The Stacks project

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. $\square$