Definition 28.6.1. A scheme $S$ is said to be *Jacobson* if its underlying topological space is Jacobson.

## 28.6 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 10.35.1.

Lemma 28.6.2. An affine scheme $\mathop{\mathrm{Spec}}(R)$ is Jacobson if and only if the ring $R$ is Jacobson.

**Proof.**
This is Algebra, Lemma 10.35.4.
$\square$

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

Lemma 28.6.3. 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 28.4.2.

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 28.6.2, 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 28.4.3. Any localization of a Jacobson ring at an element is Jacobson, see Algebra, Lemma 10.35.14. 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 28.4.1.
$\square$

Many schemes used commonly in algebraic geometry are Jacobson, see Morphisms, Lemma 29.16.10. We mention here the following interesting case.

Lemma 28.6.4. Examples of Noetherian Jacobson schemes.

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

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.

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. $\square$

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