The Stacks project

Example 15.93.5. Let $A$ be a ring and let $f \in A$. Denote $K \mapsto K^\wedge $ the derived completion with respect to $(f)$. Let $M$ be an $A$-module. Using that

\[ M^\wedge = R\mathop{\mathrm{lim}}\nolimits (M \xrightarrow {f^ n} M) \]

by Lemma 15.91.18 and using Lemma 15.87.4 we obtain

\[ H^{-1}(M^\wedge ) = \mathop{\mathrm{lim}}\nolimits M[f^ n] = T_ f(M) \]

the $f$-adic Tate module of $M$. Here the maps $M[f^ n] \to M[f^{n - 1}]$ are given by multiplication by $f$. Then there is a short exact sequence

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits M[f^ n] \to H^0(M^\wedge ) \to \mathop{\mathrm{lim}}\nolimits M/f^ n M \to 0 \]

describing $H^0(M^\wedge )$. We have $H^1(M^\wedge ) = R^1\mathop{\mathrm{lim}}\nolimits M/f^ nM = 0$ as the transition maps are surjective (Lemma 15.87.1). All the other cohomologies of $M^\wedge $ are zero for trivial reasons. Finally, for $K \in D(A)$ and $p \in \mathbf{Z}$ there is a short exact sequence

\[ 0 \to H^0(H^ p(K)^\wedge ) \to H^ p(K^\wedge ) \to T_ f(H^{p + 1}(K)) \to 0 \]

This follows from the spectral sequence of Example 15.91.22 because it degenerates at $E_2$ (as only $i = -1, 0$ give nonzero terms); the next lemma gives more information.

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