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

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

Proof

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.

Proof

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 $(A, \mathfrak m)$.

We 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 $\hat A \otimes_A M \cong \hat M$.

Note

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

Proof

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.

Proof

(⇒) 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) $a_1, \ldots, a_n$.

Example

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 Â of A. By the proposition, Â 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 $\hat A$.

Proof

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 $\hat{M}$ ?

• limsup says:

No, we don’t assume the canonical map $M\to \hat M$ is injective here.

2. Vanya says:

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