Lemma 28.6.3. 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 28.4.2.

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 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. $\square$

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