The Stacks project

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. $\square$

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Your email address will not be published.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 08M4. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 18.28.15 as pdf