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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Comment #5047 by Laurent Moret-Bailly on
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