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Ostrowski's theorem

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 Title: Ostrowski's theorem Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

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 Q is equivalent to either the usual real absolute value or a p-adic absolute value.

Contents

• Definitions 1
• Proof 2
• Case I: ∃n ∈ N   |n|∗ > 1 2.1
• Case II: ∀n ∈ N   |n|∗ ≤ 1 2.2
• Another Ostrowski's theorem 3
• References 5

Definitions

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

|x|_{\ast} = |x|^{c} \text{ for all } x \in \mathbf{K}.

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

|x|_{0} := \begin{cases} 0, & \text{if } x = 0 \\ 1, & \text{if } x \ne 0. \end{cases}

The real absolute value on the rationals Q is the standard absolute value on the reals, defined to be

|x|_\infty := \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if } 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 Q is defined as follows: any non-zero rational x, can be written uniquely as x=p^{n}\dfrac{a}{b} with a, b and p pairwise coprime and n\in\mathbf{Z} some integer; so we define

|x|_{p} := \begin{cases} 0, & \text{if } x = 0 \\ p^{-n}, & \text{if } x \ne 0. \end{cases}

Proof

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

(i) \exists{n\in\mathbf{N}},|n|_{\ast}>1
(ii) \forall{n\in\mathbf{N}},|n|_{\ast}\leq 1

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

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

and for negative rationals

|-x|_\ast=|x|_\ast=|x|^c_{\ast \ast}=|-x|^c_{\ast \ast}.

Case I: ∃n ∈ N   |n|∗ > 1

Consider the following calculation. Let a, b and n be natural numbers with a, b > 1. Expressing bn in base a yields

b^n=\sum_{i < m}c_i a^i, \qquad c_i \in \{0,1,\ldots,a-1\}, \quad m\leq n\tfrac{\log b}{\log a} +1.

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

\begin{align} |b|_\ast^n &= |b^{n}|_{\ast} \\ &\leq am\max \left \{|a|_\ast^m,1 \right \}\\ &\leq a(n\log_a b+1)\max \left \{|a|_\ast^{n\log_a b},1 \right \} \end{align}

Therefore

|b|_{\ast} \leq \left(a(n\log_a b+1)\right)^{\frac{1}{n}} \max \left \{|a|_\ast^{\log_a b},1 \right \}

However we have:

\left(a(n\log_a b+1)\right)^{\frac{1}{n}} \to 1, \quad \text{as} \quad n \to \infty

which implies:

|b|_{\ast} \leq \max \left \{|a|_\ast^{\log_{a}b},1 \right \}.

Now choose 1 < bN such that |b| > 1. Using this in the above ensures that |a| > 1 regardless of the choice of a (else |a|_\ast^{\log_a b}\leq1 implying |b|_\ast\leq 1). Thus for any choice of a, b > 1 above, we get

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

i.e.

\frac{\log|b|_{\ast}}{\log b} \leq \frac{\log|a|_{\ast}}{\log a}.

By symmetry, this inequality is an equality.

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

Case II: ∀n ∈ N   |n|∗ ≤ 1

As this valuation is non-trivial, there must be a natural number for which |n| < 1. Factorising this natural,

n = \prod_{i

yields |p| must be less than 1, for at least one of the prime factors p = pj. We claim than in fact, that 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\mathbf{N}^{+} be such that |p|_{\ast}^{e},|q|_{\ast}^{e}<\tfrac{1}{2}. By the Euclidean algorithm, there are m, nZ such that mp^{e}+nq^{e}=1. This yields

1=|1|_{\ast}\leq |m|_{\ast}|p|_\ast^{e}+|n|_{\ast}|q|_{\ast}^{e}<\frac{|m|_{\ast}+|n|_{\ast}}{2}\leq 1,

So we must have |pj| = α < 1 for some j, and |pi| = 1 for ij. Letting

c=-\tfrac{\log\alpha}{\log p},

we see that for general positive naturals

|n|_\ast= \left |\prod_{i.

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

\blacksquare

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