Lemma 10.29.2. Let $R \to S$ be a finite type ring map. Denote $X = \mathop{\mathrm{Spec}}(R)$ and $Y = \mathop{\mathrm{Spec}}(S)$. Write $f : Y \to X$ the induced map of spectra. Let $E \subset Y = \mathop{\mathrm{Spec}}(S)$ be a constructible set. If a point $\xi \in X$ is in $f(E)$, then $\overline{\{ \xi \} } \cap f(E)$ contains an open dense subset of $\overline{\{ \xi \} }$.

**Proof.**
Let $\xi \in X$ be a point of $f(E)$. Choose a point $\eta \in E$ mapping to $\xi $. Let $\mathfrak p \subset R$ be the prime corresponding to $\xi $ and let $\mathfrak q \subset S$ be the prime corresponding to $\eta $. Consider the diagram

By Lemma 10.28.2 the set $E \cap Y'$ is constructible in $Y'$. It follows that we may replace $X$ by $X'$ and $Y$ by $Y'$. Hence we may assume that $R \subset S$ is an inclusion of domains, $\xi $ is the generic point of $X$, and $\eta $ is the generic point of $Y$. By Lemma 10.29.1 combined with Chevalley's theorem (Theorem 10.28.9) we see that there exist dense opens $U \subset X$, $V \subset Y$ such that $f(V) \subset U$ and such that $f : V \to U$ maps constructible sets to constructible sets. Note that $E \cap V$ is constructible in $V$, see Topology, Lemma 5.15.4. Hence $f(E \cap V)$ is constructible in $U$ and contains $\xi $. By Topology, Lemma 5.15.15 we see that $f(E \cap V)$ contains a dense open $U' \subset U$. $\square$

## 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Your email address will not be published.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)