The Stacks project

Example 10.28.3. Let $R$ be a ring and let $S$ be a multiplicative subset of $R$. We claim that $\mathcal{F} = \{ I \subset R \mid I \cap S \not= \emptyset \}$ is an Oka family. Namely, suppose that $(I : a), (I, a) \in \mathcal{F}$ for some $a \in R$. Then pick $s \in (I, a) \cap S$ and $s' \in (I : a) \cap S$. Then $ss' \in I \cap S$ and hence $I \in \mathcal{F}$. Thus $\mathcal{F}$ is an Oka family.

Example 10.28.4. Let $R$ be a ring, $I \subset R$ an ideal, and $a \in R$. If $(I : a)$ is generated by $a_1, \ldots , a_ n$ and $(I, a)$ is generated by $a, b_1, \ldots , b_ m$ with $b_1, \ldots , b_ m \in I$, then $I$ is generated by $aa_1, \ldots , aa_ n, b_1, \ldots , b_ m$. To see this, note that if $x \in I$, then $x \in (I, a)$ is a linear combination of $a, b_1, \ldots , b_ m$, but the coefficient of $a$ must lie in $(I:a)$. As a result, we deduce that the family of finitely generated ideals is an Oka family.

Example 10.28.5. Let us show that the family of principal ideals of a ring $R$ is an Oka family. Indeed, suppose $I \subset R$ is an ideal, $a \in R$, and $(I, a)$ and $(I : a)$ are principal. Note that $(I : a) = (I : (I, a))$. Setting $J = (I, a)$, we find that $J$ is principal and $(I : J)$ is too. By Lemma 10.28.1 we have $I = J (I : J)$. Thus we find in our situation that since $J = (I, a)$ and $(I : J)$ are principal, $I$ is principal.

Example 10.28.6. Let $R$ be a ring. Let $\kappa$ be an infinite cardinal. The family of ideals which can be generated by at most $\kappa$ elements is an Oka family. The argument is analogous to the argument in Example 10.28.4 and is omitted.

Example 10.28.7. Let $A$ be a ring, $I \subset A$ an ideal, and $a \in A$ an element. There is a short exact sequence $0 \to A/(I : a) \to A/I \to A/(I, a) \to 0$ where the first arrow is given by multiplication by $a$. Thus if $P$ is a property of $A$-modules that is stable under extensions and holds for $0$, then the family of ideals $I$ such that $A/I$ has $P$ is an Oka family.

Example 10.28.14.reference 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.