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A useful version is the strong form of Hensel lemma. Let $K$ be complete for a Non-Archimedean absolute value, let $f\in {\mathcal{O}}_K[X]$, and suppose $a\in {\mathcal{O}}_K$ satisfiesThen there is a unique root $\alpha$ in the ball $|\alpha -a|<|f^{\prime }(a)|$, and $|\alpha -a|\le |f(a)/f^{\prime }(a)|$.

Put $h=|f^{\prime }(a)|$ and $\eta =|f(a)|/h^2<1$. Starting with $x_0=a$, apply Newton iteration over a valued field $x_{n+1}=x_n-f(x_n)/f^{\prime }(x_n)$. Taylor expansion over the valuation ring givesfor integral $x,u$. Inductively $|f^{\prime }(x_n)|=h$ and $|f(x_n)|/h^2\le {\eta }^{2^n}$. Indeed the Newton increment has size at most $h{\eta }^{2^n}<h$, so the derivative size is unchanged, and the new function value is bounded by the square of the increment. The increments tend to zero, and the ultrametric inequality makes the sequence Cauchy. Completeness supplies a root with the stated distance bound.

If $b,c$ are distinct points of the ball, their divided difference is $f^{\prime }(a)$ plus a term of size at most $\max (|b-a|,|c-a|)<h$, so $(f(b)-f(c))/(b-c)$ is nonzero. Two roots in that ball are impossible. This proves the stated Hensel lemma. In particular a simple root modulo the maximal ideal lifts uniquely to a root in its residue class.

An absolute value on a field is a map $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ satisfyingA Non-Archimedean absolute value satisfies the stronger ultrametric inequality $|x+y|\le \max (|x|,|y|)$. Two equivalent absolute values induce the same topology, or equivalently differ by a positive real power. The trivial absolute value on a field takes value one on every nonzero element. The rational classification and the compactness criterion below concern nontrivial absolute values on a field; the trivial exceptions are given explicitly.

Here the additive valuation has real values. For any $c>1$, the mutually inverse constructions areMultiplicativity becomes $v(xy)=v(x)+v(y)$, and the ultrametric inequality becomes $v(x+y)\ge \min (v(x),v(y))$. Equivalent real-valued valuations differ by positive scaling, so these constructions give the required bijection on equivalence classes. Changing $c$ merely rescales the valuation.

To justify the topology formulation, $|a|<1$ is equivalent to $a^n\to 0$. Thus two nontrivial absolute values on a field with the same topology give the same strict positivity relation on their additive valuations. Fix $t$ with $v_1(t)>0$. Comparing the signs of $m{v}_j(t)-n{v}_j(x)=v_j(t^m{x}^{-n})$, for integers $m$ and positive integers $n$, shows that $v_1(x)/v_1(t)$ and $v_2(x)/v_2(t)$ have identical rational cuts. They are equal, proving $v_2=c{v}_1$ for $c>0$. If one allows valuations in arbitrary ordered groups, the nontrivial classes arising this way are precisely rank-one valuations: higher-rank ordered value groups do not embed order-preservingly in $\mathbb{R}$.

For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, $|n|\le 1$ for every integer $n$, by repeatedly applying the ultrametric inequality to sums of ones. Some prime $p$ must have $|p|<1$, otherwise prime factorization and multiplicativity would make every nonzero rational have value one. There is at most one such prime: if both $|p|,|q|<1$, a Bezout identity $ap+bq=1$ contradicts the ultrametric inequality. If $p\nmid m$, another Bezout identity gives $|m|=1$. HenceThis proves the non-Archimedean part of the Ostrowski theorem. If the trivial absolute value on a field is admitted, it supplies one additional class and is not equivalent to any p-adic absolute value.

The valuation ring, its maximal ideal, and its residue field areSuppose the absolute value on a field is nontrivial and $R$ is compact. The ideal $\mathfrak{m}$ is an open additive subgroup of $R$, so $k$ is discrete; as a continuous image of a compact space it is finite. The subgroup $\mathfrak{m}$ is also closed, since all its cosets are open, and is therefore compact. The continuous function $|\cdot |$ attains a maximum $\rho$ on $\mathfrak{m}$, with $0<\rho <1$. Choose $\pi$ with $|\pi |=\rho$. Then the positive values of $v=-\log |\cdot |$ have least element $-\log \rho$. Division with remainder in this additive subgroup of $\mathbb{R}$ proves $v(K^{\times })=v(\pi )\mathbb{Z}$. Thus $K$ is a discretely valued field and $\pi$ is a uniformizer.

Conversely, normalize the discrete valuation by $v(\pi )=1$. If $k$ has $q$ elements, $R/{\pi }^{N}R$ has $q^N$ elements. Each quotient therefore supplies a finite cover by balls of radius tending to zero. The ring $R$ is a closed subset of the complete metric field $K$, hence complete and totally bounded, so it is compact. Equivalently,This is the local compactness criterion for a complete non-Archimedean field, with nontriviality understood. For the trivial absolute value on a field, $R=K$ has the discrete topology and is compact exactly when $K$ is a finite field; its value group is zero rather than a nonzero discrete cyclic group.

A number-field completion is finite. Let $F$ be the given number field and $L=F_{\mathfrak{p}}$ its completion. Normalize its absolute value on a field so that the restriction to $\mathbb{Q}$ is $|\cdot {|}_p$: if $v_{\mathfrak{p}}(p)=e$, use $|x|=p^{-v_{\mathfrak{p}}(x)/e}$. Changing the original absolute value by this positive power does not change its completion. The closure of $\mathbb{Q}$ in $L$ is an isometric copy of ${\mathbb{Q}}_p$.

Take a $\mathbb{Q}$-basis of $F$ and let $V$ be its ${\mathbb{Q}}_p$-span in $L$. Choose a maximal linearly independent subset of that finite spanning set. The compact-sphere proof of finite-dimensional non-Archimedean norm equivalence applies to this basis too, so $V$ is complete and therefore closed in $L$. Since $F\subset V$ is dense in $L$, it follows that $V=L$. Hence

Every finite extension is obtained this way. Let $E/{\mathbb{Q}}_p$ be finite. Use its extending Non-Archimedean absolute value, whose completeness and uniqueness were proved above. In characteristic zero the primitive element theorem gives $E={\mathbb{Q}}_p(\alpha )$; scaling $\alpha$ by a power of $p$, we may arrange that it is integral. Its monic minimal polynomial $f$ belongs to ${\mathbb{Z}}_p[T]$. If the degree is one, $E={\mathbb{Q}}_p$ is already the completion of $\mathbb{Q}$.

Otherwise choosewhere ${\alpha }^{\prime }$ runs over the conjugates in an algebraic closure. Approximate the coefficients of $f$ by integers, obtaining a monic $g\in \mathbb{Z}[T]$ of the same degree. Since $f^{\prime }(\alpha )\ne 0$, sufficiently accurate approximation gives $|g(\alpha )|<|g^{\prime }(\alpha ){|}^2$ and makes $|g(\alpha )/g^{\prime }(\alpha )|<\delta$. The strong form of Hensel lemma in $E$ produces a root $\beta \in E$ of $g$ with $|\beta -\alpha |<\delta$.

For completeness, the required Krasner's lemma follows from the unique extension of an absolute value to a finite extension. In a finite Galois extension containing $\alpha ,\beta$, every automorphism $\sigma$ fixing $\beta$ is an isometry, andIf $\sigma \alpha \ne \alpha$, this contradicts the choice of $\delta$. Thus every automorphism fixing $\beta$ fixes $\alpha$, and Galois correspondence gives ${\mathbb{Q}}_p(\alpha )\subseteq {\mathbb{Q}}_p(\beta )$. Since $\beta \in E$, equality follows. In particular $g$ is irreducible over ${\mathbb{Q}}_p$, and therefore over $\mathbb{Q}$.

Put $F=\mathbb{Q}(\beta )\subset E$. It is a number field, and the closure of $F$ in $E$ contains ${\mathbb{Q}}_p(\beta )=E$, so $F$ is dense. LetEvery element of the ring of integers of a number field has absolute value at most one: otherwise its highest-degree term would strictly dominate all other terms of a monic integer equation. Thus reduction into the residue field of $E$ makes $\mathfrak{p}$ a prime ideal, and $\mathfrak{p}\cap \mathbb{Z}=p\mathbb{Z}$. The restriction of the valuation to $F$ is the valuation of this prime, up to normalization. Indeed, the localized discrete valuation ring $({\mathcal{O}}_F{)}_{\mathfrak{p}}$ maps into ${\mathcal{O}}_E$, its units have absolute value one, and its uniformizer has absolute value less than one. Density and completeness giveThis proves the local realization as a completion of a number field.

The non-Archimedean part of the Ostrowski theorem says that every nontrivial Non-Archimedean absolute value on $\mathbb{Q}$ is equivalent to the p-adic absolute value for a unique prime $p$.

Indeed, $|n|\le 1$ for every integer $n$. Nontriviality gives a prime $p$ with $|p|<1$, and there cannot be two such primes because the Bezout identity would make $1$ a sum of two terms of absolute value below one. If $a$ is coprime to $p$, another Bezout identity shows $|a|=1$. Consequentlyfor every $x\in {\mathbb{Q}}^{\times }$, which is a positive real power of $|x{|}_p$.

More generally, the Non-Archimedean absolute values on a number field $K$ are indexed, up to equivalence, by the nonzero prime ideals $\mathfrak{p}\subset {\mathcal{O}}_K$. The value attached to $\mathfrak{p}$ isTo prove completeness of the list, restrict an absolute value to $\mathbb{Q}$ and obtain a rational prime $p$. Its valuation ring contains ${\mathcal{O}}_K$ away from a unique prime above $p$; equivalently, its centeris a nonzero prime ideal. Since ${\mathcal{O}}_{K,\mathfrak{p}}$ is a discrete valuation ring, every $x\in K^{\times }$ is a unit times a power of a uniformizer, so the given value is equivalent to the displayed $\mathfrak{p}$-adic value. An absolute value trivial on $\mathbb{Q}$ is trivial on the algebraic extension $K$, so no further cases occur.

One direction follows by restriction. Conversely, suppose $|\cdot |$ is a Non-Archimedean absolute value. Then $|m{|}_L=|m|\le 1$ for every integer $m$. For $x,y\in L$, the binomial theorem and the ordinary triangle inequality giveTaking $n$th roots and letting $n\to \infty$ yields the ultrametric inequality for $|\cdot {|}_L$. Thus an extension of an absolute value is non-Archimedean exactly when its restriction is.

An absolute value on a field is a map $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ satisfying $|x|=0\iff x=0$, $|xy|=|x||y|$, and $|x+y|\le |x|+|y|$. It is non-Archimedean when the stronger inequality $|x+y|\le \max (|x|,|y|)$ holds.

If $\mathrm{char}K=p>0$, thenThe ordinary triangle inequality givesLetting $r\to \infty$ proves the strong triangle inequality.

Suppose a Non-Archimedean absolute value extends uniquely to every finite extension. If $f\in {\mathcal{O}}_K[X]$ is monic and irreducible, every root has absolute value at most one: otherwise its leading power dominates all lower terms, contradicting $f(\alpha )=0$ by the ultrametric inequality. In a splitting field, let $\phi$ be the minimal polynomial of an algebraic element of the residue of one root. Lift its coefficients to ${\mathcal{O}}_K$. The lift has absolute value less than one at that root and hence, by equal absolute values of algebraic conjugates, at every root. Consequently every root residue is a zero of $\phi$. Since reduction preserves the product of the linear factors with multiplicities, every irreducible factor of $\overline{f}$ is $\phi$, proving the displayed identity for some $m\ge 1$.

A nontrivial additive valuation has rank one when its ordered value group is Archimedean and therefore embeds order-preservingly in $\mathbb{R}$. Exponentiating a real-valued representative gives a Non-Archimedean absolute value; changing the embedding rescales the valuation positively. Higher-rank ordered value groups cannot be represented by real-valued absolute values on a field.

Let $L/K$ be a finite unramified extension of complete fields with a Non-Archimedean absolute value, with separable residue extension and $[L:K]=[k_L:k_K]=n$. If $x$ is integral and its residue generates $k_L/k_K$, then $1,x,\dots ,x^{n-1}$ form a $K$-basis. Divide any nonzero combination by a coefficient of largest absolute value. Its remaining coefficients are integral, at least one is a unit, and the residue-basis independence makes the reduction nonzero. The displayed norm formula follows. In particular, the combination is integral exactly when every coefficient is integral, proving ${\mathcal{O}}_L={\mathcal{O}}_K[x]$ without requiring a discrete valuation.

The ultrametric inequality, also called the strong triangle inequality, bounds a direct distance by the larger of the two distances through an intermediate point:A metric space satisfying this rule is an ultrametric space. For a Non-Archimedean absolute value, the corresponding inequality is $|a+b|\le \max \{|a|,|b|\}$, and the associated metric is $d(x,y)=|x-y|$. The ordinary triangle inequality follows because the maximum of two nonnegative real numbers is at most their sum.

For example, the 2-adic absolute value gives the three points $0,1,8$ distancesEach distance is bounded by the maximum of the other two. The two largest distances are equal: every triangle in an ultrametric space has this property. Otherwise its unique longest side would exceed the maximum of the other two, contradicting the ultrametric inequality. The drawn lengths of a triangle on paper need not represent these distances.

The ultrametric inequality gives a useful exact rule when $|a|\ne |b|$:To see why, suppose $|a|>|b|$. The upper bound is $|a+b|\le |a|$; applying the ultrametric inequality to $a=(a+b)-b$ forces the reverse bound. Equal sizes can instead cancel: $|1{|}_2=|1{|}_2=1$, but $|1+1{|}_2=1/2$. This makes a uniquely largest term decisive in a sum, a principle used in the Newton polygon and in arguments about roots of a polynomial.

Choose ordered minimal topological generators $a_1,\dots ,a_d$ of a uniform pro-p group for odd $p$, put $b_i=a_i-1$, and choose $1/(p-1)<s<1$. Ordered monomials $b^{\alpha }=b_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ give power-series coordinates in the integral completed group algebra. The displayed Non-Archimedean absolute value filtration gives a complete normed ${\mathbb{Q}}_p$-algebra after completion of the rational group algebra. The power relation raises valuation by one, and interchanging $b_i,b_j$ adds terms of valuation at least $1+s>2s$, so multiplication does not decrease total valuation. The p-adic logarithm and p-adic exponential are inverse on the relevant small balls. In particular $\log G$ is a faithful realization of the Lie algebra of a uniform pro-p group, useful for verifying its intrinsic limit operations and the Baker--Campbell--Hausdorff formula without assuming the correspondence in advance.