Lemma 67.17.1. Let $S$ be a scheme. Let $W \subset S$ be a scheme theoretically dense open subscheme (Morphisms, Definition 29.7.1). Let $f : X \to S$ be a morphism of schemes which is flat, locally of finite presentation, and locally quasi-finite. Then $f^{-1}(W)$ is scheme theoretically dense in $X$.
Proof. We will use the characterization of Morphisms, Lemma 29.7.5. Assume $V \subset X$ is an open and $g \in \Gamma (V, \mathcal{O}_ V)$ is a function which restricts to zero on $f^{-1}(W) \cap V$. We have to show that $g = 0$. Assume $g \not= 0$ to get a contradiction. By More on Morphisms, Lemma 37.45.6 we may shrink $V$, find an open $U \subset S$ fitting into a commutative diagram
\[ \xymatrix{ V \ar[r] \ar[d]_\pi & X \ar[d]^ f \\ U \ar[r] & S, } \]
a quasi-coherent subsheaf $\mathcal{F} \subset \mathcal{O}_ U$, an integer $r > 0$, and an injective $\mathcal{O}_ U$-module map $\mathcal{F}^{\oplus r} \to \pi _*\mathcal{O}_ V$ whose image contains $g|_ V$. Say $(g_1, \ldots , g_ r) \in \Gamma (U, \mathcal{F}^{\oplus r})$ maps to $g$. Then we see that $g_ i|_{W \cap U} = 0$ because $g|_{f^{-1}W \cap V} = 0$. Hence $g_ i = 0$ because $\mathcal{F} \subset \mathcal{O}_ U$ and $W$ is scheme theoretically dense in $S$. This implies $g = 0$ which is the desired contradiction. $\square$
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Your email address will not be published.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)