Artin-Rees lemma

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 $$n \ge n_0$$, $$IM_n = M_{n+1}$$).

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 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 $$B_IR$$ 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 R.

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 $$B_IR$$ 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 $$B_IR$$ 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 $$B_IR$$ is a Noetherian ring.

Thus, if $$B_\mathcal{F}(M)$$ is a finitely generated module over $$B_IR$$, 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 $$B_IR$$.
 * Since $$R$$ is Noetherian, $$B_I(R)$$ is Noetherian.
 * The module corresponding to the induced filtration on $$N$$ is a $$B_IR$$-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.