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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