Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 136 / 2 / a / Solution

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.