proof of Artin-Rees theorem

Define the graded ring $\mathrm{Pow}_{{\mathfrak{A}}}(A)\subseteq A[X]$ , where $X$ is an indeterminate by

	$\displaystyle\mathrm{Pow}_{{\mathfrak{A}}}(A)$	$\displaystyle=$	$\displaystyle\coprod_{n\geq 0}{\mathfrak{A}}^{n}X^{n}$
		$\displaystyle=$	$\displaystyle\{z_{0}+z_{1}X+\cdots+z_{r}X^{r}\mid r\geq 0,\>z_{j}\in{\mathfrak% {A}}^{j}\}.$

Now, $M$ gives rise to a graded module, $M^{\prime}$ , over $\mathrm{Pow}_{{\mathfrak{A}}}(A)$ , namely

	$\displaystyle M^{\prime}$	$\displaystyle=$	$\displaystyle\coprod_{n\geq 0}{\mathfrak{A}}^{n}MX^{n}$
		$\displaystyle=$	$\displaystyle\{z_{0}+z_{1}X+\cdots+z_{r}X^{r}\mid r\geq 0,\>z_{j}\in{\mathfrak% {A}}^{j}M\}.$

Observe that $\mathrm{Pow}_{{\mathfrak{A}}}(A)$ is a noetherian ring. For, if $\alpha_{1},\ldots,\alpha_{q}$ generate ${\mathfrak{A}}$ in $A$ , then the elements of ${\mathfrak{A}}^{n}$ are sums of degree $n$ monomials in the $\alpha_{j}$ ’s, i.e., if $Y_{1},\ldots,Y_{q}$ are independent indeterminates the map

$A[Y_{1},\ldots,Y_{q}]\longrightarrow\mathrm{Pow}_{{\mathfrak{A}}}(A)$

via $Y_{j}\mapsto\alpha_{j}X$ is surjective, and as $A[Y_{1},\ldots,Y_{q}]$ is noetherian, so is $\mathrm{Pow}_{{\mathfrak{A}}}(A)$ .

Let $m_{1},\ldots,m_{t}$ generate $M$ over $A$ . Then, $m_{1},\ldots,m_{t}$ generate $M^{\prime}$ over $\mathrm{Pow}_{{\mathfrak{A}}}(A)$ . Therefore, $M^{\prime}$ is a noetherian module. Set

$N^{\prime}=\coprod_{n\geq 0}({\mathfrak{A}}^{n}M\cap N)X^{n}\subseteq M^{% \prime},$

a submodule of $M^{\prime}$ . Moreover, $N^{\prime}$ is a homogeneous submodule of $M$ and it is f.g. as $M^{\prime}$ is noetherian. Consequently, $N^{\prime}$ possesses a finite number of homogeneous generators: $u_{1}X^{n_{1}},\ldots,u_{s}X^{n_{s}}$ , where $u_{j}\in{\mathfrak{A}}^{n_{j}}M\cap N$ . Let $k=\max\{n_{1},\ldots,n_{s}\}$ . Given any $n\geq k$ and any $z\in{\mathfrak{A}}^{n}M\cap N$ , look at $zX^{n}\in N^{\prime}_{n}$ . We have

$zX^{n}=\sum_{l=1}^{s}a_{l}X^{n-n_{l}}u_{l}X^{n_{l}},$

where $a_{l}X^{n-n_{l}}\in\bigl{(}\mathrm{Pow}_{{\mathfrak{A}}}(A)\bigr{)}_{n-n_{l}}$ . Thus,

$a_{l}\in{\mathfrak{A}}^{n-n_{l}}={\mathfrak{A}}^{n-k}{\mathfrak{A}}^{k-n_{l}}$

and

$a_{l}u_{l}\in{\mathfrak{A}}^{n-k}({\mathfrak{A}}^{k-n_{l}}u_{l})\subseteq{% \mathfrak{A}}^{n-k}({\mathfrak{A}}^{k-n_{l}}({\mathfrak{A}}^{n_{l}}M\cap N))% \subseteq{\mathfrak{A}}^{n-k}({\mathfrak{A}}^{k}M\cap N).$

It follows that $z=\sum_{l=1}^{s}a_{l}u_{l}\in{\mathfrak{A}}^{n-k}({\mathfrak{A}}^{k}M\cap N)$ , so

${\mathfrak{A}}^{n}M\cap N\subseteq{\mathfrak{A}}^{n-k}({\mathfrak{A}}^{k}M\cap N).$

Now, it is clear that the righthand side is contained in ${\mathfrak{A}}^{n}M\cap N$ , as ${\mathfrak{A}}^{n-k}N\subseteq N$ . $\Box$

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