## 18.28 Flat modules

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

Definition 18.28.1. Let $\mathcal{C}$ be a category. Let $\mathcal{O}$ be a presheaf of rings.

1. A presheaf $\mathcal{F}$ of $\mathcal{O}$-modules is called flat if the functor

$\textit{PMod}(\mathcal{O}) \longrightarrow \textit{PMod}(\mathcal{O}), \quad \mathcal{G} \mapsto \mathcal{G} \otimes _{p, \mathcal{O}} \mathcal{F}$

is exact.

2. A map $\mathcal{O} \to \mathcal{O}'$ of presheaves of rings is called flat if $\mathcal{O}'$ is flat as a presheaf of $\mathcal{O}$-modules.

3. If $\mathcal{C}$ is a site, $\mathcal{O}$ is a sheaf of rings and $\mathcal{F}$ is a sheaf of $\mathcal{O}$-modules, then we say $\mathcal{F}$ is flat if the functor

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

is exact.

4. A map $\mathcal{O} \to \mathcal{O}'$ of sheaves of rings on a site is called flat if $\mathcal{O}'$ is flat as a sheaf of $\mathcal{O}$-modules.

The notion of a flat module or flat ring map is intrinsic (Section 18.18).

Lemma 18.28.2. Let $\mathcal{C}$ be a category. Let $\mathcal{O}$ be a presheaf of rings. Let $\mathcal{F}$ be a presheaf of $\mathcal{O}$-modules. If each $\mathcal{F}(U)$ is a flat $\mathcal{O}(U)$-module, then $\mathcal{F}$ is flat.

Proof. This is immediate from the definitions. $\square$

Lemma 18.28.3. Let $\mathcal{C}$ be a site. Let $\mathcal{O}$ be a presheaf of rings. Let $\mathcal{F}$ be a presheaf of $\mathcal{O}$-modules. If $\mathcal{F}$ is a flat $\mathcal{O}$-module, then $\mathcal{F}^\#$ is a flat $\mathcal{O}^\#$-module.

Proof. Omitted. (Hint: Sheafification is exact.) $\square$

Lemma 18.28.4. Let $\mathcal{C}$ be a site. Let $\mathcal{O}$ be a presheaf of rings. Let $\mathcal{F}$ be a presheaf of $\mathcal{O}$-modules. Assume that every object $U$ of $\mathcal{C}$ has a covering $\{ U_ i \to U\} _{i \in I}$ such that $\mathcal{F}(U_ i)$ is a flat $\mathcal{O}(U_ i)$-module. Then $\mathcal{F}^\#$ is a flat $\mathcal{O}^\#$-module.

Proof. Let $\mathcal{G} \subset \mathcal{G}'$ be an inclusion of $\mathcal{O}^\#$-modules. We have to show that

$\mathcal{G} \otimes _{\mathcal{O}^\# } \mathcal{F}^\# \longrightarrow \mathcal{G}' \otimes _{\mathcal{O}^\# } \mathcal{F}^\#$

is injective. By Lemma 18.26.1 the source of this arrow is the sheafification of the presheaf $\mathcal{G} \otimes _{p, \mathcal{O}} \mathcal{F}$ and similarly for the target. If $U$ is an object of $\mathcal{C}$ such that $\mathcal{F}(U)$ is a flat $\mathcal{O}(U)$-module, then

$(\mathcal{G} \otimes _{p, \mathcal{O}} \mathcal{F})(U) = \mathcal{G}(U) \otimes _{\mathcal{O}(U)} \mathcal{F}(U) \longrightarrow \mathcal{G}'(U) \otimes _{\mathcal{O}(U)} \mathcal{F}(U) = (\mathcal{G}' \otimes _{p, \mathcal{O}} \mathcal{F})(U)$

is injective. Hence we reduce to showing: given a map of presheaves $f : \mathcal{H} \to \mathcal{H}'$ on $\mathcal{C}$ such that every $U$ in $\mathcal{C}$ has a covering $\{ U_ i \to U\} _{i \in I}$ with $\mathcal{H}(U_ i) \to \mathcal{H}'(U_ i)$ injective, then $f^\#$ is injective. This we leave to the reader as an exercise. $\square$

Lemma 18.28.5. Colimits and tensor product.

1. A filtered colimit of flat presheaves of modules is flat. A direct sum of flat presheaves of modules is flat.

2. A filtered colimit of flat sheaves of modules is flat. A direct sum of flat sheaves of modules is flat.

Proof. Part (1) follows from Lemma 18.27.7 and Algebra, Lemma 10.8.8 by looking at sections over objects. To see part (2), use Lemma 18.27.7 and the fact that a filtered colimit of exact complexes is an exact complex (this uses that sheafification is exact and commutes with colimits). Some details omitted. $\square$

Lemma 18.28.6. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $U$ be an object of $\mathcal{C}$. If $\mathcal{F}$ is a flat $\mathcal{O}$-module, then $\mathcal{F}|_ U$ is a flat $\mathcal{O}_ U$-module.

Proof. Let $\mathcal{G}_1 \to \mathcal{G}_2 \to \mathcal{G}_3$ be an exact complex of $\mathcal{O}_ U$-modules. Since $j_{U!}$ is exact (Lemma 18.19.3) and $\mathcal{F}$ is flat as an $\mathcal{O}$-modules then we see that the complex made up of the modules

$j_{U!}(\mathcal{G}_ i \otimes _{\mathcal{O}_ U} \mathcal{F}|_ U) = j_{U!}\mathcal{G}_ i \otimes _\mathcal {O} \mathcal{F}$

(Lemma 18.27.9) is exact. We conclude that $\mathcal{G}_1 \otimes _{\mathcal{O}_ U} \mathcal{F}|_ U \to \mathcal{G}_2 \otimes _{\mathcal{O}_ U} \mathcal{F}|_ U \to \mathcal{G}_3 \otimes _{\mathcal{O}_ U} \mathcal{F}|_ U$ is exact by Lemma 18.19.4. $\square$

Lemma 18.28.7. Let $\mathcal{C}$ be a category. Let $\mathcal{O}$ be a presheaf of rings. Let $U$ be an object of $\mathcal{C}$. Consider the functor $j_ U : \mathcal{C}/U \to \mathcal{C}$.

1. The presheaf of $\mathcal{O}$-modules $j_{U!}\mathcal{O}_ U$ (see Remark 18.19.7) is flat.

2. If $\mathcal{C}$ is a site, $\mathcal{O}$ is a sheaf of rings, $j_{U!}\mathcal{O}_ U$ is a flat sheaf of $\mathcal{O}$-modules.

Proof. Proof of (1). By the discussion in Remark 18.19.7 we see that

$j_{U!}\mathcal{O}_ U(V) = \bigoplus \nolimits _{\varphi \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(V, U)} \mathcal{O}(V)$

which is a flat $\mathcal{O}(V)$-module. Hence (1) follows from Lemma 18.28.2. Then (2) follows as $j_{U!}\mathcal{O}_ U = (j_{U!}\mathcal{O}_ U)^\#$ (the first $j_{U!}$ on sheaves, the second on presheaves) and Lemma 18.28.3. $\square$

Lemma 18.28.8. Let $\mathcal{C}$ be a category. Let $\mathcal{O}$ be a presheaf of rings.

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

2. Any presheaf of $\mathcal{O}$-modules is a quotient of a flat presheaf of $\mathcal{O}$-modules.

3. If $\mathcal{C}$ is a site, $\mathcal{O}$ is a sheaf of rings, then any sheaf of $\mathcal{O}$-modules is a quotient of a direct sum $\bigoplus j_{U_ i!}\mathcal{O}_{U_ i}$.

4. If $\mathcal{C}$ is a site, $\mathcal{O}$ is a sheaf of rings, then any sheaf of $\mathcal{O}$-modules is a quotient of a flat sheaf of $\mathcal{O}$-modules.

Proof. Proof of (1). For every object $U$ of $\mathcal{C}$ 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. The source is flat by combining Lemmas 18.28.5 and 18.28.7 which proves (2). The sheaf case follows from this either by sheafifying or repeating the same argument. $\square$

Lemma 18.28.9. Let $\mathcal{C}$ be a category. Let $\mathcal{O}$ be a presheaf of rings. Let

$0 \to \mathcal{F}'' \to \mathcal{F}' \to \mathcal{F} \to 0$

be a short exact sequence of presheaves of $\mathcal{O}$-modules. Let $\mathcal{G}$ be a presheaf of $\mathcal{O}$-modules.

1. If $\mathcal{F}$ is a flat presheaf of modules, then the sequence

$0 \to \mathcal{F}'' \otimes _{p, \mathcal{O}} \mathcal{G} \to \mathcal{F}' \otimes _{p, \mathcal{O}} \mathcal{G} \to \mathcal{F} \otimes _{p, \mathcal{O}} \mathcal{G} \to 0$

is exact.

2. If $\mathcal{C}$ is a site, $\mathcal{O}$, $\mathcal{F}$, $\mathcal{F}'$, $\mathcal{F}''$, and $\mathcal{G}$ are sheaves, and $\mathcal{F}$ is flat as a sheaf of modules, then 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. Choose a flat presheaf of $\mathcal{O}$-modules $\mathcal{G}'$ which surjects onto $\mathcal{G}$. This is possible by Lemma 18.28.8. Let $\mathcal{G}'' = \mathop{\mathrm{Ker}}(\mathcal{G}' \to \mathcal{G})$. The lemma follows by applying the snake lemma to the following diagram

$\begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & \mathcal{F}'' \otimes _{p, \mathcal{O}} \mathcal{G} & \to & \mathcal{F}' \otimes _{p, \mathcal{O}} \mathcal{G} & \to & \mathcal{F} \otimes _{p, \mathcal{O}} \mathcal{G} & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ 0 & \to & \mathcal{F}'' \otimes _{p, \mathcal{O}} \mathcal{G}' & \to & \mathcal{F}' \otimes _{p, \mathcal{O}} \mathcal{G}' & \to & \mathcal{F} \otimes _{p, \mathcal{O}} \mathcal{G}' & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & \mathcal{F}'' \otimes _{p, \mathcal{O}} \mathcal{G}'' & \to & \mathcal{F}' \otimes _{p, \mathcal{O}} \mathcal{G}'' & \to & \mathcal{F} \otimes _{p, \mathcal{O}} \mathcal{G}'' & \to & 0 \\ & & & & & & \uparrow & & \\ & & & & & & 0 & & \end{matrix}$

with exact rows and columns. The middle row is exact because tensoring with the flat module $\mathcal{G}'$ is exact. The proof in the case of sheaves is exactly the same. $\square$

Lemma 18.28.10. Let $\mathcal{C}$ be a category. Let $\mathcal{O}$ be a presheaf of rings. Let

$0 \to \mathcal{F}_2 \to \mathcal{F}_1 \to \mathcal{F}_0 \to 0$

be a short exact sequence of presheaves of $\mathcal{O}$-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 $\mathcal{F}_2$.

If $\mathcal{C}$ is a site and $\mathcal{O}$ is a sheaf of rings then the same result holds in $\textit{Mod}(\mathcal{O})$.

Proof. Let $\mathcal{G}^\bullet$ be an arbitrary exact complex of presheaves of $\mathcal{O}$-modules. Assume that $\mathcal{F}_0$ is flat. By Lemma 18.28.9 we see that

$0 \to \mathcal{G}^\bullet \otimes _{p, \mathcal{O}} \mathcal{F}_2 \to \mathcal{G}^\bullet \otimes _{p, \mathcal{O}} \mathcal{F}_1 \to \mathcal{G}^\bullet \otimes _{p, \mathcal{O}} \mathcal{F}_0 \to 0$

is a short exact sequence of complexes of presheaves of $\mathcal{O}$-modules. Hence (1) and (2) follow from the snake lemma. The case of sheaves of modules is proved in the same way. $\square$

Lemma 18.28.11. Let $\mathcal{C}$ be a category. Let $\mathcal{O}$ be a presheaf of rings. Let

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

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

$\ldots \to \mathcal{F}_2 \otimes _{p, \mathcal{O}} \mathcal{G} \to \mathcal{F}_1 \otimes _{p, \mathcal{O}} \mathcal{G} \to \mathcal{F}_0 \otimes _{p, \mathcal{O}} \mathcal{G} \to \mathcal{Q} \otimes _{p, \mathcal{O}} \mathcal{G} \to 0$

is exact also. If $\mathcal{C}$ is a site and $\mathcal{O}$ is a sheaf of rings then the same result holds $\textit{Mod}(\mathcal{O})$.

Proof. Follows from Lemma 18.28.9 by splitting the complex into short exact sequences and using Lemma 18.28.10 to prove inductively that $\mathop{\mathrm{Im}}(\mathcal{F}_{i + 1} \to \mathcal{F}_ i)$ is flat. $\square$

Lemma 18.28.12. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. If $\mathcal{G}$ and $\mathcal{F}$ are flat $\mathcal{O}$-modules, then $\mathcal{G} \otimes _\mathcal {O} \mathcal{F}$ is a flat $\mathcal{O}$-module.

Proof. This is true because

$(\mathcal{G} \otimes _\mathcal {O} \mathcal{F}) \otimes _\mathcal {O} \mathcal{H} = \mathcal{G} \otimes _\mathcal {O} (\mathcal{F} \otimes _\mathcal {O} \mathcal{H})$

and a composition of exact functors is exact. $\square$

Lemma 18.28.13. Let $\mathcal{O}_1 \to \mathcal{O}_2$ be a map of sheaves of rings on a site $\mathcal{C}$. If $\mathcal{G}$ is a flat $\mathcal{O}_1$-module, then $\mathcal{G} \otimes _{\mathcal{O}_1} \mathcal{O}_2$ is a flat $\mathcal{O}_2$-module.

Proof. This is true because

$(\mathcal{G} \otimes _{\mathcal{O}_1} \mathcal{O}_2) \otimes _{\mathcal{O}_2} \mathcal{H} = \mathcal{G} \otimes _{\mathcal{O}_1} \mathcal{F}$

(as sheaves of abelian groups for example). $\square$

The following lemma is the analogue of the equational criterion of flatness (Algebra, Lemma 10.39.11).

Lemma 18.28.14. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{F}$ be an $\mathcal{O}$-module. The following are equivalent

1. $\mathcal{F}$ is a flat $\mathcal{O}$-module.

2. Let $U$ be an object of $\mathcal{C}$ 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. Then there exists a covering $\{ U_ i \to U\}$ and for each $i$ a factorization

$\mathcal{O}_{U_ i}^{\oplus n} \xrightarrow {B_ i} \mathcal{O}_{U_ i}^{\oplus l_ i} \xrightarrow {(t_{i1}, \ldots , t_{il_ i})} \mathcal{F}|_{U_ i}$

of $(s_1, \ldots , s_ n)|_{U_ i}$ such that $B_ i \circ (f_1, \ldots , f_ n)|_{U_ i} = 0$.

3. Let $U$ be an object of $\mathcal{C}$ and let

$\mathcal{O}_ U^{\oplus m} \xrightarrow {A} \mathcal{O}_ U^{\oplus n} \xrightarrow {(s_1, \ldots , s_ n)} \mathcal{F}|_ U$

be a complex of $\mathcal{O}_ U$-modules. Then there exists a covering $\{ U_ i \to U\}$ and for each $i$ a factorization

$\mathcal{O}_{U_ i}^{\oplus n} \xrightarrow {B_ i} \mathcal{O}_{U_ i}^{\oplus l_ i} \xrightarrow {(t_{i1}, \ldots , t_{il_ i})} \mathcal{F}|_{U_ i}$

of $(s_1, \ldots , s_ n)|_{U_ i}$ such that $B_ i \circ A|_{U_ i} = 0$.

Proof. Assume (1). Let $\mathcal{I} \subset \mathcal{O}_ U$ be the sheaf of ideals generated by $f_1, \ldots , f_ n$. Then $\sum f_ j \otimes s_ j$ 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 (Lemma 18.28.6) 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 a covering $\{ U_ i \to U\}$ such that $\sum f_ j|_{U_ i} \otimes s_ j|_{U_ i}$ is zero in $\mathcal{I}(U_ i) \otimes _{\mathcal{O}(U_ i)} \mathcal{F}(U_ i)$. Unwinding the definitions using Algebra, Lemma 10.107.10 we find $t_{i1}, \ldots , t_{i l_ i} \in \mathcal{F}(U_ i)$ and $a_{ijk} \in \mathcal{O}(U_ i)$ such that $\sum _ j a_{ijk}f_ j|_{U_ i} = 0$ and $s_ j|_{U_ i} = \sum _ k a_{ijk}t_{ik}$. Thus (2) holds.

Assume (2). Let $U$, $n$, $m$, $A$ and $s_1, \ldots , s_ n$ as in (3) be given. Observe that $A$ has $m$ columns. We will prove the assertion of (3) is true by induction on $m$. For the base case $m = 0$ we can use the factorization through the zero sheaf (in other words $l_ i = 0$). Let $(f_1, \ldots , f_ n)$ be the last column of $A$ and apply (2). This gives new diagrams

$\mathcal{O}_{U_ i}^{\oplus m} \xrightarrow {B_ i \circ A|_{U_ i}} \mathcal{O}_{U_ i}^{\oplus l_ i} \xrightarrow {(t_{i1}, \ldots , t_{il_ i})} \mathcal{F}|_{U_ i}$

but the first column of $A_ i = B_ i \circ A|_{U_ i}$ is zero. Hence we can apply the induction hypothesis to $U_ i$, $l_ i$, $m - 1$, the matrix consisting of the first $m - 1$ columns of $A_ i$, and $t_{i1}, \ldots , t_{il_ i}$ to get coverings $\{ U_{ij} \to U_ j\}$ and factorizations

$\mathcal{O}_{U_{ij}}^{\oplus l_ i} \xrightarrow {C_{ij}} \mathcal{O}_{U_{ij}}^{\oplus k_{ij}} \xrightarrow {(v_{ij1}, \ldots , v_{ij k_{ij}})} \mathcal{F}|_{U_{ij}}$

of $(t_{i1}, \ldots , t_{il_ i})|_{U_{ij}}$ such that $C_ i \circ B_ i|_{U_{ij}} \circ A|_{U_{ij}} = 0$. Then $\{ U_{ij} \to U\}$ is a covering and we get the desired factorizations using $B_{ij} = C_ i \circ B_ i|_{U_{ij}}$ and $v_{ija}$. In this way we see that (2) implies (3).

Assume (3). Let $\mathcal{G} \to \mathcal{H}$ be an injective homomorphism of $\mathcal{O}$-modules. We have to show that $\mathcal{G} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{H} \otimes _\mathcal {O} \mathcal{F}$ is injective. Let $U$ be an object of $\mathcal{C}$ and let $s \in (\mathcal{G} \otimes _\mathcal {O} \mathcal{F})(U)$ be a section which maps to zero in $\mathcal{H} \otimes _\mathcal {O} \mathcal{F}$. We have to show that $s$ is zero. Since $\mathcal{G} \otimes _\mathcal {O} \mathcal{F}$ is a sheaf, it suffices to find a covering $\{ U_ i \to U\} _{i \in I}$ of $\mathcal{C}$ such that $s|_{U_ i}$ is zero for all $i \in I$. Hence we may always replace $U$ by the members of a covering. In particular, since $\mathcal{G} \otimes _\mathcal {O} \mathcal{F}$ is the sheafification of $\mathcal{G} \otimes _{p, \mathcal{O}} \mathcal{F}$ we may assume that $s$ is the image of $s' \in \mathcal{G}(U) \otimes _{\mathcal{O}(U)} \mathcal{F}(U)$. Arguing similarly for $\mathcal{H} \otimes _\mathcal {O} \mathcal{F}$ we may assume that $s'$ maps to zero in $\mathcal{H}(U) \otimes _{\mathcal{O}(U)} \mathcal{F}(U)$. Write $\mathcal{F}(U) = \mathop{\mathrm{colim}}\nolimits M_\alpha$ as a filtered colimit of finitely presented $\mathcal{O}(U)$-modules $M_\alpha$ (Algebra, Lemma 10.11.3). Since tensor product commutes with filtered colimits (Algebra, Lemma 10.12.9) we can choose an $\alpha$ such that $s'$ comes from some $s'' \in \mathcal{G}(U) \otimes _{\mathcal{O}(U)} M_\alpha$ and such that $s''$ maps to zero in $\mathcal{H}(U) \otimes _{\mathcal{O}(U)} M_\alpha$. Fix $\alpha$ and $s''$. Choose a presentation

$\mathcal{O}(U)^{\oplus m} \xrightarrow {A} \mathcal{O}(U)^{\oplus n} \to M_\alpha \to 0$

We apply (3) to the corresponding complex of $\mathcal{O}_ U$-modules

$\mathcal{O}_ U^{\oplus m} \xrightarrow {A} \mathcal{O}_ U^{\oplus n} \xrightarrow {(s_1, \ldots , s_ n)} \mathcal{F}|_ U$

After replacing $U$ by the members of the covering $U_ i$ we find that the map

$M_\alpha \to \mathcal{F}(U)$

factors through a free module $\mathcal{O}(U)^{\oplus l}$ for some $l$. Since $\mathcal{G}(U) \to \mathcal{H}(U)$ is injective we conclude that

$\mathcal{G}(U) \otimes _{\mathcal{O}(U)} \mathcal{O}(U)^{\oplus l} \to \mathcal{H}(U) \otimes _{\mathcal{O}(U)} \mathcal{O}(U)^{\oplus l}$

is injective too. Hence as $s''$ maps to zero in the module on the right, it also maps to zero in the module on the left, i.e., $s$ is zero as desired. $\square$

Lemma 18.28.15. Let $\mathcal{C}$ be a site. Let $\mathcal{O}' \to \mathcal{O}$ be a surjection of sheaves of rings whose kernel $\mathcal{I}$ is an ideal of square zero. Let $\mathcal{F}'$ be an $\mathcal{O}'$-module and set $\mathcal{F} = \mathcal{F}'/\mathcal{I}\mathcal{F}'$. The following are equivalent

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

2. $\mathcal{F}$ is a flat $\mathcal{O}$-module and $\mathcal{I} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{F}'$ is injective.

Proof. If (1) holds, then $\mathcal{F} = \mathcal{F}' \otimes _{\mathcal{O}'} \mathcal{O}$ is flat over $\mathcal{O}$ by Lemma 18.28.13 and we see the map $\mathcal{I} \otimes _\mathcal {O} \mathcal{F} \to \mathcal{F}'$ is injective by applying $- \otimes _{\mathcal{O}'} \mathcal{F}'$ to the exact sequence $0 \to \mathcal{I} \to \mathcal{O}' \to \mathcal{O} \to 0$, see Lemma 18.28.9. Assume (2). In the rest of the proof we will use without further mention that $\mathcal{K} \otimes _{\mathcal{O}'} \mathcal{F}' = \mathcal{K} \otimes _\mathcal {O} \mathcal{F}$ for any $\mathcal{O}'$-module $\mathcal{K}$ annihilated by $\mathcal{I}$. Let $\alpha : \mathcal{G}' \to \mathcal{H}'$ be an injective map of $\mathcal{O}'$-modules. Let $\mathcal{G} \subset \mathcal{G}'$, resp. $\mathcal{H} \subset \mathcal{H}'$ be the subsheaf of sections annihilated by $\mathcal{I}$. Consider the diagram

$\xymatrix{ \mathcal{G} \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] \ar[d] & \mathcal{G}' \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] \ar[d] & \mathcal{G}'/\mathcal{G} \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] \ar[d] & 0 \\ \mathcal{H} \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] & \mathcal{H}' \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] & \mathcal{H}'/\mathcal{H} \otimes _{\mathcal{O}'} \mathcal{F}' \ar[r] & 0 }$

Note that $\mathcal{G}'/\mathcal{G}$ and $\mathcal{H}'/\mathcal{H}$ are annihilated by $\mathcal{I}$ and that $\mathcal{G}'/\mathcal{G} \to \mathcal{H}'/\mathcal{H}$ is injective. Thus the right vertical arrow is injective as $\mathcal{F}$ is flat over $\mathcal{O}$. The same is true for the left vertical arrow. Hence the middle vertical arrow is injective and $\mathcal{F}'$ is flat. $\square$

Lemma 18.28.16. Let $\mathcal{C}$ be a site. Let $\mathcal{O} \to \mathcal{O}'$ be a flat homorphism of sheaves of rings. Let $\mathcal{I} \subset \mathcal{O}$ be a sheaf of ideals such that the induced map $\mathcal{O}/\mathcal{I} \to \mathcal{O}'/\mathcal{I}\mathcal{O}'$ is an isomorphism. For any $\mathcal{O}$-module $\mathcal{F}$ annihilated by $\mathcal{I}^ n$ for some $n \geq 0$ the map $\text{id} \otimes 1 : \mathcal{F} \to \mathcal{F} \otimes _\mathcal {O} \mathcal{O}'$ is an isomorphism.

Proof. Omitted. Hint: See More on Algebra, Lemma 15.89.2. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).