TITLE: $A_4$ extension of $\mathbb{Q}$ ramified at one prime
QUESTION [21 upvotes]: How can one show that an $A_4$-extension of $\mathbb{Q}$ ramified at only one prime must be totally real?

REPLY [11 votes]: $\def\Z{\mathbf{Z}}$
$\def\Q{\mathbf{Q}}$
$\def\PSL{\mathrm{PSL}}$
$\def\SL{\mathrm{SL}}$
$\def\inv{\mathrm{inv}}$
$\def\Br{\mathrm{Br}}$
$\def\Gal{\mathrm{Gal}}$
$\def\F{\mathbf{F}}$
$\def\R{\mathbf{R}}$
$\def\C{\mathbf{C}}$
This is a consequence of class field theory. 
After some obvious reductions, one can assume that $p \equiv 1 \pmod 3$, and that the inertia and decomposition groups at $p$ are given by $I_p = D_p = \Z/3\Z \subset A_4$.
Note that $A_4 = \Gal(K/\Q) = \PSL_2(\F_3)$. Consider the problem of lifting such a representation to $\SL_2(\F_3)$, which is a non-split central extension by $\Z/2\Z$. Group theory says that the obstruction to such  a lifting lies in
$$H^2(G_{\Q},\Z/2\Z) = H^2(G_{\Q},\mu_2) = \Br(\Q)[2],$$
the $2$-torsion in the Brauer group of $\Q$. By the Albert-Brauer-Hasse-Noether theorem (class field theory) the Brauer group injects into the sum of local Brauer groups, so the only obstructions are local. Similarly (again by class field theory), the image of $\Br(\Q)$ in the sum of the local Brauer groups is the kernel of the invariant map $\inv$ to $\Q/\Z$, so it follows that any non-trivial class must be non-trivial in $\Br(\Q_v)$ for at least two places $v$ of $\Q$. (For $2$-torsion classes, one even sees it has to be ramified at an even number of $v$, but that is not relevant here).  
There is no obstruction to lifting away from the two places $p$ and $\infty$ because unramified representations trivially lift. There is also no obstruction at the prime $p$, because $I_p = D_p = \Z/3\Z$, and $G_{\Q_p} \rightarrow \Z/3\Z \rightarrow \PSL_2(\F_3)$ trivially lifts to $\SL_2(\F_3)$. Thus the only possible obstruction can occur at $\infty$. But, by the observation above, the image of a non-trivial element of $\Br(\Q)$ must be ramified at at least two primes. Thus there is no local obstruction at $\infty$ either.
The result is immediate: if complex conjugation was non-trivial in $\PSL_2(\F_3)$, there would be a local obstruction at $\infty$ since the lift of any order $2$ element in $\PSL_2(\F_3)$ to $\SL_2(\F_3)$ has order $4$, so a non-trivial map $\Z/2\Z = \Gal(\C/\R) \rightarrow \PSL_2(\F_3)$ cannot lift to $\SL_2(\F_3)$. Hence the $A_4$-extension is totally real, as desired.