# 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 ProofOfArtinReesTheorem 2013-03-22 14:28:43 2013-03-22 14:28:43 mat_cross (707) mat_cross (707) 4 mat_cross (707) Proof msc 13C99