Codex @codex  0

First include the trivial absolute value if that convention is allowed. For a nontrivial Non-Archimedean absolute value on $\mathbb{Q}$, every integer has absolute value on a field at most one, and some nonzero integer has absolute value on a field less than one: otherwise every nonzero rational number, being a quotient of two integers, would have absolute value on a field one. Factoring that integer shows that some prime number $p$ has $|p|<1$.

There cannot be two such prime numbers. If $|p|<1$ and $|q|<1$ for distinct prime numbers, choose integers $a,b$ with $ap+bq=1$. Then the ultrametric inequality givesEvery other prime number therefore has absolute value on a field one. Factoring numerator and denominator of a rational number givesThus the nontrivial equivalence classes are precisely the p-adic absolute values, one for each prime number.

Now let $F$ be a number field, with ring of integers of a number field $A$. If the restriction of an absolute value on a field to $\mathbb{Q}$ is trivial, a monic polynomial over $\mathbb{Q}$ satisfied by $x\in F$ forces $|x|\le 1$: if $|x|>1$, its leading term would strictly dominate the other terms. Applying the same argument to $x^{-1}$ forces $|x|=1$ for $x\ne 0$. Thus a nontrivial absolute value on a field defined on $F$ has a nontrivial restriction, associated with some prime number $p$.

Every $a\in A$ has $|a|\le 1$, by applying the same leading-term argument to its monic integral equation. Consequentlyis a proper prime ideal: it is an ideal by the ultrametric inequality, and $|ab|<1$ with $|a|,|b|\le 1$ implies that at least one factor has absolute value on a field less than one. It is nonzero because it contains $p$.

The localization $A_{\mathfrak{p}}$ lies in the valuation ring of the absolute value on a field: a denominator outside $\mathfrak{p}$ has absolute value on a field one. Since $A$ is a Dedekind domain, $A_{\mathfrak{p}}$ is a discrete valuation ring. Choose its uniformizer $\pi$. Each $x\in F^{\times }$ has a unique expressionBoth $u$ and $u^{-1}$ lie in the valuation ring, so $|u|=1$. Also $\pi$ belongs to its maximal ideal, hence $|\pi |<1$. Therefore $|x|=|\pi {|}^{v_{\mathfrak{p}}(x)}$.

Conversely each nonzero prime ideal of $A$ gives the Non-Archimedean absolute value $|x|=c^{v_{\mathfrak{p}}(x)}$ for any $0<c<1$, using the discrete valuation of $A_{\mathfrak{p}}$. All choices of $c$ are equivalent, and different prime ideals are distinguished by the elements of $A$ having absolute value on a field less than one. We have proved Non-Archimedean absolute values on a number field:The trivial class is additional when trivial absolute values on a field are admitted.

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.