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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: