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reference request – Explicit isomorphism for quaternion algebras over $mathbb{Q}$? Answer

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reference request – Explicit isomorphism for quaternion algebras over $mathbb{Q}$?

Timo Hanke has written on this drawback within the generality of cyclic algebras ( He reveals that it’s equal to the answer of a norm equation, which has an algorithmic answer over world fields. In your illustration, this could in common contain the answer of a norm equation over a biquadratic bailiwick; I do not know the way nicely this could fare in drill.

In the particular illustration of quaternion algebras that you’re inquiring about, there’s a hyperlink to quadratic varieties, and as Keith Conrad mentions, it’s equal to discover a zero of a quadratic figure in six variables. This goes advocate to Albert, who checked out this figure intimately in organize to show that there was an “honest” (non-quaternion) biquaternion algebra. reference for that is part 16 of the “Book of Involutions” or part XII.2 of Lam’s “Introduction to Quadratic Forms over Fields”. It might appear love only a reformulation of the issue, but it surely turns on the market are algorithmic strategies to seek out factors on quadrics over quantity fields which are fairly environment friendly in drill: the buzzword right here is “indefinite LLL”, and Watkins ( explains what Magma does to achieve this job.

In common, right here is an thought I kicked round as soon as which a minimum of reduces the issue to decipher norm equations over quadratic extensions (as a substitute of biquadratic extensions). This energy breathe solely of theoretic/algorithmic curiosity, however a minimum of it probably generalizes. We want to check if $A cong B$ over a world bailiwick $F$ (say of attribute not $2$ for now), and if that’s the case, to seek out an categorical isomorphism. If $A=(a,b)$ and $B=(c,d)$ and $a=c$, then there may be an isomorphism if and provided that $b/d$ is a norm from $mathbb{Q}(sqrt{a})$, and this will breathe completed algorithmically by a norm equation over this bailiwick; so it is sufficient to dwindle to this illustration. To discover a frequent subfield $Ok=mathbb{Q}(sqrt{a})$ in $A,B$, one can merely choose one (select $Ok$ such that $K_v$ is just not splinter in any respect locations $v$ ramified in $A$ and $B$, e.g., take $a=-mathrm{lcm}(ab,cd)$ if $gcd(a,b)=gcd(c,d)=1$, so this step doesn’t plane require factoring). The drawback of embedding a quadratic bailiwick $Ok$ in a quaternion algebra is equal to (checking if $Ok$ splits the algebra and so) to a norm equation (a benchmark outcome, behold e.g. Once the bailiwick is embedded, we will diagonalize the quadratic figure to dwindle to the illustration the place $a=c$.

This method most likely generalizes to cyclic algebras (of any attribute), and whether it is fascinating to you, it’s one thing I’d breathe fortunate to labor out with you.

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