Lemma 10.28.1. Let $R$ be a ring. For a principal ideal $J \subset R$, and for any ideal $I \subset J$ we have $I = J (I : J)$.
10.28 A meta-observation about prime ideals
This section is taken from the CRing project. Let $R$ be a ring and let $S \subset R$ be a multiplicative subset. A consequence of Lemma 10.17.5 is that an ideal $I \subset R$ maximal with respect to the property of not intersecting $S$ is prime. The reason is that $I = R \cap \mathfrak m$ for some maximal ideal $\mathfrak m$ of the ring $S^{-1}R$. It turns out that for many properties of ideals, the maximal ones are prime. A general method of seeing this was developed in [Lam-Reyes]. In this section, we digress to explain this phenomenon.
Let $R$ be a ring. If $I$ is an ideal of $R$ and $a \in R$, we define
\[ (I : a) = \left\{ x \in R \mid xa \in I\right\} . \]
More generally, if $J \subset R$ is an ideal, we define
\[ (I : J) = \left\{ x \in R \mid xJ \subset I\right\} . \]
Proof. Say $J = (a)$. Then $(I : J) = (I : a)$. Since $I \subset J$ we see that any $y \in I$ is of the form $y = xa$ for some $x \in (I : a)$. Hence $I \subset J (I : J)$. Conversely, if $x \in (I : a)$, then $xJ = (xa) \subset I$, which proves the other inclusion. $\square$
Let $\mathcal{F}$ be a collection of ideals of $R$. We are interested in conditions that will guarantee that the maximal elements in the complement of $\mathcal{F}$ are prime.
Definition 10.28.2. Let $R$ be a ring. Let $\mathcal{F}$ be a set of ideals of $R$. We say $\mathcal{F}$ is an Oka family if $R \in \mathcal{F}$ and whenever $I \subset R$ is an ideal and $(I : a), (I, a) \in \mathcal{F}$ for some $a \in R$, then $I \in \mathcal{F}$.
Let us give some examples of Oka families. The first example is the basic example discussed in the introduction to this section.
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.
Proposition 10.28.8. If $\mathcal{F}$ is an Oka family of ideals, then any maximal element of the complement of $\mathcal{F}$ is prime.
Proof. Suppose $I \not\in \mathcal{F}$ is maximal with respect to not being in $\mathcal{F}$ but $I$ is not prime. Note that $I \not= R$ because $R \in \mathcal{F}$. Since $I$ is not prime we can find $a, b \in R - I$ with $ab \in I$. It follows that $(I, a) \neq I$ and $(I : a)$ contains $b \not\in I$ so also $(I : a) \neq I$. Thus $(I : a), (I, a)$ both strictly contain $I$, so they must belong to $\mathcal{F}$. By the Oka condition, we have $I \in \mathcal{F}$, a contradiction. $\square$
At this point we are able to turn most of the examples above into a lemma about prime ideals in a ring.
Lemma 10.28.9. Let $R$ be a ring. Let $S$ be a multiplicative subset of $R$. An ideal $I \subset R$ which is maximal with respect to the property that $I \cap S = \emptyset $ is prime.
Proof. This is the example discussed in the introduction to this section. For an alternative proof, combine Example 10.28.3 with Proposition 10.28.8. $\square$
Lemma 10.28.10. Let $R$ be a ring.
An ideal $I \subset R$ maximal with respect to not being finitely generated is prime.
If every prime ideal of $R$ is finitely generated, then every ideal of $R$ is finitely generated1.
Proof. The first assertion is an immediate consequence of Example 10.28.4 and Proposition 10.28.8. For the second, suppose that there exists an ideal $I \subset R$ which is not finitely generated. The union of a totally ordered chain $\left\{ I_\alpha \right\} $ of ideals that are not finitely generated is not finitely generated; indeed, if $I = \bigcup I_\alpha $ were generated by $a_1, \ldots , a_ n$, then all the generators would belong to some $I_\alpha $ and would consequently generate it. By Zorn's lemma, there is an ideal maximal with respect to being not finitely generated. By the first part this ideal is prime. $\square$
Lemma 10.28.11. Let $R$ be a ring.
An ideal $I \subset R$ maximal with respect to not being principal is prime.
If every prime ideal of $R$ is principal, then every ideal of $R$ is principal.
Proof. The first part follows from Example 10.28.5 and Proposition 10.28.8. For the second, suppose that there exists an ideal $I \subset R$ which is not principal. The union of a totally ordered chain $\left\{ I_\alpha \right\} $ of ideals that not principal is not principal; indeed, if $I = \bigcup I_\alpha $ were generated by $a$, then $a$ would belong to some $I_\alpha $ and $a$ would generate it. By Zorn's lemma, there is an ideal maximal with respect to not being principal. This ideal is necessarily prime by the first part. $\square$
Lemma 10.28.12. Let $R$ be a ring.
An ideal maximal among the ideals which do not contain a nonzerodivisor is prime.
If $R$ is nonzero and every nonzero prime ideal in $R$ contains a nonzerodivisor, then $R$ is a domain.
Proof. Consider the set $S$ of nonzerodivisors. It is a multiplicative subset of $R$. Hence any ideal maximal with respect to not intersecting $S$ is prime, see Lemma 10.28.9. Thus, if every nonzero prime ideal contains a nonzerodivisor, then $(0)$ is prime, i.e., $R$ is a domain. $\square$
Remark 10.28.13. Let $R$ be a ring. Let $\kappa $ be an infinite cardinal. By applying Example 10.28.6 and Proposition 10.28.8 we see that any ideal maximal with respect to the property of not being generated by $\kappa $ elements is prime. This result is not so useful because there exists a ring for which every prime ideal of $R$ can be generated by $\aleph _0$ elements, but some ideal cannot. Namely, let $k$ be a field, let $T$ be a set whose cardinality is greater than $\aleph _0$ and let
\[ R = k[\{ x_ n\} _{n \geq 1}, \{ z_{t, n}\} _{t \in T, n \geq 0}]/ (x_ n^2, z_{t, n}^2, x_ n z_{t, n} - z_{t, n - 1}) \]
This is a local ring with unique prime ideal $\mathfrak m = (x_ n)$. But the ideal $(z_{t, n})$ cannot be generated by countably many elements.
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
\[ \mathcal{F} = \{ I \subset R \mid \sqrt{I} = \sqrt{(f_1, \ldots , f_ n)}\text{ for some }n \text{ and }f_1, \ldots , f_ n \in R\} . \]
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 10.28.10 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.
[1] Later we will say that $R$ is Noetherian.
Comments (6)
Comment #4916 by Stephan Snegirov on
Comment #5186 by Johan on
Comment #5606 by Lukas Heger on
Comment #5758 by Johan on
Comment #5791 by Lukas Heger on
Comment #5806 by Johan on