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A Non-Archimedean absolute value on a field is a function $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ with $|x|=0$ exactly for $x=0$, $|xy|=|x||y|$, and the ultrametric inequality $|x+y|\le \max (|x|,|y|)$. It is nontrivial if some nonzero element has value different from one. Equivalent absolute values induce the same topology; for nontrivial absolute values this is equivalent to $|x{|}_2=|x{|}_1^c$ for some $c>0$.

Here is also a proof of the latter equivalence in the case needed. A positive power plainly preserves the open neighborhoods of zero. Conversely, if the topologies coincide, then $|x{|}_1<1$ if and only if $x^n\to 0$ in the first topology, hence if and only if $|x{|}_2<1$. Choose $a$ with $0<|a{|}_1<1$. For every $x\ne 0$, $m\in \mathbb{Z}$ and $n\ge 1$, whether $|x^n/a^m|<1$ agrees for the two values. Comparison with all rational numbers $m/n$ therefore identifies the two ratios $\log |x{|}_j/\log |a{|}_j$. Thus $|x{|}_2=|x{|}_1^c$ with $c=\log |a{|}_2/\log |a{|}_1>0$.

Now let $|\cdot |$ be nontrivial and non-Archimedean on $\mathbb{Q}$. Since $|1|=1$ and every positive integer is a sum of ones, $|n|\le 1$ for every integer. If all nonzero integers had value one, multiplicativity would make the absolute value trivial on all rationals. Hence some integer has value less than one; its prime factorization shows that some prime $p$ has $|p|<1$.

There cannot be two distinct such primes $p,q$. A Bezout identity $ap+bq=1$ with $a,b\in \mathbb{Z}$ would giveThus every other prime has value one. Prime factorization of the numerator and denominator of $x\in {\mathbb{Q}}^{\times }$ now givesConversely $|x{|}_p=p^{-v_p(x)}$ is multiplicative and satisfies the ultrametric inequality, since $v_p(x+y)\ge \min (v_p(x),v_p(y))$. Distinct primes give inequivalent values: $p^m\to 0$ for the $p$-adic absolute value but $|p^m{|}_q=1$ for $q\ne p$. We have proved the non-Archimedean part of Ostrowski theorem:There is exactly one equivalence class for each prime.

Two absolute values on a field are equivalent absolute values when they induce the same topology. For nontrivial absolute values on a field this is equivalent tofor a fixed real number $c>0$. The trivial absolute value is equivalent only to itself.

Restriction of a Non-Archimedean absolute value plainly preserves the ultrametric inequality. Conversely, if the restriction is non-Archimedean, every integer, interpreted in either field, has absolute value on a field at most one. In particular every binomial coefficient has absolute value on a field at most one. The triangle inequality applied to the binomial expansion gives, for $x,y\in L$ and every positive integer $N$,Raise both sides to the power $1/N$ and let $N\to \infty$. Since $(N+1{)}^{1/N}\to 1$, this yieldsThis is the bounded-integer criterion for a non-Archimedean absolute value. The proof works for arbitrary field extensions, not just algebraic ones, and in positive characteristic as well.

Choose $a\in K^{\times }$ with $0<|a{|}_1<1$. For any $x\ne 0$, the powers $x^n$ tend to zero precisely when $|x{|}_1<1$. Equality of the two topologies therefore impliesIn particular $0<|a{|}_2<1$. For positive integers $n$ and arbitrary integers $m$, apply this equivalence to $x^n{a}^{-m}$. It says thatThe inequality reverses on division by either negative logarithm. The rational cuts of the two real numbers are equal, so density of $\mathbb{Q}$ makes the numbers equal. Hence the equivalent absolute values satisfyThe formula also holds at zero. Nontriviality is essential to choosing $a$; no Archimedean or non-Archimedean restriction was used in the proof.

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.

Choose a $K$-basis of $L$. Any extension $|\cdot {|}_L$ defines a norm on the finite-dimensional $K$-vector space $L$, and all norms on such a space over a complete valued field induce the same topology. Hence any two extensions induce the same topology and are equivalent absolute values. The coordinate sup norm is complete because $K$ is complete, so the equivalent topology defined by $|\cdot {|}_L$ is complete as well. This proves the unique extension of an absolute value to a finite extension and the completeness of $L$.

Completeness is essential. Give $K=\mathbb{Q}$ its $5$-adic value and take $L=\mathbb{Q}(i)$. Sincein $\mathbb{Z}[i]$, the two primes $(2+i)$ and $(2-i)$ define two inequivalent extensions. For example, $(2+i)/(2-i)$ has positive valuation for one and negative valuation for the other. This is the valuation ring need not equal the integral closure over a noncomplete field example.

The common topology determines which elements have absolute value below one, becausein that topology. It therefore also determines all comparisons: $|x{|}_i<|y{|}_i$ exactly when $|x/y{|}_i<1$.

Choose $a\in K^{\times }$ with $|a{|}_1>1$; then also $|a{|}_2>1$. For every $x\in K^{\times }$ and positive integers $m,n$, the preceding observation givesThus the two real numbershave the same rational upper cuts and are equal. Settinggives $|x{|}_1=|x{|}_2^c$ for every $x$. Hence the two absolute values are equivalent absolute values.