Lemma 85.11.1. With notation as above. The morphism $a : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D})$ is flat if and only if $a_ n : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n), \mathcal{O}_ n) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{D}), \mathcal{O}_\mathcal {D})$ is flat for $n \geq 0$.
Proof. Since $g_ n : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ n), \mathcal{O}_ n) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{total}), \mathcal{O})$ is flat, we see that if $a$ is flat, then $a_ n = a \circ g_ n$ is flat as a composition. Conversely, suppose that $a_ n$ is flat for all $n$. We have to check that $\mathcal{O}$ is flat as a sheaf of $a^{-1}\mathcal{O}_\mathcal {D}$-modules. Let $\mathcal{F} \to \mathcal{G}$ be an injective map of $a^{-1}\mathcal{O}_\mathcal {D}$-modules. We have to show that
\[ \mathcal{F} \otimes _{a^{-1}\mathcal{O}_\mathcal {D}} \mathcal{O} \to \mathcal{G} \otimes _{a^{-1}\mathcal{O}_\mathcal {D}} \mathcal{O} \]
is injective. We can check this on $\mathcal{C}_ n$, i.e., after applying $g_ n^{-1}$. Since $g_ n^* = g_ n^{-1}$ because $g_ n^{-1}\mathcal{O} = \mathcal{O}_ n$ we obtain
\[ g_ n^{-1}\mathcal{F} \otimes _{g_ n^{-1}a^{-1}\mathcal{O}_\mathcal {D}} \mathcal{O}_ n \to g_ n^{-1}\mathcal{G} \otimes _{g_ n^{-1}a^{-1}\mathcal{O}_\mathcal {D}} \mathcal{O}_ n \]
which is injective because $g_ n^{-1}a^{-1}\mathcal{O}_\mathcal {D} = a_ n^{-1}\mathcal{O}_\mathcal {D}$ and we assume $a_ n$ was flat. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: