Artin-Rees lemma

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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 \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

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 <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_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.

References