
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

$\xymatrix{ \eta \ar[r] \ar@{|->}[d] & E \cap Y' \ar[r] \ar[d] & Y' = \mathop{\mathrm{Spec}}(S/\mathfrak q) \ar[r] \ar[d] & Y \ar[d] \\ \xi \ar[r] & f(E) \cap X' \ar[r] & X' = \mathop{\mathrm{Spec}}(R/\mathfrak p) \ar[r] & X }$

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 $U' \subset U$. $\square$

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