So, we saw in the previous post that completion can be defined generally for abelian groups. Now, to specialize to rings and modules.

**Rings **

The case in which we are primarily interested comes from a ring with a descending filtration (satisfying ), which implies the are ideals; as we saw, the completion will also be a ring. Most often, there will be an ideal such that , i.e. the filtration is -adic. We have a completion functor from filtered rings to rings, sending . Given a filtered -module , there is a completion , which is also a -module; this gives a functor from filtered -modules to -modules.

We’ll also primarily restrict attention to the case when is Noetherian and finitely generated. The following is thus useful:

Theorem 1If is Noetherian and an ideal, then the -adic completion is too.

As I mentioned above, the trick is to use the above result to get a filtration , with (by the end of Sunday’s post)

But is an -algebra, and if generate as an ideal, then is generated as an -algebra by , so is Noetherian.

Given , we can consider the **initial form** defined as follows: if is such that , then is the image of in in the -th part of the direct sum defining . Note that is well-defined except when , since since a complete ring is Hausdorff. Now fix an ideal . Consider the ideal generated by ; being Noetherian, we can find a finite number of generators for .

Set . I now claim that , which will prove the theorem.

This in turn will follow from

Claim 1

Wlog, so is homogeneous of degree . Now , which means we can find , with

By taking homogeneous parts, assume that the are homogeneous of degree , so lift each to . Then , so

which implies (1).

Now, to prove , proceed as follows. If , find , with ; then find with

Repeat this to get , and note that converges for each . This proves .

The previous proof highlighted an important technique which we will use often: *successive approximation*. But first, here is a corollary.

Corollary 2If is Noetherian, the power series ring is Noetherian.

Indeed, Hilbert’s basis theorem implies the polynomial ring is Noetherian; now, as is easily checked, the power series ring is the completion of with respect to the ideal . So the previous result implies the theorem.

Incidentally, Zariski-Samuels prove this corollary by imitating the proof of Hilbert’s basis theorem itself; using completions is more general though. Even so, there is a useful structure theorem stating that in many cases, complete rings are actually of this form.

In view of the somewhat technical proof above, the question arises: Why does this all matter? Well, one of the most important examples is the -adic integers , defined as the completion of with respect to the -adic (i.e. -adic) topology, for prime. The -adic integers and their quotient field, the -adic numbers are important in number theory, as I hope to eventually discuss (when I get to class field theory). One of their important properties, shared generally by completions, is *Hensel’s lemma*: using a similar successive approximation technique, we can lift “approximate” solutions of equations to exact ones in a complete ring. Thus, it becomes much easier to check whether a polynomial has a root in a complete ring. And since “local-global” results such as the Hasse-Minkowski theorem imply that properties over can be studied by the completions, these become very useful indeed.

**Modules **

Let be a commutative ring and an ideal. If is an -module, we can consider the -adic completion If is Noetherian, then -adic completion is actually an exact functor on the category of finitely generated -modules.

Theorem 3If is a Noetherian ring and an exact sequence of f.g. -modules, then is exact.

First of all, I claim that . This follows from the Artin-Rees lemma: the projective limit is the completion of with respect to the topology from the filtration . But Artin-Rees implies this is the same topology as that induced by the -adic filtration . Since completion only depends on the topology, the claim follows.

This is one reason I like to think in terms of topologies and not merely algebraically via inverse limits, though some books don’t seem to indicate this (in my opinion, cleaner) proof.

Now by the claim, for each we have an exact sequence

When we take the inverse limit of these sequences, we get the sequence of completions. The result follows from the two general lemmas:

Lemma 4If is an inverse system of exact sequences, then

is exact.

The proof is perhaps best thought of as a general phenomenon involving adjoint functors, but there is a quick (if less enlightening) direct proof too:

Injectivity on the left can be checked directly from the definition of as a subset of some direct product. Similarly, if has each in , then in fact each and satisfies appropriate compatibility conditions, so it follows that comes from some element in .

Right-exactness is not true generally, though in some special cases it is:

Lemma 5If is an inverse system of exact sequences with the maps for surjective, then

is exact.

We need only to show that if we are given , we can lift this to some . For a preliminary approximation, choose for each some with ; however needn’t satisfy the compatibility conditions for the inverse limit.

Define the from the inductively as follows. Set . Assume are defined, satisfy the compatibility conditions, and map to . Then if is the map defining the inverse system, we have maps to zero in (draw a commutative diagram). So . Now by surjectivity, choose with

and set . Since , still maps to in , and (2) implies the compatibility condition holds. Now continue the induction.

August 27, 2009 at 10:43 am

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September 1, 2009 at 4:26 pm

[…] claim that is the completion of with respect to the -adic topology. This follows because consists of equivalence classes of […]