Remark 64.18.5. If $\mathcal{F} = \left\{ \mathcal{F}_ n\right\} _{n\geq 1}$ is a $\mathbf{Z}_\ell$-sheaf on $X$ and $\bar x$ is a geometric point then $M_ n = \left\{ \mathcal{F}_{n, \bar x}\right\}$ is an inverse system of finite $\mathbf{Z}/\ell ^ n\mathbf{Z}$-modules such that $M_{n+1}\to M_ n$ is surjective and $M_ n = M_{n+1}/\ell ^ n M_{n+1}$. It follows that

$M = \mathop{\mathrm{lim}}\nolimits _ n M_ n = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_{n, \bar x}$

is a finite $\mathbf{Z}_\ell$-module. This follows from Algebra, Lemmas 10.98.2 and 10.96.12 and the fact that $M/\ell M = M_1$ is finite over $\mathbf{F}_\ell$. Therefore, $M\cong \mathbf{Z}_\ell ^{\oplus r} \oplus \oplus _{i = 1}^ t \mathbf{Z}_\ell /\ell ^{e_ i}\mathbf{Z}_\ell$ for some $r, t\geq 0$, $e_ i\geq 1$. The module $M = \mathcal{F}_{\bar x}$ is called the stalk of $\mathcal{F}$ at $\bar x$.

Comment #8876 by ZW on

I think one can prove that such $M$ is $\ell$-adically complete, and then use induction to show that $M$ is finitely generated by $M/\ell M = M_1$. Maybe this is what is meant by Nakayama.

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