Lemma 64.17.1. 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 37.40.6 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. $\square$

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