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. $\square$

Comment #519 by Fred Rohrer on

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

Comment #6538 by Jonas Ehrhard on

Suggested slogan: "Flat maps locally of finite presentation are open".

Comment #6539 by Jonas Ehrhard on

As it turns out, this is also the content of Lemma 01UA, I missed this at first.

Comment #6590 by on

If you click on the "xx tags refer to this tag" link on the right, then you can find results that use this particular lemma. So this way you could have found Lemma 29.25.10. Hope that helps.

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