Exactness of Completion

Throughout this article, A denotes a filtered ring.

Proposition 1.

Let $0 \to N \to M \to P \to 0$ be a short exact sequence of A-modules. Suppose M is filtered, inducing filtrations on N and P. Then

$0 \longrightarrow \hat N \longrightarrow\hat M \longrightarrow \hat P \longrightarrow 0$

is also exact as $\hat A$ -modules.

Proof

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

$0 \longrightarrow \overbrace{N/(M_i \cap N)}^{N/N_i} \longrightarrow M/M_i \longrightarrow \overbrace{M/(M_i + N)}^{P/P_i} \longrightarrow 0$

since $N/(M_i \cap N) \cong (M_i + N)/M_i$ by the second isomorphism theorem. By proposition 1 here, taking (inverse) limit is left-exact so we obtain an exact sequence

$0\longrightarrow \hat N \longrightarrow \hat M \longrightarrow \hat P$ .

To show that $\hat M \to \hat P$ is surjective, we pick an element of $\hat P$ . Since $P/P_k \cong M/(M_k + N)$ , the element is represented by a sequence $(m_k)$ in M such that $m_{k+1} - m_k \in M_k + N$ . We need to show there is a sequence $(x_k)$ in M such that

$k\ge 0 \implies x_{k+1} - x_k \in M_k, x_k - m_k \in M_k + N.$

When k = 0, just pick any $x_0$ . Suppose we have $x_0, \ldots, x_k$ ; we need $x_{k+1} \in M$ such that $x_{k+1} - x_k \in M_k$ and $x_{k+1} - m_{k+1} \in M_{k+1} + N$ . But observe that $m_{k+1} - x_k = (m_{k+1} - m_k) + (m_k - x_k) \in M_k + N$ . If we write $m_{k+1} - x_k = m + n$ for $m\in M_k, n\in N$ , then $x_{k+1} := m_{k+1} - n$ works. ♦

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Completion of Completion

Lemma 1.

We have $\hat M /\hat M_n \cong M/M_n$ , where $M_n$ has the filtration induced from M.

Proof

Let $P = M/M_n$ . From proposition 1 we get an exact sequence

$0 \to \hat M_n \to \hat M \to \hat P \to 0.$

But we also have $P/P_m = M/(M_m + M_n)$ which is $M/M_n$ for all $m\ge n$ . Thus $\hat P = M/M_n$ and we are done. ♦

Hence if we let $\hat M$ take the filtration given by

$\hat M = \hat M_0 \supseteq \hat M_1 \supseteq \ldots$

then by lemma 1, the completion of $\hat M$ with respect to this filtration is still $\hat M$ .

If $m_1, m_2, \ldots \in \hat M$ is a Cauchy sequence, from the previous article we have its limit

$(\lim_{n\to \infty} m_n) \in \hat{\hat M} = \hat M$

Since the map from $\hat M$ to its completion is injective, we have $\cap_n \hat M_n = 0$ so as shown in exercise A.3 here, we can define an (ultra)metric on $\hat M$ such that the resulting topology has a basis comprising of the set of all cosets $\{m + \hat M_n\}$ . From the above, every Cauchy sequence converges in $\hat M$ . Thus:

Summary.

$\hat M$ is a complete metric space.

Furthermore, the image of $M \to \hat M$ is dense; indeed any basic open subset of $\hat M$ is of the form $m + \hat M_n$ for $m\in \hat M$ and $n\ge 0$ . Since $\hat M / \hat M_n\cong M/M_n$ , we see that m can be represented by an element of M. Thus any non-empty open subset of $\hat M$ contains an element of M.

Thus $\hat M$ is the completion of M even in the topological sense.

Note

For visualization, one can show that $\mathbb Z_2$ is homeomorphic to the Cantor set:

cantor_2-adic

E.g. the point above corresponds to a 2-adic integer ending at $(\ldots 0010)_2$ .

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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 $\mathfrak a$ :

$M_n = \mathfrak a^n M \implies \hat M = \varprojlim M/\mathfrak a^n M.$

Clearly if M is given the 𝔞-adic filtration, so is any quotient, because $\mathfrak a^n(M/N) = (\mathfrak a^n M + N)/N$ , so the induced filtration on M/N is also 𝔞-adic. On the other hand, the induced filtration on a submodule N is $\mathfrak a^n M \cap N\ne \mathfrak a^n N$ .

But the situation is salvageable when A is noetherian. Instead of the 𝔞-adic filtration, let us loosen our definition a little.

Definition.

A filtration $(M_n)$ of M is said to be 𝔞-stable if for some n, we have $M_{n+k} = \mathfrak a^k M_n$ for all $k\ge 0$ .

In other words, an 𝔞-stable filtration is “eventually 𝔞-adic”. When we take the completion, we get the same thing.

Proposition 2.

Suppose M is an A-module with an 𝔞-stable filtration. Its completion is canonically isomorphic to the 𝔞-adic completion of M.

Proof

Since $(M_n)$ is a filtration for M we have $A_i M_j \subseteq M_{i+j}$ , i.e. $\mathfrak a^i M_j \subseteq M_{i+j}$ . Now fix an n such that $M_{n+k} = \mathfrak a^k M_n$ for all $k\ge 0$ . We get

$k\ge 0 \implies M_k \supseteq \mathfrak a^k M \supseteq \mathfrak a^k M_n = M_{n+k} \supseteq \mathfrak a^{n+k}M$

and hence maps $M/\mathfrak a^{n+k}\to M/M_{n+k} \to M/\mathfrak a^k M \to M/M_k$ . Taking the inverse limit:

$\varprojlim_k M/\mathfrak a^{n+k}M \to \varprojlim_k M/M_{n+k} \to \varprojlim M/\mathfrak a^k M \to \varprojlim M/M_k$ .

By explicitly writing out elements of inverse limits, we see that the above give isomorphisms $\varprojlim_k M/M_{n+k} \cong \varprojlim M/M_k$ and $\varprojlim_k M/\mathfrak a^{n+k} \cong \varprojlim M/\mathfrak a^k$ ; thus

$\hat M \cong \varprojlim M/\mathfrak a^k M$ . ♦

Exercise A

1. Fill in the last step of the proof.

2. Show that in any category, the inverse limit of the diagram

limit_diagram_N

remains the same when we drop finitely many terms on the right.

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Artin-Rees Lemma

The main result we wish to prove is the following.

Artin-Rees Lemma.

Let A be a noetherian ring with the $\mathfrak a$ -adic filtration, and N a submodule of a finitely generated A-module M. If M has an $\mathfrak a$ -stable filtration, the induced filtration on N is also $\mathfrak a$ -stable.

Proof

Step 1: define blowup algebra and module.

Definition.

Given any filtered module M over a filtered ring A, the blowup algebra and blowup module are defined by

$B(A) := A_0 \oplus A_1 \oplus \ldots, \quad B(M) := M_0 \oplus M_1 \oplus \ldots.$

We define a product operation $A_i \times A_j \to A_{i+j}$ 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 $A_i \times M_j \to M_{i+j}$ which gives B(M) a structure of a graded B(A)-module. When A and M are given the 𝔞-adic filtration, we write $B_{\mathfrak a}(A)$ and $B_{\mathfrak a}(M)$ for their blowup algebra and module.

Step 2: if A is a noetherian ring, so is B_𝔞(A).

Since A is noetherian, we can write $\mathfrak a = x_1 A + \ldots + x_k A$ for some $x_1, \ldots, x_k \in \mathfrak a$ . It follows that $\mathfrak a^n$ is a sum of $x_1^{d_1}\ldots x_k^{d_k} A$ where $\sum_{i=1}^k d_i = n$ . Hence the map

$A[X_1, \ldots, X_k] \longrightarrow B_{\mathfrak a}(A), \quad X_i \mapsto (x_i \in A_1)$

is a surjective ring homomorphism so $B_{\mathfrak a}(A)$ 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 $B(M) = M_0 \oplus M_1 \oplus \ldots \oplus M_n \oplus \mathfrak a M_n \oplus \mathfrak a^2 M_n \oplus \ldots.$ Since M is a noetherian A-module, each $M_i$ ( $0\le i \le n$ ) is finitely generated as an A-module by, say $m_{i1}, \ldots, m_{iN}$ . Now we take the set of $m_{ij}$ , as homogeneous elements of B(M) of degree i.

generators_of_BM

In the above, each homogeneous element of $M_0, \ldots, M_n$ is an A-linear combination of these generators. Furthermore, $M_{n+k} = \mathfrak a^k M_n = A_k M_n$ so $m_{n1}, \ldots, m_{nN} \in B(M)_n$ generate (over $B_{\mathfrak a}(A)$ ) the homogeneous elements in B(M) of degree n and higher.

(⇒) Suppose $B(M)$ is finitely generated over $B_{\mathfrak a}(A)$ by homogeneous elements $x_1, \ldots, x_k$ ; let $d_i = \deg x_i$ and $N = \max d_i$ . We claim that $M_{n+1} = \mathfrak a M_n$ for all $n\ge N$ . Since M is filtered, we have $\mathfrak a M_n \subseteq M_{n+1}$

Conversely take $y\in M_{n+1}$ , regard it as an element of $B(M)_{n+1}$ and write $y = a_1 x_1 + \ldots +a_k x_k$ with $a_i \in B_{\mathfrak a}(A)$ . Since y and $x_i$ are homogeneous, we may assume $a_i$ is homogeneous of degree $e_i := n+1 - d_i > 0$ . So $a_i \in B_{\mathfrak a}(A)_{e_i} = \mathfrak a^{e_i}$ . Write

$a_i = b_{i1} c_{i1} + b_{i2} c_{i2} + \ldots + b_{ik} c_{ik}, \quad b_{ij} \in \mathfrak a, c_{ij} \in \mathfrak a^{e_i-1} \subseteq A.$

Now y is a sum of $b_{ij}c_{ij} x_i$ , with $c_{ij} x_i \in M_n$ so $y \in \mathfrak a M_n$ .

Step 4: prove the Artin-Rees lemma.

By step 2, $B_{\mathfrak a}(A)$ is a noetherian ring; since M has an $\mathfrak a$ -stable filtration, by step 3 B(M) is a noetherian $B_{\mathfrak a}(A)$ -module. And since $B(N) \subseteq B(M)$ is a $B_{\mathfrak a}(A)$ -submodule it is also noetherian. By step 3 again, this says the induced filtration on N is $\mathfrak a$ -stable. ♦

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Commutative Algebra 55

Exactness of Completion

Completion of Completion

The 𝔞-adic Filtration

Artin-Rees Lemma

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Commutative Algebra 55

Exactness of Completion

Completion of Completion

The 𝔞-adic Filtration

Artin-Rees Lemma

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