Posted by: hilbertthm90 | November 14, 2009

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.

Posted in algebra, topology | Tags: , , ,

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Responses

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

    By: Akhil Mathew on November 15, 2009
    at 7:37 am


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