TITLE: Question on units in a cubic number field
QUESTION [5 upvotes]: If $K$ is a cubic extension of $\Bbb Q$ (the rational numbers) having only one real embedding in $\Bbb R$ ( the real numbers), then why should the units in the ring of integers be of the form $\pm u^k$, where $u (>1)$ is the fundamental unit?
I know Dirichlet` theorem of units.
My question is :
1) Why should $\pm 1$ be the only roots of unity in the ring of integers in $K$.
2) Why should $u$ be an integer in $\Bbb Z$, and why should it be $>1$.
This sum is required for problem 35 of chapter 5 of marcus.
Please help. I am stuck with fundamentals...

REPLY [2 votes]: Roots of unity do not fit well inside $\mathbb{R}$: since our field $K$ has a real embedding, this tells us that $K$ only has the second roots of unity.
In the real embedding of $K$, any generator of our unit group is clearly real, and as noted in the comments, there will be a unique generator greater than 1.
However, if we are not considering a real embedding, then it's not the case that the generator must be real. For example, $K=\mathbb{Q}(2^{1/3})$ is a field with one real embedding. In the real embedding, you may check that $u = 2^{1/3} - 1$ is a fundamental unit greater than 1. If we consider this element in one of the non real embeddings we get, for example, $v = \omega 2^{1/3} - 1$ where $\omega$ is a primitive third root of unity--this is definitely not real. The only generators for the non-torsion part of the unit group are $\pm v^{\pm 1}$, none of which are real.
Implicit in the Marcus exercise is that we are considering the real embedding of $K$.