Absolute Values, Valuations and Completions

Introduction

An absolute value on a field is a real-valued size function compatible with multiplication and satisfying the triangle inequality; it turns the field into a metric space and hence into a topological field, and it makes possible the operation of completion that produces $\mathbb{R}$ from $\mathbb{Q}$ and $\mathbb{Q}_p$ from $\mathbb{Q}$. The absolute values that satisfy the stronger ultrametric inequality are the non-Archimedean ones, and these are equivalent to the valuations of commutative algebra: maps to an ordered abelian group that measure divisibility. Ostrowski's theorem states that every nontrivial absolute value on $\mathbb{Q}$ is equivalent either to the usual one or to one of the $p$-adic ones, so the completions of $\mathbb{Q}$ are exactly $\mathbb{R}$ and the fields $\mathbb{Q}_p$.

This article develops absolute values, the ultrametric geometry of the non-Archimedean case, valuations and their rings, Ostrowski's theorem, and the completion of a valued field, with $\mathbb{Q}_p$ as the central example. The general theory of topological rings and of the $I$-adic completion is from Topological Rings and Fields; the two have in common the inverse limit description of the completion, and they differ in that on a field the only $I$-adic topologies are the trivial ones, so the field case must proceed through a metric or a valuation.

Throughout, $F$ is a field, $\lvert \cdot \rvert$ an absolute value on it, and $v$ a valuation with value group $\Gamma$. The completion is written $\widehat{F}$, and $\mathbb{Q}_p$, $\mathbb{Z}_p$ and $\mathbb{F}_p$ retain their standard meanings.

Absolute Values

Definition and Examples

Definition. An absolute value on a field $F$ is a function $\lvert \cdot \rvert : F \to \mathbb{R}_{\geq 0}$ such that for all $x, y \in F$:

(AV1) $\lvert x \rvert = 0$ if and only if $x = 0$;

(AV2) $\lvert xy \rvert = \lvert x \rvert \lvert y \rvert$;

(AV3) $\lvert x + y \rvert \leq \lvert x \rvert + \lvert y \rvert$.

The absolute value is non-Archimedean if it satisfies the stronger ultrametric inequality

(AV3') $\lvert x + y \rvert \leq \max\{\lvert x \rvert, \lvert y \rvert\}$,

and Archimedean otherwise. It is trivial if $\lvert x \rvert = 1$ for all $x \neq 0$, and discrete if the subgroup $\lvert F^\times \rvert \subseteq \mathbb{R}_{>0}$ is discrete.

Proposition. Let $\lvert \cdot \rvert$ be an absolute value on $F$.

(a) $\lvert 1 \rvert = 1$ and $\lvert -x \rvert = \lvert x \rvert$; more generally $\lvert \zeta \rvert = 1$ for every root of unity $\zeta$.

(b) $\lvert x^{-1} \rvert = \lvert x \rvert^{-1}$ for $x \neq 0$.

(c) $\lvert x^n \rvert = \lvert x \rvert^n$ for $n \in \mathbb{Z}$, and $\lvert \lvert x \rvert - \lvert y \rvert \rvert \leq \lvert x - y \rvert$.

(d) $d(x,y) = \lvert x - y \rvert$ is a metric on $F$, and $F$ with the induced topology is a topological field.

(e) $\lvert \cdot \rvert$ satisfies (AV3') if and only if $\lvert n \cdot 1 \rvert \leq 1$ for every integer $n$.

Proof. (a) $\lvert 1 \rvert = \lvert 1 \rvert^2$ and $\lvert 1 \rvert \neq 0$; $\lvert -x \rvert^2 = \lvert x^2 \rvert = \lvert x \rvert^2$. (b) $1 = \lvert x x^{-1} \rvert$. (c) Induction gives the first, and the second is the usual reverse triangle inequality. (d) The metric axioms follow from (AV1)-(AV3); continuity of addition and multiplication follows from the estimates $\lvert (x+h) - (x+h') \rvert \leq \lvert h \rvert + \lvert h' \rvert$ and $\lvert xy - x'y' \rvert \leq \lvert x \rvert \lvert y - y' \rvert + \lvert y' \rvert \lvert x - x' \rvert$, and continuity of inversion from $\lvert x^{-1} - y^{-1} \rvert = \lvert x - y \rvert / \lvert xy \rvert$ with $\lvert x \rvert$ bounded below near $x \neq 0$. (e) If (AV3') holds then $\lvert n \cdot 1 \rvert \leq 1$ by induction. Conversely, if $\lvert n \cdot 1 \rvert \leq 1$ for all $n$, the binomial expansion gives $\lvert x + y \rvert^n \leq (n+1) \max\{\lvert x \rvert, \lvert y \rvert\}^n$, and taking $n$-th roots and letting $n \to \infty$ gives (AV3').

Standard Examples

Example (the usual absolute value). On $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ the usual absolute value (or modulus) is Archimedean, and $\lvert n \cdot 1 \rvert = n$, so it is not bounded on the integers.

Example ($p$-adic absolute values). Fix a prime $p$. Every $x \in \mathbb{Q}^\times$ factors uniquely as

$$ x = \pm \prod_p p^{a_p(x)}, \qquad a_p(x) \in \mathbb{Z},\ \text{finitely many nonzero}, $$

and the $p$-adic valuation is $v_p(x) = a_p(x)$, with $v_p(0) = \infty$. The $p$-adic absolute value is

$$ \lvert x \rvert_p = p^{-v_p(x)} \ (\text{for } x \neq 0), \qquad \lvert 0 \rvert_p = 0 . $$

It is non-Archimedean and discrete: $\lvert p \rvert_p = 1/p$ and $\lvert n \rvert_p = 1$ for $n$ coprime to $p$.

Example (function fields). On $k(t)$ the $t$-adic valuation $v_t(f)$ is the order of vanishing of $f$ at $t = 0$, and $\lvert f \rvert = c^{-v_t(f)}$ for a fixed $c > 1$ is a non-Archimedean absolute value; more generally, each irreducible polynomial and the "point at infinity" give an absolute value.

Example (trivial and discrete). The trivial absolute value gives the discrete topology. Every absolute value of the form $\lvert x \rvert^\alpha$ with $0 < \alpha \leq 1$ is again an absolute value, Archimedean or not accordingly.

Equivalence and the Product Formula

Definition. Two absolute values $\lvert \cdot \rvert_1$, $\lvert \cdot \rvert_2$ on $F$ are equivalent if they induce the same topology on $F$.

Theorem. Two nontrivial absolute values on a field $F$ are equivalent if and only if there is a real number $\alpha > 0$ with

$$ \lvert x \rvert_2 = \lvert x \rvert_1^{\alpha} \quad \text{for all } x \in F . $$

Proof sketch. If the relation holds, the two absolute values have the same balls up to scaling of the radius, hence the same topology. Conversely, if the topologies agree, then $\lvert x \rvert_1 < 1$ implies $\lvert x \rvert_2 < 1$ (a sequence tending to $0$ in one topology tends to $0$ in the other), and one shows that the subgroups $\{x : \lvert x \rvert_1 < 1\}$ and $\{x : \lvert x \rvert_2 < 1\}$ coincide; choosing $a$ with $\lvert a \rvert_1 \neq 0, 1$ and comparing $\lvert a^n \rvert_1$ with $\lvert x \rvert_2$ for all $x$ gives the exponent $\alpha$.

Theorem (product formula for $\mathbb{Q}$). For every $x \in \mathbb{Q}^\times$,

$$ \lvert x \rvert_\infty \prod_{p} \lvert x \rvert_p = 1, $$

where the product is over all primes and the product is finite because $v_p(x) = 0$ for almost all $p$.

Proof. Write $\lvert x \rvert_\infty = \prod_p p^{a_p(x)}$ in the factorization of $x$, which holds because $\lvert x \rvert_\infty$ collects the prime powers with sign. Then $\lvert x \rvert_p = p^{-a_p(x)}$, so the product of the $p$-adic terms is $\prod_p p^{-a_p(x)} = \lvert x \rvert_\infty^{-1}$.

Non-Archimedean Absolute Values

Ultrametric Geometry

Theorem. Let $\lvert \cdot \rvert$ be a non-Archimedean absolute value on $F$.

(a) $\lvert x + y \rvert = \max\{\lvert x \rvert, \lvert y \rvert\}$ whenever $\lvert x \rvert \neq \lvert y \rvert$.

(b) Every triangle is isosceles: for any $x,y,z$, the three numbers $\lvert x - y \rvert$, $\lvert y - z \rvert$, $\lvert z - x \rvert$ have the property that the two largest are equal.

(c) Every point of a closed ball $\{x : \lvert x - a \rvert \leq r\}$ is a centre, and the same for open balls.

(d) Any two balls are either disjoint or one contains the other; if they have the same radius and intersect, they are equal.

(e) Balls are both open and closed, and $F$ is totally disconnected.

(f) A sequence $(x_n)$ is Cauchy if and only if $\lvert x_{n+1} - x_n \rvert \to 0$, and a series $\sum a_n$ converges if and only if $a_n \to 0$.

Proof. (a) Suppose $\lvert x \rvert > \lvert y \rvert$. Then $\lvert x \rvert = \lvert (x+y) - y \rvert \leq \max\{\lvert x+y \rvert, \lvert y \rvert\}$, and since $\lvert y \rvert < \lvert x \rvert$ this forces $\lvert x + y \rvert \geq \lvert x \rvert$; combined with (AV3') gives equality. (b) Apply (a) to the identity $x - z = (x-y) + (y-z)$ and its permutations. (c) If $\lvert x - a \rvert \leq r$ and $\lvert y - a \rvert \leq r$ then $\lvert y - x \rvert \leq r$, so the ball with centre $x$ and radius $r$ is contained in the ball with centre $a$ and radius $r$, and symmetry gives equality. (d) Suppose balls $B(a,r)$ and $B(b,s)$ with $r \leq s$ intersect at $c$; then $\lvert x - b \rvert \leq \max\{\lvert x - a \rvert, \lvert a - c \rvert, \lvert c - b \rvert\} \leq s$ for $x \in B(a,r)$ after the estimates, so $B(a,r) \subseteq B(b,s)$. (e) A ball is open by definition; its complement is a union of balls of the same radius, hence open, so it is closed. Total disconnectedness follows from (d), since the only connected subsets are points. (f) If $\lvert x_{n+1} - x_n \rvert \to 0$ then for $m > n$ the telescoping estimate $\lvert x_m - x_n \rvert \leq \max_{n \leq k < m} \lvert x_{k+1} - x_k \rvert \to 0$ shows the sequence is Cauchy; the converse is immediate. For series, the same telescoping argument applies to the partial sums.

Remark (two uses of "Archimedean"). The word Archimedean occurs in two unrelated senses in this corpus. An ordered field is Archimedean when the natural numbers are unbounded in it, as in Ordered Fields; an absolute value is non-Archimedean when it satisfies the ultrametric inequality, equivalently when it is bounded on the prime subring. A field with a non-Archimedean absolute value may still carry an ordering, as $\mathbb{Q}(t)$ does, and an Archimedean ordered field may carry a non-Archimedean absolute value, as $\mathbb{Q}$ does through $\lvert \cdot \rvert_p$; the two conditions govern different structures. What is true is that the metric topology of a non-Archimedean absolute value is totally disconnected, so it can never coincide with the order topology of an order-complete field.

The Value Group

Definition. Let $\lvert \cdot \rvert$ be a non-Archimedean absolute value on $F$. The value group is the subgroup

$$ \Gamma = \lvert F^\times \rvert \subseteq \mathbb{R}_{>0}, $$

written multiplicatively, and the residue field is $k = \mathcal{O}/\mathrm{M}$ where

$$ \mathcal{O} = \{x \in F : \lvert x \rvert \leq 1\}, \qquad \mathrm{M} = \{x \in F : \lvert x \rvert < 1\}. $$

Proposition. $\mathcal{O}$ is a subring of $F$ containing $1$ and

$$ \mathcal{O} = \{x : \lvert x \rvert \leq 1\}, \qquad \mathcal{O}^\times = \{x : \lvert x \rvert = 1\}, \qquad \mathrm{M} = \{x : \lvert x \rvert < 1\} $$

is its unique maximal ideal; $\mathcal{O}$ is a local ring with residue field $k = \mathcal{O}/\mathrm{M}$, and $F = \operatorname{Frac}(\mathcal{O})$.

Proof. $\mathcal{O}$ is closed under addition by (AV3') and under multiplication by (AV2); the elements of absolute value $1$ are exactly the units, because $\lvert x \rvert = 1$ gives $\lvert x^{-1} \rvert = 1$, and every element of $\mathrm{M}$ is noninvertible since $\lvert x \rvert < 1$ forces $\lvert x^{-1} \rvert > 1$. Every proper ideal is contained in $\mathrm{M}$, because a unit generates the whole ring, so $\mathrm{M}$ is the unique maximal ideal.

Valuations

Definition

Definition. A valuation on a field $F$ is a map

$$ v : F \to \Gamma \cup \{\infty\} $$

where $\Gamma$ is a totally ordered abelian group (written additively) and $\infty$ is a symbol greater than every element of $\Gamma$, such that for all $x, y \in F$:

(V1) $v(x) = \infty$ if and only if $x = 0$;

(V2) $v(xy) = v(x) + v(y)$;

(V3) $v(x + y) \geq \min\{v(x), v(y)\}$.

A valuation is discrete if $\Gamma \cong \mathbb{Z}$, and a rank-one valuation if $\Gamma$ embeds in $\mathbb{R}$.

Theorem (correspondence). Let $F$ be a field and $c$ a real number with $c > 1$. The assignment

$$ v \mapsto \lvert x \rvert = c^{-v(x)} $$

is a bijection between the valuations of $F$ with value group $\Gamma \subseteq \mathbb{R}$ and the non-Archimedean absolute values of $F$, and this correspondence is compatible with equivalence: two valuations give equivalent absolute values exactly when they have the same valuation ring.

Proof. (V2) and (V3) become (AV2) and (AV3') under $v \mapsto c^{-v}$; conversely $\log_c \lvert x \rvert$ recovers $v$ up to the sign convention. Equivalence of absolute values preserves the unit group $\{x : \lvert x \rvert = 1\}$, hence the valuation ring, and conversely equality of valuation rings determines the unit group and the set $\{x : \lvert x \rvert < 1\}$, hence the topology.

Definition. The valuation ring of $v$ is

$$ \mathcal{O}_v = \{x \in F : v(x) \geq 0\}, $$

with maximal ideal $\mathrm{M}_v = \{x : v(x) > 0\}$ and residue field $k(v) = \mathcal{O}_v/\mathrm{M}_v$. A discrete valuation ring (DVR) is the valuation ring of a discrete valuation.

Proposition. Valuation rings are local domains with $F = \operatorname{Frac}(\mathcal{O}_v)$, and $\mathcal{O}_v$ is a maximal proper subring of $F$: if $\mathcal{O}_v \subsetneq R \subseteq F$ with $R$ a subring, then $R = F$.

Proof. The local and domain properties are proved as for $\mathcal{O}$ above. For maximality, let $R$ be a subring with $\mathcal{O}_v \subseteq R \subseteq F$ and $R \neq \mathcal{O}_v$, and let $x \in R \setminus \mathcal{O}_v$; then $v(x) < 0$, so $v(x^{-1}) > 0$ and $x^{-1} \in \mathrm{M}_v \subseteq \mathcal{O}_v \subseteq R$. Given $y \in F$, choose $m \geq 1$ with $v(y) - m\,v(x) \geq 0$, that is, $y x^{-m} \in \mathcal{O}_v \subseteq R$; then $y = (y x^{-m})\,x^{m} \in R$. Hence $R = F$, and $\mathcal{O}_v$ is a maximal proper subring of $F$.

Examples

Example ($p$-adic). On $\mathbb{Q}$, $v_p$ is a discrete valuation with $\mathcal{O} = \mathbb{Z}_{(p)} = \{a/b : p \nmid b\}$, $\mathrm{M} = p\mathbb{Z}_{(p)}$ and residue field $\mathbb{F}_p$; here $\mathbb{Z}_{(p)}$ is the localization of Localization and the Fraction Field.

Example (DVRs). $\mathbb{Z}_{(p)}$ and $k[t]_{(t)}$ are DVRs, as is $\mathbb{Z}_p$ after completion. A DVR is a principal ideal domain with a unique nonzero maximal ideal, and its nonzero ideals form the chain $\mathcal{O} \supsetneq \mathrm{M} \supsetneq \mathrm{M}^2 \supsetneq \cdots$, which is the $I$-adic filtration of Topological Rings and Fields.

Example (higher rank). The valuation on $\mathbb{C}((t))$ trivial on $\mathbb{C}$ has $\Gamma = \mathbb{Z}$, so it is discrete and of rank one. A field of generalized power series whose exponents form the lexicographically ordered group $\mathbb{Z}^2$ carries a valuation with $\Gamma = \mathbb{Z}^2$; that ordered group does not embed in $\mathbb{R}$, since $(0,1)$ is positive while $n(0,1) < (1,0)$ for every $n$, so the valuation is not of rank one and is not given by $-\log_c \lvert \cdot \rvert$ for any absolute value of that field.

Ostrowski's Theorem

Theorem (Ostrowski). Every nontrivial absolute value on $\mathbb{Q}$ is equivalent either to the usual absolute value $\lvert \cdot \rvert_\infty$ or to a $p$-adic absolute value $\lvert \cdot \rvert_p$ for exactly one prime $p$.

Proof sketch. Let $\lvert \cdot \rvert$ be nontrivial on $\mathbb{Q}$. If it is Archimedean, one shows that $\lvert \cdot \rvert = \lvert \cdot \rvert_\infty^\alpha$ for some $\alpha > 0$ by comparing the growth of $\lvert m \rvert$ and $\lvert n \rvert$ for integers, using the binomial expansion as in the proof that an Archimedean absolute value is unbounded on $\mathbb{Z}$; this is the classical argument of Ostrowski. If it is non-Archimedean, then $\lvert n \rvert \leq 1$ for all $n$, and since the absolute value is nontrivial there is a prime $p$ with $\lvert p \rvert < 1$. No second prime $q \neq p$ can satisfy $\lvert q \rvert < 1$: if it did, then from a Bézout relation $1 = ap + bq$ one gets $\lvert 1 \rvert \leq \max\{\lvert a \rvert \lvert p \rvert, \lvert b \rvert \lvert q \rvert\} < 1$, a contradiction. Hence $\lvert n \rvert = 1$ for all $n$ coprime to $p$, and $\lvert \cdot \rvert$ is determined by its value $\lvert p \rvert = c \in (0,1)$, giving $\lvert x \rvert = c^{v_p(x)}$, which is equivalent to $\lvert x \rvert_p$ by the choice $\alpha = \log_p c^{-1}$.

Corollary (the places of $\mathbb{Q}$). The places of $\mathbb{Q}$, that is, the equivalence classes of nontrivial absolute values, are the Archimedean place $\infty$ and the non-Archimedean places $p$ for primes $p$; the product formula of the previous section holds over all of them.

Corollary. The nontrivial completions of $\mathbb{Q}$ are $\mathbb{R}$ and the fields $\mathbb{Q}_p$, one for each prime $p$.

Completions

The Completion of a Valued Field

Definition. Let $F$ be a field with absolute value $\lvert \cdot \rvert$. The completion $\widehat{F}$ is the set of equivalence classes of Cauchy sequences in the metric $d(x,y) = \lvert x - y \rvert$, with the operations defined termwise and the absolute value extended by continuity:

$$ \widehat{\lvert (x_n) \rvert} = \lim_n \lvert x_n \rvert . $$

Theorem. Let $F$ be a field with absolute value $\lvert \cdot \rvert$.

(a) $\widehat{F}$ is a field, $\lvert \cdot \rvert$ extends to an absolute value on it, and $F$ embeds densely in $\widehat{F}$.

(b) $\widehat{F}$ is complete: every Cauchy sequence in $\widehat{F}$ converges.

(c) The absolute value is non-Archimedean exactly when the same is true on $F$. When it is non-Archimedean, so that the valuation ring is defined, both invariants are unchanged: $\Gamma_{\widehat{F}} = \Gamma_F$, and the natural map $\mathcal{O}_F/\mathrm{M}_F \to \mathcal{O}_{\widehat{F}}/\mathrm{M}_{\widehat{F}}$ is an isomorphism. The hypothesis is necessary; in the Archimedean case the value set can grow, since $\sqrt2$ is a value of the completion of $\mathbb{Q}$ at $\lvert \cdot \rvert_\infty$ but not a value on $\mathbb{Q}$.

(d) $\widehat{F}$ is the unique complete field containing $F$ densely, up to an isometry fixing $F$; it satisfies the universal property that every isometric embedding of $F$ into a complete field extends uniquely.

Proof sketch. The termwise operations are well defined on classes because sums and products of Cauchy sequences are Cauchy, and the only obstruction to field axioms is that an element with a Cauchy sequence tending to $0$ must be excluded, which is the definition of the equivalence. Completeness of $\widehat{F}$ is proved by diagonal extraction, and the extension of the absolute value is by continuity. Uniqueness follows by the same argument that identifies two completions of a metric space. For (c), the ultrametric inequality passes to the extension by continuity; if $x = \lim_n x_n \neq 0$ in $\widehat{F}$ then $\lvert x_n - x \rvert < \lvert x \rvert$ for all large $n$, and for those $n$ the ultrametric inequality gives $\lvert x_n \rvert = \lvert x \rvert$, so no new values occur and $\Gamma_{\widehat{F}} = \Gamma_F$; the residue field is unchanged because density supplies, for every $x \in \widehat{F}$ with $\lvert x \rvert \leq 1$, an element $y \in \mathcal{O}_F$ with $\lvert y - x \rvert < 1$.

Proposition. Suppose the topology on $F$ is induced by the $I$-adic topology of a subring $\mathcal{O} \subseteq F$, where $I$ is a proper ideal of $\mathcal{O}$ (as for $\mathbb{Z}_{(p)} \subseteq \mathbb{Q}$ with $I = (p)$, or $k[t]_{(t)} \subseteq k(t)$ with $I = (t)$). Then the metric completion of $\mathcal{O}$ is its $I$-adic completion,

$$ \widehat{\mathcal{O}} = \varprojlim_n \mathcal{O} / I^n , $$

and the completion of $F$ is the fraction field of that ring, $\widehat{F} = \operatorname{Frac}(\widehat{\mathcal{O}})$. In particular $\widehat{\mathbb{Z}}_{(p)} = \mathbb{Z}_p$ and $\widehat{\mathbb{Q}} = \mathbb{Q}_p$, while $\widehat{k[t]}_{(t)} = k[[t]]$ and $\widehat{k(t)} = k((t))$; the completion is taken on the subring $\mathcal{O}$, which has proper nonzero ideals, and not on the field $F$, which has none.

Proof. By hypothesis the two topologies on $F$ agree, so the powers $I^n$ form a fundamental system of neighbourhoods of $0$ in the metric topology on $\mathcal{O}$ as well as in the $I$-adic topology; the Cauchy sequences for the metric and for the filtration therefore coincide, and the completion of $\mathcal{O}$ is the inverse limit of the quotients $\mathcal{O}/I^n$. The absolute value of $F$ extends to the fraction field of that completion, which is therefore the completion of $F$; its valuation ring is $\widehat{\mathcal{O}}$ by the persistence of the residue field, part (c) above.

The $p$-adic Fields

Theorem. The completion $\mathbb{Q}_p$ of $\mathbb{Q}$ at $\lvert \cdot \rvert_p$ is a complete non-Archimedean valued field with

$$ \mathcal{O} = \mathbb{Z}_p, \qquad \mathrm{M} = p\mathbb{Z}_p, \qquad k = \mathbb{F}_p, \qquad \Gamma = \mathbb{Z}, $$

and the residue field is $\mathbb{F}_p$, the value group is $\mathbb{Z}$. Every $x \in \mathbb{Q}_p^\times$ has a unique Laurent expansion

$$ x = \sum_{n \geq n_0} a_n p^n, \qquad a_n \in \{0, 1, \dots, p-1\}, $$

with $n_0 = v_p(x)$; $\mathbb{Z}_p$ corresponds to $n_0 \geq 0$.

Proof. The assertions about the valuation ring, maximal ideal, residue field and value group follow from the corresponding facts for $\mathbb{Z}_{(p)}$ and the persistence of the residue field under completion (part (c) of the completion theorem); the expansion is the $p$-adic analogue of decimal expansion for fractions.

Theorem (Hensel's lemma). Let $F$ be complete with respect to a non-Archimedean absolute value, let $\mathcal{O}$ be its valuation ring, and let $f \in \mathcal{O}[x]$. If there is $a \in \mathcal{O}$ with

$$ \lvert f(a) \rvert < \lvert f'(a) \rvert^2, $$

then there is a unique $b \in \mathcal{O}$ with $f(b) = 0$ and $\lvert b - a \rvert < \lvert f'(a) \rvert$.

Proof sketch. The Newton iteration $a_{n+1} = a_n - f(a_n)/f'(a_n)$ is defined by the hypothesis and is Cauchy for the non-Archimedean absolute value, with $\lvert a_{n+1} - a_n \rvert$ decreasing quadratically; completeness gives a limit $b$, and continuity of $f$ gives $f(b) = 0$. Uniqueness is by the ultrametric estimate on $f(b) - f(b')$.

Corollary (roots of units). Let $p$ be odd and let $u \in \mathbb{Z}_p^\times$. Then $u$ is a square in $\mathbb{Z}_p$ if and only if its image in $\mathbb{F}_p^\times$ is a square.

Proof. Apply Hensel's lemma to $f(x) = x^2 - u$ at a lift $a$ of a square root of the image of $u$ in $\mathbb{F}_p$: since the image is a nonzero square, $f'(a) = 2a$ has absolute value $1$, while $\lvert f(a) \rvert \leq 1/p < 1 = \lvert f'(a) \rvert^2$, so the hypothesis of Hensel's lemma holds.

Theorem (algebraic closure of $\mathbb{Q}_p$). The algebraic closure $\overline{\mathbb{Q}_p}$ is not complete; its completion $\mathbb{C}_p$ is algebraically closed and complete, and $\mathbb{C}_p$ is the smallest algebraically closed and complete extension of $\mathbb{Q}_p$.

Proof sketch. The valuation of $\mathbb{Q}_p$ extends uniquely to $\overline{\mathbb{Q}_p}$; the value group of the algebraic closure is $\mathbb{Q}$ and its residue field is $\overline{\mathbb{F}_p}$, and by part (c) above both are unchanged when the completion is taken, so $\mathbb{C}_p$ has value group $\mathbb{Q}$ and residue field $\overline{\mathbb{F}_p}$. Krasner's lemma shows that the completion of an algebraic closure of a complete non-Archimedean field is algebraically closed.

Completions and the Number Systems

Base field	Absolute value	Completion
$\mathbb{Q}$	$\lvert \cdot \rvert_\infty$	$\mathbb{R}$
$\mathbb{Q}$	$\lvert \cdot \rvert_p$	$\mathbb{Q}_p$
$k(t)$	$t$-adic	$k((t))$
$\overline{\mathbb{Q}_p}$	$p$-adic (extended)	$\mathbb{C}_p$
$\overline{\mathbb{Q}}$	usual (restricted)	$\mathbb{C}$
$\mathbb{C}(t)$	$t$-adic	$\mathbb{C}((t))$

Remark. The completion of $\mathbb{Q}$ at the Archimedean place is $\mathbb{R}$, whose construction is the order-theoretic one of Real-Closed and Complete Ordered Fields and The Real Numbers; the completion at the non-Archimedean places gives the fields $\mathbb{Q}_p$, whose study is the beginning of algebraic number theory. The two constructions agree in making the field complete and, in the non-Archimedean case, in preserving the value group and the residue field, and they differ in that $\mathbb{R}$ is order-complete and real closed while $\mathbb{Q}_p$ is neither orderable nor algebraically closed.

Summary

An absolute value on a field is a multiplicative size function satisfying the triangle inequality; it defines a metric and makes the field a topological field. It is non-Archimedean when the ultrametric inequality $\lvert x+y \rvert \leq \max\{\lvert x \rvert, \lvert y \rvert\}$ holds, equivalently when $\lvert n \cdot 1 \rvert \leq 1$ for all integers $n$. Two nontrivial absolute values are equivalent exactly when they are positive powers of one another, and on $\mathbb{Q}$ Ostrowski's theorem says that the nontrivial absolute values are, up to equivalence, the usual one and the $p$-adic ones; the product formula $\lvert x \rvert_\infty \prod_p \lvert x \rvert_p = 1$ holds for every nonzero rational.

In the non-Archimedean case the geometry is ultrametric: $\lvert x+y \rvert = \max$ when the terms have different sizes, every triangle is isosceles, every point of a ball is a centre, balls are nested or disjoint and are open and closed, the space is totally disconnected, and a series converges exactly when its terms tend to zero. The unit ball is the valuation ring $\mathcal{O}$, a local domain with maximal ideal $\mathrm{M}$ and residue field $k = \mathcal{O}/\mathrm{M}$, and the non-Archimedean absolute values correspond to valuations into ordered abelian groups, with $\mathcal{O}$ a maximal proper subring of $F$ and $F = \operatorname{Frac}(\mathcal{O})$.

Every valued field has a completion, unique up to isometry, which is complete and contains the field densely, and which agrees with the $I$-adic completion when the topology is the $I$-adic topology of a subring; it preserves the non-Archimedean character and, in that case, the value group $\Gamma_{\widehat{F}} = \Gamma_F$ and the residue field, while in the Archimedean case the value set can grow. The completions of $\mathbb{Q}$ are $\mathbb{R}$ and the $p$-adic fields $\mathbb{Q}_p$, with valuation ring $\mathbb{Z}_p$, maximal ideal $p\mathbb{Z}_p$, residue field $\mathbb{F}_p$ and value group $\mathbb{Z}$; Hensel's lemma gives a criterion for lifting simple roots modulo $\mathrm{M}$, and the completion $\mathbb{C}_p$ of the algebraic closure of $\mathbb{Q}_p$ is algebraically closed.

Summary of Notation

Symbol	Meaning
$F$	Field with an absolute value or valuation
$\lvert \cdot \rvert$	Absolute value
$\lvert \cdot \rvert_p$, $\lvert \cdot \rvert_\infty$	$p$-adic and usual absolute values
$v$, $v_p$	Valuation, $p$-adic valuation
$\Gamma$	Value group (multiplicative for absolute values, additive for valuations)
$\Gamma_{\widehat{F}}$	Value group of the completion, equal to $\Gamma_F$ in the non-Archimedean case
$\operatorname{Frac}$	Field of fractions, $\widehat{F} = \operatorname{Frac}(\widehat{\mathcal{O}})$
$\mathcal{O}$, $\mathcal{O}_v$	Valuation ring $\{x : \lvert x \rvert \leq 1\}$
$\mathrm{M}$, $\mathrm{M}_v$	Maximal ideal $\{x : \lvert x \rvert < 1\}$
$k$, $k(v)$	Residue field $\mathcal{O}/\mathrm{M}$
$c$	Base of the exponential, $c > 1$, relating $v$ and $\lvert \cdot \rvert$
$\widehat{F}$	Completion of $F$
$\mathbb{Z}_{(p)}$, $k[t]_{(t)}$	Localizations, examples of DVRs
$\mathbb{Z}_p$, $\mathbb{Q}_p$, $\mathbb{C}_p$	$p$-adic integers, numbers, and completed algebraic closure
$k((t))$	Formal Laurent series field
$d(x,y) = \lvert x-y \rvert$	Induced metric
$f'(x)$	Formal derivative, in Hensel's lemma