Artin-Rees lemma

This article is about the statement of a simple but indispensable lemma in commutative algebra
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This article defines a result where the base ring (or one or more of the rings involved) is Noetherian
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This fact is an application of the following pivotal fact/result/idea: Hilbert basis theorem
View other applications of Hilbert basis theorem OR Read a survey article on applying Hilbert basis theorem

Statement

Suppose $A$ is a Noetherian commutative unital ring and $M$ is a finitely generated $A$ -module and $N$ a submodule of $M$ .

Suppose:

$M = M_{0} \supset M_{1} \supset M_{2} \supset \dots$

is an essentially $I$ -adic filtration (in other words, there exists $n_{0}$ such that for all $n \geq n_{0}$ , $I M_{n} = M_{n + 1}$ ).

Then the filtration of $N$ given by:

$N = N \cap M_{0} \supset N \cap M_{1} \supset N \cap M_{2} \supset \dots$

is also essentially $I$ -adic.

Proof

Proof outline

The key idea is the following:

To any filtration, find an associated module over a Noetherian ring such that the filtration being essentially $I$ -adic is equivalent to the module being finitely generated
Prove that intersecting the filtration with a submodule is equivalent to taking a submodule of the associated module.

Setting up the modules

Denote by $B_{I} R$ the blowup algebra of the ideal $I$ in $R$ ; in other words, the ring:

$R \oplus I \oplus I^{2} \oplus \dots$

where the multiplication of graded components is just the usual multiplication as elements of <mth>R</math>.

We now describe a way to associate, to any filtration of a module, an associated module over the blowup algebra. Suppose the filtration is:

$M_{0} \supset M_{1} \supset M_{2} \supset \dots$

We define the associated module over $B_{I} R$ as:

$B_{F} (M) = M_{0} \oplus M_{1} \oplus M_{2} \oplus \dots$

where the multiplication is defined in the usual way.

The crucial observation

The main step of the proof is to observe that $B_{F} (M)$ is a finitely generated module over $B_{I} R$ if and only if the filtration of $M$ is essentially $I$ -adic.

Induced filtration on submodule gives submodule on blowup

The module associated for the induced filtration on the submodule $N$ of $M$ , is clearly a submodule of the module $B_{F} (M)$ .

Applying the Hilbert basis theorem

Since $R$ is a Noetherian ring, the ideal $I$ is a finitely generated ideal. Since the blowup algebra is, by construction, generated by its elements of degree zero and one, we see that the blowup algebra is a finitely generated algebra over $R$ . Thus, the blowup algebra is a quotient of a polynomial ring over $R$ . By the Hilbert basis theorem and the fact that quotients of Noetherian rings are Noetherian, we obtain that $B_{I} R$ is a Noetherian ring.

Thus, if $B_{F} (M)$ is a finitely generated module over $B_{I} R$ , so is the submodule for the induced filtration on $N$ .

Putting the pieces together

Here's the summary of the proof:

If $F$ is an essentially $I$ -adic filtration on $M$ , then the corresponding module $B_{F} (M)$ is finitely generated over $B_{I} R$ .
Since $R$ is Noetherian, $B_{I} (R)$ is Noetherian.
The module corresponding to the induced filtration on $N$ is a $B_{I} R$ -submodule of $B_{F} (M)$ . Since the big module is finitely generated and the ring is Noetherian, the submodule is also finitely generated.
Finally, since the module corresponding to the induced filtration is finitely generated, the induced filtration itself is essentially $I$ -adic.

References

Book:Eisenbud, Page 151-152