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 $r=1$?



Are there primitive quartic CM fields whose norms of units give all totally positive units of the real quadratic subfield?
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.


