Proposition 10.41.8. Let $R \to S$ be flat and of finite presentation. Then $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is open. More generally this holds for any ring map $R \to S$ of finite presentation which satisfies going down.

**Proof.**
If $R \to S$ is flat, then $R \to S$ satisfies going down by Lemma 10.39.19. Thus to prove the lemma we may assume that $R \to S$ has finite presentation and satisfies going down.

Since the standard opens $D(g) \subset \mathop{\mathrm{Spec}}(S)$, $g \in S$ form a basis for the topology, it suffices to prove that the image of $D(g)$ is open. Recall that $\mathop{\mathrm{Spec}}(S_ g) \to \mathop{\mathrm{Spec}}(S)$ is a homeomorphism of $\mathop{\mathrm{Spec}}(S_ g)$ onto $D(g)$ (Lemma 10.17.6). Since $S \to S_ g$ satisfies going down (see above), we see that $R \to S_ g$ satisfies going down by Lemma 10.41.4. Thus after replacing $S$ by $S_ g$ we see it suffices to prove the image is open. By Chevalley's theorem (Theorem 10.29.10) the image is a constructible set $E$. And $E$ is stable under generalization because $R \to S$ satisfies going down, see Topology, Lemmas 5.19.2 and 5.19.6. Hence $E$ is open by Lemma 10.41.7. $\square$

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