Proposition 10.35.19. Let $R$ be a Jacobson ring. Let $R \to S$ be a ring map of finite type. Then

1. The ring $S$ is Jacobson.

2. The map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ transforms closed points to closed points.

3. For $\mathfrak m' \subset S$ maximal lying over $\mathfrak m \subset R$ the field extension $\kappa (\mathfrak m')/\kappa (\mathfrak m)$ is finite.

Proof. Let $\mathfrak m' \subset S$ be a maximal ideal and $R \cap \mathfrak m' = \mathfrak m$. Then $R/\mathfrak m \to S/\mathfrak m'$ satisfies the conditions of Lemma 10.35.18 by Lemma 10.35.17. Hence $R/\mathfrak m$ is a field and $\mathfrak m$ a maximal ideal and the induced residue field extension is finite. This proves (2) and (3).

If $S$ is not Jacobson, then by Lemma 10.35.5 there exists a non-maximal prime ideal $\mathfrak q$ of $S$ and an $g \in S$, $g \not\in \mathfrak q$ such that $(S/\mathfrak q)_ g$ is a field. To arrive at a contradiction we show that $\mathfrak q$ is a maximal ideal. Let $\mathfrak p = \mathfrak q \cap R$. Then $R/\mathfrak p \to (S/\mathfrak q)_ g$ satisfies the conditions of Lemma 10.35.18 by Lemma 10.35.17. Hence $R/\mathfrak p$ is a field and the field extension $\kappa (\mathfrak p) \to (S/\mathfrak q)_ g = \kappa (\mathfrak q)$ is finite, thus algebraic. Then $\mathfrak q$ is a maximal ideal of $S$ by Lemma 10.35.9. Contradiction. $\square$

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