Exactness of Completion
Throughout this article, A denotes a filtered ring.
Let be a short exact sequence of A-modules. Suppose M is filtered, inducing filtrations on N and P. Then
is also exact as -modules.
Without loss of generality, assume N is a submodule of M and P = M/N. Each term in the filtration gives a short exact sequence
since by the second isomorphism theorem. By proposition 1 here, taking (inverse) limit is left-exact so we obtain an exact sequence
To show that is surjective, we pick an element of . Since , the element is represented by a sequence in M such that . We need to show there is a sequence in M such that
When k = 0, just pick any . Suppose we have ; we need such that and . But observe that . If we write for , then works. ♦
Completion of Completion
We have , where has the filtration induced from M.
Let . From proposition 1 we get an exact sequence
But we also have which is for all . Thus and we are done. ♦
Hence if we let take the filtration given by
then by lemma 1, the completion of with respect to this filtration is still .
If is a Cauchy sequence, from the previous article we have its limit
Since the map from to its completion is injective, we have so as shown in exercise A.3 here, we can define an (ultra)metric on such that the resulting topology has a basis comprising of the set of all cosets . From the above, every Cauchy sequence converges in . Thus:
is a complete metric space.
Furthermore, the image of is dense; indeed any basic open subset of is of the form for and . Since , we see that m can be represented by an element of M. Thus any non-empty open subset of contains an element of M.
Thus is the completion of M even in the topological sense.
For visualization, one can show that is homeomorphic to the Cantor set:
E.g. the point above corresponds to a 2-adic integer ending at .
The 𝔞-adic Filtration
Now instead of arbitrary filtrations on M, we will focus our attention to the 𝔞-adic filtrations on A and M for a fixed ideal :
Clearly if M is given the 𝔞-adic filtration, so is any quotient, because , so the induced filtration on M/N is also 𝔞-adic. On the other hand, the induced filtration on a submodule N is .
But the situation is salvageable when A is noetherian. Instead of the 𝔞-adic filtration, let us loosen our definition a little.
A filtration of M is said to be 𝔞-stable if for some n, we have for all .
In other words, an 𝔞-stable filtration is “eventually 𝔞-adic”. When we take the completion, we get the same thing.
Suppose M is an A-module with an 𝔞-stable filtration. Its completion is canonically isomorphic to the 𝔞-adic completion of M.
Since is a filtration for M we have , i.e. . Now fix an n such that for all . We get
and hence maps . Taking the inverse limit:
By explicitly writing out elements of inverse limits, we see that the above give isomorphisms and ; thus
1. Fill in the last step of the proof.
2. Show that in any category, the inverse limit of the diagram
remains the same when we drop finitely many terms on the right.
The main result we wish to prove is the following.
Let A be a noetherian ring with the -adic filtration, and N a submodule of a finitely generated A-module M. If M has an -stable filtration, the induced filtration on N is also -stable.
Step 1: define blowup algebra and module.
Given any filtered module M over a filtered ring A, the blowup algebra and blowup module are defined by
We define a product operation from multiplication in A. Hence, B(A) has a canonical structure of a graded ring.
Similarly, since M is a filtered module, we obtain a product operation which gives B(M) a structure of a graded B(A)-module. When A and M are given the 𝔞-adic filtration, we write and for their blowup algebra and module.
Step 2: if A is a noetherian ring, so is B𝔞(A).
Since A is noetherian, we can write for some . It follows that is a sum of where . Hence the map
is a surjective ring homomorphism so is also noetherian.
Now we suppose A is noetherian and is given the 𝔞-adic filtration. Let M be a finitely generated filtered A-module.
Step 3: B(M) is finitely generated if and only if the filtration on M is 𝔞-stable.
(⇐) For some n we have Since M is a noetherian A-module, each () is finitely generated as an A-module by, say . Now we take the set of , as homogeneous elements of B(M) of degree i.
In the above, each homogeneous element of is an A-linear combination of these generators. Furthermore, so generate (over ) the homogeneous elements in B(M) of degree n and higher.
(⇒) Suppose is finitely generated over by homogeneous elements ; let and . We claim that for all . Since M is filtered, we have
Conversely take , regard it as an element of and write with . Since y and are homogeneous, we may assume is homogeneous of degree . So . Write
Now y is a sum of , with so .
Step 4: prove the Artin-Rees lemma.
By step 2, is a noetherian ring; since M has an -stable filtration, by step 3 B(M) is a noetherian -module. And since is a -submodule it is also noetherian. By step 3 again, this says the induced filtration on N is -stable. ♦