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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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