
Lemma 10.34.21. Let $R \to S$ be a finite type ring map of Jacobson rings. 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. Denote with a subscript ${}_0$ the set of closed points of a topological space.

1. We have $f(E)_0 = f(E_0) = X_0 \cap f(E)$.

2. A point $\xi \in X$ is in $f(E)$ if and only if $\overline{\{ \xi \} } \cap f(E_0)$ is dense in $\overline{\{ \xi \} }$.

Proof. We have a commutative diagram of continuous maps

$\xymatrix{ E \ar[r] \ar[d] & Y \ar[d] \\ f(E) \ar[r] & X }$

Suppose $x \in f(E)$ is closed in $f(E)$. Then $f^{-1}(\{ x\} )\cap E$ is nonempty and closed in $E$. Applying Topology, Lemma 5.18.5 to both inclusions

$f^{-1}(\{ x\} ) \cap E \subset E \subset Y$

we find there exists a point $y \in f^{-1}(\{ x\} ) \cap E$ which is closed in $Y$. In other words, there exists $y \in Y_0$ and $y \in E_0$ mapping to $x$. Hence $x \in f(E_0)$. This proves that $f(E)_0 \subset f(E_0)$. Proposition 10.34.19 implies that $f(E_0) \subset X_0 \cap f(E)$. The inclusion $X_0 \cap f(E) \subset f(E)_0$ is trivial. This proves the first assertion.

Suppose that $\xi \in f(E)$. According to Lemma 10.29.2 the set $f(E) \cap \overline{\{ \xi \} }$ contains a dense open subset of $\overline{\{ \xi \} }$. Since $X$ is Jacobson we conclude that $f(E) \cap \overline{\{ \xi \} }$ contains a dense set of closed points, see Topology, Lemma 5.18.5. We conclude by part (1) of the lemma.

On the other hand, suppose that $\overline{\{ \xi \} } \cap f(E_0)$ is dense in $\overline{\{ \xi \} }$. By Lemma 10.28.3 there exists a ring map $S \to S'$ of finite presentation such that $E$ is the image of $Y' := \mathop{\mathrm{Spec}}(S') \to Y$. Then $E_0$ is the image of $Y'_0$ by the first part of the lemma applied to the ring map $S \to S'$. Thus we may assume that $E = Y$ by replacing $S$ by $S'$. Suppose $\xi$ corresponds to $\mathfrak p \subset R$. Consider the diagram

$\xymatrix{ S \ar[r] & S/\mathfrak p S \\ R \ar[r] \ar[u] & R/\mathfrak p \ar[u] }$

This diagram and the density of $f(Y_0) \cap V(\mathfrak p)$ in $V(\mathfrak p)$ shows that the morphism $R/\mathfrak p \to S/\mathfrak p S$ satisfies condition (2) of Lemma 10.29.4. Hence we conclude there exists a prime $\overline{\mathfrak q} \subset S/\mathfrak pS$ mapping to $(0)$. In other words the inverse image $\mathfrak q$ of $\overline{\mathfrak q}$ in $S$ maps to $\mathfrak p$ as desired. $\square$

Comment #743 by Wei Xu on

In the proof, "Hence $f^{-1}(\{x\})\cap E$ is constructible, nonempty in $Y$." should be "Hence $f^{-1}(\{x\})\cap E$ is locally closed, nonempty in $Y$.", then everything is ok.

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