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$\newcommand{\fin}{\mathrm{fin}}$Under what hypotheses does there exist a uniform ideal $I$ on $\omega_1$ such that $P(\omega_1)/I \cong P(\omega)/\fin$? What is the consistency strength?

It follows from work of Woodin that if we have $\diamondsuit$ and a countably complete $\omega_1$-dense ideal on $\omega_1$, then such a (only finitely complete) ideal $I$ exists as above.

Note that I am not asking whether $P(\omega_1)/I$ and $P(\omega)/\fin$ are forcing-equivalent, but rather if they are literally isomorphic.

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  • $\begingroup$ Do you have a reference for that result of Woodin? Sounds very interesting. $\endgroup$
    – Will Brian
    Commented Jun 8 at 9:14
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    $\begingroup$ @WillBrian I gave a sketch of the proof in our seminar recently. ucloud.univie.ac.at/index.php/s/f89ENYQLkdg4BNo $\endgroup$
    – Monroe Eskew
    Commented Jun 8 at 14:29
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    $\begingroup$ What about taking a copy $K$ of $\omega^*$ in the set of uniform ultrafilters on $\omega_1$? The ideal $I=\{A:A^*\cap K=\emptyset\}$ is a uniform and $K$ is the Stone space of $\mathcal{P}(\omega_1)/I$. $\endgroup$
    – KP Hart
    Commented Jun 17 at 8:02

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To expand my comment into an answer: take, for each $n\in\omega$, a uniform ultrafilter $u_n$ on $\omega_1$ that contains the set $\{\lambda+n:\lambda$ is a limit or $0\}$. The set $U=\{u_n:n\in\omega\}$ is (relatively) discrete in the Čech-Stone compactification $\beta\omega_1$, hence its closure $K$ is homeomorphic to $\beta\omega$. The space $L=K\setminus U$ is homeomorphic to $\beta\omega\setminus\omega$.

Now let $I$ be the ideal of subsets of $\omega_1$ whose closure in $\beta\omega_1$ is disjoint from $L$. By Stone-duality $L$ is the Stone space of $\mathcal{P}(\omega_1)/I$, and so $\mathcal{P}(\omega_1)/I$ is isomorphic to $\mathcal{P}(\omega)/\mathit{fin}$. Because the ultrafilters $u_n$ are uniform the ideal $I$ contains all countable sets.

Alternative description of $I$: one has $A\in I$ iff $\{n:A\in u_n\}$ is finite.

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    $\begingroup$ Let me see if I understand correctly. It looks like we can give a more direct argument and skip the topological considerations. Just define the ideal $I$ as in your last line, or alternatively, against any countable sequence $u_n$ such that $u_n$ is a uniform ultrafilter on $\omega_1$ concentrating on $X_n$, where the $X_n$ form a partition of $\omega_1$. The we can just map $[A]_I \mapsto [\{ n : A \in u_n \} ]_{\mathrm{fin}}$ and verify directly that this is a boolean isomorphism. Or am I missing some subtlety? $\endgroup$
    – Monroe Eskew
    Commented Jun 24 at 16:04
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    $\begingroup$ @MonroeEskew No, there's no subtlety involved; it's just that Stone-duality makes it easier (for me) to see what you were asking. The point here is that quotients are dual to subspaces, so you were asking whether the space of uniform ultrafilters has a subspace homeomorphic to $\beta\omega\setminus\omega$. The same reasoning would provide a two-line answer to this question of yours, because the Stone space of that algebra contains a copy of $\beta\omega$, hence of $\beta\omega\setminus\omega$. $\endgroup$
    – KP Hart
    Commented Jun 24 at 16:19

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