We recall the definition of metric on a set $ X$.

Definition 16.2.1 (Metric)   A on a set $ X$ is a map

$\displaystyle d : X \times X \rightarrow \mathbf{R}

such that for all $ x,y,z\in X$,
  1. $ d(x,y)\geq 0$ and $ d(x,y)=0$ if and only if $ x=y$,
  2. $ d(x,y)=d(y,x)$, and
  3. $ d(x,z)\leq d(x,y)+d(y,z)$.

A is a sequence $ (x_n)$ in $ X$ such that for all $ \varepsilon >0$ there exists $ M$ such that for all $ n,m>M$ we have $ d(x_n,x_m)<\varepsilon $. The of $ X$ is the set of Cauchy sequences $ (x_n)$ in $ X$ modulo the equivalence relation in which two Cauchy sequences $ (x_n)$ and $ (y_n)$ are equivalent if $ \lim_{n\rightarrow \infty} d(x_n,y_n)=0$. A metric space is if every Cauchy sequence converges, and one can show that the completion of $ X$ with respect to a metric is complete.

For example, $ d(x,y)=\vert x-y\vert$ (usual archimedean absolute value) defines a metric on  $ \mathbf{Q}$. The completion of $ \mathbf{Q}$ with respect to this metric is the field $ \mathbf{R}$ of real numbers. More generally, whenever $ \left\vert \cdot \right\vert$ is a valuation on a field $ K$ that satisfies the triangle inequality, then $ d(x,y)=\left\vert x-y\right\vert$ defines a metric on $ K$. Consider for the rest of this section only valuations that satisfy the triangle inequality.

Definition 16.2.2 (Complete)   A field $ K$ is with respect to a valuation $ \left\vert \cdot \right\vert$ if given any Cauchy sequence $ a_n$, ( $ n=1,2,\ldots$), i.e., one for which

$\displaystyle \left\vert a_m - a_n\right\vert \to 0 \qquad(m,n\to \infty,\infty),

there is an $ a^*\in K$ such that

$\displaystyle a_n \to a^*$    w.r.t. $\displaystyle \left\vert \cdot \right\vert

(i.e., $ \left\vert a_n-a^*\right\vert\to 0$).

Theorem 16.2.3   Every field $ K$ with valuation $ v=\left\vert \cdot \right\vert$ can be embedded in a complete field $ K_v$ with a valuation $ \left\vert \cdot \right\vert{}$ extending the original one in such a way that $ K_v$ is the closure of $ K$ with respect to $ \left\vert \cdot \right\vert{}$. Further $ K_v$ is unique up to a unique isomorphism fixing $ K$.

Proof. Define $ K_v$ to be the completion of $ K$ with respect to the metric defined by $ \left\vert \cdot \right\vert$. Thus $ K_v$ is the set of equivalence classes of Cauchy sequences, and there is a natural injective map from $ K$ to $ K_v$ sending an element $ a\in K$ to the constant Cauchy sequence $ (a)$. Because the field operations on $ K$ are continuous, they induce well-defined field operations on equivalence classes of Cauchy sequences componentwise. Also, define a valuation on $ K_v$ by

$\displaystyle \left\vert(a_n)_{n=1}^{\infty}\right\vert = \lim_{n\to\infty} \left\vert a_n\right\vert,$

and note that this is well defined and extends the valuation on $ K$.

To see that $ K_v$ is unique up to a unique isomorphism fixing $ K$, we observe that there are no nontrivial continuous automorphisms $ K_v\to
K_v$ that fix $ K$. This is because, by denseness, a continuous automorphism $ \sigma: K_v\to K_v$ is determined by what it does to $ K$, and by assumption $ \sigma$ is the identity map on $ K$. More precisely, suppose $ a\in K_v$ and $ n$ is a positive integer. Then by continuity there is $ \delta>0$ (with $ \delta<1/n$) such that if $ a_n\in K_v$ and $ \left\vert a-a_n\right\vert<\delta$ then $ \left\vert\sigma(a)-\sigma(a_n)\right\vert<1/n$. Since $ K$ is dense in $ K_v$, we can choose the $ a_n$ above to be an element of $ K$. Then by hypothesis $ \sigma(a_n)=a_n$, so $ \left\vert\sigma(a) - a_n\right\vert < 1/n$. Thus $ \sigma(a) = \lim_{n\to\infty} a_n = a$. $ \qedsymbol$

Corollary 16.2.4   The valuation $ \left\vert \cdot \right\vert$ is non-archimedean on $ K_v$ if and only if it is so on $ K$. If $ \left\vert \cdot \right\vert$ is non-archimedean, then the set of values taken by $ \left\vert \cdot \right\vert{}$ on $ K$ and $ K_v$ are the same.

Proof. The first part follows from Lemma 15.2.10 which asserts that a valuation is non-archimedean if and only if $ \left\vert n\right\vert<1$ for all integers $ n$. Since the valuation on $ K_v$ extends the valuation on $ K$, and all $ n$ are in $ K$, the first statement follows.

For the second, suppose that $ \left\vert \cdot \right\vert{}$ is non-archimedean (but not necessarily discrete). Suppose $ b\in K_v$ with $ b\neq 0$. First I claim that there is $ c\in K$ such that $ \left\vert b-c\right\vert < \left\vert b\right\vert$. To see this, let $ c'=b-\frac{b}{a}$, where $ a$ is some element of $ K_v$ with $ \left\vert a\right\vert>1$, note that $ \left\vert b-c'\right\vert=\left\vert\frac{b}{a}\right\vert<\left\vert b\right\vert$, and choose $ c\in K$ such that $ \left\vert c-c'\right\vert < \left\vert b-c'\right\vert$, so

$\displaystyle \left\vert b-c\right\vert = \left\vert b-c' - (c-c')\right\vert
...t c-c'\right\vert\right) = \left\vert b-c'\right\vert<\left\vert b\right\vert.

Since $ \left\vert \cdot \right\vert{}$ is non-archimedean, we have

$\displaystyle \left\vert b\right\vert = \left\vert(b-c)+c\right\vert \leq \max\...
...\vert b-c\right\vert,\left\vert c\right\vert\right) = \left\vert c\right\vert,

where in the last equality we use that $ \left\vert b-c\right\vert < \left\vert b\right\vert$. Also,

$\displaystyle \left\vert c\right\vert = \left\vert b + (c-b)\right\vert \leq \m...
...\vert b\right\vert,\left\vert c-b\right\vert\right) = \left\vert b\right\vert,

so $ \left\vert b\right\vert = \left\vert c\right\vert$, which is in the set of values of $ \left\vert \cdot \right\vert{}$ on $ K$. $ \qedsymbol$

William Stein 2004-05-06