Ostrowski's Theorem

Theorem

Every non-trivial norm on the rational numbers $\Q$ is equivalent to either:

the $p$-adic norm $\norm {\, \cdot \,}_p$ for some prime $p$

or:

the absolute value, $\size {\, \cdot \,}$.

Proof

Let $\norm {\, \cdot \,}$ be a non-trivial norm on the rational numbers $\Q$.

Archimedean Norm Case

Let $\norm {\, \cdot \,}$ be an Archimedean norm.

$\exists n \in \N$ such that $\norm n > 1$

Let $n_0 = \min \set {n \in \N : \norm n > 1}$

By Norm of Unity then:

$n_0 > 1$

Let $\alpha = \dfrac {\log \norm {n_0} } {\log n_0}$

Since $n_0, \norm n_0 > 1$ then:

$\alpha > 0$

Lemma 1.1

$\forall n \in N: \norm n \le n^\alpha$

$\Box$

Lemma 1.2

$\forall n \in N: \norm n \ge n^\alpha$

$\Box$

Hence:

$\forall n \in \N: \norm n = n^\alpha = \size n^\alpha$

By Equivalent Norms on Rational Numbers then $\norm {\, \cdot \,}$ is equivalent to the absolute value $\size {\, \cdot \,}$.

$\Box$

Non-Archimedean Norm Case

Let $\norm {\, \cdot \,}$ be a non-Archimedean Norm.

$\forall n \in \N: \norm n \le 1$

Lemma 2.1

$\exists n \in \N: 0 < \norm n < 1$.

$\Box$

Let $n_0 = \min \set {n \in N : \norm n < 1}$.

Lemma 2.2

$n_0$ is a prime number.

$\Box$

Let $p = n_0$.

Let $\alpha = - \dfrac {\log \norm p } {\log p}$ then:

$\norm p = p^{-\alpha} = \paren {p^{-1}}^\alpha = \norm p_p^\alpha$

Let $b \in N$

Case 1: $p \nmid b$

Let $p \nmid b$.

$p$ and $b$ are coprime, that is, $p \perp b$
$\norm b = 1$

By the definition of the $p$-adic norm:

$\norm b_p = 1$

Hence:

$\norm b = 1 = 1^\alpha = \norm b_p^\alpha$

$\Box$

Case 2: $p \divides b$

Let $p \divides b$.

Let $\nu = \map {\nu_p} b$ where $\nu_p$ is the $p$-adic valuation on $\Z$.

Then:

$b = p^\nu c$

where $p \nmid c$

From #Case 1:

$\norm c = 1$

Hence:

 $\ds \norm b$ $=$ $\ds \norm p^\nu \norm {c}$ Non-Archimedean Norm Axiom $\text N 2$: Multiplicativity $\ds$ $=$ $\ds \norm p^\nu$ $\ds$ $=$ $\ds \norm p_p^{\alpha \nu}$ $\ds$ $=$ $\ds \paren {p^{-1} }^{\alpha \nu}$ Definition of $p$-adic norm $\ds$ $=$ $\ds p^{-\alpha \nu}$ $\ds$ $=$ $\ds \paren {p^{-\nu} }^\alpha$ $\ds$ $=$ $\ds \norm b_p^\alpha$ Definition of $p$-adic norm

$\Box$

In either case:

$\norm b = \norm b_p^\alpha$

Since $b$ was arbitrary, it has been shown:

$\forall b \in \N: \norm b = \norm b_p^\alpha$
$\norm {\, \cdot \,}$ is equivalent to the $p$-adic norm $\norm {\, \cdot \,}_p$.

$\blacksquare$

Source of Name

This entry was named for Alexander Markowich Ostrowski.

Historical Note

In the same paper as the one he presented this theorem, published in $1918$, Ostrowski also proved that, up to isomorphism, $\R$ and $\C$ are the only fields that are complete with respect to an archimedean norm.

That theorem is sometimes also referred to as Ostrowski's theorem.