Testing Equivalence of Ternary Cubics

Tom Fisher

University of Cambridge, DPMMS, Centre for Mathematical Sciences, Wilberforce Road, Cambridge CB3 0WB, UK

                        T.A.Fisher@dpmms.cam.ac.uk

Abstract. Let C be a smooth plane cubic curve with Jacobian E. We give a formula for the action of the 3-torsion of E on C, and explain how it is useful in studying the 3-Selmer group of an elliptic curve defined over a number field.

We work over a field K of characteristic zero, with algebraic closure \overline{K}.

1 The Invariants of a Ternary Cubic

Let X_{3} = A^{10} be the space of all ternary cubics

\begin{aligned}U(X, Y, Z) & = & aX^{3} + bY^{3} + cZ^{3} + a_{2}X^{2}Y + a_{3}X^{2}Z \\ & & + b_{1}XY^{2} + b_{3}Y^{2}Z + c_{1}XZ^{2} + c_{2}Y Z^{2} + mXY Z . \end{aligned}

The co-ordinate ring of X_{3} is the polynomial ring

K[X_{3}] = K[a, b, c, a_{2}, a_{3}, b_{1}, b_{3}, c_{1}, c_{2}, m] .

There is a natural action of GL_{3} on X_{3} given by

(gU)(X, Y, Z) = U(g_{11}X + g_{21}Y + g_{31}Z, …, g_{13}X + g_{23}Y + g_{33}Z) .

The ring of invariants is

K[X_{3}]^{SL^{3}} = \{F ∈ K[X_{3}] : F ∘ g = F for all g ∈ SL_{3}(\overline{K})\} .

A homogeneous invariant F satisfies

F ∘ g = χ(g)F  (1)

for all g ∈ GL_{3}(\overline{K}), for some rational character χ : GL_{3} → G_{m}. But the only rational characters of GL_{3} are of the form χ(g) = (det g)^{k} for k an integer. We say that F is an invariant of weight k. Taking g a scalar matrix in (1) shows that F has weight equal to its degree. The following facts are well known: see [1], [10], [15].

Theorem 1.1. There are invariants c_{4}, c_{6} and Δ of weights 4, 6 and 12, related by c_{4}^{3} − c_{6}^{2} = 1728Δ, with the following properties:

(i) The invariants of

U_{E}(X, Y, Z) = Y^{2}Z + a_{1}XY Z + a_{3}Y Z^{2} − X^{3} − a_{2}X^{2}Z − a_{4}XZ^{2} − a_{6}Z^{3}

are given by the standard formulae: see [14, Chap. III]. (ii) The ring of invariants is a polynomial ring in two variables, generated by c_{4} and c_{6}. (iii) A ternary cubic U is non-singular if and only if Δ(U) ≠ 0. (iv) If the plane cubic {U = 0} ⊂ P^{2} is non-singular then it has Jacobian

y^{2} = x^{3} − 27c_{4}(U)x − 54c_{6}(U) .

The Hessian of U(X, Y, Z) is

H(X, Y, Z) = (−1/2) × | \begin{matrix}\frac{∂^{2}U}{∂X^{2}} & \frac{∂^{2}U}{∂X∂Y} & \frac{∂^{2}U}{∂X∂Z} \\ \frac{∂^{2}U}{∂X∂Y} & \frac{∂^{2}U}{∂Y^{2}} & \frac{∂^{2}U}{∂Y ∂Z} \\ \frac{∂^{2}U}{∂X∂Z} & \frac{∂^{2}U}{∂Y ∂Z} & \frac{∂^{2}U}{∂Z^{2}} \end{matrix} | .

The factor −1/2, although not standard, is a choice we find convenient. The Hessian is a polynomial map H : X_{3} → X_{3} satisfying

H ∘ g = (det g)^{2}g ∘ H

for all g ∈ GL_{3}(\overline{K}). We say it is a covariant of weight 2. Putting c_{4} = c_{4}(U), c_{6} = c_{6}(U) and H = H(U) we find

H(λU + μH) = 3(c_{4}λ^{2}μ + 2c_{6}λμ^{2} + c_{4}^{2}μ^{3})U + (λ^{3} − 3c_{4}λμ^{2} − 2c_{6}μ^{3})H .

This formula is classical: see [7], [11]. It is easily verified by restricting to any family of plane cubics covering the j-line. It also gives a convenient way of computing the invariants c_{4} and c_{6}.

2 The 3-Selmer Group

Definition 2.1. Let U_{1} and U_{2} be ternary cubics over K.

(i) U_{1} and U_{2} are equivalent if U_{2} = λ(gU_{1}) for some λ ∈ K^{×} and g ∈ GL_{3}(K). (ii) U_{1} and U_{2} are properly equivalent if U_{2} = (det g)^{−1}(gU_{1}) for some g ∈ GL_{3}(K).

Lemma 2.2. Let U_{1} and U_{2} be non-singular ternary cubics over K. If U_{1} and U_{2} are properly equivalent then they have the same invariants. If K = \overline{K} then the converse is also true.

Proof. The first statement follows from the fact that a homogeneous invariant has weight equal to its degree. For the second statement we may assume

\begin{aligned}U_{1}(X, Y, Z) & = & Y^{2}Z − (X^{3} + a_{1}XZ^{2} + b_{1}Z^{3}) \\ U_{2}(X, Y, Z) & = & Y^{2}Z − (X^{3} + a_{2}XZ^{2} + b_{2}Z^{3}) \end{aligned}

for some a_{1}, b_{1}, a_{2}, b_{2} ∈ \overline{K}. Since U_{1} and U_{2} have the same invariants, it follows by Theorem 1.1(i) that U_{1} = U_{2}. ⊓ ⊔

We consider pairs (C → S, ω) where C → S is a morphism from a smooth curve of genus one C to a Brauer-Severi variety S, and ω is a regular 1-form on C. An isomorphism between (C_{1} → S_{1}, ω_{1}) and (C_{2} → S_{2}, ω_{2}) is a pair of isomorphisms ϕ : C_{1} ≅ C_{2} and ψ : S_{1} ≅ S_{2} such that ϕ^{∗}ω_{2} = ω_{1} and the diagram

\xymatrix{C_{1} \ar[r] \ar[d]_{ϕ} & S_{1} \ar[d]^{ψ} \\ C_{2} \ar[r] & S_{2}}

commutes.

Let n ≥ 2 be an integer. Let E/K be an elliptic curve with invariant differential ω_{E}. We map E → P^{n−1} via the complete linear system |n.0_{E}|. We recall that objects defined over K are called twists if they are isomorphic over \overline{K}.

Lemma 2.3. The twists of (E → P^{n−1}, ω_{E}), up to K-isomorphism, are parametrised by H^{1}(K, E[n]).

Proof. The automorphisms α of E with α^{∗}ω_{E} = ω_{E} are the translation maps. If τ_{P} : E → E is translation by P ∈ E(\overline{K}), we know that τ_{P}^{∗} (n.0_{E}) ∼ n.0_{E} if and only if nP = 0_{E}. So Aut(E → P^{n−1}, ω_{E}) ≅ E[n]. An injective map from the isomorphism classes of twists to H^{1}(K, E[n]) is given by comparing an isomorphism defined over \overline{K} with its Galois conjugates. It remains to prove surjectivity. This follows from the well known facts that the twists of E are parametrised by H^{1}(K, Isom(E)) and the twists of P^{n−1} are parametrised by H^{1}(K, PGL_{n}). ⊓ ⊔

Remark 2.4. This interpretation of H^{1}(K, E[n]) is a variant of one given in [4], [9]. If ϕ : C → E is an isomorphism of curves defined over \overline{K} with ϕ^{∗}ω_{E} = ω then we make C a torsor under E via (P, Q) ↦ ϕ^{−1}(P + ϕ(Q)). This action depends on ω but not on ϕ.

The obstruction map, defined in [9], is

\begin{aligned}Ob : & H^{1}(K, E[n]) & → & Br(K) \\ & (C → S, ω) & ↦ & [S] . \end{aligned}

In general this map is not a group homomorphism. Nevertheless we write ker(Ob) for the inverse image of the identity. We specialise to the case n = 3.

Theorem 2.5. Let U_{E} = 0 be a Weierstrass equation for E. Then the ternary cubics with the same invariants as U_{E}, up to proper K-equivalence, are parametrised by ker(Ob) ⊂ H^{1}(K, E[3]).

Proof. A ternary cubic U determines a plane cubic C = {U = 0} ⊂ P^{2} and a regular 1-form on C

ω = \frac{Z^{2}d(Y/Z)}{\frac{∂U}{∂X} (X, Y, Z)} .

Conversely, every twist (C → S, ω) of (E → P^{2}, ω_{E}) with S ≅ P^{2} arises in this way. In view of Lemmas 2.2 and 2.3 it only remains to show that ternary cubics U_{1} and U_{2} are properly equivalent if and only if they determine isomorphic pairs (C_{1} → P^{2}, ω_{1}) and (C_{2} → P^{2}, ω_{2}). This is immediate from the next lemma, or more precisely the special case of it where g ∈ GL_{3}(K). ⊓ ⊔

Lemma 2.6. Let U_{1} and U_{2} be non-singular ternary cubics, determining pairs (C_{1} → P^{2}, ω_{1}) and (C_{2} → P^{2}, ω_{2}). If gU_{1} = U_{2} for some g ∈ GL_{3}(\overline{K}) then the isomorphism induced by g, namely

γ : C_{2} → C_{1} ; (X : Y : Z) ↦ (g_{11}X + g_{21}Y + g_{31}Z : …),

satisfies γ^{∗}ω_{1} = (det g)ω_{2}.

Proof. If the lemma is true for g_{1}, g_{2} ∈ GL_{3}(\overline{K}) then it is true for g_{1}g_{2}. So it suffices to let g run over a set of generators for GL_{3}(\overline{K}). The result is already clear for matrices of the form

g = ( \begin{matrix}λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & μ & λ_{3} \end{matrix} ) .

Then for

g = ( \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} ) and g = ( \begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} )

we use the identities

\frac{Z}{Y} d (\frac{Y}{Z} ) + \frac{Y}{Z} d (\frac{Z}{Y} ) = 0

and

\frac{1}{Z^{2}} \frac{∂U}{∂X} (X, Y, Z)d (\frac{X}{Z} ) + \frac{1}{Z^{2}} \frac{∂U}{∂Y} (X, Y, Z)d (\frac{Y}{Z} ) = 0 .

⊓ ⊔

Remark 2.7. The subset ker(Ob) ⊂ H^{1}(K, E[3]) contains the identity and is closed under taking inverses. A ternary cubic U represents the identity if and only if it has a K-rational point of inflection. The inverse of U is −U.

Remark 2.8. We claim that if K is a number field then the everywhere locally soluble ternary cubics with the same invariants as U_{E}, up to proper Kequivalence, are parametrised by the 3-Selmer group S^{(3)}(E/K). It is shown in [9] that S^{(3)}(E/K) ⊂ ker(Ob), so this claim is a special case of Theorem 2.5.

This interpretation of S^{(3)}(E/K) becomes more useful if we can find algorithms for performing the following tasks. We write [U] for the proper equivalence class of U.

1. Given U test whether [U] = 0.

2. Given U_{1}, U_{2} test whether [U_{1}] = [U_{2}]. If so find the change of co-ordinates that relates them.

3. Given U_{1}, U_{2}, U_{3} test whether [U_{1}] + [U_{2}] = [U_{3}].

4. Given U_{1}, U_{2} determine whether there exists U_{3} with [U_{1}] + [U_{2}] = [U_{3}]. If so compute U_{3}.

The analogues of these problems for the 2-Selmer group have been solved in [3].

3 The Etale Algebra

Let R be the étale algebra of E[3]. It is a product of field extensions of K, one for each orbit for the action of Gal(\overline{K}/K) on E[3]. It is shown in [6], [12] that there is an injective group homomorphism

w_{1} : H^{1}(K, E[3]) → R^{×}/(R^{×})^{3} .

According to [4, Paper I, Corollary 3.12] the restriction to ker(Ob) is given by

(C → P^{2}, ω) ↦ α = det M

where M ∈ GL_{3}(R) = Map_{K}(E[3], GL_{3}(\overline{K})) describes the action of E[3] on C → P^{2}. (Recall that C is a torsor under E.)

In joint work [4] we describe a method for converting elements of ker(Ob) represented by α ∈ R^{×} to elements of ker(Ob) represented by a ternary cubic U(X, Y, Z). In this article we work in the opposite direction. We start with a ternary cubic U(X, Y, Z) and convert it to α ∈ R^{×}. We also give a formula for the matrix M ∈ GL_{3}(R). This enables us to solve the problems listed at the end of Sect. 2.

4 The Hesse Family

Let C be a smooth plane cubic with Jacobian E. Let ζ ∈ \overline{K} be a primitive cube root of unity. Let S, T be a basis for E[3] with e_{3}(S, T) = ζ, where e_{3} is the Weil pairing. Making a suitable choice of co-ordinates over \overline{K} we may assume

M_{S} = ( \begin{matrix}1 & 0 & 0 \\ 0 & ζ & 0 \\ 0 & 0 & ζ^{2} \end{matrix} ) , M_{T} = ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) .

Then C has equation

U(X, Y, Z) = a(X^{3} + Y^{3} + Z^{3}) − 3bXY Z .

The invariants of this ternary cubic are

\begin{aligned}c_{4}(a, b) & = & 3^{4}(8a^{3} + b^{3})b \\ c_{6}(a, b) & = & 3^{6}(8a^{6} + 20a^{3}b^{3} − b^{6}) \\ Δ(a, b) & = & −3^{9}a^{3}(a^{3} − b^{3})^{3} . \end{aligned}

The Hessian is

H(X, Y, Z) = 27ab^{2}(X^{3} + Y^{3} + Z^{3}) − 27(4a^{3} − b^{3})XY Z .

Taking 0_{C} = (0 : 1 : −1) the elliptic curve (C, 0_{C}) has Weierstrass equation

y^{2}z = x^{3} − 27c_{4}(a, b)xz^{2} − 54c_{6}(a, b)z^{3} .

An explicit isomorphism is given by

\begin{aligned}x & = & −27(4a^{3} − b^{3})X − 81ab^{2}(Y + Z) \\ y & = & 972a(a^{3} − b^{3})(Y − Z) \\ z & = & bX + a(Y + Z) . \end{aligned}  (2)

5 The Syzygetic Triangles

Let U(X, Y, Z) be a non-singular ternary cubic with Jacobian E. The pencil of cubics spanned by U and its Hessian is a twist of the Hesse family. So there are exactly 4 singular fibres, and each singular fibre is a triangle. The sides of each triangle are the fixed lines for the action of M_{T} on P^{2} for some 0 ≠ T ∈ E[3]. So there is a Galois equivariant bijection between the syzygetic triangles and

P(E[3]) = \frac{E[3] \ \{0\}}{\{±1\}} .

Lemma 5.1. Let U be a non-singular ternary cubic with invariants c_{4}, c_{6} and Hessian H. Let T = (x_{T} , y_{T}) be a non-zero 3-torsion point on the Jacobian

E : y^{2} = x^{3} − 27c_{4}x − 54c_{6} .

Then the syzygetic triangle corresponding to ±T has equation

T = \frac{1}{3}x_{T} U + H

and this equation satisfies H(T ) = \frac{1}{27}y_{T}^{2} T .

Proof. We may assume that U belongs to the Hesse family with T the image of (0 : ζ : −ζ^{2}) under (2). The lemma follows by direct calculation. ⊓ ⊔

Remark 5.2. The Hessian of a triangle is a non-zero multiple of the triangle. So in Lemma 5.1 we have y_{T} ≠ 0. This is no surprise, since a non-zero 3-torsion point on E cannot also be a 2-torsion point.

6 The Invariants of a Triangle

Let S_{3} act on Q[α_{1}, α_{2}, α_{3}, β_{1}, β_{2}, β_{3}] by simultaneously permuting the α_{i} and the β_{i}. The ring of invariants has Hilbert series

h(t) = \frac{1 + t^{2} + 2t^{3} + t^{4} + t^{6}}{(1 − t)^{2}(1 − t^{2})^{2}(1 − t^{3})^{2}} .

Let s_{1}, s_{2}, s_{3} (respectively t_{1}, t_{2}, t_{3}) be the elementary symmetric polynomials in the α_{i} (respectively β_{i}). According to MAGMA the primary invariants are s_{1}, s_{2}, s_{3}, t_{1}, t_{2}, t_{3}. The remaining coefficients of

T_{1}(X, Y, Z) = ∏_{i=1}^{3}(X + α_{i}Y + β_{i}Z)

are

\begin{aligned}u & = & α_{1}(β_{2} + β_{3}) + α_{2}(β_{3} + β_{1}) + α_{3}(β_{1} + β_{2}) \\ v & = & α_{1}α_{2}β_{3} + α_{1}β_{2}α_{3} + β_{1}α_{2}α_{3} \\ w & = & α_{1}β_{2}β_{3} + β_{1}α_{2}β_{3} + β_{1}β_{2}α_{3} . \end{aligned}

The secondary invariants are 1, u, v, w, u^{2}, vw. So as a Q-algebra, the ring of invariants is generated by the coefficients of T_{1}. There are 5 relations. These are obtained by writing uv, uw, v^{2}, w^{2}, u^{3} as linear combinations of the secondary invariants. In fact MAGMA can rewrite any invariant as a Q[s_{1}, s_{2}, s_{3}, t_{1}, t_{2}, t_{3}]- linear combination of the secondary invariants. For example

\begin{matrix}(α_{1} − α_{2})(α_{2} − α_{3})(α_{3} − α_{1}) | \begin{matrix}1 & α_{1} & β_{1} \\ 1 & α_{2} & β_{2} \\ 1 & α_{3} & β_{3} \end{matrix} | \\ = 2s_{1}^{2}v − s_{1}s_{2}u − 6s_{1}s_{3}t_{1} + 2s_{2}^{2}t_{1} − 6s_{2}v + 9s_{3}u . \end{matrix}

7 Formulae

Let C be a smooth plane cubic defined over K, with Jacobian E. Let L/K be any field extension. Given T ∈ E[3](L) we aim to compute M_{T} ∈ GL_{3}(L) describing the action of T on C. We start with an equation U = 0 for C. Then we construct the syzygetic triangle T = \frac{1}{3}x_{T} U + H as described in Lemma 5.1. Making a change of co-ordinates if necessary, we may assume T (1, 0, 0) ≠ 0. Then factoring over the algebraic closure gives

T (X, Y, Z) = r ∏_{i=1}^{3}(X + α_{i}Y + β_{i}Z) .  (3)

We put

P = ( \begin{matrix}1 & α_{1} & β_{1} \\ 1 & α_{2} & β_{2} \\ 1 & α_{3} & β_{3} \end{matrix} )

and ξ = α_{1} + ζα_{2} + ζ^{2}α_{3}.

Theorem 7.1. If ξ ≠ 0 then the matrix

M_{T} = rξP^{−1} ( \begin{matrix}1 & 0 & 0 \\ 0 & ζ^{2} & 0 \\ 0 & 0 & ζ \end{matrix} ) P

belongs to GL_{3}(L) and describes the action of T (or −T) on C.

Proof. The required matrix has image in PGL_{3} of order 3, and acts on P^{2} with fixed lines the sides of T = 0. So the second statement is clear. We must check that M_{T} has coefficients in L.

We write r^{−1}(det P)M_{T} = A + (ζ − ζ^{2})B where A and B are matrices with entries in Q[α_{1}, α_{2}, α_{3}, β_{1}, β_{2}, β_{3}]. We find σ(A) = sign(σ)A and σ(B) = B for all σ ∈ S_{3}. So the entries of (det P)A and B are polynomials in the coefficients of T_{1} = r^{−1}T . As discussed in Sect. 6 we can compute these polynomials using MAGMA.

By (3) we have H(T ) = −r^{2}(det P)^{2}T . Comparing with Lemma 5.1 it follows that y_{T} = ±3(ζ − ζ^{2})r det P. Therefore

(det P)^{2}M_{T} = r(det P)A ± \frac{1}{3}y_{T} B

and M_{T} has entries in L as required. ⊓ ⊔

We write

\begin{aligned}T (X, Y, Z) & = & rX^{3} + s_{1}X^{2}Y + s_{2}XY^{2} + s_{3}Y^{3} \\ & & + t_{1}X^{2}Z + t_{2}XZ^{2} + t_{3}Z^{3} \\ & & + Y Z(uX + vY + wZ) . \end{aligned}  (4)

Theorem 7.2. det(M_{T}) = \frac{1}{2}(R ± \frac{27r}{y_{T}} S) where

\begin{aligned}R & = & 2s_{1}^{3} − 9rs_{1}s_{2} + 27r^{2}s_{3} \\ S & = & 2s_{1}^{2}v − s_{1}s_{2}u − 6s_{1}s_{3}t_{1} + 2s_{2}^{2}t_{1} − 6rs_{2}v + 9rs_{3}u . \end{aligned}

Proof. Comparing coefficients in (3) and (4) we find

\begin{aligned}det(M_{T}) & = & r^{3}(α_{1} + ζα_{2} + ζ^{2}α_{3})^{3} \\ & = & \frac{1}{2}(R − 3(ζ − ζ^{2})r^{3}δ) \end{aligned}

where δ = (α_{1} − α_{2})(α_{2} − α_{3})(α_{3} − α_{1}). By the example in Sect. 6 we have r^{3}δ det P = S. Finally we recall from the proof of Theorem 7.1 that y_{T} = ±3(ζ − ζ^{2})r det P. ⊓ ⊔

Remark 7.3. The formulae of Theorems 7.1 and 7.2 sometimes fail and give zero. (The situation is analogous to the proof of Hilbert’s theorem 90 using Lagrange resolvents.) However if they fail for both T and −T then

α_{1} + ζα_{2} + ζ^{2}α_{3} = 0

and

α_{1} + ζ^{2}α_{2} + ζα_{3} = 0 .

From these we deduce det P = 0, contradicting that (3) is the equation of a syzygetic triangle. So if our formula for M_{T} fails then we can use (M_{−T})^{−1} instead.

8 Galois Actions

The formulae of Sect. 7 are slightly easier to use in the case E does not admit a rational 3-isogeny.

Lemma 8.1. If E does not admit a rational 3-isogeny and [U] ≠ 0, then we are guaranteed that T (1, 0, 0) ≠ 0.

Proof. We recall that T = \frac{1}{3}x_{T} U +H. By hypothesis x_{T} ∉ K. So if T (1, 0, 0) = 0 then U(1, 0, 0) = H(1, 0, 0) = 0. But then (1 : 0 : 0) is a K-rational point of inflection and [U] = 0. ⊓ ⊔

Let G ⊂ GL_{2}(F_{3}) ≅ Aut(E[3]) be the image of Galois.

Lemma 8.2. If E does not admit a rational 3-isogeny then −I_{2} ∈ G.

Proof. By hypothesis the image of G in PGL_{2}(F_{3}) ≅ S_{4} acts on P^{1}(F_{3}) without fixed points. If this image is A_{4} or S_{4} then G contains SL_{2}(F_{3}) by [13, IV, §3.4, Lemma 2]. Otherwise G is a 2-group, and so conjugate to a subgroup of the Sylow 2-subgroup generated by

a = ( \begin{matrix}1 & 1 \\ −1 & 1 \end{matrix} ) and b = ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) .

The only non-trivial subgroups of ⟨a, b⟩, not containing −I_{2}, are the conjugates of ⟨b⟩. These possibilities for G are again ruled out by the assumption that E does not admit a rational 3-isogeny. ⊓ ⊔

If −I_{2} ∈ G then our formula for M_{T} works if and only if our formula for M_{−T} works. According to Remark 7.3 they cannot both fail, so they must both work.

9 Applications

In our examples we take K = Q. Elliptic curves over Q are referenced by their labellings in [2].

9.1 Testing Proper Equivalence

We are given non-singular ternary cubics U_{1} and U_{2}, and must decide whether they are properly equivalent. First we check that they have the same invariants c_{4} and c_{6}. Then the plane cubics U_{1} = 0 and U_{2} = 0 each have Jacobian

E : y^{2} = x^{3} − 27c_{4}x − 54c_{6} .

We compute α_{1}, α_{2} ∈ R^{×} by using Theorem 7.2 once for each orbit for the action of Gal(\overline{K}/K) on E[3]. Then U_{1} and U_{2} are properly equivalent if and only if α_{1}/α_{2} ∈ (R^{×})^{3}.

For example the ternary cubics

\begin{aligned}U_{1}(X, Y, Z) & = & X^{3} − 180Y^{3} + 24Z^{3} + 8X^{2}Y − 3X^{2}Z \\ & & + 3XY^{2} − 148Y^{2}Z + 76XZ^{2} − 280Y Z^{2} + 59XY Z \end{aligned}

and

\begin{aligned}U_{2}(X, Y, Z) & = & 32X^{3} + 48Y^{3} + 32Z^{3} − 14X^{2}Y − 17X^{2}Z \\ & & + 14XY^{2} + 68Y^{2}Z − 34XZ^{2} + 34Y Z^{2} − 91XY Z \end{aligned}

each have invariants c_{4} = 1073512497 and c_{6} = 35173095391575. The Jacobian is

2534e2 : y^{2} + xy + y = x^{3} − x^{2} − 22364844x − 40704009937 .

A non-trivial 3-torsion point is T = (x_{T} , y_{T}) where

x_{T} = \frac{1}{12}(289u^{6} + 765u^{4} + 24567u^{2} + 22035)
y_{T} = −\frac{1}{312}(239307u^{7} + 3757u^{6} + 638911u^{5} + 9945u^{4}
+ 20357181u^{3} + 319371u^{2} + 45909405u + 286611)

and u is a root of X^{8} + 78X^{4} − 36X^{2} − 507 = 0. We have R = Q × L where L = Q(u) is a number field of degree 8. The first factor of Q may be ignored. Using Theorem 7.2 we compute

α_{1} = \frac{144}{13}(548276415600669u^{7} − 912344032067546u^{6}
+1459379319052681u^{5} − 2428439574347826u^{4}
+46650075622210203u^{3} − 77626752951639190u^{2}
+104433275464300347u − 173779291524426198)
α_{2} = \frac{1152}{13}(23737183831720776u^{7} + 38664498064205221u^{6}
+63182645951465768u^{5} + 102915548856548337u^{4}
+2019677284143385464u^{3} + 3289767129786200531u^{2}
+4521354220053126264u + 7364643132168529779) .

We find α_{1}/α_{2} = b^{3} where

\begin{aligned}b & = & \frac{1}{31499104}(−35980u^{7} + 9880u^{6} − 90181u^{5} + 294515u^{4} \\ & & −2820090u^{3} + 1603888u^{2} − 6288205u + 17147429) . \end{aligned}

It follows that U_{1} and U_{2} are properly equivalent.

Remark 9.1. Suppose we are given non-singular ternary cubics U and U′ with invariants c_{4}, c_{6} and c′_{4}, c′_{6}. To test for equivalence we first find all λ ∈ K^{×} satisfying c′_{4} = λ^{4}c_{4} and c′_{6} = λ^{6}c_{6}. Then for each such λ we test whether λU and U′ are properly equivalent.

9.2 Finding Equivalences

We continue with the example of the last subsection and find the change of coordinates relating U_{1} and U_{2}. Following the proof of Theorem 7.1 we compute matrices M_{1}, M_{2} ∈ GL_{3}(L) describing the action of T on U_{1} = 0 and U_{2} = 0. Since α_{1}/α_{2} ∈ (L^{×})^{3} we may arrange that det M_{1} = det M_{2}. We are looking for g ∈ GL_{3}(K) with U_{2} = (det g)^{−1}(gU_{1}). We must have

M_{1}g^{T} = cg^{T} M_{2}

for some c ∈ L^{×}. Taking determinants gives c^{3} = 1. Since L contains no nontrivial cube roots of unity it follows that c = 1. Solving for g by linear algebra we find

g = ( \begin{matrix}19 & −1 & 6 \\ −8 & −8 & 0 \\ 22 & −2 & −4 \end{matrix} ) .

Remark 9.2. If E[3](K) ≠ 0 then there will be more than one change of coordinates relating U_{1} and U_{2}. These will correspond to different choices for the constant c. Indeed by the Weil pairing there is an inclusion E[3](K) ⊂ μ_{3}(R).

9.3 Addition of Selmer Group Elements

The rank 0 elliptic curve

E = 4343b1 : y^{2} + y = x^{3} − 325259x − 71398995

has Tate-Shafarevich group of analytic order 9. The following two elements of S^{(3)}(E/Q) are visible in the rank 1 elliptic curves 21715a1 and 117261k1. (We will explain these calculations more fully in subsequent work. The concept of visibility was introduced in [5].)

U_{1}(X, Y, Z) = X^{3} + 15Y^{3} − 17Z^{3} − 8X^{2}Y + 4X^{2}Z
+ 15XY^{2} − 13Y^{2}Z + 32XZ^{2} + 26Y Z^{2} + 4XY Z
U_{2}(X, Y, Z) = 7X^{3} − 13Y^{3} − 17Z^{3} + 7X^{2}Y + 3X^{2}Z
− 4XY^{2} − 2Y^{2}Z + 12XZ^{2} − 15Y Z^{2} − 30XY Z

We use Theorem 7.2 to compute α_{1}, α_{2} ∈ R^{×}. We find that α_{1}, α_{2} are independent in R^{×}/(R^{×})^{3}. Applying the work of [4] to α_{1}α_{2} and α_{1}/α_{2} we obtain

U_{3}(X, Y, Z) = −5X^{3} + 12Y^{3} + 31Z^{3} + 3X^{2}Y − 5X^{2}Z
+ 5XY^{2} + 2Y^{2}Z + 4XZ^{2} + 26Y Z^{2} + 40XY Z
U_{4}(X, Y, Z) = −11X^{3} + 8Y^{3} − 13Z^{3} − 9X^{2}Y + 11X^{2}Z
− 15XY^{2} − Y^{2}Z − 16XZ^{2} − 3Y Z^{2} − 38XY Z .

Assuming the Birch Swinnerton-Dyer conjecture, we have

�(E/Q) = {0, ±[U_{1}], ±[U_{2}], ±[U_{3}], ±[U_{4}]} .

We have found these equations without the need to compute the class group or unit group of any number field.

9.4 Testing Global Solubility

We show that the ternary cubic

\begin{aligned}U_{1}(X, Y, Z) & = & 7X^{3} + 9Y^{3} + 16Z^{3} + 2X^{2}Y \\ & & + 5Y^{2}Z + 5XZ^{2} − 7Y Z^{2} − 31XY Z \end{aligned}

is insoluble over Q. The Jacobian

E = 35882a1 : y^{2} + xy = x^{3} − x^{2} − 156926x − 24991340

has Mordell-Weil group E(Q) ≅ Z/2Z × Z. A point of infinite order is P = (693, 13750). Embedding E ⊂ P^{2} via |2.0_{E} + P | we obtain

\begin{aligned}U_{2}(X, Y, Z) & = & 15Y^{3} + 1254Z^{3} + X^{2}Z − XY^{2} \\ & & + 674Y^{2}Z + 10XZ^{2} − 291Y Z^{2} + XY Z . \end{aligned}

We use Theorem 7.2 to compute α_{1}, α_{2} ∈ R^{×}. We then check that α_{1}, α_{2} are independent in R^{×}/(R^{×})^{3}. Since U_{2} is soluble over Q and E(Q)/3E(Q) ≅ Z/3Z, it follows that U_{1} is insoluble over Q.

Alternatively this could be checked using the explicit formulae for the covering map given in [1].

9.5 Reduction of Ternary Cubics

It is desirable to be able to replace an integer coefficient ternary cubic by an equivalent one with smaller coefficients. One method, explained to me by Michael Stoll, first computes a certain inner product, and then uses standard lattice reduction techniques. By an inner product on a complex vector space we mean a positive definite Hermitian form. We recall the Weyl unitary trick.

Lemma 9.3. Let V be an irreducible complex representation of a finite group G. Then (up to scalars) there is a unique G-invariant inner product ⟨ , ⟩ : V ×V → C.

Proof. Let ⟨ , ⟩_{0} be any inner product on V. Then

⟨u, v⟩ = ∑_{g∈G}⟨gu, gv⟩_{0}

is a G-invariant inner product. By Schur’s lemma the complex vector space of G-invariant sesquilinear forms on V is 1-dimensional. ⊓ ⊔

We now take C ⊂ P^{2} a smooth plane cubic defined over Q with Jacobian E. The action of E[3] on C extends to P^{2} to give χ : E[3] → PGL_{3}. Lifting to SL_{3} we obtain a diagram

[figure: 0; μ_{3}; H_{3}; E[3]; χ; 0; 0; μ_{3}; _{SL_{3}}; _{PGL_{3}}; 0 .]

The Heisenberg group H_{3} is a non-abelian group of order 27. For the reduction of ternary cubics, we use the unique Heisenberg-invariant inner product. Theorem 7.1 gives a convenient way of computing this inner product. Indeed if M_{1} ∈ SL_{3}(R) and M_{2} ∈ SL_{3}(C) generate the action of E[3] on C then the required inner product on C^{3} has Gram matrix

∑_{r=0}^{2}(\overline{M_{2}}^{r})^{T} ( ∑_{s=0}^{2}(M_{1}^{s})^{T} M_{1}^{s})M_{2}^{r} .

Acknowledgements

I would like to thank John Cremona, Cathy O’Neil and Michael Stoll for many useful discussions. All computer calculations in support of this work were performed using MAGMA [8]. The examples in Sect. 9 were prepared with the assistance of some programs written by Michael Stoll in connection with the joint work [4].

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