Suppose that $K$ is a number field, and (writing $G\_K=\mathrm{Gal}(\overline{K}/K)$), suppose that $\phi:G\_K\to \overline{\mathbb{Q}}$ is a finite order character of $G\_K$. I believe that the obstruction to taking a square root of $\phi$ (that is, the obstruction to finding some finite order $\chi:G\_K\to \overline{\mathbb{Q}}$ with $\chi^2=\phi$) can be identified with a 2-torsion element of the Brauer group of $K$, and I believe that the proof just involves taking a long exact sequence from a cunningly-chosen short exact sequence; but I cannot now reconstruct it. Can anyone remind me?



obstruction to taking the square root of a Galois character
With luck, this will be blunder-free (and if it's not, please tell me!):


Consider the short exact sequence 
$$1 \to \mu\_2 \to \overline{\mathbb Q}^{\times} \to \overline{\mathbb Q}^{\times}
\to 1,$$
with the third arrow being squaring, and with trivial $G\_k$-action. Passing to cohomology,
the sequence of $H^0$s is just this same sequence, and the sequence of higher cohomology
becomes
$$0 \to Hom(G\_K,\mu\_2) \to Hom(G\_K,\overline{\mathbb Q}^{\times}) \to
Hom(G\_K,\overline{\mathbb Q}^{\times}) \to H^2(G\_K,\mu\_2),$$
with the last arrow being the obstruction you asked about. 


Added: See Brian Conrad's comment below for a cleaner point of view, showing
that $H^2(G\_K,\mu\_2)$ is the precise obstruction space.


