THE INVARIANTS OF A GENUS ONE CURVE

TOM FISHER

Abstract. It was first pointed out by Weil [26] that we can use classical invariant theory to compute the Jacobian of a genus one curve. The invariants required for curves of degree n = 2, 3, 4 were already known to the nineteenth centuary invariant theorists. We have succeeded in extending these methods to curves of degree n = 5, where although the invariants are too large to write down as explicit polynomials, we have found a practical algorithm for evaluating them.

1. Introduction

We work throughout over a perfect field K with algebraic closure \overline{K}. In this introduction we further assume that char (K) ≠ 2, 3. Let C be a smooth curve of genus one defined over K and let D be a K-rational divisor on C of degree n. If n = 1 then C(K) ≠ ∅, and C is defined by a Weierstrass equation. If n ≥ 2 then the complete linear system |D| defines a morphism C → ℙ^{n−1}. If n ≥ 3 then this morphism is an embedding and we call the image a genus one normal curve of degree n. For n ≤ 5 the pair (C, D) is described by data of the following form.

Definition 1.1. A genus one model of degree n = 1, 2, 3, 4, 5 is

(i) if n = 1 a Weierstrass equation (ii) if n = 2 a binary quartic (iii) if n = 3 a ternary cubic (iv) if n = 4 a pair of quadrics in 4 variables

(v) if n = 5 a 5 × 5 alternating matrix of linear forms in 5 variables.

The equations defined by a genus one model of degree 5 are the 4×4 Pfaffians of the matrix. It is a classical fact that every genus one normal quintic is defined by equations of this form. An algorithm for computing these matrices is given in [12], based on the Buchsbaum-Eisenbud structure theorem [5], [6] for Gorenstein ideals of codimension 3.

We write X_{n} for the (affine) space of all genus one models of degree n and K[X_{n}] for its co-ordinate ring. In §4 we specify a linear algebraic group G_{n} acting on X_{n}. In each case we find that the ring of invariants K[X_{n}]^{G_{n}} is a polynomial ring in two variables. (This could be deduced by a theorem of Kempf [18, Theorem 2.4] if char (K) = 0.) We label the generators c_{4} and c_{6}. In the case n = 1 these are the usual polynomials defined in [24, Chapter III]. In the cases n = 2, 3, 4, 5 we find that c_{4} and c_{6} are homogeneous polynomials of degree 4n/(6 − n) and 6n/(6 − n). Moreover, as first pointed out by Weil [26] in the case n = 2, the invariants c_{4} and c_{6} give a formula for the Jacobian. The invariants for n = 2, 3, 4 have been known since the nineteenth century, and are surveyed in [1].

Date: 9th October 2006.

Our work has two main goals. The first is to describe the ring of invariants in a way that emphasises the similarities between the cases n = 1, 2, 3, 4, 5. The second is to give practical methods for evaluating the invariants. In both instances our main original contribution is in the case n = 5.

A relatively easy argument (reduction to the case n = 1) shows that if invariants c_{4} and c_{6} of the expected degrees exist then they generate the ring of invariants. In the cases n = 2, 3 the existence of these invariants is settled by writing them down as explicit polynomials. In the case n = 4 there is a classical trick for reducing to the case n = 2. But in the case n = 5 the invariants are too large to write down as explicit polynomials. This makes it difficult to show they exist.

One of the tools of classical invariant theory is the so-called symbolic notation, as described in [14]. This is an extremely compact notation for specifying invariants. For example in the case n = 3 the invariants may be written (see [22, §§220,221] or [25, §4.5])

\begin{aligned}c_{4} & = & 54 × (abc)(bcd)(cda)(dab) \\ c_{6} & = & 972 × (abc)(abd)(bce)(caf)(def)^{2}. \end{aligned}

By introducing non-commuting symbols it is possible to write down similar expressions in the case n = 5. But we have no way of showing these invariants are non-zero without expanding them as explicit polynomials. As remarked above, this is not feasible.

In principle we could use the representation theory of Lie algebras, specifically the Weyl character formula, to compute the dimension of the vector space of invariants of any given degree. For details in the case n = 3 we refer to [13, Exercise 13.20] or [25, §4.4]. Unfortunately, when we tried this approach in the case n = 5, we were again defeated by combinatorial explosion.

The plan of the paper is as follows. In §2 we explain the role played by the invariant differential in computing the Jacobian of a genus one curve. In §3 we revisit and motivate our definition of a genus one model.

Notice that we modify the definition in the case n = 2 to accommodate fields of characteristic 2.

In §4 we study the ring of invariants. We show that it is generated by invariants c_{4} and c_{6} of the expected degrees and that these invariants give a formula for the Jacobian. We also show (in all characteristics) that a genus one model defines a smooth curve of genus one if and only if its discriminant Δ = (c_{4}^{3} − c_{6}^{2})/1728 is non-zero. The proofs rely on geometric results proved in §5 and formulae recorded in §6.

In §7 we recall some classical methods for computing the invariants in the cases n = 2, 3, 4. These formulae have already been surveyed in [1], but are included here to demonstrate our preferred choice of scalings. In the case n = 5 we have found an algorithm for evaluating the invariants. Our algorithm, presented in §8, is inspired by the methods of nineteenth century invariant theory, in that we approach the invariants through the construction of certain covariants. The key step relies on a geometric “accident” satisfied by the secant variety of a genus one normal quintic. In §9 we compare our invariant-theoretic approach with some other methods for computing the Jacobian of a genus one curve.

Finally in §10 we include a brief note on the invariants in characteristics 2 and 3. We find in these cases that the invariants are insufficient to compute the Jacobian. Instead it should be possible to find a formula for the Jacobian that works in all characteristics by modifying the formulae in characteristic 0. This has been carried out by Artin, Rodriguez-Villegas and Tate [3] in the case n = 3.

The formulae and algorithms presented in §§7,8 have been contributed to MAGMA [19, Version 2.13] by the author.

2. Geometric Invariants

Let C be a smooth curve of genus one defined over K, and let ω be a non-zero regular 1-form on C, also defined over K. We say that ω is an invariant differential. Over \overline{K}, the pair (C, ω) may be put in the form

(1)  y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}

with ω = dx/(2y + a_{1}x + a_{3}).

Definition 2.1. The geometric invariants of the pair (C, ω) are

\begin{aligned}c_{4} & = & b_{2}^{2} − 24b_{4} \\ c_{6} & = & −b_{2}^{3} + 36b_{2}b_{4} − 216b_{6} \end{aligned}

where b_{2} = a_{1}^{2} + 4a_{2}, b_{4} = 2a_{4} + a_{1}a_{3} and b_{6} = a_{3}^{2} + 4a_{6}.

It is clear from the formulae in [24, Chapter III] that c_{4} and c_{6} depend only on the pair (C, ω) and not on the choice of Weierstrass equation (1). We deduce by Galois theory that c_{4}, c_{6} ∈ K.

Lemma 2.2. If (C, ω) has geometric invariants c_{4} and c_{6}, and λ ∈ K^{∗}, then (C, λ^{−1}ω) has geometric invariants λ^{4}c_{4} and λ^{6}c_{6}.

Proof: This is again clear from [24, Chapter III]. □

We show in §5.4 that if a genus one model of degree n defines a smooth curve of genus one, then it also defines an invariant differential on the curve. This enables us to construct the invariants c_{4}, c_{6} ∈ K[X_{n}]^{G_{n}} as the geometric invariants of the generic genus one model of degree n. In particular we treat the cases n = 2, 3, 4, 5 in a uniform manner, and avoid the problem of combinatorial explosion in the case n = 5.

The geometric invariants give a formula for the Jacobian.

Proposition 2.3. Assume char (K) ≠ 2, 3. If (C, ω) has geometric invariants c_{4} and c_{6} then C has Jacobian

y^{2} = x^{3} − 27c_{4}x − 54c_{6}.

The proof relies on two easy lemmas.

Lemma 2.4. Assume char (K) ≠ 2, 3. Let E be an elliptic curve defined over K with invariant differential ω. Let α be an automorphism of E. Then α is a translation map if and only if α^{∗}ω = ω.

Proof: We write τ_{P} : E → E for translation by P ∈ E. The map P ↦ τ_{P}^{∗}(ω)/ω is a morphism E → 𝔾_{m}. It must therefore be constant. Specialising to P = 0 we deduce that τ_{P}^{∗}(ω) = ω for all P ∈ E. (An alternative proof is given by writing τ_{P} as the commutator of τ_{Q} and [−1] where 2Q = P.)

Conversely if α is not a translation then α − 1 is not constant, and therefore surjective. So α has a fixed point. Conjugating by a translation, we may suppose that the fixed point is 0 ∈ E. Since char (K) ≠ 2, 3 we can put E in shorter Weierstrass form y^{2} = x^{3} + Ax + B. Then the only automorphisms of (E, 0) are of the form (x, y) ↦ (u^{2}x, u^{3}y). Since ω is a multiple of dx/y the result is now clear. □

Lemma 2.5. Assume char (K) ≠ 2, 3. Let E be an elliptic curve and C a smooth curve of genus one, both defined over K. Let ω_{E} and ω_{C} be invariant differentials on E and C, also defined over K. If there is an isomorphism α : C ≅ E defined over \overline{K} with α^{∗}ω_{E} = ω_{C} then E is the Jacobian of C.

Proof: Let ξ_{σ} = σ(α)α^{−1} for σ ∈ Gal(\overline{K}/K). Since ω_{E} and ω_{C} are both K-rational we deduce that ξ_{σ}^{∗}ω_{E} = ω_{E}. It follows by Lemma 2.4 that ξ_{σ} is a translation. So C is the twist of E by the class of {ξ_{σ}} in H^{1}(K, E). In particular C is a torsor under E, the action μ : E × C → C being given by

μ(P, Q) = α(P + α^{−1}Q).

It follows that E is the Jacobian of C. □

Proof of Proposition 2.3: We are given (C, ω) with geometric invariants c_{4} and c_{6}. Let E be the elliptic curve over K with Weierstrass equation

y^{2} = x^{3} − 27c_{4}x − 54c_{6}.

The pairs (C, ω) and (E, 3dx/y) have the same geometric invariants, and are therefore isomorphic over \overline{K}. It follows by Lemma 2.5 that E is the Jacobian of C. □

Before we can use Proposition 2.3 to compute the Jacobian of a genus one curve, we first need to compute an invariant differential on the curve. It is easy to generalise the construction of §5.4 to genus one normal curves of arbitrary degree. An alternative is the following.

Let C ⊂ ℙ^{n−1} be a genus one normal curve of degree n with hyperplane section H. We identify the Riemann-Roch space L(H) with the space of linear forms on ℙ^{n−1}. If we fix ω then there is a linear map

∧^{2}L(H) → L(2H) ; f ∧ g ↦ \frac{fdg−gdf}{ω} .

By Lemma 5.5 with d = 2 the natural map S^{2}L(H) → L(2H) is surjective. Thus there is an alternating matrix of quadrics Ω = (Ω_{ij}) with

ω = \frac{x_{j}^{2}d(x_{i}/x_{j})}{Ω_{ij}}

for all i ≠ j. This matrix has the property that

(\frac{∂f}{∂x_{1}} … \frac{∂f}{∂x_{n}} ) Ω ≡ 0 (mod I(C))

for all f ∈ I(C). Starting from generators for I(C) we can use this property to solve for Ω by linear algebra. Then Ω is the data we use to specify ω. Notice that the entries of Ω are determined only up to the addition of quadrics in I(C).

In §9 we compare our invariant-theoretic approach with some other methods for computing the geometric invariants.

3. Genus one models

Let C be a smooth curve of genus one defined over K, and let D be a K-rational divisor on C of degree n. In each of the cases n = 1, 2, 3, 4, 5 we find equations for the pair (C, D), and use the form of these equations to motivate our definition of a genus one model.

3.1. Genus one models of degree 1. If n = 1 then we pick x, y ∈ K(C) such that L(2D) and L(3D) have bases 1, x and 1, x, y. The 7 elements 1, x, y, x^{2}, xy, x^{3}, y^{2} in the 6-dimensional space L(6D) satisfy a linear dependence relation. Moreover the coefficients of x^{3} and y^{2} are non-zero. Rescaling x and y if necessary we find that C has Weierstrass equation

(2)  y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}.

A genus one model of degree 1 is a tuple ϕ = (a_{1}, a_{2}, a_{3}, a_{4}, a_{6}). We write C_{ϕ} ⊂ ℙ^{2} for the curve with Weierstrass equation (2). Genus one models ϕ and ϕ′ of degree 1 are equivalent if they are related by substitutions

\begin{aligned}x & = & u^{2}x′ + r \\ y & = & u^{3}y′ + u^{2}sx′ + t \end{aligned}

and ϕ′ = u^{−6}ϕ, with u ≠ 0. We write G_{1} for the group of all such transformations [u; r, s, t].

3.2. Genus one models of degree 2. If n = 2 then we pick x, y ∈ K(C) such that L(D) and L(2D) have bases 1, x and 1, x, y, x^{2}. The 9 elements 1, x, x^{2}, y, x^{3}, xy, x^{4}, x^{2}y, y^{2} in the 8-dimensional space L(4D) satisfy a linear dependence relation. Moreover the coefficient of y^{2} is non-zero. We find that C has equation

y^{2} + (α_{0}x^{2} + α_{1}x + α_{2})y = ax^{4} + bx^{3} + cx^{2} + dx + e.

A genus one model of degree 2 is a pair of homogeneous polynomials ϕ = (p(x, z), q(x, z)) of degrees 2 and 4. We write C_{ϕ} ⊂ ℙ(1, 1, 2) for the curve defined by

y^{2} + p(x, z)y = q(x, z).

Here the ambient space ℙ(1, 1, 2) is a weighted project plane, with degrees 1, 1, 2 assigned to the co-ordinates x, z, y. Genus one models ϕ and ϕ′ of degree 2 are equivalent if they are related by substitutions

\begin{aligned}x & = & B_{11}x′ + B_{21}z′ \\ z & = & B_{12}x′ + B_{22}z′ \\ y & = & μ^{−1}y′ + r_{0}x^{′2} + r_{1}x′z′ + r_{2}z^{′2} \end{aligned}

and ϕ′ = μ^{2}ϕ, with μ det B ≠ 0. We write G_{2} for the group of all such transformations [μ, r, B].

If char (K) ≠ 2 then by completing the square it suffices to consider models of the form (0, q(x, z)). These are the binary quartics of Definition 1.1.

If n ≥ 3 then the complete linear system |D| determines an embedding C → ℙ^{n−1}. We identify C with its image, which is called a genus one normal curve of degree n. Some basic facts about these curves are recalled in §5.1.

3.3. Genus one models of degree 3. If n = 3 then C ⊂ ℙ^{2} is a plane cubic. A genus one model of degree 3 is a single homogeneous polynomial ϕ = (f(x_{1}, x_{2}, x_{3})) of degree 3. We write C_{ϕ} ⊂ ℙ^{2} for the variety defined by f = 0. Genus one models ϕ and ϕ′ of degree 3 are equivalent if they are related by substitutions ϕ′ = μϕ and x_{j} = ∑_{i=1}^{3} B_{ij}x′_{i} with μ det B ≠ 0. We write G_{3} = 𝔾_{m} × GL_{3} for the group of all such transformations.

3.4. Genus one models of degree 4. If n = 4 then C ⊂ ℙ^{3} is the complete intersection of two quadrics. A genus one model of degree 4 is a pair of homogeneous polynomials

ϕ = (\begin{matrix}q_{1}(x_{1}, x_{2}, x_{3}, x_{4}) \\ q_{2}(x_{1}, x_{2}, x_{3}, x_{4}) \end{matrix})

of degree 2. We write C_{ϕ} ⊂ ℙ^{3} for the variety defined by q_{1} = q_{2} = 0. Genus one models ϕ and ϕ′ of degree 4 are equivalent if they are related by substitutions ϕ′ = Aϕ and x_{j} = ∑_{i=1}^{4} B_{ij}x′_{i} with det A det B ≠ 0. We write G_{4} = GL_{2} × GL_{4} for the group of all such transformations.

3.5. Genus one models of degree 5. If n = 5 then C ⊂ ℙ^{4} is defined by the 4×4 Pfaffians of a 5×5 alternating matrix of linear forms. (See for example [12] and the references cited there.) A genus one model of degree 5 is a 5 × 5 alternating matrix of linear forms in 5 variables. We write C_{ϕ} ⊂ ℙ^{4} for the variety defined by its 4 × 4 Pfaffians. Genus one models ϕ and ϕ′ of degree 5 are equivalent if they are related by substitutions ϕ′ = AϕA^{T} and x_{j} = ∑_{i=1}^{4} B_{ij}x′_{i} with det A det B ≠ 0. We write G_{5} = GL_{5} × GL_{5} for the group of all such transformations.

4. The ring of invariants

Let X_{n} be the space of genus one models of degree n. For n = 1, 2, 3, 4, 5 this is an affine space of dimension N = 5, 8, 10, 20, 50. The co-ordinate ring K[X_{n}] is a polynomial ring in N variables. For n = 3, 4, 5 we give this ring its usual grading by degree. In the cases n = 1, 2 the rings are

\begin{aligned}K[X_{1}] & = & K[a_{1}, a_{2}, a_{3}, a_{4}, a_{6}] \\ K[X_{2}] & = & K[α_{0}, α_{1}, α_{2}, a, b, c, d, e]. \end{aligned}

We assign degrees deg(a_{i}) = i, deg(α_{i}) = 1, deg(a) = … = deg(e) = 2. In §3 we defined a linear algebraic group G_{n} acting on X_{n}. We now write G_{n} for the commutator subgroup, i.e.

G_{1} = {[1; r, s, t] ∈ G_{1}}
G_{2} = {[1, r, B] ∈ G_{2} : B ∈ SL_{2}}
G_{3} = SL_{3}
G_{4} = SL_{2} × SL_{4}
G_{5} = SL_{5} × SL_{5}.

Definition 4.1. The ring of invariants is

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

The definition is extended to an integral domain R by putting

R[X_{n}]^{G_{n}} = R[X_{n}] ∩ K[X_{n}]^{G_{n}}.

where K is the field of fractions of R.

We define a rational character det : G_{n} → 𝔾_{m}

n = 1 [u; r, s, t] ↦ u^{−1}
n = 2  [μ, r, B] ↦ μ det B
n = 3  [μ, B] ↦ μ det B
n = 4  [A, B] ↦ det A det B
n = 5  [A, B] ↦ (det A)^{2} det B.

Definition 4.2. The vector space of invariants of weight k is

K[X_{n}]_{k}^{G_{n}} = \{F ∈ K[X_{n}] : F ∘ g = (det g)^{k}F for all g ∈ G_{n}(\overline{K})\}.

Lemma 4.3. Every homogeneous invariant of degree d is an invariant of weight k where

d = \{ \begin{matrix}k & if n = 1, 2, 3 \\ 2k & if n = 4 \\ 5k & if n = 5. \end{matrix}

In particular the ring of invariants is graded by weight, i.e.

K[X_{n}]^{G_{n}} = ⊕_{k≥0}K[X_{n}]_{k}^{G_{n}}.

Proof: We treat the cases n = 4, 5. Since the only rational characters of G_{n} are of the form [A, B] ↦ (det A)^{p}(det B)^{q} we have

F ∘ [A, B] = (det A)^{p}(det B)^{q}F

for some integers p, q. Considering [A, B] in the centre of G_{n} we deduce

\begin{matrix}n = 4 & \{ \begin{matrix}d & = & 2p \\ 2d & = & 4q \end{matrix} \\ n = 5 & \{ \begin{matrix}2d & = & 5p \\ d & = & 5q \end{matrix} \end{matrix}

We are done by the definition of det : G_{n} → 𝔾_{m}. The cases n = 1, 2, 3 are similar. □

We are ready to state our main theorem.

Theorem 4.4. There are invariants c_{4}, c_{6}, Δ ∈ K[X_{n}]^{G_{n}} of weights 4, 6 and 12, related by c_{4}^{3} − c_{6}^{2} = 1728Δ, such that

(i) If char (K) ≠ 2, 3 then K[X_{n}]^{G_{n}} = K[c_{4}, c_{6}]. (ii) The variety C_{ϕ} defined by ϕ ∈ X_{n} is a smooth curve of genus one if and only if Δ(ϕ) ≠ 0. (iii) If char (K) ≠ 2, 3 and ϕ ∈ X_{n} with Δ(ϕ) ≠ 0 then C_{ϕ} has Jacobian

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

The proof depends on the following geometric statements.

Proposition 4.5. Assume K = \overline{K}. Let X_{n}^{sing} be the set of all models ϕ ∈ X_{n} which do not define a smooth curve of genus one. Then X_{n}^{sing} is an irreducible Zariski closed subset of X_{n}. In particular the generic genus one model of degree n defines a smooth curve of genus one.

Proof: The cases n = 1, 2, 3 are well known. A proof for n = 3, 4, 5 is given in §5.3. □

Let ℙ(X_{n}) be the projective space determined by X_{n}. (This is a weighted projective space in the cases n = 1, 2.) We recall that elements of X_{n} are equivalent if they lie in the same G_{n}-orbit.

Proposition 4.6. Assume K = \overline{K}. Let ϕ, ϕ′ ∈ X_{n} with C_{ϕ} and C_{ϕ′} either smooth curves of genus one or rational curves with a single node. Then C_{ϕ} and C_{ϕ′} are isomorphic as curves if and only if ϕ and ϕ′ are equivalent. Moreover the stabiliser of ϕ for the action of G_{n} on ℙ(X_{n}) is finite.

Proof: The cases n = 1, 2 are straightforward. A proof for n = 3, 4, 5 is given in §5.2. □

We identify K[X_{n}]^{G_{n}} as a subring of K[X_{1}]^{G_{1}}. To do this we start with an elliptic curve E in Weierstrass form

y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}.

The complete linear system |n.0| determines a morphism E → ℙ^{n−1}. The image is described by a genus one model of degree n. In §6 we specify such a model and hence define a morphism π_{n} : X_{1} → X_{n}. The models π_{n}(ϕ) for ϕ ∈ X_{1} are called Weierstrass models. Collectively they form the Weierstrass family.

Proposition 4.7. There are morphisms π_{n} : X_{1} → X_{n} and γ_{n} : G_{1} → G_{n} with the following properties.

(i) If ϕ′ = π_{n}(ϕ) then C_{ϕ} and C_{ϕ′} are isomorphic as curves. (ii) γ_{n} is a group homomorphism. (iii) (γ_{n}g)(π_{n}ϕ) = π_{n}(gϕ) for all g ∈ G_{1} and ϕ ∈ X_{1}. (iv) det(γ_{n}g) = det(g) for all g ∈ G_{1}.

Proof: The proposition is checked by direct computation using the formulae in §6. □

The map π_{n} : X_{1} → X_{n} induces a homomorphism of polynomial rings π_{n}^{∗} : K[X_{n}] → K[X_{1}] ; F ↦ F ∘π_{n}. By Proposition 4.7 it restricts to a homomorphism of graded rings

π_{n}^{∗} : K[X_{n}]^{G_{n}} → K[X_{1}]^{G_{1}}

where the grading is by weight.

Lemma 4.8. The map π_{n}^{∗} : K[X_{n}]^{G_{n}} → K[X_{1}]^{G_{1}} is an injective homomorphism of graded rings.

Proof: We must show that π_{n}^{∗} is injective. For this we are free to assume that K is algebraically closed. If F ∈ K[X_{n}]^{G_{n}} is a homogeneous invariant vanishing on the Weierstrass family then by Propositions 4.6 and 4.7 it also vanishes at every ϕ ∈ X_{n} for which C_{ϕ} is a smooth curve of genus one. Proposition 4.5 tells us that the latter are Zariski dense in X_{n}. It follows that F is identically zero and hence π_{n}^{∗} is injective. □

Computing the ring K[X_{1}]^{G_{1}} is entirely routine. We recall that

K[X_{1}] = K[a_{1}, a_{2}, a_{3}, a_{4}, a_{6}].

Following Tate’s formulaire [24, Chapter III] we put

(3)  \begin{aligned}b_{2} & = & a_{1}^{2} + 4a_{2} \\ b_{4} & = & 2a_{4} + a_{1}a_{3} \\ b_{6} & = & a_{3}^{2} + 4a_{6} \\ b_{8} & = & a_{1}^{2}a_{6} + 4a_{2}a_{6} − a_{1}a_{3}a_{4} + a_{2}a_{3}^{2} − a_{4}^{2} \end{aligned}

and

(4)  \begin{aligned}c_{4} & = & b_{2}^{2} − 24b_{4} \\ c_{6} & = & −b_{2}^{3} + 36b_{2}b_{4} − 216b_{6} \\ Δ & = & −b_{2}^{2}b_{8} − 8b_{4}^{3} − 27b_{6}^{2} + 9b_{2}b_{4}b_{6}. \end{aligned}

It is well known that c_{4}, c_{6}, Δ ∈ ℤ[X_{1}]^{G_{1}} and c_{4}^{3} − c_{6}^{2} = 1728Δ.

Lemma 4.9. If char (K) ≠ 2, 3 then K[X_{1}]^{G_{1}} = K[c_{4}, c_{6}].

Proof: This is Theorem 4.4(i) in the case n = 1. It is an immediate consequence of the standard procedure for putting a Weierstrass equation in the shorter form y^{2} = x^{3} + Ax + B. The required isomorphism is ι^{∗} where

ι : 𝔸^{2} → X_{1} ; (c_{4}, c_{6}) ↦ (0, 0, 0, −c_{4}/48, −c_{6}/864).
□

We have reduced the proof of Theorem 4.4(i) to showing that π_{n}^{∗} is surjective. Equivalently, we must show that K[X_{n}]^{G_{n}} contains invariants of weights 4 and 6. One method would be to split into the cases n = 2, 3, 4, 5 and use the explicit constructions presented in §§7,8. This makes the theorem appear an accident, especially in the case n = 5. Instead we give a construction based on Proposition 4.6.

Lemma 4.10. Assume K = \overline{K}. Let ϕ, ϕ′ ∈ X_{n} with C_{ϕ} and C_{ϕ′} either smooth curves of genus one or rational curves with a single node. Then the Zariski closure of the G_{n}-orbit of ϕ is the zero locus of an irreducible homogeneous invariant F ∈ K[X_{n}]^{G_{n}}. Moreover F (ϕ′) = 0 if and only if ϕ and ϕ′ are equivalent.

Proof: By Proposition 4.6 the morphism G_{n} → ℙ(X_{n}); g ↦ g(ϕ) has zero-dimensional fibres. But for each n we find

dim(G_{n}) = dim(X_{n}) − 2.

So the G_{n}-orbit of ϕ in ℙ(X_{n}) has codimension 1. Moreover since G_{n} is irreducible, every G_{n}-orbit is irreducible. Therefore the Zariski closure of the orbit of ϕ is the zero locus of an irreducible homogeneous polynomial F ∈ K[X_{n}]. Since the equivalence class of ϕ determines F uniquely up to scalars, and G_{n} is the commutator subgroup of G_{n}, it follows that F is an invariant.

If ϕ and ϕ′ are equivalent then clearly F (ϕ′) = 0. For the converse we suppose F (ϕ′) = 0. Then the G_{n}-orbits of ϕ and ϕ′ in ℙ(X_{n}) have the same Zariski closure, Z say. A standard argument (see e.g. [23, Chapter I, §5.3, Theorem 6]) shows that each of these orbits contains a non-empty open subset of Z. Since Z is irreducible these open sets must intersect. It follows that ϕ and ϕ′ are equivalent. □

We restrict these invariants to the Weierstrass family.

Lemma 4.11. Assume K = \overline{K} and char (K) ≠ 2, 3. Then there are irreducible invariants F_{4}, F_{6} ∈ K[X_{n}]^{G_{n}} and integers p, q ≥ 1 such that F_{4} ∘ π_{n} = c_{4}^{p} and F_{6} ∘ π_{n} = c_{6}^{q}.

Proof: By Proposition 4.7 we can pick ϕ ∈ X_{n} with C_{ϕ} a smooth curve of genus one with j-invariant 0. Let F_{4} ∈ K[X_{n}]^{G_{n}} be the invariant constructed from ϕ in Lemma 4.10. Then F_{4} ∘ π_{n} is a homogeneous element of K[X_{1}]^{G_{1}} = K[c_{4}, c_{6}]. Rescaling F_{4} we can write

F_{4} ∘ π_{n} = c_{4}^{p} c_{6}^{q} Δ^{r} ∏_{ν=1}^{s}(c_{4}^{3} − j_{ν}Δ)

for some integers p, q, r, s ≥ 0 and constants j_{1}, … j_{s} ≠ 0, 1728.

Now let ϕ′ = π_{n}(ϕ_{1}) be a Weierstrass model with C_{ϕ′} a smooth curve of genus one. If this curve has j-invariant not equal to 0 then by Lemma 4.10 we have F_{4}(ϕ′) ≠ 0. By varying the choice of ϕ_{1} we deduce that q = s = 0. We then repeat the argument for C_{ϕ′} a Weierstrass model with a node. This shows that r = 0. The statement for c_{6} is proved similarly, starting with j-invariant 1728. □

The proof of Theorem 4.4(i) now reduces to showing that p = q = 1 in Lemma 4.11. For this we quote a geometric result whose proof uses properties of the invariant differential.

Definition 4.12. Genus one models ϕ, ϕ′ ∈ X_{n} are properly equivalent if there exists g ∈ G_{n} with gϕ = ϕ′ and det(g) = 1.

Proposition 4.13. Assume K = \overline{K} and char (K) ≠ 2, 3. Let ϕ ∈ X_{n} with C_{ϕ} a smooth curve of genus one. Then ϕ is properly equivalent to π_{n}(0, 0, 0, A, B) for some unique A, B ∈ K.

Proof: The existence is already clear from Propositions 4.6 and 4.7. We prove uniqueness in §5.4. □

Lemma 4.14. Assume K = \overline{K} and char (K) = 0. Then the map π_{n}^{∗} : K[X_{n}]^{G_{n}} → K[X_{1}]^{G_{1}} is surjective.

Proof: Let ϕ ∈ X_{n}(𝕂) be the generic model defined over the function field 𝕂 = K(X_{n}). We have assumed char (K) = 0 so that 𝕂 is perfect. Proposition 4.5 tells us that C_{ϕ} is a smooth curve of genus one. So by Proposition 4.13, ϕ is properly equivalent to π_{n}(0, 0, 0, A, B) for some unique A, B ∈ \overline{𝕂}. The uniquess statement shows that A and B are fixed by Gal(\overline{𝕂}/𝕂) and hence A, B ∈ 𝕂.

Let F_{4}, F_{6} ∈ K[X_{n}]^{G_{n}} be the irreducible invariants constructed in Lemma 4.11. Then F_{4} = f_{4}^{p} and F_{6} = f_{6}^{q} where

\begin{aligned}f_{4} & = & c_{4}(0, 0, 0, A, B) & = & −48A \\ f_{6} & = & c_{6}(0, 0, 0, A, B) & = & −864B. \end{aligned}

Since K[X_{n}] is integrally closed (in its field of fractions 𝕂) and F_{4}, F_{6} ∈ K[X_{n}] are irreducible it follows that p = q = 1. □

Applying Lemma 4.14 with K = \overline{ℚ} we learn that the invariants c_{4}, c_{6}, Δ ∈ ℤ[X_{1}]^{G_{1}} extend to invariants in \overline{ℚ}[X_{n}]^{G_{n}}. These invariants are again denoted c_{4}, c_{6}, Δ. Since π_{n}^{∗} is injective it follows by Galois theory that c_{4}, c_{6}, Δ ∈ ℚ[X_{n}]. In fact the coefficients are integers.

Lemma 4.15. c_{4}, c_{6}, Δ ∈ ℤ[X_{n}].

Proof: Let F = c_{4}, c_{6} or Δ. Let p be a prime and r ≥ 0 an integer. We suppose for a contradiction that p^{r+1}F ∈ ℤ_{p}[X_{n}] yet p^{r}F ∉ ℤ_{p}[X_{n}]. Then each coefficient of π_{n}^{∗}(p^{r+1}F) ∈ ℤ[X_{1}] is divisible by p. So if G ∈ 𝔽_{p}[X_{n}] is the reduction of p^{r+1}F mod p then π_{n}^{∗}G = 0. The injectivity established in Lemma 4.8 shows that G = 0. Therefore p^{r}F ∈ ℤ_{p}[X_{n}]. This is the required contradiction. □

Remark 4.16. Since the original c_{4}, c_{6}, Δ ∈ ℤ[X_{1}] are primitive it is clear that the new c_{4}, c_{6}, Δ ∈ ℤ[X_{n}] are also primitive. This means it is possible to specify our scalings of c_{4}, c_{6}, Δ, at least up to sign, without the need to compute their restrictions to the Weierstrass family.

We revert to working over an arbitrary perfect field K.

Proof of Theorem 4.4: Let c_{4}, c_{6}, Δ ∈ K[X_{n}] be the images of c_{4}, c_{6}, Δ ∈ ℤ[X_{n}]. These polynomials are invariants of weights 4, 6 and 12, satisfying c_{4}^{3} − c_{6}^{2} = 1728Δ. They are non-zero by Remark 4.16.

(i) If char (K) ≠ 2, 3 then by Lemmas 4.8 and 4.9 the map

π_{n}^{∗} : K[X_{n}]^{G_{n}} → K[X_{1}]^{G_{1}} = K[c_{4}, c_{6}]

is an isomorphism of graded rings. (ii) We may assume that K is algebraically closed. If ϕ ∈ X_{n} with C_{ϕ} a smooth curve of genus one then by Propositions 4.6 and 4.7 it is equivalent to a Weierstrass model. We deduce Δ(ϕ) ≠ 0. So there is an inclusion

{Δ = 0} ⊂ X_{n}^{sing}.

But Proposition 4.5 asserts that X_{n}^{sing} is closed and irreducible. So the inclusion is in fact an equality. (iii) Let ϕ ∈ X_{n} with C_{ϕ} a smooth curve of genus one. In §5.4 we use ϕ ∈ X_{n} to define an invariant differential ω_{ϕ} on C_{ϕ}. In the case n = 5 we must assume char (K) ≠ 2. We then show that c_{4}(ϕ) and c_{6}(ϕ) are the geometric invariants of the pair (C_{ϕ}, ω_{ϕ}). The formula for the Jacobian follows by Proposition 2.3. □

Theorem 4.4(iii) is proved in [1] for n = 2, 3, 4 by giving formulae for the covering map (of degree n^{2}) from a genus one curve to its Jacobian. We have extended to the case n = 5 by taking a different approach, based on properties of the invariant differential.

It turns out that the map π_{n}^{∗} : K[X_{n}]^{G_{n}} → K[X_{1}]^{G_{1}} is an isomorphism in all characteristics. The proof in characteristics 2 and 3 is given in §10.

The remaining sections of the paper may be read in any order.

5. Geometry

The aim of this section is to prove the geometric results cited in §4. We work over an algebraically closed field K. The homogeneous ideal of a projective variety X is denoted I(X).

5.1. Genus one normal curves. We recall some basic facts about genus one normal curves and rational nodal curves.

Definition 5.1. Let n ≥ 3 be an integer.

(i) A genus one normal curve C ⊂ ℙ^{n−1} is a smooth curve of genus one and degree n that spans ℙ^{n−1}. (ii) A rational nodal curve C ⊂ ℙ^{n−1} is a rational curve of degree n that spans ℙ^{n−1} and has a single node.

Remark 5.2. Equivalently, a genus one normal curve is a smooth curve of genus one embedded by a complete linear system of degree n. A rational nodal curve is the image of a morphism ℙ^{1} → ℙ^{n−1} determined by a linear system of the form

{f ∈ L(D)|f(P_{1}) = f(P_{2})}

for D a divisor on ℙ^{1} of degree n, and P_{1}, P_{2} ∈ ℙ^{1} distinct.

Proposition 5.3. Let C ⊂ ℙ^{n−1} be either a genus one normal curve or a rational nodal curve. If n ≥ 4 then the ideal I(C) is generated by a vector space of quadrics of dimension n(n − 3)/2.

This proposition is well known, at least for genus one normal curves. Our proof, based on an argument in [17], has the advantage of working for rational nodal curves at the same time. We write R = K[x_{1}, …, x_{n}] and R′ = K[x_{1}, …, x_{n−1}] for the homogeneous co-ordinate rings of ℙ^{n−1} and ℙ^{n−2}. We give each ring its usual grading by degree, say R = ⊕_{d≥0}R_{d} and R′ = ⊕_{d≥0}R′_{d}.

Lemma 5.4. Let X ⊂ ℙ^{n−2} be a set of n points in general position.

(i) The evaluation map π_{X} : R′_{d} → K^{n} is surjective for all d ≥ 2. (ii) If n ≥ 4 then the ideal I(X) ⊂ R′ is generated by quadrics.

Proof: We change co-ordinates so that X is the set of points (1 : 0 : …: 0), (0 : 1 : …: 0), …, (0 : 0 : …: 1) and (1 : 1 : …: 1). The proof is now straightforward. □

We show that the curves defined in Definition 5.1 are projectively normal.

Lemma 5.5. Let C ⊂ ℙ^{n−1} be either a genus one normal curve or a rational nodal curve. Let H be the divisor of a hyperplane section, say cut out by a linear form h ∈ R_{1}. Then the map

π_{C} : R_{d} → L(dH) ; f ↦ f/h^{d}

is surjective for all d ≥ 1.

Proof: The proof is by induction on d, the case d = 1 being clear from Riemann-Roch. For the induction step we choose a hyperplane {ξ = 0} meeting C in n distinct points disjoint from H. Again by Riemann-Roch any n − 1 distinct points on C span a hyperplane. So X = C ∩ {ξ = 0} satisfies the hypothesis of Lemma 5.4.

Let d ≥ 2. We are given f ∈ L(dH) and wish to show that it belongs to the image of π_{C}. By Lemma 5.4(i) it suffices to treat the case where f vanishes on X. But then f = (ξ/h)f′ for some f′ ∈ L((d − 1)H). Applying the induction hypothesis to f′, we deduce that f is in the image of π_{C} as required. □

Proof of Proposition 5.3: We continue with the notation of the last proof. Since C is contained in no hyperplane, the natural map

(5)  I(C) ∩ R_{2} → I(X) ∩ R′_{2}

is injective. By Lemmas 5.4 and 5.5 these spaces each have dimension n(n−3)/2. So (5) is an isomorphism. Now let f ∈ I(C)∩R_{d}. We must show that f is in the ideal generated by I(C) ∩ R_{2}. By Lemma 5.4(ii) and the surjectivity of (5) it suffices to treat the case where f vanishes on X. But then f = ξf′ for some f′ ∈ I(C) ∩ R_{d−1}. The proposition now follows by induction on d. □

We say that curves C, C′ ⊂ ℙ^{n−1} are projectively equivalent if there exists α ∈ PGL_{n} with α(C) = C′.

Lemma 5.6. (i) Genus one normal curves C, C′ ⊂ ℙ^{n−1} are projectively equivalent if and only if they have the same j-invariant. (ii) Any two rational nodal curves C, C′ ⊂ ℙ^{n−1} are projectively equivalent.

Proof: (i) Let C and C′ have hyperplane sections H and H′. If C and C′ are isomorphic as curves then composing with a translation map we can find an isomorphism α : C ≅ C′ with α^{∗}H′ ∼ H. (ii) This is clear from Remark 5.2. □

Lemma 5.7. Let C ⊂ ℙ^{n−1} be either a genus one normal curve or a rational nodal curve. Then there are only finitely many α ∈ PGL_{n} with α(C) = C.

Proof: We first treat the case C is a genus one normal curve, say with hyperplane section H. We are interested in the automorphisms α of C with α^{∗}H ∼ H. The automorphism group of C sits in an exact sequence

0 → E → Aut(C) → Aut(E, 0) → 0

where E is the Jacobian of C. The first map is P ↦ τ_{P} where τ_{P} is translation by P. Since H is a divisor of degree n we have τ_{P}^{∗}H ∼ H if and only if nP = 0. The lemma follows from the fact that E[n] and Aut(E, 0) are both finite.

If C is a rational nodal curve then without loss of generality it is the image of

ℙ^{1} ↦ ℙ^{n−1} ; (s : t) ↦ (s^{n} + t^{n} : st^{n−1} : …: s^{n−1}t).

The group of automorphisms of ℙ^{1} that extend to automorphisms of ℙ^{n−1} form a copy of the dihedral group generated by (s : t) ↦ (t : s) and (s : t) ↦ (ζs : t) for ζ an nth root of unity. □

5.2. Minimal free resolutions. We recall that a genus one model of degree n = 3, 4, 5 is a collection of homogenoeus polynomials in R = K[x_{1}, …, x_{n}]. Splitting into the cases n = 3, 4, 5 we now use ϕ ∈ X_{n} to define an ideal I_{ϕ} ⊂ R and a complex of graded free R-modules F_{•}(ϕ). We write R(d) for the graded R-module with R(d)_{e} = R_{d+e}.

If n = 3 then ϕ consists of a single polynomial f ∈ R. This polynomial generates an ideal I_{ϕ} ⊂ R and defines a complex

F_{•}(ϕ) : 0 ⟶ R(−3) ⟶^{f} R ⟶ 0.

If n = 4 then ϕ consists of polynomials q_{1}, q_{2}. These polynomials generate an ideal I_{ϕ} ⊂ R and define a complex

F_{•}(ϕ) : 0 ⟶ R(−4) \underrightarrow{(\begin{matrix}−q_{2} \\ q_{1} \end{matrix})} R(−2)^{2} \underrightarrow{(q_{1} q_{2})} R ⟶ 0.

If n = 5 then ϕ is a 5 × 5 alternating matrix of linear forms. The Pfaffian of a 4 × 4 alternating matrix is

pf (\begin{matrix}0 & a_{1} & a_{2} & a_{3} \\ & 0 & b_{3} & b_{2} \\ & & 0 & b_{1} \\ & & & 0 \end{matrix}) = a_{1}b_{1} − a_{2}b_{2} + a_{3}b_{3}.

We write ϕ^{{i}} for the submatrix of ϕ obtained by deleting the ith row and ith column. Then the vector of submaximal Pfaffians of ϕ is P = (p_{1}, …, p_{5}) where

p_{i} = (−1)^{i+1} pf(ϕ^{{i}}).

These polynomials generate an ideal I_{ϕ} ⊂ R and define a complex

F_{•}(ϕ) : 0 ⟶ R(−5) ⟶^{P^{T}} R(−3)^{5} ⟶^{ϕ} R(−2)^{5} ⟶^{P} R ⟶ 0.

In each case n = 3, 4, 5, the variety C_{ϕ} ⊂ ℙ^{n−1} is that defined by the ideal I_{ϕ} ⊂ R. We say that F_{•}(ϕ) is a minimal free resolution of R/I_{ϕ} if it is exact at every term except the final copy of R where the homology is R/I_{ϕ}.

Lemma 5.8. Let n = 3, 4, 5 and let ϕ ∈ X_{n}.

(i) Every component of C_{ϕ} has dimension at least 1. (ii) If every component of C_{ϕ} has dimension 1 then F_{•}(ϕ) is a minimal free resolution of R/I_{ϕ}.

Proof: (i) This is clear for n = 3, 4. For n = 5 we recall that the 4 × 4 Pfaffians of a generic 5 × 5 alternating matrix define the image of the Plucker embedding Gr(2, 5) → ℙ^{9}. Then C_{ϕ} is the intersection of this Grassmannian with a linear subspace ℙ^{4}. Since Gr(2, 5) has dimension 6 we are done by [15, I, Theorem 7.2]. (ii) If n = 3, 4 then our claim is that f is non-zero, respectively that q_{1}, q_{2} are coprime. This is clear. The case n = 5 is an application of the Buchsbaum-Eisenbud acyclicity criterion, for which we refer to [4, Theorem 1.4.13] or [10, Theorem 20.9]. □

We recall that if A is a finitely generated graded K-algebra, say A = ⊕_{d≥0}A_{d}, then there is a polynomial h_{A}(t), called the Hilbert polynomial, with the property that h_{A}(d) = dim(A_{d}) for all d ≫ 0.

Lemma 5.9. (i) Let n = 3, 4, 5 and let ϕ ∈ X_{n}. If the complex F_{•}(ϕ) is a minimal free resolution of R/I_{ϕ} then

h_{R/I_{ϕ}}(t) = nt.

(ii) If C ⊂ ℙ^{n−1} is a curve of arithmetic genus g and degree d then

h_{R/I(C)}(t) = dt + (1 − g).

Proof: (i) We compute the Hilbert polynomial from the minimal free resolution in the usual way. For example in the case n = 5,

h(t) = \binom{t + 4}{4} − 5\binom{t + 2}{4} + 5\binom{t + 1}{4} − \binom{t − 1}{4} = 5t.

(ii) This is a definition. See for example [15, I, §7]. □

Proposition 5.10. Let n = 3, 4, 5 and let ϕ ∈ X_{n}.

(i) If C_{ϕ} ⊂ ℙ^{n−1} is a smooth curve of genus one then it is a genus one normal curve of degree n. (ii) If C_{ϕ} ⊂ ℙ^{n−1} is a rational curve with a single node then it is a rational nodal curve of degree n.

Proof: By Lemma 5.8 the complex F_{•}(ϕ) is a minimal free resolution of R/I_{ϕ}. Since I_{ϕ} ⊂ I(C_{ϕ}), a comparison of Hilbert polynomials as described in Lemma 5.9 shows that C_{ϕ} has degree d ≤ n. If C_{ϕ} ⊂ ℙ^{n−1} spans a linear subspace of dimension m−1 it follows by Riemann-Roch that 3 ≤ m ≤ d ≤ n. We must show that m = n. In the case n = 3 this is already clear. If n = 4, 5 then C_{ϕ} is defined by quadrics. This enables us to rule out the unwanted possibilities for (m, d), with the exception of (m, d) = (4, 4) in the case n = 5. This possiblity is excluded by the following lemma. □

Lemma 5.11. Let C ⊂ ℙ^{3} be either a genus one normal curve or a rational nodal curve. Then C cannot be defined by the 4 × 4 Pfaffians of a 5 × 5 alternating matrix of linear forms on ℙ^{3}.

Proof: Let ϕ be such a matrix, with vector of submaximal Pfaffians P = (p_{1}, …, p_{5}). Let C be defined by quadrics q_{1}, q_{2}. By Proposition 5.3 we have ⟨p_{1}, …, p_{5}⟩ = ⟨q_{1}, q_{2}⟩. Replacing ϕ by A^{T} ϕA for suitable A ∈ GL_{5} we may suppose that P = (q_{1}, q_{2}, 0, 0, 0). Since P ϕ = 0, and q_{1}, q_{2} are coprime, it follows that the first two rows of ϕ are zero. But then every 4 × 4 Pfaffian of ϕ vanishes, which is a contradiction.

□

Lemma 5.12. Let n = 3, 4, 5 and let ϕ ∈ X_{n}. If C_{ϕ} ⊂ ℙ^{n−1} is either a genus one normal curve or a rational nodal curve then I_{ϕ} is a radical ideal, equivalently I(C_{ϕ}) = I_{ϕ}.

Proof: By Lemma 5.8 the complex F_{•}(ϕ) is a minimal free resolution of R/I_{ϕ}. If n = 3 then I_{ϕ} = (f) where f is an irreducible cubic. If n = 4, 5 then I_{ϕ} is generated by a vector space of quadrics of dimension d = 2, 5. Since I_{ϕ} ⊂ I(C_{ϕ}) we are done by Proposition 5.3. □

Lemma 5.13. Let n = 3, 4, 5 and let ϕ, ϕ′ ∈ X_{n}. Suppose that

(i) there exists α ∈ PGL_{n} with α(C_{ϕ}) = C_{ϕ′}, (ii) F_{•}(ϕ) and F_{•}(ϕ′) are minimal free resolutions of R/I_{ϕ} and R/I_{ϕ′}, (iii) the ideals I_{ϕ} and I_{ϕ′} are radical ideals. Then ϕ and ϕ′ are equivalent.

Proof: By (i) we may assume C_{ϕ} = C_{ϕ′}. Then (iii) gives I_{ϕ} = I_{ϕ′}. The cases n = 3, 4 are now clear. If n = 5 then there is an isomorphism of complexes

[figure: 0; R(−5); P^{T}; c; R(−3)^{5}; ϕ; B; R(−2)^{5}; P; A; R; 0; 0; R(−5); P^{′T}; R(−3)^{5}; ϕ′; R(−2)^{5}; P′; R; 0]

The matrices A, B ∈ GL_{5} are uniquely determined. Comparing this diagram with its dual gives B = cA^{−T}. So ϕ′ = [A, c^{−1}I_{5}]ϕ. □

Lemma 5.14. Let n = 3, 4, 5 and let ϕ ∈ X_{n}. Suppose that

(i) there are only finitely many α ∈ PGL_{n} with α(C_{ϕ}) = C_{ϕ}, (ii) F_{•}(ϕ) is a minimal free resolution of R/I_{ϕ}, (iii) the ideal I_{ϕ} is a radical ideal. Then the stabiliser of ϕ for the action of G_{n} on ℙ(X_{n}) is finite.

Proof: This is clear for n = 3, 4. In the case n = 5 it suffices to show that if [A, I_{5}]ϕ = λϕ for some A ∈ GL_{5} and λ ∈ K^{∗}, then A is a scalar matrix. Taking submaximal Pfaffians we obtain P adj A = λ^{2}P. By (ii) the components of P are linearly independent. It follows that adj A and hence A is a scalar matrix. □

Proof of Proposition 4.6: (For n = 3, 4, 5.) We are given ϕ, ϕ′ ∈ X_{n} with C_{ϕ} and C_{ϕ′} either smooth curves of genus one or rational curves with a single node. By Proposition 5.10 these are either genus one normal curves or rational nodal curves. The hypotheses of Lemmas 5.13 and 5.14 are satisfied by Lemmas 5.6, 5.7, 5.8 and 5.12. □

5.3. The generic model. We show that the generic genus one model of degree n = 3, 4, 5 defines a smooth curve of genus one.

Definition 5.15. Let n = 3, 4, 5. The Jacobian matrix J_{ϕ} of a genus one model ϕ ∈ X_{n} is

\begin{matrix}n = 3 & ϕ = (f) & J_{ϕ} = (\frac{∂f}{∂x_{j}} ) \\ n = 4 & ϕ = (\begin{matrix}q_{1} \\ q_{2} \end{matrix}) & J_{ϕ} = (\frac{∂q_{i}}{∂x_{j}} ) \\ n = 5 & p_{i} = (−1)^{i+1} pf(ϕ^{\{i\}}) & J_{ϕ} = (\frac{∂p_{i}}{∂x_{j}} ). \end{matrix}

Lemma 5.16. Let n = 3, 4, 5 and let ϕ ∈ X_{n}.

(i) If P ∈ C_{ϕ} then rank J_{ϕ}(P) ≤ n − 2. (ii) If rank J_{ϕ}(P) = n − 2 for every P ∈ C_{ϕ} then C_{ϕ} is a smooth curve of genus one.

Proof: (i) We saw in Lemma 5.8(i) that every component of C_{ϕ} has dimension at least 1. Therefore dim T_{P} (C_{ϕ}) ≥ 1. Since I_{ϕ} ⊂ I(C_{ϕ}) it follows that rank J_{ϕ}(P) ≤ n − 2. (ii) The argument used in (i) shows that every component of C_{ϕ} has dimension 1. So by Lemma 5.8(ii) the complex F_{•}(ϕ) is a minimal free resolution of R/I_{ϕ}. In particular R/I_{ϕ} is Cohen-Macaulay. It follows by Serre’s criterion (see [10, §18.3]) that I_{ϕ} is a prime ideal. Hence C_{ϕ} is an irreducible smooth curve and I(C_{ϕ}) = I_{ϕ}. It only remains to check that C_{ϕ} has genus 1. We do this by computing the Hilbert polynomial as described in Lemma 5.9. □

We define some “bad” subsets B_{n} ⊂ X_{n}.

Definition 5.17. (i) Let B_{3} ⊂ X_{3} consist of all models of the form

ϕ = (x_{1}f_{1}(x_{2}, x_{3}) + f_{2}(x_{2}, x_{3})) .

(ii) Let B_{4} ⊂ X_{4} consist of all models of the form

ϕ = (\begin{matrix}x_{1}x_{2} + g_{1}(x_{2}, x_{3}, x_{4}) \\ g_{2}(x_{2}, x_{3}, x_{4}) \end{matrix}) .

(iii) Let B_{5} ⊂ X_{5} consist of all models ϕ with ϕ_{ij}(1, 0, 0, 0, 0) = 0 for all {i, j} ≠ {1, 2}, and ϕ_{45}(x_{1}, …, x_{5}) ≡ 0.

Lemma 5.18. Let n = 3, 4, 5 and let ϕ ∈ X_{n}. The following are equivalent.

(i) C_{ϕ} is not a smooth curve of genus one. (ii) rank J_{ϕ}(P) < n − 2 for some P ∈ C_{ϕ}. (iii) ϕ is equivalent to a model in B_{n}.

Proof: (i) ⇒ (ii). This is a restatement of Lemma 5.16. (ii) ⇒ (i). This follows from Proposition 5.10(i), Lemma 5.12 and the Jacobian criterion for smoothness. (iii) ⇒ (ii). Without loss of generality ϕ ∈ B_{n}. Then the point P = (1 : 0 : …: 0) belongs to C_{ϕ} and rank J_{ϕ}(P) < n − 2. (ii) ⇒ (iii). This is clear for n = 3, 4. We take n = 5. Since P ∈ C_{ϕ} the 4 × 4 Pfaffians of ϕ(P) vanish. So rank ϕ(P) = 0 or 2.

If rank ϕ(P) = 2 then we may assume P = (1 : 0 : …: 0) and

ϕ = (\begin{matrix}0 & x_{1} & ϕ_{13} & ϕ_{14} & ϕ_{15} \\ & 0 & ϕ_{23} & ϕ_{24} & ϕ_{25} \\ & & 0 & ℓ_{3} & −ℓ_{2} \\ & − & & 0 & ℓ_{1} \\ & & & & 0 \end{matrix})

for some ϕ_{ij}, ℓ_{k} ∈ ⟨x_{2}, x_{3}, x_{4}, x_{5}⟩. Since rank J_{ϕ}(P) < 3 the linear forms ℓ_{1}, ℓ_{2}, ℓ_{3} are linearly dependent. Replacing ϕ by AϕA^{T} for suitable A ∈ GL_{5} we may suppose that ℓ_{1} = 0. Then ϕ ∈ B_{5} as required.

If rank ϕ(P) = 0 then we may assume

ϕ = (\begin{matrix}0 & ϕ_{12} & ϕ_{13} & ϕ_{14} & ℓ_{1} \\ & 0 & ϕ_{23} & ϕ_{24} & ℓ_{2} \\ & & 0 & ϕ_{34} & ℓ_{3} \\ & − & & 0 & ℓ_{4} \\ & & & & 0 \end{matrix})

for some ϕ_{ij} ∈ ⟨x_{2}, x_{3}, x_{4}, x_{5}⟩ and ℓ_{j} ∈ ⟨x_{3}, x_{4}, x_{5}⟩. It is clear that ℓ_{1}, …, ℓ_{4} are linearly dependent. Replacing ϕ by AϕA^{T} for suitable A ∈ GL_{5} we may suppose that ℓ_{4} = 0. Then ϕ ∈ B_{5} as required. □

Proof of Proposition 4.5: (For n = 3, 4, 5.) Let X_{n}^{sing} be the set of all models ϕ ∈ X_{n} which do not define a smooth curve of genus one. We consider the projective variety

Z_{n} = {(ϕ, P) ∈ ℙ(X_{n}) × ℙ^{n−1}|P ∈ C_{ϕ} and rank J_{ϕ}(P) < n − 2}.

Let pr_{1} : Z_{n} → ℙ(X_{n}) be the first projection. Lemma 5.18 identifies pr_{1}(Z_{n}) = ℙ(X_{n}^{sing}). Since the image of a projective variety is again projective it follows that X_{n}^{sing} ⊂ X_{n} is a Zariski closed subset.

Lemma 5.18 also identifies X_{n}^{sing} as the image of a morphism

G_{n} × B_{n} → X_{n}.

Since G_{n} and B_{n} are irreducible it follows that X_{n}^{sing} is irreducible. □

5.4. The invariant differential. We continue to work over an algebraically closed field K. In the case n = 5 we further suppose that char (K) ≠ 2.

Let ϕ ∈ X_{n} with C_{ϕ} a smooth of curve genus one. We use ϕ to define an invariant differential ω_{ϕ} on C_{ϕ}. In the cases n = 1, 2 we put

\begin{matrix}n = 1 & ϕ = (a_{1}, a_{2}, a_{3}, a_{4}, a_{6}) & ω_{ϕ} = \frac{dx}{2y+a_{1}x+a_{3}} \\ n = 2 & ϕ = (p(x, z), q(x, z)) & ω_{ϕ} = \frac{z^{2}d(x/z)}{2y+p(x,z)}. \end{matrix}

In the cases n = 3, 4, 5 we start with the complex

F_{•}(ϕ) : 0 ⟶ R ⟶^{ϕ_{n−2}} F_{n−3} ⟶ … ⟶ F_{1} ⟶^{ϕ_{1}} R ⟶ 0

defined in §5.2. We identify the maps ϕ_{i} with the matrices of homogeneous polynomials that represent them. Then we define

ω_{ϕ} = \frac{x_{1}^{2}d(x_{2}/x_{1})}{\frac{∂ϕ_{1}}{∂x_{3}} ∘ … ∘ \frac{∂ϕ_{n−2}}{∂x_{n}}}

where the partial derivative of a matrix is the matrix of partial derivatives. In the cases n = 3, 4 this formula works out as

ω_{ϕ} = \frac{x_{1}^{2}d(x_{2}/x_{1})}{\frac{∂f}{∂x_{3}}} and ω_{ϕ} = \frac{x_{1}^{2}d(x_{2}/x_{1})}{\frac{∂q_{1}}{∂x_{4}} \frac{∂q_{2}}{∂x_{3}} − \frac{∂q_{1}}{∂x_{3}} \frac{∂q_{2}}{∂x_{4}}} .

Proposition 5.19. Let ϕ ∈ X_{n} with C_{ϕ} a smooth curve of genus one. If ϕ′ = gϕ for some g ∈ G_{n} then the isomorphism γ : C_{ϕ′} ≅ C_{ϕ} determined by g satisfies

γ^{∗}ω_{ϕ} = (det g)ω_{ϕ′}.

Proof: If the proposition holds for g_{1}, g_{2} ∈ G_{n} then it holds for g_{1}g_{2}. So we only need to consider g running over a set of generators for G_{n}. Since the cases n = 1, 2 are well known we take n = 3, 4, 5. The result is clear for g of the form [1, B] with

B = (\begin{matrix}μ_{1} & λ \\ & μ_{2} \\ & & . & . & . \\ & & & & & μ_{n} \end{matrix}) .

If n = 3 and g = [μ, I_{3}] then the result is again clear. If n = 4 and g = [A, I_{4}] then there is an isomorphism of complexes

[figure: 0; R(−4); ϕ′_{2}; det A; R(−2)^{2}; ϕ′_{1}; A^{T}; R; 0; 0; R(−4); ϕ_{2}; R(−2)^{2}; ϕ_{1}; R; 0]

We deduce

γ^{∗}ω_{ϕ} = (det A)ω_{ϕ′} = (det g)ω_{ϕ′}.

If n = 5 and g = [A, I_{5}] then there is an isomorphism of complexes

[figure: 0; R(−5); ϕ′_{3}; (det A)^{2}; R(−3)^{5}; ϕ′_{2}; (det A)A^{T}; R(−2)^{5}; ϕ′_{1}; adj A; R; 0; 0; R(−5); ϕ_{3}; R(−3)^{5}; ϕ_{2}; R(−2)^{5}; ϕ_{1}; R; 0]

We deduce

γ^{∗}ω_{ϕ} = (det A)^{2}ω_{ϕ′} = (det g)ω_{ϕ′}.

It only remains to prove the proposition for g = [1, B] with B a permutation matrix. This in turn reduces to checking the result for a set of transpositions generating the symmetric group S_{n}. The symmetry (12) is already clear from the identity

x_{1}^{2}d(x_{2}/x_{1}) + x_{2}^{2}d(x_{1}/x_{2}) = 0.

Since the entries of ϕ_{1} belong to I(C_{ϕ}) we have

(6)  ∑_{i=2}^{n} \frac{∂ϕ_{1}}{∂x_{i}} d(x_{i}/x_{1}) = 0.

If n = 3 then (6) gives the symmetry (23). If n = 4 then the symmetry (34) is clear. By (6) we have

|\begin{matrix}\frac{∂q_{1}}{∂x_{2}} & \frac{∂q_{1}}{∂x_{4}} \\ \frac{∂q_{2}}{∂x_{2}} & \frac{∂q_{2}}{∂x_{4}} \end{matrix} | d(x_{2}/x_{1}) + |\begin{matrix}\frac{∂q_{1}}{∂x_{3}} & \frac{∂q_{1}}{∂x_{4}} \\ \frac{∂q_{2}}{∂x_{3}} & \frac{∂q_{2}}{∂x_{4}} \end{matrix} | d(x_{3}/x_{1}) = 0

and this establishes the symmetry (23).

If n = 5 then the symmetry (35) is clear. Differentiating ϕ_{1}ϕ_{2} = 0 and ϕ_{2}ϕ_{3} = 0 we find

(7)  \frac{∂ϕ_{1}}{∂x_{3}} \frac{∂ϕ_{2}}{∂x_{4}} \frac{∂ϕ_{3}}{∂x_{5}} + \frac{∂ϕ_{1}}{∂x_{3}} \frac{∂ϕ_{2}}{∂x_{5}} \frac{∂ϕ_{3}}{∂x_{4}} = ϕ_{1} \frac{∂ϕ_{2}}{∂x_{3}} \frac{∂^{2}ϕ_{3}}{∂x_{4}∂x_{5}} .

This establishes the symmetry (45). Using (6) we get

∑_{i=2}^{5} \frac{∂ϕ_{1}}{∂x_{i}} \frac{∂ϕ_{2}}{∂x_{4}} \frac{∂ϕ_{3}}{∂x_{5}} d(x_{i}/x_{1}) = 0.

The terms for i = 4, 5 vanish since char (K) ≠ 2 and analogous to (7) we have

2 \frac{∂ϕ_{1}}{∂x_{4}} \frac{∂ϕ_{2}}{∂x_{4}} \frac{∂ϕ_{3}}{∂x_{5}} = \frac{∂^{2}ϕ_{1}}{∂x_{4}^{2}} \frac{∂ϕ_{2}}{∂x_{5}} ϕ_{3} .

This establishes the symmetry (23). □

Lemma 5.20. Let ϕ = π_{n}(ϕ_{1}) be a Weierstrass model with C_{ϕ} a smooth curve of genus one. Then the natural isomorphism γ : C_{ϕ_{1}} ≅ C_{ϕ} satisfies γ^{∗}ω_{ϕ} = ω_{ϕ_{1}}.

Proof: We check this by direct calculation using the definition of ω_{ϕ} and the formulae of §6. □

Lemma 5.21. Let ϕ ∈ X_{n} with C_{ϕ} a smooth curve of genus one. Then ω_{ϕ} is an invariant differential on C_{ϕ}.

Proof: Our claim is that ω_{ϕ} is a non-zero regular 1-form. By Propositions 4.6, 4.7 and 5.19 it suffices to prove this for ϕ a Weierstrass model. Then Lemma 5.20 reduces us to the case n = 1, and in this case the result is well known. □

We recall from Definition 4.12 that models ϕ, ϕ′ ∈ X_{n} are properly equivalent if there exists g ∈ G_{n} with gϕ = ϕ′ and det(g) = 1.

Lemma 5.22. Let ϕ, ϕ′ ∈ X_{1} with C_{ϕ} and C_{ϕ′} smooth curves of genus one. Then ϕ and ϕ′ are properly equivalent if and only if π_{n}(ϕ) and π_{n}(ϕ′) are properly equivalent.

Proof: One implication is clear from Proposition 4.7. For the converse we suppose π_{n}(ϕ) and π_{n}(ϕ′) are properly equivalent. Then by Proposition 5.19 and Lemma 5.20 there is an isomorphism γ : C_{ϕ} ≅ C_{ϕ′} with γ^{∗}ω_{ϕ′} = ω_{ϕ}. Composing with a translation we may suppose that γ is determined by some g ∈ G_{1}. It follows that ϕ and ϕ′ are properly equivalent. □

Proof of Proposition 4.13: Let ϕ ∈ X_{n} with C_{ϕ} a smooth curve of genus one. We must show that ϕ is properly equivalent to a Weierstrass model π_{n}(0, 0, 0, A, B) for some unique A, B ∈ K. The existence is already clear from Propositions 4.6 and 4.7. To prove uniqueness we use Lemma 5.22 to reduce to the case n = 1. In this case the result is well known. □

In the proof of Theorem 4.4(iii) we used

Proposition 5.23. Let ϕ ∈ X_{n} with C_{ϕ} a smooth curve of genus one. Then the geometric invariants of (C_{ϕ}, ω_{ϕ}) are c_{4}(ϕ) and c_{6}(ϕ).

Proof: We are free to replace ϕ by any properly equivalent model. So we may assume that ϕ is a Weierstrass model. Then Lemma 5.20 reduces us to the case n = 1. In this case the result is a tautology. □

6. Weierstrass models

Let E be an elliptic curve with Weierstrass equation

y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}.

In the notation of §3 we have E = C_{ϕ} where ϕ = (a_{1}, a_{2}, a_{3}, a_{4}, a_{6}). The complete linear system |n.0| determines a morphism E → ℙ^{n−1}

\begin{matrix}n = 2 & (x, y) & ↦ & (x : 1) \\ n = 3 & (x, y) & ↦ & (1 : x : y) \\ n = 4 & (x, y) & ↦ & (1 : x : y : x^{2}) \\ n = 5 & (x, y) & ↦ & (1 : x : y : x^{2} : xy). \end{matrix}

The image is defined by a genus one model

\begin{aligned}π_{2}(ϕ) & = & (a_{1}xz + a_{3}z^{2}, x^{3}z + a_{2}x^{2}z^{2} + a_{4}xz^{3} + a_{6}z^{4}) \\ π_{3}(ϕ) & = & (y^{2}z + a_{1}xyz + a_{3}yz^{2} − x^{3} − a_{2}x^{2}z − a_{4}xz^{2} − a_{6}z^{3}) \\ π_{4}(ϕ) & = & (\begin{matrix}x_{1}x_{4} − x_{2}^{2} \\ x_{3}^{2} + a_{1}x_{2}x_{3} + a_{3}x_{1}x_{3} − x_{2}x_{4} − a_{2}x_{2}^{2} − a_{4}x_{1}x_{2} − a_{6}x_{1}^{2} \end{matrix}) \\ π_{5}(ϕ) & = & (\begin{matrix}0 & ℓ & x_{5} & x_{4} & x_{3} \\ & 0 & x_{4} & x_{3} & x_{2} \\ & & 0 & −x_{2} & 0 \\ & − & & 0 & x_{1} \\ & & & & 0 \end{matrix}) \end{aligned}

where ℓ = a_{1}x_{5} − a_{2}x_{4} + a_{3}x_{3} − a_{4}x_{2} − a_{6}x_{1}.

These formulae define a morphism π_{n} : X_{1} → X_{n}. A morphism γ_{n} : G_{1} → G_{n} with the properties specified in Proposition 4.7 is given by

\begin{aligned}γ_{2}([u; r, s, t]) & = & [u^{−3}, (0, u^{2}s, t), (\begin{matrix}u^{2} & 0 \\ r & 1 \end{matrix})] \\ γ_{3}([u; r, s, t]) & = & [u^{−6}, (\begin{matrix}1 & r & t \\ 0 & u^{2} & u^{2}s \\ 0 & 0 & u^{3} \end{matrix})] \\ γ_{4}([u; r, s, t]) & = & [(\begin{matrix}u^{−4} & 0 \\ u^{−6}r & u^{−6} \end{matrix}) , (\begin{matrix}1 & r & t & r^{2} \\ 0 & u^{2} & u^{2}s & 2u^{2}r \\ 0 & 0 & u^{3} & 0 \\ 0 & 0 & 0 & u^{4} \end{matrix})] \end{aligned}

and γ_{5}([u; r, s, t]) = [A_{5}, B_{5}] where

A_{5} = u^{−2} (\begin{matrix}1 & −s & 2r − s^{2} & rs − t & −r^{2} + rs^{2} − st \\ 0 & u & 2us & −ur & u(−2rs + t) \\ 0 & 0 & u^{2} & 0 & −u^{2}r \\ 0 & 0 & 0 & u^{3} & u^{3}s \\ 0 & 0 & 0 & 0 & u^{4} \end{matrix})

and

B_{5} = u^{−3} (\begin{matrix}1 & r & t & r^{2} & rt \\ 0 & u^{2} & u^{2}s & 2u^{2}r & u^{2}(rs + t) \\ 0 & 0 & u^{3} & 0 & u^{3}r \\ 0 & 0 & 0 & u^{4} & u^{4}s \\ 0 & 0 & 0 & 0 & u^{5} \end{matrix}) .

7. Formulae

We recall some formulae for the invariants in the cases n = 2, 3, 4. In each case we scale the invariants so as to give the usual formulae when restricted to the Weierstrass family. As noted in Remark 4.16 these are also the scalings, unique up to sign, for which c_{4}, c_{6} and Δ are primitive integer coefficient polynomials. We assume for simplicity that char (K) ≠ 2, 3.

7.1. Formulae for the invariants: case n = 2. The invariants in the case n = 2 are classical. Here is one way to compute them. We start with the binary quartic

f = ax^{4} + bx^{3}z + cx^{2}z^{2} + dxz^{3} + ez^{4}

and compute (a scalar multiple of) its Hessian

\begin{aligned}H & = & (8ac − 3b^{2})x^{4} + (24ad − 4bc)x^{3}z + (48ae + 6bd − 4c^{2})x^{2}z^{2} \\ & & +(24be − 4cd)xz^{3} + (8ce − 3d^{2})z^{4}. \end{aligned}

We then turn f into a differential operator by substituting ∂/∂z and −∂/∂x for x and z. Letting this operator act on f and H gives the invariants

\begin{aligned}c_{4} & = & 2^{4}(12ae − 3bd + c^{2}) \\ c_{6} & = & 2^{5}(72ace − 27ad^{2} − 27b^{2}e + 9bcd − 2c^{3}). \end{aligned}

The discriminant Δ = (c_{4}^{3} −c_{6}^{2})/1728 is 16 times the usual discriminant of a degree 4 polynomial. If the cross terms α_{0}, α_{1}, α_{2} are included (by computing the square) then c_{4}, c_{6} and Δ are primitive integer coefficient polynomials in α_{0}, α_{1}, α_{2}, a, b, c, d, e.

7.2. Formulae for the invariants: case n = 3. The invariants in the case n = 3 are again classical. The ternary cubic

\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}yz^{2} + mxyz \end{aligned}

has Hessian

H(U) = (−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} | .

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 [16, §II.7] or [22, §225]. It is easily verified by restricting to any family of plane cubics covering the j-line, for example the Weierstrass family defined in §6. We solve to find

c_{4} = −216abcm + 144abc_{1}c_{2} + 144acb_{1}b_{3} − 48ab_{1}c_{2}^{2} + …
… + 24a_{3}b_{1}c_{2}m − 8a_{3}b_{3}m^{2} + 16b_{1}^{2}c_{1}^{2} − 8b_{1}c_{1}m^{2} + m^{4}
c_{6} = 5832a^{2}b^{2}c^{2} − 3888a^{2}bcb_{3}c_{2} + 864a^{2}bc_{2}^{3} + 864a^{2}cb_{3}^{3} + …

… + 12a_{3}b_{3}m^{4} + 64b_{1}^{3}c_{1}^{3} − 48b_{1}^{2}c_{1}^{2}m^{2} + 12b_{1}c_{1}m^{4} − m^{6}

where the full expressions have 25 terms and 103 terms respectively. These polynomials are written out completely in [1], [9, §10.3], [22, §§220,221] and [25, §§4.4,4.5].

We may compute the discriminant as Δ = (c_{4}^{3} − c_{6}^{2})/1728. An alternative, taken from [22, §241], is the following. We compute the partial derivatives of U and H = H(U), and arrange the coefficients of these quadrics in a 6×6 matrix. Then this matrix has determinant ±1728Δ.

7.3. Formulae for the invariants: case n = 4. We identify a genus one model of degree 4 with a pair of 4×4 symmetric matrices. Explicitly

ϕ = (\begin{matrix}q_{1} \\ q_{2} \end{matrix}) ≡ (\begin{matrix}A \\ B \end{matrix})

where

q_{1}(x_{1}, …, x_{4}) = \frac{1}{2}x^{T} Ax and q_{2}(x_{1}, …, x_{4}) = \frac{1}{2}x^{T} Bx.

The invariants are found by computing the binary quartic

det(sA + tB) = as^{4} + bs^{3}t + cs^{2}t^{2} + dst^{3} + et^{4}

and then using the formulae for n = 2. The correct scalings are

\begin{aligned}c_{4} & = & 12ae − 3bd + c^{2} \\ c_{6} & = & \frac{1}{2}(72ace − 27ad^{2} − 27b^{2}e + 9bcd − 2c^{3}). \end{aligned}

Since b, d ∈ 2ℤ[X_{4}] the coefficients of c_{4} and c_{6} are indeed integers as predicted by Lemma 4.15

We may compute the discriminant as Δ = (c_{4}^{3} − c_{6}^{2})/1728. An alternative is the following. Let T_{1} and T_{2} be the symmetric matrices defined in [1], [20] by

adj(s(adj A) + t(adj B)) = a^{2}As^{3} + aT_{1}s^{2}t + eT_{2}st^{2} + e^{2}Bt^{3}.

The corresponding quadrics are

q′_{1}(x_{1}, … x_{4}) = \frac{1}{2}x^{T} T_{1}x and q′_{2}(x_{1}, … x_{4}) = \frac{1}{2}x^{T} T_{2}x.

For a permutation π ∈ S_{4} we define

Ω_{π(1),π(2)} = sign(π) (\frac{∂q_{1}}{∂x_{π(3)}} \frac{∂q_{2}}{∂x_{π(4)}} − \frac{∂q_{1}}{∂x_{π(4)}} \frac{∂q_{2}}{∂x_{π(3)}} ) .

Then we arrange the coefficients of the quadrics q_{1}, q_{2}, q′_{1}, q′_{2} and Ω_{r,s} for 1 ≤ r < s ≤ 4 in a 10 × 10 matrix. The determinant of this matrix turns out to be ±16Δ. As seen in §5.4, the quadrics Ω_{r,s} arise naturally in the construction of an invariant differential ω_{ϕ} on C_{ϕ}.

8. An evaluation algorithm

In the case n = 5 the invariants c_{4} and c_{6} are homogeneous polynomials of degrees 20 and 30 in 50 variables. They are therefore too large to write down as explicit polynomials. Nonetheless we have found a practical algorithm for evaluating them. We assume throughout this section that char (K) ≠ 2, 3, 5.

We identify X_{5} = ∧^{2}V ⊗W where V and W are 5-dimensional vector spaces. Explicitly

(ϕ_{ij}(x_{1}, …, x_{5}))_{i,j=1,...,5} ≡ ∑_{i<j}(v_{i} ∧ v_{j}) ⊗ ϕ_{ij}(x_{1}, …, x_{5})

where v_{1}, …, v_{5} and x_{1}, …, x_{5} are fixed bases for V and W. The action of G_{5} = GL(V)×GL(W) is the natural one. The commutator subgroup of G_{5} is G_{5} = SL(V) × SL(W).

Definition 8.1. Let (ρ, Y) be a rational representation of G_{5}. A covariant is a polynomial map F : ∧^{2}V ⊗ W → Y such that F ∘ g = ρ(g) ∘ F for all g ∈ G_{5}.

Notice that the invariants are the covariants in the case of the trivial representation. For a fixed representation (ρ, Y) the covariants form a module over the ring of invariants.

The 4 × 4 Pfaffians of ϕ are quadrics p_{1}, …, p_{5} satisfying

ϕ ∧ ϕ ∧ v_{i} = p_{i}(x_{1}, …, x_{5}) v_{1} ∧ … ∧ v_{5}.

We may therefore define covariants

\begin{matrix}P : ∧^{2}V ⊗ W → V^{∗} ⊗ S^{2}W ; & ϕ ↦ ∑_{i=1}^{5} v_{i}^{∗} ⊗ p_{i}(x_{1}, …, x_{5}) \\ S : ∧^{2}V ⊗ W → S^{5}W ; & ϕ ↦ det(\frac{∂p_{i}}{∂x_{j}}) \end{matrix}

where v_{1}^{∗}, …, v_{5}^{∗} is the basis for V^{∗} dual to v_{1}, …, v_{5}.

Our method for evaluating the invariants relies on the following geometric “accident”.

Lemma 8.2. Let ϕ ∈ X_{5} with C_{ϕ} a smooth curve of genus one and let p_{1}, …, p_{5} be the 4 × 4 Pfaffians of ϕ.

(i) The secant variety of C_{ϕ} is the hypersurface defined by S(ϕ) = 0. (ii) The partial derivatives \frac{∂}{∂x_{i}} S(ϕ) are quadrics in p_{1}, …, p_{5}.

Proof: (i) See [12, Lemma 6.7] or [17, VIII.2.5]. (ii) The lemma may be checked by direct computation on any family of models covering the j-line, for example the Weierstrass family defined in §6. A more illuminating proof is given in [11, Corollary 7.5]. □

Lemma 8.2(ii) is accompanied by the following uniqueness statement.

Lemma 8.3. Let ϕ ∈ X_{5} with 4 × 4 Pfaffians p_{1}, …, p_{5}. If S(ϕ) ≠ 0 then the quartics {p_{i}p_{j} : 1 ≤ i ≤ j ≤ 5} are linearly independent.

Proof: The condition S(ϕ) ≠ 0 gives that p_{1}, …, p_{5} are linearly independent. Now suppose q(v_{1}, …, v_{5}) is a quadric in 5 variables with q(p_{1}, …, p_{5}) = 0. We differentiate with respect to x_{j} to obtain

∑_{i=1}^{5} \frac{∂q}{∂v_{i}} (p_{1}, …, p_{5})\frac{∂p_{i}}{∂x_{j}} (x_{1}, …, x_{5}) = 0.

Our assumption S(ϕ) ≠ 0 then gives \frac{∂q}{∂v_{i}} (p_{1}, …, p_{5}) = 0 for all i. Since p_{1}, …, p_{5} are linearly independent, it follows that all partial derivatives of q are identically zero, and hence that q itself is identically zero. □

Lemma 8.4. There is a covariant

Q : ∧^{2}V ⊗ W → S^{2}V ⊗ W ; ϕ ↦ ∑ q_{i}(v_{1}, …, v_{5}) ⊗ x_{i}

with the property that if ϕ ∈ X_{5} with 4 × 4 Pfaffians p_{1}, …, p_{5} then

\frac{∂}{∂x_{i}} S(ϕ) = q_{i}(p_{1}, …, p_{5})

for all i. Moreover Q is uniquely determined by this property.

Proof: Let ϕ ∈ X_{5}(𝕂) be the generic model defined over the function field 𝕂 = K(X_{5}). By Proposition 4.5 we know that C_{ϕ} is a smooth curve of genus one. Then by Lemma 8.2 we can solve for quadrics q_{1}, …, q_{5} with the required property. These quadrics define a rational map

Q : ∧^{2}V ⊗ W − →S^{2}V ⊗ W.

By Lemma 8.3 the quadrics q_{1}, …, q_{5} are uniquely determined. So the covariance property is clear. We must show that Q is regular, and for this we may work over an algebraically closed field.

We first claim that Q is regular at all ϕ ∈ X_{5} with S(ϕ) ≠ 0. The coefficients of the quartics {p_{i}p_{j} : 1 ≤ i ≤ j ≤ 5} may be arranged in a 15 × 70 matrix. Let h_{1}, …, h_{N} ∈ K[X_{5}] be the 15 × 15 minors of this matrix. If ϕ ∈ X_{5} with S(ϕ) ≠ 0 then Lemma 8.3 gives h_{i}(ϕ) ≠ 0 for some i. Our claim follows since Q is regular on each of the open sets {h_{i} ≠ 0}.

Now let F ∈ K[X_{5}] be a homogeneous polynomial of least degree such that F Q is regular. Then Q is regular at ϕ if and only if F (ϕ) ≠ 0. The above claim gives F (ϕ) ≠ 0 whenever S(ϕ) ≠ 0. But we know by Lemma 8.2(i) that S(ϕ) ≠ 0 for C_{ϕ} a smooth curve of genus one. By Theorem 4.4(ii) and the irreducibility of Δ (which is inherited from the case n = 1) it follows that F is a power of Δ. To complete the proof it only remains to show that S is not divisible by Δ. Since S has degree 10 and Δ has degree 60, this is clear. □

Starting from P and Q we compute covariants M and N_{λ} taking values in S^{5}V^{∗} and S^{5}V. We then use the natural identification S^{5}V^{∗} = (S^{5}V)^{∗} to contract these covariants, and hence compute the invariants. We arrive at the following algorithm.

Algorithm 8.5. Assume char (K) ≠ 2, 3, 5. INPUT: A genus one model ϕ ∈ X_{5} = ∧^{2}V ⊗ W. OUTPUT: The invariants c_{4}(ϕ), c_{6}(ϕ), Δ(ϕ).

(1) Compute the 4 × 4 Pfaffians p_{1}, …, p_{5} of ϕ.

(2) Check that the quartics {p_{i}p_{j} : 1 ≤ i ≤ j ≤ 5} are linearly independent. If not return 0, 0, 0.

(3) Compute the secant quintic s = det(\frac{∂p_{i}}{∂x_{j}}).

(4) Solve for the auxiliary quadrics q_{1}, …, q_{5} satisfying

\frac{∂s}{∂x_{i}} = q_{i}(p_{1}, …, p_{5}).

(5) Compute the quintic M = det(∑_{k=1}^{5} \frac{∂^{2}p_{k}}{∂x_{i}∂x_{j}} v_{k}^{∗}) ∈ S^{5}V^{∗}.

(6) Compute the quintic N_{λ} = det(λ\frac{∂q_{i}}{∂v_{j}} + ∑_{k=1}^{5} \frac{∂ϕ_{jk}}{∂x_{i}} v_{k}) ∈ S^{5}V.

(7) Contract M and N_{λ} to obtain

⟨M, N_{λ}⟩ = 40c_{4}λ − 320c_{6}λ^{3} + 128c_{8}λ^{5}.

(8) Check that c_{8} = c_{4}^{2}.

(9) Return c_{4}, c_{6}, (c_{4}^{3} − c_{6}^{2})/1728.

It is easy to show that the quantities c_{4} and c_{6} computed are invariants of weights 4 and 6. By Theorem 4.4 the invariants of weights 4 and 6 each form a 1-dimensional vector space. So it only remains to check that the invariants computed are not identically zero, and that they are correctly scaled. We did this by computing their restriction to the Weierstrass family, but in fact it would suffice to compute a single numerical example.

To complete the justification of Algorithm 8.5 we must show that if the quartics in Step 2 are linearly dependent then the invariants are necessarily zero. By Lemma 8.3 we have S(ϕ) = 0. Then Δ(ϕ) = 0 by Lemma 8.2(i). Since c_{4}^{3} − c_{6}^{2} = 1728Δ it only remains to show that c_{4}(ϕ) = 0. We do this by constructing a covariant

T : ∧^{2}V ⊗ W → S^{5}W^{∗}

of degree 30 with ⟨S, T ⟩ = c_{4}^{2}. We omit the (lengthy) details, since our main interest is in applying Algorithm 8.5 in the case C_{ϕ} is a smooth curve of genus one.

An alternative method for computing the discriminant is the following. Let ϕ ∈ X_{5} with 4 × 4 Pfaffians p_{1}, …, p_{5}. For a permutation π ∈ S_{5} we define

Ω_{π(1),π(2)} = sign(π) ∑_{i,j=1}^{5} \frac{∂p_{i}}{∂x_{π(3)}} \frac{∂ϕ_{ij}}{∂x_{π(4)}} \frac{∂p_{j}}{∂x_{π(5)}} .

The calculations of §5.4 show that Ω_{r,s} is well-defined up to the addition of quadrics in the space spanned by p_{1}, …, p_{5}. We arrange the coefficients of p_{1}, …, p_{5} and Ω_{r,s} for 1 ≤ r < s ≤ 5 in a 15 × 15 matrix. Then the determinant of this matrix is an invariant of degree 60, and hence weight 12. We claim it is ±32Δ. Since the invariants of weight 12 form a 2-dimensional vector space, our claim is verified by computing two (suitably chosen) numerical examples. This method for computing the discriminant is in practice much faster than using Algorithm 8.5.

9. Computing the geometric invariants

Let C ⊂ ℙ^{n−1} be a genus one normal curve of degree n ≥ 3, and let ω be an invariant differential on C, both defined over a field K. The geometric invariants c_{4} and c_{6} of the pair (C, ω) were defined in §2. We are interested in computing geometric invariants for the following two reasons.

Computing the Jacobian. Given equations defining a genus one normal curve C ⊂ ℙ^{n−1} of degree n, we aim to compute a Weierstrass equation for its Jacobian. The first step is to compute an invariant differential ω on C. We can do this using either the method of §2 or the method of §5.4. Proposition 2.3 then reduces the problem of computing the Jacobian to that of computing the geometric invariants.

Minimisation. Let K be a local field with discrete valuation ord : K^{∗} → ℤ. If n ≤ 5 then by minimisation we mean the task of finding an integer coefficient genus one model equivalent to a given one, with ord(Δ) minimal. We refer to [8] for a treatment of this problem in the case n = 2. In general the same question can be asked provided we have a notion of genus one model with the following properties:

• a (non-singular) genus one model defines a pair (C, ω),

• it is possible to decide whether a genus one model has integer

coefficients. We will not discuss the possible definitions of genus one model for n > 5, but merely note that if we are to keep track of our progress in minimising, we must be able to compute geometric invariants.

We have compiled the following list of methods for computing geometric invariants. By Lemma 2.2 we are free to rescale ω at any stage (provided we keep track of the scalars).

9.1. The invariants method. We assume that C has degree n ≤ 5. The first step is to compute a genus one model ϕ ∈ X_{n} with C = C_{ϕ}. For n ≤ 4 this is trivial. For n = 5 we use the algorithm described in [12]. Then the formulae and algorithms of §§7,8 are used to compute c_{4}(ϕ) and c_{6}(ϕ). By Proposition 5.23 these are the geometric invariants of (C_{ϕ}, ω_{ϕ}).

The main disadvantage of the invariants method is that we are currently restricted to n ≤ 5.

9.2. The projection method. Extending our field (if necessary) we first find a rational point P ∈ C(K). For instance we might find P by intersecting our curve with a random hyperplane, or by taking the generic point defined over the function field. Then we project away from P to obtain a genus one normal curve C_{P} ⊂ ℙ^{n−2} of degree n − 1. Explicitly, we change co-ordinates on ℙ^{n−1} so that P = (0 : 0 : …: 0 : 1) and the tangent line at P is x_{1} = … = x_{n−2} = 0. Then the projection map

ℙ^{n−1}− →ℙ^{n−2} ; (x_{1} : …: x_{n−1} : x_{n}) ↦ (x_{1} : …: x_{n−1})

restricts to an isomorphism π : C ≅ C_{P} with π(P) = (0 : 0 : …: 0 : 1). We eliminate x_{n} from the quadrics generating I(C) by linear algebra. If n ≥ 5 then by Proposition 5.3 the remaining quadrics are sufficient to generate I(C_{P}). The invariant differential ω on C is specified by an n × n matrix of quadrics, as described in §2. The corresponding invariant differential on C_{P} is obtained by deleting the last row and column of this matrix. We eliminate x_{n} from the remaining entries by subtracting suitable elements of I(C).

At this stage we may either project away from π(P) or switch to another method. If we keep projecting away from a rational point, then eventually we obtain a curve in Weierstrass form. (The final stages of this process are described in [7, §8].) Alternatively if a method for computing Riemann-Roch spaces is available, then we may pass directly to a Weierstrass equation by computing L(mP) for m = 1, 2, 3.

The main disadvantage of the projection method is that it requires a field extension.

9.3. The covering method. Suppose we are given pairs (C_{1}, ω_{1}) and (C_{2}, ω_{2}), and a morphism π : C_{1} → C_{2}. Further suppose that π is a twist of the multiplication-by-m map on an elliptic curve. Then (C_{1}, π^{∗}ω_{2}) and (C_{2}, mω_{2}) have the same geometric invariants. This enables us to compute the geometric invariants of (C_{1}, ω_{1}) from those of (C_{2}, ω_{2}).

The main disadvantage of the covering method is that we need to know a suitable map π : C_{1} → C_{2}. However if the curve C_{1} is found by a descent calculation then it is likely that such a map will be known. In this setting we already know the Jacobian, and the application we have in mind is minimisation.

9.4. The Wronskian method. The invariant differential ω determines a derivation f ↦ df/ω on the function field K(C). Anderson [2] gives a formula in terms of Wronskian determinants for the covering map of degree n^{2} from C to its Jacobian. From this data it is easy to read off the geometric invariants.

The main disadvantage of the Wronskian method is that it requires extensive calculations in the function field.

An example. Wuthrich [27] has constructed an element of order 5 in the Tate-Shafarevich group of an elliptic curve E over ℚ, where the elliptic curve E does not admit any rational 5-isogenies. Written as a genus one normal quintic his example has equations

p_{1} = 3x_{1}^{2} + x_{1}x_{5} − x_{2}x_{4} − x_{3}^{2}
p_{2} = 17x_{1}^{2} − 10x_{1}x_{3} + 7x_{1}x_{5} − 7x_{2}x_{4} − 4x_{2}x_{5} + 4x_{3}x_{4}
p_{3} = 215x_{1}^{2} − 16x_{1}x_{2} − 80x_{1}x_{3} + 16x_{1}x_{4} + 81x_{1}x_{5} − 49x_{2}x_{4}
−28x_{2}x_{5} − 16x_{3}x_{5} − 16x_{4}^{2}
p_{4} = 60x_{1}^{2} + 48x_{1}x_{2} − 34x_{1}x_{3} − 24x_{1}x_{4} + 20x_{1}x_{5} − 8x_{2}^{2} − 5x_{2}x_{3}
−12x_{2}x_{4} + 16x_{2}x_{5} − 14x_{3}x_{5} − 8x_{4}x_{5}
p_{5} = 18x_{1}^{2} + 9x_{1}x_{3} − 4x_{1}x_{4} − 4x_{1}x_{5} − 4x_{2}x_{3} − 8x_{2}x_{4} − 6x_{2}x_{5}
+8x_{3}x_{5} − 4x_{5}^{2}.

We use the algorithm in [12] to write these quadrics as the 4×4 Pfaffians of a matrix of linear forms:

(\begin{matrix}0 & 310x_{1} + 3x_{2} + 162x_{5} & −34x_{1} − 5x_{2} − 14x_{5} & 10x_{1} + 28x_{4} + 16x_{5} & 80x_{1} − 32x_{4} \\ & 0 & 6x_{1} + 3x_{2} + 2x_{5} & −6x_{1} + 7x_{3} − 4x_{4} & −14x_{2} − 8x_{3} \\ & & 0 & −x_{3} & 2x_{2} \\ & − & & 0 & −4x_{1} \\ & & & & 0 \end{matrix})

Algorithm 8.5 then computes the invariants

c_{4} = 2^{44} × 151009, c_{6} = −2^{66} × 34871057.

Thus the Jacobian is the elliptic curve of conductor 1 289 106 508 910 with minimal Weierstrass equation

y^{2} + xy + y = x^{3} + x^{2} − 3146x + 39049.

According to MAGMA [19] this elliptic curve has rank 0 and the analytic order of its Tate-Shafarevich group is 25. It is also the only elliptic curve in its isogeny class.

We were also able to compute this example using the projection and Wronskian methods. In our current implementation (written in MAGMA [19], and available from the author’s website) the invariants method is slightly faster than the projection method, each taking around a second. The Wronskian method is much slower, taking around 30 seconds in this case, but has the advantage of giving equations for the covering map. These timings are of course heavily dependent on details of the implementation we have not described here.

10. Invariants in characteristics 2 and 3

In §4 we showed that there is an injective homomorphism of graded rings

π_{n}^{∗} : K[X_{n}]^{G_{n}} → K[X_{1}]^{G_{1}}.

We also recalled the usual formulae for b_{2}, b_{4}, b_{6}, b_{8} and c_{4}, c_{6}, Δ as polynomials in

K[X_{1}] = K[a_{1}, a_{2}, a_{3}, a_{4}, a_{6}].

In Lemma 4.9 we saw that if char (K) ≠ 2, 3 then K[X_{1}]^{G_{1}} = K[c_{4}, c_{6}]. The analogue of this result in characteristics 2 and 3 is the following.

Lemma 10.1. The ring of invariants is

K[X_{1}]^{G_{1}} = \{ \begin{matrix}K[a_{1}, Δ] & if char (K) = 2 \\ K[b_{2}, Δ] & if char (K) = 3. \end{matrix}

Proof: It is easy to show that a_{1} and Δ, respectively b_{2} and Δ, are invariants. We must show that they generate the ring of invariants. As in the proof of Lemma 4.9, this is deduced from the existence of a suitable normal form. Case char (K) = 2. We start with the general Weierstrass equation

y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}.

One easily computes j = a_{1}^{12}/Δ. We assume j ≠ 0 and following [24, Appendix A] make substitutions x = x′ + r and y = y′ + t so that a_{3} = a_{4} = 0. We are free to suppose that K is algebraically closed. Then a further substitution y = y′ + sx′ gives a_{2} = 0. We arrive at the normal form

y^{2} + a_{1}xy = x^{3} + a_{6}

with Δ = a_{1}^{6}a_{6}. It follows that every invariant is a polynomial in a_{1} and Δ/a_{1}^{6}. We are done since a_{1} does not divide Δ. Case char (K) = 3. We start with a general Weierstrass equation and complete the square to obtain

y^{2} = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}.

One easily computes j = a_{2}^{6}/Δ. We assume j ≠ 0 and following [24, Appendix A] make a substitution x = x′ + r so that a_{4} = 0. We arrive at the normal form

y^{2} = x^{3} + a_{2}x^{2} + a_{6}

with b_{2} = a_{2} and Δ = −a_{2}^{3}a_{6}. It follows that every invariant is a polynomial in b_{2} and Δ/b_{2}^{3}. We are done since b_{2} does not divide Δ. □

Theorem 10.2. Let n = 2, 3, 4, 5. Then the map π_{n}^{∗} : K[X_{n}]^{G_{n}} → K[X_{1}]^{G_{1}} is an isomorphism in all characteristics.

Proof: In §4 we saw that c_{4}, c_{6}, Δ ∈ K[X_{1}]^{G_{1}} extend to invariants c_{4}, c_{6}, Δ ∈ K[X_{n}]^{G_{n}}. So it only remains to show that in characteristics 2 and 3 there are invariants in K[X_{n}]^{G_{n}} of weights 1 and 2.

If char (K) = 2 or 3 then c_{4}^{3} − c_{6}^{2} = 1728Δ = 0. So an invariant of weight 2 exists by unique factorization in K[X_{n}].

We now take char (K) = 2 and split into the cases n = 2, 3, 4, 5. In the cases n = 2, 3 the coefficient of xyz is an invariant of weight 1. In the case n = 4 we write

\begin{aligned}q_{1}(x_{1}, …, x_{4}) & = & ∑_{i≤j} a_{ij}x_{i}x_{j} \\ q_{2}(x_{1}, …, x_{4}) & = & ∑_{i≤j} b_{ij}x_{i}x_{j} \end{aligned}

and find

a_{1} = a_{12}b_{34} + a_{13}b_{24} + a_{14}b_{23} + a_{23}b_{14} + a_{24}b_{13} + a_{34}b_{12}.

If n = 5 then our genus one model is a matrix of linear forms, say ϕ = (ϕ_{ij}(x_{1}, …, x_{5})). Let T be a set of left coset representatives for D_{5} = ⟨(12345), (25)(34)⟩ as a subgroup of S_{5}. Then a_{1} is the coefficient of ∏_{i=1}^{5} x_{i} in ∑_{σ∈T} ∏_{i=1}^{5} ϕ_{σ(i) σ(i+1)}. □

Remark 10.3. If char (K) = 2 or 3 then the invariants do not suffice to compute the Jacobian. For example the elliptic curves y^{2} +xy = x^{3} +1 and y^{2} + xy = x^{3} + x^{2} + 1 over 𝔽_{2} have invariants a_{1} = Δ = 1, but are not isomorphic. Similarly the elliptic curves y^{2} = x^{3} − x ± 1 over 𝔽_{3} have invariants b_{2} = 0 and Δ = 1, but are not isomorphic. These examples should be seen as a consequence of the failure of Lemma 2.4 in characteristics 2 and 3.

As we noted in the introduction, it should instead be possible to find a formula for the Jacobian that works in all characteristics by modifying the formulae in characteristic 0. This has been carried out by Artin, Rodriguez-Villegas and Tate [3] in the case n = 3.

Acknowledgements. I would like to thank Nick Shepherd-Barron for introducing me to this problem. The computer calculations in support of this work were performed using MAGMA [19] and PARI [21].

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[27] C. Wuthrich, Une quintique de genre 1 qui contredit le principe de Hasse, Enseign. Math. (2) 47 (2001), no. 1-2, 161–172.

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

  E-mail address: T.A.Fisher@dpmms.cam.ac.uk