proof of Artin-Rees theorem


Define the graded ringMathworldPlanetmath Pow𝔄(A)A[X], where X is an indeterminate by

Pow𝔄(A) = n0𝔄nXn
= {z0+z1X++zrXrr0,zj𝔄j}.

Now, M gives rise to a graded moduleMathworldPlanetmath, M, over Pow𝔄(A), namely

M = n0𝔄nMXn
= {z0+z1X++zrXrr0,zj𝔄jM}.

Observe that Pow𝔄(A) is a noetherian ringMathworldPlanetmath. For, if α1,,αq generate 𝔄 in A, then the elements of 𝔄n are sums of degree n monomialsPlanetmathPlanetmathPlanetmath in the αj’s, i.e., if Y1,,Yq are independent indeterminates the map

A[Y1,,Yq]Pow𝔄(A)

via YjαjX is surjectivePlanetmathPlanetmath, and as A[Y1,,Yq] is noetherian, so is Pow𝔄(A).

Let m1,,mt generate M over A. Then, m1,,mt generate M over Pow𝔄(A). Therefore, M is a noetherian module. Set

N=n0(𝔄nMN)XnM,

a submoduleMathworldPlanetmath of M. Moreover, N is a homogeneous submodule of M and it is f.g. as M is noetherian. Consequently, N possesses a finite number of homogeneousPlanetmathPlanetmathPlanetmath generatorsPlanetmathPlanetmath: u1Xn1,,usXns, where uj𝔄njMN. Let k=max{n1,,ns}. Given any nk and any z𝔄nMN, look at zXnNn. We have

zXn=l=1salXn-nlulXnl,

where alXn-nl(Pow𝔄(A))n-nl. Thus,

al𝔄n-nl=𝔄n-k𝔄k-nl

and

alul𝔄n-k(𝔄k-nlul)𝔄n-k(𝔄k-nl(𝔄nlMN))𝔄n-k(𝔄kMN).

It follows that z=l=1salul𝔄n-k(𝔄kMN), so

𝔄nMN𝔄n-k(𝔄kMN).

Now, it is clear that the righthand side is contained in 𝔄nMN, as 𝔄n-kNN.

Title proof of Artin-Rees theorem
Canonical name ProofOfArtinReesTheorem
Date of creation 2013-03-22 14:28:43
Last modified on 2013-03-22 14:28:43
Owner mat_cross (707)
Last modified by mat_cross (707)
Numerical id 4
Author mat_cross (707)
Entry type Proof
Classification msc 13C99