Commutative Algebra 55

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 PM/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 B𝔞(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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