Character of normal ultrafilters The character of an ultrafilter $U$, denoted $\chi(U)$, is the minimal size of an $A \subseteq U$ such that $(\forall x \in U ) (\exists y \in A) y \subseteq x$.  This cardinal characteristic has been studied for ultrafilters on $\omega$.  For all nonprincipal $U$, $\omega_1 \leq \chi(U)$, and it is known that, consistently, there exists nonprincipal $U$ with $\chi(U) < 2^\omega$.  My question is, for normal ultrafilters $U$ on a measurable cardinal $\kappa$, is it possible that $\chi(U) < 2^\kappa$?
Here are some relevant papers about ultrafilters on $\omega$, by Shelah, Brendle, and Hart:
MR1686797, MR0987317, MR2365799, MR2847327
(available through http://www.ams.org/mathscinet/)
 A: To get the ball rolling...
One can show easily that $\chi(U)$ must be at least $\kappa^+$, since otherwise one can take the diagonal intersection of a $\kappa$-sized family and find a single set that supposedly generates $U$, which is impossible. 
Thus, your situation would require that $\kappa$ is measurable and $2^\kappa\gt\kappa^+$, a situation already whose consistency strength strictly exceeds that of a measurable cardinal. Silver showed that this is consistent relative to a $\kappa^+$-supercompact cardinal, but it is now known to be equiconsistent with a cardinal $\kappa$ that is $\kappa^{++}$-tall. But it follows from this that one cannot prove that your situation is consistent if one starts only from the assumption that measurable cardinals are consistent (unless that assumption is inconsistent). We need to use a stronger hypothesis. 
But it is conceivable that we might hope to mimic the methods on $\omega$ higher up, if we start with a supercompactness assumption on $\kappa$....
