The Stacks project

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)

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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 054Y. Beware of the difference between the letter 'O' and the digit '0'.