We proved earlier that every valuation $\abs{\cdot}$ is equivalent to one that satisfies the triangle inequality. Thus $\abs{\cdot}$ gives $K$ the structure of a metric space by letting $\rho(x, y) = \abs{x - y}^r$.
Theorem Two valuations $\abs{\cdot}_1$ and $\abs{\cdot}_2$ induce the same topology on $K$ iff they are equivalent.
Proof From the definition of the topology, it is clear that if $\abs{x}_1^r = \abs{x}_2$, then $\abs{\cdot}_1$ and $\abs{\cdot}_2$ induce the same topology on $K$.
Now suppose $\abs{\cdot}_1$ and $\abs{\cdot}_2$ are nontrivial and induce the same topology on $K$ There exists an $x_0$ such that $\abs{x_0}_1 > 1$, and let $r = \frac{\log\abs{x_0}_2}{\log\abs{x_0}_1}$. I claim that this is the desired $r$. Since $\abs{x_0^{-n}}_1\to 0$, $x_0^{-n}\to 0$, so $\abs{x_0^{-n}}_2 \to 0$, which means that $\abs{x_0}_2 > 1$ as well. Thus $r > 0$.
Fix another $x\in K^\times$, and let $\epsilon = \log\abs{x}_2 - r\log\abs{x}_1$. Suppose for the sake of contradiction that $\epsilon \neq 0$. Since $\abs{\cdot}_1$ and $\abs{\cdot}_2$ are equivalent,
Fix another $x\in K^\times$, and let $\epsilon = \log\abs{x}_2 - r\log\abs{x}_1$. Suppose for the sake of contradiction that $\epsilon \neq 0$. Since $\abs{\cdot}_1$ and $\abs{\cdot}_2$ are equivalent,
\[ \abs{x_0^{e_i} x^{f_i}}_1\to 0 \iff \abs{x_0^{e_i} x^{f_i}}_2\to 0, \] or equivalently, \[ e_i\log\abs{x_0}_1 + f_i\log \abs{x}_1\to -\infty \iff r(e_i\log\abs{x_0}_1 + f_i \log\abs{x}_1) + \epsilon f_i\to -\infty.\] Let $f_n = -n\cdot \operatorname{sgn}(\epsilon)$, $e_n = \left\lceil \frac{-f_n\log\abs{x}_1}{\log \abs{x_0}_1} \right\rceil$. Then
\[ 0 \le e_i\log\abs{x_0}_1 + f_i\log \abs{x}_1 \le \abs{x_0}_1. \] However, \[ \epsilon f_i \le r(e_i\log\abs{x_0}_1 + f_i \log\abs{x}_1) + \epsilon f_i \le \epsilon f_i + r\abs{x_0}_1, \] so the first series does not converge to $-\infty$ while the second does, contradiction. Thus for all $x\in K^\times$, $\log\abs{x}_2 - r\log\abs{x}_1 = 0$. $\boxed{}$
Given a valuation $v$ on $K$, we can complete $K$ with respect to the topology induced by $v$ to get a new metric space $K_v$.
Proposition There is a unique way to give $K_v$ the structure of a field such that addition and multiplication are continuous with respect to the valuation $v$.
Proof Note that multiplication and addition are clearly continuous on $K$, so they extend uniquely to multiplication and addition on $K$. Thus $K_v$ already has a ring structure. It suffices to show that $K_v$ has inverses. This is also not hard to show and is left as an exercise to the reader.$\boxed{}$
Proposition Suppose $v$ is a ((non-)Archimedian) valuation on $K$. Then $v$ extends uniquely to a ((non-)Archimedian) valuation on $K_v$.
Proof Obvious by the continuity of $v$. The details are left as an exercise to the reader. $\boxed{}$.
Theorem (Big Ostrowski, Part II) Suppose $v$ is an Archimedian valuation on the number field $K$. Then there exist an embedding $\sigma: K\to \BC$ and a real number $r > 0$ such that $\abs{x}_v = \abs{\sigma(x)}_{\BC}^r$.
Proof Let $K_v$ be the completion of $K$ with respect to $K$. It suffices to show that there exists a valuation-preserving isomorphism from either $K_v\to \BR$ or $K_v\to \BC$.
Note that since we have an embedding $\BQ\to K$ and the restriction of $v$ onto $\BQ$ must be the absolute value norm on $\BQ$, there exists an embedding $\BQ\to K_v$ that extends uniquely to an embedding $\BR\simeq \BQ_v \to K_v$. Thus it suffices to show that every element of $K_v$ has degree at most 2 over $\BR$.
Let $x\in K_v$, and define the function $f_x: \BC\to \BR$ as $z\mapsto \abs{x^2 - (z + \bar{z})x + z\cdot \bar{z}}_{\BC}$. Note that $x$ having degree at most 2 over $\BR$ is equivalent to $f_x(z) = 0$ for some $z\in \BC$, so suppose for the sake of contradiction that $f_x$ never hits 0. Note that the region $\{z: \abs{f_x(z)}_{\BC} < R\}$ is bounded for any $R$. Thus $f_x$ attains a minimum $r>0$ at $z_0$. Since $f_x^{-1}(r)$ is closed and bounded, it is compact. Thus we can pick $z_0$ such that $\abs{z_0}_{\BC}$ is maximal.
Let $0 < \epsilon < r$ be a real number, and let $z_1$ be a root of $z^2 - (z_0 + \bar{z}_0)z+ z_0\cdot \bar{z}_0 + \epsilon = 0$. Since this is also a real quadratic, we have $\abs{z_1}_{\BC}^2 = z_1\cdot \bar{z_1} = z_0\cdot \bar{z}_0 + \epsilon > \abs{z_0}_{\BC}^2$, so $f_x(z_1) > r$.
Now consider $g_n(t) = (t^2 - (z_0 + \bar{z}_0) t+ z_0\cdot \bar{z}_0)^n - (-\epsilon)^n\in \BR[t]$. Note that $g_n(t)$ can be factored uniquely into (real) linear terms and irreducible quadratic terms, so if $t_1, \ldots, t_{2n}\in \BC$ are the roots of $g_n$, then \[ g_n(x)^2 = g_n(t)\cdot \overline{g_n(t)} = \prod_{i = 1}^{2n} (t^2 - (t_i + \bar{t_i})t + t_i\cdot \bar{t_i}) \] Furthermore, $z_1$ is always a root of $g_n(t)$, so without loss of generality $z_1 = t_1$. Thus \[ \abs{g_n(x)}^2_v = \prod_{i = 1}^{2n} \abs{x^2 - (t_i + \bar{t_i})x + t_i\cdot \bar{t_i}}_v\ge f_x(z_1)\cdot r^{2n - 1} \] However, $g_n(x) = r^n - (-\epsilon)^n$. Thus $f_x(z_1)\le \frac{(r^n + (-\epsilon)^n)^2}{r^{2n - 1}}$, so taking $n\to\infty$ gives $f_x(z_1) \le r$, contradiction.$\boxed{}$
Given a valuation $v$ on $K$, we can complete $K$ with respect to the topology induced by $v$ to get a new metric space $K_v$.
Proposition There is a unique way to give $K_v$ the structure of a field such that addition and multiplication are continuous with respect to the valuation $v$.
Proof Note that multiplication and addition are clearly continuous on $K$, so they extend uniquely to multiplication and addition on $K$. Thus $K_v$ already has a ring structure. It suffices to show that $K_v$ has inverses. This is also not hard to show and is left as an exercise to the reader.$\boxed{}$
Proposition Suppose $v$ is a ((non-)Archimedian) valuation on $K$. Then $v$ extends uniquely to a ((non-)Archimedian) valuation on $K_v$.
Proof Obvious by the continuity of $v$. The details are left as an exercise to the reader. $\boxed{}$.
Theorem (Big Ostrowski, Part II) Suppose $v$ is an Archimedian valuation on the number field $K$. Then there exist an embedding $\sigma: K\to \BC$ and a real number $r > 0$ such that $\abs{x}_v = \abs{\sigma(x)}_{\BC}^r$.
Proof Let $K_v$ be the completion of $K$ with respect to $K$. It suffices to show that there exists a valuation-preserving isomorphism from either $K_v\to \BR$ or $K_v\to \BC$.
Note that since we have an embedding $\BQ\to K$ and the restriction of $v$ onto $\BQ$ must be the absolute value norm on $\BQ$, there exists an embedding $\BQ\to K_v$ that extends uniquely to an embedding $\BR\simeq \BQ_v \to K_v$. Thus it suffices to show that every element of $K_v$ has degree at most 2 over $\BR$.
Let $x\in K_v$, and define the function $f_x: \BC\to \BR$ as $z\mapsto \abs{x^2 - (z + \bar{z})x + z\cdot \bar{z}}_{\BC}$. Note that $x$ having degree at most 2 over $\BR$ is equivalent to $f_x(z) = 0$ for some $z\in \BC$, so suppose for the sake of contradiction that $f_x$ never hits 0. Note that the region $\{z: \abs{f_x(z)}_{\BC} < R\}$ is bounded for any $R$. Thus $f_x$ attains a minimum $r>0$ at $z_0$. Since $f_x^{-1}(r)$ is closed and bounded, it is compact. Thus we can pick $z_0$ such that $\abs{z_0}_{\BC}$ is maximal.
Let $0 < \epsilon < r$ be a real number, and let $z_1$ be a root of $z^2 - (z_0 + \bar{z}_0)z+ z_0\cdot \bar{z}_0 + \epsilon = 0$. Since this is also a real quadratic, we have $\abs{z_1}_{\BC}^2 = z_1\cdot \bar{z_1} = z_0\cdot \bar{z}_0 + \epsilon > \abs{z_0}_{\BC}^2$, so $f_x(z_1) > r$.
Now consider $g_n(t) = (t^2 - (z_0 + \bar{z}_0) t+ z_0\cdot \bar{z}_0)^n - (-\epsilon)^n\in \BR[t]$. Note that $g_n(t)$ can be factored uniquely into (real) linear terms and irreducible quadratic terms, so if $t_1, \ldots, t_{2n}\in \BC$ are the roots of $g_n$, then \[ g_n(x)^2 = g_n(t)\cdot \overline{g_n(t)} = \prod_{i = 1}^{2n} (t^2 - (t_i + \bar{t_i})t + t_i\cdot \bar{t_i}) \] Furthermore, $z_1$ is always a root of $g_n(t)$, so without loss of generality $z_1 = t_1$. Thus \[ \abs{g_n(x)}^2_v = \prod_{i = 1}^{2n} \abs{x^2 - (t_i + \bar{t_i})x + t_i\cdot \bar{t_i}}_v\ge f_x(z_1)\cdot r^{2n - 1} \] However, $g_n(x) = r^n - (-\epsilon)^n$. Thus $f_x(z_1)\le \frac{(r^n + (-\epsilon)^n)^2}{r^{2n - 1}}$, so taking $n\to\infty$ gives $f_x(z_1) \le r$, contradiction.$\boxed{}$
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