# Is every non-archimedean absolute value on a number field equivalent to a $|\cdot|_{\mathfrak{p}}$?

Let $K$ be an algebraic number field, i.e. a finite field extension of $\Bbb{Q}$. I would like to prove that every non-archimedean absolute value on $K$ is equivalent to $$|x|_{\mathfrak{p}} := N_K(\mathfrak{p})^{-\text{ord}_{\mathfrak{p}}(x)}$$ for some prime ideal $\mathfrak{p}$ of $\mathcal{O}_K$, where $|0|_\mathfrak{p} = 0$ by convention and otherwise $\text{ord}_{\mathfrak{p}}(x)$ is the exponent of $\mathfrak{p}$ in the prime factorization of $(x)$, and where $N_K(\mathfrak{p}) = |\mathcal{O}_K/\mathfrak{p}|$ is the absolute norm of $\mathfrak{p}$.

Preliminary question: Is this actually true?

If so, how can I prove this? I know that the completion of $K$ with respect to a non-archimedean absolute value is isomorphic to a finite extension of $\Bbb{Q}_p$ for some prime number $p$, and that $$|x|_p = |N_{L/\Bbb{Q}_p}(x)|_p^{1/[L:\Bbb{Q}_p]}$$ is the unique extension of $|\cdot|_p$ to a finite extension $L$ of $\Bbb{Q}_p$, but I don't know how to conclude from here.

Yes, this is true. The case for $\mathbb{Q}$ is sometimes called Ostrowski's Theorem, although related results also go by this name. The proof for number fields is extremely similar (and sometimes also called Ostrowski's Theorem).
• I checked those references. Sadly, (just like every other reference I was able to find) they mention only the version of Ostrowski's theorem for $\Bbb{Q}$. I don't see any immediate way to derive the version for $K \supset \Bbb{Q}$ from that one, though. – A.P. Jun 19 '15 at 10:43