Lemma 37.26.4. Let f : X \to Y be a morphism of schemes. Assume
Y is irreducible with generic point \eta ,
X_\eta is geometrically reduced, and
f is of finite type.
Then there exists a nonempty open subscheme V \subset Y such that X_ V \to V has geometrically reduced fibres.
Proof.
Let Y' \subset Y be the reduction of Y. Let X' \to Y' be the base change of f. Note that Y' \to Y induces a bijection on points and that X' \to X identifies fibres. Hence we may assume that Y' is reduced, i.e., integral, see Properties, Lemma 28.3.4. We may also replace Y by an affine open. Hence we may assume that Y = \mathop{\mathrm{Spec}}(A) with A a domain. Denote K the fraction field of A. After shrinking Y a bit we may also assume that X \to Y is flat and of finite presentation, see Morphisms, Proposition 29.27.1.
As X_\eta is geometrically reduced there exists an open dense subset V \subset X_\eta such that V \to \mathop{\mathrm{Spec}}(K) is smooth, see Varieties, Lemma 33.25.7. Let U \subset X be the set of points where f is smooth. By Morphisms, Lemma 29.34.15 we see that V \subset U_\eta . Thus the generic fibre of U is dense in the generic fibre of X. Since X_\eta is reduced, it follows that U_\eta is scheme theoretically dense in X_\eta , see Morphisms, Lemma 29.7.8. We note that as U \to Y is smooth all the fibres of U \to Y are geometrically reduced. Thus it suffices to show that, after shrinking Y, for all y \in Y the scheme U_ y is scheme theoretically dense in X_ y, see Morphisms, Lemma 29.7.9. This follows from Lemma 37.24.4.
\square
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