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I wanted to compute any maximal order in the non-split quaternion algebra $\left(\frac{21, -7}{\mathbf{Q}(\sqrt{-3})}\right)$, so I did the following in MAGMA:

_<x> := PolynomialRing(Rationals());
F<t> := NumberField(x^2 + 3);
A := QuaternionAlgebra<F | 21, -7>;
O := MaximalOrder(A);
PseudoBasis(O);

I got some output that was hard to understand:

[
    <Principal Ideal
    Generator:
        $.1, 1>,
    <Fractional Ideal
    Two element generators:
        1/2*$.1
        -1/6*$.1 - 1/6*$.2, 3 + i>,
    <Principal Ideal
    Generator:
        -$.1, j>,
    <Fractional Ideal
    Two element generators:
        1/14*$.1
        -1/42*$.1 - 1/42*$.2, 21*j + k>
]

Surprisingly, it was difficult for me to get a straight answer from the documentation about the meaning of this (and if anyone can point me to where it is explained that would be wonderful). Here is my interpretation of it (after checking using IntegralBasis(F) what the generators \$.1 and \$.2 are and thinking about what the structure theorem for finitely generated torsion-free modules over a Dedekind domain says): this maximal order is $$\mathcal{O} = \mathcal{O}_F + \left(\frac{1}{2}\mathcal{O}_F + \frac{3 + \sqrt{-3}}{12}\mathcal{O}_F\right)(3 + i) + \mathcal{O}_F j + \left(\frac{1}{14}\mathcal{O}_F + \frac{3 + \sqrt{-3}}{84}\mathcal{O}_F\right)(21 j + k)$$ where $\mathcal{O}_F = \mathbf{Z}[\frac{1 + \sqrt{-3}}{2}]$ is the maximal order of $F = \mathbf{Q}(\sqrt{-3})$ and $1, i, j, k$ are the usual basis of the quaternion algebra.

I have two questions:

(1) Is the above interpretation of MAGMA's output correct ?

(2) Of course the point of the pseudobasis stuff is that MAGMA is supposed to work even when $\mathcal{O}_F$ is not a PID. But in this case it is. In this particular case, we can rewrite $$\mathcal{O} = \mathcal{O}_F + \frac{1}{6}\sqrt{-3}\mathcal{O}_F(3 + i) + \mathcal{O}_F j + \frac{1}{42}\sqrt{-3}\mathcal{O}_F(21j + k).$$

Why did MAGMA choose to give such complicated expressions for the ideals involved ? Could we have forced it to choose the minimum possible number of generators (in this case 1) for the ideals that are part of the data of the pseudobasis here ?

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1 Answer 1

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(1) Yes, your interpretation is correct.

(2) Computing a single generator (when it even exists) of a fractional ideal in a number field can be very costly (not polynomial time) and yield a very large generator (not polynomial size) in general, contrary to the two-generators representation. This is enough for computer algebra software to usually choose not to compute them by default. You could then argue that for fields that are small enough they could choose to do it (imaginary quadratic fields are special in this respect). But then it is not clear which threshold to choose, the behaviour would depend on the field in a way that could be impredictable to the user, and the software would be performing an extra computation even if the user does not need it. It is better to let the user do the extra computation if they want to.

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  • $\begingroup$ Thank you ! Sounds good to me. $\endgroup$ Commented Oct 15, 2024 at 19:35

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