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Suppose $P$ is the set of all subsets of a set $X$ and $P$ is a ring. Let $p$ be an element in $P$ (so that $p$ is a subset of $X$). What does it mean to say "an ideal generated by $p$"? And suppose there is some $q\in P$ then what does it mean to say "an ideal generated by (p,q)"? 

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Usually sets don't have ideals... are you talking about rings? –  Ted Feb 15 '12 at 17:15
    
@Ted: The set of all subsets of set X forms a ring, you are right. –  eugene Feb 15 '12 at 17:29
    
Or are you talking about ideals in the order-theoretic sense, viewing $P$ as a partially ordered set under inclusion? –  Arturo Magidin Feb 15 '12 at 17:36
    
@ArturoMagidin: I am not familiar with the entity you have described, so I guess it isn't that. I have edited the question. P is supposed to be a ring. –  eugene Feb 15 '12 at 17:39
    
@eugene If you give a reference to the text that you seek to comprehend, then we can use this to help supply an optimal answer. Lacking such, we are forced to guess the context, which may result in non-optimal replies. –  Math Gems Feb 15 '12 at 18:12

2 Answers 2

$P$ is a Boolean algebra (with $\cup = \lor$ and $\cap = \land$) and there is a notion of ideal there: an ideal $I$ of elements from $P$ obeys:

(1) $\forall p,q \in I : p \lor q \in I$

(2) $\forall p \in I, \forall q \in P: q \le p \rightarrow q \in I$

Which translates in your case as: an ideal is a collection of subsets closed under finite unions and subsets.

So the ideal generated by an element $p$ must contain all subsets of $p$, and then note that the collection of subsets of $p$ is closed under unions. So the ideal generated by $p$ equals $\{ q \mid q \subset p \}$.

The ideal generated by $p$ and $q$ must contain $p \cup q$ and all of its subsets, and one checks easily that it indeed is an ideal. So the ideal generated by $(p,q)$ equals $\{ (r \in P \mid r \subset p \cup q \}$

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Thanks, Henno, what I don't understand is I thought an ideal of a ring has to satisfy the strong closure property, so for all $ x \in (p) $ and $y \in P$ we must have $x\cdot y\in (p)$, where $\cdot$ is some binary operation combining the 2 elements, which I don't know what it is since the elements of $P$ are sets –  eugene Feb 15 '12 at 17:35
    
$P$ is a ring, btw. Sorry for leaving that out. –  eugene Feb 15 '12 at 17:40
    
Well, this condition is true for the operation = intersection; and $P$ is a (algebraic) ring with operations $\cdot$ intersection and $+$ is symmetric difference. And the above notion of ideal is also a ring theoretic ideal in that sense with these operations. –  Henno Brandsma Feb 15 '12 at 18:14

It could be that $P$ is considered as a Boolean ring (or algebra). In this case the ideals are the same as the order ideals of $P$. These are subsets of $P$ (collections of subsets of $X$) closed under finite unions and under intersections with arbitrary elements of $P$ (the latter menas they are lower sets: with every part of $X$ that is in the ideal, all smaller parts are also in it). An ideal generated by a finite number of sets in $P$ is the collection of all the subsets of their union; Henno Bransdma described what this means for ideals generated by one or two elements of $P$.

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