On the non existence of non real embeddings Let $K/\Bbb Q$ be a finite Galois extension.
Denote by $U_K$ the units group of $\mathcal O_K$.
I am studying the impact of the finiteness of $U_K/(U_K\cap \Bbb R)$ on the existence of non real embeddings of $K$ in $\Bbb C$.
I have the following guess:
$U_K/(U_K\cap \Bbb R)$ finite $\Leftrightarrow$ there are no non real embeddings of $K$ in $\Bbb C$
If my guess is correct, I don't see how to prove it from the following manipulation:
By the Dirichlet Unit theorem
$U_K\cong T\times \Bbb Z^{r+s-1}$ with $r,s$ respectively the number of real, complex pair embeddings and $T$ the subgroup of roots of unity in $K$.
Then:
$U_K/(U_K\cap \Bbb R) \cong (T\times \Bbb Z^{r+s-1})/((T\cap \Bbb R)\times (\Bbb Z^{r+s-1}\cap \Bbb R)$    (step 1)
$\cong (T\times \Bbb Z^{r+s-1})/((T\cap \Bbb R)\times \Bbb Z^{r+s-1})$ (step 2)
$\cong T/(T\cap \Bbb R)$ (step 3).
I am not sure if something is incorrect in the steps above or otherwise, how to go further.
Thank you for any help to see this more clearly. 
 A: Let $K/K^{+}$ be any non-trivial extension of number fields. Then the natural statement is that $U_K/U_{K^+}$ is infinite unless $K^{+}$ is totally real and $K$ is a totally complex quadratic extension, i.e. a CM field. In particular, if $K$ is Galois over $\mathbf{Q}$, then every conjugate of complex conjugation fixes $K^{+}$ and so must lie in $\mathrm{Gal}(K/K^{+})$, which is thus normal and so central. Let's prove this.
Suppose that $K^{+}$ has signature $(r,s)$, and that $K/K^{+}$ has degree $d \ge 2$. The complex places $s$ of $K^{+}$ split into $ds$ complex places of $K$.  $m \le r$ of the real places of $K^{+}$ remain real, and so give $dm$ real places of $K$, and the other $(r-m)$ real places split, and thus give $d/2(r-m)$ complex places of $K$ (the real places can only split when $d$ is even). It follows that $K$ has signature $(R,S) = (dm,ds + d/2(r-m))$. For the unit quotient to be finite, the ranks of the groups must be the same, and thus by Dirichlet's theorem $R+S=r+s$. But
$$R+S-r-s = dm + ds + d/2(r-m) - r- s = \frac{1}{2}
\left(d(m+s) + (d-2)(r+s)\right).$$
This can only be zero if $d = 2$ and $m+s=0$, or if $(r,s) = (r,0)$, and $(R,S) = (0,2r)$, or if $K^{+}$ is totally real and $K$ is totally complex.

For your actual question, you can deduce the following:
Let $\sigma$ be an embedding of $K$ into the complex numbers, let $K^{+}$ denote the intersection of $\sigma(K)$ with $\mathbf{R}$, so $U_{K^{+}} = U_K \cap \mathbf{R}$. Then either $K = K^{+}$, or $K$ is a CM extension.
Note that $K \ne K^{+}$ exactly when $\sigma$ is not a real embedding, so one can rephrase this as:
If $\sigma$ is a complex (non-real) embedding of $K$ and $U_K/U_K \cap \mathbf{R}$ is finite, then $K$ is CM. 
Of course you can't say much more than that, because if $\sigma$ is a real embedding then $K^{+}=K$ and the condition you give is trivial. Perhaps you have the condition for all $\sigma$, which would imply that either $K$ is totally real (and so complex conjugation is trivial) or $K$ is CM.
