Is every boolean algebra isomorphic to the quotient of a powerset algebra? Let $\mathbb{B}$ be a boolean algebra. Then we know that $\mathbb{B}$ is isomorphic to an algebra of sets, viz. ${h:\mathbb{B}\longrightarrow \mathrm{Clopen}(\mathrm{Ult}(\mathbb{B}))\subseteq\wp(\mathrm{Ult}(\mathbb{B}))}$ via $h(b)=U_{b}=\{\mathfrak{F}\in\mathrm{Ult}(\mathbb{B})\mid b\in\mathfrak{F}\}$.
I want to know, whether there is a powerset algebra $\wp(X)$ and an epimorphism, ${g:\wp(X)\longrightarrow\mathbb{B}}$. This is equivalent to the question: is there a powerset algebra $\wp(X)$ and an ideal $\mathcal{I}\subseteq \wp(X)$, such that $\mathbb{B}\cong\wp(X)/\mathcal{I}$.
I am also aware of the Loomis-Sikorski Theorem: Every $\sigma$-complete Boolean algebra is isomorphic to the a quotient $\mathbf{F}/\mathcal{I}$, where $\mathbf{F}$ is a $σ$-field of sets and $\mathcal{I}\subseteq\mathbf{F}$ is a $\sigma$-ideal. Is there an equivalent theorem for complete boolean algebras?

UPDATE: Answered by @Keith Kearnes here [https://mathoverflow.net/questions/306578/is-every-complete-boolean-algebra-isomorphic-to-the-quotient-of-a-powerset-algeb].
 A: A free algebra on an infinite set is not complete (exercise). If it would be a quotient of a powerset it would be a retract of it (exercise), and a retract of a complete algebra is complete (exercise).
Practically the same argument shows that the same algebra is also not a quotient of any countably complete one.
Another argument uses the fact that images of atoms under surjective homomorphisms are either 0 or atoms themselves. 
Since there is some controversy in comments, let me try to supply an argument for that. Take $\pi:P\twoheadrightarrow B$ and suppose $a\in P$ is an atom. Take any $x\leqslant\pi(a)$. Choose $y\in P$ with $\pi(y)=x$. Then also $\pi(a\land y)=x$ (since $\pi(a\land y)=\pi(a)\land\pi(y)=\pi(a)\land x=x$).
Now since $a$ is an atom, either $a\land y=a$ or $a\land y=0$, so that either $x=\pi(a)$ or $x=0$.
However to the contrary of what I stated, this does not imply that an atomless algebra cannot be quotient of a powerset; thanks to JDH for noticing that.
In fact for complete algebras the answer seems to be positive. In fact there is a theorem by Sikorski that injective objects in Boolean algebras are precisely the complete ones. Thus any embedding $i:B\hookrightarrow P$ of a complete Boolean algebra $B$ as a subalgebra of any Boolean algebra $P$ admits a retraction $\pi:P\twoheadrightarrow B$ (i. e. $\pi i$ is identity). Thus in fact any complete Boolean algebra is a retract of a powerset.
A proof can be found on planetmath. Note that it uses AC; that planetmath page claims that it is not known whether AC is unavoidable.
I believe also that not every quotient of a powerset is complete, i. e. retract - e. g. I think quotient by the ideal of finite subsets is not complete.
