regularity of ultrafilters An ultrafilter $U$ is $(\mu,\kappa)$-regular if there is a sequence $\langle X_\alpha : \alpha < \kappa \rangle \subseteq U$ such that for all $y \in [\kappa]^\mu$, $\bigcap_{\alpha \in y} X_\alpha = \emptyset$.  Countable incompleteness is equivalent to $(\omega,\omega)$-regularity, and for every $\kappa$, there is always an $(\omega,\kappa)$-regular ultrafilter.  The failure of $(\omega,\omega_1)$-regularity for an ultrafilter on $\omega_1$ is consistent relative to large cardinals, and carries some large cardinal strength.
In the 1970s, Kanamori proved that if $\lambda$ is singular, then every ultrafilter on $\lambda^+$ is $(\lambda,\lambda^+)$-regular.  Question: Is it known whether Kanamori's result can be improved to, "If $\lambda$ is singular, then every ultrafilter on $\lambda^+$ is $(\kappa,\lambda^+)$-regular for some $\kappa < \lambda$"?  What about in the case $\lambda=\aleph_\omega$?
 A: Assuming large cardinals, the answer is no even for $\aleph_\omega$.
Given a supercompact cardinal, Ben-David and Magidor constructed a model in which $\aleph_{\omega+1}$ carries a uniform indecomposable ultrafilter $U$, which means that $U$ is $\theta$-indecomposable for $\aleph_0<\theta<\aleph_\omega$.
For regular $\theta$, $\theta$-indecomposability means that the ultrafilter is closed  closed under intersections of decreasing sequences of length $\theta$, and for ultrafilters this is equivalent to the failure of $(\theta,\theta)$-regularity.
[due to Kanamori or Ketonen --- I'll need to check.]
Said another way, this means that any point-$<\theta$ family of sets from the ultrafilter has size $<\theta$, and hence certainly $(\theta,\aleph_{\omega+1})$-regularity fails.
The ultrafilter in the Ben-David/Magidor paper is $\aleph_n$-indecomposable whenever $0<n<\omega$, hence it cannot be $(\kappa,\aleph_{\omega+1})$ regular for any $\kappa<\aleph_\omega$.
[Note: In the other direction, the answer is surely yes if $V=L$ and the proof ought to go by showing that a counterexample implies some variety of stationary reflection, but I haven't had time to investigate.]
