Lemma 10.34.4. A ring $R$ is Jacobson if and only if $\mathop{\mathrm{Spec}}(R)$ is Jacobson, see Topology, Definition 5.18.1.

**Proof.**
Suppose $R$ is Jacobson. Let $Z \subset \mathop{\mathrm{Spec}}(R)$ be a closed subset. We have to show that the set of closed points in $Z$ is dense in $Z$. Let $U \subset \mathop{\mathrm{Spec}}(R)$ be an open such that $U \cap Z$ is nonempty. We have to show $Z \cap U$ contains a closed point of $\mathop{\mathrm{Spec}}(R)$. We may assume $U = D(f)$ as standard opens form a basis for the topology on $\mathop{\mathrm{Spec}}(R)$. According to Lemma 10.16.2 we may assume that $Z = V(I)$, where $I$ is a radical ideal. We see also that $f \not\in I$. By assumption, there exists a maximal ideal $\mathfrak m \subset R$ such that $I \subset \mathfrak m$ but $f \not\in \mathfrak m$. Hence $\mathfrak m \in D(f) \cap V(I) = U \cap Z$ as desired.

Conversely, suppose that $\mathop{\mathrm{Spec}}(R)$ is Jacobson. Let $I \subset R$ be a radical ideal. Let $J = \cap _{I \subset \mathfrak m} \mathfrak m$ be the intersection of the maximal ideals containing $I$. Clearly $J$ is a radical ideal, $V(J) \subset V(I)$, and $V(J)$ is the smallest closed subset of $V(I)$ containing all the closed points of $V(I)$. By assumption we see that $V(J) = V(I)$. But Lemma 10.16.2 shows there is a bijection between Zariski closed sets and radical ideals, hence $I = J$ as desired. $\square$

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