Lemma 37.28.3. Let $f : X \to Y$, $s : Y \to X$ be as in Situation 37.28.1. Assume $f$ of finite type. Let $y \in Y$ be a point. Then there exists a nonempty open $V \subset \overline{\{ y\} }$ such that the inverse image of $X^0$ in the base change $X_ V$ is open and closed in $X_ V$.

**Proof.**
Let $Z \subset Y$ be the induced reduced closed subscheme structure on $\overline{\{ y\} }$. Let $f_ Z : X_ Z \to Z$ and $s_ Z : Z \to X_ Z$ be the base changes of $f$ and $s$. By Lemma 37.28.2 we have $(X_ Z)^0 = (X^0)_ Z$. Hence it suffices to prove the lemma for the morphism $X_ Z \to Z$ and the point $x \in X_ Z$ which maps to the generic point of $Z$. In other words we have reduced the problem to the case where $Y$ is an integral scheme (see Properties, Lemma 28.3.4) with generic point $\eta $. Our goal is to show that after shrinking $Y$ the subset $X^0$ becomes an open and closed subset of $X$.

Note that the scheme $X_\eta $ is of finite type over a field, hence Noetherian. Thus its connected components are open as well as closed. Hence we may write $X_\eta = X_\eta ^0 \amalg T_\eta $ for some open and closed subset $T_\eta $ of $X_\eta $. Next, let $T \subset X$ be the closure of $T_\eta $ and let $X^{00} \subset X$ be the closure of $X_\eta ^0$. Note that $T_\eta $, resp. $X^0_\eta $ is the generic fibre of $T$, resp. $X^{00}$, see discussion preceding Lemma 37.23.5. Moreover, that lemma implies that after shrinking $Y$ we may assume that $X = X^{00} \cup T$ (set theoretically). Note that $(T \cap X^{00})_\eta = T_\eta \cap X^0_\eta = \emptyset $. Hence after shrinking $Y$ we may assume that $T \cap X^{00} = \emptyset $, see Lemma 37.23.1. In particular $X^{00}$ is open in $X$. Note that $X^0_\eta $ is connected and has a rational point, namely $s(\eta )$, hence it is geometrically connected, see Varieties, Lemma 33.7.14. Thus after shrinking $Y$ we may assume that all fibres of $X^{00} \to Y$ are geometrically connected, see Lemma 37.27.4. At this point it follows that the fibres $X^{00}_ y$ are open, closed, and connected subsets of $X_ y$ containing $\sigma (y)$. It follows that $X^0 = X^{00}$ and we win. $\square$

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