Lemma 15.92.10 (091V)—The Stacks project

Derived completions along finitely generated ideals exist, and can be computed by a Čech procedure.

Lemma 15.92.10. 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 ¹.

Let $f \in A$. By Lemma 15.92.9 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 15.92.1.

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 $D(A)$. $\square$

[1] Namely, if $E$ is derived complete and $a : K \to E$ is a map, then the commutative diagram

\[ \xymatrix{ K \ar[r]_ a \ar[d] & E \ar[d]^{\cong } \\ K^\wedge \ar[r]^{a^\wedge } & E^\wedge } \]

shows that any map into $E$ factors through $K^\wedge $. Choose a distinguished triangle $K \to K^\wedge \to C$ and apply ${}^\wedge $. We find that $C^\wedge = 0$. By the above this means that $\mathop{\mathrm{Hom}}\nolimits (C, E) = 0$ for $E$ derived complete and this proves that $\mathop{\mathrm{Hom}}\nolimits (K, E) = \mathop{\mathrm{Hom}}\nolimits (K^\wedge , E)$ as desired.

Comments (3)

Comment #855 by Bhargav Bhatt on July 26, 2014 at 16:22

Suggested slogan: Derived completions along finitely generated ideals exist, and can be computed by a Cech procedure.

Comment #10852 by nkym on September 27, 2025 at 17:20

I thought we still needed to show that the completion of $K\to \hat{K}$ and the completion for $\hat{K}$ coincide, but I could be wrong

Comment #10877 by Stacks Project on September 28, 2025 at 16:09

OK, I added the argument in a footnote. I think it's what you had in mind but since your comment was very short, I can't be sure. See this commit.

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