The Stacks project

[Exposé V, Corollary 1.7.1, SGA4]

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05VD. Beware of the difference between the letter 'O' and the digit '0'.