The Stacks project

Proof. Proof of (1). Let $\mathcal{J}^\bullet $ be a K-injective complex of $\mathcal{O}_ f$-modules representing $N$. By Cohomology on Sites, Lemma 21.20.10 it follows that $\mathcal{J}^\bullet $ is a K-injective complex of $\mathcal{O}$-modules as well. Let $\mathcal{F}^\bullet $ be a complex of $\mathcal{O}_ f$-modules representing $L$. Then

by Modules on Sites, Lemma 18.11.4 because $\mathcal{J}^\bullet $ is a K-injective complex of $\mathcal{O}$ and of $\mathcal{O}_ f$-modules.

Proof of (2). Let $\mathcal{I}^\bullet $ be a K-injective complex of $\mathcal{O}$-modules representing $K$. Then $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, K)$ is represented by $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, \mathcal{I}^\bullet )$ which is a K-injective complex of $\mathcal{O}_ f$-modules and of $\mathcal{O}$-modules by Cohomology on Sites, Lemmas 21.20.11 and 21.20.10. Let $\mathcal{F}^\bullet $ be a complex of $\mathcal{O}_ f$-modules representing $L$. Then

by Modules on Sites, Lemma 18.27.8 and because $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, \mathcal{I}^\bullet )$ is a K-injective complex of $\mathcal{O}_ f$-modules.

Proof of (3). This follows from the fact that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ g, \mathcal{I}^\bullet )$ is K-injective as a complex of $\mathcal{O}$-modules and the fact that $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ g, \mathcal{H})) = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_{gf}, \mathcal{H})$ for all sheaves of $\mathcal{O}$-modules $\mathcal{H}$. $\square$

Proof. It is clear that (2) implies (1). The implication (1) $\Rightarrow $ (2) follows from Lemma 52.6.1. A free resolution of the $\mathcal{O}$-module $\mathcal{O}_ f$ is given by

where the first map sends a local section $(x_0, x_1, \ldots )$ to $(x_0, x_1 - fx_0, x_2 - fx_1, \ldots )$ and the second map sends $(x_0, x_1, \ldots )$ to $x_0 + x_1/f + x_2/f^2 + \ldots $. Applying $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(-, \mathcal{I}^\bullet )$ where $\mathcal{I}^\bullet $ is a K-injective complex of $\mathcal{O}$-modules representing $K$ we get a short exact sequence of complexes

because $\mathcal{I}^ n$ is an injective $\mathcal{O}$-module. The products are products in $D(\mathcal{O})$, see Injectives, Lemma 19.13.4. This means that the object $T(K, f)$ is a representative of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, K)$ in $D(\mathcal{O})$. Thus the equivalence of (1) and (3). $\square$

Proof. We will use the results of Lemma 52.6.2 without further mention. If $f \in \mathcal{I}(U)$, and $g \in \mathcal{O}(U)$, then $\mathcal{O}_{U, gf}$ is an $\mathcal{O}_{U, f}$-module hence $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_{U, gf}, K|_ U) = 0$, hence $gf \in \mathcal{I}(U)$. Suppose $f, g \in \mathcal{O}(U)$. Then there is a short exact sequence

because $f, g$ generate the unit ideal in $\mathcal{O}(U)_{f + g}$. This follows from Algebra, Lemma 10.24.2 and the easy fact that the last arrow is surjective. Because $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}( - , K|_ U)$ is an exact functor of triangulated categories the vanishing of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, f(f + g)}, K|_ U)$, $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, g(f + g)}, K|_ U)$, and $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, gf(f + g)}, K|_ U)$, implies the vanishing of $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, f + g}, K|_ U)$. We omit the verification of the sheaf condition. $\square$

Proof. Let $U$ be an object of $\mathcal{C}$ and let $f \in \mathcal{I}(U)$. Recall that

by Cohomology on Sites, Lemma 21.35.2. The right hand side is zero by Lemma 52.6.2 and the relationship between internal hom and actual hom, see Cohomology on Sites, Lemma 21.35.1. The same vanishing holds for all $U'/U$. Thus the object $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, f}, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(K, L)|_ U)$ of $D(\mathcal{O}_ U)$ has vanishing $0$th cohomology sheaf (by locus citatus). Similarly for the other cohomology sheaves, i.e., $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, f}, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(K, L)|_ U)$ is zero in $D(\mathcal{O}_ U)$. By Lemma 52.6.2 we conclude. $\square$

Proof. Using Lemma 52.6.3 we see that $K' \in D(\mathcal{O}')$ is in $D_{comp}(\mathcal{O}', \mathcal{I}\mathcal{O}')$ if and only if $T(K'|_ U, f)$ is zero for every local section $f \in \mathcal{I}(U)$. Observe that the cohomology sheaves of $T(K'|_ U, f)$ are computed in the category of abelian sheaves, so it doesn't matter whether we think of $f$ as a section of $\mathcal{O}$ or take the image of $f$ as a section of $\mathcal{O}'$. The lemma follows immediately from this and the definition of derived complete objects. $\square$

Proof. We may assume $f$ is given by a morphism of ringed sites corresponding to a continuous functor $\mathcal{C} \to \mathcal{D}$ (Modules on Sites, Lemma 18.7.2 ). Let $U$ be an object of $\mathcal{C}$ and let $g$ be a section of $\mathcal{I}$ over $U$. We have to show that $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{O}_ U)}(\mathcal{O}_{U, g}, Rf_*K|_ U) = 0$ whenever $K$ is derived complete with respect to $\mathcal{I}'$. Namely, by Cohomology on Sites, Lemma 21.35.1 this, applied to all objects over $U$ and all shifts of $K$, will imply that $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(\mathcal{O}_{U, g}, Rf_*K|_ U)$ is zero, which implies that $T(Rf_*K|_ U, g)$ is zero (Lemma 52.6.2) which is what we have to show (Definition 52.6.4). Let $V$ in $\mathcal{D}$ be the image of $U$. Then

where $g' = f^\sharp (g) \in \mathcal{I}'(V)$. The second equality because $K$ is derived complete and the first equality because the derived pullback of $\mathcal{O}_{U, g}$ is $\mathcal{O}'_{V, g'}$ and Cohomology on Sites, Lemma 21.19.1. $\square$

Proof. Define $K^\wedge $ by the last displayed formula of the lemma. There is a map of complexes

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 $f$ be a global section of $\mathcal{O}$. By Lemma 52.6.1 the object $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, K^\wedge )$ is equal to

If $f = f_ i$ for some $i$, then $f_1, \ldots , f_ r$ generate the unit ideal in $\mathcal{O}_ f$, hence the extended alternating Čech complex

is zero (even homotopic to zero). In this way we see that $K^\wedge $ is derived complete.

If $K$ is derived complete, then $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(\mathcal{O}_ f, K)$ is zero for all $f = f_{i_0} \ldots f_{i_ p}$, $p \geq 0$. Thus $K \to K^\wedge $ is an isomorphism in $D(\mathcal{O})$. $\square$

Proof. In More on Algebra, Lemma 15.29.6 we have seen that the extended alternating Čech complex

is a colimit of the Koszul complexes $K^ n = K(\mathcal{O}, f_1^ n, \ldots , f_ r^ n)$ sitting in degrees $0, \ldots , r$. Note that $K^ n$ is a finite chain complex of finite free $\mathcal{O}$-modules with dual $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(K^ n, \mathcal{O}) = K_ n$ where $K_ n$ is the Koszul cochain complex sitting in degrees $-r, \ldots , 0$ (as usual). By Lemma 52.6.8 the functor $E \mapsto E^\wedge $ is gotten by taking $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits $ from the extended alternating Čech complex into $E$:

This is equal to $R\mathop{\mathrm{lim}}\nolimits (E \otimes _\mathcal {O}^\mathbf {L} K_ n)$ by Cohomology on Sites, Lemma 21.48.8. $\square$

Proof. Let $S$ be the set of rings $A_0$ of the form $A_0 = \mathbf{Z}[x_1, \ldots , x_ n]/J$. Every finite type $\mathbf{Z}$-algebra is isomorphic to an element of $S$. Let $\mathcal{A}_0$ be the category whose objects are pairs $(A_0, I_0)$ where $A_0 \in S$ and $I_0 \subset A_0$ is an ideal and whose morphisms $(A_0, I_0) \to (B_0, J_0)$ are ring maps $\varphi : A_0 \to B_0$ such that $J_0 = \varphi (I_0)B_0$.

Suppose we can construct $K(A_0, I_0) \to A_0$ functorially for objects of $\mathcal{A}_0$ having properties (a), (b), (c), and (d). Then we take

where the colimit is over ring maps $\varphi : A_0 \to A$ such that $\varphi (I_0)A = I$ with $(A_0, I_0)$ in $\mathcal{A}_0$. A morphism between $(A_0, I_0) \to (A, I)$ and $(A_0', I_0') \to (A, I)$ are given by maps $(A_0, I_0) \to (A_0', I_0')$ in $\mathcal{A}_0$ commuting with maps to $A$. The category of these $(A_0, I_0) \to (A, I)$ is filtered (details omitted). Moreover, $\mathop{\mathrm{colim}}\nolimits _{\varphi : (A_0, I_0) \to (A, I)} A_0 = A$ so that $K(A, I)$ is a complex of $A$-modules. Finally, given $\varphi : A \to B$ and $I \subset A$ for every $(A_0, I_0) \to (A, I)$ in the colimit, the composition $(A_0, I_0) \to (B, IB)$ lives in the colimit for $(B, IB)$. In this way we get a map on colimits. Properties (a), (b), (c), and (d) follow readily from this and the corresponding properties of the complexes $K(A_0, I_0)$.

Endow $\mathcal{C}_0 = \mathcal{A}_0^{opp}$ with the chaotic topology. We equip $\mathcal{C}_0$ with the sheaf of rings $\mathcal{O} : (A, I) \mapsto A$. The ideals $I$ fit together to give a sheaf of ideals $\mathcal{I} \subset \mathcal{O}$. Choose an injective resolution $\mathcal{O} \to \mathcal{J}^\bullet $. Consider the object

Let $U = (A, I) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_0)$. Since the topology in $\mathcal{C}_0$ is chaotic, the value $\mathcal{J}^\bullet (U)$ is a resolution of $A$ by injective $A$-modules. Hence the value $\mathcal{F}^\bullet (U)$ is an object of $D(A)$ representing the image of $R\Gamma _ I(A)$ in $D(A)$, see Dualizing Complexes, Section 47.9. Choose a complex of $\mathcal{O}$-modules $\mathcal{K}^\bullet $ and a commutative diagram

where the horizontal arrows are quasi-isomorphisms. This is possible by the construction of the derived category $D(\mathcal{O})$. Set $K(A, I) = \mathcal{K}^\bullet (U)$ where $U = (A, I)$. Properties (a) and (b) are clear and properties (c) and (d) follow from Dualizing Complexes, Lemmas 47.10.2 and 47.10.3. $\square$

Proof. Let $\mathcal{C}' \subset \mathcal{C}$ be the full subcategory of objects $U$ such that $\mathcal{I}|_ U$ is generated by finitely many sections. Then $\mathcal{C}' \to \mathcal{C}$ is a special cocontinuous functor (Sites, Definition 7.29.2). Hence it suffices to work with $\mathcal{C}'$, see Sites, Lemma 7.29.1. In other words we may assume that for every object $U$ of $\mathcal{C}$ there exists a finitely generated ideal $I \subset \mathcal{I}(U)$ such that $\mathcal{I}|_ U = \mathop{\mathrm{Im}}(I \otimes \mathcal{O}_ U \to \mathcal{O}_ U)$. We will say that $I$ generates $\mathcal{I}|_ U$. Warning: We do not know that $\mathcal{I}(U)$ is a finitely generated ideal in $\mathcal{O}(U)$.

Let $U$ be an object and $I \subset \mathcal{O}(U)$ a finitely generated ideal which generates $\mathcal{I}|_ U$. On the category $\mathcal{C}/U$ consider the complex of presheaves

with $K(-, -)$ as in Lemma 52.6.10. We claim that the sheafification of this is independent of the choice of $I$. Indeed, if $I' \subset \mathcal{O}(U)$ is a finitely generated ideal which also generates $\mathcal{I}|_ U$, then there exists a covering $\{ U_ j \to U\} $ such that $I\mathcal{O}(U_ j) = I'\mathcal{O}(U_ j)$. (Hint: this works because both $I$ and $I'$ are finitely generated and generate $\mathcal{I}|_ U$.) Hence $K_{U, I}^\bullet $ and $K_{U, I'}^\bullet $ are the same for any object lying over one of the $U_ j$. The statement on sheafifications follows. Denote $K_ U^\bullet $ the common value.

The independence of choice of $I$ also shows that $K_ U^\bullet |_{\mathcal{C}/U'} = K_{U'}^\bullet $ whenever we are given a morphism $U' \to U$ and hence a localization morphism $\mathcal{C}/U' \to \mathcal{C}/U$. Thus the complexes $K_ U^\bullet $ glue to give a single well defined complex $K^\bullet $ of $\mathcal{O}$-modules. The existence of the map $K^\bullet \to \mathcal{O}$ and the quasi-isomorphism of the lemma follow immediately from the corresponding properties of the complexes $K(-, -)$ in Lemma 52.6.10. $\square$

Proof. Let $K \to \mathcal{O}$ in $D(\mathcal{O})$ be as constructed in Lemma 52.6.11. Let $E \in D(\mathcal{O})$. Then $E^\wedge = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, E)$ together with the map $E \to E^\wedge $ will do the job. Namely, locally on the site $\mathcal{C}$ we recover the adjoint of Lemma 52.6.8. This shows that $E^\wedge $ is always derived complete and that $E \to E^\wedge $ is an isomorphism if $E$ is derived complete. $\square$

Proof. Observe that the second complex is the tensor product of the first complex with $\mathcal{O}'$. We can write the first extended alternating Čech complex as a colimit of the Koszul complexes $K_ n = K(\mathcal{O}, f_1^ n, \ldots , f_ r^ n)$, see More on Algebra, Lemma 15.29.6. Hence it suffices to prove $K_ n \to K_ n \otimes _\mathcal {O} \mathcal{O}'$ is a quasi-isomorphism. Since $\mathcal{O} \to \mathcal{O}'$ is flat it suffices to show that $H^ i \to H^ i \otimes _\mathcal {O} \mathcal{O}'$ is an isomorphism where $H^ i$ is the $i$th cohomology sheaf $H^ i = H^ i(K_ n)$. These sheaves are annihilated by $f_1^ n, \ldots , f_ r^ n$, see More on Algebra, Lemma 15.28.6. Hence these sheaves are annihilated by $(f_1, \ldots , f_ r)^ m$ for some $m \gg 0$. Thus $H^ i \to H^ i \otimes _\mathcal {O} \mathcal{O}'$ is an isomorphism by Modules on Sites, Lemma 18.28.16. $\square$

Proof. Lemma 52.6.7 implies restriction $r : D(\mathcal{O}') \to D(\mathcal{O})$ sends $D_{comp}(\mathcal{O}', \mathcal{I}\mathcal{O}')$ into $D_{comp}(\mathcal{O}, \mathcal{I})$. We will construct a quasi-inverse $E \mapsto E'$.

Let $K \to \mathcal{O}$ be the morphism of $D(\mathcal{O})$ constructed in Lemma 52.6.11. Set $K' = K \otimes _\mathcal {O}^\mathbf {L} \mathcal{O}'$ in $D(\mathcal{O}')$. Then $K' \to \mathcal{O}'$ is a map in $D(\mathcal{O}')$ which satisfies the conclusions of Lemma 52.6.11 with respect to $\mathcal{I}' = \mathcal{I}\mathcal{O}'$. The map $K \to r(K')$ is a quasi-isomorphism by Lemma 52.6.16. Now, for $E \in D_{comp}(\mathcal{O}, \mathcal{I})$ we set

viewed as an object in $D(\mathcal{O}')$ using the $\mathcal{O}'$-module structure on $K'$. Since $E$ is derived complete we have $E = R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _\mathcal {O}(K, E)$, see proof of Proposition 52.6.12. On the other hand, since $K \to r(K')$ is an isomorphism in we see that there is an isomorphism $E \to r(E')$ in $D(\mathcal{O})$. To finish the proof we have to show that, if $E = r(M')$ for an object $M'$ of $D_{comp}(\mathcal{O}', \mathcal{I}')$, then $E' \cong M'$. To get a map we use

where the second arrow uses the map $K' \to \mathcal{O}'$. To see that this is an isomorphism, one shows that $r$ applied to this arrow is the same as the isomorphism $E \to r(E')$ above. Details omitted. $\square$

Proof. The first statement we have seen in Lemma 52.6.7. Note that the second statement makes sense as we have a derived completion functor $D(\mathcal{O}') \to D_{comp}(\mathcal{O}', \mathcal{I}')$ by Proposition 52.6.12. OK, so now let $K \in D_{comp}(\mathcal{O}, \mathcal{I})$ and $M \in D_{comp}(\mathcal{O}', \mathcal{I}')$. Then we have

by the universal property of derived completion. $\square$

Proof. By Proposition 52.6.12 the derived completion functors exist. By Lemma 52.6.7 the object $Rf_*(K^\wedge )$ is derived complete, and hence we obtain a canonical map $(Rf_*K)^\wedge \to Rf_*(K^\wedge )$ by the universal property of derived completion. We may check this map is an isomorphism locally on $\mathcal{C}$. Thus, since derived completion commutes with localization (Remark 52.6.14) we may assume that $\mathcal{I}$ is generated by global sections $f_1, \ldots , f_ r$. Then $\mathcal{I}'$ is generated by $g_ i = f^\sharp (f_ i)$. By Lemma 52.6.9 we have to prove that

Because $Rf_*$ commutes with $R\mathop{\mathrm{lim}}\nolimits $ (Cohomology on Sites, Lemma 21.23.3) it suffices to prove that

This follows from the projection formula (Cohomology on Sites, Lemma 21.50.1) and the fact that $Lf^*K(\mathcal{O}, f_1^ n, \ldots , f_ r^ n) = K(\mathcal{O}', g_1^ n, \ldots , g_ r^ n)$. $\square$

Proof. Apply Lemma 52.6.19 to the morphism of ringed topoi $(\mathcal{C}, \mathcal{O}) \to (pt, A)$ and take cohomology to get the first statement. The second spectral sequence is the second spectral sequence of Derived Categories, Lemma 13.21.3. The first spectral sequence is the spectral sequence of More on Algebra, Example 15.91.22 applied to $R\Gamma (\mathcal{C}, \mathcal{F})^\wedge $. $\square$