The Stacks project

Proof. Let $y' \in Y'$ with image $y \in Y$. We may think of $X^0_ y$ as a closed subscheme of $X_ y$, see for example Morphisms, Definition 29.26.3. As $s(y) \in X^0_ y$ we conclude from Varieties, Lemma 33.7.14 that $X_ y^0$ is a geometrically connected scheme over $\kappa (y)$. Hence $X_ y^0 \times _ y y' \to X'_{y'}$ is a connected closed subscheme which contains $s'(y')$. Thus $X_ y^0 \times _ y y' \subset (X'_{y'})^0$. The other inclusion $X_ y^0 \times _ y y' \supset (X'_{y'})^0$ is clear as the image of $(X'_{y'})^0$ in $X_ y$ is a connected subset of $X_ y$ which contains $s(y)$. $\square$

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 37.29.2 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 37.24.5. 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 37.24.1. 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 37.28.4. 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. $\square$

Proof. Let $x \in X$. We have to show that there exists an open neighbourhood $U$ of $x$ such that $X^0 \cap U$ is constructible in $U$. This reduces us to the case where $Y$ is affine. Write $Y = \mathop{\mathrm{Spec}}(A)$ and $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as a directed limit of finite type $\mathbf{Z}$-algebras. By Limits, Lemma 32.10.1 we can find an $i$ and a morphism $f_ i : X_ i \to \mathop{\mathrm{Spec}}(A_ i)$ of finite presentation, endowed with a section $s_ i : \mathop{\mathrm{Spec}}(A_ i) \to X_ i$ whose base change to $Y$ recovers $f$ and the section $s$. By Lemma 37.29.2 it suffices to prove the lemma for $f_ i, s_ i$. Thus we reduce to the case where $Y$ is the spectrum of a Noetherian ring.

Assume $Y$ is a Noetherian affine scheme. Since $f$ is of finite presentation, i.e., of finite type, we see that $X$ is a Noetherian scheme too, see Morphisms, Lemma 29.15.6. In order to prove the lemma in this case it suffices to show that for every irreducible closed subset $Z \subset X$ the intersection $Z \cap X^0$ either contains a nonempty open of $Z$ or is not dense in $Z$, see Topology, Lemma 5.16.3. Let $x \in Z$ be the generic point, and let $y = f(x)$. By Lemma 37.29.3 there exists a nonempty open subset $V \subset \overline{\{ y\} }$ such that $X^0 \cap X_ V$ is open and closed in $X_ V$. Since $f(Z) \subset \overline{\{ y\} }$ and $f(x) = y \in V$ we see that $W = f^{-1}(V) \cap Z$ is a nonempty open subset of $Z$. It follows that $X^0 \cap W$ is open and closed in $W$. Since $W$ is irreducible we see that $X^0 \cap W$ is either empty or equal to $W$. This proves the lemma. $\square$

Proof. We may replace $Y$ with an affine open neighbourhood of $y$. Write $Y = \mathop{\mathrm{Spec}}(A)$ and $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as a directed limit of finite type $\mathbf{Z}$-algebras. By Limits, Lemma 32.10.1 we can find an $i$ and a morphism $f_ i : X_ i \to \mathop{\mathrm{Spec}}(A_ i)$ of finite presentation, endowed with a section $s_ i : \mathop{\mathrm{Spec}}(A_ i) \to X_ i$ whose base change to $Y$ recovers $f$ and the section $s$. After possibly increasing $i$ we may also assume that $f_ i$ is flat, see Limits, Lemma 32.8.7. Let $y_ i$ be the image of $y$ in $Y_ i$. Note that $X_ y = (X_{i, y_ i}) \times _{y_ i} y$. Hence $X_{i, y_ i}$ is geometrically reduced, see Varieties, Lemma 33.6.6. By Lemma 37.29.2 it suffices to prove the lemma for the system $f_ i, s_ i, y_ i \in Y_ i$. Thus we reduce to the case where $Y$ is the spectrum of a Noetherian ring.

Assume $Y$ is the spectrum of a Noetherian ring. Since $f$ is of finite presentation, i.e., of finite type, we see that $X$ is a Noetherian scheme too, see Morphisms, Lemma 29.15.6. Let $x \in X^0$ be a point lying over $y$. By Topology, Lemma 5.16.4 it suffices to prove that for any irreducible closed $Z \subset X$ passing through $x$ the intersection $X^0 \cap Z$ is dense in $Z$. In particular it suffices to prove that the generic point $x' \in Z$ is in $X^0$. By Properties, Lemma 28.5.10 we can find a discrete valuation ring $R$ and a morphism $\mathop{\mathrm{Spec}}(R) \to X$ which maps the special point to $x$ and the generic point to $x'$. We are going to think of $\mathop{\mathrm{Spec}}(R)$ as a scheme over $Y$ via the composition $\mathop{\mathrm{Spec}}(R) \to X \to Y$. By Lemma 37.29.2 we have that $(X_ R)^0$ is the inverse image of $X^0$. By construction we have a second section $t : \mathop{\mathrm{Spec}}(R) \to X_ R$ (besides the base change $s_ R$ of $s$) of the structure morphism $X_ R \to \mathop{\mathrm{Spec}}(R)$ such that $t(\eta _ R)$ is a point of $X_ R$ which maps to $x'$ and $t(0_ R)$ is a point of $X_ R$ which maps to $x$. Note that $t(0_ R)$ is in $(X_ R)^0$ and that $t(\eta _ R) \leadsto t(0_ R)$. Thus it suffices to prove that this implies that $t(\eta _ R) \in (X_ R)^0$. Hence it suffices to prove the lemma in the case where $Y$ is the spectrum of a discrete valuation ring and $y$ its closed point.

Assume $Y$ is the spectrum of a discrete valuation ring and $y$ is its closed point. Our goal is to prove that $X^0$ is a neighbourhood of $X^0_ y$. Note that $X^0_ y$ is open and closed in $X_ y$ as $X_ y$ has finitely many irreducible components. Hence the complement $C = X_ y \setminus X_ y^0$ is closed in $X$. Thus $U = X \setminus C$ is an open neighbourhood of $X^0_ y$ and $U^0 = X^0$. Hence it suffices to prove the result for the morphism $U \to Y$. In other words, we may assume that $X_ y$ is connected. Suppose that $X$ is disconnected, say $X = X_1 \amalg \ldots \amalg X_ n$ is a decomposition into connected components. Then $s(Y)$ is completely contained in one of the $X_ i$. Say $s(Y) \subset X_1$. Then $X^0 \subset X_1$. Hence we may replace $X$ by $X_1$ and assume that $X$ is connected. At this point Lemma 37.28.7 implies that $X_\eta $ is connected, i.e., $X^0 = X$ and we win. $\square$

Proof. This is an immediate consequence of Lemma 37.29.5. $\square$