In the course of doing some calculations I have found myself wanting to answer the following question:

Let $D/\mathbb{Q}$ be a quaternion algebra ramified at a prime $p$ and at $\infty$ and let $\mathcal{O}$ be a maximal order.

Is it possible to find $\lambda,\mu\in\mathcal{O}$ such that:

$1. N(\lambda) = p-1$

$2. N(\mu) = p$

$3. Tr(\lambda\overline{\mu})=0$

If the answer is no I would consider the weaker question of whether we are always able to choose $\mathcal{O}$ such that the above is possible.

I have definitely found examples of this occurrence for $p=2,3,5,7,11$ (cases where I have needed elements with these properties for certain applications) and can set congruence conditions on $p$ where it is possible etc, but I am wondering whether the result is true in general.

I have considered the idea of using the classification of such quaternion algebras and searching for obvious integral elements such as $\alpha + \beta i + \gamma j + \delta k$ with $\alpha,\beta,\gamma,\delta\in\mathbb{Z}$ with the required properties but it seems to be tough to do this way (even if we make simple choices for $\mu$ such as $\mu = j$ when the quaternion algebra is nicely presented).


1 Answer 1


First of all, it is not even true that every such order $\mathcal{O}$ contains an element $\mu \in \mathcal{O}$ with $\mathrm{nrd}(\mu)=p$. The first counterexample is $p=37$, and this corresponds to the fact that this is the first prime $p$ such that there is a supersingular elliptic curve with $j$-invariant $\in \mathbb{F}_{p^2} \setminus \mathbb{F}_p$. So we should consider the weaker question.

In that case, viewed in the language of quadratic forms, you are asking whether or not the norm form $Q$ on some maximal order represents the binary quadratic form $R=pu^2 + (p-1)v^2$. Then the integral local-global (Hasse-Minowski) principle for lattices applies: an integral quadratic form $R$ is represented by some form in the genus of an integral quadratic form $Q$ if and only if $R_v$ is represented by $Q_v$ for all places $v$ of $\mathbb{Q}$. Now check the latter: at any place $v \neq p,\infty$, the norm form $Q$ is the determinant and check that this represents any binary quadratic form, in particular $R$. The place $v=\infty$ is no problem since the form is positive definite. At $v=p$ with $p \neq 2$, the norm form is $t^2 - ex^2 + py^2 - pez^2$ where $e$ is any quadratic nonresidue modulo $p$, and we are again OK: the regular form $t^2-ex^2-(p-1)w^2$ has a zero. (Check the case $p=2$ separately, which you have done.)

Here is Magma code (run at http://magma.maths.usyd.edu.au/calc/) that produces all solutions for each $p$, up to conjugation:

    for p in PrimesUpTo(30) do
        B := QuaternionAlgebra(p);
        a,b := StandardForm(B);
        Omax := MaximalOrder(B);
        Os := ConjugacyClasses(Omax);
        for O in Os do
            munufound := [* *];
            L := LatticeWithGram(GramMatrix(O));
            vs := ShortVectors(L, 2*p, 2*p);
            ws := ShortVectors(L, 2*(p-1), 2*(p-1));
            for v in vs do
                mu := O!Eltseq(v[1]);
                for w in ws do
                    nu := O!Eltseq(w[1]);
                    if Trace(mu*Conjugate(nu)) eq 0 then
                        Append(~munufound, [B!mu,B!nu]);
                    end if;
                end for;
            end for;
            print p, <a,b>, munufound;
        end for;
    end for;
  • $\begingroup$ Thanks! Lovely answer, I knew it would be possible but couldn't do it using elementary calculations. $\endgroup$
    – fretty
    Mar 10, 2014 at 18:12

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