Lemma 10.59.3 (00K6)—The Stacks project

Processing math: 100%

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.59: Noetherian local rings
Lemma 10.59.3 (cite)

numberstags

Lemma 10.59.3. Suppose that $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of finite $R$ -modules. Then there exists a submodule $N \subset M'$ with finite colength $l$ and $c \geq 0$ such that

$\chi _{I, M}(n) = \chi _{I, M''}(n) + \chi _{I, N}(n - c) + l$

and

$\varphi _{I, M}(n) = \varphi _{I, M''}(n) + \varphi _{I, N}(n - c)$

for all $n \geq c$ .

Proof. Note that $M/I^ nM \to M''/I^ nM''$ is surjective with kernel $M' / M' \cap I^ nM$ . By the Artin-Rees Lemma 10.51.2 there exists a constant $c$ such that $M' \cap I^ nM = I^{n - c}(M' \cap I^ cM)$ . Denote $N = M' \cap I^ cM$ . Note that $I^ c M' \subset N \subset M'$ . Hence $\text{length}_ R(M' / M' \cap I^ nM) = \text{length}_ R(M'/N) + \text{length}_ R(N/I^{n - c}N)$ for $n \geq c$ . From the short exact sequence

$0 \to M' / M' \cap I^ nM \to M/I^ nM \to M''/I^ nM'' \to 0$

and additivity of lengths (Lemma 10.52.3) we obtain the equality

$\chi _{I, M}(n - 1) = \chi _{I, M''}(n - 1) + \chi _{I, N}(n - c - 1) + \text{length}_ R(M'/N)$

for $n \geq c$ . We have $\varphi _{I, M}(n) = \chi _{I, M}(n) - \chi _{I, M}(n - 1)$ and similarly for the modules $M''$ and $N$ . Hence we get $\varphi _{I, M}(n) = \varphi _{I, M''}(n) + \varphi _{I, N}(n-c)$ for $n \geq c$ . $\square$

Comments (0)

There are also:

1 comment(s) on Section 10.59: Noetherian local rings

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).

Comments (0)

Post a comment