We suggest looking at Cohomology, Sections 20.36 and 20.39 first.
Lemma 52.3.1. Let $X$ be a scheme. Let $f \in \Gamma (X, \mathcal{O}_ X)$. Let
\[ \ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1 \]
be an inverse system of quasi-coherent $\mathcal{O}_ X$-modules. The following are equivalent
for all $n \geq 1$ the map $f : \mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1}$ factors through $\mathcal{F}_{n + 1} \to \mathcal{F}_ n$ to give a short exact sequence $0 \to \mathcal{F}_ n \to \mathcal{F}_{n + 1} \to \mathcal{F}_1 \to 0$,
for all $n \geq 1$ the map $f^ n : \mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1}$ factors through $\mathcal{F}_{n + 1} \to \mathcal{F}_1$ to give a short exact sequence $0 \to \mathcal{F}_1 \to \mathcal{F}_{n + 1} \to \mathcal{F}_ n \to 0$
there exists an $\mathcal{O}_ X$-module $\mathcal{G}$ which is $f$-divisible such that $\mathcal{F}_ n = \mathcal{G}[f^ n]$.
there exists an $\mathcal{O}_ X$-module $\mathcal{F}$ which is $f$-torsion free such that $\mathcal{F}_ n = \mathcal{F}/f^ n\mathcal{F}$.
Proof.
The equivalence of (1), (2), (3) and the implication (4) $\Rightarrow $ (1) are proven in Cohomology, Lemma 20.36.1. Assume (1) holds. Set $\mathcal{F} = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. By Lemma 52.2.1 part (4) we have $\mathcal{F}_ n = \mathcal{F}/f^ n\mathcal{F}$. Let $U \subset X$ be open and $s = (s_ n) \in \mathcal{F}(U) = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n(U)$. Choose $n \geq 1$. If $fs = 0$, then $s_{n + 1}$ is in the kernel of $\mathcal{F}_{n + 1} \to \mathcal{F}_ n$ by condition (1). Hence $s_ n = 0$. Since $n$ was arbitrary, we see $s = 0$. Thus $\mathcal{F}$ is $f$-torsion free.
$\square$
reference
Lemma 52.3.2. Let $A$ be a ring and $f \in A$. Let $X$ be a scheme over $A$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume that $\mathcal{F}[f^ n] = \mathop{\mathrm{Ker}}(f^ n : \mathcal{F} \to \mathcal{F})$ stabilizes. Then
\[ R\Gamma (X, \mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n\mathcal{F}) = R\Gamma (X, \mathcal{F})^\wedge \]
where the right hand side indicates the derived completion with respect to the ideal $(f) \subset A$. Consequently, for $p \in \mathbf{Z}$ we obtain a commutative diagram
\[ \xymatrix{ & 0 & 0 \\ 0 \ar[r] & \widehat{H^ p(X, \mathcal{F})} \ar[r] \ar[u] & \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}/f^ n\mathcal{F}) \ar[r] \ar[u] & T_ f(H^{p + 1}(X, \mathcal{F})) \ar[r] & 0 \\ 0 \ar[r] & H^0(H^ p(X, \mathcal{F})^\wedge ) \ar[r] \ar[u] & H^ p(X, \mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n\mathcal{F}) \ar[r] \ar[u] & T_ f(H^{p + 1}(X, \mathcal{F})) \ar[r] \ar@{=}[u] & 0 \\ & R^1\mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F})[f^ n] \ar[u] \ar[r]^\cong & R^1\mathop{\mathrm{lim}}\nolimits H^{p - 1}(X, \mathcal{F}/f^ n\mathcal{F}) \ar[u] \\ & 0 \ar[u] & 0 \ar[u] } \]
with exact rows and columns where $\widehat{H^ p(X, \mathcal{F})} = \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F})/f^ n H^ p(X, \mathcal{F})$ is the usual $f$-adic completion and $T_ f(-)$ denotes the $f$-adic Tate module as in More on Algebra, Example 15.93.5.
Proof.
By Lemma 52.2.1 we have $\mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n\mathcal{F} = R\mathop{\mathrm{lim}}\nolimits \mathcal{F}/f^ n \mathcal{F}$. Everything else follows from Cohomology, Example 20.39.3.
$\square$
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