ESSENTIAL DIMENSION OF CUBICS

Grégory Berhuy & Giordano Favi

March 2003

Abstract. In this paper, we compute the essential dimension of cubics in three variables, when the base field has characteristic different from 2 and 3 and contains a primitive third root of unity. For this, we use canonical pencils of cubics, Galois descent techniques, and the material introduced in[BeF].

§0 Introduction

Let C be a polynomial in n variables with coefficients somewhere, say in a ring or a field. A question one may ask is whether is it possible, by changing linearly the coordinates, to drop some of its coefficients or make it “nicer”. For instance, the quadratic polynomial X^{2} + bX + c can always be brought to the form X^{2} + d as soon as one can divide by 2. Similarly the cubic polynomial X^{3}+aX^{2}+bX +c can be reduced to X^{3}+dX +d when \frac{1}{3} makes sense. In both cases one feels that “only one parameter is needed” to describe these polynomials. We shall say that in these cases the essential dimension is 1. Essential dimension makes precise the notion of “how many parameters are needed to describe a given structure” in some general context. It turns out that this number is not always easy to compute. One has to carry out some tools in order to estimate it. The aim of this paper is to use techniques, previously developed in [BeF], for the computation of the essential dimension of homogenous cubic polynomials in three variables. The authors would thank warmly Philippe Chabloz, Manuel Ojanguren, Zinovy Reichstein and Armin Rigo for helpful conversation. The first named author also gratefully acknowledges support from the Swiss National Science Fundation, grant No 2100-065128.01/1 (Project leader: E.Bayer-Fluckiger).

§1 Essential dimension of functors: some definition and results

Let k be a field. We denote by ℭ_{k} the category of field extensions of k, i.e. the category whose objects are field extensions K over k and whose morphisms are field homomorphisms which fix k. We write 𝔉_{k} for the category of all covariant functors from ℭ_{k} to the category of sets. If F is such a functor in 𝔉_{k} and K/k ⟶ L/k is a morphism in ℭ_{k}, for every element a ∈ F(K/k) its image under the map F(K/k) ⟶ F(L/k) will be denoted by a_{L}. When no confusion is possible, we will write F(K) instead of F(K/k). By k_{s} we will always mean a separable closure of k. If k has characteristic different from 3, we will denote by ε ∈ k_{s} a primitive third root of unity.

Typeset by AMS-TEX

We recall the definition of the essential dimension of a functor F : ℭ_{k} ⟶ Sets as introduced in [BeF].

Definition 1.1. Let F be an object of 𝔉_{k}, K/k a field extension and a ∈ F(K). For n ∈ ℕ, we say that the essential dimension of a is ≤ n (and we write ed(a) ≤ n), if there exists a subextension E/k of K/k such that :

i) trdeg(E : k) ≤ n, ii) the element a is in the image of the map F(E) ⟶ F(K). We say that ed(a) = n if ed(a) ≤ n and ed(a) ≰ n − 1. The essential dimension of F is the supremum of ed(a) for all a ∈ F(K) and for all K/k. The essential dimension of F will be denoted by ed_{k}(F).

For a group scheme G of finite type over k the essential dimension of the Galois cohomology functor H^{1}(_{−}, G) will be denoted by ed_{k}(G).

Let us recall some results proved in [BeF]:

For any field extension k′/k, any functor F : ℭ_{k} ⟶ Sets can be considered as an element of 𝔉_{k′}. We denote by ed_{k′}(F) its essential dimension. It is easily checked that the inequality ed_{k′}(F) ≤ ed_{k}(F) holds. We will often use this fact. For example to give lower bounds of the essential dimension of a functor one can suppose k algebraically closed.

We shall say that a morphism of functors f : F ⟶ F′ is a surjection if, for every L/k, the corresponding map of sets f_{L} : F(L) ⟶ F′(L) is a surjection.

Lemma 1.1. Let \xymatrix{F \ar@{->>}[r] & F′} be a surjection between functors. Then ed_{k}(F) ≥ ed_{k}(F′).

Proof. See [BeF].

One case of special interest, though not used extensively in these notes, is when one of the functors is a scheme over k. Indeed take X a k-scheme of finite type. One can view it as a functor simply saying X(L) = Hom(Spec(L), X) for L/k. Its essential dimension is easily computed as ed_{k}(X) = dim(X). Now we shall say that X is a classifying scheme for a functor F if there is a surjection \xymatrix{X \ar@{->>}[r] & F} . In this case the above lemma tells us that ed_{k}(F) ≤ dim(X).

Here is a new result which happens to be very useful for our purpose.

Let i : k → k′ an object of ℭ_{k}. We will describe a construction which will give rise to a functor i^{∗} : 𝔉_{k′} ⟶ 𝔉_{k}. Let F be any functor on ℭ_{k′}. For any object L/k of ℭ_{k} we set

(i^{∗}F)(L/k) = ∐_{Hom_{k}(k′,L)}F(L/k′).

This means, more precisely, that for every k-linear map f : k′/k → L/k (if there is any) we take a copy of the set F(L/k′) where L is considered as an object of ℭ_{k′} via f. In other words elements of (i^{∗}F)(L/k) are elements of F(L/k) labelled by k-morphisms from k′/k to L/k. We may write elements in (i^{∗}F)(L/k) as pairs (f, a) where f : k′ → L is a morphism and a ∈ F(L/k′). If now φ : L/k → L′/k is a morphism in ℭ_{k} we define

i^{∗}φ : ∐_{Hom_{k}(k′,L)}F(L/k′) ⟶ ∐_{Hom_{k}(k′,L′)}F(L′/k′)

to be the map which sends an element a ∈ F(L/k′) (labelled by the morphism f : k′/k → L/k) to the element F(φ)(a)(L′/L/k′) (labelled by the morphism φ ∘ f : k′/k → L/k). That is the pair (f, a) goes to (φ ∘ f, a_{L′}). The functoriality is left to the reader.

Take now the representable functor h_{i} : ℭ_{k} ⟶ Sets defined by h_{i}(K/k) = Hom_{k}(k′/k, K/k). One easily checks that i^{∗}(1) = h_{i} where 1 denotes the one-point functor over ℭ_{k′} sending each object to a one-point set. Moreover one computes that ed_{k}(h_{i}) = trdeg(k′ : k). The following lemma generalizes this fact.

Lemma 1.2. Let i : k → k′ be any morphism and F be an object of 𝔉_{k′}. Then

ed_{k}(i^{∗}F) = ed_{k′}(F) + trdeg(k′:k).

Proof. Take an element (f, a) ∈ i^{∗}F(L/k) for some extension L/k where f : k′/k → L/k and a ∈ F(L/k′). Take now a′ ∈ F(L′/k′) for some f′ = L′/k′ and some k′-morphism L′ → L such that a′_{L} = a, where trdeg(L′/k′) is minimal. The element (f, a) now comes from the element (f′, a′) ∈ i^{∗}F(L′/k′/k). It follows that

ed(f, a) ≤ trdeg(L′ : k) = trdeg(L′ : k′) + trdeg(k′ : k) ≤ ed_{k′}(F) + trdeg(k′ : k).

Consequently ed_{k}(i^{∗}F) ≤ ed_{k′}(F) + trdeg(k′ : k). For the reverse inequality take a ∈ F(L/k′) for some extension L/k′. This defines an element (f, a) ∈ i^{∗}F(L/k′/k) where f denotes the extension L/k′. Take an extension L′/k, a k-morphism φ : L′ → L and an element (f′, a′) ∈ i^{∗}F(L′/k) such that (f′, a′)_{L} = (f, a), where trdeg(L′/k) is minimal. This means that φ ∘ f′ = f hence φ : L′/k′ → L/k′ is k′-linear. Consequently a is defined over L′/k′ and we have

ed(a) ≤ trdeg(L′ : k′) = trdeg(L′ : k) − trdeg(k′ : k) ≤ ed_{k}(i^{∗}F) − trdeg(k′ : k).

It follows that ed_{k}(F) ≤ ed_{k}(i^{∗}F) − trdeg(k′ : k).

Remark. The previous things are almost trivial. The only real problem is notation. It is an (tedious) exercice left to the reader to show that i^{∗} : 𝔉_{k′} → 𝔉_{k} is a functor. One could go a step further and prove that the association k ↦ 𝔉_{k} is functorial. We will not use such things and leave it to the courageous reader.

The following result will be useful for our purpose. It relates the essential dimension of an algebraic group to that of a closed subgroup.

Proposition 1.1. Let G be an algebraic group defined over k, and let H be a closed subgroup. Then

ed_{k}(H) + dim(H) ≤ ed_{k}(G) + dim(G).

Proof. See [BeF].

Proposition 1.2. Let G be an algebraic group over k acting linearly on an affine space 𝔸(V). Assume that there exists a non-empty G-stable open subset U of 𝔸(V) such that:

1) the quotient U/G exists

2) for every L/k the stabilizer of each element of U(L_{s}) under G(L_{s}) is trivial. Then U/G is a classifying scheme for H^{1}(_{−}, G). In particular, we have

ed_{k}(G) ≤ dim(V) − dim(G).

Proof. See [BeF].

We now give an application of this last proposition which we will use later.

Proposition 1.3. Let G be a finite constant closed subgroup of PGL_{n} defined over k, and let G̃ be the inverse image of G under the canonical projection π : GL_{n} ⟶ PGL_{n}. Then ed_{k}(G̃) ≤ n − 1.

Proof. The inclusion G̃ ⊂ GL_{n} induces a natural action of G̃ on 𝔸^{n}. The idea is to find an open subset of 𝔸^{n} for which the hypotheses of Proposition 1.2 are fulfilled. Notice first that the quotient U/G̃ exists when one takes U = 𝔸^{n} \ {0}. Indeed it is easily seen that this quotient is the same as the quotient of ℙ^{n−1} by the finite group G. So we only have to worry about the condition of stabilizers. Actually one only has to cut out of 𝔸^{n} a bad closed set those points have non trivial stabilizer. We will now go into all the details.

By definition G̃(K) = π_{K}^{−1}(G(K)) for any field extension K/k and dim G̃ = 1. Moreover, the map π_{K} induces a group isomorphism G̃(K)/K^{×} ≃ G(K). For any element of G(k), choose a preimage in G̃(k). We choose I_{n} for the preimage of \overline{I}_{n}. Denote by S the set of these preimages. Since G is a constant group scheme we have G(K) = G(k) = G for all field extension K/k. Hence the previous isomorphism shows that

G̃(K) = {μg | μ ∈ K^{×}, g ∈ S}.

Take g ∈ S − {I_{n}} and write it g = (m_{ij}). Let I_{g} be the ideal of k[X_{1}, …, X_{n}] generated by the polynomials

X_{i}( ∑_{j} m_{1j}X_{j}) − X_{1}( ∑_{j} m_{ij}X_{j}) i = 1, …, n.

Let F = ⋃ V(I_{g}). This is a closed subset of 𝔸^{n} since S − {I_{n}} is finite. Notice that, by construction, F (K) = {v ∈ K^{n} | there exists g ∈ S − {I_{n}} and λ ∈ K such that gv = λv}. Let U = 𝔸^{n} \ F. This is a dense open subset of 𝔸^{n} which does not contain 0.

We first show that U(L_{s}) is G̃(L_{s})-stable for every L/k.

Let v ∈ U(L_{s}), and μg ∈ G̃(L_{s}). Assume that (μg)v ∉ U(L_{s}), that is (μg)v ∈ F (L_{s}). Then there exists λ ∈ L_{s} and g′ ∈ S − {I_{n}}, such that g′(μgv) = λ(μgv). Since G̃(K) acts linearly on K^{n} and μ ≠ 0, we get g′gv = λgv, hence (g^{−1}g′g)v = λv. Let g′′ ∈ S which represents π_{k}(g^{−1}g′g). Then there exists γ ∈ k^{×} such that g^{−1}g′g = γg′′. Then g′′v = γ^{−1}λv. Since v ∈ U(L_{s}), this implies that g′′ = I_{n}, so π_{k}(g^{−1}g′g) = \overline{I}_{n}. Then we get π_{k}(g′) = \overline{I}_{n}. By construction of S, this implies that g′ = I_{n}, which is a contradiction.

We now check that the stabilizer of any element of U(L_{s}) is trivial.

Let v ∈ U(L_{s}), and let μg ∈ G̃(L_{s}) such that μgv = v. We then have gv = μ^{−1}v, hence g = I_{n} by hypothesis on v. Since v ∈ U(L_{s}), we have v ≠ 0, hence μ = 1. This implies that μg = I_{n}.

Thus the action of G̃ on 𝔸_{k}^{n} satisfy the conditions of Proposition 1.2. Hence

ed(G̃) ≤ dim(𝔸^{n}) − dim(G̃) = n − 1.

§2 Degree d curves and specialization

Let k be a field and let d ≥ 2, n ≥ 1 be two integers. We consider C_{d,n} the functor of nonzero homogeneous polynomials of degree d in n variables up to a scalar. Elements of C_{d,n} are called degree d curves in n variables. We will often use the same notation for a curve and for one polynomial which defines it. We also will have to consider non-singular curves in the sequel. Let’s denote by C_{d,n}^{+} the functor of non-singular degree d curves in n variables.

We want to discuss the following general question. Take C a degree d curve in n variables and write it down C = ∑ a_{i_{1},... ,i_{n}}X_{1}^{i_{1}} ⋯ X_{n}^{i_{n}} (where i_{1} +⋯+i_{n} = d) for some coefficients a_{i_{1},... ,i_{n}}

in a field extension of k. In general it has \binom{d+n−1}{n−1} coefficients. But as soon as one makes a linear change of coordinates some of these coefficients may drop or become equal. Hence we would like to know how many parameters are needed to describe the curve C as soon as we allow ourselves to change a little the equation defining it.

The group GL_{n} acts on C_{d,n} as described above by linear change of coordinates. More precisely, if C ∈ C_{d,n}(L) and φ ∈ GL_{n}(L), define φ(C) to be the curve defined by C ∘ φ. Since scalar matrices do nothing on curves this action induces an action of PGL_{n} on C_{d,n}.

We denote by F_{d,n} the functor of curves up to this action, and sometimes by [C] the class of C ∈ C_{d,n}(L). The action of GL_{n} clearly restricts to C_{d,n}^{+}. We then denote by F_{d,n}^{+} the functor C_{d,n}^{+}/GL_{n}. These are exactely the functors we are interested in (at least for small values of d and n) since we would like to count the minimal number of parameters needed to describe a degree d curve up to change of coordinates. In other words we would like to compute its essential dimension.

At this point there is a useful remark to be made. In order to compute the essential dimension of F_{d,n} one sees that it is sufficient to minimize the number of parameters appearing in the most general polynomial, that is C_{0} = ∑ t_{I}X^{I} where the t_{I}’s are algebraically independent variables over k. This C_{0} is called the generic polynomial of degree d in n variables. We will show in the detail that ed(F_{d,n}) = ed([C_{0}]) = ed(F_{d,n}^{+}). In the sequel we will often use this fact.

So let C_{0} = ∑ t_{I}X^{I} be the generic homogeneous polynomial of degree d in n variables, where the t_{I}’s are independent indeterminates over k (with obvious notation). Set t = (t_{I}) and K_{0} = k(t).

We begin by a technical lemma, which says in particular that the generic cubic [C_{0}] can be specialized in almost any cubic [C].

Technical Lemma. Let P ∈ k[t]. Let φ ∈ GL_{n}(K_{0}), L/k be a field extension and C be a homogeneous polynomial of degree d in n variables with coefficients in L. If L is infinite, there exists ψ ∈ GL_{n}(L) and a specialization t → b = (b_{I}) such that:

1) φ(b) is well-defined and φ(b) ∈ GL_{n}(L),

2) P (b) ≠ 0,

3) C_{0}(b) ∘ φ(b) = C ∘ ψ ∘ φ(b).

Proof. Write φ = (φ_{ij}(t)) = (\frac{P_{ij}(t)}{Q_{ij}(t)}) and C_{0} ∘ φ = ∑_{I} F_{I}(t)X^{I}.

Clearly there exists some polynomials with coefficients in K_{0} in n^{2} variables T_{ij}, say R_{I}(T_{ij}, t), such that F_{I}(t) = R_{I}(φ_{ij}(t), t). In particular, the set of poles of these F_{I}’s is contained is the set of zeros of the Q_{ij}’s. Hence F_{I}(b) is well-defined if Q_{ij}(b) ≠ 0 for all i, j. Set det φ = \frac{D_{1}(t)}{D_{2}(t)}. The set of poles of det φ is contained in the set of zeros of the Q_{ij}’s, so φ(b) is well-defined and invertible if D_{1}(b) ≠ 0 and Q_{ij}(b) ≠ 0 for all i, j.

Let ψ = (ψ_{ij}) ∈ M_{n}(L), where ψ_{ij} have to be determined, and write C ∘ ψ = ∑ b_{I}X^{I}.

Clearly C ∘ψ∘φ = ∑ R_{I}(φ_{ij}(t), b)X^{I}. Since the b_{I}’s are polynomials in ψ_{ij} (and in coefficients of C), we can write b_{I} = S_{I}(ψ_{ij}) for some polynomials S_{I}(T_{ij}) ∈ L[T_{ij}]. Set S = (S_{I}) and let U(T_{ij}) be the product of the polynomials Q_{ij}(S), D_{1}(S), P (S) and det(T_{ij}). The polynomial U with coefficients in L is non zero, hence there exist an element (ψ_{11}, ⋯, ψ_{nn}) of L^{n^{2}} such that U(ψ_{ij}) ≠ 0, since L is infinite. The specialization t_{I} ↦ b_{I} = S_{I}(ψ_{ij}) then satisfies the required conditions.

Definition 2.1. Let [C] ∈ F_{d,n}(L). We say that [C] is isotropic if the equation C = 0 has a non trivial solution in L^{n}. Clearly, this does not depends on the choice of C.

Corollary 2.1. Using the above notation the following holds:

1) The class of the cubic [C_{0}] is anisotropic, hence non singular.

2) One has ed_{k}(F_{d,n}) = ed_{k}([C_{0}]) and ed_{k}(F_{d,n}^{+}) = ed_{k}^{+}([C_{0}]), where ed_{k}^{+}([C_{0}]) denotes the essential dimension of [C_{0}] viewed as an element of F_{d,n}^{+}.

Proof. 1) Assume that [C_{0}] is isotropic, and let (P_{1}(t), ⋯, P_{n}(t)) be a nontrivial solution of the equation C_{0} = 0. Let P the product of all P_{i}’s which are non zero, let L = k(s_{1}, ⋯, s_{n}), where the s_{i}’s are independent indeterminates over k and let C = s_{1}X_{1}^{d} + ⋯ + s_{n}X_{n}^{d}. By the Technical Lemma, there is a specialization of the t_{I}’s such that [C_{0}] maps to [C], and such that the specialization of P is non zero. Consequently, the non zero P_{i}’s map to non zero elements of L and [C] is then isotropic over L. This is not the case, by [Re], Theorem 3.2, hence we get a contradiction.

2) We only show the first equality. The proof of the second one is similar. The inequality ed([C_{0}]) ≤ ed_{k}(F_{d,n}) follows from the definition. Let L/k be a field extension, and take [C] ∈ F_{d,n}(L). We have to show that ed([C]) ≤ ed([C_{0}]). If L is finite, then L/k is algebraic, hence ed([C]) = 0 ≤ ed([C_{0}]). Now assume that L is infinite. Let K ⊂ K_{0} be a field extension such that [C_{0}] = [C′_{0}]_{K_{0}} for some [C′_{0}] ∈ F_{d,n}(K) with trdeg(K : k) = ed([C_{0}]). By definition, there exists λ ∈ K_{0}^{×} and φ ∈ GL_{n}(K_{0}) such that λC_{0} ∘ φ = C′_{0}. Write C′_{0} = ∑ F_{I}(t)X^{I}. Notice that C′_{0} is defined over k(F_{I}(t)). By minimality of the transcendence degree of K, we have

ed_{k}([C_{0}]) = trdeg(K : k) = trdeg(k(F_{I}(t)) : k).

Write λ = \frac{P_{1}}{P_{2}} . By the Technical Lemma, there exists a specialization t → b such that C_{0} ∘ φ maps to to a polynomial C′ equivalent to C and P_{1}P_{2} maps to a non zero element of L. It follows that λ maps to a non zero element μ of L, hence C′_{0} maps to μC′. In particular, μC′ = ∑ F (b)X^{I}, so [μC′] = [C] is defined over k(F_{I}(b)). Since

ed([C]) ≤ trdeg(k(F_{I}(b)) : k) ≤ trdeg(k(F_{I}(t)) : k) = ed([C_{0}])

we then get the result.

Corollary 2.2. We have ed_{k}(F_{d,n}) = ed_{k}(F_{d,n}^{+}).

Proof. This follows from part 2) of the above corollary and the easy fact (left to the reader) that the essential dimension of C_{0} in F_{d,n} is the same that its essential dimension in F_{d,n}^{+}.

§3 Some considerations on cubics

0. Warm-up

Let us come back to our problem. For d = 3, elements of C_{d,n} are called cubics, and the functor F_{d,n} (resp. F_{d,n}^{+}) is simply denoted by Cub_{n} (resp. by Cub_{n}^{+}).

We begin with the two variables case which can be handled without any extra tool.

Proposition 3.1. Let k be a field. If char(k) ≠ 3, then ed_{k}(Cub_{2}) = 1

Proof. We first show that ed_{k}(Cub_{2}) ≤ 1. Let L/k be a field extension, and let C_{0} be the generic cubic polynomial in 2 variables. Write C_{0} = t_{1}X^{3} + t_{2}X^{2}Y + t_{3}XY^{2} + t_{4}Y^{3} We have to show that, up to a linear change of coordinates and to a scalar, C_{0} is defined over an extension of k of transcendence degree at most 1.

Since t_{1} ≠ 0 one can divide by t_{1} and obtain X^{3} + s_{2}X^{2}Y + s_{3}XY^{2} + s_{4}Y^{3} where s_{i} = \frac{t_{i}}{t_{1}} .

Let φ = ( \begin{matrix}1 & −\frac{1}{3}s_{2} \\ 0 & 1 \end{matrix} ). Then C ∘φ = X^{3} +uXY^{2} +vY^{3} for some some u, v ∈ k(s_{2}, s_{3}, s_{4}) which

are easily computed to be non zero. Let now φ′ = ( \begin{matrix}\frac{v}{u} & 0 \\ 0 & 1 \end{matrix} ). Then

\frac{u^{3}}{v^{3}} C ∘ φ′ = X^{3} + \frac{u^{3}}{v^{2}} XY^{2} + \frac{u^{3}}{v^{2}} Y^{3},

so [C] is defined over k(\frac{u^{3}}{v^{2}} ), which has transcendence degree at most 1 over k.

It remains to show that ed_{k}(Cub_{2}) ≥ 1. One can assume that k is algebraically closed. Let L = k(t), where t is an indeterminate over k and C = X^{3} − tY^{3}. Assume that ed([C]) = 0. This means that [C] is defined over k, since k(t)/k is purely transcendental. Hence there exists λ ∈ k(t)^{×}, φ ∈ GL(k(t)) and a polynomial C′ with coefficients in k, such that C = λC′ ∘ φ. In this case, C′ would be isotropic over k (since k is algebraically closed), hence over k(t). Consequently, C is also isotropic over k(t). But this is clearly not the case, since t ∉ k(t)^{×3}. Hence ed_{k}(Cub_{2}) ≥ ed([C]) = 1. This concludes the proof of the statement.

1. Basic facts about cubics in three variables

From now on we will consider the case n = 3. Assume until the end of this section that char(k) ≠ 3.

For any field extension L/k and any λ ∈ L, let C_{λ} = X_{1}^{3} + X_{2}^{3} + X_{3}^{3} − 3λX_{1}X_{2}X_{3}. We also define C_{∞} = X_{1}X_{2}X_{3}. It is easy to see that C_{λ} for λ ∈ L is non-singular if and only if λ is not a 3rd root of unity.

We recall some well-known facts about cubics in 3 variables.

We first begin with the Hessian group G_{216}. It plays a crucial role in our work. We follow [BK].

The Hessian group G_{216} is the group of special affinities SA_{2}(𝔽_{3}), which is generated by the translations of the plane 𝔽_{3}^{2} and the elements of SL_{2}(𝔽_{3}). One can view this group as a subgroup of PGL_{3}(k_{s}) as follows:

Let x_{00}, ⋯, x_{22} the nine points of ℙ_{k(ε)}^{2} defined by:

x_{00} = (0, −1, 1), x_{01} = (0, −ε, 1), x_{02} = (0, −ε^{2}, 1)

x_{10} = (1, 0, −1), x_{11} = (1, 0, −ε), x_{12} = (1, 0, −ε^{2})

x_{20} = (−1, 1, 0), x_{21} = (−ε, 1, 0), x_{22} = (−ε^{2}, 1, 0).

If g ∈ SA_{2}(𝔽_{3}), then g induces a permutation σ_{g} of these nine points as follows:

If g(ā, b̄) = (c̄, d̄) (where a, b, c, d ∈ {0, 1, 2}), then set σ_{g}(x_{ab}) = x_{cd}.

Computation then shows that there exists a unique element \overline{M}_{g} ∈ PGL_{3}(k_{s}) which induces the permutation σ_{g} on the points x_{ab} (the image of the point x_{ab} is computed by left multiplication by x_{ab}, since we use the row convention).

The two translations T_{(20)} and T_{(02)} then correspond respectively to \overline{A} and \overline{C}, where

A = ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) and C = ( \begin{matrix}1 & 0 & 0 \\ 0 & ε & 0 \\ 0 & 0 & ε^{2} \end{matrix} ).

The three generators ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ), ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) and ( \begin{matrix}2 & 1 \\ 1 & 1 \end{matrix} ) correspond to \overline{D}, \overline{E} and \overline{E′}, where

D = ( \begin{matrix}1 & 0 & 0 \\ 0 & ε & 0 \\ 0 & 0 & ε \end{matrix} ), E = ( \begin{matrix}1 & 1 & 1 \\ 1 & ε & ε^{2} \\ 1 & ε^{2} & ε \end{matrix} ) and E′ = ( \begin{matrix}ε^{2} & 1 & 1 \\ ε & 1 & ε \\ ε & ε & 1 \end{matrix} ).

Notice that the set of generators for SA_{2}(𝔽_{3}) in [BK] is not completely correct. Indeed, the 2-Sylow subgroup of G_{216} is the quaternion group, so it is generated by 2 elements of order 4, but the element ( \begin{matrix}2 & 0 \\ 0 & 2 \end{matrix} ), which corresponds to the class of B = ( \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} ), has order 2.

Notice that G_{216} is in fact a subgroup of PGL_{3}(k(ε)).

When talking about cubics in three variables it’s hard not to mention the so called j-invariant. For our purpose we will need only few things about it. First of all we have to know that it exists. That is for a non-singular cubic C = ∑_{i_{1}+i_{2}+i_{3}=3} a_{i_{1},i_{2},i_{3}}X_{1}^{i_{1}}X^{i_{2}}X_{3}^{i_{3}} with coefficients in

a field K there is a rational expression of the coefficients, denoted by j(C), which lies in the ground field K and which does not depend on the class of the cubic. For a non-singular cubic of the form C_{λ} one has j(C_{λ}) = \frac{λ^{3}(λ^{3}+8)^{3}}{(λ^{3}−1)^{3}} (see [BK] p.301-302).

We now recall some results proved in [BK], p.292-298:

Lemma 3.1. Assume that k = k_{s}. Then:

1) Every non-singular cubic C can be mapped to some C_{λ} for some λ ∈ k. Moreover nonsingular cubics are classified by their j-invariant, that is two non-singular cubics are equivalent if and only if they have same j-invariant.

2) Let λ ∈ k ∪ \{∞\}. For any \overline{φ} ∈ PGL_{3}(k), \overline{φ} maps C_{λ} to some C_{μ} if and only if \overline{φ} ∈ G_{216}.

3) Let λ ∈ k ∪ \{∞\}. For any \overline{φ} ∈ PGL_{3}(k), \overline{φ} maps the cubic C_{λ} to itself if and only if \overline{φ} belongs to the subgroup H = ⟨\overline{A}, \overline{B}, \overline{C}⟩.

The two first statements are proved in the case where k is the field of complex numbers, but it is easy to check that they are still true when k is a separably closed field of characteristic different from 3. The third one is only mentionned without proof, but can be obtained by easy computation. Notice that in the two last statements, C_{λ} is not supposed to be non-singular.

2. Canonical pencils of cubics

If C is a cubic polynomial in 3 variables with coefficients in L, let H_{C} = det (\frac{∂^{2}C}{∂X_{i}∂X_{j}} ), and let

F_{C} be the set of cubics of the form αC + βH_{C}, for some α, β ∈ L. The set F_{C} is called the canonical pencil associated to C. Since H_{αC} = α^{3}H_{C} for any α ∈ L^{×}, this set does only depend on the cubic defined by C.

Let 𝔓(L) denote the set {F_{C} | C ∈ C_{3,3}(L)}. For a cubic C over a field L and for any kmorphism L → L′ we define a map 𝔓(L) ⟶ 𝔓(L′) by sending the pencil F_{C} to the pencil F_{C_{L′}}. We then obtain a functor 𝔓 : ℭ_{k} ⟶ Sets. The association C ↦ F_{C} gives rise to a surjective map of functors \xymatrix{C_{3,3} \ar@{->>}[r] & 𝔓} . Let now act the group GL_{3} naturally on 𝔓 as follows: for φ ∈ GL_{3}(L) and C ∈ C_{3,3}(L) we set φ(F_{C}) = F_{φ(C)}.

We say that F_{C} and F_{C′} are isomorphic over L if they are in the same orbit under this action. We denote by [F_{C}] the isomorphism class of F_{C} and we denote by Pen_{3} the functor of isomorphism classes of such pencils.

Lemma 3.2. Let C be a cubic in three variables with coefficients in L. Then sending the class of C to the class of its pencil F_{C} induces a well defined morphism of functors Cub_{3} ⟶ Pen_{3}.

Proof. The statement follows from the formula H_{C∘φ} = (det φ)^{2}H_{C} ∘ φ. The proof of this formula is left to the reader.

Lemma 3.1 tells us that, over a separably closed field, one can bring every non-singular cubic to some canonical form depending on one parameter. However, unlike quadratic forms, there are several cubics defined over L which are not isomorphic over L_{s}. Hence one cannot classify cubics using Galois cohomology like in the quadratic form case. However the next lemma shows that one can do something for pencils of cubics.

Lemma 3.3. Assume that chark ≠ 2, 3 and let L/k be a field extension. For any λ ∈ L, λ^{3} ≠ 1, we have

F_{C_{λ}} = {C_{μ} | μ ∈ L ∪ {∞}}.

In particular, for all C, C′ ∈ Cub_{3}^{+}(L_{s}), the pencils F_{C} and F_{C′} are isomorphic.

Proof. It is easy to see that H_{C_{λ}} = −54λ^{2}(X_{1}^{3} + X_{2}^{3} + X_{3}^{3}) − 3(18λ^{3} − 72)X_{1}X_{2}X_{3}, hence we get

αC_{λ} + βH_{C_{λ}} = (α − 54λ^{2}β)(X_{1}^{3} + X_{2}^{3} + X_{3}^{3}) − 3(αλ + 18λ^{3}β − 2β)X_{1}X_{2}X_{3}.

Let μ ∈ L. If μ = λ, take α = 1 and β = 0. Assume now that μ ≠ λ. Take β = 1 and

α = \frac{72 − 54λ^{2}μ − 18λ^{3}}{λ − μ}.

We claim that α−54λ^{2} ≠ 0. Indeed, assume the contrary. Then we easily get that 72(1−λ^{3}) = 1. Since char(k) ≠ 2, 3, this implies that λ^{3} ≠ 1, which is not the case.

Thus, with these choices of α and β, we get αC_{λ} +βH_{C_{λ}} = (α−54λ^{2})C_{μ}, hence the polynomials αC_{λ} + βH_{C_{λ}} and C_{μ} belong to the same class.

If μ = ∞, take α = −\frac{λ^{2}}{4(λ^{3} − 1)} and β = −\frac{1}{216(λ^{3} − 1)}.

Remark 3.1. If λ^{3} = 1, the lemma is not true. Indeed, it is easy to see that in this case F_{C_{λ}} = {C_{λ}}. Since we want to apply Galois descent to pencils of cubics, we have to restrict ourselves to pencils of non-singular cubics.

We will denote by 𝔓^{+} and Pen_{3}^{+} the corresponding functors. This little restriction does not matter for the computation of essential dimension for we have seen that ed(Cub_{3}) = ed(Cub_{3}^{+}).

Lemma 3.4. Let L/k be a field extension and let C ∈ Cub_{3}^{+}(L). Then

ed([F_{C}]) ≤ ed([C]) ≤ ed([F_{C}]) + 1.

Proof. Let K/k such that trdeg(K : k) = ed_{k}([C]). Then clearly F_{C} is defined over K, hence ed([F_{C}]) ≤ ed([C]). Assume now that ed([F_{C}]) = n. Then there exists a field extension E/k of transcendence degree equal to n, and C′ ∈ C_{3,3}(E) such that F_{C} = F_{C′_{K}}. By definition, there exists φ ∈ GL_{3}(K) such that F_{φ(C)} = F_{C′_{K}}. In particular, there exists α, β ∈ K such that the polynomials C ∘φ and αC′ +βH_{C′} are proportional. Hence [C] = [αC′ +βH_{C′}]. Since α or β is non zero, C is then defined over E(\frac{α}{β}) or E(\frac{β}{α}). Thus C is defined over a field of transcendence degree at most n + 1.

§4 Galois descent for functors. Applications to cubics

We just dealt with pencils of cubics and saw how all pencils become isomorphic over a separably closed field. A natural idea is then to classify them using Galois cohomology set. The problem is that the objects we want to classify are not standard “algebraic structures”. In this section, we prove a Galois descent lemma for reasonable functors which is a slight generalization of [BOI], Proposition (29.1). This lemma will apply to our situation.

Let k be any field, and let F be an object of 𝔉_{k}. We denote by Aut(F) the functor defined by

Aut(F)(L) = {η : F_{L} ⟶ F_{L} | η is an isomorphism of functors}

for any L/k. Notice that for any extension L/k, the action of the absolute Galois group Γ_{L} on L_{s} induces an action on F(L_{s}) by functoriality.

Let G be a group-valuated functor and ρ : G ⟶ Aut(F) be a morphism of group-valuated functors which is Γ-equivariant. For each E/k we define an equivalence relation on F (E) saying that b, b′ ∈ F (E) are equivalent if there exists g ∈ G(E) such that ρ_{E}(g)(b) = b′. We note this by b ∼_{E} b′. Let k′/k be a field extension, and a ∈ F(K). For every extension L/k′ set

X(L) = {b ∈ F(L) | b ∼_{L_{s}} a}.

Denote by Stab_{G}(a) the subfunctor of G defined by

Stab_{G}(a)(L) = {g ∈ G(L) | ρ_{L}(g)(a_{L}) = a_{L}}

for any extension L/k′. This is a group valuated subfunctor of G_{K}.

Finally, we denote by F_{a}(L) the set of equivalence classes of elements of X(L) under the relation b ∼_{L} b′. This defines an object of 𝔉_{k′}, denoted by F_{a}.

We now state the Galois descent lemma:

Galois Descent Lemma. Let ρ : G ⟶ Aut(F) as above. Assume that for any L ∈ ℭ_{k}, the following conditions hold:

1) H^{1}(L, G(L_{s})) = 1

2) F(L_{s})^{Γ_{L}} = F(L) and G(L_{s})^{Γ_{L}} = G(L). Then for any k′/k and for any a ∈ F(k′), there is a natural isomorphism of functors of 𝔉_{k′}

F_{a} ˜ ⟶H^{1}(_{−}, Stab_{G}(a)).

Moreover, this isomorphism maps the class of a_{L} to the base point of H^{1}(L, Stab_{G}(a)(L_{s})).

Proof. We fix once for all an extension k′/k and an element a ∈ F(k′). Let L/k′ an extension of k′. For the proof we will denote by Γ instead of Γ_{L} the Galois group of L. We set A = Stab_{G}(a)(L_{s}) and B = G(L_{s}). It is well-known that there is a natural bijection between ker(H^{1}(L, A) ⟶ H^{1}(L, B)) and the orbit set of the group B^{Γ} in (B/A)^{Γ} (see [BOI], Corollary 28.2 for example).

Since the group G(L_{s}) acts transitively on X(L_{s}), the Γ-set X(L_{s}) can be identified with the set of left cosets of G(L_{s}) modulo Stab_{G}(a)(L_{s}), hence B/A ≃ X(L_{s}). By assumption on F, the set (B/A)^{Γ} is then equal to X(L). Moreover, B^{Γ} = G(L_{s})^{Γ} = G(L). It follows that the orbit set of B^{Γ} in (B/A)^{Γ} is precisely X(L). Since H^{1}(L, G(L_{s})) is trivial, we then obtain is a natural a bijection of pointed sets between H^{1}(L, Stab_{G}(a)(L_{s})) and F_{a}(L). The functoriality is left to the reader.

Example 4.1. Assume that chark ≠ 2, 3. Take F = 𝔓^{+} and let the group G = GL_{3} act on 𝔓^{+}. Take λ ∈ k with λ^{3} ≠ 1 and set a = F_{C_{λ}}. Then Lemma 3.3 tells us that F_{a}(L) = Pen_{3}^{+}(L) for any extension L/k.

We now determine the stabilizer of the pencil F_{C_{λ}}. Notice that the functors Cub_{3}, Cub_{3}^{+}, Pen_{3} and Pen_{3}^{+} can be naturally extended to k-algebras. The functor Stab_{G}(a) is then the stabilizer of a point of the Grassmanian, hence is a representable functor. This shows that Stab_{G}(a) is an algebraic group scheme defined over k.

We first compute the image of this group scheme by the natural projection π : GL_{3} → PGL_{3}. We have π(Stab_{G}(a))(k_{s}) = \{\overline{φ} ∈ PGL_{3}(k_{s}) | φ(F_{C_{λ}}) = F_{C_{λ}}\}.

Since over k_{s} the pencil F_{C_{λ}} is equal to {C_{μ} | μ ∈ k_{s}∪{∞}} it follows that any φ in Stab_{G}(a)(k_{s}) maps C_{λ} to some C_{μ}. So the same holds for \overline{φ} and hence \overline{φ} belongs to G_{216} ⊆ PGL_{3}(k(ε)) by Lemma 3.1. Conversely, if \overline{φ} ∈ G_{216}, then it is clear that \overline{φ} ∈ π(Stab_{G}(a))(k_{s}) by Lemma 3.1 again. Hence π(Stab_{G}(a))(k_{s}) = G_{216}. Since Γ_{k} acts continuously on G_{216}, the group π(Stab_{G}(a)) is then the étale group scheme G_{216,ét} (see [BOI], Proposition 20.16). Thus

Stab_{G}(F_{C_{λ}}) = π^{−1}(G_{216,ét}).

We denote this last group by G̃_{216}. Since the hypotheses of the Galois Descent Lemma are clearly fulfilled, we get

Pen_{3}^{+} ≃ H^{1}(_{−}, G̃_{216}).

In particular, ed_{k}(Pen_{3}^{+}) = ed_{k}(G̃_{216}).

Example 4.2. Assume that chark ≠ 3. Take F = C_{3,3}^{+} and let G = GL_{3} act on it as usual. Let k′/k be a field extension and take a = C_{λ} for some λ ∈ k′ with λ^{3} ≠ 1. Then F_{a}(L) is the set of cubics in L which are equivalent to C_{λ} over L_{s}. Arguing as previously, one can see that Stab_{G}(a) is the algebraic group k′-scheme π^{−1}(H_{ét}), where H is the subgroup of G_{216} described in Lemma 3.1. We will denote it by H̃. Hence, for any field extension k′/k, for any λ ∈ k′, λ^{3} ≠ 1, and for any field extension L/k′, we have a one-to-one correspondence

F_{a}(L) = {[C] ∈ Cub_{3}^{+}(L) | C ∼_{L_{s}} C_{λ}} ≃ H^{1}(L, H̃).

Hence ed_{k′}(F_{a}) = ed_{k′}(H̃). Again we have classified cubics which become isomorphic to a fixed cubic by a Galois cohomology set.

§5 Essential dimension of cubics

We can finally state and prove our main result:

Theorem 5.1. Let k be a field. Assume that char(k) ≠ 2, 3. If k contains ε, then

ed_{k}(Cub_{3}) = 3.

In particular, ed_{k}(Cub_{3}) ≥ 3 for any field k.

This section is devoted to the proof of the statement. We will prove the first part of the statement, the second one will follow from the fact that ed_{k(ε)}(F) ≤ ed_{k}(F). We will restrict ourselves to the functor Cub_{3}^{+} since we know by §2 that ed(Cub_{3}) = ed(Cub_{3}^{+}).

By Example 4.1, we have ed_{k}(Pen_{3}^{+}) = ed_{k}(G̃_{216}). Since ε ∈ k, the group Γ_{k} acts trivially on G_{216,ét}(k_{s}), hence G_{216,ét} is the constant algebraic group G_{216}. Applying Proposition 1.3

with G = G_{216} and n = 3 then gives ed_{k}(G̃_{216}) ≤ 2. Lemma 3.4 implies in particular that ed_{k}(Cub_{3}^{+}) ≤ ed_{k}(Pen_{3}^{+}) + 1, hence ed_{k}(Cub_{3}^{+}) ≤ 3. Notice that G̃_{216} contains the constant subgroup (ℤ/3Z)^{3} generated by εI_{3}, C and D, so using Proposition 1.1, we get ed_{k}(G̃_{216}) = ed_{k}(Pen_{3}^{+}) = 2.

The hard part is to show the converse inequality. We will proceed in several steps.

Let k′/k be a field extension and λ ∈ k′ with λ^{3} ≠ 1. We define an object F_{λ} of 𝔉_{k′} as follows. If L/k′ is a field extension, set

F_{λ}(L/k′) = {[C′] ∈ Cub_{3}^{+}(L) | there exists E/L such that C′ ∼_{E} C_{λ}}.

Notice that F_{λ}(L/k′) = {[C′] ∈ Cub_{3}^{+}(L) | C′ ∼_{L_{s}} C_{λ}}. Indeed, let [C′] ∈ Cub_{3}^{+}(L) such that C′ ∼_{E} C_{λ}, for some E/L. We then have j(C′_{E}) = j(C_{λ}_{E}), hence j([C′]) = j(C_{λ}_{L}). Thus j(C′_{L_{s}}) = j(C_{λ}_{L_{s}}), so C′ ∼_{L_{s}} C_{λ}. The reverse inclusion is clear. Example 4.2 then shows that F_{λ} ≃ H^{1}(_{−}, H̃) and thus

ed_{k′}(F_{λ}) = ed_{k′}(H̃).

This means in particular that the essential dimension of F_{λ} does not depend on λ.

Our next task is to compute the essential dimension of F_{λ}, that is the essential dimension of H̃. Precisely, we will show the following result:

Proposition 5.1. Let k′ be a field of characteristic different from 2 and 3 containing μ_{3}. Then ed_{k′}(H̃) = 2.

Let S be the constant subgroup of PGL_{3} isomorphic to (ℤ/3ℤ)^{2} generated by \overline{A} and \overline{C}, and let S̃ = π^{−1}(S). Clearly, we have the following exact sequence of group schemes:

1 ⟶ 𝔾_{m} ⟶ S̃ ⟶ S ⟶ 1,

hence dim(S̃) = 1. Similarly, we have dim(H̃) = 1. Applying Proposition 1.3 we get ed_{k′}(H̃) ≤ 2. We now prove the reverse inequality.

By Proposition 1.1. we have ed_{k′}(H̃) ≥ ed_{k′}(S̃). Since μ_{3} ∈ k′ ⊆ L, we can identify the algebraic group S with μ_{3} × μ_{3}, where the identification is given on the L-points by mapping \overline{A}^{m}\overline{C}^{n} ∈ S(L) to (ε^{m}, ε^{n}) ∈ μ_{3}(L) × μ_{3}(L). We then have

1 ⟶ 𝔾_{m} ⟶ S̃ ⟶ μ_{3} × μ_{3} ⟶ 1,

where the second map is given by

S̃(L) ⟶ μ_{3}(L) × μ_{3}(L)
λA^{m}C^{n} ↦ (ε^{n}, ε^{m}).

Moreover, for any field extension L/k′, the above exact sequence induces the following exact sequence in cohomology:

H^{1}(L, S̃) ⟶ H^{1}(L, μ_{3}) × H^{1}(L, μ_{3}) ⟶ H^{2}(L, 𝔾_{m}).

Recall that H^{1}(L, μ_{3}) is in one-to-one correspondence with L^{×}/L^{×3} as follows:

For aL^{×3} ∈ L^{×}/L^{×3}, let α ∈ L_{s} such that α^{3} = a. Then the map c_{a} : Γ_{L} → μ_{3} defined by σ ↦ \frac{σ(α)}{α} is a 1-cocycle, and its cohomology class does not depend on the choice of α and a. We will write (a)_{3} the class of the correponding cocycle.

Lemma 5.1. For any field L, the connecting map

∂ : H^{1}(L, μ_{3}) × H^{1}(L, μ_{3}) → H^{2}(L, 𝔾_{m})

is defined by ∂((a)_{3}, (b)_{3}) = −(a)_{3} ∪ (b)_{3}, where ∪ is the cup-product associated to the natural pairing μ_{3}(L) × μ_{3}(L) → μ_{3}(L) defined by (ε^{m}, ε^{n}) ↦ ε^{mn}.

Proof. Let (a)_{3}, (b)_{3} ∈ H^{1}(L, μ_{3}). Let a, b ∈ L^{×}. If σ ∈ Γ_{L}, write c_{a}(σ) = ε^{m_{σ}} and c_{b}(σ) = ε^{n_{σ}} for some m_{σ}, n_{σ} ∈ {0, 1, 2}. Since Γ_{L} acts trivially on μ_{3} × μ_{3}, the element ∂((a)_{3}, (b)_{3}) is the class of the 2-cocycle

α : σ, τ ∈ Γ_{L} × Γ_{L} ↦ β_{σ}σβ_{τ} β_{στ}^{−1},

where β_{σ} is any preimage of (c_{a}(σ), c_{b}(σ)).

If (c_{a}(σ), c_{b}(σ)) = (ε^{m_{σ}}, ε^{n_{σ}}), we choose β_{σ} = A^{m_{σ}}C^{n_{σ}}. Notice that we have σβ_{τ} = β_{τ} for any σ, τ ∈ Γ_{L}. We then have α_{σ,τ} = A^{m_{σ}}C^{n_{σ}}A^{m_{τ}} C^{n_{τ}} C^{−n_{στ}} A^{−m_{στ}}. Since CA = εAC, we get

α(σ, τ) = ε^{n_{σ}m_{τ}} A^{m_{σ}+m_{τ}} C^{n_{σ}+n_{τ} −n_{στ}} A^{−m_{στ}} .

The fact that c_{a} and c_{b} are cocycles and that Γ_{L} acts trivially on μ_{3}(L_{s}) implies that m_{σ} + m_{τ} − m_{στ} and n_{σ} + n_{τ} − n_{στ} are divisible by 3. Hence we get

α : (σ, τ) ∈ Γ_{L} × Γ_{L} ↦ ε^{m_{σ}n_{τ}} ∈ μ_{3}(L),

which is precisely a cocycle representing (b)_{3} ∪ (a)_{3} since Γ_{L} acts trivially on μ_{3}(L_{s}). The conclusion then follows from the equality (b)_{3} ∪ (a)_{3} = −(a)_{3} ∪ (b)_{3}.

We then have a surjection of functors ∂ : H^{1}(_{−}, S̃) → N, where N is the object of 𝔉_{k′} defined by

N(L) = {((a)_{3}, (b)_{3}) ∈ H^{1}(L, μ_{3}) × H^{1}(L, μ_{3}) | (a)_{3} ∪ (b)_{3} = 0}

for any field extension L/k′. Hence, by Lemma 1.1 we get

ed_{k′}(S̃) ≥ ed_{k′}(N).

To conclude the proof of Proposition 5.1, it suffices to prove the following:

Lemma 5.2. We have ed_{k′}(N) ≥ 2.

Proof. It suffices to show the inequality when k′ is algebraically closed.

Let ((a)_{3}, (b)_{3}) ∈ N(L) for some L/k′. Let’s consider the cubic in 4 variables

C_{a,b} = X^{3} + aY^{3} + bZ^{3} + abT^{3}.

The equivalence class of the cubic C_{a,b} does not depend on the choice of representatives of (a)_{3} and (b)_{3}. Moreover, this assignment is functorial in L. Hence we have a morphism of functors N ⟶ Cub_{4}. Notice that ((a)_{3}, (b)_{3}) ∈ N(L) if and only if a is a norm of the extension L(β)/L, where β^{3} = b. Now let s, t, u be independent inderminates over k′, and set b = t, a = 1 + ts^{3} + t^{2}u^{3} − 3stu. Set L = k′(s, t, u). Then a = N_{L(β)/L}(1 + sβ + uβ^{2}), hence ((a)_{3}, (b)_{3}) ∈ N(L).

Now assume that ((a)_{3}, (b)_{3}) is defined over a field L′/k′ of transcendence degree at most 1 over k′, then so is [C_{a,b}]. Then L′ is a C_{1} field (since k′ is algebraically closed), hence [C_{a,b}] is isotropic over L′, hence over L.

To get a contradiction, it remains to show that C_{a,b} is anisotropic over L.

Lemma 5.3. The polynomial C_{a,b}, with a = 1 + ts^{3} + t^{2}u^{3} − 3stu and b = t is anisotropic over k′(s, t, u).

Proof. Assume the contrary. Then there exists P_{1}, ⋯, P_{4} ∈ k′[s, t, u] not all zero, such that P_{1}^{3} + aP_{2}^{3} + bP_{3}^{3} + abP_{4}^{3} = 0. Consider P_{1}, ⋯, P_{4} and a as elements of k′(s, u)[t] , and write P_{i} = a^{n_{i}}Q_{i} where Q_{i} is not divisible by a as soon as Q_{i} ≠ 0. If n_{1} ≤ n_{2}, n_{3}, n_{4}, then one gets

Q_{1}^{3} + a^{3(n_{2}−n_{1})+1}Q_{2}^{3} + ta^{3(n_{3}−n_{1})}Q_{3}^{3} + ta^{3(n_{4}−n_{1})+1}Q_{4}^{3} = 0.

Hence Q_{1}^{3} + ta^{3(n_{3}−n_{1})}Q_{3}^{3} = 0 in E = k′(s, u)[T]/(a).

Assume first that Q_{3} = 0. Then the previous equation shows that Q_{1} is divisible by a, which implies that Q_{1} = 0 by choice of the Q_{i}’s. Hence Q_{1} = Q_{3} = 0, so Q_{2} and Q_{4} are both non zero, and a^{3(n_{2}−n_{1})+1}Q_{2}^{3} + ta^{3(n_{4}−n_{1})+1}Q_{4}^{3} = 0. If n_{2} − n_{1} ≠ n_{4} − n_{1}, it would imply that Q_{1} or Q_{3} is divisible by a, which gives a contradiction since Q_{1} and Q_{2} are non zero polynomials. Hence n_{2} − n_{1} = n_{4} − n_{1}, and t is then a cube in E. Assume now Q_{3} is non zero. In this case Q_{3} is not divisible by a, so t is a cube in E. The remaining cases also give that t is a cube in E.

By definition, we have E = k′(s, u)(\sqrt{Δ}), where Δ = (s^{3}−3su)^{2}−4u^{3}, and t = \frac{3su − s^{3} ± \sqrt{Δ}}{2u^{3}}.

Then, if t is a cube in E then 12su − 4s^{3} ± 4\sqrt{Δ} is a cube in E. Let denote by τ the unique non trivial k′(s, u)-automorphism of E. An element λ ∈ E is a cube if and only if τ(λ) is a cube. So λ = 12su − 4s^{3} + 4\sqrt{Δ} is a cube.

Write λ = μ^{3}, with μ = \frac{R_{1}+R_{2}\sqrt{Δ}}{R_{3}}, where R_{1}, R_{2}, R_{3} ∈ k′[s, u]. One can assume that R_{1}, R_{2}, R_{3} are relatively prime. We have N_{E/k(s,u)}(λ) = 16(3su − s^{3})^{2} − 16Δ = 64u^{3}, hence

64R_{3}^{6}u^{3} = (R_{1}^{2} − R_{2}^{2}Δ)^{3}.

We also have (12su − 4s^{3} + 4\sqrt{Δ})R_{3}^{3} = R_{1}^{3} + 3R_{1}R_{2}^{2}Δ + (R_{2}^{3}Δ + 3R_{1}^{2}R_{2})\sqrt{Δ}, hence

4R_{3}^{3} = R_{2}^{3}Δ + 3R_{1}^{2}R_{2} and (12su − 4s^{3})R_{3}^{3} = R_{1}^{3} + 3R_{1}R_{2}^{2}Δ.

We now show that R_{3} is a constant. Assume that S is an irreducible divisor of R_{3}. Then S divides R_{1}^{2} − R_{2}^{2}Δ by the first equation, hence S divides 3R_{1}^{3} − 3R_{1}R_{2}^{2}Δ. The third equation implies that S divides R_{1}^{3} + 3R_{1}R_{2}^{2}Δ, hence S divides R_{1}^{3}, so S divides R_{1}. Hence S^{3} divides (12su − 4s^{2})R_{3}^{3} − R_{1}^{3}, so S^{2} divides 3R_{2}^{2}Δ. Consequently, S divides R_{2}^{2} (even if S = Δ), so S divides R_{2}. This is impossible since R_{1}, R_{2}, R_{3} are relatively prime. Hence, one can assume that R_{3} = 1.

We then have the equations

64u^{3} = (R_{1}^{2} − R_{2}^{2}Δ)^{3}, 4 = R_{2}^{3}Δ + 3R_{1}^{2}R_{2}, 12su − 4s^{3} = R_{1}^{3} + 3R_{1}R2^{2}Δ.

The second equation then implies that R_{2} and R_{2}^{2}Δ + 3R_{1}^{2} are non zero constant polynomials. In particular R_{1}^{2} and Δ has same degree in u. This gives a contradiction since these degrees don’t have the same parity.

We are now able to finish the proof of Theorem 5.1 using Proposition 5.1 and Lemma 1.2.

Let t be an indeterminate over k, let \overline{k(t)} be an algebraic closure of k(t). Let i be the composite k → k(t) → \overline{k(t)}, where the first map is the natural inclusion, and k(t) → \overline{k(t)} is a fixed k-linear morphism which maps t to itself. In the sequel we set k′ = \overline{k(t)}. Let λ ∈ k′ such that j(C_{λ}) = t and consider the functor F_{λ}. By Lemma 1.2, we have

ed_{k}(i^{∗}F_{λ}) = ed_{k′}(F_{λ}) + trdeg(k′ : k).

By Proposition 5.1, we have ed_{k′}(F_{λ}) = 2. Moreover we have trdeg(k′ : k) = 1, so we get

ed_{k}(i^{∗}F_{λ}) = 3.

Thus there exists a field extension L/k with trdeg(L : k) = 3 and and an element x = (f, [C]) ∈ (i^{∗}F_{λ})(L/k) which can not be defined over a subextension of L of a smaller transcendence degree. By definition of i^{∗}F_{λ}(L/k), the extension L/k is the composite f ∘i. Since [C] ∈ F_{λ}(L/k′), then in particular [C] ∈ Cub_{3}^{+}(L/k′), so [C] can be viewed as an element of Cub_{3}^{+}(L/k) via i.

We now proceed to show that ed([C]) = 3 in Cub_{3}^{+}.

Assume that there exists a subextension K′/k of L/k with trdeg(K′ : k) ≤ 2 and [C′] ∈ Cub_{3}^{+}(L) such that [C′]_{L} = [C]. Let φ : k ⟶ L, ψ : k ⟶ K′ and θ : K′ ⟶ L. Since K′/k is a subextension of L/k, we then have θ ∘ ψ = φ. Since [C] ∈ F_{λ}(L/k′), we have j(C′_{L}) = j(C) = j(C_{λ}_{L}) = f(t), hence j(C′_{L}) is transcendental over k. Consequently, j(C′) ∈ K′ is transcendental over k and we can define a morphism of k-extensions β : k(t) → K′ by β(t) = j(C′) and β_{|k} = ψ.

We now check that the composite maps η_{1} : k(t) ⟶ k′ ⟶ L and η_{2} : k(t) ⟶ K′ ⟶ L are the same. Let α ∈ k. We have η_{1}(α) = f(α) and η_{2}(α) = θ(ψ(α)) = φ(α). By definition of L/k, we have φ = f ∘ i (since L/k factors through i). Hence η_{2}(α) = f(i(α)) = f(α). Moreover, since we have η_{1}(t) = f(t) and η_{2}(t) = ψ(j(C′)) = j(C′_{L}), the maps η_{1} and η_{2} coincide.

Hence we can define the compositum E/k(t) of k′/k(t) and K′/k(t) in L/k(t).

By definition of E/k(t), the composite map K′ ⟶ E ⟶ L is equal to θ, so we have ([C′]_{E})_{L} = [C′]_{L} = [C]. Now consider the extension E/k defined by k ⟶ k′ ⟶ E. By construction, this extension factors through i. Let g : k′ ⟶ E. We then have (g, [C′]_{E}) ∈ (i^{∗}F_{λ})(E/k). Moreover, by definition of E/k(t) again, the map k′ ⟶ E ⟶ L is precisely f, hence (g, [C′]_{E})_{L} = (f, [C]). Since trdeg(K′ : k) ≤ 2 and trdeg(k(t) : k) = 1, we have trdeg(K′ : k(t)) ≤ 1. Moreover, trdeg(k′ : k(t)) = 0, hence trdeg(E : k(t)) ≤ 1, so trdeg(E : k) ≤ 2. Consequently, x = (f, [C]) is defined over a subextension of L/k of transcendence degree at most 2, which is impossible by choice of x.

We then get ed_{k}(Cub_{3}^{+}) ≥ ed([C]) = 3, which concludes the proof of Theorem 5.1.

REFERENCES

[BeF]G.Berhuy, G. Favi Essential dimension: a functorial point of view. Preprint (2003) Available on ...

[BK] E. Brieskorn, H. Knörrer, Plane Algebraic Curves, Birkhaüser Verlag, Basel, Boston,Stuttgart (1986)

[BOI] M.-A. Knus, A.S. Merkurjev, M. Rost, J.-P. Tignol, The Book of Involutions, AMS Coll. Pub. vol. 44 (1998)

[Re] Z. Reichstein, On a Theorem of Hermite and Joubert. Canadian J. of Mathematics, 51(1), 69-95 (1999)

Grégory Berhuy Ecole Polytechnique Fédérale de Lausanne MA-Ecublens CH-1015 Lausanne SWITZERLAND

Email:gregory.berhuy@epfl.ch

Giordano Favi Institut de Mathématiques Université de Lausanne Dorigny CH-1015 Lausanne SWITZERLAND

Email: giordano.favi@ima.unil.ch