The Stacks project

Lemma 20.36.3. Let $A$ be a Noetherian ring complete with respect to a principal ideal $(f)$. Let $X$ be a topological space. Let

\[ \ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1 \]

be an inverse system of sheaves of $A$-modules. Assume

  1. $\Gamma (X, \mathcal{F}_1)$ is a finite $A$-module,

  2. $X$, $f$, $(\mathcal{F}_ n)$ satisfy condition (1) of Lemma 20.36.1.


\[ M = \mathop{\mathrm{lim}}\nolimits \Gamma (X, \mathcal{F}_ n) \]

is a finite $A$-module, $f$ is a nonzerodivisor on $M$, and $M/fM$ is the image of $M$ in $\Gamma (X, \mathcal{F}_1)$.

Proof. By Lemma 20.36.2 we have $M/fM \subset H^0(X, \mathcal{F}_1)$. From (1) and the Noetherian property of $A$ we get that $M/fM$ is a finite $A$-module. Observe that $\bigcap f^ nM = 0$ as $f^ nM$ maps to zero in $H^0(X, \mathcal{F}_ n)$. By Algebra, Lemma 10.96.12 we conclude that $M$ is finite over $A$. Finally, suppose $s = (s_ n) \in M = \mathop{\mathrm{lim}}\nolimits H^0(X, \mathcal{F}_ n)$ satisfies $fs = 0$. Then $s_{n + 1}$ is in the kernel of $\mathcal{F}_{n + 1} \to \mathcal{F}_ n$ by condition (1) of Lemma 20.36.1. Hence $s_ n = 0$. Since $n$ was arbitrary, we see $s = 0$. Thus $f$ is a nonzerodivisor on $M$. $\square$

Comments (0)

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Your email address will not be published.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