Lemma 39.19.4. Assumptions and notation as in Lemma 39.19.3. There exists an $R$-invariant open $V \subset U$ and a quasi-compact open $W'$ such that $W \subset V \subset W' \subset U$.

Proof. Set $E = t(s^{-1}(W))$. Recall that $E$ is set-theoretically $R$-invariant (Lemma 39.19.2). By Lemma 39.19.3 there exists a quasi-compact open $W'$ containing $E$. Let $Z = U \setminus W'$ and consider $T = t(s^{-1}(Z))$. Observe that $Z \subset T$ and that $E \cap T = \emptyset$ because $s^{-1}(E) = t^{-1}(E)$ is disjoint from $s^{-1}(Z)$. Since $T$ is the image of the closed subset $s^{-1}(Z) \subset R$ under the quasi-compact morphism $t : R \to U$ we see that any point $\xi$ in the closure $\overline{T}$ is the specialization of a point of $T$, see Morphisms, Lemma 29.6.5 (and Morphisms, Lemma 29.6.3 to see that the scheme theoretic image is the closure of the image). Say $\xi ' \leadsto \xi$ with $\xi ' \in T$. Suppose that $r \in R$ and $s(r) = \xi$. Since $s$ is flat we can find a specialization $r' \leadsto r$ in $R$ such that $s(r') = \xi '$ (Morphisms, Lemma 29.25.9). Then $t(r') \leadsto t(r)$. We conclude that $t(r') \in T$ as $T$ is set-theoretically invariant by Lemma 39.19.2. Thus $\overline{T}$ is a set-theoretically $R$-invariant closed subset and $V = U \setminus \overline{T}$ is the open we are looking for. It is contained in $W'$ which finishes the proof. $\square$

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