The Stacks project

Lemma 10.59.2. Suppose that $M' \subset M$ are finite $R$-modules with finite length quotient. Then there exists a constants $c_1, c_2$ such that for all $n \geq c_2$ we have

\[ c_1 + \chi _{I, M'}(n - c_2) \leq \chi _{I, M}(n) \leq c_1 + \chi _{I, M'}(n) \]

Proof. Since $M/M'$ has finite length there is a $c_2 \geq 0$ such that $I^{c_2}M \subset M'$. Let $c_1 = \text{length}_ R(M/M')$. For $n \geq c_2$ we have

\begin{eqnarray*} \chi _{I, M}(n) & = & \text{length}_ R(M/I^{n + 1}M) \\ & = & c_1 + \text{length}_ R(M'/I^{n + 1}M) \\ & \leq & c_1 + \text{length}_ R(M'/I^{n + 1}M') \\ & = & c_1 + \chi _{I, M'}(n) \end{eqnarray*}

On the other hand, since $I^{c_2}M \subset M'$, we have $I^ nM \subset I^{n - c_2}M'$ for $n \geq c_2$. Thus for $n \geq c_2$ we get

\begin{eqnarray*} \chi _{I, M}(n) & = & \text{length}_ R(M/I^{n + 1}M) \\ & = & c_1 + \text{length}_ R(M'/I^{n + 1}M) \\ & \geq & c_1 + \text{length}_ R(M'/I^{n + 1 - c_2}M') \\ & = & c_1 + \chi _{I, M'}(n - c_2) \end{eqnarray*}

which finishes the proof. $\square$


Comments (0)

There are also:

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

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Your email address will not be published.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 00K5. Beware of the difference between the letter 'O' and the digit '0'.