= Solution
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> gives
$$
1=|1|\leq\max(|a||p|,|b||q|)<1.
$$
Every other <prime number> therefore has <absolute value on a field> one. Factoring numerator and denominator of a <rational number> gives
$$
|x|=|p|^{v_p(x)}=|x|_p^c,\qquad
c=\frac{-\log |p|}{\log p}>0.
$$
Thus 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 $|x|\leq1$: if $|x|>1$, its leading term would strictly dominate the other terms. Applying the same argument to $x^{-1}$ forces $|x|=1$ for $x\ne0$. 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 $|a|\leq1$, by applying the same leading-term argument to its monic integral equation. Consequently
$$
\mathfrak p=\{a\in A:|a|<1\}
$$
is a proper <prime ideal>: it is an ideal by the <ultrametric inequality>, and $|ab|<1$ with $|a|,|b|\leq1$ 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 expression
$$
x=\pi^m u,\qquad m=v_{\mathfrak p}(x),\quad u\in A_{\mathfrak p}^{\times}.
$$
Both $u$ 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>:
$$
\boxed{\{\text{nontrivial equivalence classes on }F\}
\longleftrightarrow
\{\text{nonzero prime ideals of }\mathcal O_F\}.}
$$
The trivial class is additional when trivial <absolute values on a field> are admitted.
Back to article page
