Completeness

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

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

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

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

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

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

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

(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

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

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

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

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

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

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

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

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 $K$. $\qedsymbol$