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 ${R}$.)