# Exactness of Completion

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. ♦

# 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:

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

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

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

# 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.

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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