I$ -adic completion). Let $I \subseteq R$ be an ideal ( $R$ a ring). The $I$ -adic completion of $R$ is the

\hat{R} : = \lim_{\overset{[}{R}] \leftarrow} ∕ I^{n}

where $R ∕ I^{n + 1} \to R ∕ I^{n}$ is the natural projection.

Note that there exists a natural map $i : R \to \hat{R}$ by the Universal property of inverse limits (there exist maps $R \to R ∕ I^{n}$ ). We say $R$ is $I$ -adically complete if it is an isomorphism.

Fact: $\ker (i : R \to \hat{R}) = ⋂_{n = 1}^{\infty} I^{n}$ .

Let $(K, | ∙ |)$ be a non-archimedean valued fieldand $π \in O_{K}$ such that $| π | < 1$ .

Proposition 3.4. Assuming that:

$K$ is complete with respect to $| ∙ |$

Then

(i) Then $O_{K} ≅ \lim_{\overset{[}{n}] \leftarrow} O_{K} ∕ π^{n} O_{K}$ ( $O_{K}$ is $π$ -adically complete)
(ii) Every $x \in O_{K}$ can be written uniquely as $x = \sum_{i = 0}^{n} a_{i} π^{i}$ , $a_{i} \in A$ , where $A \subseteq O_{K}$ is a set of coset representatives for $O_{K} ∕ π O_{K}$ .

Proof.

(i) $K$ is complete and $O_{K}$ is closed, so $O_{K}$ is complete.
$x \in ⋂_{n = 1}^{\infty} π^{n} O_{K}$ impies $v (x) \geq n v (π)$ for all $n$ , and hence $x = 0$ . Hence $O_{K} \to {\lim_{\overset{[}{O}] \leftarrow}}_{K} ∕ π^{n} O_{K}$ is injective.

Let ${(x_{n})}_{n = 1}^{\infty} \in {\lim_{\overset{[}{O}] \leftarrow}}_{K} ∕ π^{n} O_{K}$ and for each $n$ , let $y_{n} \in O_{K}$ be a lifting of $x_{n} \in O_{K} ∕ π^{n} O_{K}$ . Then $y_{n} - y_{n + 1} \in π^{n} O_{K}$ so that $v (y_{n} - y_{n + 1}) \geq n v (π)$ .

Thus ${(y_{n})}_{n = 1}^{\infty}$ is a Cauchy sequence in $O_{K}$ . Let $y_{n} \to y \in O_{K}$ . Then $y$ maps to ${(x_{n})}_{n = 1}^{\infty}$ in the $\lim_{\overset{[}{n}] \leftarrow} O_{K} ∕ π^{n} O_{K}$ . Thus $O_{K} \to \lim_{\overset{[}{n}] \leftarrow} O_{K} ∕ π^{n} O_{K}$ is surjective.
(ii) Exercise on Example Sheet 1. □

Corollary 3.5.

(i) $ℤ_{p} ≅ \lim_{\overset{[}{ℤ}] \leftarrow} ∕ p^{n} ℤ$ .
(ii) Every element $x \in ℚ_{p}$ can be written uniquely as
$x = \sum_{i = n}^{\infty} a_{i} p^{i},$

with $n \in ℤ$ , $a_{i} \in {0, 1, \dots, - 1}$ .

Proof.

(i) It suffices by Proposition 3.4 to show that
$ℤ_{p} ∕ p^{n} ℤ_{p} ≅ ℤ ∕ p^{n} ℤ .$

Let $f_{n} : ℤ \to ℤ_{p} ∕ p^{n} ℤ_{p}$ be the natural map
$\ker (f_{n}) = {x \in ℤ | {| x |}_{p} \leq p^{- n}} = p^{n} ℤ,$

hence $ℤ ∕ p^{n} ℤ \to ℤ_{p} ∕ p^{n} ℤ_{p}$ is injective.

Let $τ \in ℤ_{p} ∕ p^{n} ℤ_{p}$ and let $c \in ℤ_{p}$ be a lift. Since $ℤ$ is dense in $ℤ_{p}$ , there exists $x \in ℤ$ such that $x \in c + p^{n} ℤ_{p}$ is open in $ℤ_{p}$ . Then $f_{n} (x) = τ$ , hence $ℤ ∕ p^{n} ℤ \to ℤ_{p} ∕ p^{n} ℤ_{p}$ is surjective.
(ii) It follows from Proposition 3.4(ii) to $p^{- n} x \in ℤ_{p}$ for some $n \in ℤ$ □

Example.

\frac{1}{1 - p} = 1 + p + p^{2} + p^{3} + \dots

3 The p-adic numbers

3 The $p$ -adic numbers