trisection for supersingular genus 2 curves in ... · pdf fileunicamp.jpg ubiobio.jpg...
TRANSCRIPT
Unicamp.jpgUBiobio.jpg
Trisection for supersingular genus 2 curves in characteristic 2
Josep Miret1, Jordi Pujolas1, and Nicolas Theriault2
1Departament de Matematica, Universitat de Lleida, Espana.
2Departamento de Matematica, Universidad del Bıo-Bıo, Chile.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 1 / 25
Unicamp.jpgUBiobio.jpg
1 Introduction
2 3-torsions
3 TrisectionsGeneral caseSpecial cases
4 Structure of Jac(C)[3∞](F2m )Structure of Jac(C)[3](F2m )Structure of Jac(C)[3k ](F2m ), k > 1
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 2 / 25
Unicamp.jpgUBiobio.jpg
Supersingular Hyperelliptic Curves of Genus 2
Let Fq be a finite field of characteristic 2, then a super singular curve of genus 2 can begiven by an equation of the form
y 2 + h0y = x5 + f3x3 + f1x + f0
with h0 6= 0.
From a hyper elliptic curve, we can define the divisor class group: the quotient group ofthe divisors of degree zero modulo the principal divisors.
The reduced divisors in each class group can be given in Mumford representation,
D = [u(x), v(x)],
with u monic of degree at most 2, deg(v) < deg(u), and satisfying u|v 2 + h0v + f .
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 3 / 25
Unicamp.jpgUBiobio.jpg
Trisections
Given a reduced divisor D3 = [u3(x), v3(x)], we want to find all the possible reduceddivisors D1 = [u1(x), v1(x)] such that
D3 = 3D1,
the trisections of D3.
From the group structure, we know that
Jac(C)[3] ≡ (Z/3Z)4 .
Over Fq, the problem consists in computing the 81 trisections of D3.
Over Fq, we have to identify which of the 81 trisections are defined over Fq.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 4 / 25
Unicamp.jpgUBiobio.jpg
3-torsions
We are looking for the non-zero trisections of the divisor class 0 (Mumfordrepresentation: [1, 0]), i.e. divisors D such that 3D ≡ 0.
We first observe that for curves of genus 2, D must have weight 2, soD = [u(x), v(x)] = [x2 + u1x + u0, v1x + v0].
Rather than solve 3D = 0, we will solve 2D ≡ −D = [x2 + u1x + u0, v1x + (v0 + h0)].
To do this we look for an auxiliary polynomial k(x) such that v(x) in 2D = [u(x)2, v(x)]is of the form v(x) = v(x) + k(x)u(x).
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 5 / 25
Unicamp.jpgUBiobio.jpg
3-torsions
The coefficients of x0, x1, x2, and x3 in
k21u(x)2 =
v 2(x) + h0v(x) + f (x)
u(x)+ h0k(x) + k2(x)u(x).
give us 4 equations in {u1, u0, v1, v0, k1, k0}, and the divisibility condition
v(x)2 + h0v(x) + f (x) ≡ 0 mod u(x)
gives us two more, completing the system.
In the system, u1, u0, and v1 can be written in terms of k1, k0 and v0.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 6 / 25
Unicamp.jpgUBiobio.jpg
3-torsions
k1 satisfies the degree 5 equation
h0k15 + f3k1
4 + 1 = 0,
k0 satisfies a degree 8 equation (depending on k1)
k41 k0
8 + h20k
101 k0
4 + h20k
81 k0
2 + h30k
121 k0 + f 2
1 k121 + f3h
20k
101 + h2
0k61 = 0,
and v0 satisfies a degree 2 equation (depending on k1 and k0)
k21 v0
2 + h0k21 v0 + k2
1 f0 + h0k30 = 0.
Theorem
There is a bijection between triples of solutions (k1, k0, v0) ∈ F2m × F2m × F2m satisfyingthe three equations above, and the set of divisors of order 3 in Jac(C)(F2m ).
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 7 / 25
Unicamp.jpgUBiobio.jpg
Factorization types
Pu1 (X ) is the translate of a 2-linearized polynomial=⇒ The number of roots is 0, 1, 2, 4, 8 in any field extension.
Proposition
The possible factorization types of Pu1 (X ):
[1, 1, 1, 1, 2, 2], [1, 1, 2, 4], [2, 2, 2, 2], [4, 4] if m is odd.
[1, 1, 1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 2, 2], [1, 1, 3, 3], [2, 2, 2, 2], [2, 6], [4, 4] if m is even.
Direct consequences:
Corollary
if 3-torsion subgroup of rank 4, then m is even.
if Pu1 (X ) has 8 linear factors, then m is even.
In general, restricts the possible extension degrees where the 3-torsion divisors are defined.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 8 / 25
Unicamp.jpgUBiobio.jpg
Reducing 3D1 (Cantor)
Suppose that we obtain 3D1 = [u31(x), v(x)] using Cantor’s composition algorithm (D1 is
not of order 3), then the simple reduction of Cantor works as follows (assuming v(x) hasdegree 5):
ua(x) = Monic
(v 2 + hv + f
u31
)va(x) = v + h mod ua(x)
u3(x) = Monic
(v 2a + hva + f
ua
)v3(x) = va + h mod u3(x).
We refer to this as double linear reduction.
Special cases:
D1 of weight 1
3D1 = [u31(x), v(x)] with v(x) of degree ≤ 4 (simple quadratic reduction)
D3 of weight 1 (va(x) has degree ≤ 2).
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 9 / 25
Unicamp.jpgUBiobio.jpg
De-reduction
Our goal is to go from D3 = [u3, v3] to 3D1 = [u31(x), v(x)], or rather to D1 = [u1, v1]
with v1 = v mod u1, undoing the reduction algorithm.
There should be 81 valid solutions over Fq, and either 0 or 3r solutions over Fq where r isthe 3-rank of the curve over Fq.
Writing va(x) and v(x) as
va(x) = v3(x) + (k1x + k0)u3(x) + h(x)
v(x) = va(x) + (k3x + k2)ua(x) + h(x)
where k1 · k3 6= 0.
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, hencethe name double linear de-reduction.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 10 / 25
Unicamp.jpgUBiobio.jpg
General case
Using the representations of va(x) and v(x) into the equations for the u(x)-terms, weobtain
ua(x) =1
k21
(v 2
3 + v3h + f
u3+ (k1x + k0)2u3 + (k1x + k0)h
)and then
u1(x)3 =1
k31
(k2
1u3 + (k3x + k2)2ua + (k3x + k2)h).
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
t1 = 1/k1
t3 = 1/k3
z = t3k2 = k2/k3
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 11 / 25
Unicamp.jpgUBiobio.jpg
General case
From the coefficients of x5 and x4 we get:
u11 = u31 + t21
u10 = u30 + u231 + z2 + k2
0 t21 + u31t
21 + t4
1
and then the coefficients of x2 and x3 can be combined to find k0:
k0 =1
h0u231t
41
t14
1 + u31t121 + t6
1z4 + u2
31t61z
2 + u431t
61 + f3u31t
81
+u31t41z
4 + h0u31t71 + u2
31t21z
4 + u431t
21z
2
+u231v
231t
41 + u3
31u30t41 + f3u31t
41z
2 + u230t
61 + f 2
3 t61
+h0u31t31z
2 + h0u331t
31 + u31u
230t
41 + f3u31u30t
41
+u231u
230t
21 + h0u31u30t
31 + h2
0t41 + u2
31t23
Remark
The case u31 = 0 must be handled separately and is easier than the case u31 6= 0.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 12 / 25
Unicamp.jpgUBiobio.jpg
General case
The coefficients in x2 and x3 also leave us an equation
0 = t281 + u2
31t241 + t12
1 z8 + u431t
121 z4 + u8
31t121 + f 2
3 u231t
161 + u2
31t81z
8
+ h20u
231t
141 + h2
0u331t
121 + u4
31t41z
8 + u831t
41z
4 + u431v
431t
81 + u6
31u230t
81
+ f 23 u
231t
81z
4 + h20u
431t
101 + u4
30t121 + f 4
3 t121 + h2
0u331t
81z
2 + h20u
231t
61z
4
+ h20u
431t
61z
2 + u231u
430t
81 + f 2
3 u231u
230t
81 + h2
0u331u30t
81 + f3h
20u
331t
81
+ h30u
331t
71 + u4
31u430t
41 + h2
0u231u
230t
61 + h4
0t81 + u4
31t43
and substituting k0 in the coefficients of x0 and x1 gives us two more equations in z , t1
and t3.
After some careful combinations, we can write t3 in terms of t1 and z , and obtain
p1(t1, z) = 0
p2(t1, z) = 0
with p1 of degree 33 in t1 and degree 12 in z , and p2 of degree 15 in t1 and degree 2 in z .
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 13 / 25
Unicamp.jpgUBiobio.jpg
General case
Taking the resultant (in z) of p1 and p2 and cleaning out “parasitic” factors, we find
Resz(p1(t1, z), p2(t1, z)) = t91 · (t5
1 + (u231 + f3)t1 + h0)12 · qD3 (t1)
where qD3 (t1) is a polynomial of degree 81 (except in some special cases).
Proposition
Each non-zero root t1 of qD3 (X ) = 0 corresponds to a trisection D1 of D3 via doublelinear de-reduction.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 14 / 25
Unicamp.jpgUBiobio.jpg
Weight-1 trisections
If D1 = [x + u10, v10], the reduction of 3D takes only one simple step.Setting v(x) = v3(x) + k0u3(x) + h(x), we obtain the equality
(x + u10)3 =v3(x)2 + v3(x)h0 + f (x)
u3(x)+ k2
0u3(x) + k0h0,
which turns into 3 equations, first
u10 = k20 + u31,
and then
s1(k0) = k40 + k2
0u31 + f3 + u30
s2(k0) =(u2
31 + f3)k2
0 + h0k0 + v 231 + u31f3
Proposition
D3 admits a trisection of weight 1 if and only if
f 63 + f 4
3 u230 + f 3
3 h20u31 + f 2
3 h20u
331 + f 2
3 h20u31u30 + f 2
3 u231v
431 + f3h
40
+f3h20u
531 + h4
0u30 + h20u
531u30 + h2
0u31v431 + u8
31u230 + u6
31v431 + v 8
31 = 0
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 15 / 25
Unicamp.jpgUBiobio.jpg
Weight-1 trisectees
If D3 = [x + u31, v31] (D3 has weight 1), the process is similar to the general case (doublelinear de-reduction), except that no special case can occur. (all trisections have weight 2and all de-reductions are linear).
ua =v 2
3 + v3h + f
u3+ (k1x + k0)2u3 + (k1x + k0)h
(x2 + u11x + u10)3 =u3
k23
+(k3x + k2)2ua
k23
+(k3x + k2)h
k23
From which we get u11, u10 and k0 in terms of k1 and t0 = k2k3
:
u11 = k21 + u30
u10 = k41 + k2
1u30 + f3 + t20
k20 = k6
1 + k21 f3 + k1h0
with the final three equations in terms of k1, t0 and t1 = 1k3
.
Since one equation does not depend on t1, solving the can be reduced (via tworesultants) to a degree 81 polynomial in k1.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 16 / 25
Unicamp.jpgUBiobio.jpg
Simple quadratic de-reduction
In some cases, D1 and D3 both have weight 2, but v in 3D1 = [u31 , v ] is of degree ≤ 4.
In this case, the reduction requires only one step, and we can write
v(x) = v3(x) + (k2x2 + k1x + k0)u3 + h(x)
Proposition
D3 = [x2 + u31x + u30, v31x + v30] admits a simple quadratic de-reduction only if
u531 + f 2
3 u31 + h20 = 0
and the associated D1 has u11 = u31.
working out the equations, we obtain a degree 9 trisection polynomial
pD3 (t) = t9 + u431h
20t
3 + (u1231 + u6
31v431 + u8
31f2
3 + u531h
20u30)t + u6
31h30.
and u10 = u30 + t2/u31
Remark
If u531 + f 2
3 u31 + h20 = 0, then the trisection polynomial for double linear de-reduction has
degree 72 after removing 9 (false) roots t1 = 0.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 17 / 25
Unicamp.jpgUBiobio.jpg
Goals
@ supersingular curves of genus 2 over any field F2m with a 3-torsion subgroup ofrank 3
the exponents of Jac(C)[3∞](F2m ) are equal
Jac(C)[3k ](F2m ) has a distinguished basis for every k
For curves with coefficients in F2, the first two results follow from C. Xing’sOn supersingular abelian varieties of dimension two over finite fields (1996).
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 18 / 25
Unicamp.jpgUBiobio.jpg
Structure of the group
Supersingular curves have simplified equation.
D = [x2 + u1x + u0, v1x + v0]D ′ = [x2 + u1x + u′0, v
′1x + v ′0]
}⇒ u1(±(D + D ′) = u1(±(D − D ′))
by direct computation + supersingularity.
Proposition
If Pu1 (X ) has more than 4 roots (type [1, . . . , 1]), then rank(Jac(C)[3]) 6= 3:
if 3-torsion has 3 generators D1,D2,D3 with the same u1, there must be 5 other divisorsin 〈D1,D2,D3〉 with same u1, but they go together in pairs by property above!
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 19 / 25
Unicamp.jpgUBiobio.jpg
Rank 3
Theorem
There are no supersingular C in characteristic 2 such that Jac(C)[3] has rank 3
We can assume factorization type is not [1, . . . , 1]
for each u1 obtain at most 4 u0’s, so need the 5 u1’s over F2m .
40− 13 = 27 remaining pairs (u1, u0) to be found over extension of degree 2, 3, 6because of factorization types.
But we need 26/2 = 13 first coordinates over F2m . This leads to contradiction forevery m = 2, 3, 6.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 20 / 25
Unicamp.jpgUBiobio.jpg
Distinguished Bases
Corollary (B)
There exists a basis of Jac(C)[3] with elements having the same u1.
In rank 4, there must exist a root u1 leading 8 u0’s. From the divisors with thesefirst components x2 + u1x + u0 there exists a basis of Jac(C)[3]
In rank 2, it follows from the factorization types of Pu1 (x)
Call such divisors “distinguished”
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 21 / 25
Unicamp.jpgUBiobio.jpg
Quadratic De-reductions
Consider the situation:
D = [x2 + u1x + u0, v1x + v0] a divisor such that u1 satisfies
u51 + f 2
3 u1 + h20 = 0
pD(t) = t9 + ...+ u631h
30
the degree 9 quadratic-dereduction polynomial
qD(t1) = (...)t721 + ...+ h24
0
the degree 72 twice-linear-dereduction polynomial.
Proposition (D)
If qD(t1) has a root α1 ∈ F2m then pD(t) has a root β1 ∈ F2m .
That is, if D has a trisection by twice-linear-dereduction D only if it has a trisection byquadratic-dereduction
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 22 / 25
Unicamp.jpgUBiobio.jpg
Group Structure
Theorem
The nonzero exponents of Jac(C)(F2m )[3∞] are equal.
Corollary
For every k, there is a distinguished basis of Jac(C)[3k ](F2m ).
Sketch:
By Corollary (B), we can take a distinguished basis in Jac(C)[3]
By Proposition (D), if a distinguished divisor has a trisection, it has a distinguishedtrisection (this is not trivial in rank 2)
Prove by induction: For D = [x2 + u31x + u30, v31x + v30] ∈ Jac(C)[3k ](F2m ), Ddistinguished of order 3k , =⇒ pD(x) does not depend on undistinguished coefficientsu30, v31, v30 but only on u31 and k.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 23 / 25
Unicamp.jpgUBiobio.jpg
Group Structure
k = 1: in 3-torsion the only “wrong” coefficients satisfy
h20u30 + v 4
31u31 = f 23 u
331 + u7
31
troubled coefficients in next level pD′(x) are
h20u11 + u31v
411 = h2
0u30 + u31v431 +
(h2
0
u31t2 + (u31 + 1)4t4 +
(f3 + u231)4
h20
t8
).
By induction hypothesis there exists a formula by which h20u30 + u31v
431 depends only
on u31, and t (a root of pD(x) = pku31
(x)), then clearly pD′(x) = pk+1u31
(x) for any D ′.
Hence pD(x) is the same for all distinguished D of order 3k : pD(x) = pku31
(x).
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 24 / 25
Unicamp.jpgUBiobio.jpg
Group Structure
It only remains to show: if 13D 6= ∅ for some undistinguished D,
does there exist a distinguished D of the same order such that 13D 6= ∅?
By induction, suppose there exists a distinguished D ∈ Jac(C)[3k ](F2m ) such that3k−1D equals any distinguished basis elements in Jac(C)[3](F2m ).
Let D ∈ Jac(C)[3k ](F2m ) be undistinguished of order 3k such that 13D 6= ∅.
By definition of k, Jac(C)[3k ] must also contain a distinguished D such that3k−1D = 3k−1D.
=⇒ there exists an E of order 3s with s < k such that D = D + E .Therefore 1
3E 6= ∅, hence 1
3D 6= ∅ and our proof is complete.
Miret, Pujolas y Theriault (U.Lleida & Ubiobio) Trisection for supersingular genus 2 curves 25 / 25