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

## 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.

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:

The scheme $X$ is Jacobson.

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

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

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

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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