Archimedian Valuations on Number Fields

Definition A valuation $\abs{\cdot}$ is non-Archimedian if $\abs{x + y} \le \max(\abs{x}, \abs{y})$. Otherwise, we say that the valuation is Archimedian.

Remark It's not hard to see that if $\abs{\cdot}$ is Archimedian, then so is $\abs{\cdot}^r$ for any real number $r$.

Example(s) If $K = \BQ$, for any prime $p$, $\abs{\cdot}_p$ is non-Archimedian, while $\abs{\cdot}_\infty$ is Archimedian.

Example Let $K$ be a number field, and suppose $\mf{p}$ is a prime ideal of $K$. For each $x$, let $v_{\mf{p}}(x)$ be the greatest integer $n$ such that $x\in \mf{p}^n$. Then the function given by $\abs{x} = \mc{N}(\mf{p})^{-v_{\mf{p}}(x)}$ is a non-Archimedian norm. (This is called the $\mf{p}$-adic valuation on $K$ and is denoted $\abs{\cdot}_{\mf{p}}$.)

Proposition A valuation $\abs{\cdot}$ is non-Archimedian if and only if there exists a real number $M$ such that $\abs{m} < M$ for all $m\in \BZ$. (i.e. bounded on the rational integers)

Proof Note that the proposition is unchanged by replacing $\abs{\cdot}$ with $\abs{\cdot}^r$ for any real number $r > 0$, so we may assume that $\abs{\cdot}$ satisfies the triangle inequality.

For one direction, if $\abs{\cdot}$ is non-Archimedian, then $\abs{m}\le \abs{1}$, so the boundedness condition clearly holds.

Now we prove the other direction. Suppose $\abs{m} < M$ for all $m\in \BZ$. Then
\[ \begin{align*} \abs{x + y}^n &= \abs{(x + y)^n} \\ &= \abs{\sum_{i = 0}^n \binom{n}{i} x^k y^{n-i}} \\ &\le \sum_{i = 0}^n \abs{\binom{n}{i}}\cdot \abs{x^k y^{n-i}}\\ &\le M(n + 1)\max(\abs{x}^n, \abs{y}^n) \end{align*} \] Thus $\abs{x + y}\le (M(n + 1))^{1 / n}\max(\abs{x}, \abs{y})$ for all positive integers $n$, so taking $n\to\infty$ gives $\abs{x + y}\le \max(\abs{x}, \abs{y})$. $\boxed{}$.

For the remainder of this post, suppose $K$ is a number field (i.e. a finite extension of $\BQ$). Let $\mathcal{O}_K$ denote the ring of integers in $K$.

Lemma Suppose $\abs{\cdot}$ is a non-Archimedian valuation on $K$. Then for all $x\in \mc{O}_K$, $\abs{x} \le 1$.

Proof Let the minimal polynomial of $x$ be $x^n + a_{n-1}x^{n-1} + \ldots + a_0$, where $a_0, \ldots, a_{n-1}\in \BZ$. Then $\abs{a_i}\le i$ for all $i$, so \[ \begin{align*} \abs{x}^n = \abs{x^n} &= \abs{\sum_{i = 0}^{n - 1}a_i x^i}\\ &\le \max_{0\le i\le n - 1} \abs{a_i x^i}\\ &\le \max_{0\le i\le n - 1} \abs{x}^i\\ &= \max(1, \abs{x}^{n-1}) \end{align*} \] This is only possible if $\abs{x} \le 1$. $\boxed{}$.

Theorem (Big Ostrowski, Part I) Suppose $\abs{\cdot}$ is a (nontrivial) non-Archimedian norm on $K$. Then $\abs{\cdot}$ is equivalent to $\abs{\cdot}_{\mf{p}}$ for some prime ideal $\mf{p}\subseteq \mc{O}_K$.

Proof The proof of this theorem very closely resembles the proof of the non-Archimedian case of Little Ostrowski's Theorem. However, instead of rational primes, we have prime ideals, and we have to make a little tweak with the class group. In any case....

Let $\abs{\cdot}$ be a non-Archimedian norm on $K$, let $I = \{x\in \mc{O}_K : \abs{x} < 1\}$. By the previous lemma, $I$ has to be a proper ideal of $\mc{O}_K$. Since $\abs{\cdot}$ is nontrivial, $I$ is also nonzero. Finally, if $xy\in I$, then either $x\in I$ or $y\in I$, so $I = \mf{p}$ for some prime ideal $\mf{p}$. It remains to show that $\abs{x} = c^{v_\mf{p}(x)}$ for some real number $0 < c < 1$.

Let $n = \abs{\mc{Cl}_K}$ be the size of the class group of $K$. Then for each prime ideal $\mf{q}$, there exists an $x_{\mf{q}}\in \mc{O}_K$ such that $\mf{q}^n = (x_{\mf{q}})$. In addition, we have \[\abs{x_{\mf{q}}} < 1\iff x_{\mf{q}}\in \mf{p} \iff \mf{q} = \mf{p}.\]Given an $x\in K$, we have that  $(x) = \prod_{\mf{q}} \mf{q}^{v_{\mf{q}}(x)}$ is the unique factorization of the fractional ideal $K$ into prime ideals. Then \[(x^n) = \prod_{\mf{q}} \mf{q}^{nv_{\mf{q}}(x)} = \prod_{\mf{q}} (x_{\mf{q}})^{v_{\mf{q}}(x)}\implies x^n = \prod_{\mf{q}} x_{\mf{q}}^{v_{\mf{q}}(x)}.\] Thus \[\abs{x} = \abs{x^n}^{1/n} = \prod_{\mf{q}} \abs{x_{\mf{q}}}^{v_{\mf{q}}(x)} = \abs{x_{\mf{p}}}^{v_{\mf{q}}(x)/n},\] as desired. $\boxed{}$