I have just read that a selective ultrafilter must necessary be an ultrafilter. Is this also true for q-filters?
Im not sure if using CH for instance, we can follow the construction of a q-point but this time by fixing a set and ensuring neither the fixed set nor its complement should be in the filter.
Yes, exactly what you suggest works. Constructing by transfinite induction an increasing family of countably generated filters $F_\xi$ such that the evens and the odds are both $F_\xi$ positive for all $\xi$ you are confronted at stage $\xi$ with a partition $P$ into finite sets. Choosing by finite induction a selector for $P$ you notice that given any infinite set $X$ it is possible to extend your partial selector to intersect $X$ (this is not possible in the selective construcution). So you choose your selector meeting all sets in $F_\xi$ and their intersections with either the evens or the odds and let $F_{\xi+1}$ be generated by $F_\xi$ and this new selector.
I think you can also produce a counterexample from the assumption that there are two non RK equivalent selectives, simply by putting one of these selectives on the evens and the other on the odds.
Another way to get a Q-filter that is not an ultrafilter is to destroy a Q-point using an $\omega^\omega$-bounding forcing, e.g., Grigorieff forcing with a selective ultrafilter.
Grigorieff forcing destroys the selective ultrafilter, but because Grigorieff forcing with a P-filter is $\omega^\omega$-bounding, the (formerly selective) filter will remain a Q-filter.
