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

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

the complex $S$ is exact, and

$f' = f \bmod I^{c + 1}M$ 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.
$\square$

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