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:
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:
Identifying M with its image in ,
In particular, if M is finitely generated, so is .
We also have:
For any ideal and A-module M
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.
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 .
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
For any finitely generated M, we have a natural isomorphism .
In short, for finitely generated module M, taking its completion is the same as taking the induced Â-module of M.
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:
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.
(⇒) 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. ♦
is a flat A-algebra.
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
But also has a ring structure! Indeed by definition it is the completion obtained from the -adic filtration as an A-module
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:
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:
Suppose is an ideal generated by . Then the completion of is the quotient of by the ideal generated by (the images of) .
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.
For each , is invertible in .
In particular, (by proposition 4 here) is contained in the Jacobson radical of .
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 .