Lemma 58.16.3. Let $f : X \to S$ be a proper morphism with geometrically connected fibres. Let $s' \leadsto s$ be a specialization of points of $S$ and let $sp : \pi _1(X_{\overline{s}'}) \to \pi _1(X_{\overline{s}})$ be a specialization map. Then there exists a strictly henselian valuation ring $R$ over $S$ with algebraically closed fraction field such that $sp$ is isomorphic to $sp_ R$ defined above.
Proof. Let $\mathcal{O}_{S, s} \to A$ be the strict henselization constructed using $\kappa (s) \to \kappa (\overline{s})$. Let $A \to \kappa (\overline{s}')$ be the map used to construct $sp$. Let $R \subset \kappa (\overline{s}')$ be a valuation ring with fraction field $\kappa (\overline{s}')$ dominating the image of $A$. See Algebra, Lemma 10.50.2. Observe that $R$ is strictly henselian for example by Lemma 58.12.2 and Algebra, Lemma 10.50.3. Then the lemma is clear. $\square$
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Your email address will not be published.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)