Ostrowski's Theorem/Non-Archimedean Norm

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

Then $\norm {\, \cdot \,}$ is equivalent to the $p$-adic Norm $\norm {\, \cdot \,}_p$ for some prime $p$.

Proof
By Characterisation of Non-Archimedean Division Ring Norms then:
 * $\forall n \in \N: \norm {n} \le 1$

Let $z \in \Z$, then $\size z \in \N$.

By Norm of Negative then:
 * $\norm {z} = \norm {\size{z}} \le 1$

Hence:
 * $\forall z \in \Z: \norm {z} \le 1$

Lemma 2.1
Let $n_0 = \min \set {n \in N : \norm {n} < 1}$.

Lemma 2.2
Let $p = n_0$.

Let $q$ be a prime number such that $q \neq p$.

By Prime not Divisor implies Coprime then:
 * $p \perp q$

Aiming for a contradiction suppose that $\norm {q} < 1$.

Since $p \perp q$, by Bézout's Identity then:
 * $\exists n, m \in \Z : m p + n q = 1$

It follows that:

Similarly $\norm {n q} \lt 1$.

Then:

This is a contradiction.

Hence $\norm {q} = 1$.

By the Fundamental Theorem of Arithmetic, any positive integer $a$ can be factored into prime divisors: $a = p_1^{b_1} p_2^{b_2} \dots p_r^{b_r}$.

Then:


 * $\left \Vert {a}\right \Vert = \left \Vert {p_1}\right \Vert ^{b_1} \left \Vert {p_2}\right \Vert ^{b_2} \dots \left \Vert {p_r}\right \Vert^{b_r}$

But $\left \Vert {p_i}\right \Vert$ will be different from $1$ only if $p_i = p$.

Its corresponding $b_i$ will be $\nu_p^\Z \left({a}\right)$, where $\nu_p^\Z$ is the $p$-adic valuation on $\Z$.

Hence, if we let $\rho = \left \Vert {p}\right \Vert < 1$, we have:


 * $\left \Vert {a}\right \Vert = \rho^{\nu_p^\Z \left({a}\right)}$

By the properties of norms, this same formula holds with any nonzero rational number in place of $a$.