# 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 $\mathrm{\left|}m\mathrm{\right|}\mathrm{\le}\mathrm{max}{\mathrm{(}\mathrm{1}\mathrm{,}\mathrm{\left|}n\mathrm{\right|}\mathrm{)}}^{\mathrm{log}\mathit{}m\mathrm{/}\mathrm{log}\mathit{}n}$.

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

$$\left|{a}_{i}\right|=\left|\underset{{a}_{i}}{\underset{\u23df}{1+\mathrm{\cdots}+1}}\right|\le {a}_{i}\left|1\right|={a}_{i}\le n$$ |

by the triangle inequality^{}; also, $$.

Thus

$\left|m\right|$ | $=|{a}_{0}+{a}_{1}n+\mathrm{\cdots}+{a}_{r}{n}^{r}|\le (r+1)n\mathrm{max}{(1,\left|n\right|)}^{r}$ | ||

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

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

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

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, $\left|m\right|\le {\left|n\right|}^{\mathrm{log}m/\mathrm{log}n}$, so that for every $m,n$ we have

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

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

If instead, for some $n>1$ we have $$, then by the lemma, for every $m$, $\left|m\right|\le 1$. Thus the absolute value is nonarchimedean. Define $A=\{x\in \mathbb{Q}\mid \left|x\right|\le 1\}$ and let $\U0001d52a\subset A$ be the (unique) maximal ideal defined by $$. Then $\mathbb{Z}\subset A$ since $\left|m\right|\le 1$ for every $m$, and $\U0001d52a\cap \mathbb{Z}$ is nonzero since otherwise the valuation would be trivial (we would have $\left|m\right|=1$ for every $m$). Thus $\U0001d52a\cap \mathbb{Z}$ is prime since $\U0001d52a$ 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 |