
## 15.4 A comment on the Artin-Rees property

Some of this material is taken from . 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 10.50.3. 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 $g'(L) \cap I^ nM \subset g'(I^{n - c}L)$ 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 $(g')^{-1}(I^ nM) \subset f'(L) + I^{n - c}M$ 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 10.50.5. $\square$

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. $\square$

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