Lemma 10.51.2 (Artin-Rees). Suppose that $R$ is Noetherian, $I \subset R$ an ideal. Let $N \subset M$ be finite $R$-modules. There exists a constant $c > 0$ such that $I^ n M \cap N = I^{n-c}(I^ cM \cap N)$ for all $n \geq c$.

Proof. Consider the ring $S = R \oplus I \oplus I^2 \oplus \ldots = \bigoplus _{n \geq 0} I^ n$. Convention: $I^0 = R$. Multiplication maps $I^ n \times I^ m$ into $I^{n + m}$ by multiplication in $R$. Note that if $I = (f_1, \ldots , f_ t)$ then $S$ is a quotient of the Noetherian ring $R[X_1, \ldots , X_ t]$. The map just sends the monomial $X_1^{e_1}\ldots X_ t^{e_ t}$ to $f_1^{e_1}\ldots f_ t^{e_ t}$. Thus $S$ is Noetherian. Similarly, consider the module $M \oplus IM \oplus I^2M \oplus \ldots = \bigoplus _{n \geq 0} I^ nM$. This is a finitely generated $S$-module. Namely, if $x_1, \ldots , x_ r$ generate $M$ over $R$, then they also generate $\bigoplus _{n \geq 0} I^ nM$ over $S$. Next, consider the submodule $\bigoplus _{n \geq 0} I^ nM \cap N$. This is an $S$-submodule, as is easily verified. By Lemma 10.51.1 it is finitely generated as an $S$-module, say by $\xi _ j \in \bigoplus _{n \geq 0} I^ nM \cap N$, $j = 1, \ldots , s$. We may assume by decomposing each $\xi _ j$ into its homogeneous pieces that each $\xi _ j \in I^{d_ j}M \cap N$ for some $d_ j$. Set $c = \max \{ d_ j\}$. Then for all $n \geq c$ every element in $I^ nM \cap N$ is of the form $\sum h_ j \xi _ j$ with $h_ j \in I^{n - d_ j}$. The lemma now follows from this and the trivial observation that $I^{n-d_ j}(I^{d_ j}M \cap N) \subset I^{n-c}(I^ cM \cap N)$. $\square$

Comment #944 by correction_bot on

In the statement of the lemma, it should say for all $n \geq c$.

Comment #2097 by on

More generally, the conclusion holds if $R$ is arbitrary and $M$ is noetherian (via reduction to the noetherian ring $R/(0:_RM)$).

Comment #2125 by on

@#2097: Yes, this is true, but I am going leave it as is for the moment.

Comment #8349 by Hao Xiao on

Very minor comment: Since $R$ is Noetherian, it suffices to let $M$ be finite. $N$ being finite is then automatic.

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