## 15.84 Derived Completion

Some references for the material in this section are , , [PSY], [dag12] (especially Chapter 4). Our exposition follows [BS]. The analogue (or “dual”) of this section for torsion modules is Dualizing Complexes, Section 47.9. The relationship between the derived category of complexes with torsion cohomology and derived complete complexes can be found in Dualizing Complexes, Section 47.12.

Let $K \in D(A)$. Let $f \in A$. We denote $T(K, f)$ a derived limit of the system

$\ldots \to K \xrightarrow {f} K \xrightarrow {f} K$

in $D(A)$.

Lemma 15.84.1. Let $A$ be a ring. Let $f \in A$. Let $K \in D(A)$. The following are equivalent

1. $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_ f, K) = 0$ for all $n$,

2. $\mathop{\mathrm{Hom}}\nolimits _{D(A)}(E, K) = 0$ for all $E$ in $D(A_ f)$,

3. $T(K, f) = 0$,

4. for every $p \in \mathbf{Z}$ we have $T(H^ p(K), f) = 0$,

5. for every $p \in \mathbf{Z}$ we have $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, H^ p(K)) = 0$ and $\mathop{\mathrm{Ext}}\nolimits ^1_ A(A_ f, H^ p(K)) = 0$,

6. $R\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, K) = 0$,

7. the map $\prod _{n \geq 0} K \to \prod _{n \geq 0} K$, $(x_0, x_1, \ldots ) \mapsto (x_0 - fx_1, x_1 - fx_2, \ldots )$ is an isomorphism in $D(A)$, and

Proof. It is clear that (2) implies (1) and that (1) is equivalent to (6). Assume (1). Let $I^\bullet$ be a K-injective complex of $A$-modules representing $K$. Condition (1) signifies that $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, I^\bullet )$ is acyclic. Let $M^\bullet$ be a complex of $A_ f$-modules representing $E$. Then

$\mathop{\mathrm{Hom}}\nolimits _{D(A)}(E, K) = \mathop{\mathrm{Hom}}\nolimits _{K(A)}(M^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(A_ f)}(M^\bullet , \mathop{\mathrm{Hom}}\nolimits _ A(A_ f, I^\bullet ))$

by Algebra, Lemma 10.13.4. As $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, I^\bullet )$ is a K-injective complex of $A_ f$-modules by Lemma 15.55.3 the fact that it is acyclic implies that it is homotopy equivalent to zero (Derived Categories, Lemma 13.31.2). Thus we get (2).

A free resolution of the $A$-module $A_ f$ is given by

$0 \to \bigoplus \nolimits _{n \in \mathbf{N}} A \to \bigoplus \nolimits _{n \in \mathbf{N}} A \to A_ f \to 0$

where the first map sends the $(a_0, a_1, a_2, \ldots )$ to $(a_0, a_1 - fa_0, a_2 - fa_1, \ldots )$ and the second map sends $(a_0, a_1, a_2, \ldots )$ to $a_0 + a_1/f + a_2/f^2 + \ldots$. Applying $\mathop{\mathrm{Hom}}\nolimits _ A(-, I^\bullet )$ we get

$0 \to \mathop{\mathrm{Hom}}\nolimits _ A(A_ f, I^\bullet ) \to \prod I^\bullet \to \prod I^\bullet \to 0$

Since $\prod I^\bullet$ represents $\prod _{n \geq 0} K$ this proves the equivalence of (1) and (7). On the other hand, by construction of derived limits in Derived Categories, Section 13.34 the displayed exact sequence shows the object $T(K, f)$ is a representative of $R\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, K)$ in $D(A)$. Thus the equivalence of (1) and (3).

There is a spectral sequence

$E_2^{p, q} = \mathop{\mathrm{Ext}}\nolimits ^ q_ A(A_ f, H^ p(K)) \Rightarrow \mathop{\mathrm{Ext}}\nolimits ^{p + q}_ A(A_ f, K)$

(details omitted). This spectral sequence degenerates at $E_2$ because $A_ f$ has a length $1$ resolution by projective $A$-modules (see above) hence the $E_2$-page has only 2 nonzero rows. Thus we obtain short exact sequences

$0 \to \mathop{\mathrm{Ext}}\nolimits ^1_ A(A_ f, H^{p - 1}(K)) \to \text{Ext}^ p_ A(A_ f, K) \to \mathop{\mathrm{Hom}}\nolimits _ A(A_ f, H^ p(K)) \to 0$

This proves (4) and (5) are equivalent to (1). $\square$

Lemma 15.84.2. Let $A$ be a ring. Let $K \in D(A)$. The set $I$ of $f \in A$ such that $T(K, f) = 0$ is a radical ideal of $A$.

Proof. We will use the results of Lemma 15.84.1 without further mention. If $f \in I$, and $g \in A$, then $A_{gf}$ is an $A_ f$-module hence $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_{gf}, K) = 0$ for all $n$, hence $gf \in I$. Suppose $f, g \in I$. Then there is a short exact sequence

$0 \to A_{f + g} \to A_{f(f + g)} \oplus A_{g(f + g)} \to A_{gf(f + g)} \to 0$

because $f, g$ generate the unit ideal in $A_{f + g}$. This follows from Algebra, Lemma 10.23.2 and the easy fact that the last arrow is surjective. From the long exact sequence of $\mathop{\mathrm{Ext}}\nolimits$ and the vanishing of $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_{f(f + g)}, K)$, $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_{g(f + g)}, K)$, and $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_{gf(f + g)}, K)$ for all $n$ we deduce the vanishing of $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_{f + g}, K)$ for all $n$. Finally, if $f^ n \in I$ for some $n > 0$, then $f \in I$ because $T(K, f) = T(K, f^ n)$ or because $A_ f \cong A_{f^ n}$. $\square$

Lemma 15.84.3. Let $A$ be a ring. Let $I \subset A$ be an ideal. Let $M$ be an $A$-module.

1. If $M$ is $I$-adically complete, then $T(M, f) = 0$ for all $f \in I$.

2. Conversely, if $T(M, f) = 0$ for all $f \in I$ and $I$ is finitely generated, then $M \to \mathop{\mathrm{lim}}\nolimits M/I^ nM$ is surjective.

Proof. Proof of (1). Assume $M$ is $I$-adically complete. By Lemma 15.84.1 it suffices to prove $\mathop{\mathrm{Ext}}\nolimits ^1_ A(A_ f, M) = 0$ and $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, M) = 0$. Since $M = \mathop{\mathrm{lim}}\nolimits M/I^ nM$ and since $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, M/I^ nM) = 0$ it follows that $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, M) = 0$. Suppose we have an extension

$0 \to M \to E \to A_ f \to 0$

For $n \geq 0$ pick $e_ n \in E$ mapping to $1/f^ n$. Set $\delta _ n = fe_{n + 1} - e_ n \in M$ for $n \geq 0$. Replace $e_ n$ by

$e'_ n = e_ n + \delta _ n + f\delta _{n + 1} + f^2 \delta _{n + 2} + \ldots$

The infinite sum exists as $M$ is complete with respect to $I$ and $f \in I$. A simple calculation shows that $fe'_{n + 1} = e'_ n$. Thus we get a splitting of the extension by mapping $1/f^ n$ to $e'_ n$.

Proof of (2). Assume that $I = (f_1, \ldots , f_ r)$ and that $T(M, f_ i) = 0$ for $i = 1, \ldots , r$. By Algebra, Lemma 10.95.7 we may assume $I = (f)$ and $T(M, f) = 0$. Let $x_ n \in M$ for $n \geq 0$. Consider the extension

$0 \to M \to E \to A_ f \to 0$

given by

$E = M \oplus \bigoplus Ae_ n\Big/\langle x_ n - fe_{n + 1} + e_ n\rangle$

mapping $e_ n$ to $1/f^ n$ in $A_ f$ (see above). By assumption and Lemma 15.84.1 this extension is split, hence we obtain an element $x + e_0$ which generates a copy of $A_ f$ in $E$. Then

$x + e_0 = x - x_0 + fe_1 = x - x_0 - x_1 + f^2 e_2 = \ldots$

Since $M/f^ nM = E/f^ nE$ by the snake lemma, we see that $x = x_0 + fx_1 + \ldots + f^{n - 1}x_{n - 1}$ modulo $f^ nM$. In other words, the map $M \to \mathop{\mathrm{lim}}\nolimits M/f^ nM$ is surjective as desired. $\square$

Motivated by the results above we make the following definition.

Definition 15.84.4. Let $A$ be a ring. Let $K \in D(A)$. Let $I \subset A$ be an ideal. We say $K$ is derived complete with respect to $I$ if for every $f \in I$ we have $T(K, f) = 0$. If $M$ is an $A$-module, then we say $M$ is derived complete with respect to $I$ if $M \in D(A)$ is derived complete with respect to $I$.

The full subcategory $D_{comp}(A) = D_{comp}(A, I) \subset D(A)$ consisting of derived complete objects is a strictly full, saturated triangulated subcategory, see Derived Categories, Definitions 13.3.4 and 13.6.1. By Lemma 15.84.2 the subcategory $D_{comp}(A, I)$ depends only on the radical $\sqrt{I}$ of $I$, in other words it depends only on the closed subset $Z = V(I)$ of $\mathop{\mathrm{Spec}}(A)$. The subcategory $D_{comp}(A, I)$ is preserved under products and homotopy limits in $D(A)$. But it is not preserved under countable direct sums in general. We will often simply say $M$ is a derived complete module if the choice of the ideal $I$ is clear from the context.

Proposition 15.84.5. Let $I \subset A$ be a finitely generated ideal of a ring $A$. Let $M$ be an $A$-module. The following are equivalent

1. $M$ is $I$-adically complete, and

2. $M$ is derived complete with respect to $I$ and $\bigcap I^ nM = 0$.

Proof. This is clear from the results of Lemma 15.84.3. $\square$

The next lemma shows that the category $\mathcal{C}$ of derived complete modules is abelian. It turns out that $\mathcal{C}$ is not a Grothendieck abelian category, see Examples, Section 107.10.

Lemma 15.84.6. Let $I$ be an ideal of a ring $A$.

1. The derived complete $A$-modules form a weak Serre subcategory $\mathcal{C}$ of $\text{Mod}_ A$.

2. $D_\mathcal {C}(A) \subset D(A)$ is the full subcategory of derived complete objects.

Proof. Part (2) is immediate from Lemma 15.84.1 and the definitions. For part (1), suppose that $M \to N$ is a map of derived complete modules. Denote $K = (M \to N)$ the corresponding object of $D(A)$. Pick $f \in I$. Then $\mathop{\mathrm{Ext}}\nolimits _ A^ n(A_ f, K)$ is zero for all $n$ because $\mathop{\mathrm{Ext}}\nolimits _ A^ n(A_ f, M)$ and $\text{Ext}_ A^ n(A_ f, N)$ are zero for all $n$. Hence $K$ is derived complete. By (2) we see that $\mathop{\mathrm{Ker}}(M \to N)$ and $\mathop{\mathrm{Coker}}(M \to N)$ are objects of $\mathcal{C}$. Finally, suppose that $0 \to M_1 \to M_2 \to M_3 \to 0$ is a short exact sequence of $A$-modules and $M_1$, $M_3$ are derived complete. Then it follows from the long exact sequence of $\mathop{\mathrm{Ext}}\nolimits$'s that $M_2$ is derived complete. Thus $\mathcal{C}$ is a weak Serre subcategory by Homology, Lemma 12.10.3. $\square$

Lemma 15.84.7. Let $I$ be a finitely generated ideal of a ring $A$. Let $M$ be a derived complete $A$-module. If $M/IM = 0$, then $M = 0$.

Proof. Assume that $M/IM$ is zero. Let $I = (f_1, \ldots , f_ r)$. Let $i < r$ be the largest integer such that $N = M/(f_1, \ldots , f_ i)M$ is nonzero. If $i$ does not exist, then $M = 0$ which is what we want to show. Then $N$ is derived complete as a cokernel of a map between derived complete modules, see Lemma 15.84.6. By our choice of $i$ we have that $f_{i + 1} : N \to N$ is surjective. Hence

$\mathop{\mathrm{lim}}\nolimits (\ldots \to N \xrightarrow {f_{i + 1}} N \xrightarrow {f_{i + 1}} N)$

is nonzero, contradicting the derived completeness of $N$. $\square$

If the ring is $I$-adically complete, then one obtains an ample supply of derived complete complexes.

Lemma 15.84.8. Let $A$ be a ring and $I \subset A$ an ideal. If $A$ is $I$-adically complete then any pseudo-coherent object of $D(A)$ is derived complete.

Proof. Let $K$ be a pseudo-coherent object of $D(A)$. By definition this means $K$ is represented by a bounded above complex $K^\bullet$ of finite free $A$-modules. Since $A$ is $I$-adically complete, it is derived complete (Lemma 15.84.3). It follows that $H^ n(K)$ is derived complete for all $n$, by part (1) of Lemma 15.84.6. This in turn implies that $K$ is derived complete by part (2) of the same lemma. $\square$

Lemma 15.84.9. Let $A$ be a ring. Let $f, g \in A$. Then for $K \in D(A)$ we have $R\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, R\mathop{\mathrm{Hom}}\nolimits _ A(A_ g, K)) = R\mathop{\mathrm{Hom}}\nolimits _ A(A_{fg}, K)$.

Proof. This follows from Lemma 15.69.1. $\square$

Lemma 15.84.10. 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 15.84.9 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 15.28.13. (In fact, if $f = f_ i$ for some $i$, then this complex is homotopic to zero; 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 15.84.1.

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 $D(A)$. $\square$

Lemma 15.84.11. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $K^\bullet$ be a complex of $A$-modules such that $f : K^\bullet \to K^\bullet$ is an isomorphism for some $f \in I$, i.e., $K^\bullet$ is a complex of $A_ f$-modules. Then the derived completion of $K^\bullet$ is zero.

Proof. Indeed, in this case the $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ is zero for any derived complete complex $L$, see Lemma 15.84.1. Hence $K^\wedge$ is zero by the universal property in Lemma 15.84.10. $\square$

Lemma 15.84.12. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $K, L \in D(A)$. Then

$R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)^\wedge = R\mathop{\mathrm{Hom}}\nolimits _ A(K, L^\wedge ) = R\mathop{\mathrm{Hom}}\nolimits _ A(K^\wedge , L^\wedge )$

Proof. By Lemma 15.84.10 we know that derived completion is given by $R\mathop{\mathrm{Hom}}\nolimits _ A(C, -)$ for some $C \in D(A)$. Then

\begin{align*} R\mathop{\mathrm{Hom}}\nolimits _ A(C, R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)) & = R\mathop{\mathrm{Hom}}\nolimits _ A(C \otimes _ A^\mathbf {L} K, L) \\ & = R\mathop{\mathrm{Hom}}\nolimits _ A(K, R\mathop{\mathrm{Hom}}\nolimits _ A(C, L)) \end{align*}

by Lemma 15.69.1. This proves the first equation. The map $K \to K^\wedge$ induces a map

$R\mathop{\mathrm{Hom}}\nolimits _ A(K^\wedge , L^\wedge ) \to R\mathop{\mathrm{Hom}}\nolimits _ A(K, L^\wedge )$

which is an isomorphism in $D(A)$ by definition of the derived completion as the left adjoint to the inclusion functor. $\square$

Lemma 15.84.13. Let $A$ be a ring and let $I \subset A$ be an ideal. Let $(K_ n)$ be an inverse system of objects of $D(A)$ such that for all $f \in I$ and $n$ there exists an $e = e(n, f)$ such that $f^ e$ is zero on $K_ n$. Then for $K \in D(A)$ the object $K' = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} K_ n)$ is derived complete with respect to $I$.

Proof. Since the category of derived complete objects is preserved under $R\mathop{\mathrm{lim}}\nolimits$ it suffices to show that each $K \otimes _ A^\mathbf {L} K_ n$ is derived complete. By assumption for all $f \in I$ there is an $e$ such that $f^ e$ is zero on $K \otimes _ A^\mathbf {L} K_ n$. Of course this implies that $T(K \otimes _ A^\mathbf {L} K_ n, f) = 0$ and we win. $\square$

Situation 15.84.14. Let $A$ be a ring. Let $I = (f_1, \ldots , f_ r) \subset A$. Let $K_ n^\bullet = K_\bullet (A, f_1^ n, \ldots , f_ r^ n)$ be the Koszul complex on $f_1^ n, \ldots , f_ r^ n$ viewed as a cochain complex in degrees $-r, -r + 1, \ldots , 0$. Using the functoriality of Lemma 15.28.3 we obtain an inverse system

$\ldots \to K_3^\bullet \to K_2^\bullet \to K_1^\bullet$

compatible with the inverse system $H^0(K_ n^\bullet ) = A/(f_1^ n, \ldots , f_ r^ n)$ and compatible with the maps $A \to K_ n^\bullet$.

A key feature of the discussion below will use that for $m > n$ the map

$K_ m^{-p} = \wedge ^ p(A^{\oplus r}) \to \wedge ^ p(A^{\oplus r}) = K_ n^{-p}$

is given by multiplication by $f_{i_1}^{m - n} \ldots f_{i_ p}^{m - n}$ on the basis element $e_{i_1} \wedge \ldots \wedge e_{i_ p}$.

Lemma 15.84.15. In Situation 15.84.14. For $K \in D(A)$ the object $K' = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} K_ n^\bullet )$ is derived complete with respect to $I$.

Proof. This is a special case of Lemma 15.84.13 because $f_ i^ n$ acts by an endomorphism of $K_ n^\bullet$ which is homotopic to zero by Lemma 15.28.6. $\square$

Lemma 15.84.16. In Situation 15.84.14. Let $K \in D(A)$. The following are equivalent

1. $K$ is derived complete with respect to $I$, and

2. the canonical map $K \to R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} K_ n^\bullet )$ is an isomorphism of $D(A)$.

Proof. If (2) holds, then $K$ is derived complete with respect to $I$ by Lemma 15.84.15. Conversely, assume that $K$ is derived complete with respect to $I$. Consider the filtrations

$K_ n^\bullet \supset \sigma _{\geq -r + 1}K_ n^\bullet \supset \sigma _{\geq -r + 2}K_ n^\bullet \supset \ldots \supset \sigma _{\geq -1}K_ n^\bullet \supset \sigma _{\geq 0}K_ n^\bullet = A$

by stupid truncations (Homology, Section 12.15). Because the construction $R\mathop{\mathrm{lim}}\nolimits (K \otimes E)$ is exact in the second variable (Lemma 15.80.11) we see that it suffices to show

$R\mathop{\mathrm{lim}}\nolimits \left( K \otimes _ A^\mathbf {L} (\sigma _{\geq p}K_ n^\bullet / \sigma _{\geq p + 1}K_ n^\bullet ) \right) = 0$

for $p < 0$. The explicit description of the Koszul complexes above shows that

$R\mathop{\mathrm{lim}}\nolimits \left( K \otimes _ A^\mathbf {L} (\sigma _{\geq p}K_ n^\bullet / \sigma _{\geq p + 1}K_ n^\bullet ) \right) = \bigoplus \nolimits _{i_1, \ldots , i_{-p}} T(K, f_{i_1}\ldots f_{i_{-p}})$

which is zero for $p < 0$ by assumption on $K$. $\square$

Lemma 15.84.17. In Situation 15.84.14. The functor which sends $K \in D(A)$ to the derived limit $K' = R\mathop{\mathrm{lim}}\nolimits ( K \otimes _ A^\mathbf {L} K_ n^\bullet )$ is the left adjoint to the inclusion functor $D_{comp}(A) \to D(A)$ constructed in Lemma 15.84.10.

First proof. The assignment $K \leadsto K'$ is a functor and $K'$ is derived complete with respect to $I$ by Lemma 15.84.15. By a formal argument (omitted) we see that it suffices to show $K \to K'$ is an isomorphism if $K$ is derived complete with respect to $I$. This is Lemma 15.84.16. $\square$

Second proof. Denote $K \mapsto K^\wedge$ the adjoint constructed in Lemma 15.84.10. By that lemma we have

$K^\wedge = R\mathop{\mathrm{Hom}}\nolimits \left((A \to \prod \nolimits _{i_0} A_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_ r}), K\right)$

In Lemma 15.28.13 we have seen that the extended alternating Čech complex

$A \to \prod \nolimits _{i_0} A_{f_{i_0}} \to \prod \nolimits _{i_0 < i_1} A_{f_{i_0}f_{i_1}} \to \ldots \to A_{f_1\ldots f_ r}$

is a colimit of the Koszul complexes $K^ n = K(A, f_1^ n, \ldots , f_ r^ n)$ sitting in degrees $0, \ldots , r$. Note that $K^ n$ is a finite chain complex of finite free $A$-modules with dual (as in Lemma 15.70.14) $R\mathop{\mathrm{Hom}}\nolimits _ A(K^ n, A) = K_ n$ where $K_ n$ is the Koszul cochain complex sitting in degrees $-r, \ldots , 0$ (as usual). Thus it suffices to show that

$R\mathop{\mathrm{Hom}}\nolimits _ A(\text{hocolim} K^ n, K) = R\mathop{\mathrm{lim}}\nolimits (K \otimes _ A^\mathbf {L} K_ n)$

This follows from Lemma 15.70.15. $\square$

As an application of the relationship with the Koszul complex we obtain that derived completion has finite cohomological dimension.

Lemma 15.84.18. Let $A$ be a ring and let $I \subset A$ be an ideal which can be generated by $r$ elements. Then derived completion has finite cohomological dimension:

1. Let $K \to L$ be a morphism in $D(A)$ such that $H^ i(K) \to H^ i(L)$ is an isomorphism for $i \geq 1$ and surjective for $i = 0$. Then $H^ i(K^\wedge ) \to H^ i(L^\wedge )$ is an isomorphism for $i \geq 1$ and surjective for $i = 0$.

2. Let $K \to L$ be a morphism of $D(A)$ such that $H^ i(K) \to H^ i(L)$ is an isomorphism for $i \leq -1$ and injective for $i = 0$. Then $H^ i(K^\wedge ) \to H^ i(L^\wedge )$ is an isomorphism for $i \leq -r - 1$ and injective for $i = -r$.

Proof. Say $I$ is generated by $f_1, \ldots , f_ r$. For any $K \in D(A)$ by Lemma 15.84.17 we have $K^\wedge = R\mathop{\mathrm{lim}}\nolimits K \otimes _ A^\mathbf {L} K_ n$ where $K_ n$ is the Koszul complex on $f_1^ n, \ldots , f_ r^ n$ and hence we obtain a short exact sequence

$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{i - 1}(K \otimes _ A^\mathbf {L} K_ n) \to H^ i(K^\wedge ) \to \mathop{\mathrm{lim}}\nolimits H^ i(K \otimes _ A^\mathbf {L} K_ n) \to 0$

by Lemma 15.80.4.

Proof of (1). Pick a distinguished triangle $K \to L \to C \to K$. Then $H^ i(C) = 0$ for $i \geq 0$. Since $K_ n$ is sitting in degrees $\leq 0$ we see that $H^ i(C \otimes _ A^\mathbf {L} K_ n) = 0$ for $i \geq 0$ and that $H^{-1}(C \otimes _ A^\mathbf {L} K_ n) = H^{-1}(C) \otimes _ A A/(f_1^ n, \ldots , f_ r^ n)$ is a system with surjective transition maps. The displayed equation above shows that $H^ i(C^\wedge ) = 0$ for $i \geq 0$. Applying the distinguished triangle $K^\wedge \to L^\wedge \to C^\wedge \to K^\wedge $ we get (1).

Proof of (2). Pick a distinguished triangle $K \to L \to C \to K$. Then $H^ i(C) = 0$ for $i < 0$. Since $K_ n$ is sitting in degrees $-r, \ldots , 0$ we see that $H^ i(C \otimes _ A^\mathbf {L} K_ n) = 0$ for $i < -r$. The displayed equation above shows that $H^ i(C^\wedge ) = 0$ for $i < r$. Applying the distinguished triangle $K^\wedge \to L^\wedge \to C^\wedge \to K^\wedge $ we get (2). $\square$

Lemma 15.84.19. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $K^\bullet$ be a filtered complex of $A$-modules. There exists a canonical spectral sequence $(E_ r, \text{d}_ r)_{r \geq 1}$ of bigraded derived complete $A$-modules with $d_ r$ of bidegree $(r, -r + 1)$ and with

$E_1^{p, q} = H^{p + q}((\text{gr}^ pK^\bullet )^\wedge )$

If the filtration on each $K^ n$ is finite, then the spectral sequence is bounded and converges to $H^*((K^\bullet )^\wedge )$.

Proof. By Lemma 15.84.10 we know that derived completion is given by $R\mathop{\mathrm{Hom}}\nolimits _ A(C, -)$ for some $C \in D^ b(A)$. By Lemmas 15.84.18 and 15.65.2 we see that $C$ has finite projective dimension. Thus we may choose a bounded complex of projective modules $P^\bullet$ representing $C$. Then

$M^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet , K^\bullet )$

is a complex of $A$-modules representing $(K^\bullet )^\wedge$. It comes with a filtration given by $F^ pM^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet , F^ pK^\bullet )$. We see that $F^ pM^\bullet$ represents $(F^ pK^\bullet )^\wedge$ and hence $\text{gr}^ pM^\bullet$ represents $(\text{gr}K^\bullet )^\wedge$. Thus we find our spectral sequence by taking the spectral sequence of the filtered complex $M^\bullet$, see Homology, Section 12.24. If the filtration on each $K^ n$ is finite, then the filtration on each $M^ n$ is finite because $P^\bullet$ is a bounded complex. Hence the final statement follows from Homology, Lemma 12.24.11. $\square$

Example 15.84.20. Let $A$ be a ring and let $I \subset A$ be a finitely generated ideal. Let $K^\bullet$ be a complex of $A$-modules. We can apply Lemma 15.84.19 with $F^ pK^\bullet = \tau _{\leq -p}K^\bullet$. Then we get a bounded spectral sequence

$E_1^{p, q} = H^{p + q}(H^{-p}(K^\bullet )^\wedge [p]) = H^{2p + q}(H^{-p}(K^\bullet )^\wedge )$

converging to $H^{p + q}((K^\bullet )^\wedge )$. After renumbering $p = -j$ and $q = i + 2j$ we find that for any $K \in D(A)$ there is a bounded spectral sequence $(E'_ r, d'_ r)_{r \geq 2}$ of bigraded derived complete modules with $d'_ r$ of bidegree $(r, -r + 1)$, with

$(E'_2)^{i, j} = H^ i(H^ j(K)^\wedge )$

and converging to $H^{i + j}(K^\wedge )$.

Lemma 15.84.21. Let $A \to B$ be a ring map. Let $I \subset A$ be an ideal. The inverse image of $D_{comp}(A, I)$ under the restriction functor $D(B) \to D(A)$ is $D_{comp}(B, IB)$.

Proof. Using Lemma 15.84.2 we see that $L \in D(B)$ is in $D_{comp}(B, IB)$ if and only if $T(L, f)$ is zero for every local section $f \in I$. Observe that the cohomology of $T(L, f)$ is computed in the category of abelian groups, so it doesn't matter whether we think of $f$ as an element of $A$ or take the image of $f$ in $B$. The lemma follows immediately from this and the definition of derived complete objects. $\square$

Lemma 15.84.22. Let $A \to B$ be a ring map. Let $I \subset A$ be a finitely generated ideal. If $A \to B$ is flat and $A/I \cong B/IB$, then the restriction functor $D(B) \to D(A)$ induces an equivalence $D_{comp}(B, IB) \to D_{comp}(A, I)$.

Proof. Choose generators $f_1, \ldots , f_ r$ of $I$. Denote $\check{\mathcal{C}}^\bullet _ A \to \check{\mathcal{C}}^\bullet _ B$ the quasi-isomorphism of extended alternating Čech complexes of Lemma 15.82.4. Let $K \in D_{comp}(A, I)$. Let $I^\bullet$ be a K-injective complex of $A$-modules representing $K$. Since $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_ f, K)$ and $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(B_ f, K)$ are zero for all $f \in I$ and $n \in \mathbf{Z}$ (Lemma 15.84.1) we conclude that $\check{\mathcal{C}}^\bullet _ A \to A$ and $\check{\mathcal{C}}^\bullet _ B \to B$ induce quasi-isomorphisms

$I^\bullet = \mathop{\mathrm{Hom}}\nolimits _ A(A, I^\bullet ) \longrightarrow \text{Tot}(\mathop{\mathrm{Hom}}\nolimits _ A(\check{\mathcal{C}}^\bullet _ A, I^\bullet ))$

and

$\mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet ) \longrightarrow \text{Tot}(\mathop{\mathrm{Hom}}\nolimits _ A(\check{\mathcal{C}}^\bullet _ B, I^\bullet ))$

Some details omitted. Since $\check{\mathcal{C}}^\bullet _ A \to \check{\mathcal{C}}^\bullet _ B$ is a quasi-isomorphism and $I^\bullet$ is K-injective we conclude that $\mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet ) \to I^\bullet$ is a quasi-isomorphism. As the complex $\mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )$ is a complex of $B$-modules we conclude that $K$ is in the image of the restriction map, i.e., the functor is essentially surjective

In fact, the argument shows that $F : D_{comp}(A, I) \to D_{comp}(B, IB)$, $K \mapsto \mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )$ is a left inverse to restriction. Finally, suppose that $L \in D_{comp}(B, IB)$. Represent $L$ by a K-injective complex $J^\bullet$ of $B$-modules. Then $J^\bullet$ is also K-injective as a complex of $A$-modules (Lemma 15.55.1) hence $F(\text{restriction of }L) = \mathop{\mathrm{Hom}}\nolimits _ A(B, J^\bullet )$. There is a map $J^\bullet \to \mathop{\mathrm{Hom}}\nolimits _ A(B, J^\bullet )$ of complexes of $B$-modules, whose composition with $\mathop{\mathrm{Hom}}\nolimits _ A(B, J^\bullet ) \to J^\bullet$ is the identity. We conclude that $F$ is also a right inverse to restriction and the proof is finished. $\square$

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