# The Stacks Project

## Tag 06Y7

### 20.27. Flat resolutions

A reference for the material in this section is [Spaltenstein]. 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 13.16.5 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 20.27.1. 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 15.49.1 and 15.49.2. $\square$

Definition 20.27.2. 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 20.27.3. 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 20.27.1 and the fact that quasi-isomorphisms are characterized by having acyclic cones. $\square$

Lemma 20.27.4. 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 $\mathcal{O}_{X, x}$ is K-flat (More on Algebra, Definition 15.49.3).

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 20.27.5. 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 20.27.6. 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 20.27.1 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 20.27.7. 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 20.27.4. Hence this follows from Sheaves, Lemma 6.26.4 and More on Algebra, Lemma 15.49.5. $\square$

Lemma 20.27.8. 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 20.27.4. Thus this lemma follows from Modules, Lemma 17.16.2 and More on Algebra, Lemma 15.49.8. $\square$

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

Lemma 20.27.9. 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 $\mathop{\rm colim}\nolimits_i \mathcal{K}_i^\bullet$ is K-flat.

Proof. Because we are taking termwise colimits it is clear that $$\mathop{\rm colim}\nolimits_i \text{Tot}( \mathcal{F}^\bullet \otimes_{\mathcal{O}_X} \mathcal{K}_i^\bullet) = \text{Tot}(\mathcal{F}^\bullet \otimes_{\mathcal{O}_X} \mathop{\rm colim}\nolimits_i \mathcal{K}_i^\bullet)$$ Hence the lemma follows from the fact that filtered colimits are exact. $\square$

Lemma 20.27.10. 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 $\mathop{\rm colim}\nolimits \mathcal{K}_n^\bullet \to \mathcal{G}^\bullet$ 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 13.28.1. The induced map $\mathop{\rm colim}\nolimits \mathcal{K}_n^\bullet \to \mathcal{G}^\bullet$ is a quasi-isomorphism because filtered colimits are exact. $\square$

Lemma 20.27.11. 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 $\mathcal{K}^\bullet \to \mathcal{G}^\bullet$.

Proof. Choose a diagram as in Lemma 20.27.10. 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 20.27.8. The induced map $\mathop{\rm colim}\nolimits \mathcal{K}_n^\bullet \to \mathcal{G}^\bullet$ is a quasi-isomorphism by construction. Since $\mathop{\rm colim}\nolimits \mathcal{K}_n^\bullet$ is K-flat by Lemma 20.27.9 we win. $\square$

Lemma 20.27.12. 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 20.27.11. 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 20.27.3 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 20.27.11. By Lemma 20.27.1 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 20.27.3 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 20.27.12 the resulting functor (up to isomorphism) does not depend on the choice of the K-flat resolution.

Definition 20.27.13. 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 $D(\mathcal{O}_X)$. 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 20.27.14. 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 20.27.15. 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. $\square$

The code snippet corresponding to this tag is a part of the file cohomology.tex and is located in lines 5390–5791 (see updates for more information).

\section{Flat resolutions}
\label{section-flat}

\noindent
A reference for the material in this section is \cite{Spaltenstein}.
Let $(X, \mathcal{O}_X)$ be a ringed space. By
Modules, Lemma \ref{modules-lemma-module-quotient-flat}
any $\mathcal{O}_X$-module is a quotient of a flat $\mathcal{O}_X$-module.
By
Derived Categories, Lemma \ref{derived-lemma-subcategory-left-resolution}
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.

\begin{lemma}
\label{lemma-derived-tor-exact}
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.
\end{lemma}

\begin{proof}
Omitted. Hint: See
More on Algebra, Lemmas \ref{more-algebra-lemma-derived-tor-homotopy} and
\ref{more-algebra-lemma-derived-tor-exact}.
\end{proof}

\begin{definition}
\label{definition-K-flat}
Let $(X, \mathcal{O}_X)$ be a ringed space.
A complex $\mathcal{K}^\bullet$ of $\mathcal{O}_X$-modules is
called {\it 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.
\end{definition}

\begin{lemma}
\label{lemma-K-flat-quasi-isomorphism}
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.
\end{lemma}

\begin{proof}
Follows from
Lemma \ref{lemma-derived-tor-exact}
and the fact that quasi-isomorphisms are characterized by having
acyclic cones.
\end{proof}

\begin{lemma}
\label{lemma-check-K-flat-stalks}
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 $\mathcal{O}_{X, x}$ is K-flat
(More on Algebra, Definition \ref{more-algebra-definition-K-flat}).
\end{lemma}

\begin{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 \ref{modules-lemma-abelian}.
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.
\end{proof}

\begin{lemma}
\label{lemma-tensor-product-K-flat}
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.
\end{lemma}

\begin{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.
\end{proof}

\begin{lemma}
\label{lemma-K-flat-two-out-of-three}
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.
\end{lemma}

\begin{proof}
Follows from
Lemma \ref{lemma-derived-tor-exact}
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.
\end{proof}

\begin{lemma}
\label{lemma-pullback-K-flat}
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.
\end{lemma}

\begin{proof}
We can check this on stalks, see
Lemma \ref{lemma-check-K-flat-stalks}.
Hence this follows from
Sheaves, Lemma \ref{sheaves-lemma-stalk-pullback-modules}
and
More on Algebra, Lemma \ref{more-algebra-lemma-base-change-K-flat}.
\end{proof}

\begin{lemma}
\label{lemma-bounded-flat-K-flat}
Let $(X, \mathcal{O}_X)$ be a ringed space. A bounded above complex
of flat $\mathcal{O}_X$-modules is K-flat.
\end{lemma}

\begin{proof}
We can check this on stalks, see
Lemma \ref{lemma-check-K-flat-stalks}.
Thus this lemma follows from
Modules, Lemma \ref{modules-lemma-flat-stalks-flat}
and
More on Algebra, Lemma \ref{more-algebra-lemma-derived-tor-quasi-isomorphism}.
\end{proof}

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

\begin{lemma}
\label{lemma-colimit-K-flat}
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 $\colim_i \mathcal{K}_i^\bullet$ is K-flat.
\end{lemma}

\begin{proof}
Because we are taking termwise colimits it is clear that
$$\colim_i \text{Tot}( \mathcal{F}^\bullet \otimes_{\mathcal{O}_X} \mathcal{K}_i^\bullet) = \text{Tot}(\mathcal{F}^\bullet \otimes_{\mathcal{O}_X} \colim_i \mathcal{K}_i^\bullet)$$
Hence the lemma follows from the fact that filtered colimits are
exact.
\end{proof}

\begin{lemma}
\label{lemma-resolution-by-direct-sums-extensions-by-zero}
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
$\colim \mathcal{K}_n^\bullet \to \mathcal{G}^\bullet$ is a quasi-isomorphism.
\end{lemma}

\begin{proof}
The existence of the diagram and properties (1), (2), (3) follows immediately
from
Modules, Lemma \ref{modules-lemma-module-quotient-flat}
and
Derived Categories, Lemma \ref{derived-lemma-special-direct-system}.
The induced map
$\colim \mathcal{K}_n^\bullet \to \mathcal{G}^\bullet$
is a quasi-isomorphism because filtered colimits are exact.
\end{proof}

\begin{lemma}
\label{lemma-K-flat-resolution}
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
$\mathcal{K}^\bullet \to \mathcal{G}^\bullet$.
\end{lemma}

\begin{proof}
Choose a diagram as in
Lemma \ref{lemma-resolution-by-direct-sums-extensions-by-zero}.
Each complex $\mathcal{K}_n^\bullet$ is a bounded
above complex of flat modules, see
Modules, Lemma \ref{modules-lemma-j-shriek-flat}.
Hence $\mathcal{K}_n^\bullet$ is K-flat by
Lemma \ref{lemma-bounded-flat-K-flat}.
The induced map
$\colim \mathcal{K}_n^\bullet \to \mathcal{G}^\bullet$
is a quasi-isomorphism by construction. Since
$\colim \mathcal{K}_n^\bullet$ is K-flat by
Lemma \ref{lemma-colimit-K-flat}
we win.
\end{proof}

\begin{lemma}
\label{lemma-derived-tor-quasi-isomorphism-other-side}
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.
\end{lemma}

\begin{proof}
Choose a quasi-isomorphism $\mathcal{K}^\bullet \to \mathcal{F}^\bullet$
with $\mathcal{K}^\bullet$ a K-flat complex, see
Lemma \ref{lemma-K-flat-resolution}.
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 \ref{lemma-K-flat-quasi-isomorphism}
the vertical arrows and the top horizontal arrow are quasi-isomorphisms.
\end{proof}

\noindent
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 \ref{lemma-K-flat-resolution}.
By
Lemma \ref{lemma-derived-tor-exact}
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 \ref{lemma-K-flat-quasi-isomorphism}
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 \ref{lemma-derived-tor-quasi-isomorphism-other-side}
the resulting functor (up to isomorphism)
does not depend on the choice of the K-flat resolution.

\begin{definition}
\label{definition-derived-tor}
Let $(X, \mathcal{O}_X)$ be a ringed space.
Let $\mathcal{F}^\bullet$ be an object of $D(\mathcal{O}_X)$.
The {\it 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.
\end{definition}

\noindent
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 $D(\mathcal{O}_X)$.
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.

\begin{definition}
\label{definition-tor}
Let $(X, \mathcal{O}_X)$ be a ringed space.
Let $\mathcal{F}$, $\mathcal{G}$ be $\mathcal{O}_X$-modules.
The {\it 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.
\end{definition}

\noindent
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.

\begin{lemma}
\label{lemma-flat-tor-zero}
Let $(X, \mathcal{O}_X)$ be a ringed space.
Let $\mathcal{F}$ be an $\mathcal{O}_X$-module.
The following are equivalent
\begin{enumerate}
\item $\mathcal{F}$ is a flat $\mathcal{O}_X$-module, and
\item $\text{Tor}_1^{\mathcal{O}_X}(\mathcal{F}, \mathcal{G}) = 0$
for every $\mathcal{O}_X$-module $\mathcal{G}$.
\end{enumerate}
\end{lemma}

\begin{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.
\end{proof}

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