# obstruction to taking the square root of a Galois character

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?

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.
• The last term can also be taken to be $\mathrm{Ext}^1(G_K^{ab},\mu_2)$. – David Corwin Aug 25 '10 at 4:13
• Matt, use $0 \rightarrow \mathbf{Z}/(n) \rightarrow \mathbf{Q}/\mathbf{Z} \rightarrow \mathbf{Q}/\mathbf{Z} \rightarrow 0$ since ${\rm{H}}^2(K,\mathbf{Q}/\mathbf{Z}) = 0$ for global $K$ (Tate), so ${\rm{H}}^2(K,\mathbf{Z}/(n))$ is exact obstruction space (for $n$th roots). Is there local-to-global problem for $n$th root of finite-order char? For $n = 2$ OK: $\mathbf{Z}/(2) = \mu_2$ + Brauer gp. In general, Tate global duality + Kummer thy implies at most "one" bad char. & occurs exactly in Grun.-Wang cases (so $8|n$, char. 0). For $n=8$, $K = \mathbf{Q}(\sqrt{7})$, what order & ramification?!? – BCnrd Aug 25 '10 at 4:22