Throughout this article, *A* denotes a noetherian ring and 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

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 is a submodule, then can be identified as a submodule of .

2. If are submodules, then

.

3. If is a map of *A*-modules, then satisfies

.

In particular, for a fixed , take the *A*-linear map . Taking the -adic completion gives as well, where is the canonical map. Hence

From property 2, we obtain, for ,

Thus we have shown:

Proposition 2.Identifying M with its image in ,

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

We also have:

Corollary 1.For any ideal and A-module M

.

**Proof**

By proposition 2, . ♦

# Krull’s Intersection Theorem

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

Krull’s Intersection Theorem.Suppose is local and noetherian. If M is a finitely generated A-module, then .

In particular, the canonical map is injective where is the -adic completion of M.

**Proof**

Let , a submodule of *M*. By the Artin-Rees lemma, the -adic filtration on *M* induces a -stable filtration on *N* so for some *n*,

by Nakayama’s lemma. ♦

In particular, is an injective ring homorphism when we take the -adic completion of a local ring .

We give an example where Krull’s intersection theorem fails when *A* is not noetherian. Take the set of all infinitely differentiable functions , where is an open interval containing 0; let *A* be the set of equivalence classes under the relation: and are equivalent if for some containing 0.

Now *A* is a ring with addition and product given by pointwise addition and product. Its unique maximal ideal is . Then since it contains .

**Exercise A**

1. Find a noetherian ring *A* and a proper ideal such that .

2. Prove that if *A* is a noetherian integral domain, then any proper ideal satisfies . [ 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 .

**Note**

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

**Proof**

Since is an -module with a canonical *A*-linear , by universal property of induced modules we have a map 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 . 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 is exact when restricted to the category of *finitely generated A-modules*. To see that 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 , the resulting is also injective.

**Proof**

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

By the given condition, is injective for each . By proposition 3 here, taking the direct limit gives an injective

By exercise B.4 here, the LHS is isomorphic to Likewise the RHS is isomorphic to so is injective. ♦

Corollary 2.is a flat A-algebra.

**Exercise B**

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

- For each ideal , is injective.
- For each
*finitely generated*ideal , is injective.

# Completion and Quotients

Recall that for any submodule we have . In particular if is an ideal then

as -modules.

But also has a ring structure! Indeed by definition it is the completion obtained from the -adic filtration as an *A*-module

, where

which is also a filtration of *as a ring* since . The construction which gives us as *A*-modules also gives us the inverse limit as rings. One easily verifies that is a ring homomorphism so:

Proposition 4.We have an isomorphism of rings

,

where is its -adic completion as a ring.

Furthermore, by proposition 2, if is generated (as an ideal) by , then is generated by the images of in . Thus we have shown:

Proposition 5.Suppose is an ideal generated by . Then the completion of is the quotient of by the ideal generated by (the images of) .

**Example**

Take the example with from an earlier example; we wish to compute the -adic completion *Â* of *A*. By the proposition, *Â* is the quotient of (the -adic completion) by . But we clearly have so

as we had claimed. In the next article, we will show that this ring is isomorphic to .

## Completion of Completion

As a special case, we have

,

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

Hence, the -adic completion of is isomorphic to . We also have the following.

Lemma 2.For each , is invertible in .

In particular, (by proposition 4 here) is contained in the Jacobson radical of .

**Proof**

Since , we can take the infinite sum

.

Then so in . ♦

Does Proposition 2 assume that in order to identify with ?

No, we don’t assume the canonical map is injective here.

Sorry “in order to identify M with its image in .