Let $f: X\to Y$ be a finite morphism of schemes, let $y\in Y, Y(\bar{y}):={\rm Spec}(\mathscr{O}_{Y, \bar{y}}), X_{\bar{y}}:=X\times_Y{\rm Spec}(\kappa_y^s)=X_y\otimes_{\kappa_y}\kappa_y^s, X({\bar{y}}):=X\times_YY(\bar{y})$, here $\kappa_y^s$ is a separable closure of the residue field $\kappa_y$. By some results about strict henselian local rings I can see $$X({\bar{y}})\cong\coprod_{a\in X_{\bar{y}}}{\rm Spec}(\mathscr{O}_{X({\bar{y}}), a})=\coprod_{x\in X_y}\coprod_{a\in(X_{\bar{y}}\to X_y)^{-1}(x)}{\rm Spec}(\mathscr{O}_{X({\bar{y}}), a}).$$ I believe $\mathscr{O}_{X(\bar{y}), a}=\mathscr{O}_{X, \bar{x}}$ if $a\in(X_{\bar{y}}\to X_y)^{-1}(x)$ but can't prove it. Can anyone prove it in some detail or give a correct form of $\mathscr{O}_{X(\bar{y}), a}$?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ I think it's correct, see e.g. Tag 05WR. In your setup, $R \to S$ is a finite ring map, and $R_{\mathfrak{p}}^{sh} \otimes _{R_{\mathfrak{p}}} S_{\mathfrak{q}}$ is a localization of $R_{\mathfrak{p}}^{sh} \otimes _{R_{\mathfrak{p}}} (R \setminus \mathfrak{p})^{-1}S$, which is finite over $R_{\mathfrak{p}}^{sh}$, hence a finite product of (strictly henselian) local rings. $\endgroup$– Minseon ShinCommented Nov 7, 2019 at 21:48
-
$\begingroup$ I can confirm it's correct now by passing to limit using [Milne / Étale Cohomology, Chapter II, Lemma 3.3], which will give another decomposition of $X({\bar{y}})$ of the same form, but the terms in the coproduct are now $\mathscr{O}_{X, \bar{x}}$, forcing $\mathscr{O}_{X(\bar{y}), a}\cong\mathscr{O}_{X, \bar{x}}$. $\endgroup$– Lao-tzuCommented Nov 8, 2019 at 7:34
-
$\begingroup$ Anyway, this argument is rather indirect, there should be a more direct way to show it. I will just leave it there in case others may give a direct proof. $\endgroup$– Lao-tzuCommented Nov 8, 2019 at 7:35
Add a comment
|