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 $r=1$?

  • $\begingroup$ a) Yes, or even ${\bf Q}(\sqrt{t^2+1})$ (cubing the fundamental unit if necessary); but that's begging the question if you're given only the discriminant of $K_0$ (e.g. for ${\bf Q}(\sqrt{193})$ the first $t$ is $1761432$). I think there's still no full criterion known for the existence of a negative unit, only some partial results (most easily: yes if the discriminant is prime, no if it has a factor $\equiv 3 \bmod 4$). $\endgroup$ Jun 28, 2017 at 15:50
  • $\begingroup$ @Noam D. Elkies : Thanks! I am no expert in this, but what do you mean by: "there is no criterion known for the existence of a negative unit"? I was reading the article "On real quadratic fields containing units with norm -1" by Yokoi and it says $\mathbb{Q}(\sqrt{D})$ has a unit of negative norm if and only if $D-4$ is a square. Or did I misunderstand something? What do you mean by "even $\mathbb{Q}(\sqrt{t^2+1})$"? Does it mean all fields I get using $\sqrt{t^2+4}$ I can get using $\sqrt{t_0^2+1}$ for some $t_0\neq t$? $\endgroup$
    – Bernie
    Jun 29, 2017 at 8:57

1 Answer 1


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.


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