The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.50.2 (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 10.50.1 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 (3)

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.


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