3. Torsion on Jacobians of Curves 67
3.2. Solving Norm Equations
3.2. Solving Norm Equations
In this section we relate the norm of elements in the function field of a curve with torsion divisors on the jacobian of the curve defined overQ. This connection of these two objects plays a central role in all constructions of torsion points of large order on jacobians of curves. After proving this connection, we make direct use of it to explicitly describe a one-dimensional family of hyperelliptic curves of genus two with a five-torsion divisor and an example of a hyperelliptic curve of genus two with a seven-torsion divisor.
Since points in the jacobian of a curve can be described as divisors of degree zero modulo principal divisors, we get a direct connection between elements in the function field of the curve and divisors which are equivalent to zero in the jacobian. We are especially interested in functionsf ∈K(C), where C is a curve over the field K, such thatdiv(f) =N D for some degree zero divisor Dand some integer N. The existence of such a functionf is equivalent with the existence of a torsion point of order dividingN. This observation we make explicit in the following two lemmas.
Lemma 3.3. Assume C/Q:Y2 = F(X) is a hyperelliptic curve of genus two with a Q-rational Weierstrasspoint. Assume further that there is a point D∈Jac(C)[N](Q) for a given natural number N. Then there exists a functionf ∈Q(C) such that
NQ(C)/Q(x)(f) =εu(x)N
for some polynomial u∈Q[X]\Q with deg(u)≤2 andε∈Q∗.
Proof : LetC/Q:Y2 =F(X) be a hyperelliptic curve of genus two with deg(F) = 5.
Now we know that every pointD∈Jac(C)(Q)can be represented by two points Q1 6=Q2 ∈C(Q)∪ {P∞} orQ1, Q2=Qσ1 ∈C(K)
for some quadratic extension Q⊂ K, where σ is the non-trivial automorphism ofK overQ. So write D=Q1+Q2−2P∞. The assumption that Dis of order dividingN translates to the fact that
O=N D=N Q1+N Q2−2N P∞. We have to look at two different cases.
1. caseP∞6∈ {Q1, Q2}:
Thus there exists a function f ∈Q(C) which has a pole of order 2N atP∞ and nowhere else and a zero of orderN at the two pointsQ1 andQ2. So we can deduce
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thatf ∈ L(2N P∞) and f can be written in the form f =a(x) +b(x)y
with polynomialsa, b∈Q[X]by Lemma 1.8. SinceQ1 andQ2 are assumed to be N-fold zeros of f, we get
f(Qi) =a(x(Qi)) +y(Qi)b(x(Qi)) = 0 fori= 1,2. This implies that
a(x(Qi))2−b(x(Qi))2F(x(Qi)) = 0.
Therefore, we get that (X−x(Qi))N dividesa2−b2F for i= 1,2. By comparing degrees, we obtain
a(x)2−b(x)2F(x) = N
Q(C)/Q(x)(f) =ε(x−x(Q1))N(x−x(Q2))N and the lemma is proven since(X−x(Q1))(X−x(Q2))∈Q[X]by assumption.
2. caseP∞∈ {Q1, Q2}:
We can assume without loss of generality thatQ2 =P∞, so we getD=Q1−P∞
andQ1∈C(Q). Hence there exists a function f ∈ L(N P∞) with aN-fold zero at Q1 and no zeros elsewhere. With the same arguments as above we get
a2−b2F = NQ(C)/Q(x)(f) =ε(x−x(Q1))N for someε∈Q∗.
Remark. The polynomial u in Lemma 3.3 gives the first coordinate of the Mumford representation of the point D ∈ Jac(C). Obviously the assertion in the lemma can also be formulated for curves of arbitrary genus and for curves without a Q-rational Weierstrasspoint. The proofs are completely analogous to the given one, but we would have to consider a lot more cases. Since this gives no further insights, we restricted ourselves to the most simple case.
The next lemma states that the converse of Lemma 3.3 is also true.
Lemma 3.4. Let C/Qbe a hyperelliptic curve of genus two with a Q-rational Weier-strass point and D ∈ Jac(C)(Q) having first coordinate in Mumford represen-tation equal to the polynomial u with 1 ≤ deg(u) ≤ 2. If there exists a function
f ∈ L(deg(u)N P∞) such that N
Q(C)/Q(x)(f) =εuN for some ε∈Q∗, then ord(D)|N.
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3.2. Solving Norm Equations
Proof : AssumeC/Qis a hyperelliptic curve of genus two and there exists a function f ∈ L(deg(u)N P∞) such that
NQ(C)/Q(x)(f) =εuN. Then the principal divisor of f is
div(f) =N Q1+N Q2−2N P∞
for some pointsQ1, Q2∈C such thatu(x(Qi)) = 0. Therefore, N(Q1+Q2−2P∞)
has to be the identity in the jacobian. Hence, the divisor(Q1+Q2−2P∞)has to be of order dividingN.
Remark. Again, this can easily be generalized to curves with arbitrary genus.
We use these two lemmas to produce families of hyperelliptic curves admitting a point of certain order on their jacobians. The strategy is to take a parametrized familyC of hyperelliptic curves and assume the existence of a function f ∈Q(C) with a norm equal to
NQ(C)/Q(x)(f) =εuN
forC∈ C. Then, by comparing coefficients, we get equations in the parameters of the familyC, the parameters of the functionf and theu-coordinate of the potential point of orderN. If we further start with a prime N, we can be sure that on every constructed curve there is a point of exact orderN on the jacobian.
We compute an example of this construction for five-torsion. We start with the universal hyperelliptic curve of genus two. Since the equations we are dealing with are rather complicated, we restrict the functionsf ∈ L(10P∞) to be in the subvectorspace spanned by{1, x, x2, x3, x4, x5, y}.
Set
f5 :=− 3
128λ5+ 5
16λ3µ− 15
8 λµ2+ 2η, f4 :=−25
512λ6+ 85
128λ4µ−135
32 λ2µ2+ 5λη−5 8µ3, f3 :=− 25
1024λ7+ 75
256λ5µ−115
64 λ3µ2+ 15
4 λ2η−95
16λµ3+ 5µη, f2 := 25
16384λ8− 75
1024λ6µ+ 375
512λ4µ2+5
8λ3η−415
64 λ2µ3+15
2 λµη−95 64µ4,
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f1 :=−5
64λ4η+15
8 λ2µη−5λµ4+15 4 µ2η, f0 :=η2−µ5.
Proposition 3.5. The family
Cλ,µ,η :y2 =f5x5+f4x4+f3x3+f2x2+f1x+f0, generically admits a Q-rational point of order 5 on its jacobian.
The u-coordinate of this point in Mumford-representation is given by the polynomial u:=x2+λx+µ.
Proof : Let
C:Y2 =f5X5+f4X4+f3X3+f2X2+f1X+f0 =:F(X)
be the universal hyperelliptic curve of genus two with one rationalWeierstrass-Point at infinity. Assumex2+λx+µto be the first coordinate of a Q-rational point
D:=P1+P2−2P∞∈Jac(C)[5](Q)
of order five on the jacobianJac(C)inMumford-representation. Since we have assumed D∈Jac(C)[5], we havel(5D)≥1. This means that there exists a function f ∈Q(C)of normε(x2+λx+µ)5 by Lemma 3.3. So there is a solution to the equation
a2−F b2 =ε(X2+λX+µ)5,
whereaandbare polynomials inX. By assumingb= 1, we can simplify the equation to a2−F = (X2+λX+µ)5,
whereahas to be monic of degree five. Setting
a:=η+a1X+a2X2+a3X3+a4X4+X5
and by comparing coefficients of both sides of the equation, we get the equations for f0, . . . , f5.
The familyCλ,µ,η in Proposition 3.5 is one-dimensional as we can see with the following considerations. First we consider the subfamily given byλ:= 1andµ:= 0and show that
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3.2. Solving Norm Equations
this family is one-dimensional. This is done by computing the absoluteIgusainvariants of this family. These are given by
α(C1,0,η) =
17
24335η4−21017335η3+2139921335η2−2283532η+21753834
(η2−237η+ 2142532)2
β(C1,0,η) = h
(η2− 237η+ 2142532)3 γ(C1,0,η) = η5(η− 3
27)5(η− 3
28)2(η2− 3
27η+2116) (η2−237η+2142532)5 , whereh is a polynomial of degree eight inη.
Since these invariants are non-constant rational functions in the parameterη, we get that there are infinitely many pairwise non-isomorphic curves in this family. So the familyC1,0,η has to be one-dimensional. Therefore, the dimension of Cλ,µ,η has to be at least one.
The next step is to show that for every choice of λ, µ, η ∈ Q there exists a η0 ∈ Q such that Cλ,µ,η ∼=C1,0,η0. To show this we look at the differences of the absoluteIgusa invariants of Cλ,µ,η and C1,0,η0 and compute the greatest common divisor d of their nominators. This is given byd=d2η02+d1η0+d0, where
d2 :=λ10−20λ8µ+ 160λ6µ2−640λ4µ3+ 1280λ2µ4−1024µ5, d1 :=24µ5−30λ2µ4+ 15λ4µ3−15
4 λ6µ2+15
32λ8µ− 3 128λ10, d0 :=15
8 λµ2η− 9
64µ5−η2−45
64λ2µ4− 5
16λ3µη+ 3
128λ5η− 25 1024λ6µ2 + 105
512λ4µ3+ 15 16384λ8µ.
So for any given λ, µ, η ∈ Q we can find η0 ∈ Q such that d = 0. Therefore, Cλ,µ,η ∼=C1,0,η0 overQ. We now check under what conditions this isomorphism is defined over the rational numbers. This is done by computing the roots of the polynomialdand check when these are defined overQ(λ, µ, η). Sincedis a polynomial of degree two, we can easily determine its roots. These roots are rational if and only if
d21 4d22 −d0
d2 = 32(λ5−403λ3µ+ 80λµ2− 2563 η)2 2562(λ2−4µ)5
is a square in Q. This is true if and only if λ2−4µis a square in Q. This answers the question of the field of definition of the isomorphism described above.
We also have tried to construct a family of curves defined over Qadmitting a seven-torsion point on their jacobians. But the restriction to functions in h1, y, x, . . . , x7iQ
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just leads to the empty family. Allowing functions in the whole spaceL(14P∞) gives equations which define a four-dimensional parameter space. Since these equations are rather large, we do not state them here. The following example is the fiber above a rational point in the mentioned parameter space.
Example 3.2. Let C :Y2 =F(X) with F =X5−39
4 X4+ 65X3−146X2+ 198X−127.
Then there exists a non-trivial point D inJac(C)[7](Q).
The attempt to construct in this direct approach a family of curves with a seven-torsion point on their jacobians shows that in order to construct large torsion orders one has to be able to findQ-rational points in affine varieties which are given by complicated polynomials in the parameters.
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