The Stacks project

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

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

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

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

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. $\square$

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. $\square$

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

(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. $\square$

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

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. $\square$

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

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. $\square$

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

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. $\square$

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

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. $\square$

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. $\square$

Proof. This is true because

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

Proof. This is true because

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

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

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

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

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

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

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

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. $\square$

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

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. $\square$

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