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.

Then

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

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 $M$. $\square$

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).