Let $A$ be a ring (commutative unital), and $\mathcal I$ be a nonempty family of proper ideals of $A$.

I will say that $\mathcal I$ has property $\dagger$ if for any $\mathfrak a\in\mathcal I$ and any $xy\in \mathfrak a$, one of $\mathfrak a+(x),\mathfrak a+(y)$ is in $\mathcal I$.

In particular, any maximal (w.r.t. inclusion) element of $\mathcal I$ is prime.

Does $\dagger$ (or maybe $\dagger$ + hypothesis of Zorn's lemma) have a name (a prime family, perhaps, as I suggest in the title)? As a side question, are there some interesting criteria for checking that a given family of ideals has property $\dagger$?

It seems to me that posets with property $\dagger$ are rather abundant in commutative algebra (e.g. in proof of Cohen's characterization of Noetherian rings), but I've yet to see $\dagger$ discussed on its own, or any name for it.


$\newcommand{\F}{\mathcal{F}}$ $\newcommand{\I}{\mathcal{I}}$ $\newcommand{\a}{\mathfrak{a}}$ In a paper titled A Prime Ideal Principle in commutative algebra, T.Y. Lam and I defined a family $\F$ of ideals of $A$ to be an Ako family if $A \in \F$ and, for all ideals $I$ of $A$ and all $x,y \in A$, the following implication holds: $$I+(x),I+(y) \in \F \implies I+(xy) \in \F.$$ We were interested in these for precisely the kind of observation you make: an ideal that is maximal with respect to not being an element of $\F$ is prime. (This is part of the "Prime Ideal Principle" in the title of the paper.)

Now I will show that a (possibly empty) family $\I$ satisfies your property $\dagger$ if and only if the complement $\F = \I^c$ (taken in the set of all ideals of $A$) is an Ako family. (Notice that your property that $\I$ consist of proper ideals ensures that $A \in \I^c = \F$.)

First, suppose that $\I$ satisfies $\dagger$. To prove $\F = \I^c$ is Ako, let $I$ be an ideal and $x,y \in A$ such that $I+(x), I+(y) \in \F$. Assume for contradiction that the ideal $\a = I+(xy)$ is not in $\F$, so that $\mathfrak{a} \in \mathcal{I}$. By property $\dagger$, one of $\mathfrak{a}+(x) = I + (xy) + (x) = I+(x)$ or $\mathfrak{a}+(y) = I+(xy)+(y) = I+(y)$ is an element of $I = \mathcal{F}^c$, contradicting the assumptions. Thus $\mathcal{F}$ is an Ako family.

Conversely, suppose that $\F = \I^c$ is an Ako family. To prove that $\I$ satisfies $\dagger$, let $\a\in\I$ and let $x,y \in A$ be elements with $xy \in \a$. Assume for contradiction that neither of $\a+(x),\a+(y)$ is an element of $\I$. Then $\a+(x),\a+(y) \in \F$, and the Ako property implies that $\a+(xy) \in \F$. But $xy \in \a$, so $\a = \a+(xy) \in \F = \I^c$, a contradiction.

You're right that these are very common families in commutative algebra. We investigated a related kind of ideal family, which we called an Oka family in honor of a the mathematician Kiyoshi Oka, who worked in several complex variables. (In fact, the term "Ako" was playfully chosen as the revers of the word "Oka.") Oka families seem to be even more ubiquitous in commutative algebra, since they are rather easy to construct if one thinks module-theoretically. Families of ideals that are Oka but not Ako are rather easy to come by. On the other hand, it seems to require much more care to construct an example of an Ako family that is not Oka; this was accomplished in a subsequent paper, Oka and Ako ideal families in commutative rings.

  • $\begingroup$ See also Magidin's answer here. $\endgroup$ – user26857 Feb 15 '13 at 22:11

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