Commutative Algebra 54

Filtered Rings


Let A be a ring. A filtration on A is a sequence of additive subgroups

A = A_0 \supseteq A_1 \supseteq A_2 \supseteq \ldots

such that A_i A_j \subseteq A_{i+j} for any i, j\ge 0. A filtered ring is a ring with a designated filtration.


Since A\cdot A_i = A_0 \cdot A_i \subseteq A_i, in fact each A_i is an ideal of A.


1. If A = \oplus_{i=0}^\infty A_i is a grading, we can form a filtration by taking n \mapsto A_n \oplus A_{n+1} \oplus \ldots (where A_n, A_{n+1}, \ldots refers to the grading).

2. Let \mathfrak a\subseteq A be an ideal. The \mathfrak aadic filtration is given by A_i = \mathfrak a^i. E.g. we can take A = \mathbb Z and \mathfrak a = 2\mathbb Z, then A_i is the set of integers divisible by 2^i. More generally, in a dvr with uniformizer \pi, we can take A_i = (\pi^i).


Suppose A is a filtered ring. A filtration on an A-module M is a sequence of additive subgroups

M = M_0 \supseteq M_1 \supseteq M_2 \supseteq \ldots

such that A_i M_j \subseteq M_{i+j} for any i, j\ge 0. A filtered module is a module with a designated filtration.


Since A\cdot M_i = A_0 M_i \subseteq M_i, each M_i 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.


Again, for an ideal \mathfrak a \subseteq A, we obtain the \mathfrak aadic filtration of M, where M_i = \mathfrak a^i M.


Induced Filtrations

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


Let M, N be filtered A-modules. A linear map f:M\to N is said to be filtered if f(M_i) \subseteq N_i for each i.

We also have:


Let f:M\to N be a linear map of A-modules.

  • If (M_i) is a filtration of M, the induced filtration on N via f is given by N_i = f(M_i).
  • If (N_i) is a filtration of N, the induced filtration on M via f is given by M_i = f^{-1}(N_i).


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

A_i N_j = A_i f(M_j) = f(A_i M_j) \subseteq f(M_{i+j}) = N_{i+j}.

And in the second,

f(A_i M_j) = A_i\cdot f(M_j) \subseteq A_i N_j \subseteq N_{i+j} \implies A_i M_j \subseteq M_{i+j}.

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

N_i = M_i \cap N, \quad (M/N)_i = (M_i + N)/N.

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

warningIf f:M\to N is a filtered A-linear map, then M/\mathrm{ker} f and \mathrm{im }f have induced filtrations via quotient module of M and submodule of N. Although M/\mathrm{ker } f \cong \mathrm{im }f 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

(M / \mathrm{ker } f)_i = (M_i + \mathrm{ker } f)/\mathrm{ker } f, \quad (\mathrm{im } f)_i = (\mathrm{im} f) \cap N_i

which gives

(M_i + \mathrm{ker } f)/\mathrm{ker } f \cong M_i / (M_i \cap \mathrm{ker } f) = M_i / \mathrm{ker} (f|_{M_i}) \cong \mathrm{im} (f|_{M_i})

which is not isomorphic to (\mathrm{im} f) \cap N_i in general.




Let M be a filtered A-module, the completion of M is the following (inverse) limit of A-modules:

\hat M := \varprojlim M/M_n.


By the description of inverse limits in the category of A-modules, \hat M comprises of the set of all (\ldots, x_n, \ldots, x_2, x_1) \in \prod_{i=1}^\infty (M/M_i) such that for each i, x_{i+1} maps to x_i under the canonical map M/M_{i+1}\to M/M_i.

The canonical maps M\to M/M_n induce an A-linear map M \to \hat M by the universal property of inverse limits. This can be described as follows: for each m\in M, let m_i be its image in M/M_i; then the map takes (m \in M) \mapsto (\ldots, m_n, \ldots, m_2, m_1) \in \hat M. From this we see that:

Lemma 1.

The map i:M\to \hat M is injective if and only if \cap_n M_n = 0, in which case we say the filtration is Hausdorff.

Next, by setting MA, we also have \hat A = \varprojlim A/A_n. Although we defined \hat A as an A-module, one sees by the explicit construction that \hat A has a ring structure. To be specific,

(a_n)_{n=1}^\infty, (b_n)_{n=1}^\infty \in \hat A \implies (a_n) \times (b_n) := (a_n b_n)_{n=1}^\infty.

Exercise A

1. Prove that \hat M has a canonical structure as an \hat A-module.

2. Prove that \hat A is the inverse limit of A/A_n in the category of rings. In particular, the canonical map i : A\to \hat A is a ring homomorphism.

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

d(x, y) := 2^{-|x-y|}, where |z| := \sup\{ n : z\in M_n\},

such that the collection of all cosets \{ m + M_n : m\in M, n \ge 0\} forms a basis for the resulting topology. In fact, d is an ultrametric.


Limits in Completion

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


Let m_1, m_2, \ldots \in M be a sequence in a filtered module M. We say the sequence is Cauchy if: for any i, there exists n such that

m_n \equiv m_{n+1} \equiv m_{n+2} \equiv \ldots \pmod {M_i}.

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


The limit of a Cauchy sequence m_1, m_2, \ldots \in M is defined as follows. For each i, pick n such that m_n \equiv m_{n+1} \equiv \ldots modulo M_i, with image y_i \in M/M_i. Now define:

\lim_{n\to\infty} m_n := (\ldots, y_n, \ldots, y_2, y_1) \in \hat M.

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

\begin{aligned}\lim_{n\to\infty} (m_n + m_n') &= (\lim_{n\to\infty} m_n) + (\lim_{n\to\infty} m_n') \in \hat M, \\ \lim_{n\to\infty} (a_n m_n) &= (\lim_{n\to\infty} a_n) (\lim_{n\to\infty} m_n) \in \hat M.\end{aligned}

Example: 𝔪-adic Completion

Pick a ring A with maximal ideal \mathfrak m; we will take the \mathfrak m-adic filtration A_n = \mathfrak m^n. Given a sequence x_i \in A_i for i=0, 1, \ldots, let

y_i := x_0 + \ldots + x_{i-1} \pmod {\mathfrak m^i}, an element of A/A_i.

Then (y_i) is a Cauchy sequence in A and we write:

\sum_{i=0}^\infty x_i := \lim_{n\to\infty} y_n \in \hat A.

Arithmetic Example.

Let A = \mathbb Z and A_n = p^n \mathbb Z for a prime p. The completion \hat A is called the ring of p-adic integers and denoted by \mathbb Z_p. Let us take p = 2 and the element:

y = 1 + 2^2 + 2^4 + 2^6 + \ldots \in \mathbb Z_2.

Then 3y = 3 + 3(2^2) + 3(2^4) + \ldots is congruent to -1 modulo any 2^n. Thus y = -\frac 1 3. In general, an element of \mathbb Z_p can be regarded as having an infinite base-p expansion. Thus the above y \in \mathbb Z_2 would have base-2 expansion (\ldots 1010101)_2. One easily checks that three times this value is (\ldots 1111)_2, which is -1.

Geometric Example

Let A be a ring, B = A[X, Y] and B_n = \mathfrak m^n where \mathfrak m = (X, Y). Again, we can take the infinite sum \alpha = 1 + XY + (XY)^2 + \ldots and check that \alpha(1-XY) = 1. Note that \hat B \cong A[[X,Y]], the ring of formal power series with coefficients in A.


Let A be any ring. A formal power series in X with coefficients in A is an expression

f(X) = a_0 + a_1 X + a_2 X^2 + \ldots, where a_i \in A.

Unlike polynomials, we allow infinitely many a_i to be non-zero. Addition and multiplication of formal power series are defined as follows. For f(X) = \sum_{i=0}^\infty a_n X^n and g(X) = \sum_{j=0}^\infty b_n X^m,

\begin{aligned} f(X) + g(X) &=(a_0 + b_0) + (a_1 + b_1)X + (a_2 + b_2)X^2 + \ldots \\ f(X)\times g(X) &= (a_0 b_0) + (a_0 b_1 + a_1 b_0)X + (a_0 b_2 + a_1 b_1 + a_2 b_0)X^2 + \ldots \end{aligned}

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

A[[X_1, \ldots, X_n]] := (A[[X_1, \ldots, X_{n-1}]])[[X_n]]].

As another example, let us take the \mathfrak m-adic completion for A = \mathbb C[X, Y]/(Y^2 - X^3 + X) and \mathfrak m = (X, Y). We will prove later that

\hat A \cong \mathbb C[[X, Y]]/(Y^2 - X^3 + X)

and that the map \mathbb C[[Y]] \to \hat A is an isomorphism. Geometrically, this means when we project E: Y^2 = X^3 - X to the Y-axis, the map is locally invertible at the origin.


[ Image edited from GeoGebra plot. ]

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

Exercise B

1. Prove that the canonical map \mathbb C[[X]] \to \mathbb C[[X, Y]]/(Y^2 - X^3 + X) is not an isomorphism.

2. Let A be a filtered ring. Prove that if f(X) \in A[[X]] is a_0 + a_1 X + a_2 X^2 + \ldots, then f defines a map

A_1 \to \hat A, \quad f(\alpha) := \sum_{n=0}^\infty a_n \alpha^n \in \hat A.

Prove that if we fix \alpha \in A_1, we get a ring homomorphism A[[X]] \to \hat A, f\mapsto f(\alpha).


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5 Responses to Commutative Algebra 54

  1. Vanya says:

    In the first definition in the section “Limits in completion” it is mentioned that “Let m_1, m_2, \ldots \in M, but in the congruences immediately following that sentence x_n \equiv x_{n+1} \dots \pmod{M_i}.

  2. Vanya says:

    The scalar multiplication just before \mathfrak{m}-adic completion has a typo.

  3. Vanya says:

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

    • limsup says:

      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 y^2 = x^3 - x^2 or y^2 = x^3 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 A = \mathbb C[X, Y]/(Y^2 - X^3 + X) at the origin is in fact isomorphic to \mathbb C[[Y]] yet the localization of A at the origin is not the localization of a polynomial ring.

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