The Stacks project

27.6 Jacobson schemes

Recall that a space is said to be Jacobson if the closed points are dense in every closed subset, see Topology, Section 5.18.

Definition 27.6.1. A scheme $S$ is said to be Jacobson if its underlying topological space is Jacobson.

Recall that a ring $R$ is Jacobson if every radical ideal of $R$ is the intersection of maximal ideals, see Algebra, Definition 10.34.1.

Lemma 27.6.2. An affine scheme $\mathop{\mathrm{Spec}}(R)$ is Jacobson if and only if the ring $R$ is Jacobson.

Proof. This is Algebra, Lemma 10.34.4. $\square$

Here is the standard result characterizing Jacobson schemes. Intuitively it claims that Jacobson $\Leftrightarrow $ locally Jacobson.

Lemma 27.6.3. Let $X$ be a scheme. The following are equivalent:

  1. The scheme $X$ is Jacobson.

  2. The scheme $X$ is “locally Jacobson” in the sense of Definition 27.4.2.

  3. For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is Jacobson.

  4. There exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is Jacobson.

  5. There exists an open covering $X = \bigcup X_ j$ such that each open subscheme $X_ j$ is Jacobson.

Moreover, if $X$ is Jacobson then every open subscheme is Jacobson.

Proof. The final assertion of the lemma holds by Topology, Lemma 5.18.5. The equivalence of (5) and (1) is Topology, Lemma 5.18.4. Hence, using Lemma 27.6.2, we see that (1) $\Leftrightarrow $ (2). To finish proving the lemma it suffices to show that “Jacobson” is a local property of rings, see Lemma 27.4.3. Any localization of a Jacobson ring at an element is Jacobson, see Algebra, Lemma 10.34.14. Suppose $R$ is a ring, $f_1, \ldots , f_ n \in R$ generate the unit ideal and each $R_{f_ i}$ is Jacobson. Then we see that $\mathop{\mathrm{Spec}}(R) = \bigcup D(f_ i)$ is a union of open subsets which are all Jacobson, and hence $\mathop{\mathrm{Spec}}(R)$ is Jacobson by Topology, Lemma 5.18.4 again. This proves the second property of Definition 27.4.1. $\square$

Many schemes used commonly in algebraic geometry are Jacobson, see Morphisms, Lemma 28.15.10. We mention here the following interesting case.

Lemma 27.6.4. Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak m$. In this case the scheme $S = \mathop{\mathrm{Spec}}(R) \setminus \{ \mathfrak m\} $ is Jacobson.

Proof. Since $\mathop{\mathrm{Spec}}(R)$ is a Noetherian scheme, $S$ is a Noetherian scheme (Lemma 27.5.6). Hence $S$ is a sober, Noetherian topological space (use Schemes, Lemma 25.11.1). 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 10.60.1. $\square$


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