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61.20 Derived completion in the constant Noetherian case

We continue the discussion started in Algebraic and Formal Geometry, Section 52.6; we assume the reader has read at least some of that section.

Let $\mathcal{C}$ be a site. Let $\Lambda $ be a Noetherian ring and let $I \subset \Lambda $ be an ideal. Recall from Modules on Sites, Lemma 18.42.4 that

\[ \underline{\Lambda }^\wedge = \mathop{\mathrm{lim}}\nolimits \underline{\Lambda /I^ n} \]

is a flat $\underline{\Lambda }$-algebra and that the map $\underline{\Lambda } \to \underline{\Lambda }^\wedge $ identifies quotients by $I$. Hence Algebraic and Formal Geometry, Lemma 52.6.17 tells us that

\[ D_{comp}(\mathcal{C}, \Lambda ) = D_{comp}(\mathcal{C}, \underline{\Lambda }^\wedge ) \]

In particular the cohomology sheaves $H^ i(K)$ of an object $K$ of $D_{comp}(\mathcal{C}, \Lambda )$ are sheaves of $\underline{\Lambda }^\wedge $-modules. For notational convenience we often work with $D_{comp}(\mathcal{C}, \Lambda )$.

Lemma 61.20.1. Let $\mathcal{C}$ be a site. Let $\Lambda $ be a Noetherian ring and let $I \subset \Lambda $ be an ideal. The left adjoint to the inclusion functor $D_{comp}(\mathcal{C}, \Lambda ) \to D(\mathcal{C}, \Lambda )$ of Algebraic and Formal Geometry, Proposition 52.6.12 sends $K$ to

\[ K^\wedge = R\mathop{\mathrm{lim}}\nolimits (K \otimes _\Lambda ^\mathbf {L} \underline{\Lambda /I^ n}) \]

In particular, $K$ is derived complete if and only if $K = R\mathop{\mathrm{lim}}\nolimits (K \otimes _\Lambda ^\mathbf {L} \underline{\Lambda /I^ n})$.

Proof. Choose generators $f_1, \ldots , f_ r$ of $I$. By Algebraic and Formal Geometry, Lemma 52.6.9 we have

\[ K^\wedge = R\mathop{\mathrm{lim}}\nolimits (K \otimes _\Lambda ^\mathbf {L} \underline{K_ n}) \]

where $K_ n = K(\Lambda , f_1^ n, \ldots , f_ r^ n)$. In More on Algebra, Lemma 15.94.1 we have seen that the pro-systems $\{ K_ n\} $ and $\{ \Lambda /I^ n\} $ of $D(\Lambda )$ are isomorphic. Thus the lemma follows. $\square$

Lemma 61.20.2. Let $\Lambda $ be a Noetherian ring. Let $I \subset \Lambda $ be an ideal. Let $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ be a morphism of topoi. Then

  1. $Rf_*$ sends $D_{comp}(\mathcal{D}, \Lambda )$ into $D_{comp}(\mathcal{C}, \Lambda )$,

  2. the map $Rf_* : D_{comp}(\mathcal{D}, \Lambda ) \to D_{comp}(\mathcal{C}, \Lambda )$ has a left adjoint $Lf_{comp}^* : D_{comp}(\mathcal{C}, \Lambda ) \to D_{comp}(\mathcal{D}, \Lambda )$ which is $Lf^*$ followed by derived completion,

  3. $Rf_*$ commutes with derived completion,

  4. for $K$ in $D_{comp}(\mathcal{D}, \Lambda )$ we have $Rf_*K = R\mathop{\mathrm{lim}}\nolimits Rf_*(K \otimes ^\mathbf {L}_\Lambda \underline{\Lambda /I^ n})$.

  5. for $M$ in $D_{comp}(\mathcal{C}, \Lambda )$ we have $Lf^*_{comp}M = R\mathop{\mathrm{lim}}\nolimits Lf^*(M \otimes ^\mathbf {L}_\Lambda \underline{\Lambda /I^ n})$.

Proof. We have seen (1) and (2) in Algebraic and Formal Geometry, Lemma 52.6.18. Part (3) follows from Algebraic and Formal Geometry, Lemma 52.6.19. For (4) let $K$ be derived complete. Then

\[ Rf_*K = Rf_*( R\mathop{\mathrm{lim}}\nolimits K \otimes ^\mathbf {L}_\Lambda \underline{\Lambda /I^ n}) = R\mathop{\mathrm{lim}}\nolimits Rf_*(K \otimes ^\mathbf {L}_\Lambda \underline{\Lambda /I^ n}) \]

the first equality by Lemma 61.20.1 and the second because $Rf_*$ commutes with $R\mathop{\mathrm{lim}}\nolimits $ (Cohomology on Sites, Lemma 21.23.3). This proves (4). To prove (5), by Lemma 61.20.1 we have

\[ Lf_{comp}^*M = R\mathop{\mathrm{lim}}\nolimits ( Lf^*M \otimes _\Lambda ^\mathbf {L} \underline{\Lambda /I^ n}) \]

Since $Lf^*$ commutes with derived tensor product by Cohomology on Sites, Lemma 21.18.4 and since $Lf^*\underline{\Lambda /I^ n} = \underline{\Lambda /I^ n}$ we get (5). $\square$

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