Lemma 15.91.10. Let $I$ be a finitely generated ideal of a ring $A$. The inclusion functor $D_{comp}(A, I) \to D(A)$ has a left adjoint, i.e., given any object $K$ of $D(A)$ there exists a map $K \to K^\wedge $ of $K$ into a derived complete object of $D(A)$ such that the map
\[ \mathop{\mathrm{Hom}}\nolimits _{D(A)}(K^\wedge , E) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{D(A)}(K, E) \]
is bijective whenever $E$ is a derived complete object of $D(A)$. In fact, if $I$ is generated by $f_1, \ldots , f_ r \in A$, then we have
\[ K^\wedge = R\mathop{\mathrm{Hom}}\nolimits \left((A \to \prod \nolimits _{i_0} A_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_ r}), K\right) \]
functorially in $K$.
Proof.
Define $K^\wedge $ by the last displayed formula of the lemma. There is a map of complexes
\[ (A \to \prod \nolimits _{i_0} A_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_ r}) \longrightarrow A \]
which induces a map $K \to K^\wedge $. It suffices to prove that $K^\wedge $ is derived complete and that $K \to K^\wedge $ is an isomorphism if $K$ is derived complete.
Let $f \in A$. By Lemma 15.91.9 the object $R\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, K^\wedge )$ is equal to
\[ R\mathop{\mathrm{Hom}}\nolimits \left((A_ f \to \prod \nolimits _{i_0} A_{ff_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{ff_{i_0}f_{i_1}} \to \ldots \to A_{ff_1\ldots f_ r}), K\right) \]
If $f \in I$, then $f_1, \ldots , f_ r$ generate the unit ideal in $A_ f$, hence the extended alternating Čech complex
\[ A_ f \to \prod \nolimits _{i_0} A_{ff_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{ff_{i_0}f_{i_1}} \to \ldots \to A_{ff_1\ldots f_ r} \]
is zero in $D(A)$ by Lemma 15.29.5. (In fact, if $f = f_ i$ for some $i$, then this complex is homotopic to zero by Lemma 15.29.4; this is the only case we need.) Hence $R\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, K^\wedge ) = 0$ and we conclude that $K^\wedge $ is derived complete by Lemma 15.91.1.
Conversely, if $K$ is derived complete, then $R\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, K)$ is zero for all $f = f_{i_0} \ldots f_{i_ p}$, $p \geq 0$. Thus $K \to K^\wedge $ is an isomorphism in $D(A)$.
$\square$
Comments (1)
Comment #855 by Bhargav Bhatt on
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