= Solution
An <absolute value on a field> is a map $|\cdot|:K\to\mathbb R_{\geq0}$ satisfying
$$
|x|=0\Longleftrightarrow x=0,\qquad |xy|=|x||y|,\qquad |x+y|\leq|x|+|y|.
$$
A <Non-Archimedean absolute value> satisfies the stronger <ultrametric inequality> $|x+y|\leq\max(|x|,|y|)$. 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 are
$$
v(x)=-\log_c|x|,\quad v(0)=+\infty,\qquad |x|=c^{-v(x)}.
$$
Multiplicativity becomes $v(xy)=v(x)+v(y)$, and the <ultrametric inequality> becomes $v(x+y)\geq\min(v(x),v(y))$. Equivalent real-valued <valuations> differ by positive scaling, so these constructions give the required \b[bijection on equivalence classes]. Changing $c$ merely rescales the <valuation>.
To justify the topology formulation, $|a|<1$ is equivalent to $a^n\to0$. 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 $\mathbb Q$, $|n|\leq1$ 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> $a p+b q=1$ contradicts the <ultrametric inequality>. If $p\nmid m$, another <Bezout identity> gives $|m|=1$. Hence
$$
\boxed{|x|=|p|^{v_p(x)}=|x|_p^{\alpha},\qquad \alpha=-\frac{\log|p|}{\log p}>0.}
$$
This 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> are
$$
R=\{x:|x|\leq1\},\qquad\mathfrak m=\{x:|x|<1\},\qquad k=R/\mathfrak m.
$$
Suppose 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,
$$
\boxed{R\cong\varprojlim_N R/\pi^N R,\qquad R\text{ compact}\Longleftrightarrow v(K^\times)\text{ discrete and }|k|<\infty.}
$$
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.
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