Lemma 28.6.4. Examples of Noetherian Jacobson schemes.

1. If $(R, \mathfrak m)$ is a Noetherian local ring, then the punctured spectrum $\mathop{\mathrm{Spec}}(R) \setminus \{ \mathfrak m\}$ is a Jacobson scheme.

2. If $R$ is a Noetherian ring with Jacobson radical $\text{rad}(R)$ then $\mathop{\mathrm{Spec}}(R) \setminus V(\text{rad}(R))$ is a Jacobson scheme.

3. If $(R, I)$ is a Zariski pair (More on Algebra, Definition 15.10.1) with $R$ Noetherian, then $\mathop{\mathrm{Spec}}(R) \setminus V(I)$ is a Jacobson scheme.

Proof. Proof of (3). Observe that $\mathop{\mathrm{Spec}}(R) - V(I)$ has a covering by the affine opens $\mathop{\mathrm{Spec}}(R_ f)$ for $f \in I$. The rings $R_ f$ are Jacobson by More on Algebra, Lemma 15.10.5. Hence $\mathop{\mathrm{Spec}}(R) \setminus V(I)$ is Jacobson by Lemma 28.6.3. Parts (1) and (2) are special cases of (3).

Direct proof of case (1). Since $\mathop{\mathrm{Spec}}(R)$ is a Noetherian scheme, $S$ is a Noetherian scheme (Lemma 28.5.6). Hence $S$ is a sober, Noetherian topological space (use Schemes, Lemma 26.11.1). Assume $S$ is not Jacobson to get a contradiction. By Topology, Lemma 5.18.3 there exists some non-closed point $\xi \in S$ such that $\{ \xi \}$ is locally closed. This corresponds to a prime $\mathfrak p \subset R$ such that (1) there exists a prime $\mathfrak q$, $\mathfrak p \subset \mathfrak q \subset \mathfrak m$ with both inclusions strict, and (2) $\{ \mathfrak p\}$ is open in $\mathop{\mathrm{Spec}}(R/\mathfrak p)$. This is impossible by Algebra, Lemma 10.61.1. $\square$

Comment #2435 by Matthew Emerton on

A minor comment: In the first sentence of the proof, having both Since and hence is redundant.

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