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
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Statement

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

Suppose:

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

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

Then the filtration of $N$ given by:

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

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:

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

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:

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

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

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

where the multiplication is defined in the usual way.

The crucial observation

The main step of the proof is to observe that $B_{\mathcal {F}}(M)$ 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 $B_{\mathcal {F}}(M)$.

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 $B_{\mathcal {F}}(M)$ 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 ${\mathcal {F}}$ is an essentially $I$-adic filtration on $M$, then the corresponding module $B_{\mathcal {F}}(M)$ 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 $B_{\mathcal {F}}(M)$. 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