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. Assume that $R \to S$ has finite presentation and satisfies going down. It suffices to prove that the image of a standard open $D(f)$ is open. Since $S \to S_ f$ satisfies going down as well, we see that $R \to S_ f$ satisfies going down. Thus after replacing $S$ by $S_ f$ 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. $\square$

Comment #519 by Fred Rohrer on

In the statement of the result the map between the spectra goes the wrong way.

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