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 theelements 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 indeterminatesthe 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, $M^{\\prime}$ isa 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 homogeneousgenerators: $u_{1}X^{n_{1}},\\ldots,u_{s}X^{n_{s}}$ , 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\\capN).

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