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
$\forall n \in \N: \left \Vert {n}\right \Vert \le 1$:

Let $n_0$ be the least positive integer such that $\left \Vert {n}\right \Vert < 1$.

Such a number exists because we have assumed $\left \Vert {*}\right \Vert$ is non-trivial.

If $n_0$ is composite, there are positive integers $n_1, n_2 < n_0$ such that $n_0 = n_1 n_2$.

Then:


 * $\left \Vert {n_1}\right \Vert = \left \Vert {n_2}\right \Vert = 1$

by the choice of $n_0$, and so:


 * $\left \Vert {n_0}\right \Vert = \left \Vert {n_1}\right \Vert \left \Vert {n_2}\right \Vert = 1$

which is a contradiction. So $n_0$ must be prime.

Let $n_0 = p$.

Claim: $\left \Vert {q}\right \Vert = 1$ if $q$ is a prime not equal to $p$.

Suppose not; then $\left \Vert {q}\right \Vert < 1$, and for some large $N$ we have:


 * $\left \Vert {q^N}\right \Vert = \left \Vert {q} \right \Vert ^N < \frac 1 2$

Also, for some large $M$ we have $\left \Vert {p}\right \Vert^M < \frac 1 2$.

Since $p^M, q^N$ are relatively prime, by Bézout's Identity we can find integers $n, m$ such that $m p^M + n q^N = 1$.

But then:

Now, $\left \Vert {k}\right \Vert \le 1$ for any integer $k$: If $k=0$, then $\left \Vert {k}\right \Vert=0$. If $k\ne 0$, then either $k$ or $-k$ is positive and since $\left \Vert {k}\right \Vert=\left \Vert {-k}\right \Vert$ for any norm, it follows by assumption.

Thus $\left \Vert {m}\right \Vert, \left \Vert {n}\right \Vert \le 1$, so that:


 * $1 \le \left \Vert {m}\right \Vert \left \Vert {p^M}\right \Vert + \left \Vert {n}\right \Vert \left \Vert {q^N}\right \Vert \le \left \Vert {p^M}\right \Vert + \left \Vert {q^N}\right \Vert < \frac 1 2 + \frac 1 2 = 1$

which is a contradiction.

Hence $\left \Vert {q}\right \Vert = 1$.

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 $a$.