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 functors
and
are exact functors of triangulated categories.
A reference for the material in this section is [Spaltenstein]. Let $(X, \mathcal{O}_ X)$ be a ringed space. By Modules, Lemma 17.17.6 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 functors
and
are exact functors of triangulated categories.
Proof. This follows from Derived Categories, Remark 13.10.9. $\square$
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
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
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 15.59.1).
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 $x \in X$ and let $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
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 $(X, \mathcal{O}_ X)$ be a ringed space. Let $0 \to \mathcal{K}_1^\bullet \to \mathcal{K}_2^\bullet \to \mathcal{K}_3^\bullet \to 0$ be a short exact sequence of complexes such that the terms of $\mathcal{K}_3^\bullet $ are flat $\mathcal{O}_ X$-modules. If two out of three of $\mathcal{K}_ i^\bullet $ are K-flat, so is the third.
Proof. By Modules, Lemma 17.17.7 for every complex $\mathcal{L}^\bullet $ we obtain a short exact sequence
of complexes. Hence the lemma follows from the long exact sequence of cohomology sheaves and the definition of K-flat complexes. $\square$
Lemma 20.26.8. 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 15.59.3. $\square$
Lemma 20.26.9. 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 17.17.2 and More on Algebra, Lemma 15.59.7. $\square$
In the following lemma by a colimit of a system of complexes we mean the termwise colimit.
Lemma 20.26.10. 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
Hence the lemma follows from the fact that filtered colimits are exact. $\square$
Lemma 20.26.11. 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
with the following properties: (1) the vertical arrows are quasi-isomorphisms and termwise surjective, (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 17.17.6 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 20.26.12. Let $(X, \mathcal{O}_ X)$ be a ringed space. For any complex $\mathcal{G}^\bullet $ there exists a $K$-flat complex $\mathcal{K}^\bullet $ whose terms are flat $\mathcal{O}_ X$-modules and a quasi-isomorphism $\mathcal{K}^\bullet \to \mathcal{G}^\bullet $ which is termwise surjective.
Proof. Choose a diagram as in Lemma 20.26.11. Each complex $\mathcal{K}_ n^\bullet $ is a bounded above complex of flat modules, see Modules, Lemma 17.17.5. Hence $\mathcal{K}_ n^\bullet $ is K-flat by Lemma 20.26.9. Thus $\mathop{\mathrm{colim}}\nolimits \mathcal{K}_ n^\bullet $ is K-flat by Lemma 20.26.10. The induced map $\mathop{\mathrm{colim}}\nolimits \mathcal{K}_ n^\bullet \to \mathcal{G}^\bullet $ is a quasi-isomorphism and termwise surjective by construction. Property (3) of Lemma 20.26.11 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 20.26.13. 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
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 20.26.12. Consider the commutative diagram
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 20.26.12. By Lemma 20.26.1 we obtain an exact functor of triangulated categories
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 20.26.13 the resulting functor (up to isomorphism) does not depend on the choice of the K-flat resolution.
Definition 20.26.14. 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
is the exact functor of triangulated categories described above.
It is clear from our explicit constructions that there is a canonical isomorphism
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 15.72. 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 20.26.15. 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
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
for every $\mathcal{O}_ X$-module $\mathcal{G}$. This will be called the long exact sequence of $\text{Tor}$ associated to the situation.
Lemma 20.26.16.slogan Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. The following are equivalent
$\mathcal{F}$ is a flat $\mathcal{O}_ X$-module, and
$\text{Tor}_1^{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G}) = 0$ 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. $\square$
Lemma 20.26.17. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $a : \mathcal{K}^\bullet \to \mathcal{L}^\bullet $ be a map of complexes of $\mathcal{O}_ X$-modules. If $\mathcal{K}^\bullet $ is K-flat, then there exist a complex $\mathcal{N}^\bullet $ and maps of complexes $b : \mathcal{K}^\bullet \to \mathcal{N}^\bullet $ and $c : \mathcal{N}^\bullet \to \mathcal{L}^\bullet $ such that
$\mathcal{N}^\bullet $ is K-flat,
$c$ is a quasi-isomorphism,
$a$ is homotopic to $c \circ b$.
If the terms of $\mathcal{K}^\bullet $ are flat, then we may choose $\mathcal{N}^\bullet $, $b$, and $c$ such that the same is true for $\mathcal{N}^\bullet $.
Proof. We will use that the homotopy category $K(\textit{Mod}(\mathcal{O}_ X))$ is a triangulated category, see Derived Categories, Proposition 13.10.3. Choose a distinguished triangle $\mathcal{K}^\bullet \to \mathcal{L}^\bullet \to \mathcal{C}^\bullet \to \mathcal{K}^\bullet [1]$. Choose a quasi-isomorphism $\mathcal{M}^\bullet \to \mathcal{C}^\bullet $ with $\mathcal{M}^\bullet $ K-flat with flat terms, see Lemma 20.26.12. By the axioms of triangulated categories, we may fit the composition $\mathcal{M}^\bullet \to \mathcal{C}^\bullet \to \mathcal{K}^\bullet [1]$ into a distinguished triangle $\mathcal{K}^\bullet \to \mathcal{N}^\bullet \to \mathcal{M}^\bullet \to \mathcal{K}^\bullet [1]$. By Lemma 20.26.6 we see that $\mathcal{N}^\bullet $ is K-flat. Again using the axioms of triangulated categories, we can choose a map $\mathcal{N}^\bullet \to \mathcal{L}^\bullet $ fitting into the following morphism of distinghuised triangles
Since two out of three of the arrows are quasi-isomorphisms, so is the third arrow $\mathcal{N}^\bullet \to \mathcal{L}^\bullet $ by the long exact sequences of cohomology associated to these distinguished triangles (or you can look at the image of this diagram in $D(\mathcal{O}_ X)$ and use Derived Categories, Lemma 13.4.3 if you like). This finishes the proof of (1), (2), and (3). To prove the final assertion, we may choose $\mathcal{N}^\bullet $ such that $\mathcal{N}^ n \cong \mathcal{M}^ n \oplus \mathcal{K}^ n$, see Derived Categories, Lemma 13.10.7. Hence we get the desired flatness if the terms of $\mathcal{K}^\bullet $ are flat. $\square$
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