Lemma 51.17.3 (0EBX)—The Stacks project

Processing math: 100%

Table of contents
Part 3: Topics in Scheme Theory
Chapter 51: Local Cohomology
Section 51.17: Frobenius action
Lemma 51.17.3 (cite)

numberstags

See [Lech-inequalities] and [Lemma 2 page 300, MatCA].

Lemma 51.17.3. Let $A$ be a ring. If $f_1, \ldots , f_{r - 1}, f_ rg_ r$ are independent and if the $A$ -module $A/(f_1, \ldots , f_{r - 1}, f_ rg_ r)$ has finite length, then

$\begin{align*} & \text{length}_ A(A/(f_1, \ldots , f_{r - 1}, f_ rg_ r)) \\ & = \text{length}_ A(A/(f_1, \ldots , f_{r - 1}, f_ r)) + \text{length}_ A(A/(f_1, \ldots , f_{r - 1}, g_ r)) \end{align*}$

Proof. We claim there is an exact sequence

$0 \to A/(f_1, \ldots , f_{r - 1}, g_ r) \xrightarrow {f_ r} A/(f_1, \ldots , f_{r - 1}, f_ rg_ r) \to A/(f_1, \ldots , f_{r - 1}, f_ r) \to 0$

Namely, if $a f_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$ , then $\sum _{i < r} a_ i f_ i + (a + bg_ r)f_ r = 0$ for some $b, a_ i \in A$ . Hence $\sum _{i < r} a_ i g_ r f_ i + (a + bg_ r)g_ rf_ r = 0$ which implies $a + bg_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$ which means that $a$ maps to zero in $A/(f_1, \ldots , f_{r - 1}, g_ r)$ . This proves the claim. To finish use additivity of lengths (Algebra, Lemma 10.52.3). $\square$

Comments (0)

There are also:

6 comment(s) on Section 51.17: Frobenius action

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