London Mathematical Society ISSN 1461–1570

COMPUTING THE RANK OF ELLIPTIC CURVES OVER NUMBER

FIELDS PREPRINT VERSION FOR ANTS IV

DENIS SIMON

Abstract

We describe an algorithm of 2-descent for computing the rank of an elliptic curve without 2-torsion, defined over a general number field. This allows us in practice to deal with fields of degree from 1 to 5.

Introduction

Computing the rank of the Mordell–Weil group of an elliptic curve E defined over ℚ has now become a classical task. This is currently achieved by using 2-descent (see [6]). This method can provably compute only the 2-Selmer group of the curve. However, in many cases the Tate–Shafarevich group happens to have no 2-torsion, in which case E(ℚ)/2E(ℚ) can be identified with the 2-Selmer group. From this, we can easily deduce the rank of E(ℚ).

Our goal is to describe completely a 2-descent method for elliptic curves defined over a number field K. Such a description has already been given for some real quadratic fields (see [8]). We dispense here with any assumption on the field K, and we give an example of an elliptic curve defined over an imaginary quadratic field with nontrivial class group. With the help of the package PARI/GP (see [3]), we have been able to treat a great number of curves defined over quadratic fields, and some over larger degree fields (up to degree 5). To make things easier, we will restrict to elliptic curves without 2-torsion over K. The basis of our method can be found in [2] and in [16].

The present method is a direct consequence of the arithmetic of invariants of quartics as described by J. Cremona in [10]. The new ingredients introduced in the present work are mainly the solution of Legendre equations over number fields, a method for minimizing some quartics constructed, over general number fields, and the implementation of the whole method.

If we are specifically interested in the Selmer group, we refer to [13], which describes a more general method for computing the Selmer group of the Jacobian of a hyperelliptic curve.

1. Description of the Method

1.1. Definition of the Morphism

Let K be a number field and E be an elliptic curve defined over K, given by an equation of the form

y^{2} = P (x) ,

. 1991 Mathematics Subject Classification 11D25, 11R, 11Y50 \begin{matrix}c \\ ◯ \end{matrix} ????, Denis SIMON

LMS J. Comput. Math. ?? (????) 1–10

Computing the Rank of Elliptic Curves over Number Fields preprint version for ANTS IV

where P (x) = x^{3} + Ax^{2} + Bx + C, with A, B and C ∈ K. The group of points of E with coordinates in K will be denoted by E(K).

Without loss of generality, we can make the assumption that A, B and C are in the ring of algebraic integers ℤ_{K} of K. We make the further assumption that the polynomial P is irreducible over K. This is equivalent to saying that E(K) has no 2-torsion.

Let θ be a root of P in ℂ. This algebraic integer generates a cubic extension L = K(θ) of K. It is not difficult to prove the following result (see [2]):

Proposition 1.1. The map

ϕ : E(K) → L^{∗}/L^{∗2}
0  ↦ 1
(x, y) ↦ x − θ

defines a group homomorphism with exact kernel 2E(K).

Thanks to this homomorphism, it is possible to obtain the rank of E(K). Indeed, if we know the image of ϕ, we also know the group E(K)/2E(K), whose cardinality is equal to 2^{r}, where r is precisely the rank of E(K). Note also the following proposition:

Proposition 1.2. The image of ϕ is a subgroup of the kernel of the norm map N_{L/K} from L^{∗}/L^{∗2} to K^{∗}/K^{∗2}.

The proof is an immediate consequence of the relation N_{L/K}(x − θ) = P (x) = y^{2}. In [13], we find an identification of this kernel with the cohomology group H^{1}(K, E(K̄)[2]).

1.2. Reduction to a Finite Set

The next two propositions show that Im ϕ is contained in some easily computable finite subgroup of the infinite group L^{∗}/L^{∗2}.

Let S be a finite set of places of L containing the infinite places. Let

L(S, 2) = \{\overline{δ} ∈ L^{∗}/L^{∗2}, ∀𝔓 ∉ S v_{𝔓}(δ) ≡ 0 (mod 2)\} .

In order to describe this group, we make use of the S-units of L, denoted by 𝕌_{L,S}, and of the S-class group Cl_{S}(L). Recall that Dirichlet’s S-unit theorem (see [11]) asserts that 𝕌_{L,S} is a finitely generated Abelian group, whose torsion part contains exactly the roots of unity in L, and whose free part has rank |S| − 1. The S-class group Cl_{S}(L) is isomorphic to the quotient of the ordinary class group Cl(L) by the subgroup generated by the classes of all prime ideals in S. In [14], we prove the following result:

Proposition 1.3. There exists an isomorphism

L(S, 2) ≃ \frac{𝕌_{L,S}}{𝕌_{L,S}^{2}} × Cl_{S}(L)[2] .

If we denote by h_{2}(S) the 2-rank of Cl_{S}(L), we have |L(S, 2)| = 2^{|S|+h_{2}(S)}. Thus, if we adjoin to S a set of h_{2}(S) prime ideals which generate exactly Cl_{S}(L)[2], we obtain a new set S′ such that

L(S, 2) ≃ L(S′, 2) ≃ 𝕌_{L,S′}/𝕌_{L,S′}^{2} .

Proposition 1.4. Let S be the set of all infinite places of L together with the prime ideals 𝔓 of L above 𝔭 in K such that 𝔓|P′(θ) and 𝔭^{2}| Disc P. We have

Im ϕ ⊂ L(S, 2) ∩ Ker N_{L/K} .

Remark 1.5. The norm in L/K of the number P′(θ) is precisely equal to the discriminant of the polynomial P. The discriminant of the elliptic curve is equal to 16 times that of the polynomial. As a consequence of this proposition, it will be enough to consider the prime ideals whose square divides Disc P. Thus, contrary to what is often stated, it is not necessary to consider all prime factors of 2 Disc P.

Proof. Let (x, y) be a nontrivial point on E(K). We shall first prove that the valuation v_{𝔓}(x − θ) is even at all primes not dividing P′(θ). Assume first that this valuation is negative. Since θ is integral, we have v_{𝔓}(x − θ) = v_{𝔓}(x). The relation y^{2} = x^{3} + ... implies that 2v_{𝔓}(y) = 3v_{𝔓}(x), which proves our assertion in this case. Assume now that v_{𝔓}(x − θ) is nonnegative and odd. In the relation

y^{2} = P (x) = (x − θ) (P′(θ) + (x − θ)\frac{P′′(θ)}{2} + (x − θ)^{2}) ,

all the elements that occur are integral at 𝔓. If we compare the valuations of both sides of the equality, we see that 𝔓 divides the second factor and therefore divides P′(θ).

It now remains to prove that if v_{𝔓}(x − θ) is odd, then 𝔭^{2} divides Disc P. Since 𝔓 divides P′(θ), we know that 𝔭 divides Disc P. Assume that 𝔭^{2} does not divide Disc P. Since we know that v_{𝔭}( Disc P) ⩾ e(𝔓/𝔭) − 1, where e(𝔓/𝔭) is the relative ramification index, it follows that 𝔭 factors as 𝔓_{1}^{2}𝔓_{2} in L/K, with 𝔓_{2} not dividing P′(θ) (since 𝔭^{2} does not divide Disc P). If we set α_{i} = v_{𝔓_{i}}(x − θ), we know that α_{2} is even and that 2v_{𝔭}(y) = v_{𝔭}(N_{L/K}(x − θ)) = α_{1} + α_{2} is also even. It follows that α_{1} = v_{𝔓}(x − θ) is even, as claimed.

1.3. Legendre Equations

All the formulas of this section (1.3) and of the next (1.4) are byproducts of the theory developed in [10].

What we have just proved allows us to embed the group Im ϕ into the finite computable group L(S, 2) ∩ Ker N_{L/K} for an explicit finite set S. It now remains to decide whether a given element \overline{δ} of L(S, 2) ∩ Ker N_{L/K} belongs to Im ϕ. Explicitly, such an element belongs to Im ϕ if it can be written in the form \overline{x − θ} with x ∈ K, or equivalently if we can find z ∈ L^{∗} such that δz^{2} = x−θ. After writing δ = a−bθ+cθ^{2} with N_{L/K}(δ) = r^{2}, and z = u + vθ + wθ^{2}, we get

δ′ = δz^{2} = q_{0}(u, v, w) − q_{1}(u, v, w)θ + q_{2}(u, v, w)θ^{2} ,

where q_{0}, q_{1} and q_{2} are three quadratic forms in the variables u, v and w. Their discriminants are respectively N_{L/K}((θ^{2} + Aθ + B)δ) = C^{2}r^{2}, N_{L/K}((θ + A)δ) = (C − AB)r^{2} and N_{L/K}(δ) = r^{2}, where A, B, and C are the coefficients of the polynomial P defining E. Since we have assumed that the polynomial P is irreducible, the quantities C and C − AB cannot vanish. We have to solve simultaneously q_{2} = 0 and q_{1} = 1. We first look for a particular solution of q_{2} = 0. We have:

q_{2}(u, v, w) = cu^{2} + (A^{2}c + Ab − Bc + a)v^{2}

+ (A^{4}c + A^{3}b − 3A^{2}Bc + A^{2}a − 2ABb + 2ACc + B^{2}c − Ba + Cb)w^{2}

− 2(b + Ac)uv + 2(A^{2}c + Ab − Bc + a)uw
+ 2(−A^{3}c − A^{2}b + 2ABc − Aa + Bb − Cc)vw

Using the auxiliary variables α = Abc + Bc^{2} − ac + b^{2} and β = A^{2}bc + ABc^{2} + Ab^{2} − Cc^{2} + ab, we make the linear change of variables

\begin{matrix}U = & cu − (b + Ac)v + (a + Ab + A^{2}c − Bc)w \\ V = & αv − βw \\ W = & rw \end{matrix}

so that we obtain −αcq_{2}(u, v, w) = Q_{2}(U, V, W) with

Q_{2}(U, V, W) = V^{2} − αU^{2} − cW^{2} . (1)

Thus, we are reduced to solving the Legendre equation Q_{2} = 0. The above is valid only if αc does not vanish. However, if c = 0, there is a trivial solution (there is essentially nothing to do), and if α = 0, we observe that N_{L/K}(δ)/δ = (r/δ)^{2}δ is equivalent to δ and has no term in θ^{2}, since this term is precisely given by α.

When αc ≠ 0, Equation (1) defines a conic. For such a curve, we know that the Hasse principle applies: the existence of a solution in K is equivalent to the existence of a local solution at every place of K. In fact, it is only necessary to test the infinite places and the places dividing 2αc. This local solubility is given by the Hilbert symbol.

In the classical situation where we look for rational points on elliptic curves defined over ℚ, this leads to Legendre equations with coefficients in ℤ. In this case, we have at our disposal an efficient algorithm to solve this equation (see [7]). This algorithm makes a crucial use of Euclidean division in ℤ. For this reason, it is difficult to write down a direct generalization valid over general number fields, and usually it is even not possible at all (see [14]). We must proceed in another way.

Let us consider the quadratic extension K(\sqrt{α})/K = F/K. Equation (1) is equivalent to

N_{F/K} (\frac{V + \sqrt{α}U}{W} ) = c . (2)

Note that the two numbers α and c play a symmetrical role. The question is whether or not c is a norm for the extension F/K, and if this is the case, to find an explicit element of norm c.

Before giving an answer to this question, we need some notation. The extension F/K is quadratic, hence cyclic. Denote by G its Galois group. Let S_{K} be a finite set of primes of K, and let S_{F} be the set of primes of F above those in S_{K}. By abuse of notation, we will write S for both. As before, we denote by 𝕌_{K,S} the S-units of K, and by 𝕌_{F,S} the S-units of F. If Cl_{S}(F) is the S-class group of F, Cl_{S}(F)^{G} is the subgroup of invariant classes under the action of G, and Cl_{S}(I^{G}) the subgroup of classes of invariant ideals under this action. We have Cl_{S}(I^{G}) ⊂ Cl_{S}(F)^{G} ⊂ Cl_{S}(F).

With this notation, we have:

Proposition 1.6. Let F/K be a cyclic extension of number fields with Galois group G. There exists an isomorphism

\frac{Cl_{S}(F)^{G}}{Cl_{S}(I^{G})} ≃ \frac{N_{F/K}(F^{∗}) ∩ 𝕌_{K,S}}{N_{F/K}(𝕌_{F,S})} .

In particular, if S generates the class group Cl(F), we have

N_{F/K}(F^{∗}) ∩ 𝕌_{K,S} = N_{F/K}(𝕌_{F,S}) .

This is essentially Chevalley’s ambiguous classes theorem, and a proof of the statement as given above and more complete results about norm equations in non-Galois extensions can be found in [15].

Thanks to this result, we know in which case the S-units that are norms for a cyclic extension are simply norms of S-units (this is not always the case!). Thus, a solution of (2) will be given by an S-unit of F, where S contains all prime factors of c, and generators of the 2-part of Cl(F). Since there is only a finite number of S-units modulo squares, there is only a finite number of potential solutions. If this construction does not give a solution, then there is no solution at all.

The above discussion gives one solution to q_{2} = 0 or proves that it is impossible. Using this solution, we can construct an element z ∈ L^{∗} such that δ′ = δz^{2} = a − bθ.

1.4. Construction of the Quartic

Assume now that δ is of the form δ = a − bθ with a and b algebraic integers in ℤ_{K}. We have N_{L/K}(δ) = r^{2}. We want to find δ′ with \overline{δ′} = \overline{δ} and of the form δ′ = x − θ. This is equivalent to looking for z = u+vθ +wθ^{2} ∈ L^{∗} such that δ′ = δ ·z^{2} satisfies both q_{2} = 0 and q_{1} = 1. The previous discussion is not applicable anymore since we are precisely in the case c = 0. These equations can be writen as follows:

q_{2}(u, v, w) = (Ab + a)v^{2} + (A^{3}b + A^{2}a − 2ABb − Ba + Cb)w^{2}
− 2buv + 2(Ab + a)uw + 2(−A^{2}b − Aa + Bb)vw
q_{1}(u, v, w) = bu^{2} − Bbv^{2} + (−ABa − A^{2}Bb + ACb + B^{2}b + Ca)w^{2}
+ 2(ABb + Ba − Cb)vw − 2Bbuw − 2auv .

Here, it is important to note that b ≠ 0 if δ ≠ 1. Indeed, otherwise we would have δ = a and a^{3} = N_{L/K}(δ) = r^{2}, and therefore a = δ would be a square, which can only occur for the trivial case δ = 1, contradiction. The case δ = 1 is of course of no interest. We can thus make the linear change of variables

\begin{matrix}W = & rw \\ V = & (a + Ab)w − bv \\ U = & (a^{2} + 2Bb^{2} − A^{2}b^{2})w + (Ab^{2} + ab)v − 2b^{2}u \end{matrix}

and we get the identity b^{2}q_{2}(u, v, w) = W^{2} − UV. The solutions of q_{2} = 0 are given by (U, V, W) = (λ^{2}/y, μ^{2}/y, λμ/y), where (λ, μ, y) belongs to K × K × K^{∗}. We can express u, v and w in terms of y,λ and μ. The equation q_{1} = 1 becomes:

4b^{3}y^{2} = λ^{4} − 2(Ab + 3a)λ^{2}μ^{2} + 8rλμ^{3} + (A^{2}b^{2} − 2Aab − 4Bb^{2} − 3a^{2})μ^{4} = Q(λ, μ) .

The discriminant of this fourth degree polynomial in λ and μ is equal to 2^{12}b^{6} Disc P.

Now recall that a quartic has two polynomial invariants I and J of total degree 2 and 3 and weight 4 and 6 in the coefficients of the quartic, such that the ring of invariants of the quartic is the free polynomial algebra generated by I and J. For example, the usual discriminant of the quartic is equal to (4I^{3} − J^{2})/27. It is easy to check that the I and J invariants of Q are related to the I and J invariants of P (considered as a degree 4 polynomial with zero leading term) by the relations I(Q) = 2^{4}b^{2}I(P) and J(Q) = 2^{6}b^{3}J(P). They are also related to the c_{4} and c_{6} invariants of the elliptic curve by I(Q) = b^{2}c_{4} and J(Q) = 2b^{3}c_{6}.

1.5. Minimization of the Quartic

In classical 2-descent over ℚ, we only get quartics with discriminant Disc P or 2^{12} Disc P (see for example [6]). This comparison suggests that the factor b^{6} can be removed and the quartic minimized. Such a minimization can be achieved as follows, using the fact that the factorization of Q mod b is formally given by a^{4}Q = (aλ − rμ)^{3}(aλ + 3rμ).

(an algorithm for finding such u and v, valid over general number fields, is given in [4]).

Proposition 1.7. Let l, a_{1}, r_{1}, u, and v be five integers in ℤ_{K}, such that

f = (lr + br_{1})v − (la + ba_{1})u ≠ 0 .

The quartic 4b^{3}y^{2} = Q(λ, μ) can be minimized to 4y^{2} = Q′(λ′, μ′) such that - Q′ has integral coefficients

- Disc Q′ = 2^{12}f^{12} Disc P

- I(Q′) = 2^{4}f^{4}I(P) and J(Q′) = 2^{6}f^{6}J(P).

Proof. Consider the linear change of variables

\begin{matrix}λ = & buλ′ + (lr + br_{1})μ′ \\ μ = & bvλ′ + (la + ba_{1})μ′. \end{matrix}

Applied to Q, this linear change of variables defines a new polynomial which is formally divisible by b^{3}. Letting Q(λ, μ) = b^{3}Q′(λ′, μ′), our quartic becomes

4y^{2} = Q′(λ′, μ′) .

The discriminant of Q′ is now equal to 2^{12}f^{12} Disc P and we obtain the conclusion of the proposition. Note that the I and J invariants are also reduced in the same way.

Remark 1.8. It is quite common that a and b are coprime. In this case, we can find l and β such that la + βb = 1 (an algorithm for finding such l and β, valid over general number fields, is given in [4]). A suitable change of variables is

\begin{matrix}λ = & bλ′ + (lr + br_{1})μ′ \\ μ = & μ′ \end{matrix}

since it gives f = 1.

I do not claim that these choices will always lead to a Q′ with small coefficients, and a better choice should be sought after. In [9], J. Cremona describes a very efficient reduction algorithm for the coefficients of a real quartic: in some sense it gives the smallest possible reduction.

1.6. Solubility of the Quartic

Before looking for a solution of the quartic over K, we can look at its local solubility. Since the quartic defines a curve of genus 1, we know that there exist local solutions at all places not dividing 2 Disc Q = 2b^{3} Disc P (see for example [8]).

The local solubility at the bad primes is tested by using a classical Hensel lift (see [8]). For the Archimedean real places, it is equivalent to knowing whether the polynomial bQ or Q′ can take negative values.

As soon as we know that the quartic is everywhere locally soluble, we can hope to find a global point. For this, we can test all values of x up to a given height, using more or less powerful methods. If we are lucky, we find a point on the quartic, from which it is not difficult to recover a point on the elliptic curve by using the formulas of this paper. If we do not know where the quartic comes from (for example if it has been constructed by another method, see [9]), it is still possible to recover the point on the elliptic curve by using a syzygy (see [10]). But since we do not have any bound for the height of a point on the quartic, it may happen that the smallest one is too large to be found by such a naive search, and hence we can miss it. In fact it is not always true that such a point will exist at all.

The set of all \overline{δ} which lead to everywhere locally soluble quartics forms a subgroup of L(S, 2) ∩ Ker N_{L/K}: this is the Selmer group S^{2}(E/K) (see [13]). Recall the exact sequence

1 → E(K)/2E(K) → S^{2}(E/K) → � (E/K)[2] → 1 .

Since the group � (E/K)[2] of elements of order 2 in the Tate–Shafarevich group is not always trivial, this implies the well-known fact that the local-global Hasse principle does not hold for genus 1 curves, in other words, the existence of local solutions does not imply the existence of a global one (over K).

Remark 1.9. As it has been pointed out to me by J. Cremona and M. Stoll, it is not necessary to have the explicit equation of the quartic in order to know its local solubility. Indeed, we can use a criteria of Siksek (in [12]) to test the local solubility directly on the system q2 = 0, q1 = 1. Using the notation of [12], it is only necessary to test the local solubility at the primes dividing 2 or ∂(q_{2}, q_{1} − 1) = r^{12} Disc P. The main advantage of this idea is that it allows to compute directly the Selmer group, and we spare the time needed for solving all Legendre equations which lead to non soluble quartics.

2. Examples

Using the PARI/GP package (see [3]), we have implemented this algorithm and computed the Mordell–Weil group of various elliptic curves for several number fields with degrees ranging from 1 to 5, and having a priori any class number. However we need a large amount of work for large degrees (we have to compute class groups and units in degree 15 fields), and for such degrees, the computation was possible only for a few well-chosen examples.

2.1. A Complete Example

For the quadratic case, the computations are quite fast (less than one second for the easy cases). The following example consists of an elliptic curve defined over an imaginary quadratic field with nontrivial class number (as opposed to the situation in [8]).

Let K = ℚ(ε), with ε^{2}−ε+4 = 0. This is the imaginary quadratic field of discriminant −15. Its class group is of order 2.

Consider the elliptic curve E given by the equation

y^{2} = P (x) = x^{3} + εx − 1 .

The discriminant of P is equal to 12ε − 11 and is not divisible by the square of a nontrivial ideal in K. The field L is generated by θ such that P (θ) = 0 = 1−θ+4θ^{2} −2θ^{3} +θ^{4} +θ^{6}. The class group of L is of order 2, and is generated by the prime ideal 𝔓_{2} above 2 such that 𝔓_{2}^{2} = (1 − θ)ℤ_{L}, and N_{L/K}(1 − θ) = ε. The units of L are generated by −1, θ and −2 + εθ + θ^{2}, with respective relative norms −1, 1 and 1.

Following Proposition 1.4, we set S = {∞_{1}, ∞_{2}, ∞_{3}} ∪ {𝔓_{2}}. For this choice, the group L(S, 2) ∩ Ker N_{L/K} is of order 4, and is generated by the classes of θ and −2 + εθ + θ^{2}. From this, we already know that the rank of E(K) is at most equal to 2. We have

L(S, 2) ∩ Ker N_{L/K} = {1, θ, −2 + εθ + θ^{2}, 1 − (ε + 2)θ + εθ^{2}} .

Let δ = δ_{1} = θ. Since δ has no coefficient in θ^{2}, we are already at the final step of the method, and we have to find a K-rational point on the quartic

−4y^{2} = λ^{4} + 8λμ^{3} − 4εμ^{4} .

We check that this quartic has no local solution at 𝔓_{2}, hence no global solution.

Let δ = δ_{2} = −2 + εθ + θ^{2}. Here, we have c = 1, hence the Legendre equation V^{2} − αU^{2} = cW^{2} is satisfied by (U, V, W) = (0, 1, 1). From this, we find the solution (u, v, w) = (ε, 2 − θ, 2) for q_{2} = 0, which gives δ′ = δ(u + vθ + wθ^{2})^{2} = ε(1 − θ), with N_{L/K}(δ′) = ε^{4}. We then have to find a K-rational point on the quartic

4ε^{3}y^{2} = λ^{4} − 6ελ^{2}μ^{2} + 8(ε − 4)λμ^{3} + (9ε + 28)μ^{4} .

Once again, we check that this quartic has no local solution at 𝔓_{2}, hence no global solution.

Let δ = δ_{3} = 1 − (ε + 2)θ + εθ^{2}. Here, we have α = ε^{2}, hence the Legendre equation V^{2} − αU^{2} = cW^{2} has the solution (U, V, W) = (1, ε, 0). From this, we find the solution (u, v, w) = (−5+ε, −2, 0) for q_{2} = 0, which gives δ′ = δ(u+vθ +wθ^{2})^{2} = (29+3ε)− (10 + 14ε)θ, with N_{L/K}(δ′) = (93 − 41ε)^{2}. We then have to find a K-rational point on the quartic

4(10 + 14ε)^{3}y^{2} = λ^{4} − 6(29 + 3ε)λ^{2}μ^{2} + 8(93 − 41ε)λμ^{3} + (5201 + 283ε)μ^{4} .

As in Remark 1.8, this quartic is minimized and reduced by the linear change of variables λ = bλ′ +(l′lr+br_{1})μ′ and μ′ = μ with l = (1−ε)^{−1}, l′ = 33−4ε, and r_{1} = −48+19ε. Thus, the quartic is equivalent to

4y^{2} = (14ε + 10)λ′^{4} + (−12ε + 68)λ′^{3}μ′ − 24ελ′^{2}μ′^{2} + (2ε − 16)λ′μ′^{3} + εμ′^{4} .

By a naive search, we find the solution λ = ε, μ = 2 and y = 4−2ε, which gives the point

(−\frac{ε}{3} + 1, \frac{ε}{9} + \frac{4}{9}) ∈ E(K) .

We have therefore proved that E(K) has rank 1 and that the group � (E/K)[2] is trivial.

2.2. A Collection of Quadratic Examples

In [5], we find a collection of elliptic curves defined over imaginary quadratic fields with class number 1. Using the theory of L functions, J. Cremona and E. Whitley computed the analytic rank and some other data associated with these elliptic curves. In each case, our 2-descent method is able to prove the observations that they made from their computations. We present the results in a table.

The number field K is given by the square root of its discriminant and by the minimal polynomial of its generator ε. We use here the standard notation for elliptic curves: its Weierstrass model is given by y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}, its conductor is N. The result of the 2-descent is given by the quantities #S^{2} (the order of the Selmer group), r (the rank), #� [2] (the order of the 2-part of the Tate–Shafarevich group) and a set of independent points on the curve.

N  a_{1}  a_{2}  a_{3}  a_{4}  a_{5}  #S^{2} r #� [2]  P
K = ℚ(\sqrt{−3}) ε^{2} − ε + 1 = 0
19  0  −ε  1  −1 + ε  0  2 1  1  (1, 0)
19  0  −ε  1  9 − 9ε  −15  2 1  1  (2ε, −1 + ε)
19  0  −ε  1 769 − 769ε  −8470  2 1  1  ( \frac{801}{49} ε, − \frac{171+ε}{343} )
19  0  −ε  1  −31 − 99ε 74 − 498ε  2 1  1  (−4 − 3ε, 2 − ε)
19  0  −ε  1  99 + 31ε −424 + 498ε 2 1  1  (3 − 3ε, 5 + 10ε)
K = ℚ(\sqrt{−4}) ε^{2} + 1 = 0
13 + 8ε 0 1 − ε ε −ε 0 2 1 1 (0, 0)
K = ℚ(\sqrt{−8}) ε^{2} + 2 = 0
13 + 8ε 0 −1 + ε 1  1 − ε  ε  1 0  1
13 + 8ε 0 −1 + ε 1  11 − 11ε  −27 + 3ε  1 0  1
13 + 8ε 0 −1 + ε 1  −9 + 39ε −210 − 39ε  1 0  1
13 + 8ε 0 1 − ε  1  2 − 2ε  1  1 0  1
K = ℚ(\sqrt{−19}) ε^{2} − ε + 5 = 0
2ε  ε 1 − ε  1  −1  0  1 0  1
2ε  ε 1 − ε  1  9 − 5ε  −24 − 2ε  1 0  1
2ε  ε 1 − ε  1  −31 − 15ε −60 − 54ε  1 0  1
−3 + 6ε 0  1  1  20  −32  4 0  4
−3 + 6ε 0  1  1  −4390  −113432  4 0  4
−3 + 6ε 0  −1  1  −2  2  4 2  1  (2, 1)
(1 + 4ε, 15 − 12ε)
K = ℚ(\sqrt{−43}) ε^{2} − ε + 11 = 0
13 + ε 0 −ε 1 + ε −2 1 − ε 2 1 1 (−1, −2)
K = ℚ(\sqrt{−67}) ε^{2} − ε + 17 = 0
3ε  ε 1 + ε  ε  −1 − 3ε  16 − 5ε  2 1  1  (−ε, −4)

Remark 2.1. A word has to be said about the two curves with nontrivial Tate–Shafarevich group. These curves are in fact defined over ℚ. It is well known that

rank(E(ℚ(\sqrt{−19}))) = rank(E(ℚ)) + rank(E^{(−19)}(ℚ)) ,

where E^{(−19)} is the twist of E by −19. From this, it is not difficult to prove that the rank must be equal to 0 over ℚ(\sqrt{−19}). On the other hand, our 2-descent produces three nontrivial everywhere locally soluble quartics which certainly represent nontrivial elements in the Tate–Shafarevich group. To be more precise, we give these three quartics for the first curve:

y^{2} = − 103x^{4} − 160x^{3} − 134x^{2} − 56x + 5
y^{2} = (27ε − 56)x^{4} + (20ε − 36)x^{3} + (−48ε + 22)x^{2} + (−4ε + 32)x + (ε + 7)
y^{2} = (50ε − 229)x^{4} + (76ε + 612)x^{3} + (−306ε − 782)x^{2}

+ (236ε + 520)x + (−53ε − 134)

2.3. A Degree 5 Example

Consider the number field K = ℚ(ε), with ε^{5} − ε^{3} − ε^{2} + ε + 1 = 0. This field has a single real embedding, and its discriminant is equal to 1609. Using 2-descent, we have computed that the curve

E : y^{2} = x^{3} + ε

has rank 1 and that the group � (E/K)[2] is trivial. The point (ε^{4} − ε, ε + 1) has infinite order in E(K). We do not go into the cumbersome details.

References

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Denis SIMON Denis.Simon@math.u-bordeaux.fr

UFR Mathématiques Université Bordeaux I - France 351, Cours de la Libération 33405 Talence France