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Suppose $D$ is an explicitly given rank 9 central simple algebra over ${\mathbb Q}$ (or a number field). For example it could be specified by two cubic subfields ${\mathbb Q}(a), {\mathbb Q}(b)\subset D$ and one relation $ba=\sum a_i b_j$. What is the best way to determine whether $D$ is a matrix algebra or a division algebra?

Any ideas or references would be highly appreciated! I simply tried to look for random elements whose minimal polynomial is reducible (sort of like Parker's meataxe for group representations), but this is not terribly efficient, and I don't know how to do this cleverly. Perhaps one could also take many cubic subfields and record which primes split in them - if the union of split primes over all subfields seems to cover all prime numbers, this suggests that $D$ is a matrix algebra; again, I don't know how to make this into an actual algorithm.

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  • $\begingroup$ What is your application? E.g., are you interested in whether or not an abelian variety is absolutely simple? I've thought about this problem in the context of endomorphism rings of modular abelian varieties; there one can tell whether or not the algebra is a matrix algebra using results of Ribet and Lario (?), but finding an explicit representation as a matrix algebra is harder. $\endgroup$
    – William Stein
    Commented Mar 5, 2011 at 21:28
  • $\begingroup$ Rachel Newton is doing explicit computations with cup products in class field theory, and one of the things she wants to test boils down to just this. So these are really just central simple algebras, honestly given by generators and relations. $\endgroup$
    – Tim Dokchitser
    Commented Mar 5, 2011 at 22:24
  • $\begingroup$ Not efficient, but an idea: assume it is a matrix algebra, i.e. there are matrices $A$ and $B$, satisfying cubics which you know, and a product relation. Let the coefficients of $A$ and $B$ be 9 variables each. From the relations you get $9+9+9$ equations, that give an ideal in $\mathbb{Q}[x_1,...,x_{18}]$. Now you can try to use Groebner bases to find a solution, or prove that one doesn't exist. $\endgroup$
    – Dror Speiser
    Commented Mar 5, 2011 at 23:32
  • $\begingroup$ @Dror: Thank you! It looks like this is way too many variables and equations for the Groebner basis machinery to handle, but I'll think a bit more about this... $\endgroup$
    – Tim Dokchitser
    Commented Mar 6, 2011 at 8:48
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    $\begingroup$ Have you tried using the Galois cohomology routines in Magma? Assuming you can get a 2-cocycle representing your algebra, have a look at SUnitCohomologyProcess and IsGloballySplit, which will test an explicit 2-cocycle to see whether it's a coboundary. I must admit I've only tried it for quaternion algebras. There is a paper by Claus Fieker describing the algorithm (essentially an S-unit calculation, but being cunning to reduce S as much as possible). Alternatively there might be something in Cremona, Fisher, O'Neil, Simon, Stoll's papers on explicit descent. (You could ask Tom.) $\endgroup$
    – Martin Bright
    Commented Mar 7, 2011 at 14:26

3 Answers 3

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This may be repeating what others have said as it essentially follows the maximal order approach, but have you looked at Nebe, Gabriele; Steel, Allan, Recognition of division algebras, J. Algebra 322 (2009), no. 3, 903–909? http://dx.doi.org/10.1016/j.jalgebra.2009.04.026

Preprint version and magma code available here: http://www.math.rwth-aachen.de/~nebe/pl.html

(I know this should really be a comment, but I'm afraid that I don't have enough reputation yet.)

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Tim, this is in general difficult (as the suggestions show). What I would do in/with Magma is - compute an explicit co-cycle parametrizing your central simple algebra. For this to work, you need a matrix representation over a number field that is normal over the centre of the algebra. (Basically compute the Galois action as conjugating matrices) - from the co-cycle you can compute the Schur-indices as the exponent in the Brauer group (either globally or locally) - comparing dimensions you should be able to see if you have a division algebra (skew-field) or not.

The MeatAxe approach can also be useful - Allan Steel developed this systematically further. However, in difficult cases, currently, one needs to revert to the Galois cohomology as above.

PS.: an improvement, while using the same core idea is to compute a maximal order of the algebra. The discriminant gives then enough information to derive the Schur indices, thus avoiding the Galois cohomology.

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Let $k$ be the base field, and $K$ be a splitting field of your algebra. Then you can calculate the relative Brauer group as the second cohomology of the (simplicial) Amitsur complex $k^\times \to K^\times \to K\otimes_k K^\times \to K\otimes_k K\otimes_k K^\times \to\dots$. Now, suppose that you know how to associate an element of $M\in K\otimes_k K^\times$ to your algebra. Then it is easy to determine whether its splits (i.e. whether $M$ equals $x\otimes x^{-1}$ for some $x\in K^\times$); in order to recover such an $x$ it suffices to project $K\otimes_k K$ onto $K$ by the linear map $a\otimes b\mapsto a\cdot tr_{K/k}(cb)$ for some (fixed) $c\in K^{\times}$.

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  • $\begingroup$ Thank you! This approach is new to me, so could I perhaps ask you to add a few more details? I can certainly find a (non-Galois) splitting field $K$ by taking a subfield of $D$; how do I can compute the class of $D$ in $(K\otimes K)^\times$ and how does this trace map help exactly? $\endgroup$
    – Tim Dokchitser
    Commented Mar 6, 2011 at 8:44
  • $\begingroup$ Actually, trace is not really relevant. You just need any $k$-linear map $f:K\to k$ such that $f(x^{-1})\neq 0$. Then we have $id_K\otimes f(x\otimes x^{-1})=x\otimes ff(x^{-1})$; so you can recover $x$ (up to a $k$-multiplier that does not matter anyway). Probably the method that associates a 2-cocycle to $D$ could be found in the original paper of Amitsur: ams.org/journals/tran/1959-090-01/S0002-9947-1959-0101265-7/… $\endgroup$
    – Mikhail Bondarko
    Commented Mar 6, 2011 at 11:13

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