 Solution

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.