10
$\begingroup$

It is known that the isomorphism class of a quaternion algebra $A=\binom{a,b}{K}$ over a number field $K$ is determined by the finite set of places $v$ of $K$ where $A\otimes_K K_v$ is a division algebra, equivalently at which the projective curve $ax^2+by^2-z^2$ fails to have a $K_v$-rational point.

Given two quaternion algebras $\binom{a,b}{\mathbb{Q}}$ and $\binom{c,d}{\mathbb{Q}}$ over $\mathbb{Q}$ that are ramified at the same places, is there a known algorithm to construct an explicit isomorphism? (I.e. to find $I,J\in \binom{a,b}{\mathbb{Q}}$ such that $I^2=c, J^2=d, IJ=-JI$?)

If so, I would love some references.

$\endgroup$
1
  • 5
    $\begingroup$ Two quaternion algebras over $\mathbf Q$ are isomorphic iff the norm forms on their pure quaternion (trace zero) subspaces are isometric quadratic forms. So this question is tantamount to asking how to construct an isometry between $-ax^2 - by^2 + abz^2$ and $-cx^2 - dy^2 + cdz^2$ if you are told they are isometric. (I'm just pointing out that the question can be phrased equivalently at the level of quadratic form equivalence.) $\endgroup$
    – KConrad
    Mar 10, 2014 at 20:48

1 Answer 1

9
$\begingroup$

Timo Hanke has written on this problem in the generality of cyclic algebras (http://arxiv.org/abs/math/0702681). He shows that it is equivalent to the solution of a norm equation, which has an algorithmic solution over global fields. In your case, this would in general involve the solution of a norm equation over a biquadratic field; I don't know how well this would fare in practice.

In the special case of quaternion algebras that you are inquiring about, there is a link to quadratic forms, and as Keith Conrad mentions, it is equivalent to find a zero of a quadratic form in six variables. This goes back to Albert, who looked at this form in detail in order to prove that there was an "honest" (non-quaternion) biquaternion algebra. A good reference for this is section 16 of the "Book of Involutions" or section XII.2 of Lam's "Introduction to Quadratic Forms over Fields". It may seem like just a reformulation of the problem, but it turns out there are algorithmic methods to find points on quadrics over number fields that are quite efficient in practice: the buzzword here is "indefinite LLL", and Watkins (http://magma.maths.usyd.edu.au/~watkins/papers/illl.pdf) explains what Magma does to accomplish this task.

In general, here is an idea I kicked around once which at least reduces the problem to solve norm equations over quadratic extensions (instead of biquadratic extensions). This might be only of theoretic/algorithmic interest, but at least it potentially generalizes. We wish to test if $A \cong B$ over a global field $F$ (say of characteristic not $2$ for now), and if so, to find an explicit isomorphism. If $A=(a,b)$ and $B=(c,d)$ and $a=c$, then there is an isomorphism if and only if $b/d$ is a norm from $\mathbb{Q}(\sqrt{a})$, and this can be accomplished algorithmically by a norm equation over this field; so it is enough to reduce to this case. To find a common subfield $K=\mathbb{Q}(\sqrt{a})$ in $A,B$, one can simply pick one (choose $K$ such that $K_v$ is not split at all places $v$ ramified in $A$ and $B$, e.g., take $a=-\mathrm{lcm}(ab,cd)$ if $\gcd(a,b)=\gcd(c,d)=1$, so this step does not even require factoring). The problem of embedding a quadratic field $K$ in a quaternion algebra is equivalent to (checking if $K$ splits the algebra and so) to a norm equation (a standard result, see e.g. http://www.math.dartmouth.edu/~jvoight/articles/quatalgs-060513.pdf). Once the field is embedded, we can diagonalize the quadratic form to reduce to the case where $a=c$.

This approach probably generalizes to cyclic algebras (of any characteristic), and if it is interesting to you, it is something I would be happy to work out with you.

$\endgroup$
1
  • $\begingroup$ (There is just a typo: one should read "there is an isomorphism if and only if đť‘Ź/đť‘‘ is a norm" ; this is for instance proved in theorem 5.1 of K. Conrad's notes on Quaternion algebras, and exercise 16 p. 80 in your book as well). $\endgroup$
    – Watson
    Jan 13, 2021 at 8:25

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.