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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.

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 intended statement uses a nontrivial Non-Archimedean absolute value. Under that convention choose $t$ with $0<|t|<1$. A compact neighborhood $C$ of zero contains an open ball of some radius $r>0$. For sufficiently large $N$, $t^N{\mathcal{O}}_K$ is contained in that ball. The valuation ring is closed, so $t^N{\mathcal{O}}_K$ is a closed subset of $C$ and is compact. Scaling shows that ${\mathcal{O}}_K$ itself is compact.

Its maximal ideal $\mathfrak{m}=\{x:|x|<1\}$ is open. The residue field $k={\mathcal{O}}_K/\mathfrak{m}$ is therefore discrete and compact, hence finite: the open cover by its singleton sets has a finite subcover. Since $\mathfrak{m}$ has finite index, its complement is a finite union of open cosets. Thus $\mathfrak{m}$ is also closed and compact.

The continuous absolute value on a field attains its maximum on $\mathfrak{m}$. Nontriviality supplies a nonzero element there, so this maximum is $\rho =|\pi |$ for some $\pi \ne 0$, with $0<\rho <1$. If $x\in \mathfrak{m}$, then $|x/\pi |\le 1$, so $\mathfrak{m}=\pi {\mathcal{O}}_K$.

For arbitrary $x\ne 0$, choose the integer $n$ for which ${\rho }^{n+1}<|x|\le {\rho }^n$. Then $\rho <|x/{\pi }^n|\le 1$. This element cannot lie in $\mathfrak{m}$, by maximality of $\rho$, so its absolute value on a field is one. Therefore $|x|={\rho }^n$. This proves discrete valuation from nontrivial local compactness:

The nontrivial valuation in the local compactness criterion is essential. If the trivial absolute value is admitted, the displayed source claim is false: take $\mathbb{Q}$ with that absolute value on a field. It is complete and discrete, hence locally compact, but its valuation ring and residue field are both $\mathbb{Q}$, which is infinite. The intended nontrivial case has been proved above.

Every Non-Archimedean absolute value on $\mathbb{Q}$ is either the trivial absolute value, or $|\cdot {|}_p^c$ for a unique prime $p$ and some $c>0$. Here $|p{|}_p=p^{-1}$.

To prove this classification, the ultrametric inequality gives $|n|\le 1$ for all integers $n$. If every nonzero integer has absolute value one, multiplicativity makes the absolute value on a field trivial on all rational numbers. Otherwise some positive integer has absolute value less than one, and its prime factorization gives a prime $p$ with $0<|p|<1$. There cannot be two such primes: if $p\ne q$ and $ap+bq=1$ with integers $a,b$, then $1\le \max (|p|,|q|)<1$, a contradiction. Every other prime consequently has absolute value one. Unique prime factorization now givesConversely these are Non-Archimedean absolute values, by the valuation inequality $v_p(x+y)\ge \min (v_p(x),v_p(y))$. Thus $\boxed{|\cdot |=|\cdot {|}_p^c(c>0)\text{ or the trivial value}}$. This is the non-Archimedean part of Ostrowski theorem, including the trivial case.