Commutative Algebra 56

Throughout this article, A denotes a noetherian ring and \mathfrak a \subseteq A is a fixed ideal. All A-modules are finitely generated.

Consequences of Artin-Rees Lemma

Suppose we have an exact sequence of finitely generated A-modules

0 \longrightarrow N \longrightarrow M \longrightarrow P \longrightarrow 0.

Let M be given the 𝔞-adic filtration; the induced filtration on P is 𝔞-adic so its completion is the 𝔞-adic completion. By the Artin-Rees lemma, the induced filtration on N is 𝔞-stable and by proposition 2 here its completion is also the 𝔞-adic completion. Hence we have shown:

Proposition 1.

The following functor is exact:


From general properties of exact functors, this has the following properties.

1. If N \subseteq M is a submodule, then \hat N can be identified as a submodule of \hat M.

2. If N_1, N_2 \subseteq M are submodules, then

(N_1 \cap N_2)\hat{} \cong \hat N_1 \cap \hat N_2, \quad (N_1 + N_2)\hat{} \cong \hat N_1 + \hat N_2.

3. If f:N\to M is a map of A-modules, then \hat f : \hat N \to \hat M satisfies

\mathrm{ker } \hat f = (\mathrm{ker } f)\hat{}, \quad \mathrm{im } \hat f = (\mathrm{im } f)\hat{}.

In particular, for a fixed m\in M, take the A-linear map f : A\to M, 1 \mapsto m. Taking the \mathfrak a-adic completion gives \hat f : \hat A \to \hat M, 1 \mapsto i(m) as well, where i : M\to \hat M is the canonical map. Hence

\hat A \cdot i(m) = \mathrm{im } \hat f = (\mathrm{im } f)\hat{} = (Am)\hat{}.

From property 2, we obtain, for m_1, \ldots, m_n \in M,

\hat A \cdot i(m_1) + \ldots + \hat A \cdot i(m_n) = (Am_1 + \ldots + Am_n)\hat{}.

Thus we have shown:

Proposition 2.

Identifying M with its image in \hat M,

\hat A \cdot M = \hat M.

In particular, if M is finitely generated, so is \hat M.

We also have:

Corollary 1.

For any ideal \mathfrak b\subseteq A and A-module M

(\mathfrak b M)\hat{} = \hat{\mathfrak b} \hat M.


By proposition 2, \hat {\mathfrak b}\hat M = (\hat A \mathfrak b)(\hat A M) = \hat A(\mathfrak b M) = (\mathfrak b M)\hat{}. ♦


Krull’s Intersection Theorem

Another interesting consequence of the Artin-Rees lemma is as follows.

Krull’s Intersection Theorem. 

Suppose (A,\mathfrak m) is local and noetherian. If M is a finitely generated A-module, then \cap_n \mathfrak m^n M = 0.

In particular, the canonical map M \to \hat M is injective where \hat M is the \mathfrak m-adic completion of M.


Let N = \cap_n \mathfrak m^n M, a submodule of M. By the Artin-Rees lemma, the \mathfrak m-adic filtration on M induces a \mathfrak m-stable filtration on N so for some n,

\mathfrak m (N \cap \mathfrak m^n M) = N \cap \mathfrak m^{n+1} M \implies \mathfrak m N = N \implies N = 0

by Nakayama’s lemma. ♦

In particular, A \to \hat A is an injective ring homorphism when we take the \mathfrak m-adic completion of a local ring (A, \mathfrak m).

warningWe give an example where Krull’s intersection theorem fails when A is not noetherian. Take the set of all infinitely differentiable functions f : I\to \mathbb R, where I is an open interval containing 0; let A be the set of equivalence classes under the relation: f : I \to \mathbb R and g : I' \to \mathbb R are equivalent if f|_J = g|_J for some J\subseteq I\cap I' containing 0.

Now A is a ring with addition and product given by pointwise addition and product. Its unique maximal ideal is \mathfrak m = \{f \in A : f(0) = 0\}. Then \cap_n \mathfrak m^n \ne 0 since it contains \exp(-\frac 1 {x^2}).

Exercise A

1. Find a noetherian ring A and a proper ideal \mathfrak a \subsetneq A such that \cap_n \mathfrak a^n \ne 0.

2. Prove that if A is a noetherian integral domain, then any proper ideal \mathfrak a\subsetneq A satisfies \cap_n \mathfrak a^n = 0. [ Hint: follow the proof of Krull’s Intersection Theorem; use the “adjugate matrix” trick. ]


Tensoring with Â

Proposition 3.

For any finitely generated M, we have a natural isomorphism \hat A \otimes_A M \cong \hat M.


In short, for finitely generated module M, taking its completion is the same as taking the induced Â-module of M.


Since \hat M is an \hat A-module with a canonical A-linear M\to \hat M, by universal property of induced modules we have a map \hat A \otimes_A M \to \hat M which is natural in M. And since M is a noetherian module, it is finitely presented so we can find an exact sequence of the form A^m \to A^n \to M \to 0. This gives a commutative diagram of maps:


where the top row is exact because tensor product is right-exact and the bottom row is exact from proposition 1. Since the first two vertical maps are isomorphisms, so is the third one. ♦

Hence the functor \hat A \otimes_A - is exact when restricted to the category of finitely generated A-modules. To see that \hat A is A-flat, we apply:

Lemma 1.

Let A be any ring (possibly non-noetherian) and M be an A-module.

M is A-flat if and only if for any injective map of finitely generated A-modules N_1 \to N_2, the resulting N_1 \otimes_A M \to N_2 \otimes_A M is also injective.


(⇒) Obvious. (⇐) Let P\subseteq Q be a submodule of any module. We need to show that P\otimes_A M\to Q\otimes_A M is injective. Let \Sigma be the set of all pairs (N_1, N_2) where N_2 \subseteq Q and N_1 \subseteq P\cap N_2 are finitely generated A-submodules, ordered by inclusion (in both terms). Clearly \Sigma is a directed set; since N_1 runs through all finitely generated submodules of P, we have direct limits

\varinjlim_{(N_1, N_2) \in \Sigma} N_1 \cong P, \quad \varinjlim_{(N_1, N_2)\in \Sigma} N_2 \cong Q.

By the given condition, N_1\otimes_A M \to N_2 \otimes_A M is injective for each (N_1, N_2) \in \Sigma. By proposition 3 here, taking the direct limit gives an injective

\varinjlim_{(N_1, N_2)} (N_1 \otimes_A M) \to \varinjlim_{(N_1, N_2)} (N_2 \otimes_A M).

By exercise B.4 here, the LHS is isomorphic to (\varinjlim_{(N_1, N_2)} N_1) \otimes_A M \cong P\otimes_A M. Likewise the RHS is isomorphic to Q\otimes_A M so P\otimes_A M \to Q\otimes_A M is injective. ♦

Corollary 2.

\hat A is a flat A-algebra.

Exercise B

Prove that in lemma 1, we can weaken the flatness condition to:

  1. For each ideal \mathfrak a\subseteq A, \mathfrak a\otimes_A M \to A\otimes_A M \cong M is injective.
  2. For each finitely generated ideal \mathfrak a\subseteq A, \mathfrak a\otimes_A M \to A\otimes_A M \cong M is injective.


Completion and Quotients

Recall that for any submodule N\subseteq M we have (M/N)\hat{} \cong \hat M / \hat N. In particular if \mathfrak b\subseteq A is an ideal then

\hat A / \hat {\mathfrak b} \cong (A/\mathfrak b)\hat{} as \hat A-modules.

But (A/\mathfrak b)\hat{} also has a ring structure! Indeed by definition it is the completion obtained from the \mathfrak a-adic filtration as an A-module

A/\mathfrak b = A_0' \supseteq A_1' \supseteq \ldots, where A'_n := (\mathfrak a^n + \mathfrak b)/\mathfrak b

which is also a filtration of A/\mathfrak b as a ring since A_i' A_j' \subseteq A_{i+j}'. The construction which gives us (A/\mathfrak b)\hat{} = \varprojlim [(A/\mathfrak b)/A_n'] as A-modules also gives us the inverse limit as rings. One easily verifies that \hat A \to (A/\mathfrak b)\hat{} is a ring homomorphism so:

Proposition 4.

We have an isomorphism of rings

\hat A / \hat {\mathfrak b} \cong (A/\mathfrak b)\hat{},

where (A/\mathfrak b)\hat{} is its (\mathfrak a + \mathfrak b)/\mathfrak b-adic completion as a ring.

Furthermore, by proposition 2, if \mathfrak b is generated (as an ideal) by a_1, \ldots, a_n, then \hat{\mathfrak b} is generated by the images of a_i in \hat A. Thus we have shown:

Proposition 5.

Suppose \mathfrak b \subseteq A is an ideal generated by a_1, \ldots, a_n. Then the completion of A/\mathfrak b is the quotient of \hat A by the ideal generated by (the images of) a_1, \ldots, a_n.


Take the example A = \mathbb C[X, Y]/(Y^2 - X^3 + X) with \mathfrak m = (X, Y) from an earlier example; we wish to compute the \mathfrak m-adic completion  of A. By the proposition,  is the quotient of \mathbb C[X, Y]^\wedge (the (X, Y)-adic completion) by (Y^2 - X^3 + X). But we clearly have \mathbb C[X, Y]^\wedge \cong \mathbb C[[X, Y]] so

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

as we had claimed. In the next article, we will show that this ring is isomorphic to \mathbb C[[Y]].

Completion of Completion

As a special case, we have

\hat A / (\hat {\mathfrak a})^n = \hat A / (\mathfrak a^n)\hat{} \cong A / \mathfrak a^n,

where the equality is from corollary 1 and the isomorphism from lemma 1 here.

Hence, the \hat a-adic completion of \hat A is isomorphic to \hat A. We also have the following.

Lemma 2.

For each x \in \hat{\mathfrak a}, 1-x is invertible in \hat A.

In particular, (by proposition 4 here) \hat{\mathfrak a} is contained in the Jacobson radical of \hat A.


Since x^n \in \hat{\mathfrak a}^n, we can take the infinite sum

y = 1 + x + x^2 + \ldots \in \hat A.

Then (1-x)y \in \cap_n \hat{\mathfrak a}^n so (1-x)y = 0 in \hat A. ♦


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3 Responses to Commutative Algebra 56

  1. Vanya says:

    Does Proposition 2 assume that \cap_n M_n = 0 in order to identify M with \hat{M} ?

  2. Vanya says:

    Sorry “in order to identify M with its image in \hat{M}.

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