Descent via (3,3)-isogeny on Jacobians of genus 2 curves

0.22 MB·

by  Nils Bruin

DESCENT VIA (3,3)-ISOGENY ON JACOBIANS OF GENUS 2 CURVES
N.BRUIN,E.V. FLYNN,AND D.TESTA
Abstract. We give parametrisation of curves C of genus 2 with a maximal isotropic (Z/3)2 in
J[3], whereJ is theJacobian varietyof C, and develop thetheory required toperform descent via
(3,3)-isogeny. Weapplythistoseveralexamples,whereitcanshownthatnon-reducibleJacobians
4
1 havenontrivial 3-part of theTate-Shafarevich group.
0
2

n
a
J 1. Introduction

3
In this article we consider curves C of genus 2 over a field k of characteristic not 2 or 3, that

] have special structure in the 3-torsion of their Jacobians J. In particular, we consider the situation
T
where J[3](k) contains a group Σ(k) of order 9. As we show in Section 2, such a curve C can be
N
given by a model of the form
.
h
t (1) y2 = F(x) = G(x)2+λH(x)3,
a
m
′ ′
where G(x) is cubic and H(x) is quadratic in x. The divisor D = [(x,G(x)) + (x,G(x )) κ],
[ ′ −
where x,x are the roots of H(x) and κ is a canonical divisor, represents a point in J[3](k), as can

1 be seen from the fact that 3D is linearly equivalent to the divisor of the function y G(x).
v −
The curves of interest to us can be expressed as such a model in several ways. As we show
0
8 in Lemma 5, the Weil pairing of the 3-torsion points can be easily expressed in terms of the
5 correspondingpolynomials G(x) and H(x). This allows us to fully describe the genus 2 curves that
0
have a subgroup Σ(k) J[3](k) of size 9 on which the Weil pairing is trivial.
1. In Section 3 we ph⊂rase the question of classification of such curves in terms of partial level
0
structures on principally polarized abelian surfaces. The relevant moduli space is (Σ), with
4 2
A
1 Σ = Z/3 Z/3. In Section 4 we determine a genus 2 curve rst over k(r,s,t) such that a suffi-
× C
: ciently general curve of the type we are interested in, can be obtained by specializing r,s,t. Our
v
Xi construction identifies k(r,s,t) with the function field of the moduli space A2(Σ), thereby giving
a particularly explicit proof of its rationality. Furthermore, we observe that if J is equipped with
ar a Σ-level structure, then the quadratic twist J(d) is naturally equipped with a Σ(d) level structure.
Thus we find that (Σ) and (Σ(d)) are naturally isomorphic.
2 2
A A
Since Σ J[3] is maximally isotropic with respect to the Weil pairing, we have that J = J/Σ
⊂ ∨
is also principally polarized and in fact has a Σ -level structure. In Section 5 we identify a curve
such that its Jacobian is /Σ. In fact, we observe that Σ∨ = Σ(−3) and henceethat the
rst rst rst
C (−3) J J ′ ′ ′
quadratic twist Crst ≃ Cr′s′t′, where (r ,s,t) = ψ0(r,s,t) is a birational transformation that we
eexplicitly determine. Whileethe final formulas we find are quite manageable, the proof that they
are correct requieres some significant computation.
Date: 1 January, 2014.
1991 Mathematics Subject Classification. Primary 11G30; Secondary 11G10, 14H40.
Key words and phrases. Higher GenusCurves, Jacobians, Shafarevich-Tate Group, Class Groups.
ThefirstauthorispartiallysupportedbyanNSERCgrant. Thesecondandthirdauthorsarepartiallysupported
by EPSRCgrant EP/F060661/1.
1
In Theorem 13 and Corollary 14 we observe some remarkable relations between the models for
and . We also identify the natural action of PGL (F ) on (Σ), as well as the involution
rst rst 2 3 2
C (C−3) A
, as automorphisms of k(r,s,t).
Crst → Crst
In Secteion 6 we get to the original motivation of this paper. If we take J to be a Jacobian
of a geneus 2 curve with a Σ level structure over a number field k, then it is particularly easy to
compute interesting information about J(k) and X(J/k)[3] via (3,3)-isogeny descent, using ideas
from [12] and [5]. Thisallows us to give examples of various absolutely simpleabelian surfaces with
interesting structures in X(J/k)[3], see Examples 19, 20, 21.
2. Three-torsion on genus two Jacobians
Let k be a field of characteristic different from 2,3 and let C be a smooth projective curve of
genus 2 over k given by an affine model
C: y2 = F(x),
where F(x) k[x] is of degree 6 (this only (mildly) restricts the admissible C if k is a finite field
∈
of 5 elements). Let J be the Jacobian of C. The group J(k) is isomorphic to the group of divisor
classes of k-rational degree 0 divisors on C. Since C is of genus 2, every degree 0 class contains a
representative of the form
D κ,
−
where D is an effective divisor of degree 2 and κ is an effective canonical divisor. Furthermore,
for any non-principal class, the divisor D is unique. For κ we have choice, since effective canonical
divisors are exactly the fibers of the hyperelliptic double cover x: C P1. We write ι: C C for
→ →
the hyperelliptic involution, i.e., ι(x,y) = (x, y).
−
Lemma 1. Let Σ Pic(C/k)[3] be a subgroup of size 9. Then Σ can be generated by a pair of
⊂
divisors
D κ ,D κ
1 1 2 2
− −
where D ,D are effective divisors of degree 2 and κ ,κ are effective canonical divisors, with the
1 2 1 2
supports of D1,D2,κ1,κ2 pairwise disjoint. A fortiori, we can ensure that x∗(D1) and x∗(D2) are
disjoint.
Proof. First, note that we can choose κ and κ to be any fiber of x: P1 over points in P1(k).
1 2
C →
Sincek has characteristic 0or at least 5, we know that#P1(k) 6. Sincex∗(D1+D2) is supported
≥
on at most 4 rational points, we can choose κ ,κ with disjoint support from D ,D .
1 2 1 2
It remains to show that we can choose D ,D with disjoint support. Since these divisors are
1 2
uniquely determined, we lose no generality by assuming that k is algebraically closed. Therefore,
we assume that Σ is generated by the classes
T = [P +Q κ ] and T = [P +Q κ ],
1 1 1 1 2 2 2 2
− −
where P ,Q ,P ,Q C(k). We want to ensure that P ,ιP ,Q ,ιQ and P ,ιP ,Q ,ιQ are
1 1 2 2 1 1 1 1 2 2 2 2
∈ { } { }
disjoint. If they are not, we can assume that P = P and it follows that
1 2
T = T T = [Q +ιQ κ].
3 1 2 1 2
− −
A straightforward computation shows that if T =T +T = [P +Q κ] has P = P or P = ιP
4 1 2 4 4 4 1 4 1
−
then either T = [2P κ] and T = [2Q κ], so that choosing T ,T works, or T ,T is not of
4 1 3 1 3 4 1 2
size 9. Therefore, one−of the choices (T ,T−) or (T ,T ) or (T ,T ) satisfies our crihteria. i (cid:3)
1 2 1 4 3 4
Example 2. Atthispointitmaybeworthnotingthatratherexceptional configurationsofsupport
for 3-torsion do occur. For instance, for
C: y2 = x6+rx3+1
2
we see that T = [(0,1) + + κ] and T = [(0,1) + − κ] are 3-torsion points. In fact, it
1 2
∞ − ∞ −
is straightforward to check that any genus 2 Jacobian with two independent 3-torsion points such
that the group generated by those 3-torsion points is supported on only 4 points of the curve must
be isomorphic to a curve of this form.
Lemma 3. Let C be a curve of genus 2. Then Pic(C/k)[3] has a subgroup Σ of size 9 if and only
if C admits a model of the form
C: y2 = F(x) = G (x)2 +λ H (x)3 = G (x)2+λ H (x)3,
1 1 1 2 1 2
×
where H ,H ,G ,G ,F k[x] and λ ,λ k and H ,H are of degree 2 and gcd(H ,H ) = 1.
1 2 1 2 1 2 1 2 1 2
∈ ∈
Proof. Suppose that T Pic(C/k)[3] is non-trivial. We assume that T = [D κ∞], where κ∞ is
∈ −
the fiber above x = and D is an effective divisor with support disjoint from κ∞. Then x∗(D)
∞
can be described by H(x) = 0, where H(x) k[x] is a quadratic monic polynomial. Since 3T is
∈
the principal class, there is a function g k(C) such that
∈
div(g) = 3D 3κ∞
−
and it is straightforward to check that we must have g = y G(x) for some G(x) k[x], with
− ∈
deg(G) 3. It follows that
≤
y2 = F(x) = G(x)2+λH(x)3,
and conversely, that any such decomposition of F(x) gives rise to a 3-torsion point T = [D κ],
−
where D is the effective degree 2 divisor described by the vanishing of y G(x),H(x) . The class
{ − }
2T = T is then described by the vanishing of y+G(x),H(x) .
− { }
The existence of Σ as stated in the lemma would lead to 4 decompositions of the type described
and simple combinatorics shows that not all quadratic polynomials H(x) featuring in them can be
equal. This proves the lemma. (cid:3)
The torsion subgroup scheme J[3] comes equipped with a non-degenerate, bilinear, alternating
Weil pairing
e : J[3] J[3] µ ,
3 3
× →
where µ is the group scheme representing the cube roots of unity.
3
We say that a subgroup Σ J[3] is isotropic if e restricts to the trivial pairing on Σ. If Σ
3
⊂
is of degree 9 then Σ is maximal isotropic, meaning Σ is not properly contained in an isotropic
∨
subgroup. The nondegeneracy of e then induces an isomorphism J[3]/Σ Σ = Hom(Σ,µ ). In
3 3
∨ →
fact, we have a direct sum decomposition J[3] = Σ Σ .
In particular, if Σ= Z/3 Z/3 then we have Σ∨×= µ µ .
3 3
× ×
Lemma 4. Let
C: y2 = F(x) = G (x)2+λ H (x)3 = G (x)2+λ H (x)3
1 1 1 2 2 2
be a genus 2 curve with H ,H quadratic monic polynomials and H = H . For i 1,2 , let
1 2 1 2
6 ∈ { }
D = y G (x),H (x) and let T = [D κ] Pic(C/k)[3]. Then
i i i i i
{ − } − ∈
λ Res(G G ,H )
2 2 1 2
e (T ,T )= −
3 1 2
λ Res(G G ,H )
1 1 2 1
−
Proof. We choose canonical divisors κ and κ above x = r and x = r respectively, such that the
1 2 1 2
divisors
D κ and D κ
1 1 2 2
− −
have disjoint support. We have the functions
y G
i
g = − with div(g )= 3D 3κ .
i (x r )3 i i − i
i
−
3
We can compute the pairing via
g (D κ )
1 2 2
e (T ,T )= − ,
3 1 2
g (D κ )
2 1 1
−
whereevaluating afunction on adivisor is definedas g( n P)= g(P)nP. Evaluating y G (x)
P 1
−
atD means evaluating G (x) G (x)at theroots of H (x), yieldingRes(G G ,H ). Evaluating
2 2 1 2 2 1 2
− P Q −
(x r ) at D yields H (r ). Noting that κ = (r , F(r ))+(r , F(r )), we see that
1 2 2 1 2 2 2 2 2
− −
G1(r2)2 F(rp2) H1(r2)3p
g (κ ) = − = λ −
1 2 (r r )6 1(r r )6
2 1 2 1
− −
and hence that
Res(G G ,H )(r r )6
2 1 2 2 1
g (D κ )= − − .
1 2− 2 λ H (r )3H (r )3
1 1 2 2 1
−
Symmetry yields the result stated in the lemma. (cid:3)
We can characterize when e (T ,T ) = 1 in terms of the polynomials G ,H in the following way.
3 1 2 i i
First note that
G2 G2 = λ H3 λ H3.
2− 1 1 1 − 2 2
Writing α ,α ,α for the cube roots of λ /λ in an algebraic closure of k, we find that
1 2 3 2 1
(G G )(G +G )= λ (H α H )(H α H )(H α H ).
2 1 2 1 1 1 1 2 1 2 2 1 3 2
− − − −
It follows that the roots of the quadratic polynomials H α H are the same as the roots of
1 i 2
−
G G and G +G . The way in which they distribute determines the Weil pairing.
2 1 2 1
−
Lemma 5. The pairing e (T ,T ) = 1 if and only if none of the polynomials (H α H ) divide
3 1 2 1 i 2
−
G G .
2 1
−
Proof. First suppose that G G = L (H α H ). Then
2 1 1 1 1 2
− −
λ Res(G G ,H ) λ Res((H α H )L ,H ) λ Res(L ,H )
2 2 1 2 2 1 1 2 1 2 2 1 2
e (T ,T ) = − = − = .
3 1 2 λ Res(G G ,H ) λ Res((H α H )L ,H ) λ α2Res(L ,H )
1 1 − 2 1 1 1− 1 2 1 1 1 1 1 1
Observe that L must divide one of the other factors (H α H ), say for j = 2. Therefore,
1 1 j 2
−
Res(L ,H ) = Res(L ,α H ). Since deg(L ) = 1 and λ /λ = α3 we obtain
1 1 1 2 2 1 2 1 1
Res(L ,H ) Res(L ,H ) α
1 2 1 2 1
e (T ,T )= α = α = ,
3 1 2 1 1
Res(L ,H ) Res(L ,α H ) α
1 1 1 2 2 2
which is indeed a primitive cube root of unity.
In the remaining situation we have G G = L L L , where, for i 1,2,3 , the polynomial
2 1 1 2 3
− ∈ { }
L divides H α H , and hence Res(L ,H )= Res(L ,α H ). We obtain that
i 1 i 2 i 1 i i 2
−
λ
2
Res(G G ,H )= α α α Res(L L L ,H )= Res(G G ,H ),
2 1 1 1 2 3 1 2 3 2 2 1 2
− λ −
1
which indeed implies that e (T ,T ) = 1. (cid:3)
3 1 2
3. Level structure on principally polarized abelian surfaces
Let n be a positive integer. We need the analogues of the modular curves Y(n), Y (n) and Y (n)
1 0
for genus 2 Jacobians. The theory is most conveniently stated in terms of slightly more general
objects, namely principally polarized abelian surfaces (PPAS). These include direct products of
elliptic curves, equipped with the product polarization. The advantage is that the category of
PPAS is closed under polarized isogenies.
Let k be a field of characteristic not dividing 6n. We write (1) for the moduli space of
2
A
isomorphismclassesofPPASoverk. Thecoarsemodulispace (1)isa3-dimensionalvariety. Note
2
A
4
however that, for any extension L of k, the set (1)(L) of L-rational points of (1) corresponds
2 2
A A
to L-rational isomorphism classes that need not contain an L-rational abelian surface; similarly
two abelian surfaces defined over L that are isomorphic over k need not be isomorphic over L.
Let J be a principally polarized abelian surface over k. Then J[n](k) has a non-degenerate
bilinear alternating Galois covariant Weil-pairing J[n] J[n] µ . Apartial level n structure on J
n
× →
consists of afinite´etale groupschemeΣ withapairingΣ Σ µ andaninjective homomorphism
n
× →
Σ J[n] that is compatible with the pairings. An isomorphism between (J,Σ J[n]) and
′→ ′ ′ →
(J ,Σ J [n]) is an isomorphism φ : J J of PPAS such that the composition of Σ J[n]
→ ′ → →
with φ yields Σ J . We write (Σ) for the moduli space of PPAS equiped with a partial level
2
→ A
n-structure involving Σ. If we take for example a sample abelian surface J and set Σ = J [n] then
0 0
(Σ) is the moduli space of PPAS with full level n structure.
2
A
We will work with the case n = 3 and Σ = Z/3 Z/3 with trivial pairing (i.e., Σ is isotropic
×
with respect to its pairing). Note that the automorphism group of Σ is the full GL (F ). However,
2 3
since on every abelian surface J the involution 1: J J induces the automorphism 1 on level
− → −
structures, we deduce that (J,Σ J) and (J, Σ J) are isomorphic as level n structures on J.
→ − →
Therefore we obtain an action of PGL (F ) on (Σ).
2 3 2
A ∨
In our case Σ J[3] is maximal isotropic, so the nondegeneracy of e yields that J[3]/Σ Σ =
3
⊂ ≃
Hom(Σ,µ ). The fact that Σ is maximal isotropic also means that the principal polarization on
3
J induces a principal polarization on the isogenous abelian surface J/Σ. Furthermore, the Weil
∨
pairing determines an injection Σ (J/Σ)[3]. Thus we see that our maximal isotropic level
→ ∨
structure leads to an isogeny J J/Σ, inducing an isomorphism (Σ) (Σ ), whose inverse
2 2
→ A → A
is induced by the dual isogeny, using the principal polarizations to identify J and J/Σ with their
duals.
Thenegation automorphism on J also gives rise to a quadratic twisting operation. We write J(d)
for the twist of J by the quadratic character of discriminant d. A level structure under twisting
gives rise to a twisted level structureΣ(d) J(d), whereΣ(d) is the quadratic twist of Σ. This gives
→
rise to an isomorphism (Σ) (Σ(d)).
2 2
A → A
We finish by making some observations about the covering degrees of the various moduli spaces
of level structures wehave introduced. Letus write (3) for thespace correspondingto fulllevel 3
2
A
data (say, for Z/3 Z/3 µ µ with obvious pairing). This has a fullPSp (F ) acting on it. The
3 3 4 3
× × ×
subgroups fixing a maximal isotropic space are all conjugate and have the group structure (Z/3)3.
There are 40 of them. We see that (3) (Σ) is finite of degree 27. Thecover (Σ) (1)
2 2 2 2
A → A A → A
is of degree 40 24, determined by the choice of isotropic space times the size of PGL (F ).
2 3
·
The variety (3) is very well-known. Its completion is the Burkhardt quartic, defined by the
2
A
homogeneous equation
(2) t4 t(w3+x3+y3+z3)+3wxyz = 0,
−
as is described in [6,8]. In particular, it, and its finite quotients, are absolutely irreducible.
Note that most PPAS are Jacobians of genus 2 curves in the sense that outside a proper closed
subvariety of (1), any point can be represented by the Jacobian of a curve of genus 2. In what
2
A
follows we will determine a genus 2 curve over the function field of (Σ) together with a level
rst 2
C A
structure on its Jacobian that makes it correspond to the generic point on (Σ).
2
A
4. Parametrisation of genus 2 curves with a maximal isotropic (Z/3)2 in J[3]
Let k be a field of characteristic different from 2,3. In this section we derive a genus 2 curve
rst
C
over k(r,s,t) with two non-trivial divisor classes T ,T Pic( /k(r,s,t))[3] with T = T and
1 2 rst 1 2
∈ C 6 ±
e (T ,T ) = 1. This specifies a Σ-level structure on the Jacobian of . In the process we
3 1 2 rst rst
J C
will see that any sufficiently general Jacobian with a Σ-level structure over k can be obtained via
5
specialization from . This identifies k(r,s,t) with the function field of (Σ), verifying that
rst 2
J A
this moduli space is indeed rational.
We use the notation from Lemmas 3 and 5. We consider the algebra k[α] = k[t]/(t3 λ /λ ),
2 1
−
which is only a field if λ /λ is not a cube in k, but at least will always be an ´etale algebra because
2 1
λ ,λ = 0. We write Nm = Norm . We have
1 2 k[x,α]/k[x]
6
(G G )(G +G ) = λ Nm(H αH ).
2 1 2 1 1 1 2
− −
From Lemma 5 it follows that
(3) H αH = LM for some L,M k[α,x]
1 2
− ∈
×
and that for some c k we have
∈
G G = 1Nm(M), G = 1(cλ Nm(L) 1Nm(M)),
2− 1 c and 1 2 1 − c
G +G = cλ Nm(L), G = 1(cλ Nm(L)+ 1Nm(M)).
2 1 1 2 2 1 c
We observe that
(cy)2 = (cG )2+(c2λ )H3 = (cG )2+(c2λ )H3,
1 1 1 2 2 2
so by adjusting the values of λ ,λ we can assume c = 1. This shows that the isomorphism class
1 2
of is determined by λ ,λ ,L,M.
1 2
C
Furthermore, if k has sufficiently many elements we can ensure that L does not vanish at x =
∞
and that L is monic, so that
L = x (l +αl +α2l ).
0 1 2
−
The fractional linear transformation
l x+tl2 l l
x 1 2 − 0 1,
7→ l x+l2 l l
2 1 − 0 2
with determinant l3 tl3 sends l +l α+l α2 to α. One can check that if l3 tl3 = 0 then either
1 − 2 0 1 2 1− 2
F has a repeated root, and hence our data does not specify a genus 2 curve, or l = l = 0. In the
1 2
latter case, L is already defined over k, so via x x l we can move its root to 0. This shows
0
7→ −
that via a fractional linear transformation, we can assume that
(4) L = x uα and M = (c +c α+c α2)x (m +m α+m α2),
0 1 2 0 1 2
− −
whereu= 0correspondstothecasewherel = l = 0. InorderforLM tobeoftheformH αH ,
1 2 1 2
−
with H ,H k[x], we need
1 2
∈
(5) c = 0, m u = 0, m = c u.
2 1 2 1
−
We set λ = s,λ = st and observe that c2F(x) is homogeneous with respect to the following
1 2
gradings.
s t c c c m m m u x c
0 1 2 0 1 2
weights 3 0 1 1 1 0 0 0 1 1 3
− − − −
0 3 0 1 2 0 1 2 1 0 0
− − − − −
0 0 0 0 0 1 1 1 1 1 3
−
We can solve (5) via either u = 0 or via m = 0. For u = 0 we find that H and H both have a
1 1 2
root at x = 0. By Lemma 1 we know we can avoid this case by changing the basis for the 3-torsion
subgroup. Thus we see that any PGL (F )-orbit has a representative that avoids this locus.
2 3
For the other case we take the affine open described by
(s,t,c ,c ,c ,m ,m ,m ,u) = (s,t,1, 1,0, r,0,0,1),
0 1 2 0 1 2
− −
leading to
H = x2+rx+t, H = x2+x+r, λ = s, λ = st.
1 2 1 2
6
It is instructive to record which cases are excluded by the choices that we make here. We use
the gradings to scale c = 1,c = 1,u = 1, so any curves that require any of these parameters
0 1
−
to be 0 are ruled out. The gradings immediately show that setting any of c ,c ,u to 0 yields at
0 1
most a 2-dimensional family of curves. For u = 0, we have seen that H ,H have a common root.
1 2
Furthermore, the gradings show that this forms at most a 2-dimensional family up to isomorphy.
For c = 0, we see that H has a root at and for c = 0 we see that H has a root at infinity.
0 1 1 2
∞
Note that if u = 0, we have applied a linear transformation to normalize the form of L, so has
6 ∞
geometric meaning. For instance, Example 2 describes a 1-dimensional family of curves that lie in
this locus.
To summarize, we have established the following theorem, where we have scaled s by 4 to avoid
some denominators in coefficients.
Theorem6. Letk beafieldofcharacteristic differentfrom2,3 andsuppose that (C,T ,T )consists
1 2
of a genus 2 curve C over k and T ,T Pic( /k)[3] such that # T ,T = 9 and e (T ,T ) = 1.
1 2 1 2 3 1 2
∈ C h i
If the specified data is sufficiently general then (C,T ,T ) is isomorphic to a suitable specialization
1 2
of r,s,t in the family described by the following data.
H = x2+rx+t
1
λ = 4s
1
G = (s st 1)x3+3s(r t)x2+3sr(r t)x st2+sr3+t
1
− − − − −
H = x2+x+r
2
λ = 4st
2
G = (s st+1)x3+3s(r t)x2+3sr(r t)x st2+sr3 t
2
− − − − −
H = sx2+(2sr st 1)x+sr2
3
− −
λ = 4t/(st+1)2
3
G = (s2t2 s2t+2st+s+1)x3+(3s2t2 3s2tr+3st+3sr)x2
3
− −
+((cid:0)3s2t2r 3s2tr2+3str+3sr2)x+s2t3 s2tr3+2st2+sr3+t /(st+1)
− −
H = (str st sr2+sr+r)x2+(st2 str st sr3+2sr2+t)x(cid:1)+st2 str2 str+sr3+t
4
− − − − − − −
λ = 4st/ st2 3str+st+sr3+t 2
4
−
G4 = (s2t(cid:0)3 3s2t2r s2tr3+6s2t(cid:1)r2 3s2tr+s2t s2r3+2st2 3str+2st sr3+t)x3
− − − − − −
+((cid:0)3s2t3 6s2t2r2 3s2t2+9s2tr3 3s2tr2+3s2tr 3s2r4+3st2 6str2+3str)x2
− − − − −
+( 3s2t3r+6s2t3 9s2t2r+3s2tr4+3s2tr3+3s2tr2 3s2r5 3st2r+6st2 3str2)x
− − − − −
s2t4+3s2t3r+s2t3 6s2t2r2+3s2tr4+s2tr3 s2r6 2st3+3st2r+st2 2str3 t2 /
− − − − − −
(st2 3str+st+sr3+t). (cid:1)
−
Here : y2 = F (x) = G2 +λ H3 for i = 1,2,3,4 and T = [ H (x) = 0,y G (x) = 0 κ]
Crst rst i i i i { i − i }−
and T = T +T and T = T T .
3 1 2 4 1 2
−
For future reference we note that
(6) Disc(F ) = 21236δ3δ3δ δ3δ δ3δ3,
rst − 1 2 3 4 5 6 7
7
where
δ = s
1
δ = t
2
δ = st+1
3
δ = r3 3rt+t2+t
(7) 4
−
δ = r3s 3rst+st2+st+t
5
−
δ = r3s2 3rs2t 3rs+s2t2+s2t+2st+s+1
6
− −
δ = r3s2t+r3s 3rs2t2 3rst+s2t3+s2t2+2st2+t.
7
− −
Remark 7. Note that (d): dy2 = F (x) has exactly the same property for Σ(d)-level structure.
Crst rst
5. The isogeny
We consider the curve as defined in Theorem 6 and its Jacobian . In this section we
rst rst
C J
determine a curve whose Jacobian is isogenous to via the isogeny /Σ. We
rst rst rst rst rst
C J J J → J
do so by determining the corresponding map between their Kummer surfaces.
e e
5.1. Isogenies and the quartic model of the Kummer surface. Let J be a principally po-
larized abelian surface with theta divisor Θ and suppose that Σ J[3] is maximal isotropic. We
J
⊂
consider J = J/Σ and the isogeny φ: J J. By [10, Proposition 16.8] there exists a principal
→
∗
polarization on J with theta divisor ΘJesuch that φ (ΘJe)= 3ΘJ.
Theclaessicaltheoryofthetadivisorsgivesuesthat (2Θ )is4-dimensionalandthattheinduced
J J
O
map P3 yieelds a quartic model of the Kummer surface = J/ 1 . We write ˜ for the
J → K h− i K
Kummer surface of J. Similarly, the linear system e(2Θe) provides a quartic model of ˜.
OJ J K
Remark 8. Note that this construction requires that we choose 2Θ to be defined over the base
e J
field k. If k is not algebraically closed, then there might not exist a k-rational divisor Θ in its
J
class. For an abelian surface, however, 2Θ is always linearly equivalent to a k-rational divisor.
J
Theisogeny induces a map K → K˜ in the following way. First note that φ∗OJe(2ΘJe) ⊂ OJ(6ΘJ).
The involution 1: J J induces a linear map on (6Θ ). We write (6Θ )+ for the fixed
J J J J
− → O O
subspace. Similarly, the translation action of Σ on J induces a linear action on the same space.
We write (6Θ )Σ for its fixed space. It is straightforward to check that
J J
O
φ∗OJe(2ΘJe) = OJ(6ΘJ)Σ ∩OJ(6ΘJ)+.
If ξ ,ξ ,ξ ,ξ forms a basis for (2Θ ) then (6Θ )+ is generated by the cubic forms in
0 1 2 3 J J J J
O O
ξ ,ξ ,ξ ,ξ . Thus we see that the isogeny φ: J J induces map ˜ which, between the
0 1 2 3
→ K → K
quartic models, is given by cubic forms.
e
5.2. Choice of model for Kummer surfaces. Let C be a curve of genus 2 given by a model
C: y2 = f x6+f x5+ +f ,
6 5 0
···
with Jacobian J. We follow [7, p.17] and choose a particular basis for (Θ ). We describe ξ =
J J
O
ξ(D) = ξ ,...,ξ as functions on J in terms of a divisor class D = [(x ,y )+(x ,y ) κ] on C as
0 3 1 1 2 2
−
follows.
Φ(ξ ,ξ ,ξ ) 2y y
0 1 2 1 2
(8) ξ = 1, ξ = x +x , ξ = x x , ξ = − , where
0 1 1 2 2 1 2 3 ξ2 4ξ ξ
1 − 0 2
Φ(ξ ,ξ ,ξ ) = 2f ξ3+f ξ2ξ +2f ξ2ξ +f ξ ξ ξ +2f ξ ξ2+f ξ2ξ +2f ξ3.
0 1 2 0 0 1 0 1 2 0 2 3 0 1 2 4 0 2 5 2 1 6 2
8
Note that for a Mumford representation D = [ x2 ξ x+ξ ,y g g x κ] we have y y =
1 2 0 1 1 2
{ − − − }−
g2+g g ξ +g2ξ , so one can compute these coordinates readily from such a representation.
0 0 1 1 1 2
The quartic equation for the model of arising from these coordinates has the shape
K
: (ξ2 4ξ ξ )ξ2+Φ(ξ ,ξ ,ξ )ξ +Ψ(ξ ,ξ ,ξ ) = 0,
K 1 − 0 2 3 0 1 2 3 0 1 2
where Ψ(ξ ,ξ ,ξ ) is a quartic form we do not need explicitly here. The important observation is
0 1 2
that one can read off the coefficients f ,...,f directly from Φ and thus recover C from it.
0 6
In order to produce Σ-invariant forms on , we use biquadratic forms from [7, p.23], arising
K ′ ′
from the addition structure on J. For i,j = 0,...,3 we have forms B k[ξ ,...,ξ ,ξ ,...,ξ ],
′ ′ i,j ∈ 0 3 0 3
biquadratic in (ξ ,...,ξ ) and (ξ ,...,ξ ) such that for points D ,D on J we have, as projective
0 3 0 3 1 2
matrices,
(9) ξ (D +D )ξ (D D )+ξ (D D )ξ (D +D ) = B (ξ(D ),ξ(D )) .
i 1 2 j 1 2 i 1 2 j 1 2 ij 1 2
− −
(cid:16) (cid:17) (cid:16) (cid:17)
We fix two points T ,T J[3] that generate Σ, write ξ(T ),ξ(T ) for the coordinate vectors of
1 2 1 2
∈
their images on the quartic model of and define
K
(10) R ξ ,...,ξ = B ξ ,...,ξ ,ξ(T ) and S ξ ,...,ξ = B ξ ,...,ξ ,ξ(T ) .
ij 0 3 ij 0 3 1 ij 0 3 ij 0 3 2
We see that(cid:0)the cubic(cid:1)forms (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
R = ξ R +ξ R +ξ R with i,j,k 1,...,4
ijk i jk j ki k ij
∈ { }
are invariant under translation by T and similarly that the forms
1
S = ξ S +ξ S +ξ S with i,j,k 1,...,4
ijk i jk j ki k ij
∈{ }
are invariant under translation by T . For C = the R and S each generate spaces of
2 rst ijk ijk
C
dimension 8 that intersect in a space of dimension 4. This intersection provides us with an explicit
∗
description of φ ( e(2Θe)).
OJ J
Generally, we expect J to be the Jacobian of a curve of genus 2, say . We can try to find a
basis ξ˜0,...,ξ˜3 for φ∗(OJe(2ΘJe)) that is the pullback of a basis of the typCe described by (8). We
can then read off the cureve , at least up to quadratic twist, from the reesulting equation for ˜.
C K
The basis choice can largely be characterized by the order of vanishing of each ξ at the identity
i
element. This leads us to coneclude that, up to scalar multiples, we should take the basis choice
ξ˜ = (1ξ +0ξ +0ξ )ξ2+ ,
0 0 1 2 3 ···
ξ˜ = (0ξ +1ξ +0ξ )ξ2+ ,
1 0 1 2 3 ···
ξ˜ = (0ξ +0ξ +1ξ )ξ2+ .
2 0 1 2 3 ···
Thedeterminationofξ˜ isalittlemoreinvolved. Theresultingformsfor = aretoovoluminous
3 rst
C C
to reproduce here, but we have made them available electronically at [4]. Via interpolation we find
tentatively the following result. e e
Theorem 9. Let be as described by Theorem 6. Then = /Σ is the Jacobian of the
rst rst rst
C J J
genus 2 curve
: 3y2 = G2+λ˜ H3,e
Crst − 4 4 4
with
e e e
G = ∆ (s st 1)x3 +3s(r t)x2+3rs(r t)x+(r3s st2 t) ,
4
− − − − − −
H4 = (r(cid:0) 1)(rs st 1)x2 +(r3s 2r2s+rst+r st2+st t)x(cid:1) (r2 t)(rs st 1),
e − − − − − − − − − −
λ˜ = 4∆st,
4
e
9
where
∆ = r6s2 6r4s2t 3r4s+2r3s2t2+2r3s2t+3r3st+r3s+r3+9r2s2t2+6r2st
(11) − −
6rs2t3 6rs2t2 9rst2 3rst 3rt+s2t4+2s2t3+s2t2+2st3+3st2+t2+t
− − − − −
5.3. Proof of Theorem 9. Since we have completely explicit descriptions of and , we can
rst rst
C C
write down explicit quartic models
e
: Q = (ξ2 4ξ ξ )ξ2+Φ(ξ ,ξ ,ξ )ξ +Ψ(ξ ,ξ ,ξ ) = 0
K 1 − 0 2 3 0 1 2 3 0 1 2
˜: Q˜ = (ξ˜2 4ξ˜ ξ˜)ξ˜2+Φ˜(ξ˜ ,ξ˜,ξ˜)ξ˜ +Ψ˜(ξ˜ ,ξ˜,ξ˜) = 0.
K 1 − 0 2 3 0 1 2 3 0 1 2
Furthermore, we have an explicit description at [4] of the map between them by expressions that
give ξ˜,...,ξ˜ as cubic forms in ξ ,...,ξ . We already know that ˜ is irreducible, because it is the
0 3 0 3
Kummer surface of a Jacobian. Therefore, to check if ˜ is indeedKthe image of , we only need to
substitute the cubic forms into the equation for ˜ andKcheck that the resultingKdegree 12 equation
K
is divisible by the quartic equation for . This is doable for specific specializations of r,s,t in Q,
K
but the computers at our disposal were not able to do this directly.
Wenote thatQandQ˜ (after substitutionof thecubicforms), arepolynomials inr,s,t,ξ ,...,ξ ,
0 3
of degrees 10 and 2 in ξ respectively. Hence, long division yields unique polynomials σ
3
∈
Q[r,s,t,ξ ,...,ξ ] and ρ ,ρ Q[r,s,t,ξ ,ξ ,ξ ] such that
0 3 0 1 0 1 2
∈
(ξ2 4ξ ξ )9Q˜ =σQ+ρ ξ +ρ .
1 − 0 2 1 3 0
We want to prove that ρ and ρ are identically zero. To this end, we analyse the appropriate
1 0
Newton polygons (or do the required computation using polynomials with coefficients truncated
to the appropriate leading terms) to verify that ρ ,ρ are of degrees at most 102,67,36 in r,s,t.
0 1
Hence, if we check that Q indeed divides Q˜ for a grid of 103 68 37 values for (r,s,t) then
× ×
a straightforward interpolation argument shows that ρ ,ρ must indeed be identically 0. This is
0 1
something that can easily be verified by a computer in less than 3 hours.
This computation shows that ˜ is indeed the Kummer surface of J = J/Σ and hence that
rst
K ∨ C
is correct up to quadratic twist. Recall from Section 3 that J/Σ comes equipped with a Σ -level
structure. In our case, we have that Σ = (Z/3)2, so Σ∨ = (µ )2 = eΣ(−3). Thus, it follows theat
3
(−3)
shouldhave a Σ-level structureitself such that theisogeny correspondingto it brings usback
Jrst (−3)
to .
Jrst
e
(−3)
Lemma 10. Let be as in Theorem 6, let be as in Theorem 9, and let be the quadratic
Crst Crst Crst
twist of by 3, using the notation in Remark 7. Define ψ by
rst 0
C −
e e
s(r 1)(r2 t)(δ r) (rs st 1)3δ2 s2(r 1)3(r2 t)3
eψ (r,s,t) = − − − 5 − , − − 4, − − .
0 (rs st 1)2δ st(r 1)3∆ (rs st 1)3δ2
(cid:16) − − 4 − − − 4 (cid:17)
(−3) ′ ′ ′
Then Cr′s′t′ is birationally equivalent to Crst , where (r ,s,t) = ψ0(r,s,t). Furthermore, as a
rational map we have ψ (ψ (r,s,t)) = (r,s,t). The Σ(−3) level structure induced on determines
0 0 rst
J
the kernel of the dual isogeny e
rst rst
J → J
e
(−3)
Proof. OnecancheckdirectlytheatCr′s′t′ isbirationallyequivalenttoCrst underthetransformation
(r3 3rt+t2+t)(rs st 1) r t ∆t(rs st 1)3(r3 3rt+t2+t)2
θ :(x,y) − − − − x+ − , − e − − y .
0 7→ (r2 t)(r 1)2s r 1 s2(r 1)3(r2 t)3
(cid:16) − − − − − (cid:17)
This naturally marks some Σ(−3) level structure on . Note that it even does so over Q, where
rst
J (−3)
we have no primitive cube root of unity. The Weil pairing implies that on , any two Σ level
Jrst
10 e
e