# PFA and saturated ideals

Does PFA imply that there is an $\omega_2$-saturated ideal on $\omega_1$? All I know is that MM implies that $NS_{\omega_1}$ is saturated.

Thanks!

• PFA does not imply that $NS_{\omega_1}$ is saturated. This doesn't quite answer your question, but it's worth mentioning. See this paper of Velickovic for some details: logique.jussieu.fr/~boban/pdf/PFA_and_NS.pdf Sep 19, 2017 at 20:52

No. It is a theorem of Shelah that PFA is consistent with the failure of weak Chang's conjecture: there exists $\langle f_i\in \omega_1^{\omega_1}: i < \omega_2+1\rangle$ that is increasing mod $NS_{\omega_1}$.
We show the existence of $\omega_2$-saturated ideal on $\omega_1$ is incompatible with this fact. Let $g=f_{\omega_2}$. For each $i<j<\omega_2$, let $E_{i,j}$ be a club in $\omega_1$ such that $\forall \xi \in E_{i,j}$, $f_i(\xi)<f_j(\xi)$ and $F_j$ be a club such that for all $\xi \in F_j$, $f_j(\xi)<g(\xi)$. Now force with $P(\omega_1)/I$, we know that $\omega_1^V\in \bigcap j(E_{i,j})\cap \bigcap j(F_j)$. But now $\{j(f_i)(\omega_1^V): i<\omega_2^V=\omega_1^{V[G]}\}$ is an uncountable increasing sequence bounded by $j(g)(\omega_1^V)\in \omega_1^{V[G]}$ which is countable. So impossible.