# Filtered Rings

Definition.Let A be a ring. A

filtrationon A is a sequence of additive subgroupssuch that for any . A

filtered ringis a ring with a designated filtration.

**Note**

Since , in fact each is an ideal of *A*.

**Examples**

1. If is a grading, we can form a filtration by taking (where refers to the grading).

2. Let be an ideal. The –**adic** **filtration** is given by . E.g. we can take and , then is the set of integers divisible by . More generally, in a dvr with uniformizer , we can take .

Definition.Suppose A is a filtered ring. A

filtrationon an A-module M is a sequence of additive subgroupssuch that for any . A

filtered moduleis a module with a designated filtration.

**Note**

Since , each is an *A*-submodule of *M*.

*Also, we need a fixed filtration on the base ring A before we can talk about filtrations on A-modules.*

**Example**

Again, for an ideal , we obtain the –**adic filtration** of *M*, where .

# Induced Filtrations

For the rest of this article, *A* denotes a filtered ring.

Definition.Let M, N be filtered A-modules. A linear map is said to be

filteredif for each i.

We also have:

Definition.Let be a linear map of A-modules.

- If is a filtration of M, the
induced filtrationon N via f is given by .- If is a filtration of N, the
induced filtrationon M via f is given by .

**Note**

Let us show that the induced filtrations are legitimate. In the first case,

.

And in the second,

In particular, if *M* is a filtered *A*-module and is a submodule, the induced filtrations on *N* and *M*/*N* are given by:

So far everything is natural, but beneath all this a danger lurks.

If is a filtered *A*-linear map, then and have induced filtrations via quotient module of *M* and submodule of *N*. Although as *A*-modules, it is not an isomorphism in the category of filtered *A*-modules!

Let us write everything out explicitly. The filtration on the LHS and RHS are given respectively by

which gives

which is not isomorphic to in general.

# Completion

Definition.Let M be a filtered A-module, the

completionof M is the following (inverse) limit of A-modules:

By the description of inverse limits in the category of *A*-modules, comprises of the set of all such that for each *i*, maps to under the canonical map .

The canonical maps induce an *A*-linear map by the universal property of inverse limits. This can be described as follows: for each , let be its image in ; then the map takes . From this we see that:

Lemma 1.The map is injective if and only if , in which case we say the filtration is

Hausdorff.

Next, by setting *M* = *A*, we also have . Although we defined as an *A*-module, one sees by the explicit construction that has a ring structure. To be specific,

**Exercise A**

1. Prove that has a canonical structure as an -module.

2. Prove that is the inverse limit of in the category of rings. In particular, the canonical map is a ring homomorphism.

3. Prove that if the filtration on *M* is Hausdorff, we can define a metric on *M* via:

, where

such that the collection of all cosets forms a basis for the resulting topology. In fact, *d* is an ultrametric.

# Limits in Completion

The completion enables us to take limits of infinite sequences and sums of infinite series in modules.

Definition.Let be a sequence in a filtered module M. We say the sequence is

Cauchyif: for any i, there exists n such that.

The astute reader would note that if the filtration is Hausdorff, then is Cauchy here if and only if it is Cauchy with respect to the metric of exercise A.3.

Definition.The

limitof a Cauchy sequence is defined as follows. For each i, pick n such that modulo , with image . Now define:.

From the definition, it is clear that if (resp. ) are Cauchy sequences in *M* (resp. in *A*), then and are also Cauchy and we have

## Example: 𝔪-adic Completion

Pick a ring *A* with maximal ideal ; we will take the -adic filtration . Given a sequence for , let

, an element of .

Then is a Cauchy sequence in *A* and we write:

.

**Arithmetic Example**.

Let and for a prime *p*. The completion is called the ring of **p****-adic integers** and denoted by . Let us take *p* = 2 and the element:

Then is congruent to -1 modulo any . Thus . In general, an element of can be regarded as having an infinite base-*p* expansion. Thus the above would have base-2 expansion . One easily checks that three times this value is , which is -1.

**Geometric Example**

Let *A* be a ring, and where . Again, we can take the infinite sum and check that . Note that , the ring of *formal power series* with coefficients in *A*.

Definition.Let A be any ring. A

formal power seriesin X with coefficients in A is an expression, where .

Unlike polynomials, we allow infinitely many to be non-zero. Addition and multiplication of formal power series are defined as follows. For and ,

This gives a ring structure on the set of formal power series. To define rings of formal power series in multiple variables, we set recursively

As another example, let us take the -adic completion for and . We will prove later that

and that the map is an isomorphism. Geometrically, this means when we project to the *Y*-axis, the map is locally invertible at the origin.

[ Image edited from GeoGebra plot. ]

Functorially, the ring behaves quite differently from , because as an *A*-module, it is a countably infinite *direct product* of copies of *A*, unlike which is a direct sum. If we follow the earlier guideline, it is (for example) generally false that for any *A*-algebra *B*.

**Exercise B**

1. Prove that the canonical map is *not* an isomorphism.

2. Let *A* be a filtered ring. Prove that if is , then *f* defines a map

Prove that if we fix , we get a ring homomorphism , .

In the first definition in the section “Limits in completion” it is mentioned that “Let , but in the congruences immediately following that sentence .

The scalar multiplication just before -adic completion has a typo.

Thanks! Corrected this error and the earlier one.

What does “locally invertible at the origin” in the diagram mean?

This one should be regarded as just a form of geometric intuition. If we “zoom in” the graph at the origin, eventually it resembles a straight line. For a non-example, if we “zoom in” the graph of or at the origin, you will see a singularity there.

This sounds familiar: you might have seen this earlier for localization, but the completion takes the intuition one step further. For example, the completion of at the origin is in fact isomorphic to yet the localization of at the origin is not the localization of a polynomial ring.