Lemma 15.10.5. Let $(A, I)$ be a Zariski pair with $A$ Noetherian. Let $f \in I$. Then $A_ f$ is a Jacobson ring.

**Proof.**
We will use the criterion of Algebra, Lemma 10.61.4. Let $\mathfrak p \subset A$ be a prime ideal such that $\mathfrak p_ f = \mathfrak p A_ f$ is prime and not maximal. We have to show that $A_ f/\mathfrak p_ f = (A/\mathfrak p)_ f$ has infinitely many prime ideals. After replacing $A$ by $A/\mathfrak p$ we may assume $A$ is a domain, $\dim A_ f > 0$, and our goal is to show that $\mathop{\mathrm{Spec}}(A_ f)$ is infinite. Since $\dim A_ f > 0$ we can find a nonzero prime ideal $\mathfrak q \subset A$ not containing $f$. Choose a maximal ideal $\mathfrak m \subset A$ containing $\mathfrak q$. Since $(A, I)$ is a Zariski pair, we see $I \subset \mathfrak m$. Hence $\mathfrak m \not= \mathfrak q$ and $\dim (A_\mathfrak m) > 1$. Hence $\mathop{\mathrm{Spec}}((A_\mathfrak m)_ f) \subset \mathop{\mathrm{Spec}}(A_ f)$ is infinite by Algebra, Lemma 10.61.1 and we win.
$\square$

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