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:
 * $\forall n \in \N: \norm {n} \le 1$

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

Lemma 2.2
Let $p = n_0$.

Let $\alpha = - \dfrac { \log {\norm p } } { \log p }$ then:
 * $\norm p = p^{-\alpha} = \paren {\dfrac 1 p}^\alpha = \norm p_p^\alpha$

Let $b$ an integer such that $p \nmid b$.

Then $p$ and $b$ are coprime, $p \perp b$.

By Corollary 5 of Three Points in Ultrametric Space has Two Equal Distances then:
 * $\norm b = 1$

By the definition of the $p$-adic norm,
 * $\norm b_p = 1$

Hence:
 * $\norm b = 1 = 1^\alpha = \norm b_p^\alpha$

Let $b$ an integer such that $p \divides b$.

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 $a$.