The Stacks project

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 $I^{n-d_ j}(I^{d_ j}M \cap N) \subset I^{n-c}(I^ cM \cap N)$. $\square$

Comments (4)

Comment #944 by correction_bot on

In the statement of the lemma, it should say for all .

Comment #2097 by on

More generally, the conclusion holds if is arbitrary and is noetherian (via reduction to the noetherian ring ).

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 is Noetherian, it suffices to let be finite. being finite is then automatic.

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00IN. Beware of the difference between the letter 'O' and the digit '0'.