ZFC proves every infinite Boolean algebra has infinitely many ultrafilters. If every ultrafilter over $\omega$ is principal, then $\mathcal{P}(\omega)/\mathrm{fin}$ has no ultrafilter.
Is it consistent with ZF that there is an infinite Boolean algebra with a unique ultrafilter? Thanks for any help.
Take the Boolean algebra $B=\mathcal{P}(\omega)/\mathrm{fin}\times \{0,1\}$. An ultrafilter on $B$ has the form either $U\times\{0,1\}$ or $\mathcal{P}(\omega)/\mathrm{fin}\times U$ where $U$ is an ultrafilter on the respective coordinate. So if $\mathcal{P}(\omega)/\mathrm{fin}$ has no ultrafilters, then the only ultrafilter on $B$ is $\mathcal{P}(\omega)/\mathrm{fin}\times\{1\}$.