Example 10.28.14. Let $R$ be a ring and $X = \mathop{\mathrm{Spec}}(R)$. Since closed subsets of $X$ correspond to radical ideas of $R$ (Lemma 10.17.2) 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

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

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 10.28.10 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.

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