Nonreal units in totally imaginary number fields Suppose we are given a totally imaginary number field $L$ of degree greater $2$, is it possible that all units of $L$ lie in $\mathbb R$? This is true for almost all quadratic imaginary number fields, but i am not sure how to proceed in the case of higher degrees. I'd also be glad if you could point out some literature on this subject.
 A: Let $L$ be a number field with a chosen complex embedding, and let $K$ be its real subfield.
If $L$ has $a$ real embeddings and $2b$ complex embeddings, then its unit rank is $a+b-1$.
If $K$ has $c$ real embeddings and $2d$ complex embeddings, then its unit rank is $c+d-1$.
Your hypothesis would require $a+b-1 = c+d-1$. However, as $[L:K] \geq 2$, we also have $a+2b \geq 2(c + 2d)$. If I've solved correctly, this implies $a=d=0$, and $b=c$.
So, the possible situation is that $K$ is a totally real number field of degree $n$, and that $L$ is a totally imaginary quadratic extension of $K$. Both of these fields have unit rank $n-1$, and so it's plausible that they have equal unit groups.
I'd imagine the conjecture would be true for nearly any example you constructed satisfying this condition.

I'm now at a computer where I can easily use sage. The very first thing I tried worked, just as I predicted:
K.<a> = NumberField(x^2 - 7)
R.<y> = PolynomialRing(K)
L.<b> = NumberField(y^2 - (a-3))
G = L.unit_group()
G <= K

prints True. Sage reports the fundamental unit of L is 3*a - 8.
The field K is $\mathbb{Q}(\sqrt{7})$, and the field L is $\mathbb{Q}(\sqrt{\sqrt{7} - 3})$. The fundamental unit of $L$ is $3 \sqrt{7} - 8$.
A: This is a comment to Hurkyl's answer, which is too long to fit in the comment box.
Let $K$ be a CM field, let $E$ be the group of units in $\mathcal{O}_K$, let $K^+$ be the maximal real subfield of $K$, and let $E^+$ be the unit group in $\mathcal{O}_{K^+}$. Let $W$ be the group of roots of unity in $K$. Then, $[E:WE^+]=1$ or $2$.
You can find a proof in Washington's "Introduction to Cyclotomic Fields", Theorem 4.12. 
In your case, you want $E=E^+$, so you need $W\subseteq E^+$ (and so $W=\{\pm 1\}$). In general all one can say is that if $W=\{\pm 1\}$, then $[E:E^+]=1$ or $2$.
