So again, we’re back to completions, though we’re going to go through it quickly. Except this time we have a field {F} with an absolute value {\left \lvert . \right \rvert} like the rationals with the usual absolute value.

 Completions   

Definition 1 The completion {\hat{F}} of {F} is defined as the set of equivalence classes of Cauchy sequences: 

  • A Cauchy sequence {\{x_n\}} satisfies {\left \lvert {x_n-x_m} \right \rvert \rightarrow 0} as {n,m \rightarrow \infty}.
  • Two Cauchy sequences {\{x_n\},\{y_n\}} are equivalent if {\left \lvert {x_n-y_m} \right \rvert \rightarrow 0} as {n \rightarrow \infty}. 

 

First off, {\hat{F}} is a field, since we can add or multiply Cauchy sequences termwise; division is also allowable if the sequence stays away from zero. There is a bit of justification to check here, but it is straighforward. Also, {F} had an absolute value, so we want to put on on {\hat{F}}. If {\{x_n\} \in \hat{F}}, define {\left \lvert\{ x_n\} \right \lvert := \lim \left \lvert x_n \right \rvert}. Third, there is a natural map {F \rightarrow \hat{F}} and the image of {F} is dense.

There are several important examples of this, of which the most basic are:

Example 1 The completion of {\mathbb{Q}} with respect to the usual absolute value is the real numbers.  

Example 2 The completion of {\mathbb{Q}} with respect to the {p}-adic absolute value is the {p}-adic numbers {\mathbb{Q}_p}.  

The second case is more representative of what we care about: completions with respect to nonarchimedean (especially discrete) valuations. By the general criterion (testing that integers have absolute value at most 1), completions preserve nonarchimedeanness. Next, here is a frequently used lemma about nonarchimedean fields: 

Lemma 2 Let {F} be a field with a nonarchimedean absolute value {\left \lvert . \right \lvert }. Then if {x, y \in F} and {\left \lvert{x-y} \right \lvert < \left \lvert x \right \lvert }, then {\left \lvert x \right \lvert = \left \lvert y \right \lvert }.

(Two elements very close together have the same absolute value. Or, any disk in a nonarchimedean field has each interior point as a center.)

Indeed, {\left \lvert y \right \lvert = \left \lvert(y-x) + x \right \lvert \leq \max( \left \lvert x-y \right \lvert , \left \lvert x \right \lvert ) = \left \lvert x \right \lvert }. Similarly {\left \lvert x \right \rvert = \left \lvert(x-y) + y \right \lvert \leq \max( \left \lvert x-y \right \rvert , \left \lvert y \right \lvert )}, and since {\left \lvert x \right \rvert > \left \lvert x-y \right \rvert} we can write in the second {\left \lvert x \right \rvert \leq \left \lvert y \right \rvert}.

Corollary 3 If {\left \lvert . \right \rvert} is discrete on {F}, it is discrete on {\hat{F}}.

Indeed, if {V \subset \mathbb{R}^+} is the value group (absolute values of nonzero elments) of {F}, then it is the value group of {\hat{F}} since {F} is dense in {\hat{F}}. 

Completions of Rings  

Now, time to connect this idea of completion with the previous one.

Take a field {F} with a discrete valuation {\left \lvert . \right \rvert} and its completion {\hat{F}}. We can take the ring of integers {R \subset F} and {\hat{R} \subset \hat{F}}, and their maximal ideals {\mathfrak{m}, \hat{\mathfrak{m}}}.

I claim that {\hat{R}} is the completion of {R} with respect to the {\mathfrak{m}}-adic topology. This follows because {\hat{R}} consists of equivalence classes of sequences {\{x_n\}} of elements of {F}, the limit of whose absolute values {\left \lvert x_n \right \rvert} is {\leq 1}. This means from some point on, the {\left \lvert x_n \right \rvert \leq 1} by discreteness, so wlog all the {x_n \in R}. This is just the definition of an element of the completion of {R}. I leave the remaining details to the reader.

(To avoid discreteness, for sequences {\{x_n\}} with {\left \lvert x_n \right \rvert \rightarrow 1} that do not go into {R}, replace it by the equivalent {\{ x_n/\left \lvert x_n \right \lvert \}}—this way one replaces it with a sequence that lies in {R}.)