
Lemma 10.60.4. Noetherian Jacobson rings.

1. Any Noetherian domain $R$ of dimension $1$ with infinitely many primes is Jacobson.

2. Any Noetherian ring such that every prime $\mathfrak p$ is either maximal or contained in infinitely many prime ideals is Jacobson.

Proof. Part (1) is a reformulation of Lemma 10.34.6.

Let $R$ be a Noetherian ring such that every non-maximal prime $\mathfrak p$ is contained in infinitely many prime ideals. Assume $\mathop{\mathrm{Spec}}(R)$ is not Jacobson to get a contradiction. By Lemmas 10.25.1 and 10.30.5 we see that $\mathop{\mathrm{Spec}}(R)$ is a sober, Noetherian topological space. By Topology, Lemma 5.18.3 we see that there exists a non-maximal ideal $\mathfrak p \subset R$ such that $\{ \mathfrak p\}$ is a locally closed subset of $\mathop{\mathrm{Spec}}(R)$. In other words, $\mathfrak p$ is not maximal and $\{ \mathfrak p\}$ is an open subset of $V(\mathfrak p)$. Consider a prime $\mathfrak q \subset R$ with $\mathfrak p \subset \mathfrak q$. Recall that the topology on the spectrum of $(R/\mathfrak p)_{\mathfrak q} = R_{\mathfrak q}/\mathfrak pR_{\mathfrak q}$ is induced from that of $\mathop{\mathrm{Spec}}(R)$, see Lemmas 10.16.5 and 10.16.7. Hence we see that $\{ (0)\}$ is a locally closed subset of $\mathop{\mathrm{Spec}}((R/\mathfrak p)_{\mathfrak q})$. By Lemma 10.60.1 we conclude that $\dim ((R/\mathfrak p)_{\mathfrak q}) = 1$. Since this holds for every $\mathfrak q \supset \mathfrak p$ we conclude that $\dim (R/\mathfrak p) = 1$. At this point we use the assumption that $\mathfrak p$ is contained in infinitely many primes to see that $\mathop{\mathrm{Spec}}(R/\mathfrak p)$ is infinite. Hence by part (1) of the lemma we see that $V(\mathfrak p) \cong \mathop{\mathrm{Spec}}(R/\mathfrak p)$ is the closure of its closed points. This is the desired contradiction since it means that $\{ \mathfrak p\} \subset V(\mathfrak p)$ cannot be open. $\square$

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