Unit in abelian extension of rationals I am studying the Dirichlet's Unit Theorem from Marcus's book, and I am stuck with this problem.
Let $K$ be an abelian extension of the rationals, let $u$ be a unit in the ring of intergers of $K$. Prove that $u$ is the product of a real number and a root of 1, with the factors either in $K$ or in an extension of degree 2 over $K$.
I know that since $K$ is an abelian extension, complex conjugate is in the center of the Galois group. Thus, by a familiar result, for all unit $u$ that is an algebraic integer, $u^k$ is real for some $k$. Thus, I can write $u$ as $u^k u^{1-k}$ where $u^k$ is real. The trouble is I don't know how I can find the root of unity in the required factorization of $u$. I am trying to show $u^{1-k}$, but currently I am stuck. Where does the extension of $K$ of degree 2 come into play?
Any help is appreciated.
 A: Let $G = \mathrm{Gal}(K/\mathbf{Q})$ be the Galois group. Since $G$ is abelian, complex conjugation $c \in G$ doesn't depend on any choice of embedding of $K$ into $\mathbf{C}$. It follows that $v = u \cdot cu$ is a totally real positive unit, that is, a unit in $K$ which is real and positive for every embedding of $K$ into $\mathbf{C}$, since the image under any embedding will just be $|u|^2$ for the image of $u$ in that embedding. So now write
$$u^2 = v \cdot \zeta = (u \cdot c u) \cdot \frac{u}{c u}.$$
We have seen that $v \in K$ is a totally positive real unit. It follows that $\zeta$ is an algebraic integer. On the other hand, the absolute value of $\zeta$ for every embedding of $K$ into $\mathbf{C}$ satisfies
$$|\zeta|^2 = \zeta \cdot c \zeta = \frac{u}{c u} \cdot \frac{cu}{u} = 1.$$
A result of Kronecker then says that $\zeta$ is a root of unity. Finally, $u = \sqrt{v} \cdot \zeta^{1/2}$. Since $v$ was a totally real and positive unit in $K$, $\sqrt{v}$ is a totally real unit in (at most) a quadratic extension of $K$. Similarly, $\sqrt{\zeta}$ is the square root of a root of unity which is also a root of unity.
