## 20.26 Flat resolutions

A reference for the material in this section is . Let $(X, \mathcal{O}_ X)$ be a ringed space. By Modules, Lemma 17.16.6 any $\mathcal{O}_ X$-module is a quotient of a flat $\mathcal{O}_ X$-module. By Derived Categories, Lemma 13.15.4 any bounded above complex of $\mathcal{O}_ X$-modules has a left resolution by a bounded above complex of flat $\mathcal{O}_ X$-modules. However, for unbounded complexes, it turns out that flat resolutions aren't good enough.

Lemma 20.26.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{G}^\bullet$ be a complex of $\mathcal{O}_ X$-modules. The functor

$K(\textit{Mod}(\mathcal{O}_ X)) \longrightarrow K(\textit{Mod}(\mathcal{O}_ X)), \quad \mathcal{F}^\bullet \longmapsto \text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{G}^\bullet )$

is an exact functor of triangulated categories.

Proof. Omitted. Hint: See More on Algebra, Lemmas 15.57.1 and 15.57.2. $\square$

Definition 20.26.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. A complex $\mathcal{K}^\bullet$ of $\mathcal{O}_ X$-modules is called K-flat if for every acyclic complex $\mathcal{F}^\bullet$ of $\mathcal{O}_ X$-modules the complex

$\text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{K}^\bullet )$

is acyclic.

Lemma 20.26.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{K}^\bullet$ be a K-flat complex. Then the functor

$K(\textit{Mod}(\mathcal{O}_ X)) \longrightarrow K(\textit{Mod}(\mathcal{O}_ X)), \quad \mathcal{F}^\bullet \longmapsto \text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{K}^\bullet )$

transforms quasi-isomorphisms into quasi-isomorphisms.

Proof. Follows from Lemma 20.26.1 and the fact that quasi-isomorphisms are characterized by having acyclic cones. $\square$

Lemma 20.26.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{K}^\bullet$ be a complex of $\mathcal{O}_ X$-modules. Then $\mathcal{K}^\bullet$ is K-flat if and only if for all $x \in X$ the complex $\mathcal{K}_ x^\bullet$ of $\mathcal{O}_{X, x}$-modules is K-flat (More on Algebra, Definition 15.57.3).

Proof. If $\mathcal{K}_ x^\bullet$ is K-flat for all $x \in X$ then we see that $\mathcal{K}^\bullet$ is K-flat because $\otimes$ and direct sums commute with taking stalks and because we can check exactness at stalks, see Modules, Lemma 17.3.1. Conversely, assume $\mathcal{K}^\bullet$ is K-flat. Pick $x \in X$ $M^\bullet$ be an acyclic complex of $\mathcal{O}_{X, x}$-modules. Then $i_{x, *}M^\bullet$ is an acyclic complex of $\mathcal{O}_ X$-modules. Thus $\text{Tot}(i_{x, *}M^\bullet \otimes _{\mathcal{O}_ X} \mathcal{K}^\bullet )$ is acyclic. Taking stalks at $x$ shows that $\text{Tot}(M^\bullet \otimes _{\mathcal{O}_{X, x}} \mathcal{K}_ x^\bullet )$ is acyclic. $\square$

Lemma 20.26.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. If $\mathcal{K}^\bullet$, $\mathcal{L}^\bullet$ are K-flat complexes of $\mathcal{O}_ X$-modules, then $\text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{L}^\bullet )$ is a K-flat complex of $\mathcal{O}_ X$-modules.

Proof. Follows from the isomorphism

$\text{Tot}(\mathcal{M}^\bullet \otimes _{\mathcal{O}_ X} \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{L}^\bullet )) = \text{Tot}(\text{Tot}(\mathcal{M}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{K}^\bullet ) \otimes _{\mathcal{O}_ X} \mathcal{L}^\bullet )$

and the definition. $\square$

Lemma 20.26.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(\mathcal{K}_1^\bullet , \mathcal{K}_2^\bullet , \mathcal{K}_3^\bullet )$ be a distinguished triangle in $K(\textit{Mod}(\mathcal{O}_ X))$. If two out of three of $\mathcal{K}_ i^\bullet$ are K-flat, so is the third.

Proof. Follows from Lemma 20.26.1 and the fact that in a distinguished triangle in $K(\textit{Mod}(\mathcal{O}_ X))$ if two out of three are acyclic, so is the third. $\square$

Lemma 20.26.7. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. The pullback of a K-flat complex of $\mathcal{O}_ Y$-modules is a K-flat complex of $\mathcal{O}_ X$-modules.

Proof. We can check this on stalks, see Lemma 20.26.4. Hence this follows from Sheaves, Lemma 6.26.4 and More on Algebra, Lemma 15.57.5. $\square$

Lemma 20.26.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. A bounded above complex of flat $\mathcal{O}_ X$-modules is K-flat.

Proof. We can check this on stalks, see Lemma 20.26.4. Thus this lemma follows from Modules, Lemma 17.16.2 and More on Algebra, Lemma 15.57.9. $\square$

In the following lemma by a colimit of a system of complexes we mean the termwise colimit.

Lemma 20.26.9. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{K}_1^\bullet \to \mathcal{K}_2^\bullet \to \ldots$ be a system of K-flat complexes. Then $\mathop{\mathrm{colim}}\nolimits _ i \mathcal{K}_ i^\bullet$ is K-flat.

Proof. Because we are taking termwise colimits it is clear that

$\mathop{\mathrm{colim}}\nolimits _ i \text{Tot}( \mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{K}_ i^\bullet ) = \text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathop{\mathrm{colim}}\nolimits _ i \mathcal{K}_ i^\bullet )$

Hence the lemma follows from the fact that filtered colimits are exact. $\square$

Lemma 20.26.10. Let $(X, \mathcal{O}_ X)$ be a ringed space. For any complex $\mathcal{G}^\bullet$ of $\mathcal{O}_ X$-modules there exists a commutative diagram of complexes of $\mathcal{O}_ X$-modules

$\xymatrix{ \mathcal{K}_1^\bullet \ar[d] \ar[r] & \mathcal{K}_2^\bullet \ar[d] \ar[r] & \ldots \\ \tau _{\leq 1}\mathcal{G}^\bullet \ar[r] & \tau _{\leq 2}\mathcal{G}^\bullet \ar[r] & \ldots }$

with the following properties: (1) the vertical arrows are quasi-isomorphisms, (2) each $\mathcal{K}_ n^\bullet$ is a bounded above complex whose terms are direct sums of $\mathcal{O}_ X$-modules of the form $j_{U!}\mathcal{O}_ U$, and (3) the maps $\mathcal{K}_ n^\bullet \to \mathcal{K}_{n + 1}^\bullet$ are termwise split injections whose cokernels are direct sums of $\mathcal{O}_ X$-modules of the form $j_{U!}\mathcal{O}_ U$. Moreover, the map $\mathop{\mathrm{colim}}\nolimits \mathcal{K}_ n^\bullet \to \mathcal{G}^\bullet$ is a quasi-isomorphism.

Proof. The existence of the diagram and properties (1), (2), (3) follows immediately from Modules, Lemma 17.16.6 and Derived Categories, Lemma 13.29.1. The induced map $\mathop{\mathrm{colim}}\nolimits \mathcal{K}_ n^\bullet \to \mathcal{G}^\bullet$ is a quasi-isomorphism because filtered colimits are exact. $\square$

Lemma 20.26.11. Let $(X, \mathcal{O}_ X)$ be a ringed space. For any complex $\mathcal{G}^\bullet$ there exists a $K$-flat complex $\mathcal{K}^\bullet$ and a quasi-isomorphism $\mathcal{K}^\bullet \to \mathcal{G}^\bullet$. Moreover, each $\mathcal{K}^ n$ is a flat $\mathcal{O}_ X$-module.

Proof. Choose a diagram as in Lemma 20.26.10. Each complex $\mathcal{K}_ n^\bullet$ is a bounded above complex of flat modules, see Modules, Lemma 17.16.5. Hence $\mathcal{K}_ n^\bullet$ is K-flat by Lemma 20.26.8. The induced map $\mathop{\mathrm{colim}}\nolimits \mathcal{K}_ n^\bullet \to \mathcal{G}^\bullet$ is a quasi-isomorphism by construction. Thus $\mathop{\mathrm{colim}}\nolimits \mathcal{K}_ n^\bullet$ is K-flat by Lemma 20.26.9. Property (3) of Lemma 20.26.10 shows that $\mathop{\mathrm{colim}}\nolimits \mathcal{K}_ n^ m$ is a direct sum of flat modules and hence flat which proves the final assertion. $\square$

Lemma 20.26.12. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\alpha : \mathcal{P}^\bullet \to \mathcal{Q}^\bullet$ be a quasi-isomorphism of K-flat complexes of $\mathcal{O}_ X$-modules. For every complex $\mathcal{F}^\bullet$ of $\mathcal{O}_ X$-modules the induced map

$\text{Tot}(\text{id}_{\mathcal{F}^\bullet } \otimes \alpha ) : \text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{P}^\bullet ) \longrightarrow \text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{Q}^\bullet )$

is a quasi-isomorphism.

Proof. Choose a quasi-isomorphism $\mathcal{K}^\bullet \to \mathcal{F}^\bullet$ with $\mathcal{K}^\bullet$ a K-flat complex, see Lemma 20.26.11. Consider the commutative diagram

$\xymatrix{ \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{P}^\bullet ) \ar[r] \ar[d] & \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{Q}^\bullet ) \ar[d] \\ \text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{P}^\bullet ) \ar[r] & \text{Tot}(\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{Q}^\bullet ) }$

The result follows as by Lemma 20.26.3 the vertical arrows and the top horizontal arrow are quasi-isomorphisms. $\square$

Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}^\bullet$ be an object of $D(\mathcal{O}_ X)$. Choose a K-flat resolution $\mathcal{K}^\bullet \to \mathcal{F}^\bullet$, see Lemma 20.26.11. By Lemma 20.26.1 we obtain an exact functor of triangulated categories

$K(\mathcal{O}_ X) \longrightarrow K(\mathcal{O}_ X), \quad \mathcal{G}^\bullet \longmapsto \text{Tot}(\mathcal{G}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{K}^\bullet )$

By Lemma 20.26.3 this functor induces a functor $D(\mathcal{O}_ X) \to D(\mathcal{O}_ X)$ simply because $D(\mathcal{O}_ X)$ is the localization of $K(\mathcal{O}_ X)$ at quasi-isomorphisms. By Lemma 20.26.12 the resulting functor (up to isomorphism) does not depend on the choice of the K-flat resolution.

Definition 20.26.13. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}^\bullet$ be an object of $D(\mathcal{O}_ X)$. The derived tensor product

$- \otimes _{\mathcal{O}_ X}^{\mathbf{L}} \mathcal{F}^\bullet : D(\mathcal{O}_ X) \longrightarrow D(\mathcal{O}_ X)$

is the exact functor of triangulated categories described above.

It is clear from our explicit constructions that there is a canonical isomorphism

$\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X}^{\mathbf{L}} \mathcal{G}^\bullet \cong \mathcal{G}^\bullet \otimes _{\mathcal{O}_ X}^{\mathbf{L}} \mathcal{F}^\bullet$

for $\mathcal{G}^\bullet$ and $\mathcal{F}^\bullet$ in $D(\mathcal{O}_ X)$. Here we use sign rules as given in More on Algebra, Section 15.68. Hence when we write $\mathcal{F}^\bullet \otimes _{\mathcal{O}_ X}^{\mathbf{L}} \mathcal{G}^\bullet$ we will usually be agnostic about which variable we are using to define the derived tensor product with.

Definition 20.26.14. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_ X$-modules. The Tor's of $\mathcal{F}$ and $\mathcal{G}$ are define by the formula

$\text{Tor}_ p^{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) = H^{-p}(\mathcal{F} \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{G})$

with derived tensor product as defined above.

This definition implies that for every short exact sequence of $\mathcal{O}_ X$-modules $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ we have a long exact cohomology sequence

$\xymatrix{ \mathcal{F}_1 \otimes _{\mathcal{O}_ X} \mathcal{G} \ar[r] & \mathcal{F}_2 \otimes _{\mathcal{O}_ X} \mathcal{G} \ar[r] & \mathcal{F}_3 \otimes _{\mathcal{O}_ X} \mathcal{G} \ar[r] & 0 \\ \text{Tor}_1^{\mathcal{O}_ X}(\mathcal{F}_1, \mathcal{G}) \ar[r] & \text{Tor}_1^{\mathcal{O}_ X}(\mathcal{F}_2, \mathcal{G}) \ar[r] & \text{Tor}_1^{\mathcal{O}_ X}(\mathcal{F}_3, \mathcal{G}) \ar[ull] }$

for every $\mathcal{O}_ X$-module $\mathcal{G}$. This will be called the long exact sequence of $\text{Tor}$ associated to the situation.

Lemma 20.26.15. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. The following are equivalent

1. $\mathcal{F}$ is a flat $\mathcal{O}_ X$-module, and

2. $\text{Tor}_1^{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) = 0$ for every $\mathcal{O}_ X$-module $\mathcal{G}$.

Proof. If $\mathcal{F}$ is flat, then $\mathcal{F} \otimes _{\mathcal{O}_ X} -$ is an exact functor and the satellites vanish. Conversely assume (2) holds. Then if $\mathcal{G} \to \mathcal{H}$ is injective with cokernel $\mathcal{Q}$, the long exact sequence of $\text{Tor}$ shows that the kernel of $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G} \to \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{H}$ is a quotient of $\text{Tor}_1^{\mathcal{O}_ X}(\mathcal{F}, \mathcal{Q})$ which is zero by assumption. Hence $\mathcal{F}$ is flat. $\square$

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