
Lemma 10.34.5. Let $R$ be a ring. If $R$ is not Jacobson there exist a prime $\mathfrak p \subset R$, an element $f \in R$ such that the following hold

1. $\mathfrak p$ is not a maximal ideal,

2. $f \not\in \mathfrak p$,

3. $V(\mathfrak p) \cap D(f) = \{ \mathfrak p\}$, and

4. $(R/\mathfrak p)_ f$ is a field.

On the other hand, if $R$ is Jacobson, then for any pair $(\mathfrak p, f)$ such that (1) and (2) hold the set $V(\mathfrak p) \cap D(f)$ is infinite.

Proof. Assume $R$ is not Jacobson. By Lemma 10.34.4 this means there exists an closed subset $T \subset \mathop{\mathrm{Spec}}(R)$ whose set $T_0 \subset T$ of closed points is not dense in $T$. Choose an $f \in R$ such that $T_0 \subset V(f)$ but $T \not\subset V(f)$. Note that $T \cap D(f)$ is homeomorphic to $\mathop{\mathrm{Spec}}((R/I)_ f)$ if $T = V(I)$, see Lemmas 10.16.7 and 10.16.6. As any ring has a maximal ideal (Lemma 10.16.2) we can choose a closed point $t$ of space $T \cap D(f)$. Then $t$ corresponds to a prime ideal $\mathfrak p \subset R$ which is not maximal (as $t \not\in T_0$). Thus (1) holds. By construction $f \not\in \mathfrak p$, hence (2). As $t$ is a closed point of $T \cap D(f)$ we see that $V(\mathfrak p) \cap D(f) = \{ \mathfrak p\}$, i.e., (3) holds. Hence we conclude that $(R/\mathfrak p)_ f$ is a domain whose spectrum has one point, hence (4) holds (for example combine Lemmas 10.17.2 and 10.24.1).

Conversely, suppose that $R$ is Jacobson and $(\mathfrak p, f)$ satisfy (1) and (2). If $V(\mathfrak p) \cap V(f) = \{ \mathfrak p, \mathfrak q_1, \ldots , \mathfrak q_ t\}$ then $\mathfrak p \not= \mathfrak q_ i$ implies there exists an element $g \in R$ such that $g \not\in \mathfrak p$ but $g \in \mathfrak q_ i$ for all $i$. Hence $V(\mathfrak p) \cap D(fg) = \{ \mathfrak p\}$ which is impossible since each locally closed subset of $\mathop{\mathrm{Spec}}(R)$ contains at least one closed point as $\mathop{\mathrm{Spec}}(R)$ is a Jacobson topological space. $\square$

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