The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

17.16 Flat modules

We can define flat modules exactly as in the case of modules over rings.

Definition 17.16.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. An $\mathcal{O}_ X$-module $\mathcal{F}$ is flat if the functor

\[ \textit{Mod}(\mathcal{O}_ X) \longrightarrow \textit{Mod}(\mathcal{O}_ X), \quad \mathcal{G} \mapsto \mathcal{G} \otimes _\mathcal {O} \mathcal{F} \]

is exact.

We can characterize flatness by looking at the stalks.

Lemma 17.16.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. An $\mathcal{O}_ X$-module $\mathcal{F}$ is flat if and only if the stalk $\mathcal{F}_ x$ is a flat $\mathcal{O}_{X, x}$-module for all $x \in X$.

Proof. Assume $\mathcal{F}_ x$ is a flat $\mathcal{O}_{X, x}$-module for all $x \in X$. In this case, if $\mathcal{G} \to \mathcal{H} \to \mathcal{K}$ is exact, then also $\mathcal{G} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{H} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{K} \otimes _{\mathcal{O}_ X} \mathcal{F}$ is exact because we can check exactness at stalks and because tensor product commutes with taking stalks, see Lemma 17.15.1. Conversely, suppose that $\mathcal{F}$ is flat, and let $x \in X$. Consider the skyscraper sheaves $i_{x, *} M$ where $M$ is a $\mathcal{O}_{X, x}$-module. Note that

\[ M \otimes _{\mathcal{O}_{X, x}} \mathcal{F}_ x = \left(i_{x, *} M \otimes _{\mathcal{O}_ X} \mathcal{F}\right)_ x \]

again by Lemma 17.15.1. Since $i_{x, *}$ is exact, we see that the fact that $\mathcal{F}$ is flat implies that $M \mapsto M \otimes _{\mathcal{O}_{X, x}} \mathcal{F}_ x$ is exact. Hence $\mathcal{F}_ x$ is a flat $\mathcal{O}_{X, x}$-module. $\square$

Thus the following definition makes sense.

Definition 17.16.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $x \in X$. An $\mathcal{O}_ X$-module $\mathcal{F}$ is flat at $x$ if $\mathcal{F}_ x$ is a flat $\mathcal{O}_{X, x}$-module.

Hence we see that $\mathcal{F}$ is a flat $\mathcal{O}_ X$-module if and only if it is flat at every point.

Lemma 17.16.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. A filtered colimit of flat $\mathcal{O}_ X$-modules is flat. A direct sum of flat $\mathcal{O}_ X$-modules is flat.

Proof. This follows from Lemma 17.15.6, Lemma 17.15.1, Algebra, Lemma 10.8.8, and the fact that we can check exactness at stalks. $\square$

Lemma 17.16.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $U \subset X$ be open. The sheaf $j_{U!}\mathcal{O}_ U$ is a flat sheaf of $\mathcal{O}_ X$-modules.

Proof. The stalks of $j_{U!}\mathcal{O}_ U$ are either zero or equal to $\mathcal{O}_{X, x}$. Apply Lemma 17.16.2. $\square$

Lemma 17.16.6. Let $(X, \mathcal{O}_ X)$ be a ringed space.

  1. Any sheaf of $\mathcal{O}_ X$-modules is a quotient of a direct sum $\bigoplus j_{U_ i!}\mathcal{O}_{U_ i}$.

  2. Any $\mathcal{O}_ X$-module is a quotient of a flat $\mathcal{O}_ X$-module.

Proof. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. For every open $U \subset X$ and every $s \in \mathcal{F}(U)$ we get a morphism $j_{U!}\mathcal{O}_ U \to \mathcal{F}$, namely the adjoint to the morphism $\mathcal{O}_ U \to \mathcal{F}|_ U$, $1 \mapsto s$. Clearly the map

\[ \bigoplus \nolimits _{(U, s)} j_{U!}\mathcal{O}_ U \longrightarrow \mathcal{F} \]

is surjective, and the source is flat by combining Lemmas 17.16.4 and 17.16.5. $\square$

Lemma 17.16.7. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let

\[ 0 \to \mathcal{F}'' \to \mathcal{F}' \to \mathcal{F} \to 0 \]

be a short exact sequence of $\mathcal{O}_ X$-modules. Assume $\mathcal{F}$ is flat. Then for any $\mathcal{O}_ X$-module $\mathcal{G}$ the sequence

\[ 0 \to \mathcal{F}'' \otimes _\mathcal {O} \mathcal{G} \to \mathcal{F}' \otimes _\mathcal {O} \mathcal{G} \to \mathcal{F} \otimes _\mathcal {O} \mathcal{G} \to 0 \]

is exact.

Proof. Using that $\mathcal{F}_ x$ is a flat $\mathcal{O}_{X, x}$-module for every $x \in X$ and that exactness can be checked on stalks, this follows from Algebra, Lemma 10.38.12. $\square$

slogan

Lemma 17.16.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let

\[ 0 \to \mathcal{F}_2 \to \mathcal{F}_1 \to \mathcal{F}_0 \to 0 \]

be a short exact sequence of $\mathcal{O}_ X$-modules.

  1. If $\mathcal{F}_2$ and $\mathcal{F}_0$ are flat so is $\mathcal{F}_1$.

  2. If $\mathcal{F}_1$ and $\mathcal{F}_0$ are flat so is $\mathcal{F}_2$.

Proof. Since exactness and flatness may be checked at the level of stalks this follows from Algebra, Lemma 10.38.13. $\square$

Lemma 17.16.9. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let

\[ \ldots \to \mathcal{F}_2 \to \mathcal{F}_1 \to \mathcal{F}_0 \to \mathcal{Q} \to 0 \]

be an exact complex of $\mathcal{O}_ X$-modules. If $\mathcal{Q}$ and all $\mathcal{F}_ i$ are flat $\mathcal{O}_ X$-modules, then for any $\mathcal{O}_ X$-module $\mathcal{G}$ the complex

\[ \ldots \to \mathcal{F}_2 \otimes _{\mathcal{O}_ X} \mathcal{G} \to \mathcal{F}_1 \otimes _{\mathcal{O}_ X} \mathcal{G} \to \mathcal{F}_0 \otimes _{\mathcal{O}_ X} \mathcal{G} \to \mathcal{Q} \otimes _{\mathcal{O}_ X} \mathcal{G} \to 0 \]

is exact also.

Proof. Follows from Lemma 17.16.7 by splitting the complex into short exact sequences and using Lemma 17.16.8 to prove inductively that $\mathop{\mathrm{Im}}(\mathcal{F}_{i + 1} \to \mathcal{F}_ i)$ is flat. $\square$

The following lemma gives one direction of the equational criterion of flatness (Algebra, Lemma 10.38.11).

Lemma 17.16.10. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a flat $\mathcal{O}_ X$-module. Let $U \subset X$ be open and let

\[ \mathcal{O}_ U \xrightarrow {(f_1, \ldots , f_ n)} \mathcal{O}_ U^{\oplus n} \xrightarrow {(s_1, \ldots , s_ n)} \mathcal{F}|_ U \]

be a complex of $\mathcal{O}_ U$-modules. For every $x \in U$ there exists an open neighbourhood $V \subset U$ of $x$ and a factorization

\[ \mathcal{O}_ V^{\oplus n} \xrightarrow {A} \mathcal{O}_ V^{\oplus m} \xrightarrow {(t_1, \ldots , t_ m)} \mathcal{F}|_ V \]

of $(s_1, \ldots , s_ n)|_ V$ such that $A \circ (f_1, \ldots , f_ n)|_ V = 0$.

Proof. Let $\mathcal{I} \subset \mathcal{O}_ U$ be the sheaf of ideals generated by $f_1, \ldots , f_ n$. Then $\sum f_ i \otimes s_ i$ is a section of $\mathcal{I} \otimes _{\mathcal{O}_ U} \mathcal{F}|_ U$ which maps to zero in $\mathcal{F}|_ U$. As $\mathcal{F}|_ U$ is flat the map $\mathcal{I} \otimes _{\mathcal{O}_ U} \mathcal{F}|_ U \to \mathcal{F}|_ U$ is injective. Since $\mathcal{I} \otimes _{\mathcal{O}_ U} \mathcal{F}|_ U$ is the sheaf associated to the presheaf tensor product, we see there exists an open neighbourhood $V \subset U$ of $x$ such that $\sum f_ i|_ V \otimes s_ i|_ V$ is zero in $\mathcal{I}(V) \otimes _{\mathcal{O}(V)} \mathcal{F}(V)$. Unwinding the definitions using Algebra, Lemma 10.106.10 we find $t_1, \ldots , t_ m \in \mathcal{F}(V)$ and $a_{ij} \in \mathcal{O}(V)$ such that $\sum a_{ij}f_ i|_ V = 0$ and $s_ i|_ V = \sum a_{ij}t_ j$. $\square$

Lemma 17.16.11. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be locally of finite presentation and flat. Then $\mathcal{F}$ is locally a direct summand of a finite free $\mathcal{O}_ X$-module.

Proof. After replacing $X$ by the members of an open covering, we may assume there exists a presentation

\[ \mathcal{O}_ X^{\oplus r} \to \mathcal{O}_ X^{\oplus n} \to \mathcal{F} \to 0 \]

Let $x \in X$. By Lemma 17.16.10 we can, after shrinking $X$ to an open neighbourhood of $x$, assume there exists a factorization

\[ \mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_1} \to \mathcal{F} \]

such that the composition $\mathcal{O}_ X^{\oplus r} \to \mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_1}$ annihilates the first summand of $\mathcal{O}_ X^{\oplus r}$. Repeating this argument $r - 1$ more times we obtain a factorization

\[ \mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_ r} \to \mathcal{F} \]

such that the composition $\mathcal{O}_ X^{\oplus r} \to \mathcal{O}_ X^{\oplus n} \to \mathcal{O}_ X^{\oplus n_ r}$ is zero. This means that the surjection $\mathcal{O}_ X^{\oplus n_ r} \to \mathcal{F}$ has a section and we win. $\square$


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