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.
Comments (0)
There are also: