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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.