
Example 15.83.4. 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 15.82.17 and using Lemma 15.78.4 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 15.78.1). All the other cohomologies of $M^\wedge$ are zero for trivial reasons. We claim that for $K \in D(A)$ there are short exact sequences

$0 \to H^0(H^ n(K)^\wedge ) \to H^ n(K^\wedge ) \to T_ f(H^{n + 1}(K)) \to 0$

Namely this follows from the spectral sequence of Example 15.82.20 because it degenerates at $E_2$ (as only $i = -1, 0$ give nonzero terms).

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).