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Let $F/{\mathbb Q}$ be an imaginary quartic extension (i.e. the degree $[K:{\mathbb Q}]=4$ and no embedding of $K$ in ${\mathbb C}$ has its image inside the real numbers). Then the unit group of the integer ring ${\mathcal O}_K$ is infinite cyclic up to the roots of unity in $K$ and one can pick a generator $\varepsilon_F$ with absolute value $>1$ (say we have chosen an embedding into ${\mathbb C}$).

I am interested in the asymptotics of $|\varepsilon_F|$; more precisely my question is the following: if we fix an imaginary quadratic extension $F_D={\mathbb Q}(\sqrt{-D})$ of ${\mathbb Q}$ then for any fundamental discriminant $d$ in the ring of integers ${\mathcal O}_D$, $F=F_D(\sqrt d)/{\mathbb Q}$ is an imaginary quartic extension, and it is known that $|\varepsilon_F|$ tends to infinity as $d$ does. I would be interested in knowing whether this convergence is uniform in $D$ or not, that is whether if given $M>1$ there is a $N\ge 0$ such that for any $D\in{\mathbb Z}_{>0}$ there are at most $N$ fundamental discriminants $d\in {\mathcal O}_D$ such that $|\varepsilon_F|\le M$. If this turns out not to be the case then I would be interested in the asymptotics of the numbers of $d$ with $|\varepsilon_F|\le M$ as (square-free) $D\to +\infty$.

It is well-known that one can reformulate this in terms of Pell-like equations: the units in such a $F$ are given by $1/2(t+u\sqrt d)$ where $(t,u)$ is an integer solution of $t^2-u^2d=4$.

My motivation for asking this question comes from geometry: the norms of fundamental units in quadratic extensions of $F_D$ correspond to the lengths of closed geodesics on the associated Bianchi orbifold, and the question amounts to asking if the number of such lengths which are less than $e^M$ is bounded when $D$ varies.

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  • $\begingroup$ What do you mean by "fundamental discriminant"? Usually, it's an integer, so that would mean that always all fd are in ${\cal O}_D$. $\endgroup$ – user1688 Jul 22 '13 at 14:40
  • $\begingroup$ I believe that "fundamental discriminant" is the standard term for the set of these integers which are quadratic residues mod 4 but not squares in ${\mathcal O}_D$ (I added "in ${\mathcal O}_D$" to put emphasis on the dependancy on $D$). $\endgroup$ – Jean Raimbault Jul 22 '13 at 15:38
  • $\begingroup$ Yes, but then $\varepsilon_F$ depends only on $d$ and not on $D$? $\endgroup$ – user1688 Jul 22 '13 at 16:39
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    $\begingroup$ I think it might be better to call these fields quartic rather than biquadratic: quartic definitely means "degree 4", while biquadratic suggests a specific type of quartic extension, namely a composite of two quadratic extensions (so of the form ${\mathbf Q}(\sqrt{a},\sqrt{b})$ with rational $a$ and $b$). $\endgroup$ – KConrad Jul 23 '13 at 2:56
  • $\begingroup$ @anton: yes, $|\varepsilon_F|$ depends only on $d$; $D$ determines the range of values of $d$. $\endgroup$ – Jean Raimbault Jul 23 '13 at 7:43
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There's certainly some uniform bound, as a special case of the theorem that for each $n$ and $M$ there are only finitely many algebraic integers $\epsilon$ of degree $n$ each of whose conjugates has absolute value at most $M$. Here $n=4$, and since $\epsilon$ is a unit conjugate to $\pm\epsilon^{-1}$ (once $D \lt -4$), the proof leads to the estimate $N = O(M^2)$ with an effective (and reasonably small) implied constant.

Write the minimal equation of $\epsilon$ as $0 = (x-\epsilon) (x\mp\epsilon^{-1}) = x^2 + ax \pm 1$ when $\epsilon$ is an algebraic number of degree $2$ with norm $\pm 1$, and as $$ 0 = (x-\epsilon) (x-\bar\epsilon) (x\mp\epsilon^{-1}) (x\mp\bar\epsilon^{-1}) $$ $$ = (x^2 - 2{\rm Re}(\epsilon) + 1) (x^2 \mp 2{\rm Re}(\epsilon^{-1}) + 1) $$ $$ = x^4 + ax^3 + bx^2 \pm ax + 1 $$ when $\epsilon$ has degree $4$. In each case the coefficients $a$ or $a,b$ are integers of size $O(M)$. Thus there are $O(M^2)$ possibilities in all, so you can take $N = O(M^2)$ as claimed.

As with Pell equations, we expect that the actual count is asymptotically smaller, but $N$ must still grow as some power of $M$ because of families such as $d = 4(t^2 \mp 1)$, $\epsilon = t + \sqrt{t^2 \mp 1}$.

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  • $\begingroup$ Thank you very much. If I understand correctly your answer, it means in particular that if I fix $M>1$ then for $D$ large enough and $d\in{\mathcal O}_D$, $d\not\in{\mathbb Z}$ a fundamental discriminant the fundamental unit $\varepsilon_F$ of $F=F_D(\sqrt d)$ has norm $>M$? $\endgroup$ – Jean Raimbault Jul 23 '13 at 12:21
  • $\begingroup$ You're welcome, and yes, I think so: for you must make sure that $u^2 d$ can't be integral, but since $d$ is fundamental that should be the case once $D \lt 4$ [for the Gaussian numbers you must watch out for $d = ib$, $u = (1-i)y$]. $\endgroup$ – Noam D. Elkies Jul 23 '13 at 13:18

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