The Stacks project

Lemma 10.35.21. 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 10.35.19 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 10.30.2 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 10.29.4 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 10.30.4. 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. $\square$

Comments (2)

Comment #743 by Wei Xu on

In the proof, "Hence is constructible, nonempty in ." should be "Hence is locally closed, nonempty in .", then everything is ok.

There are also:

  • 7 comment(s) on Section 10.35: Jacobson rings

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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00GD. Beware of the difference between the letter 'O' and the digit '0'.