8
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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, 2013 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 $D$). $\endgroup$ Jul 22, 2013 at 15:38
  • $\begingroup$ Yes, but then $\varepsilon_F$ depends only on $d$ and not on $D$? $\endgroup$
    – user1688
    Jul 22, 2013 at 16:39
  • 1
    $\begingroup$ 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, 2013 at 2:56
  • $\begingroup$ @anton: yes, $|\varepsilon_F|$ depends only on $d$; $D$ determines the range of values of $d$. $\endgroup$ Jul 23, 2013 at 7:43

1 Answer 1

8
$\begingroup$

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 $d = 4(t^2 \mp 1)$, $\epsilon = t + \sqrt{t^2 \mp 1}$.

$\endgroup$
2
  • $\begingroup$ 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 $>M$? $\endgroup$ Jul 23, 2013 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 $d = ib$, $u = (1-i)y$]. $\endgroup$ Jul 23, 2013 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.