Lemma 10.35.6. The ring $\mathbf{Z}$ is a Jacobson ring. More generally, let $R$ be a ring such that
$R$ is a domain,
$R$ is Noetherian,
any nonzero prime ideal is a maximal ideal, and
$R$ has infinitely many maximal ideals.
Then $R$ is a Jacobson ring.
Proof. Let $R$ satisfy (1), (2), (3) and (4). The statement means that $(0) = \bigcap _{\mathfrak m \subset R} \mathfrak m$. Since $R$ has infinitely many maximal ideals it suffices to show that any nonzero $x \in R$ is contained in at most finitely many maximal ideals, in other words that $V(x)$ is finite. By Lemma 10.17.7 we see that $V(x)$ is homeomorphic to $\mathop{\mathrm{Spec}}(R/xR)$. By assumption (3) every prime of $R/xR$ is minimal and hence corresponds to an irreducible component of $\mathop{\mathrm{Spec}}(R/xR)$ (Lemma 10.26.1). As $R/xR$ is Noetherian, the topological space $\mathop{\mathrm{Spec}}(R/xR)$ is Noetherian (Lemma 10.31.5) and has finitely many irreducible components (Topology, Lemma 5.9.2). Thus $V(x)$ is finite as desired. $\square$
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