Lemma 37.30.1. 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 37.24.1 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 37.24.2. Hence we win. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)