Trisection for supersingular genus 2 curves in characteristic 2

By reversing reduction in divisor class arithmetic we 
provide efficient trisection algorithms for supersingular Jacobians of genus $2$ 
curves over finite fields of characteristic $2$. With our 
technique we obtain new results for these Jacobians: we show how to 
find their $3$-torsion subgroup, we prove there is none with 
$3$-torsion subgroup of rank $3$ and we prove 
 
that the maximal $3$-power order subgroup is isomorphic to 
either $\mathbb{Z}/3^{v}\mathbb{Z}$ or $(\mathbb{Z}/3^{\frac v2}\mathbb{Z})^2$ or $(\mathbb{Z}/3^{\frac v4}\mathbb{Z})^4$, where $v$ is the $3$-adic 
 
valuation $v_{3}$(#Jac(C)$(\mathbb{F}_{2^m})$). Ours are the first trisection formulae available in literature.

To do this we look for an auxiliary polynomial k(x) such that ṽ (x) in 2D = [u(x) 2 , ṽ (x)] is of the form ṽ (x) = v (x) + k(x)u(x).
In the system, u1, u0, and v1 can be written in terms of k1, k0 and v0.3-torsions k1 satisfies the degree 5 equation k0 satisfies a degree 8 equation (depending on k1) and v0 satisfies a degree 2 equation (depending on k1 and k0)

Theorem
There is a bijection between triples of solutions (k1, k0, v0) ∈ F2m × F2m × F2m satisfying the three equations above, and the set of divisors of order 3 in Jac(C)(F2m ).
There should be 81 valid solutions over Fq, and either 0 or 3 r solutions over Fq where r is the 3-rank of the curve over Fq.
Then de-reducing comes down to finding k1x + k0 and k3x + k2 which connect [u3, v3] and [u1, v1] (via [ua, va]).This is similar to the de-reduction technique used for bisections in genus 2 curves, hence the name double linear de-reduction.

General case
Using the representations of va(x) and ṽ (x) into the equations for the u(x)-terms, we obtain Expanding this last equality according to the degrees in x, we obtain 6 equations en u11, u10, k0, k1, k2, and k3.
To simplify the equations, we make the substitutions

General case
From the coefficients of x 5 and x 4 we get: and then the coefficients of x 2 and x 3 can be combined to find k0:

Remark
The case u31 = 0 must be handled separately and is easier than the case u31 = 0.
Weight-1 trisectees , v31] (D3 has weight 1), the process is similar to the general case (double linear de-reduction), except that no special case can occur.(all trisections have weight 2 and all de-reductions are linear).
From which we get u11, u10 and k0 in terms of k1 and t0 = k 2 k 3 : with the final three equations in terms of k1, t0 and t1 = 1 k 3 .
Since one equation does not depend on t1, solving the can be reduced (via two resultants) to a degree 81 polynomial in k1.
Group Structure k = 1: in 3-torsion the only "wrong" coefficients satisfy By induction hypothesis there exists a formula by which h 2 0 u30 + u31v 4 31 depends only on u31, and t (a root of p D (x) = p k u 31 (x)), then clearly p D (x) = p k+1 u 31 (x) for any D .Hence p D (x) is the same for all distinguished D of order 3 k : p D (x) = p k u 31 (x).