# A prime poset of ideals

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.