Let $K$ be a real quadratic number field and $\epsilon$ be a fundamental unit. Then apparently the narrow Hilbert class field of $K$ equals the Hilbert class field if and only if $N_{K/\Bbb Q}(\epsilon)=-1$. Why is this the case?
Follow-up question: Does this also hold for complex cubic fields?
The narrow Hilbert class field is equal to the usual Hilbert class field if and only if the narrow class group is equal to the usual class group, which is the case if and only if the norm of the fundamental unit is $-1$.
For cubic fields, this is clearly not true since $N(-\varepsilon) = - N(\varepsilon)$.