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. $\square$

There are also:

• 7 comment(s) on Section 10.41: Going up and going down

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).