TITLE: Are there primitive quartic CM fields whose norms of units give all totally positive units of the real quadratic subfield?
QUESTION [5 upvotes]: Let $K$ be a primitive (i.e. not biquadratic) quartic CM-field. That is we have $[K:\mathbb{Q}]=4$ and let $K_0=\mathbb{Q}(\sqrt{d})$ be the totally real quadratic subfield, here $d> 1$, that is we have $[K:K_0]=[K_0:\mathbb{Q}]=2$.
Denote by $O_K^{\times}$ the set of units in $K$ and by $(O_{K_0}^{\times})^{+}$ the set of totally positive units in $K_0$. Then $N_{K/K_0}(O_K^{\times})$ is a subgroup of $(O_{K_0}^{\times})^{+}$
$\textbf{Question:}$ Can we classify all such fields $K$ with the property $r=[(O_{K_0}^{\times})^{+}:N_{K/K_0}(O_K^{\times})]=1$?
a) If $\epsilon_0$ denotes the fundamental unit of $K_0$ and we have $N_{K_0/\mathbb{Q}}(\epsilon_0)=-1$ then $r=1$. So these are all $K$ with the property that $K_0$ has a (fundamental) unit of norm -1. Are these exactly the fields $\mathbb{Q}(\sqrt{t^2+4})$ for some $t\in \mathbb{Z}$?
b)What happens if $N_{K_0/\mathbb{Q}}(\epsilon_0)=1$? Is it possible that there is a unit $\epsilon\in K$ such that $\epsilon_0=N_{K/K_0}(\epsilon)$? Then we also should get $r=1$. So are the there such fields with a unit $\epsilon$ such that $N_{K/K_0}(\epsilon)$ is a fundamental unit of $K_0$ with norm 1? Can we classify all such $K$?
c) Are there any other $K$ with $r=1$?

REPLY [2 votes]: Except for the finitely many fields where nontrivial roots of unity are present, the fundamental unit $\varepsilon$  of the quadratic subfield $k$ is the norm of a unit from $K$ if and only if $K = k(\sqrt{\pm \varepsilon})$. If $N(\varepsilon) = +1$, then $K$ is biquadratic, and if $N(\varepsilon) = -1$, then $K$ is not CM (by simple Galois theory). 
Thus the fundamental unit in $K$ is the same as in $k$ in your cases, and you have $r = 1$ if and only if $N(\varepsilon) = -1$. In this case, $k$ has the form $k = {\mathbb Q}(\sqrt{t^2+4})$ for trivial reasons: write $\varepsilon  = (T + U \sqrt{m})/2$; then $T^2 - mU^2 = -4$, hence $T^2 + 4 = mU^2$ and ${\mathbb Q}(\sqrt{m}\,) = {\mathbb Q}(\sqrt{mU^2}\,) = {\mathbb Q}(\sqrt{T^2+4}\,)$. But I would not regard this as a classification.