Lemma 37.23.5. Let $f : X \to Y$ be a finite type morphism of schemes. Assume $Y$ irreducible with generic point $\eta$. Let $X_\eta = Z_{1, \eta } \cup \ldots \cup Z_{n, \eta }$ be a covering of the generic fibre by closed subsets of $X_\eta$. Let $Z_ i$ be the closure of $Z_{i, \eta }$ in $X$ (see discussion above). Then there exists a nonempty open $V \subset Y$ such that $X_ y = Z_{1, y} \cup \ldots \cup Z_{n, y}$ for all $y \in V$.

Proof. If $Y$ is Noetherian then $U = X \setminus (Z_1 \cup \ldots \cup Z_ n)$ is of finite type over $Y$ and we can directly apply Lemma 37.23.1 to get that $U_ V = \emptyset$ for a nonempty open $V \subset Y$. In general we argue as follows. As the question is topological we may replace $Y$ by its reduction. Thus $Y$ is integral, see Properties, Lemma 28.3.4. After shrinking $Y$ we may assume that $X \to Y$ is flat, see Morphisms, Proposition 29.27.1. In this case every point $x$ in $X_ y$ is a specialization of a point $x' \in X_\eta$ by Morphisms, Lemma 29.25.9. As the $Z_ i$ are closed in $X$ and cover the generic fibre this implies that $X_ y = \bigcup Z_{i, y}$ for $y \in Y$ as desired. $\square$

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