Lemma 10.27.1. Let $R$ be a ring. For a principal ideal $J \subset R$, and for any ideal $I \subset J$ we have $I = J (I : J)$.

## 10.27 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 10.16.5 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

More generally, if $J \subset R$ is an ideal, we define

**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 10.27.2. 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 10.27.3. 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 10.27.4. 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 10.27.5. 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 10.27.1 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 10.27.6. 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 10.27.4 and is omitted.

Proposition 10.27.7. 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 10.27.8. 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 10.27.3 with Proposition 10.27.7.
$\square$

Lemma 10.27.9. 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 generated

^{1}.

**Proof.**
The first assertion is an immediate consequence of Example 10.27.4 and Proposition 10.27.7. 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 10.27.10. 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 10.27.5 and Proposition 10.27.7. 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 10.27.11. 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 10.27.8. Thus, if every nonzero prime ideal contains a nonzerodivisor, then $(0)$ is prime, i.e., $R$ is a domain.
$\square$

Remark 10.27.12. Let $R$ be a ring. Let $\kappa $ be an infinite cardinal. By applying Example 10.27.6 and Proposition 10.27.7 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

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.

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## Comments (1)

Comment #4916 by Stephan Snegirov on