1. ## Principal Ideal Ring

Let X be a set then $(\matchal{P}(X),\Delta,\cap)$ forms a commutative ring with unity X. Although not a principal ideal domain if X has at least two members, I was wondering if it was necessarily a principal ideal ring.
Namely if I was an ideal whether $I=(\bigcup I)$ necessarily.
I think it's obvious if X is a finite set, but for infinite X I can't decide as it's closed only under finite symmetric differences.

This is just for personal curiosity.

2. ## Re: Principal Ideal Ring

In fact, it is true for finite X. $A,B\in I \Rightarrow (A\Delta B)\Delta(A\cap B) = A\cup B \in I$. And since there can only be a finite number of incomparable sets for any subset I of the poset $(\mathcal{P}(X),\subseteq)$ we will eventually reach $\bigcup I$

3. ## Re: Principal Ideal Ring

To any interested, I've worked it out I believe. We always have $\mathcal{I}\subseteq\big(\bigcup \mathcal{I}\big)$ at least. Also for any set S in the powerset, the ideal it generates is precisely the family of its subsets.
$(S)=\{Z\in\mathcal{P}(X):Z\subseteq S\}$
So let's have X be infinite. And consider an arbitrary ideal I. Let Y be in $\big(\bigcup \mathcal{I}\big)$ to try to determine if it is in I. Since Y is contained in the union, there is a family that is a subset of I such that each member of the family has non-empty intersection with Y. That is

$\exists\mathcal{F}\subseteq\mathcal{I}\forall F\in\mathcal{F}(F\cap Y \neq \varnothing)$
And since I is closed under intersection from any subset of X, we discover that any arbitrarily large (but finite) subset of Y is in I. So all the finite subsets of the ideal generated by the arbitrary union of I are in I. But not necessarily any subset of that principal ideal.
To construct a counterexample, let X be the natural numbers $\omega$. Consider the ideal generated by all finite subsets of the natural numbers. It is closed under symmetric difference and intersections from all subsets of the natural numbers. But it is not equal to the whole powerset.
$\mathcal{I}=\{Z\in\mathcal{P}(\omega):Fin(Z)\}$
On the other hand, the union of all finite subsets of the natural numbers is the natural numbers, that is the entire set X. And the ideal generated by X is the entire powerset.
Thus $\mathcal{I} \neq \big(\bigcup\mathcal{I}\big)$

And the answer to my question is, this ring need not have the principal ideal property if X is infinite.

4. ## Re: Principal Ideal Ring

I realized that I didn't fully prove that the ring above isn't a principal ideal ring. I just demonstrated that the arbitrary union of the ideal fails to generate the ideal. I should have shown that any element of the ideal fails to generate it. But that's easy to fix. Suppose there was some set that generated I, call it S. It would have to be in I then. Since it's in I, it's finite. Since it's finite, it has a largest element. Call it M. We also know that the singleton of the successor of M is finite and hence in I. Since I is an ideal we have that S \delta {M+} is in I. But this set is not in (S) as it is not a subset of S. Thus S fails to generate I.