# Explicit isomorphism for quaternion algebras over $\mathbb{Q}$?

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.

• 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.) – KConrad Mar 10 '14 at 20:48

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 $$a=c$$.