Lemma 77.2.4. In Situation 77.2.1. Let $(Y_1, y_1) \to (Y, y)$ be a morphism of pointed algebraic spaces over $S$. Assume $Y_1 \to Y$ is flat at $y_1$. If $(T \to Y, t' \leadsto t, \xi )$ is an impurity of $\mathcal{F}$ above $y$, then there exists an impurity $(T_1 \to Y_1, t_1' \leadsto t_1, \xi _1)$ of the pullback $\mathcal{F}_1$ of $\mathcal{F}$ to $X_1 = Y_1 \times _ Y X$ over $y_1$ such that $T_1$ is étale over $Y_1 \times _ Y T$.
Proof. Choose an étale morphism $T_1 \to Y_1 \times _ Y T$ where $T_1$ is a scheme and let $t_1 \in T_1$ be a point mapping to $y_1$ and $t$. It is possible to find a pair $(T_1, t_1)$ like this by Properties of Spaces, Lemma 66.4.3. The morphism of schemes $T_1 \to T$ is flat at $t_1$ (use Morphisms of Spaces, Lemma 67.30.4 and the definition of flat morphisms of algebraic spaces) there exists a specialization $t'_1 \leadsto t_1$ lying over $t' \leadsto t$, see Morphisms, Lemma 29.25.9. Choose a point $\xi _1 \in |X_{T_1}|$ mapping to $t'_1$ and $\xi $ with $\xi _1 \in \text{Ass}_{X_{T_1}/T_1}(\mathcal{F}_{T_1})$. point of $\mathop{\mathrm{Spec}}(\kappa (t'_1) \otimes _{\kappa (t')} \kappa (\xi ))$. This is possible by Divisors on Spaces, Lemma 71.4.7. As the closure $Z_1$ of $\{ \xi _1\} $ in $|X_{T_1}|$ maps into the closure of $\{ \xi \} $ in $|X_ T|$ we conclude that the image of $Z_1$ in $|T_1|$ cannot contain $t_1$. Hence $(T_1 \to Y_1, t'_1 \leadsto t_1, \xi _1)$ is an impurity of $\mathcal{F}_1$ above $Y_1$. $\square$
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Your email address will not be published.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)