The Stacks project

Comment by Lukas Heger of November 12, 2020.

Example 10.27.14. Let $R$ be a ring and $X = \mathop{\mathrm{Spec}}(R)$. Since closed subsets of $X$ correspond to radical ideas of $R$ (Lemma 10.16.2) we see that $X$ is a Noetherian topological space if and only if we have ACC for radical ideals. This holds if and only if every radical ideal is the radical of a finitely generated ideal (details omitted). Let

\[ \mathcal{F} = \{ I \subset R \mid \sqrt{I} = \sqrt{(f_1, \ldots , f_ n)}\text{ for some }n \text{ and }f_1, \ldots , f_ n \in R\} . \]

The reader can show that $\mathcal{F}$ is an Oka family by using the identity

\[ \sqrt{I} = \sqrt{(I, a)(I : a)} \]

which holds for any ideal $I \subset R$ and any element $a \in R$. On the other hand, if we have a totally ordered chain of ideals $\{ I_\alpha \} $ none of which are in $\mathcal{F}$, then the union $I = \bigcup I_\alpha $ cannot be in $\mathcal{F}$ either. Otherwise $\sqrt{I} = \sqrt{(f_1, \ldots , f_ n)}$, then $f_ i^ e \in I$ for some $e$, then $f_ i^ e \in I_\alpha $ for some $\alpha $ independent of $i$, then $\sqrt{I_\alpha } = \sqrt{(f_1, \ldots , f_ n)}$, contradiction. Thus if the set of ideals not in $\mathcal{F}$ is nonempty, then it has maximal elements and exactly as in Lemma 10.27.10 we conclude that $X$ is a Noetherian topological space if and only if every prime ideal of $R$ is equal to $\sqrt{(f_1, \ldots , f_ n)}$ for some $f_1, \ldots , f_ n \in R$. If we ever need this result we will carefully state and prove this result here.


Comments (0)

There are also:

  • 6 comment(s) on Section 10.27: A meta-observation about prime ideals

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Your email address will not be published.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 0G2Z. Beware of the difference between the letter 'O' and the digit '0'.