The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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

\[ (I : a) = \left\{ x \in R \mid xa \in I\right\} . \]

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

\[ (I : J) = \left\{ x \in R \mid xJ \subset I\right\} . \]

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 $I = J (I : J)$.

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.

  1. An ideal $I \subset R$ maximal with respect to not being finitely generated is prime.

  2. 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 10.27.4 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.

  1. An ideal $I \subset R$ maximal with respect to not being principal is prime.

  2. 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.

  1. An ideal maximal among the ideals which do not contain a nonzerodivisor is prime.

  2. 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

\[ R = k[\{ x_ n\} _{n \geq 1}, \{ z_{t, n}\} _{t \in T, n \geq 0}]/ (x_ n^2, z_{t, n}^2, x_ n z_{t, n} - z_{t, n - 1}) \]

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.

[1] Later we will say that $R$ is Noetherian.

Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05K7. Beware of the difference between the letter 'O' and the digit '0'.