The Stacks project

Lemma 91.7.8. Let $t : (S, \mathcal{O}_ S) \to (S', \mathcal{O}_{S'})$, $\mathcal{J} = \mathop{\mathrm{Ker}}(t^\sharp )$, $f : (X, \mathcal{O}_ X) \to (S, \mathcal{O}_ S)$, $\mathcal{G}$, and $c : \mathcal{J} \to \mathcal{G}$ be as in ( Denote $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ S}(\mathop{N\! L}\nolimits _{S/S'}, \mathcal{J})$ the element corresponding to the extension $\mathcal{O}_{S'}$ of $\mathcal{O}_ S$ by $\mathcal{J}$ via Lemma 91.7.4. The set of isomorphism classes of solutions is canonically bijective to the fibre of

\[ \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S'}, \mathcal{G}) \to \mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(Lf^*\mathop{N\! L}\nolimits _{S/S'}, \mathcal{G}) \]

over the image of $\xi $.

Proof. By Lemma 91.7.4 applied to $X \to S'$ and the $\mathcal{O}_ X$-module $\mathcal{G}$ we see that elements $\zeta $ of $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_ X}(\mathop{N\! L}\nolimits _{X/S'}, \mathcal{G})$ parametrize extensions $0 \to \mathcal{G} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0$ of $f^{-1}\mathcal{O}_{S'}$-algebras. By Lemma 91.7.6 applied to $X \to S \to S'$ and $c : \mathcal{J} \to \mathcal{G}$ we see that there is an $S'$-morphism $X' \to S'$ compatible with $c$ and $f : X \to S$ if and only if $\zeta $ maps to $\xi $. Of course this is the same thing as saying $\mathcal{O}_{X'}$ is a solution of ( $\square$

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Your email address will not be published.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 0GQ3. Beware of the difference between the letter 'O' and the digit '0'.