proof of Ostrowski’s valuation theorem

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

Theorem 1.

(Ostrowski) Over $\mathbb{Q}$ , every nontrivial absolute value is equivalent either to $\left\lvert\cdot\right\rvert_{p}$ for some prime $p$ , or to the usual absolute value $\left\lvert\cdot\right\rvert_{\infty}$ .

We start with an estimation lemma:

Lemma 2.

If $m,n>1$ are integers and $\left\lvert\cdot\right\rvert$ any nontrivial absolute value on $\mathbb{Q}$ , then $\left\lvert m\right\rvert\leq\max(1,\left\lvert n\right\rvert)^{\log m/\log n}$ .

Proof. Write $m=a_{0}+a_{1}n+\cdots+a_{r}n^{r}$ for $a_{i}\in\mathbb{Z},0\leq a_{i}\leq n-1$ , and with $a_{r}\neq 0$ . Then clearly

$\left\lvert a_{i}\right\rvert=\left\lvert\underset{a_{i}}{\underbrace{1+\cdots% +1}}\right\rvert\leq a_{i}\left\lvert 1\right\rvert=a_{i}\leq n$

by the triangle inequality; also, $r<\frac{\log m}{\log n}$ .

Thus

	$\displaystyle\left\lvert m\right\rvert$	$\displaystyle=\left\lvert a_{0}+a_{1}n+\cdots+a_{r}n^{r}\right\rvert\leq(r+1)n% \max(1,\left\lvert n\right\rvert)^{r}$
		$\displaystyle\leq\left(1+\frac{\log m}{\log n}\right)n\max(1,\left\lvert n% \right\rvert)^{\log m/\log n}$

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

$\left\lvert m\right\rvert\leq\left(1+t\frac{\log m}{\log n}\right)^{1/t}n^{1/t% }\max(1,\left\lvert n\right\rvert)^{\log m/\log n}$

Now let $t\to\infty$ ; 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 $\left\lvert n\right\rvert>1$ . Then by the lemma, $\left\lvert m\right\rvert\leq\left\lvert n\right\rvert^{\log m/\log n}$ , so that for every $m, n$ we have

$\left\lvert m\right\rvert^{1/\log m}\leq\left\lvert n\right\rvert^{1/\log n}$

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 $m$ , $\left\lvert m\right\rvert^{1/\log m}=c$ and $\left\lvert m\right\rvert=c^{\log m}$ for some constant $c$ ; this absolute value is obviously equivalent to $\left\lvert m\right\rvert_{\infty}=e^{\log m}$ .

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

$\left\lvert x\right\rvert=\frac{\left\lvert a\right\rvert\cdot\left\lvert p% \right\rvert^{t}}{\left\lvert b\right\rvert}=\left\lvert p\right\rvert^{t}$

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