Ostrowski's theorem

In number theory, Ostrowski's theorem, due to Alexander Ostrowski (1916), states that every non-trivial absolute value on the rational numbers $\mathbb {Q}$ is equivalent to either the usual real absolute value or a p-adic absolute value.

Definitions

Raising an absolute value to a power less than 1 always results in another absolute value. Two absolute values $|\cdot |$ and $|\cdot |_{*}$ on a field K are defined to be equivalent if there exists a real number c > 0 such that

$\forall x\in K:\quad |x|_{*}=|x|^{c}.$

The trivial absolute value on any field K is defined to be

$|x|_{0}:={\begin{cases}0&x=0,\\1&x\neq 0.\end{cases}}$

The real absolute value on the rationals $\mathbb {Q}$ is the standard absolute value on the reals, defined to be

$|x|_{\infty }:={\begin{cases}x&x\geq 0,\\-x&x<0.\end{cases}}$

This is sometimes written with a subscript 1 instead of infinity.

For a prime number p, the p-adic absolute value on $\mathbb {Q}$ is defined as follows: any non-zero rational x can be written uniquely as $x=p^{n}{\tfrac {a}{b}}$, where a and b are coprime integers not divisible by p, and n is an integer; so we define

$|x|_{p}:={\begin{cases}0&x=0,\\p^{-n}&x\neq 0.\end{cases}}$

Proof

Consider a non-trivial absolute value on the rationals $(\mathbb {Q} ,|\cdot |_{*})$. We consider two cases:

${\begin{aligned}(1)\quad \exists n\in \mathbb {N} \qquad |n|_{*}&>1,\\(2)\quad \forall n\in \mathbb {N} \qquad |n|_{*}&\leq 1.\end{aligned}}$

It suffices for us to consider the valuation of integers greater than one. For, if we find $c\in \mathbb {R} _{+}$ for which $|n|_{*}=|n|_{**}^{c}$ for all naturals greater than one, then this relation trivially holds for 0 and 1, and for positive rationals

$\left|{\frac {m}{n}}\right|_{*}={\frac {|m|_{*}}{|n|_{*}}}={\frac {|m|_{**}^{c}}{|n|_{**}^{c}}}=\left({\frac {|m|_{**}}{|n|_{**}}}\right)^{c}=\left|{\frac {m}{n}}\right|_{**}^{c},$

and for negative rationals

$|-x|_{*}=|x|_{*}=|x|_{**}^{c}=|-x|_{**}^{c}.$

Case (1)

This case implies that there exists $b\in \mathbb {N}$ such that $|b|_{*}>1.$ By the properties of an absolute value, $|0|_{*}=0$ and $|1|_{*}^{2}=|1|_{*}$, so $|1|_{*}=1$ (it cannot be zero). It therefore follows that b > 1.

Now, let $a,n\in \mathbb {N}$ with a > 1. Express bⁿ in base a:

$b^{n}=\sum _{i<m}c_{i}a^{i},\qquad c_{i}\in \{0,1,\ldots ,a-1\},\quad c_{m-1}>0.$

Hence

$b^{n}\geq a^{m-1},\quad$ so $\quad m\leq n\,{\frac {\log b}{\log a}}+1.$

Then we see, by the properties of an absolute value:

$|b|_{*}^{n}=|b^{n}|_{*}\leq \sum _{i<m}|c_{i}a^{i}|_{*}$

As each of the terms in the sum is smaller than $|c_{i}|_{*}\max \left\{|a|_{*}^{m-1},1\right\}$, it follows:

${\begin{aligned}|b|_{*}^{n}&\leq a\,m\max \left\{|a|_{*}^{m-1},1\right\}\\&\leq a(n\log _{a}\!b+1)\max \left\{|a|_{*}^{n\log _{a}\!b},1\right\}\end{aligned}}$

Therefore,

$|b|_{*}\leq \left(a(n\log _{a}\!b+1)\right)^{\frac {1}{n}}\max \left\{|a|_{*}^{\log _{a}\!b},1\right\}.$

However, as $n\to \infty$, we have

$(a(n\log _{a}\!b+1))^{\frac {1}{n}}\to 1,$

which implies

$|b|_{*}\leq \max \left\{|a|_{*}^{\log _{a}\!b},1\right\}.$

Together with the condition $|b|_{*}>1,$ the above argument shows that $|a|_{*}>1$ regardless of the choice of a > 1 (otherwise $|a|_{*}^{\log _{a}\!b}\leq 1$, implying $|b|_{*}\leq 1$). As a result, the initial condition above must be satisfied by any b > 1.

Thus for any choice of natural numbers a, b > 1, we get

$|b|_{*}\leq |a|_{*}^{\frac {\log b}{\log a}},$

i.e.

${\frac {\log |b|_{*}}{\log b}}\leq {\frac {\log |a|_{*}}{\log a}}.$

By symmetry, this inequality is an equality.

Since a, b were arbitrary, there is a constant $\lambda \in \mathbb {R} _{+}$ for which $\log |n|_{*}=\lambda \log n$, i.e. $|n|_{*}=n^{\lambda }=|n|_{\infty }^{\lambda }$ for all naturals n > 1. As per the above remarks, we easily see that $|x|_{*}=|x|_{\infty }^{\lambda }$ for all rationals, thus demonstrating equivalence to the real absolute value.

Case (2)

As this valuation is non-trivial, there must be a natural number for which $|n|_{*}<1.$ Factoring into primes:

$n=\prod _{i<r}p_{i}^{e_{i}}$

yields that there exists $j$ such that $|p_{j}|_{*}<1.$ We claim that in fact this is so for only one.

Suppose per contra that p, q are distinct primes with absolute value less than 1. First, let $e\in \mathbb {N}$ be such that $|p|_{*}^{e},|q|_{*}^{e}<{\tfrac {1}{2}}$. Since $p^{e}$ and $q^{e}$ are coprime, there are $k,l\in \mathbb {Z}$ such that $kp^{e}+lq^{e}=1.$ This yields

$1=|1|_{*}\leq |k|_{*}|p|_{*}^{e}+|l|_{*}|q|_{*}^{e}<{\frac {|k|_{*}+|l|_{*}}{2}}\leq 1,$

a contradiction.

So we must have $|p_{j}|_{*}=\alpha <1$ for some j, and $|p_{i}|_{*}=1$ for i ≠ j. Letting

$c=-{\frac {\log \alpha }{\log p}},$

we see that for general positive naturals

$|n|_{*}=\left|\prod _{i<r}p_{i}^{e_{i}}\right|_{*}=\prod _{i<r}\left|p_{i}\right|_{*}^{e_{i}}=\left|p_{j}\right|_{*}^{e_{j}}=(p^{-e_{j}})^{c}=|n|_{p}^{c}.$

As per the above remarks, we see that $|x|_{*}=|x|_{p}^{c}$ for all rationals, implying that the absolute value is equivalent to the p-adic one. $\blacksquare$

One can also show a stronger conclusion, namely, that $|\cdot |_{*}:\mathbb {Q} \to \mathbb {R}$ is a nontrivial absolute value if and only if either $|\cdot |_{*}=|\cdot |_{\infty }^{c}$ for some $c\in (0,1]$ or $|\cdot |_{*}=|\cdot |_{p}^{c}$ for some $c\in (0,\infty ),\ p\in \mathbf {P}$.

Another Ostrowski's theorem

Another theorem states that any field, complete with respect to an Archimedean absolute value, is (algebraically and topologically) isomorphic to either the real numbers or the complex numbers. This is sometimes also referred to as Ostrowski's theorem.

See also

• Valuation (algebra)

References

• Cassels, J. W. S. (1986). Local Fields. London Mathematical Society Student Texts. Vol. 3. Cambridge University Press. ISBN 0-521-31525-5. Zbl 0595.12006.
• Janusz, Gerald J. (1996). Algebraic Number Fields (2nd ed.). American Mathematical Society. ISBN 0-8218-0429-4.
• Jacobson, Nathan (1989). Basic algebra II (2nd ed.). W H Freeman. ISBN 0-7167-1933-9.
• Ostrowski, Alexander (1916). "Über einige Lösungen der Funktionalgleichung φ(x)·φ(y)=φ(xy)". Acta Mathematica (2nd ed.). 41 (1): 271–284. doi:10.1007/BF02422947. ISSN 0001-5962.