The Stacks project

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GED. Beware of the difference between the letter 'O' and the digit '0'.