Lemma 91.5.6. Let $(f, f')$ be a morphism of first order thickenings as in Situation 91.3.1. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Assume $(f, f')$ is a strict morphism of thickenings and $\mathcal{F}$ flat over $S$. If there exists a pair $(\mathcal{F}', \alpha )$ consisting of an $\mathcal{O}_{X'}$-module $\mathcal{F}'$ flat over $S'$ and an isomorphism $\alpha : i^*\mathcal{F}' \to \mathcal{F}$, then the set of isomorphism classes of such pairs is principal homogeneous under $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}( \mathcal{F}, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F})$.
Proof. If we assume there exists one such module, then the canonical map
\[ f^*\mathcal{J} \otimes _{\mathcal{O}_ X} \mathcal{F} \to \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F} \]
is an isomorphism by Lemma 91.5.2. Apply Lemma 91.4.3 with $\mathcal{K} = \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{F}$ and $c = 1$. By Lemma 91.5.2 the corresponding extensions $\mathcal{F}'$ are all flat over $S'$. $\square$
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Your email address will not be published.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)