Lemma 37.24.5. 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 37.24.1 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
Comments (0)