Suppose $K$ is a quadratic imaginary field, and $\phi:G_K\rightarrow \overline{\mathbb{Q}}^{\times}$ is a finite order Galois character satisfying $\phi\phi^c=1$ (where $c$ is a complex conjugation). Can we necessarily write $\phi=\psi^c/\psi$ for some finite order character $\psi:G_K\rightarrow \overline{\mathbb{Q}}^{\times}$? What about for a CM field?

cyclicgroups (& divisibility of $D$). Abutment in degree 2 vanishes, so $E_2^{1,1}=0$ provided $E_2^{0,3}=0$. But ${\rm{H}}^3(K,D)=0$ since $K$ tot. complex. $\endgroup$ – BCnrd Aug 26 '10 at 4:11