The Stacks project

15.4 A comment on the Artin-Rees property

Some of this material is taken from [conrad-dejong]. A general discussion with additional references can be found in [Section 1, Eis].

Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Given a homomorphism $f : M \to N$ of finite $A$-modules there exists a $c \geq 0$ such that

\[ f(M) \cap I^ nN \subset f(I^{n - c}M) \]

for all $n \geq c$, see Algebra, Lemma 10.51.3. In this situation we will say $c$ works for $f$ in the Artin-Rees lemma.

Lemma 15.4.1. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal contained in the Jacobson radical of $A$. Let

\[ S : L \xrightarrow {f} M \xrightarrow {g} N \quad \text{and}\quad S' : L \xrightarrow {f'} M \xrightarrow {g'} N \]

be two complexes of finite $A$-modules as shown. Assume that

  1. $c$ works in the Artin-Rees lemma for $f$ and $g$,

  2. the complex $S$ is exact, and

  3. $f' = f \bmod I^{c + 1}M$ and $g' = g \bmod I^{c + 1}N$.

Then $c$ works in the Artin-Rees lemma for $g'$ and the complex $S'$ is exact.

Proof. We first show that $g'(M) \cap I^ nN \subset g'(I^{n - c}M)$ for $n \geq c$. Let $a$ be an element of $M$ such that $g'(a) \in I^ nN$. We want to adjust $a$ by an element of $f'(L)$, i.e, without changing $g'(a)$, so that $a \in I^{n-c}M$. Assume that $a \in I^ rM$, where $r < n - c$. Then

\[ g(a) = g'(a) + (g - g')(a) \in I^ n N + I^{r + c + 1}N = I^{r + c + 1}N. \]

By Artin-Rees for $g$ we have $g(a) \in g(I^{r + 1}M)$. Say $g(a) = g(a_1)$ with $a_1 \in I^{r + 1}M$. Since the sequence $S$ is exact, $a - a_1 \in f(L)$. Accordingly, we write $a = f(b) + a_1$ for some $b \in L$. Then $f(b) = a - a_1 \in I^ rM$. Artin-Rees for $f$ shows that if $r \geq c$, we may replace $b$ by an element of $I^{r - c}L$. Then in all cases, $a = f'(b) + a_2$, where $a_2 = (f - f')(b) + a_1 \in I^{r + 1}M$. (Namely, either $c \geq r$ and $(f - f')(b) \in I^{r + 1}M$ by assumption, or $c < r$ and $b \in I^{r - c}$, whence again $(f - f')(b) \in I^{c + 1} I^{r - c} M = I^{r + 1}M$.) So we can adjust $a$ by the element $f'(b) \in f'(L)$ to increase $r$ by $1$.

In fact, the argument above shows that $(g')^{-1}(I^ nN) \subset f'(L) + I^{n - c}M$ for all $n \geq c$. Hence $S'$ is exact because

\[ (g')^{-1}(0) = (g')^{-1}(\bigcap I^ nN) \subset \bigcap f'(L) + I^{n - c}M = f'(L) \]

as $I$ is contained in the Jacobson radical of $A$, see Algebra, Lemma 10.51.5. $\square$

Given an ideal $I \subset A$ of a ring $A$ and an $A$-module $M$ we set

\[ \text{Gr}_ I(M) = \bigoplus I^ nM/I^{n + 1}M. \]

We think of this as a graded $\text{Gr}_ I(A)$-module.

Lemma 15.4.2. Assumptions as in Lemma 15.4.1. Let $Q = \mathop{\mathrm{Coker}}(g)$ and $Q' = \mathop{\mathrm{Coker}}(g')$. Then $\text{Gr}_ I(Q) \cong \text{Gr}_ I(Q')$ as graded $\text{Gr}_ I(A)$-modules.

Proof. In degree $n$ we have $\text{Gr}_ I(Q)_ n = I^ nN/(I^{n + 1}N + g(M) \cap I^ nN)$ and similarly for $Q'$. We claim that

\[ g(M) \cap I^ nN \subset I^{n + 1}N + g'(M) \cap I^ nN. \]

By symmetry (the proof of the claim will only use that $c$ works for $g$ which also holds for $g'$ by the lemma) this will imply that

\[ I^{n + 1}N + g(M) \cap I^ nN = I^{n + 1}N + g'(M) \cap I^ nN \]

whence $\text{Gr}_ I(Q)_ n$ and $\text{Gr}_ I(Q')_ n$ agree as subquotients of $N$, implying the lemma. Observe that the claim is clear for $n \leq c$ as $f = f' \bmod I^{c + 1}N$. If $n > c$, then suppose $b \in g(M) \cap I^ nN$. Write $b = g(a)$ for $a \in I^{n - c}M$. Set $b' = g'(a)$. We have $b - b' = (g - g')(a) \in I^{n + 1}N$ as desired. $\square$

Lemma 15.4.3. Let $A \to B$ be a flat map of Noetherian rings. Let $I \subset A$ be an ideal. Let $f : M \to N$ be a homomorphism of finite $A$-modules. Assume that $c$ works for $f$ in the Artin-Rees lemma. Then $c$ works for $f \otimes 1 : M \otimes _ A B \to N \otimes _ A B$ in the Artin-Rees lemma for the ideal $IB$.

Proof. Note that

\[ (f \otimes 1)(M) \cap I^ n N \otimes _ A B = (f \otimes 1)\left((f \otimes 1)^{-1}(I^ n N \otimes _ A B)\right) \]

On the other hand,

\begin{align*} (f \otimes 1)^{-1}(I^ n N \otimes _ A B) & = \mathop{\mathrm{Ker}}(M \otimes _ A B \to N \otimes _ A B/(I^ n N \otimes _ A B)) \\ & = \mathop{\mathrm{Ker}}(M \otimes _ A B \to (N/I^ nN) \otimes _ A B) \end{align*}

As $A \to B$ is flat taking kernels and cokernels commutes with tensoring with $B$, whence this is equal to $f^{-1}(I^ nN) \otimes _ A B$. By assumption $f^{-1}(I^ nN)$ is contained in $\mathop{\mathrm{Ker}}(f) + I^{n - c}M$. Thus the lemma holds. $\square$

Comments (2)

Comment #7737 by Mingchen on

The terminology "c works for f in the Artin-Rees lemma" seems ambiguous, as the property itself depends on I as well.

Comment #7986 by on

Yes, it is ambiguous but I think it is OK here. Hopefully the reader understands we have a given pair in mind.

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 07VD. Beware of the difference between the letter 'O' and the digit '0'.