3
$\begingroup$

Let $V$ be a discrete valuation ring with the maximal ideal $\mathfrak{m}$ and let $T$ be a prime element of $V$. Assume that we have a subfield $k\subseteq V$ such that the induced map $ k \to V/\mathfrak{m}$ is an isomorphism. Prove that we have an isomorphism $k[[T]] \cong \widehat{V}$ where $\widehat{V}$ denotes the $\mathfrak{m}$-adic completion of $V$.

What is the definition of/meant by $\mathfrak{m}$-adic completion here?

$\endgroup$
  • $\begingroup$ It's not simple to explain, you should look up the wiki page on the $p$-adics. Basically it's analogous to completion of $\mathbb Q$ to get $\mathbb R$, but with respect to a different ``absolute value" than the usual one. $\endgroup$ – Gregory Grant Apr 23 '15 at 20:20
  • $\begingroup$ @GregoryGrant this actually cannot happen for $p$-adic style things because having a copy of your residue field from a $p$-adic field is impossible (different characteristics) $\endgroup$ – Adam Hughes Apr 23 '15 at 20:42
  • $\begingroup$ @AdamHughes What are you referring to exactly by "this" in "this actually cannot happen"? $\endgroup$ – Gregory Grant Apr 23 '15 at 20:51
  • $\begingroup$ @GregoryGrant the context of the op's question disqualifies the exact nature of this problem having the same behavior as the $p$-adics, but I now see that you probably just were referring to the topological question, in which case disregard my comments, please. :-) $\endgroup$ – Adam Hughes Apr 23 '15 at 20:55
0
$\begingroup$

Edit: I found an even easier way to explain it, so I have revised the answer.

The definition of $\mathfrak{m}$-adic completion is always completion (i.e. convergent sequences, since this is a metric space) relative to the $\mathfrak{m}$-adic topology or, more simply, the $\mathfrak{m}$-adic metric on $V$. Recall that this topology is defined by having $\mathfrak{m}^k, k\in\Bbb N$ be a fundamental system of neighborhoods around $0\in V$, or--in metric terms--by having $d(x,y)=|x-y|_{\mathfrak{m}}$, whichever you find more comfortable to think about.

So an element of the completion is a convergent sequence, just like in all metric spaces, so let's examine how we get at these convergent sequences in our contest.

First and foremost note that we are in a topological ring--a place where we have the all-important operation of addition available to us. Since the metric is given by $d(x,y)=|x-y|_\mathfrak{m}$ uses this operation of addition to define it, we are able to reduce the study of convergent sequences to the study of convergent series in the following way:

Let $\{a_n\}$ be a convergent sequence, then it can be given instead by the (convergent) series

$$\begin{cases} S_1=a_1 \\ S_N=a_1+\sum_{i=2}^N (a_i-a_{i-1}), N>1 \end{cases}$$

Since our field is non-archimedean, we know that a series converges iff the terms go to $0$ in absolute value. If we denote by $\kappa$ be the residue field for $V$, we know that such series with terms going to $0$ all have tails which are elements of $\kappa[[T]]$.

Now, since $k=V/\mathfrak{m}$ is just things of the form

$$x\in V, x=a_{-N}T^{-N}+\ldots +a_{-1}T^{-1}+a_0, \quad a_i\in\kappa $$

We now note that elements of $V$ are either in the maximal ideal, or they aren't. If they are not, then they are units, since $\langle a\rangle$ is an ideal of $V$, and all proper ideals are contained in the (unique) maximal ideal. So this means $N=0$ always for elements of $V/\mathfrak{m}$, and so $k=\kappa$.

This follows from the standard theory and definition of the residue field, and the definition of $V/\mathfrak{m}$. So we have that the completion is

$$k[\kappa[[T]]]=\kappa[\kappa[[T]]]=k[[T]]$$

as desired.

$\endgroup$
0
$\begingroup$

$\newcommand{\a}{\mathfrak a} \newcommand{\N}{\mathbf N}$ Consider an arbitrary ring $R$ and an ideal $\a \unlhd R$. (I assume $R$ to be commutative, but I don't think it is needed, otherwise you just need to be a little more careful about which side you multiply on.)

If you know what an inverse limit is, then the completion $\hat R$ of $R$ with respect to $\a$ is the inverse limit of $R/\a^n$ with $n\in \N$ and the natural bonding maps.

If you don't, consider the ring $R^\N$ of all sequences of elements of $R$, and $\bar \a :=\prod_{n\in \N} \a^n\unlhd R^\N$ ($\bar a$ is the ideal of sequences whose $n$-th element is in $\a^n$). Then the completion $\hat R$ of $R$ with respect to $\a$ is the subring of $R^\N/\bar \a$ which consists of those sequences $(r_n+\a^n)_{n\in \N}$ such that $\a r_n-r_{n+1}\in \a^{n+1}$.

We have a natural homomorphism $R\to \hat R$ (take $r$ to the class of a constant sequence), and whenever $\a$ has the property that $\bigcap_{n\in \N} \a^n=\{0\}$ (such as when $\a$ is the maximal ideal of a DVR), this is $1-1$.

Completions are discussed at some length in Atiyah-Macdonald (Commutative Algebra), so you might want to check it out.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy