## 18.39 Pullbacks of flat modules

The pullback of a flat module along a morphism of ringed topoi is flat. This is a bit tricky to prove.

Lemma 18.39.1. Let $(f, f^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}_\mathcal {C}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D})$ be a morphism of ringed topoi or ringed sites. Then $f^*\mathcal{F}$ is a flat $\mathcal{O}_\mathcal {C}$-module whenever $\mathcal{F}$ is a flat $\mathcal{O}_\mathcal {D}$-module.

Proof. Choose a diagram as in Lemma 18.7.2. Recall that being a flat module is intrinsic (see Section 18.18 and Definition 18.28.1). Hence it suffices to prove the lemma for the morphism $(h, h^\sharp ) : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}_{\mathcal{C}'}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}'), \mathcal{O}_{\mathcal{D}'})$. In other words, we may assume that our sites $\mathcal{C}$ and $\mathcal{D}$ have all finite limits and that $f$ is a morphism of sites induced by a continuous functor $u : \mathcal{D} \to \mathcal{C}$ which commutes with finite limits.

Recall that $f^*\mathcal{F} = \mathcal{O}_\mathcal {C} \otimes _{f^{-1}\mathcal{O}_\mathcal {D}} f^{-1}\mathcal{F}$ (Definition 18.13.1). By Lemma 18.28.13 it suffices to prove that $f^{-1}\mathcal{F}$ is a flat $f^{-1}\mathcal{O}_\mathcal {D}$-module. Combined with the previous paragraph this reduces us to the situation of the next paragraph.

Assume $\mathcal{C}$ and $\mathcal{D}$ are sites which have all finite limits and that $u : \mathcal{D} \to \mathcal{C}$ is a continuous functor which commutes with finite limits. Let $\mathcal{O}$ be a sheaf of rings on $\mathcal{D}$ and let $\mathcal{F}$ be a flat $\mathcal{O}$-module. Then $u$ defines a morphism of sites $f : \mathcal{C} \to \mathcal{D}$ (Sites, Proposition 7.14.7). To show: $f^{-1}\mathcal{F}$ is a flat $f^{-1}\mathcal{O}$-module. Let $U$ be an object of $\mathcal{C}$ and let

$f^{-1}\mathcal{O}|_ U \xrightarrow {(f_1, \ldots , f_ n)} f^{-1}\mathcal{O}|_ U^{\oplus n} \xrightarrow {(s_1, \ldots , s_ n)} f^{-1}\mathcal{F}|_ U$

be a complex of $f^{-1}\mathcal{O}|_ U$-modules. Our goal is to construct a factorization of $(s_1, \ldots , s_ n)$ on the members of a covering of $U$ as in Lemma 18.28.14 part (2). Consider the elements $s_ a \in f^{-1}\mathcal{F}(U)$ and $f_ a \in f^{-1}\mathcal{O}(U)$. Since $f^{-1}\mathcal{F}$, resp. $f^{-1}\mathcal{O}$ is the sheafification of $u_ p\mathcal{F}$ we may, after replacing $U$ by the members of a covering, assume that $s_ a$ is the image of an element $s'_ a \in u_ p\mathcal{F}(U)$ and $f_ a$ is the image of an element $f'_ a \in u_ p\mathcal{O}(U)$. Then after another replacement of $U$ by the members of a covering we may assume that $\sum f'_ as'_ a$ is zero in $u_ p\mathcal{F}(U)$. Recall that the category $(\mathcal{I}_ U^ u)^{opp}$ is directed (Sites, Lemma 7.5.2) and that $u_ p\mathcal{F}(U) = \mathop{\mathrm{colim}}\nolimits _{(\mathcal{I}_ U^ u)^{opp}} \mathcal{F}(V)$ and $u_ p\mathcal{O}(U) = \mathop{\mathrm{colim}}\nolimits _{(\mathcal{I}_ U^ u)^{opp}} \mathcal{O}(V)$. Hence we may assume there is a pair $(V, \phi ) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I}_ U^ u)$ where $V$ is an object of $\mathcal{D}$ and $\phi$ is a morphism $\phi : U \to u(V)$ of $\mathcal{D}$ and elements $s''_ a \in \mathcal{F}(V)$ and $f''_ a \in \mathcal{O}(V)$ whose images in $u_ p\mathcal{F}(U)$ and $u_ p\mathcal{O}(U)$ are equal to $s'_ a$ and $f'_ a$ and such that $\sum f''_ a s''_ a = 0$ in $\mathcal{F}(V)$. Then we obtain a complex

$\mathcal{O}|_ V \xrightarrow {(f''_1, \ldots , f''_ n)} \mathcal{O}|_ V^{\oplus n} \xrightarrow {(s''_1, \ldots , s''_ n)} \mathcal{F}|_ V$

and we can apply the other direction of Lemma 18.28.14 to see there exists a covering $\{ V_ i \to V\}$ of $\mathcal{D}$ and for each $i$ a factorization

$\mathcal{O}|_{V_ i}^{\oplus n} \xrightarrow {B''_ i} \mathcal{O}|_{V_ i}^{\oplus l_ i} \xrightarrow {(t''_{i1}, \ldots , t''_{il_ i})} \mathcal{F}|_{V_ i}$

of $(s''_1, \ldots , s''_ n)|_{V_ i}$ such that $B_ i \circ (f''_1, \ldots , f''_ n)|_{V_ i} = 0$. Set $U_ i = U \times _{\phi , u(V)} u(V_ i)$, denote $B_ i \in \text{Mat}(l_ i \times n, f^{-1}\mathcal{O}(U_ i))$ the image of $B''_ i$, and denote $t_{ij} \in f^{-1}\mathcal{F}(U_ i)$ the image of $t''_{ij}$. Then we get a factorization

$f^{-1}\mathcal{O}|_{U_ i}^{\oplus n} \xrightarrow {B_ i} f^{-1}\mathcal{O}|_{U_ i}^{\oplus l_ i} \xrightarrow {(t_{i1}, \ldots , t_{il_ i})} \mathcal{F}|_{U_ i}$

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

Lemma 18.39.2. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $p$ be a point of $\mathcal{C}$. If $\mathcal{F}$ is a flat $\mathcal{O}$-module, then $\mathcal{F}_ p$ is a flat $\mathcal{O}_ p$-module.

Proof. In Section 18.37 we have seen that we can think of $p$ as a morphism of ringed topoi

$(p, \text{id}_{\mathcal{O}_ p}) : (\mathop{\mathit{Sh}}\nolimits (pt), \mathcal{O}_ p) \longrightarrow (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}).$

such that the pullback functor $p^* : \textit{Mod}(\mathcal{O}) \to \textit{Mod}(\mathcal{O}_ p)$ equals the stalk functor. Thus the lemma follows from Lemma 18.39.1. $\square$

Lemma 18.39.3. Let $(\mathcal{C}, \mathcal{O})$ be a ringed site. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}$-modules. Let $\{ p_ i\} _{i \in I}$ be a conservative family of points of $\mathcal{C}$. Then $\mathcal{F}$ is flat if and only if $\mathcal{F}_{p_ i}$ is a flat $\mathcal{O}_{p_ i}$-module for all $i \in I$.

Proof. By Lemma 18.39.2 we see one of the implications. For the converse, use that $(\mathcal{F} \otimes _\mathcal {O} \mathcal{G})_ p = \mathcal{F}_ p \otimes _{\mathcal{O}_ p} \mathcal{G}_ p$ by Lemma 18.26.2 (as taking stalks at $p$ is given by $p^{-1}$) and Lemma 18.14.4. $\square$

Lemma 18.39.4. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}') \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O})$ be a morphism of ringed topoi. Let $0 \to \mathcal{F} \to \mathcal{G} \to \mathcal{H} \to 0$ be a short exact sequence of $\mathcal{O}$-modules with $\mathcal{H}$ a flat $\mathcal{O}$-module. Then the sequence $0 \to f^*\mathcal{F} \to f^*\mathcal{G} \to f^*\mathcal{H} \to 0$ is exact as well.

Proof. Since $f^{-1}$ is exact we have the short exact sequence $0 \to f^{-1}\mathcal{F} \to f^{-1}\mathcal{G} \to f^{-1}\mathcal{H} \to 0$ of $f^{-1}\mathcal{O}$-modules. By Lemma 18.39.1 the $f^{-1}\mathcal{O}$-module $f^{-1}\mathcal{H}$ is flat. By Lemma 18.28.9 this implies that tensoring the sequence over $f^{-1}\mathcal{O}$ with $\mathcal{O}'$ the sequence remains exact. Since $f^*\mathcal{F} = f^{-1}\mathcal{F} \otimes _{f^{-1}\mathcal{O}} \mathcal{O}'$ and similarly for $\mathcal{G}$ and $\mathcal{H}$ we conclude. $\square$

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