Ideals and filters

The notions of a filter and an ideal on a poset make intuitive sense to me, and I can understand why they are dual:

A subset $I\subset P$ of a poset $P$ is an ideal if:

• for all $x\in I$, $y\leq x$ implies $y\in I$
• for all $x,y\in I$ there exists $z\in I$ with $x\leq z$ and $y\leq z$

and a filter is the same thing with all inequalities reversed.

I feel like this should correspond to the notion of a ring ideal, where for a ring $R$ we have $I\subset R$ being a ring ideal if:

• for all $x,y\in I$ we have $x+y\in I$
• for all $x\in I$ and $r\in R$ we have $rx\in I$ and $xr\in I$

but I would like some clarification on this point. Following on, my main question is: is there a corresponding notion of a 'ring filter' which is dual to the notion of a ring ideal in the same way that a filter in a poset is dual to an ideal? Or is there no relation at all except for a coincidence in naming?

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The naming isn't a coincidence. An ideal in a Boolean ring is the same thing as an order ideal in the associated Boolean poset, where you can define the order relation by $a \le b$ if $a = ab$. But I don't see any reason to expect a well-behaved associated notion of filter for general rings.
They are also the possible "kernels" of homomorphisms between lattices with $0$ elements, so in that sense, they generalize the notion for rings. But yes, I doubt there is a notion of filter on a general ring. – Thomas Andrews May 25 '11 at 22:24
@Chris: okay, so I guess you can define a "ring filter" to be an ideal of the opposite ring $R^{op}$ (by analogy with a filter being an ideal of the opposite poset), but I don't really see a point to doing this, especially since you're talking about two-sided ideals: then a "ring filter" is still just a two-sided ideal. If you talk about left ideals, then a "left ring filter" is just a right ideal. – Qiaochu Yuan May 26 '11 at 9:18