# Completions II

We will call a topological group complete if $\phi: G\to \hat{G}$ is an isomorphism.

The case that we are particularly concerned with is when our group is a ring $R$ and we take for our inverse system some ideal $\frak{a}\subset R$ and $G_n=\frak{a}^n$. The topology that this determines is the “$\frak{a}$-adic topology”. This makes $R$ into a topological ring.

If we take the completion $\displaystyle \hat{R}=\lim_{\longleftarrow} R/G_n$, then the continuous ring homomorphism $\phi: R\to \hat{R}$ has kernel $\cap \frak{a}^n$.

Now we can also do all this with $R$-modules by taking the group to be $M$ and the inverse system $G_n=\frak{a}^nM$. The topology determined by this system is called the $\frak{a}$-topology on M. If we take the completion with respect to this topology (i.e. w.r.t this system), we get $\hat{M}$ which is a topological $\hat{R}$-module meaning the $\hat{R}$ action is continuous.

Rephrasing the motivating example from last time in this language we see that the $p$-adic integers are formed as the completion of the ring $\mathbb{Z}$ with respect to the $\frak{a}$-topology where $\frak{a}$ is the ideal $\frak{a}=(p)$.

The other really important example is to form the completion of $k[x]$ with respect to the $(x)$-adic topology. The completion is $\displaystyle \widehat{k[x]}=\lim_{\longleftarrow} k[x]/(x^n)=k[[x]]$ the ring of formal power series. Recall that by definition the inverse limit are all sequences $(a_0, \ldots, a_n, \ldots)$ such that $a_{n+1} \mod x^{n+1}\equiv a_n$. This just says that each $a_i$ is a polynomial, and it has to agree with the one before it up to the $x^i$ coefficient. So we can write each sequence $b_0+b_1x+\cdots +b_nx^n+\cdots$ where $b_i$ is the coefficient on the $x^i$ of the polynomial $a_i$. And for any power series we get a sequence in this way.

Recall our notion of $\frak{a}$-filtrations. We had a chain $M=M_0\supset M_1\supset \cdots$ such that $\frak{a}M_n\subset M_{n+1}$, and if equality held for all large $n$, then we called the filtration stable. Well, in our new language, these filtrations are inverse systems of modules, and hence determine a topology on $M$. A few posts ago we used the fact that any stable $\frak{a}$-filtrations have bounded difference. In this new language, this says precisely that all stable $\frak{a}$-filtrations determine the same topology on M, moreover this is the $\frak{a}$-topology.

Lastly, if we convert the Artin-Rees Lemma to this language, we get that if $R$ is Noetherian, $\frak{a}$ an ideal, $M$ a f.g. $R$-module, and $M'$ a submodule of $M$, then the $\frak{a}$-topology on $M'$ is actually just the subspace topology from the $\frak{a}$-topology on $M$.

We should probably do some properties of completions next time.

About these ads

### Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

### One thought on “Completions II”

1. I personally find it very difficult to think of Artin-Rees any other way.