Filters in a boolean ring?

Question

It is well known that boolean algebras are equivalent to boolean rings, and that furthermore order theoretic ideals in a boolean algebra $B$ correspond to ideals in $B$ seen as a boolean ring. It is natural to then ask what do order theoretic filters correspond to, when seeing them from a ring theory perspective.

A simple answer is "ideals in the ring $B^{op}$". After unravelling that, we see that a filter $F$ in a boolean ring $B$ is a non-empty set such that:

1. $a,b \in F \implies 1+a+b \in F$. (subgroup condition)
2. $a \in F, b \in B \implies a + ab + b \in F$. (absorption condition)

Is this definition relevant in general? Can this be restated in a nice/intuitive way using common ring theory concepts? A related question is, are there any interesting relations between $B$ and $B^{op}$ from a ring theory point of view?

Answer 1

A filter in a Boolean ring is exactly a saturated multiplicatively closed set.

Every filter $F$ is a multiplicatively closed set, since $\top = 1\in F$ and if $a,b\in F$, then $a\wedge b = ab \in F$. Indeed, a filter is exactly an upwards-closed multiplicative set. So it remains to show that a multiplicatively closed set is saturated if and only if it is upwards closed.

Suppose $F$ is upwards closed. If $ab\in F$, then since $ab\leq a$, $a\in F$, and similarly $b\in F$. So $F$ is saturated. Conversely, suppose $F$ is a saturated. If $a\in F$ and $a\leq b$, then $ab = a \in F$, so $b\in F$ by saturation. Thus $F$ is upwards closed.

Answer 2

If you interpret it directly in terms of $B$ first, it seems like some species of multiplicatively closed sets.

The first condition $a,b\in F\implies ab\in F$ makes it multiplicative, of course.

For the second condition, $a\leq r\implies r\in F$: it can be interpreted as $ar=a\implies r\in F$, which I guess means $F$ contains everything that acts like an identity on one of its elements.

A saturated multiplicatively closed set of $B$ would satisfy this, for example.

Neither of the conditions as you stated them ring any bells for me, ring theoretically.

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I erased an earlier comment talking about partially ordered rings because I realized that the order in boolean algebras given by $a\leq b \iff ab=a$ isn't compatible with multiplication the way it has to be for a partially ordered ring. But nevertheless it is interesting to talk about rings with a preorder on elements given by $a\leq b\iff (a)\subseteq (b)$.

I not sure if there are any tricks for defining a filter on a preorder. This is what I was looking at. But if one can talk about a filter $F$ in this preorder (commutative rings only), then we are lucky that $(a)\cap (b)=(ab)$ so that $F$ is multiplicatively closed (a.k.a. closed under meets, in this case). Secondly, the condition that $a\leq r\implies r\in F$ can be rewritten as $rs\in F\implies r\in F$, which is the condition for being saturated.

So it really seems that they are related to saturated multiplicative sets. But maybe I have overlooked something subtle about the preorder or the potential difficulties with the definition of a filter for preorders. I look forward to feedback.