The Stacks project

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

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EBX. Beware of the difference between the letter 'O' and the digit '0'.