The Stacks project

Lemma 10.41.2. Let $R \to S$ be a ring map. If the induced map $\varphi : \mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is open, then $R \to S$ satisfies going down.

Proof. Suppose that $\mathfrak p \subset \mathfrak p' \subset R$ and $\mathfrak q' \subset S$ lies over $\mathfrak p'$. As $\varphi $ is open, for every $g \in S$, $g \not\in \mathfrak q'$ we see that $\mathfrak p$ is in the image of $D(g) \subset \mathop{\mathrm{Spec}}(S)$. In other words $S_ g \otimes _ R \kappa (\mathfrak p)$ is not zero. Since $S_{\mathfrak q'}$ is the directed colimit of these $S_ g$ this implies that $S_{\mathfrak q'} \otimes _ R \kappa (\mathfrak p)$ is not zero, see Lemmas 10.9.9 and 10.12.9. Hence $\mathfrak p$ is in the image of $\mathop{\mathrm{Spec}}(S_{\mathfrak q'}) \to \mathop{\mathrm{Spec}}(R)$ as desired. $\square$

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