Cohen algebra and $\mathcal P(\omega)/ \mathrm{fin}$

Question

Let $C$ be the Cohen algebra, the boolean completion of the partial order of finite partial functions from $\omega$ to 2, ordered by reverse inclusion. Does there exist an ideal $I$ on $C$ such that $C/I \cong \mathcal P(\omega)/ \mathrm{fin}$?

Answer 1

Let $p=\{a_{n}|n\in\omega\}$ be a countable partition of $C$. Let $p^{*}=\{\bigvee R|R\subseteq p\}$. Then $p^{*}$ is a Boolean algebra and there is an isomorphism $i:p^{*}\rightarrow P(\omega)$. By the Sikorski extension theorem (one does not need the full Sikorski extension theorem here, but just the strictly weaker fact that every filter extends to an ultrafilter), the mapping $i$ extends to a Boolean algebra homomorphism $f:C\rightarrow P(\omega)$ and the homomorphism $f$ is surjective. Let $\pi:P(\omega)\rightarrow P(\omega)/\textrm{fin}$ be the quotient mapping. Then $\pi\circ f:C\rightarrow P(\omega)/\textrm{fin}$ is a surjective mapping, so $C/\ker(\pi\circ f)\simeq P(\omega)/\textrm{fin}$.

In fact, if $B,C$ are complete Boolean algebras, $|B|\leq|C|$, and $J$ is an ideal on $B$, then there is an ideal $I$ on $C$ such that $C/I\simeq B/J$. To prove this more general fact, one could use the Sikorski extension theorem along with the theorem by Balcar and Franek which states that every infinite complete Boolean algebra $B$ has a free subalgebra of cardinality $|B|$.