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.
$\square$

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