Lemma 37.29.1. Let $f : X \to Y$ be a morphism of schemes. Assume $Y$ irreducible with generic point $\eta $ and $f$ of finite type. If $X_\eta $ has dimension $n$, then there exists a nonempty open $V \subset Y$ such that for all $y \in V$ the fibre $X_ y$ has dimension $n$.

**Proof.**
Let $Z = \{ x \in X \mid \dim _ x(X_{f(x)}) > n \} $. By Morphisms, Lemma 29.28.4 this is a closed subset of $X$. By assumption $Z_\eta = \emptyset $. Hence by Lemma 37.23.1 we may shrink $Y$ and assume that $Z = \emptyset $. Let $Z' = \{ x \in X \mid \dim _ x(X_{f(x)}) > n - 1 \} = \{ x \in X \mid \dim _ x(X_{f(x)}) = n\} $. As before this is a closed subset of $X$. By assumption we have $Z'_\eta \not= \emptyset $. Hence after shrinking $Y$ we may assume that $Z' \to Y$ is surjective, see Lemma 37.23.2. Hence we win.
$\square$

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