proof of Ostrowski’s valuation theorem

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

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

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

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 ProofOfOstrowskisValuationTheorem 2013-03-22 17:58:26 2013-03-22 17:58:26 rm50 (10146) rm50 (10146) 4 rm50 (10146) Proof msc 13A18