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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.

For $a\in k$ and each $n$, choose a lift $x_n\in {\mathcal{O}}_K$ of the unique $p^n$th root $a^{p^{-n}}$, which exists because $k$ is a perfect field. DefineChanging $x_n$ by an element of the maximal ideal changes its $p^n$th power by an element whose valuation tends to infinity, so the limit exists and is independent of all choices. In characteristic $p$, the Frobenius endomorphism satisfies $(x+y{)}^{p^n}=x^{p^n}+y^{p^n}$, making $[\cdot ]$ a ring homomorphism lifting the identity on $k$. If $s:k\to {\mathcal{O}}_K$ is any other such section, thenfor every $n$, and the same limiting construction forces $s(a)=[a]$. This proves uniqueness of the Teichmuller lift.

Choose a uniformizer $t$. Repeatedly subtracting the lift of the residue and dividing by $t$ gives every $x\in {\mathcal{O}}_K$ a unique convergent Teichmuller expansionBecause the lift is a ring map, this identifies ${\mathcal{O}}_K$ with $k[[t]]$ and its fraction field with the Laurent series field $k((t))$. This is the equal-characteristic complete discretely valued field classification.

If $K$ is locally compact, its compact valuation ring has only finitely many disjoint residue-class balls. Thus $k$ is finite, as also follows from the local compactness criterion for a complete non-Archimedean field.

Suppose first that the valuation is discrete and the residue field $k$ is finite. For a uniformizer $\pi$, every quotient ${\mathcal{O}}_K/{\pi }^n{\mathcal{O}}_K$ is finite, and completeness givesThis inverse limit is compact. Since ${\mathcal{O}}_K$ is a compact neighborhood of zero, $K$ is locally compact.

Conversely, local compactness gives a compact ball about zero, which can be rescaled to make ${\mathcal{O}}_K$ compact. Its distinct residue classes are disjoint open balls of radius below one, so compactness forces the residue field to be finite. Cover ${\mathcal{O}}_K$ by finitely many balls of some radius $r<1$. Applying the ultrametric inequality to centers lying in the maximal ideal produces $\rho <1$ such that every nonunit has absolute value at most $\rho$. Hence the value group has a largest value below one, and the valuation is discrete. This proves the local compactness criterion for a complete non-Archimedean field.

An algebraically closed valued field has an $n$th root of every element. If its valuation were discrete and $\pi$ were a uniformizer, then $v(\sqrt[n]{\pi })=1/n$ would contradict discreteness. Therefore an algebraically closed non-Archimedean field cannot be locally compact.