proof of Ostrowski’s valuation theorem

This article proves Ostrowski’s theorem on valuations of $\mathbb{Q}$, which states:

Theorem 1.

(Ostrowski) Over $\mathrm{Q}$, every nontrivial absolute value is equivalent either to ${\mathrm{|}\mathrm{\cdot }\mathrm{|}}_p$ for some prime $p$, or to the usual absolute value ${\mathrm{|}\mathrm{\cdot }\mathrm{|}}_{\mathrm{\infty }}$.

We start with an estimation lemma:

Lemma 2.

If $m\mathrm{,}n\mathrm{>}\mathrm{1}$ are integers and $\mathrm{|}\mathrm{\cdot }\mathrm{|}$ any nontrivial absolute value on $\mathrm{Q}$, then $\mathrm{\left|}m\mathrm{\right|}\mathrm{\le }\max {\mathrm{(}\mathrm{1}\mathrm{,}\mathrm{\left|}n\mathrm{\right|}\mathrm{)}}^{\log \mathit{}m\mathrm{/}\log \mathit{}n}$.

Proof. Write $m=a_0+a_{1}n+\mathrm{\cdots }+a_r{n}^r$ for $a_i\in \mathbb{Z},0\le a_i\le n-1$, and with $a_r\ne 0$. Then clearly

$$\left|a_i\right|=\left|\underset{a_i}{\underbrace{1+\mathrm{\cdots }+1}}\right|\le a_i\left|1\right|=a_i\le n$$

by the triangle inequality; also, .

Thus

$\left|m\right|$ $=|a_0+a_{1}n+\mathrm{\cdots }+a_r{n}^r|\le (r+1)n\max {(1,\left|n\right|)}^r$

$\le (1+{\displaystyle \frac{\log m}{\log n}})n\max {(1,\left|n\right|)}^{\log m/\log n}$

Replace $m$ by $m^t$ for $t$ a positive integer, and take $t^{\mathrm{th}}$ roots of the resulting inequality, to get

$$\left|m\right|\le {(1+t\frac{\log m}{\log n})}^{1/t}{n}^{1/t}\max {(1,\left|n\right|)}^{\log m/\log n}$$

Now let $t\to \mathrm{\infty }$; the first two factors each approach $1$, and the lemma follows.

Proof of Ostrowski’s theorem:
First assume that for every $n>1$ we have $\left|n\right|>1$. Then by the lemma, $\left|m\right|\le {\left|n\right|}^{\log m/\log n}$, so that for every $m,n$ we have

$${\left|m\right|}^{1/\log m}\le {\left|n\right|}^{1/\log n}$$

Since this holds for every $m,n>0$, after reversing the roles of $m,n$, we see that in fact equality holds, so that for every $m$, ${\left|m\right|}^{1/\log m}=c$ and $\left|m\right|=c^{\log m}$ for some constant $c$; this absolute value is obviously equivalent to ${\left|m\right|}_{\mathrm{\infty }}=e^{\log m}$.

If instead, for some $n>1$ we have , then by the lemma, for every $m$, $\left|m\right|\le 1$. Thus the absolute value is nonarchimedean. Define $A=\{x\in \mathbb{Q}\mid \left|x\right|\le 1\}$ and let $𝔪\subset A$ be the (unique) maximal ideal defined by . Then $\mathbb{Z}\subset A$ since $\left|m\right|\le 1$ for every $m$, and $𝔪\cap \mathbb{Z}$ is nonzero since otherwise the valuation would be trivial (we would have $\left|m\right|=1$ for every $m$). Thus $𝔪\cap \mathbb{Z}$ is prime since $𝔪$ is, so is equal to $(p)$ for some rational prime $p$. Now, if $p\nmid a$ for an integer $a$, then $\left|a\right|$ cannot be strictly less than $1$ (else it would be in $(p)$), so $\left|a\right|=1$ and $a\in A^{\star }$. But given any $x\in \mathbb{Q}$, we can write $x=\frac{a{p}^t}{b}$ with $a,b$ prime to $p$, so that

$$\left|x\right|=\frac{\left|a\right|\cdot {\left|p\right|}^t}{\left|b\right|}={\left|p\right|}^t$$

so that the valuation is obviously equivalent to the $p$-adic valuation.

Title	proof of Ostrowski’s valuation theorem
Canonical name	ProofOfOstrowskisValuationTheorem
Date of creation	2013-03-22 17:58:26
Last modified on	2013-03-22 17:58:26
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	4
Author	rm50 (10146)
Entry type	Proof
Classification	msc 13A18