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?
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5$\begingroup$ Yes, & yes. Let $K_0$ be max. tot. real subfield, $D := \mathbf{Q}/\mathbf{Z}$. Want ${\rm{H}}^1(K/K_0,{\rm{H}}^1(K,D))=0$. Consider cyclic ext'n $K/K_0$ of # fields with $K$ tot. complex, and sp. seq. $E_2^{i,j} = {\rm{H}}^i(K/K_0,{\rm{H}}^j(K,D)) \Rightarrow {\rm{H}}^{i+j}(K_0,D)$. By Tate, ${\rm{H}}^2(F,D) = 0$ for global $F$, so $E_2^{0,2}=0$. Also, $E_2^{2,0} = {\rm{H}}^2(K/K_0,D)$ vanishes by double periodicity of Tate cohom. for cyclic groups (& 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. $\endgroup$– BCnrdCommented Aug 26, 2010 at 4:11
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2$\begingroup$ Just to clarify, for general number fields $F$, ${\rm{H}}^i(F,\cdot) = \prod_{v|\infty} {\rm{H}}^i(F_v,\cdot)$ for $i \ge 3$. In particular, this vanishes when $F$ is totally complex. I'm sure this is explained somewhere in Milne's book on arithmetic duality theorems, along with Tate's vanishing result for ${\rm{H}}^2(F,\mathbf{Q}/\mathbf{Z})$ for global $F$ (though I learned this latter fact from Serre's Durham survey article on weight-1 forms and Galois representations, where it is proved in the number field case). $\endgroup$– BCnrdCommented Aug 26, 2010 at 6:23
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