%\magnification=1200

\input amstex
\input xypic
\xyoption{all}

\documentstyle{amsppt}

\font\ptitgot=eufm10 at 10pt
\font\grogot=eufm10 at 13pt
\font\groscr=cmsy10 at 12pt
\font\ptitrm=cmr10 at 8pt

\def\Ak{\hbox{\grogot A}_k}
\def\Ck{\hbox{\grogot C}_k}
\def\Cko{\hbox{\grogot C}_{k'}}
\def\m{\hbox{\ptitgot m}}
\def\Sk{\hbox{\grogot F}_k}
\def\SK{\hbox{\grogot F}_K}
\def\Sko{\hbox{\grogot F}_{k'}}

\def\freccia{\longrightarrow}
\def\cofreccia{\longleftarrow}
\def\ds{\displaystyle}
\def\E{\varepsilon}
\def\cool{{\Cal S}_n}
\def\Fp{\hbox{\groscr F}}

\def\FFF#1{\Fp_{\hskip-1pt\scriptscriptstyle #1}}
\def\ttt{\hbox{\bf t}}
\def\bbb{\hbox{\bf b}}
\def\hhh{\hbox{\bf h}}

\def\A{{\Bbb A}}
\def\C{{\Bbb C}}
\def\F{{\Bbb F}}

\def\G{{\Bbb G}}
\def\Gm{{\Bbb G}_{\roman m}}
\def\Ga{{\Bbb G}_{\roman a}}
\def\GmL{{\Bbb G}_{{\roman m},L}}
\def\JJ{\hbox{\bf J}}
\def\NN{\hbox{\bf N}}
\def\N{{\Bbb N}}
\def\P{{\Bbb P}}
\def\Q{{\Bbb Q}}
\def\R{{\Bbb R}}
\def\Z{{\Bbb Z}}


%Tout ce qui est gras...

\def\Alg{\hbox{\bf Alg}}
\def\CC{\hbox{\bf C}}
\def\Cub{\hbox{\bf Cub}}
\def\Et{\hbox{\bf \' Et}}
\def\FF{\hbox{\bf F}}
\def\GG{\hbox{\bf G}}
\def\Groups{\hbox{\bf Groups}}
\def\GL{\hbox{\bf GL}}
\def\O{\hbox{\bf O}}
\def\Pen{\hbox{\bf Pen}}
\def\PGL{\hbox{\bf PGL}}
\def\Pol{\hbox{\bf Pol}}
\def\QQ{\hbox{\bf Q}}
\def\Sets{\hbox{\bf Sets}}
\def\Stab{\hbox{\bf Stab}}
\def\Tors{\hbox{\bf Tors}}


%Tout ce qui est droit...

\def\Aut{\roman{Aut}}
\def\Bij{\hbox{\rm Bij}}
\def\Br{\roman{Br}}
\def\cd{\roman{cd}}
\def\char{\roman{char}}
\def\Cr#1{\roman{(C}_{#1}\roman{)}}
\def\disc{\roman{disc}}
\def\dim{\roman{dim}}
\def\ed{\roman{ed}}
\def\Gal{\roman{Gal}}
\def\GLrm{\roman{GL}}
\def\Hom{\roman{Hom}}
\def\im{\roman{im}}
\def\Ind{\roman{Ind}}

\def\Isom{\hbox{\rm Isom}}
\def\PGLrm{\roman{PGL}}
\def\Prm{\hbox{\rm P}}
\def\Pprm{\hbox{\ptitrm P}}
\def\pr{\roman{pr}}
\def\rank{\roman{rank}}
\def\SA{\roman{SA}}
\def\SL{\roman{SL}}
\def\Spec{\roman{Spec}}
\def\Tr{\roman{Tr}}
\def\trdeg{\roman{trdeg}}
\def\V{\roman{V}}

%\nologo
\NoBlackBoxes
%\CompileMatrices

\parindent=0pt
\hsize=14.5cm


\topmatter
%\centerline{Una roba rapida sulla dimensione essenziale}

\vskip 10mm

\title Essential dimension of cubics\endtitle

\author Gr\'{e}gory Berhuy \& Giordano Favi \endauthor

\date March 2003 \enddate

\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\cite{BeF}.
\endabstract





\endtopmatter

\document



\heading \S0 Introduction \endheading

\bigskip

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 ${1\over3}$ 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 
{\it 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 \cite{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).
\bigskip
\bigskip

\heading \S1 Essential dimension of functors: some definition and 
results \endheading

\bigskip


\bigskip

Let $k$ be a field. We denote by $\Ck$ 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 $\Sk$ for the category of all {\it covariant} 
functors from $\Ck$ to the category of sets. If $\FF$ is such a 
functor in $\Sk$ and $K/k\freccia L/k$ is a morphism in $\Ck$, for every element 
$a\in \FF(K/k)$ its image under the map $\FF(K/k)\freccia \FF(L/k)$ will be 
denoted by $a_{L}$. When no confusion is possible, we will write $\FF(K)$ instead of $\FF(K/k)$.

By $k_s$ we will always mean a separable closure of $k$. If $k$ has 
characteristic different from $3$, we will denote by $\E\in k_s$ a 
primitive third root of unity.

\bigskip

We recall the definition of the essential dimension of a functor 
$\FF:\Ck\freccia\Sets$ as introduced in \cite{BeF}.

\bigskip

\proclaim{Definition 1.1}
Let $\FF$ be an object of $\Sk$, $K/k$ a field extension and 
$a\in\FF(K)$. For $n\in\N$, we say that the {\bf essential dimension 
of $a$ is $\leq n$} (and we write
$\ed(a)\leq n$), if there exists a subextension $E/k$ of
$K/k$ such that~:
\smallskip
i) $\trdeg(E:k)\leq n$,
\smallskip
ii) the element $a$ is in the image of the map $\FF(E)\freccia \FF(K)$.
\smallskip
We say that $\ed(a)=n$ if $\ed(a)\leq n$ and $\ed(a)\not\leq n-1$. 
The {\bf essential dimension of} $\FF$ is the supremum of $\ed(a)$ 
for all $a\in \FF(K)$ and for all $K/k$.
The essential dimension of $\FF$ will be denoted by $\ed_k(\FF)$.
\endproclaim

\bigskip

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 $\ed_k(G)$.

\bigskip

Let us recall some results proved in \cite{BeF}:

\bigskip

For any field extension $k'/k$, any functor $\FF:\Ck\freccia\Sets$ 
can be considered as an element of $\Sko$. We denote by 
$\ed_{k'}(\FF)$ its essential dimension. It is easily checked that 
the inequality $\ed_{k'}(\FF)\leq\ed_k(\FF)$ 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.


\bigskip


We shall say that a morphism of functors $f:\FF\freccia\FF'$ is a 
{\bf surjection} if, for every $L/k$, the corresponding map of sets 
$f_L:\FF(L)\freccia\FF'(L)$ is a surjection.

\bigskip

\proclaim{Lemma 1.1} Let $\xymatrix{\FF\ar@{->>}[r]&\FF'}$ be a 
surjection between functors. Then $\ed_k(\FF)\geq\ed_k(\FF')$.
\endproclaim

\demo{Proof} See \cite{BeF}.
\enddemo

\bigskip

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 
{\bf classifying scheme} for a functor $\FF$ if there is a surjection 
$\xymatrix{X\ar@{->>}[r]&\FF}$. In this case the above lemma tells us 
that $\ed_k(\FF)\leq\dim(X)$.

\bigskip

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

\bigskip

Let $i:k\to k'$ an object of $\Ck$. We will 
describe a construction which will give rise to a functor 
$i^*:\Sko\freccia\Sk$. 

Let $\FF$ be any functor on $\Cko$. For any 
object $L/k$ of $\Ck$ we 
set

$$(i^*\FF)(L/k)=\ds\coprod_{\Hom_k(k',L)}\hskip-5pt\FF(L/k').$$

This 
means, more precisely, that for every $k$-linear map $f:k'/k\to L/k$ (if there is any)
we take a copy of the set $\FF(L/k')$ where $L$ is considered as an 
object of $\Cko$ via $f$. In other words elements of $(i^*\FF)(L/k)$ 
are elements of $\FF(L/k)$ labelled by $k$-morphisms from $k'/k$ to 
$L/k$. We may write elements in $(i^*\FF)(L/k)$ as pairs $(f,a)$ 
where $f:k'\to L$ is a morphism and $a\in\FF(L/k')$. If now 
$\varphi:L/k\to L'/k$ is a morphism in $\Ck$ we 
define

$$i^*\varphi:\ds\coprod_{\Hom_k(k',L)}\hskip-5pt\FF(L/k')\freccia\ds\coprod_{\Hom_k(k',L')}\hskip-5pt\FF(L'/k')$$

to 
be the map which sends an element $a\in\FF(L/k')$ (labelled by the 
morphism $f:k'/k\to L/k$) to the element $\FF(\varphi)(a)(L'/L/k')$ 
(labelled by the morphism $\varphi\circ f:k'/k\to L/k$). That is the 
pair $(f,a)$ goes to $(\varphi\circ f, a_{L'})$. The functoriality is 
left to the reader.

\bigskip

Take now the representable functor 
$\hhh_i:\Ck\freccia\Sets$ defined by $\hhh_i(K/k)=\Hom_k(k'/k,K/k)$. 
One easily checks that $i^*(\text{\bf 1})=\hhh_i$ where {\bf 1} 
denotes the one-point functor over $\Cko$ sending each object to a 
one-point set. Moreover one computes that 
$\ed_k(\hhh_i)=\trdeg(k':k)$. The following lemma generalizes this 
fact.

\bigskip

\proclaim{Lemma 1.2} Let $i:k\to k'$ be any morphism 
and $\FF$ be an object of $\Sko$. Then 
$$\ed_k(i^*\FF)=\ed_{k'}(\FF)+\trdeg(k'\!:\!k).$$
\endproclaim

\demo{Proof}
Take an element $(f,a)\in i^*\FF(L/k)$ for 
some extension $L/k$ where $f:k'/k\to L/k$ and $a\in\FF(L/k')$. Take 
now $a'\in\FF(L'/k')$ for some $f'=L'/k'$ and some $k'$-morphism 
$L'\to L$ such that $a'_L=a$, where $\trdeg(L'/k')$ is minimal. The element $(f,a)$ now comes from the 
element $(f',a')\in i^*\FF(L'/k'/k)$. It follows that 
$$\ed(f,a)\leq\trdeg(L':k)=\trdeg(L':k')+\trdeg(k':k)\leq\ed_{k'}(\FF)+\trdeg(k':k).$$ 
Consequently 
$\ed_k(i^*\FF)\leq\ed_{k'}(\FF)+\trdeg(k':k)$.

\smallskip

For the 
reverse inequality take $a\in\FF(L/k')$ for some extension $L/k'$. 
This defines an element $(f,a)\in i^*\FF(L/k'/k)$ where $f$ denotes 
the extension $L/k'$. Take an extension $L'/k$, a $k$-morphism 
$\varphi:L'\to L$ and an element $(f',a')\in i^*\FF(L'/k)$ such that 
$(f',a')_L=(f,a)$, where $\trdeg(L^\prime/k)$ is minimal. This means that $\varphi\circ f'=f$ hence 
$\varphi:L'/k'\to L/k'$ is $k'$-linear. Consequently $a$ is defined over $L'/k'$ and we have
$$\ed(a)\leq\trdeg(L':k')=\trdeg(L':k)-\trdeg(k':k)\leq\ed_k(i^*\FF)-\trdeg(k':k).$$ 
It follows that $\ed_k(\FF)\leq\ed_k(i^*\FF)-\trdeg(k':k)$.
\enddemo

\bigskip

{\bf 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 $i^*:\Sko\to\Sk$ is a functor. One could go a 
step further and prove that the association $k\mapsto\Sk$ is 
functorial. We will not use such things and leave it to the courageous 
reader.



\bigskip

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

\bigskip



\proclaim{Proposition 1.1}Let $G$ be an algebraic group defined over 
$k$, and let $H$ be a closed subgroup. Then
$$\ed_k(H)+\dim(H)\leq\ed_k(G)+\dim(G).$$
\endproclaim

\demo{Proof} See \cite{BeF}. \enddemo

%mettre un petit blabla sur les quotients??????????????????????????????????

\bigskip

\proclaim{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:

{\parindent.5cm
\smallskip
\item{1)} the quotient $U/G$ exists
\smallskip
\item{2)} for every $L/k$ the stabilizer of each element of $U(L_s)$ 
under $G(L_s)$ is trivial. }
\smallskip
Then $U/G$ is a classifying scheme for $H^1(_-,G)$. In particular, we 
have $$\ed_k(G)\leq\dim(V)-\dim(G).$$
\endproclaim


\demo{Proof} See \cite{BeF}.\enddemo

\bigskip

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

\bigskip

\proclaim{Proposition 1.3} Let $G$ be a finite constant closed 
subgroup of $\PGL_n$ defined over $k$, and let $\ds\widetilde{G}$ be 
the inverse image of $G$ under the canonical projection 
$\pi:\GL_n\freccia\PGL_n$. Then $\ed_k(\ds\widetilde{G})\leq n-1$.
\endproclaim

\demo{Proof} The inclusion $\ds\widetilde{G}\subset\GL_n$ induces a 
natural action of $\ds\widetilde{G}$ 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 $U/\ds\widetilde{G}$ 
exists when one takes $U=\A^n\setminus\{0\}$. 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 {\it bad} 
closed set those points have non trivial stabilizer. We will now go 
into all the details.

\bigskip

By definition $\widetilde{G}(K)=\pi^{-1}_K(G(K))$ for any field 
extension $K/k$ and $\dim\,\widetilde{G}=1$. Moreover, the map 
$\pi_{K}$ induces a group isomorphism 
$\ds\widetilde{G}(K)/K^\times\simeq G(K).$  For any element of 
$G(k)$, choose a preimage in $\ds\widetilde{G}(k)$. 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 $$\ds\widetilde{G}(K)=\{ \mu g\ |\ \mu\in 
K^\times, g\in S \}.$$

Take $g\in S-\{I_n\}$ and write it $g=(m_{ij})$. Let $I_g$ be the 
ideal of $k[X_1,\ldots,X_n]$ generated by the polynomials
$$X_i\Big(\sum_jm_{1j}X_j\Big)-X_1\Big(\sum_jm_{ij}X_j\Big)\hskip.8cm 
i=1,\ldots, n.$$

Let $F=\ds\bigcup\V(I_g)$. This is a closed subset of $\A^n$ since 
$S-\{I_n\}$ is finite. Notice that, by construction, $F(K)=\{ v\in 
K^n\ \vert\hbox{ \rm there
exists }g\in S-\{I_n\} \hbox{ and } \lambda\in K \hbox{ such that } 
gv=\lambda v \}$.

Let $U=\A^n\setminus F$. This is a dense open subset of $\A^n$ which 
does not contain 0.

\medskip

We first show that $U(L_s)$ is $\ds\widetilde{G}(L_s)$-stable for every $L/k$.

\medskip

Let $v\in U(L_s)$, and $\mu g\in \ds\widetilde{G}(L_s)$. Assume that 
$(\mu g) v\notin U(L_s)$, that is $(\mu g) v\in F(L_s)$.
Then there exists $\lambda\in L_s$ and $g'\in S-\{I_n\},$ such that 
$g'(\mu g v)=\lambda(\mu g v)$. Since $\ds\widetilde{G}(K)$ acts 
linearly on $K^n$ and
$\mu\neq 0$, we get $g' g v=\lambda gv$, hence $(g^{-1}g' g)v=\lambda 
v$. Let $g^{\prime\prime}\in S$ which represents $\pi_k(g^{-1}g'
g)$. Then there exists $\gamma\in k^\times$ such that $g^{-1}g' 
g=\gamma g^{\prime\prime}$. Then 
$g^{\prime\prime}v=\gamma^{-1}\lambda v$. Since $v\in U(L_s)$, this
implies that $g^{\prime\prime}=I_n$, so $\pi_k(g^{-1}g' 
g)=\overline{I}_n$. Then we get $\pi_k(g')=\overline{I}_n$. By 
construction of $S$, this implies that $g'=I_n$, which is a
contradiction.

\medskip

We now check that the stabilizer of any element of $U(L_s)$ is trivial.

\medskip

Let $v\in U(L_s)$, and let $\mu g\in\ds\widetilde{G}(L_s)$ such that 
$\mu g v= v$. We then have $g v=\mu^{-1} v$, hence $g=I_n$ by 
hypothesis
on $v$. Since $v\in U(L_s)$, we have $v\neq 0$, hence $\mu=1$. This 
implies that $\mu g=I_n$.

\medskip


Thus the action of $\ds\widetilde{G}$ on $\A^n_k$ satisfy the 
conditions of Proposition 1.2. Hence 
$$\ed(\ds\widetilde{G})\leq\dim(\A^n)-\dim(\ds\widetilde{G})=n-1.$$
\enddemo

\bigskip

\heading \S2 Degree $d$ curves and specialization \endheading

\bigskip

Let $k$ be a field and let $d\geq2,n\geq 1$ be two integers. We 
consider $\CC_{d,n}$ the functor of nonzero homogeneous polynomials 
of degree
$d$ in $n$ variables up to a scalar. Elements of $\CC_{d,n}$ are 
called {\bf 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 $\CC_{d,n}^+$ the functor of non-singular degree $d$ 
curves in $n$ variables.

\bigskip

We want to discuss the following general question. Take $C$ a degree 
$d$ curve in $n$ variables and write it down
$\ds C=\sum a_{i_1,\ldots,i_n}X_1^{i_1}\cdots X_n^{i_n}$ (where 
$i_1+\cdots+i_n=d$) for some coefficients $a_{i_1,\ldots,i_n}$ in a 
field
extension of $k$. In general it has $\left({d+n-1\atop n-1}\right)$ 
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.

\bigskip

The group $\GL_n$  acts on $\CC_{d,n}$ as described above by linear 
change of coordinates. More precisely, if $C\in\CC_{d,n}(L)$ and
$\varphi\in\GLrm_n(L)$, define $\varphi(C)$ to be the curve defined 
by $C\circ \varphi$. Since scalar matrices do nothing on curves this
action induces an action of $\PGL_n$ on $\CC_{d,n}$.

\bigskip

We denote by $\FF_{d,n}$ the functor of curves up to this action, and 
sometimes by $[C]$ the class of $C\in\CC_{d,n}(L)$. The action of
$\GL_n$ clearly restricts to $\CC_{d,n}^+$. We then denote by 
$\FF_{d,n}^+$ the functor $\CC_{d,n}^+/\GL_n$. 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.

\bigskip

At this point there is a useful remark to be made. In order to 
compute the essential dimension of $\FF_{d,n}$ one sees that it is 
sufficient to minimize the number of parameters appearing in the most 
general polynomial, that is $C_0=\ds\sum t_IX^I$ where the $t_I$'s 
are algebraically independent variables over $k$. This $C_0$ is 
called the {\bf generic polynomial} of degree $d$ in $n$ variables. 
We will show in the detail that 
$\ed(\FF_{d,n})=\ed([C_0])=\ed(\FF_{d,n}^+)$. In the sequel we will 
often use this fact.

\bigskip

So let $C_0=\ds\sum t_I X^I$ 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 $\ttt=(t_I)$ and 
$K_0=k(\ttt)$.

\bigskip

We begin by a technical lemma, which says in particular that the 
generic cubic $[C_0]$ can be specialized in almost any cubic $[C]$.

\bigskip

\proclaim{Technical Lemma} Let $P\in k[\ttt]$. Let 
$\varphi\in\GL_n(K_0)$, $L/k$ 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 $\psi\in\GL_n(L)$ and a specialization $\ttt\to 
\bbb=(b_I)$ such that:
\smallskip
{\parindent.5cm

\item{1)} $\varphi(\bbb)$ is well-defined and $\varphi(\bbb)\in\GL_n(L)$,

\medskip

\item{2)} $P(\bbb)\neq 0$,

\medskip

\item{3)} $C_0(\bbb)\circ \varphi(\bbb)=C\circ\psi\circ\varphi(\bbb)$.}

\endproclaim

\demo{Proof} Write 
$\varphi=(\varphi_{ij}(\ttt))=\left(\frac{P_{ij}(\ttt)}{Q_{ij}(\ttt)}\right)$ 
and $C_0 \circ \varphi=\displaystyle\sum_I F_I(\hbox{\bf t})X^I$.

Clearly there exists some polynomials with coefficients in $K_0$ in 
$n^2$ variables $T_{ij}$, say $R_I(T_{ij},\ttt)$, such that 
$F_I(\ttt)=R_I(\varphi_{ij}(\ttt), \ttt)$.
In particular, the set of poles of these $F_I$'s is contained is the 
set of zeros of the $Q_{ij}$'s. Hence $F_I(\bbb)$ is well-defined if 
$Q_{ij}(\bbb)\neq 0$ for all $i,j$. Set 
$\det\varphi=\frac{D_1(\ttt)}{D_2(\ttt)}$. The set of poles of 
$\det\varphi$ is contained in the set of zeros of the $Q_{ij}$'s, so 
$\varphi(\bbb)$ is well-defined and invertible if $D_1(\bbb)\neq 0$ 
and $Q_{ij}(\bbb)\neq 0$ for all $i,j$.

\medskip

Let $\psi=(\psi_{ij})\in M_n(L)$, where $\psi_{ij}$ have to be 
determined, and write $C\circ\psi=\displaystyle\sum b_I X^I$.

Clearly $C\circ\psi\circ\varphi=\displaystyle\sum 
R_I(\varphi_{ij}(\ttt), \bbb)X^I$. Since the $b_I$'s are polynomials 
in $\psi_{ij}$ (and in coefficients of $C$), we can write 
$b_I=S_I(\psi_{ij})$ for some polynomials $S_I(T_{ij})\in L[T_{ij}]$.

Set $\hbox{\bf S}=(S_I)$ and let $U(T_{ij})$ be the product of the 
polynomials $Q_{ij}(\hbox{\bf S}), D_1(\hbox{\bf S}), P(\hbox{\bf 
S})$ and $\det(T_{ij})$.
The polynomial $U$ with coefficients in $L$ is non zero, hence there 
exist an element $(\psi_{11},\cdots,\psi_{nn})$ of $L^{n^2}$ such 
that $U(\psi_{ij})\neq 0$, since $L$ is infinite. The specialization 
$t_I\mapsto b_I=S_I(\psi_{ij})$ then satisfies the required 
conditions.
\enddemo

\bigskip

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

\bigskip

\proclaim{Corollary 2.1} Using the above notation the following holds:

\smallskip

{\parindent.5cm

\item{1)} The class of the cubic $[C_0]$ is anisotropic, hence non singular.

\medskip

\item{2)} One has $\ed_k(\FF_{d,n})=\ed_k([C_0])$ and 
$\ed_k(\FF_{d,n}^+)=\ed_k^+([C_0])$, where $\ed_k^+([C_0])$ denotes
the essential dimension of $[C_0]$ viewed as an element of $\FF_{d,n}^+$.

}

\endproclaim

\demo{Proof} 1) Assume that $[C_0]$ is isotropic, and let 
$(P_1(\ttt),\cdots, P_n(\ttt))$ be a nontrivial solution of the 
equation $C_0=0$.
Let $P$ the product of all $P_i$'s which are non zero, let 
$L=k(s_1,\cdots,s_n)$, where the $s_i$'s are independent 
indeterminates over $k$  and let
$C=s_1 X_1^d+\cdots+s_n X_n^d$.
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 \cite{Re}, Theorem 3.2, hence we get a contradiction.

\medskip

2) We only show the first equality. The proof of the second one is 
similar. The inequality $\ed([C_0])\leq\ed_k(\FF_{d,n})$ follows from 
the definition.
Let $L/k$ be a field extension, and take $[C]\in\FF_{d,n}(L)$. We 
have to show that $\ed([C])\leq\ed([C_0])$.

If $L$ is finite, then $L/k$ is algebraic, hence $\ed([C])=0\leq\ed([C_0])$.

Now assume that $L$ is infinite.
Let $K\subset K_0$ be a field extension such that 
$[C_0]=[C'_0]_{K_0}$ for some $ [C_0^\prime]\in\FF_{d,n}(K)$ with 
$\trdeg(K:k)=\ed([C_0])$.
By definition, there exists $\lambda\in K_0^\times$ and 
$\varphi\in\GL_n(K_0)$ such that $\lambda C_0\circ\varphi=C'_0$.
Write $C_0^\prime=\ds\sum F_I(\ttt)X^I$. Notice that $C'_0$ is 
defined over $k(F_I(\ttt))$. By minimality of the transcendence 
degree of $K$, we have 
$$\ed_k([C_0])=\trdeg(K:k)=\trdeg(k(F_I(\ttt)):k).$$ Write 
$\lambda=\frac{P_1}{P_2}$. By the Technical Lemma, there exists a 
specialization $\ttt\to \bbb$ such that $C_0\circ\varphi$ maps to to 
a polynomial $C^\prime$ equivalent to $C$ and $P_1P_2$ maps to a non 
zero element of $L$.
It follows that $\lambda$ maps to a non zero element $\mu$ of $L$, 
hence $C'_0$ maps to $\mu C'$. In particular, $\mu C^\prime=\ds\sum 
F(\bbb)X^I$, so $[\mu C^\prime]=[C]$ is defined over $k(F_I(\bbb))$. 
Since 
$$\ed([C])\leq\trdeg(k(F_I(\bbb)):k)\leq\trdeg(k(F_I(\ttt)):k)=\ed([C_0])$$ 
we then get the result.
\enddemo

\bigskip

\proclaim{Corollary 2.2} We have $\ed_k(\FF_{d,n})=\ed_k(\FF_{d,n}^+)$.
\endproclaim

\demo{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 $\FF_{d,n}$ is the same that its 
essential dimension in $\FF^+_{d,n}$.
\enddemo

\bigskip

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\heading \S3 Some considerations on cubics \endheading

\bigskip


{\bf 0. Warm-up}

\bigskip

Let us come back to our problem. For $d=3$, elements of $\CC_{d,n}$ 
are called {\bf cubics}, and the functor $\FF_{d,n}$ (resp. 
$\FF_{d,n}^+$) is simply denoted by
$\Cub_n$ (resp. by $\Cub_n^+$).

\bigskip

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

\bigskip

\proclaim{Proposition 3.1} Let $k$ be a field. If $\char(k)\neq 3$, 
then $\ed_k(\Cub_2)=1$
\endproclaim


\demo{Proof}
We first show that $\ed_k(\Cub_2)\leq 1$. Let $L/k$ be a field 
extension, and let $C_0$ be the generic cubic polynomial in $2$ 
variables. Write $C_0=t_1X^3+t_2X^2Y+t_3XY^2+t_4Y^3$

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~$1$.


\medskip


Since $t_1\not=0$ one can divide by $t_1$ and obtain 
$X^3+s_2X^2Y+s_3XY^2+s_4Y^3$ where $s_i={t_i\over t_1}$.

Let $\varphi=\pmatrix 1 &-\frac{1}{3}s_2\cr 0& 1\endpmatrix$. Then 
$C\circ\varphi=X^3+ uXY^2+v Y^3$ for some some $u,v\in 
k(s_2,s_3,s_4)$ which are easily computed to be non zero. Let now 
$\varphi'=\pmatrix \frac{v}{u} & 0\cr 0&1\endpmatrix$. Then
$$\frac{u^3}{v^3}C\circ \varphi'=X^3+\frac{u^3}{v^2}XY^2+\frac{u^3}{v^2}Y^3,$$
so $[C]$ is defined over $ k(\frac{u^3}{v^2})$, which has 
transcendence degree at most $1$ over $k$.

\medskip

It remains to show that $\ed_k(\Cub_2)\geq 1$. 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 $\lambda\in k(t)^\times,\varphi\in\GLrm(k(t))$ and a 
polynomial $C'$ with coefficients in $k$, such that
$C=\lambda C'\circ\varphi$. 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 $t\notin k(t)^{\times 3}$. Hence 
$\ed_k(\Cub_2)\geq\ed([C])=1$.
This concludes the proof of the statement.
\enddemo

\bigskip

{\bf 1. Basic facts about cubics in three variables}

\bigskip

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

\medskip

For any field extension $L/k$  and any $\lambda\in L$, let 
$C_\lambda=X_1^3+X_2^3+X_3^3-3\lambda X_1X_2X_3$. We also define 
$C_\infty=X_1 X_2 X_3$. It is easy to see that $C_\lambda$ for 
$\lambda\in L$ is non-singular if and only if $\lambda$ is not a 3rd 
root of unity.

\medskip

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

\medskip

We first begin with the Hessian group $G_{216}$. It plays a crucial 
role in our work. We follow~\cite{BK}.

\medskip

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 $\PGLrm_3(k_s)$ as follows:

\bigskip

Let $x_{00},\cdots,x_{22}$ the nine points of $\P^{2}_{k(\E)}$ defined by:

$$x_{00}=(0,-1,1),\ x_{01}=(0,-\E,1),\ x_{02}=(0,-\E^2,1)$$


$$x_{10}=(1,0,-1),\ x_{11}=(1,0,-\E),\ x_{12}=(1,0,-\E^2)$$


$$x_{20}=(-1,1,0),\ x_{21}=(-\E,1,0),\ x_{22}=(-\E^2,1,0).$$


If $g\in\SA_2(\F_3)$, then $g$ induces a permutation $\sigma_g$ of 
these nine points as follows:

\medskip

If $g(\bar{a},\bar{b})=(\bar{c},\bar{d})$ (where $a,b,c,d\in \{ 0, 1, 
2 \} $), then set $\sigma_g(x_{ab})=x_{cd}$.

\medskip

Computation then shows that there exists a unique element
$\overline{M}_g\in\PGLrm_3(k_s)$ which induces the permutation
$\sigma_g$ 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).

\medskip

The two translations $T_{(2 0)}$ and $T_{(0 2)}$ then correspond
respectively to $\overline{A}$ and $\overline{C}$, where
\medskip
   $A=\pmatrix 0&0&1\cr 1&0&0\cr 0&1&0\endpmatrix $ and  $C=\pmatrix
1&0&0\cr 0&\E&0\cr 0&0&\E^2\endpmatrix$.
\medskip

The three generators $\pmatrix 1&0\cr 1&1\endpmatrix$, $\pmatrix
0&1\cr -1&0\endpmatrix$ and $\pmatrix 2&1\cr 1&1\endpmatrix$
correspond to $\overline{D},\overline{E}$ and $\overline{E'}$, where
\medskip
$D=\pmatrix 1&0&0\cr 0&\E&0\cr 0&0&\E\endpmatrix$, $E=\pmatrix
1&1&1\cr 1&\E&\E^2\cr 1&\E^2&\E\endpmatrix$ and $E'=\pmatrix
\E^2&1&1\cr \E&1&\E\cr\E&\E&1\endpmatrix$.
\medskip
Notice that the set of generators for $\SA_2(\F_3)$ in \cite{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 $4$,
\medskip
but the element $\pmatrix 2&0\cr0&2\endpmatrix$, which corresponds to 
the class of $B=\pmatrix 1&0&0\cr 0&0&1\cr0&1&0\endpmatrix$, has 
order $2$.

Notice that $G_{216}$ is in fact a subgroup of $\PGLrm_3(k(\E))$.

\bigskip

%ZONGOZONGOZONGOZONGOZONGOZONGOZONGOZONGOZONGOZONGOZONGOZONGO

When talking about cubics in three variables it's hard not to mention 
the so called {\bf $j$-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 
$C=\ds\sum_{i_1+i_2+i_3=3}a_{i_1,i_2,i_3}X_1^{i_1}X^{i_2} X_3^{i_3}$ 
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 $C_\lambda$ one has 
$j(C_\lambda)={\lambda^3(\lambda^3+8)^3\over(\lambda^3-1)^3}$ (see 
\cite{BK} p.301-302).

\bigskip

We now recall some results proved in \cite{BK}, p.292-298:

\bigskip

\proclaim{Lemma 3.1} Assume that $k=k_s$. Then:

\medskip

{\parindent.7cm

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

\bigskip

\item{2)} Let $\lambda\in k\cup\{\infty\}$. For any
$\overline{\varphi}\in\PGLrm_3(k)$,
$\overline{\varphi}$ maps $C_\lambda$ to some $C_\mu$ if and only if
$\overline{\varphi}\in G_{216}$.

\bigskip

\item{3)}  Let $\lambda\in k\cup\{\infty\}$. For any 
$\overline{\varphi}\in\PGLrm_3(k)$,
$\overline{\varphi}$ maps the cubic $C_\lambda$ to itself if and only if
$\overline{\varphi}$ belongs to the subgroup  $H=\langle 
\overline{A},\overline{B},\overline{C}\rangle$.

  }


\endproclaim


\bigskip

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, $C_\lambda$ is not 
supposed to be non-singular.

\bigskip


{\bf 2. Canonical pencils of cubics}

\bigskip

If $C$ is a cubic polynomial in $3$ variables with coefficients in 
$L$, let $H_C=\det\left(\frac{\partial^2 C}{\partial X_i\partial 
X_j}\right)$,
and let $\FFF{C}$ be the set of cubics of the form $\alpha C+\beta 
H_C$, for some $\alpha,\beta\in L$. The set $\FFF{C}$ is called {\bf 
the canonical pencil associated to $C$}.
Since $H_{\alpha C}=\alpha^3 H_C$ for any $\alpha\in L^\times$, this 
set does only depend on the cubic defined by $C$.

\medskip

Let $\hbox{\ptitgot P}(L)$ denote the set $\{ \FFF{C}\ |\ C\in\CC_{3,3}(L) 
\}.$ For a cubic $C$ over a field $L$ and for any $k$-morphism $L\to 
L'$ we define a map $\hbox{\ptitgot P}(L)\freccia\hbox{\ptitgot 
P}(L')$ by sending the pencil $\FFF{C}$ to the pencil $\FFF{C_{L'}}$. 
We then obtain a functor $\hbox{\ptitgot P}:\Ck\freccia\Sets$.
The association $C\mapsto\FFF{C}$ gives rise to a surjective map of 
functors $\xymatrix{\CC_{3,3}\ar@{->>}[r]&\hbox{\ptitgot P}}$. Let 
now act the group
$\GL_3$ naturally on $\hbox{\ptitgot P}$ as follows: for 
$\varphi\in\GLrm_3(L)$ and $C\in\CC_{3,3}(L)$ we set 
$\varphi(\FFF{C})=\FFF{\varphi(C)}$.

\bigskip

We say that $\FFF{C}$ and $\FFF{C'}$ are {\bf isomorphic over} $L$ if 
they are in the same orbit under this action. We denote by
$[\FFF{C}]$ the isomorphism class of $\FFF{C}$ and we denote by 
$\Pen_3$ the functor of isomorphism classes of such pencils.

\bigskip

\proclaim{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 $\FFF{C}$ induces a well defined morphism of functors 
$\Cub_3\freccia\Pen_3$.
\endproclaim

\demo{Proof} The statement follows from the formula $H_{C\circ 
\varphi}=(\det \varphi)^2 H_C\circ \varphi$. The proof of this 
formula is left to the reader. \enddemo

\bigskip

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.

\bigskip

\proclaim{Lemma 3.3} Assume that char$k\neq 2,3$ and
let $L/k$ be a field extension. For any $\lambda\in L,\lambda^3\neq 
1$, we have $$\FFF{C_\lambda}=\big\{ C_\mu\ |\ \mu\in L\cup\{\infty\} 
\big\}.$$
In particular, for all $C,C'\in\Cub_3^+(L_s)$, the pencils $\FFF{C}$ 
and $\FFF{C'}$ are isomorphic.
\endproclaim

\demo{Proof} It is easy to see that 
$H_{C_\lambda}=-54\lambda^2(X_1^3+X_2^3+X_3^3)-3(18\lambda^3-72)X_1X_2X_3$, 

hence we get $$\alpha C_\lambda+\beta 
H_{C_\lambda}=(\alpha-54\lambda^2\beta)(X_1^3+X_2^3+X_3^3)-3(\alpha\lambda+18\lambda^3\beta-2\beta)X_1X_2X_3.$$
Let $\mu\in L$. If $\mu=\lambda$, take $\alpha=1$ and $\beta=0$. 
Assume now that $\mu\neq\lambda$. Take $\beta=1$ and
$\alpha=\ds\frac{72-54\lambda^2\mu-18\lambda^3}{\lambda-\mu}$.

We claim that $\alpha-54\lambda^2\neq 0$. Indeed, assume the 
contrary. Then we easily get that $72(1-\lambda^3)=1$. Since 
$\char(k)\neq 2,3$, this implies that $\lambda^3\neq 1$, which is not 
the case.

\medskip

Thus, with these choices of $\alpha$ and $\beta$, we get $\alpha 
C_\lambda+\beta H_{C_\lambda}=(\alpha-54\lambda^2)C_\mu$, hence
the polynomials $\alpha C_\lambda+\beta H_{C_\lambda}$ and $C_\mu$ 
belong to the same class.

If $\mu=\infty$, take $\alpha=-\ds\frac{\lambda^2}{4(\lambda^3-1)}$ 
and $\beta=-\ds\frac{1}{216(\lambda^3-1)}$.
\enddemo

\bigskip

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

\bigskip

We will denote by $\hbox{\ptitgot P}^+$ and $\Pen_3^+$ the 
corresponding functors. This little restriction does not matter for 
the computation of
essential dimension for we have seen that $\ed(\Cub_3)=\ed(\Cub^+_3)$.

\bigskip

\proclaim{Lemma 3.4}Let $L/k$ be a field extension and let 
$C\in\Cub_3^+(L).$ Then 
$$\ed([\FFF{C}])\leq\ed([C])\leq\ed([\FFF{C}])+1.$$
\endproclaim

\demo{Proof} Let $K/k$ such that $\trdeg(K:k)=\ed_k([C])$. Then 
clearly $\FFF{C}$ is defined over $K$, hence 
$\ed([\FFF{C}])\leq\ed([C])$.
Assume now that $\ed([\FFF{C}])=n$. Then there exists a field 
extension $E/k$ of transcendence degree equal to $n$, and 
$C'\in\CC_{3,3}(E)$
such that $\FFF{C}=\FFF{C'_K}$. By definition, there exists 
$\varphi\in\GL_3(K)$ such that $\FFF{\varphi(C)}=\FFF{C'_K}$.
In particular, there exists $\alpha,\beta\in K$ such that the 
polynomials $C\circ \varphi$ and $\alpha C'+\beta H_{C'}$ are 
proportional.
Hence $[C]=[\alpha C'+\beta H_{C'}]$. Since $\alpha$ or $\beta$ is 
non zero, $C$ is then defined over $ E(\frac{\alpha}{\beta})$ or 
$E(\frac{\beta}{\alpha})$.
Thus $C$ is defined over a field of transcendence degree at most $n+1$.
\enddemo

\bigskip

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\heading \S4 Galois descent for functors. Applications to cubics \endheading

\bigskip

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 \cite{BOI}, Proposition (29.1). This lemma 
will apply to our situation.

\bigskip

Let $k$ be any field, and let $\FF$ be an object of $\Sk$. We denote by $\Aut(\FF)$ the functor 
defined by $$\Aut(\FF)(L)=\{\eta:\FF_L\freccia\FF_L\ |\ \eta \hbox{ 
is an isomorphism of functors}\}$$ for any $L/k$. Notice that for any 
extension $L/k$, the action of the absolute Galois group $\Gamma_L$ 
on $L_s$ induces an action on $\FF(L_s)$ by functoriality.

\bigskip

Let $G$ be a group-valuated functor and $\rho:G\freccia\Aut(\FF)$ be 
a morphism of group-valuated functors which is $\Gamma$-equivariant.
For each $E/k$ we define an equivalence relation on $F(E)$ saying 
that $b,b'\in F(E)$ are equivalent if there exists $g\in G(E)$ such 
that $\rho_E(g)(b)=b'$. We note this by $b\sim_E b'$.

Let $k'/k$ be a field extension, and $a\in\FF(K)$. For every extension 
$L/k'$ set $$X(L)=\{ b\in\FF(L)\ |\ b\sim_{L_s}a\}.$$

Denote by $\Stab_G(a)$ the subfunctor of $G$ defined by 
$$\Stab_G(a)(L)=\{ g\in G(L)\ \vert\ \rho_L(g)(a_L)=a_L \}$$ for any
extension $L/k'$. This is a group valuated subfunctor of $G_K$.

\bigskip

Finally, we denote by $\FF_a(L)$ the set of equivalence classes of 
elements of $X(L)$ under the relation $b\sim_L b'$. This defines an
object of $\Sko$, denoted by $\FF_a$.


\bigskip

We now state the Galois descent lemma:

\bigskip

\proclaim{Galois Descent Lemma} Let $\rho:G\freccia\Aut(\FF)$ as 
above. Assume that for any $L\in \Ck$, the following conditions hold:

{\parindent.5cm
\smallskip
\item{1)} $H^1(L,G(L_s))=1$
\smallskip
\item{2)} $\FF(L_s)^{\Gamma_L}=\FF(L)$ and $G(L_s)^{\Gamma_L}=G(L)$.}
\smallskip

Then for any $k'/k$ and for any $a\in\FF(k')$, there is a natural
isomorphism of functors of $\Sko$

$$\FF_a{\ds\tilde{\freccia}} H^1(_-,\Stab_G(a)).$$

Moreover, this isomorphism maps the class of $a_L$ to the base point 
of $H^1(L,\Stab_G(a)(L_s))$.
\endproclaim

\demo{Proof} We fix once for all an extension $k'/k$ and an element 
$a\in \FF(k')$.

Let $L/k'$ an extension of $k'$. For the proof we will denote by 
$\Gamma$ instead of $\Gamma_L$ 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 
$\ker(H^1(L, A)\freccia H^1(L,B))$ and the orbit set of the group 
$B^{\Gamma}$ in $(B/A)^{\Gamma}$ (see \cite{BOI}, Corollary 28.2 for 
example).

Since the group $G(L_s)$ acts transitively on $X(L_s)$, the 
$\Gamma$-set $X(L_s)$ can be identified with the set of left cosets 
of $G(L_s)$ modulo $\Stab_G(a)(L_s)$, hence $B/A\simeq X(L_s)$. By 
assumption on $\FF$, the set $(B/A)^{\Gamma}$ is then equal to 
$X(L)$. Moreover, $B^{\Gamma}=G(L_s)^{\Gamma}=G(L)$. It follows that 
the
orbit set of $B^{\Gamma}$ in $(B/A)^{\Gamma}$ 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 
$\FF_a(L)$. The functoriality is left to the reader.

\enddemo

\bigskip

{\bf Example 4.1. }Assume that char$k\neq 2,3$. Take $\FF=\hbox{\ptitgot P}^+$ and let the group 
$G=\GL_3$ act on $\hbox{\ptitgot P}^+$. Take $\lambda\in k$ 
with $\lambda^3\neq 1$ and set $a=\FFF{C_\lambda}$. Then Lemma 3.3 
tells us that $\FF_a(L)=\Pen^+_3(L)$ for any extension $L/k$.

\medskip

We now determine the stabilizer of the pencil $\FFF{C_\lambda}$. 
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 $k$.

\medskip

We first compute the image of this group scheme by the natural 
projection $\pi:\GL_3\to \PGL_3$. We have $\pi(\Stab_G(a))(k_s)=\{ 
\overline{\varphi}\in\PGL_3(k_s)\ |\ 
\varphi(\FFF{C_\lambda})=\FFF{C_\lambda} \}$.

\medskip

Since over $k_s$ the pencil $\FFF{C_\lambda}$ is equal to $\{ C_\mu\ 
|\ \mu\in k_s\cup\{\infty\} \}$ it follows that any $\varphi$ in 
$\Stab_G(a)(k_s)$ maps $C_\lambda$ to some $C_\mu$. So the same holds 
for $\overline{\varphi}$ and hence $\overline{\varphi}$ belongs to 
$G_{216}\subseteq\PGLrm_3(k(\E))$ by Lemma 3.1. Conversely, if 
$\overline{\varphi}\in G_{216}$, then it is clear that 
$\overline{\varphi}\in\pi(\Stab_G(a))(k_s)$ by Lemma 3.1 again.
Hence $\pi(\Stab_G(a))(k_s)=G_{216}$. Since $\Gamma_k$ acts 
continuously on $G_{216}$, the group $\pi(\Stab_G(a))$ is
then the \'{e}tale group scheme $G_{216,\hbox{\ptitrm \'{e}t}}$ (see 
\cite{BOI}, Proposition 20.16). Thus 
$$\Stab_G(\FFF{C_\lambda})=\pi^{-1}(G_{216,\hbox{\ptitrm \'{e}t}}).$$

\medskip

We denote this last group by $\ds\widetilde{G}_{216}$.
Since the hypotheses of the Galois Descent Lemma are clearly fulfilled, we get

$$\Pen^+_3\simeq H^1(_-,\ds\widetilde{G}_{216}).$$
In particular, $\ed_k(\Pen_3^+)=\ed_k(\ds \widetilde{G}_{216})$.

\bigskip


{\bf Example 4.2.} Assume that char$k\neq 3$. Take $\FF=\CC_{3,3}^+$ and let $G=\GL_3$ act on it 
as usual. Let $k'/k$ be a field extension and take $a=C_\lambda$ for 
some $\lambda\in k'$ with $\lambda^3\neq 1$. Then $\FF_a(L)$ is 
the set of cubics in $L$ which are equivalent to $C_\lambda$ over 
$L_s$.
Arguing as previously, one can see that $\Stab_G(a)$ is the algebraic 
group $k'$-scheme $\pi^{-1}(H_{\hbox{\ptitrm \'{e}t}})$,
where $H$ is the subgroup of $G_{216}$ described in Lemma 3.1. We 
will denote it by $\ds\widetilde{H}$.
Hence, for any field extension $k'/k$, for any $\lambda\in k',\lambda^3\neq 1$, and for 
any field extension $L/k'$, we have a one-to-one correspondence
$$\FF_a(L)=\{ [C]\in\Cub_3^+(L)\ |\  C\sim_{L_s} C_{\lambda}\}\simeq 
H^1(L,\ds\widetilde{H}).$$

Hence $\ed_{k'}(\FF_a)=\ed_{k'}(\ds\widetilde{H})$. Again we have 
classified cubics which become isomorphic to a fixed cubic by a 
Galois cohomology set.

\bigskip

\heading \S5 Essential dimension of cubics\endheading

\bigskip

We can finally state and prove our main result:

\bigskip

\proclaim{Theorem 5.1}Let $k$ be a field. Assume that $\char(k)\neq 
2,3$. If $k$ contains $\E$, then
$$\ed_k(\Cub_3)=3.$$ In particular, $\ed_k(\Cub_3)\geq 3$ for any field $k$.
\endproclaim


\bigskip

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 $\ed_{k(\E)}(\FF)\leq\ed_k(\FF)$. We will restrict 
ourselves to the functor $\Cub^+_3$ since we know by \S2 that 
$\ed(\Cub_3)=\ed(\Cub^+_3)$.


\bigskip

By Example 4.1, we have 
$\ed_k(\Pen^+_3)=\ed_k(\ds\widetilde{G}_{216})$. Since $\E\in k$, the 
group $\Gamma_k$ acts trivially on
$G_{216,\hbox{\rm \'{e}t}}(k_s)$, hence $G_{216,\hbox{\rm \'{e}t}}$ 
is the constant algebraic group $G_{216}$. Applying Proposition 1.3 
with $G=G_{216}$ and $n=3$ then gives
$\ed_k(\ds\widetilde{G}_{216})\leq 2$. Lemma 3.4 implies in 
particular that $\ed_k(\Cub^+_3)\leq\ed_k(\Pen_3^+)+1$, hence 
$\ed_k(\Cub_3^+)\leq 3$.
Notice that $\ds\widetilde{G}_{216}$ contains the constant subgroup $(\Z/3Z)^3$ generated by $\E I_3,C$ and $D$, so 
using Proposition 1.1, we get $\ed_k(\ds\widetilde{G}_{216})=\ed_k(\Pen_3^+)=2$. 
\bigskip

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

\bigskip

Let $k'/k$ be a field extension and $\lambda\in k'$ with 
$\lambda^3\neq 1$. We define an object $\FF_\lambda$ of $\Sko$ as 
follows.
If $L/k'$ is a field extension, set $$\FF_\lambda(L/k')=\{ 
[C']\in\Cub_3^+(L)\ \vert \hbox{ there exists }E/L\hbox{ such that } 
C'\ds\sim_E C_\lambda \}.$$
Notice that $\FF_\lambda(L/k')=\{ [C']\in\Cub_3^+(L)\ \vert\ C'\ds\sim_{L_s} 
C_\lambda \}$.
Indeed, let $[C']\in\Cub_3^+(L)$ such that $C'\ds\sim_E C_\lambda$, 
for some $E/L$.
We then have $j(C'_E)=j({C_{\lambda}}_E)$, hence 
$j([C^\prime])=j({C_{\lambda}}_L)$. Thus 
$j(C'_{L_s})=j({C_\lambda}_{L_s})$, so $ C'\ds\sim_{L_s} C_\lambda$.
The reverse inclusion is clear. Example 4.2 then shows that 
$\FF_\lambda\simeq H^1(_-,\ds\widetilde{H})$ and thus 
$$\ed_{k'}(\FF_\lambda)=\ed_{k'}(\ds\widetilde{H}).$$ This means in 
particular that the essential dimension of $\FF_\lambda$ does not 
depend on $\lambda$.

\bigskip

Our next task is to compute the essential dimension of $\FF_\lambda$, 
that is the essential dimension of $\ds\widetilde{H}$.
Precisely, we will show the following result:

\medskip

\proclaim{Proposition 5.1}
Let $k'$ be a field of characteristic different from $2$ and $3$ containing 
$\mu_3$. Then $\ed_{k'}(\ds\widetilde{H})=2$.
\endproclaim

Let $S$ be the constant subgroup of $\PGL_3$ isomorphic to 
$(\Z/3\Z)^2$ generated by $\overline{A}$ and $\overline{C}$, and let 
$\ds\widetilde{S}=\pi^{-1}(S)$.

Clearly, we have the following exact sequence of group schemes:

$$1\freccia\Gm\freccia\ds\widetilde{S}\freccia S\freccia 1,$$
hence $\dim(\ds\widetilde{S})=1$. Similarly, we have 
$\dim(\ds\widetilde{H})=1$. Applying Proposition 1.3 we get 
$\ed_{k'}(\ds\widetilde{H})\leq 2$.
We now prove the reverse inequality.

By Proposition 1.1. we have 
$\ed_{k'}(\ds\widetilde{H})\geq\ed_{k'}(\ds\widetilde{S})$. Since 
$\mu_3\in k'\subseteq L$,  we can identify the algebraic group $S$ 
with $\mu_3\times\mu_3$, where the identification is given on the 
$L$-points by mapping $\overline{A}^m\overline{C}^n\in S(L)$ to 
$(\E^m,\E^n)\in\mu_3(L)\times\mu_3(L)$. We then have
$$1\freccia\Gm\freccia\ds\widetilde{S}\freccia \mu_3\times\mu_3\freccia 1,$$
where the second map is given by
$$\ \ds\widetilde{S}(L)\freccia\mu_3(L)\times\mu_3(L)$$
$$\hskip-30pt\lambda A^m C^n\mapsto (\E^n,\E^m).$$
Moreover, for any field extension $L/k'$, the above exact sequence 
induces the following exact sequence in cohomology:
$$H^1(L,\ds\widetilde{S})\freccia H^1(L,\mu_3)\times 
H^1(L,\mu_3)\freccia H^2(L,\Gm).$$

Recall that $H^1(L,\mu_3)$ is in one-to-one correspondence with 
$L^\times/L^{\times 3}$ as follows:

\medskip

For $aL^{\times 3}\in L^\times/L^{\times 3}$, let $\alpha\in L_s$ 
such that $\alpha^3=a$. Then the map $c_a:\Gamma_L\to\mu_3$ defined 
by  $\sigma\mapsto\frac{\sigma(\alpha)}{\alpha}$ is a 1-cocycle, and 
its cohomology class does not depend on the choice of $\alpha$ and 
$a$. We will write $(a)_3$ the class of the correponding cocycle.

\bigskip

\proclaim{Lemma 5.1} For any field $L$, the connecting map 
$$\partial:H^1(L,\mu_3)\times H^1(L,\mu_3)\to H^2(L,\Gm)$$ is defined 
by
$\partial((a)_3,(b)_3)=-(a)_3\cup (b)_3$, where $\cup$ is the 
cup-product associated to the natural pairing 
$\mu_3(L)\times\mu_3(L)\to\mu_3(L)$ defined by
$(\E^m,\E^n)\mapsto\E^{mn}$.
\endproclaim

\demo{Proof} Let $(a)_3,(b)_3\in H^1(L,\mu_3)$. Let $a,b\in L^\times$.
If $\sigma\in \Gamma_L$, write $c_a(\sigma)=\E^{m_\sigma}$ and 
$c_b(\sigma)=\E^{n_\sigma}$ for some $m_\sigma,n_\sigma\in \{ 0,1,2 
\}$.
Since $\Gamma_L$ acts trivially on $\mu_3\times\mu_3$, the element 
$\partial((a)_3,(b)_3)$ is the class of the $2$-cocycle
$$\alpha:\sigma,\tau\in\Gamma_L\times\Gamma_L\mapsto\beta_\sigma 
\sigma\beta_\tau\beta_{\sigma\tau}^{-1},$$ where $\beta_\sigma$ is 
any preimage of
$(c_a(\sigma),c_b(\sigma))$.

\medskip

If $(c_a(\sigma),c_b(\sigma))=(\E^{m_\sigma},\E^{n_\sigma})$, we 
choose $\beta_\sigma=A^{m_\sigma}C^{n_\sigma}$. Notice that we have 
$\sigma\beta_\tau=\beta_{\tau}$ for any
$\sigma,\tau\in\Gamma_L.$
We then have 
$\alpha_{\sigma,\tau}=A^{m_\sigma}C^{n_\sigma}A^{m_\tau}C^{n_\tau}C^{-n_{\sigma\tau}}A^{-m_{\sigma\tau}}$. 
Since $CA=\E AC$, we get $$\alpha(\sigma,\tau)=\E^{n_\sigma 
m_\tau}A^{m_\sigma+m_\tau}C^{n_\sigma+n_\tau-n_{\sigma\tau}}A^{-m_{\sigma\tau}}.$$

The fact that $c_a$ and $c_b$ are cocycles and that $\Gamma_L$ acts 
trivially on $\mu_3(L_s)$ implies that 
$m_\sigma+m_\tau-m_{\sigma\tau}$ and $n_\sigma+n_\tau-n_{\sigma\tau}$ 
are divisible by $3$.
Hence we get 
$$\alpha:(\sigma,\tau)\in\Gamma_L\times\Gamma_L\mapsto\E^{m_\sigma 
n_\tau}\in\mu_3(L),$$ which is precisely a cocycle representing
$(b)_3\cup (a)_3$ since $\Gamma_L$ acts trivially on $\mu_3(L_s)$. 
The conclusion then follows from the equality $(b)_3\cup 
(a)_3=-(a)_3\cup(b)_3$.
\enddemo

\medskip

We then have a surjection of functors 
$\partial:H^1(_-,\ds\widetilde{S})\to\NN$, where $\NN$ is the object 
of $\Sko$ defined by
$$\NN(L)=\left\{ \big((a)_3,(b)_3\big)\in H^1(L,\mu_3)\times 
H^1(L,\mu_3)\ \vert\  (a)_3\cup (b)_3=0 \right\}$$ for any field 
extension $L/k'$.
Hence, by Lemma 1.1 we get $$\ed_{k'}(\ds\widetilde{S})\geq\ed_{k'}(\NN).$$
To conclude the proof of Proposition 5.1, it suffices to prove the following:

\bigskip

\proclaim{Lemma 5.2} We have $\ed_{k'}(\NN)\geq 2$.
\endproclaim

\demo{Proof} It suffices to show the inequality when $k'$ is 
algebraically closed.

\medskip

Let $((a)_3,(b)_3)\in\NN(L)$ for some $L/k'$. Let's consider the 
cubic in $4$ variables $$C_{a,b}=X^3+a Y^3+ b Z^3+ ab T^3.$$  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 
$\NN\freccia\Cub_4$.
Notice that $((a)_3,(b)_3)\in\NN(L)$ if and only if $a$ is a norm of 
the extension $L(\beta)/L$, where $\beta^3=b$.
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 $a=N_{L(\beta)/L}(1+s\beta+u\beta^2)$, hence 
$((a)_3,(b)_3)\in\NN(L)$.

\medskip

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 $L$.

\medskip

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

\bigskip

\proclaim{Lemma 5.3} The polynomial $C_{a,b}$, with 
$a=1+ts^3+t^2u^3-3stu$ and $b=t$ is anisotropic over $k'(s,t,u)$.
\endproclaim

\demo{Proof} Assume the contrary. Then there exists 
$P_1,\cdots,P_4\in k'[s,t,u]$ not all zero, such that $P_1^3+a 
P_2^3+b P_3^3+ab P_4^3=0$.
Consider $P_1,\cdots,P_4$ 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 
$Q_i\neq 0$.
If $n_1\leq n_2,n_3,n_4$, 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 $Q_1^3+t a^{3(n_3-n_1)}Q_3^3=0$ in $E=k'(s,u)[T]/(a)$.

\medskip

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 $n_2-n_1\neq n_4-n_1$, 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~$E$.

\medskip

By definition, we have $E=k'(s,u)(\sqrt{\Delta})$, where 
$\Delta=(s^3-3su)^2-4u^3$, and 
$t=\ds\frac{3su-s^3\pm\sqrt{\Delta}}{2u^3}$.
Then, if $t$ is a cube in $E$ then $12su-4s^3\pm 4\sqrt{\Delta}$ is a 
cube in $E$.
Let denote by $\tau$ the unique non trivial $k'(s,u)$-automorphism of 
$E$. An element $\lambda\in E$ is a cube if and only if 
$\tau(\lambda)$ is a cube.
So $\lambda=12su-4s^3+4\sqrt{\Delta}$ is a cube.

\medskip

Write $\lambda=\mu^3$, with $\mu=\frac{R_1+R_2\sqrt{\Delta}}{R_3}$, 
where $R_1,R_2,R_3\in k'[s,u]$. One can assume that $R_1,R_2,R_3$ are 
relatively prime.
We have $N_{E/k(s,u)}(\lambda)=16(3su-s^3)^2-16\Delta=64u^3$, hence 
$$64 R_3^6 u^3=(R_1^2-R_2^2\Delta)^3.$$

\medskip

We also have $(12su-4s^3+4\sqrt{\Delta})R_3^3= R_1^3 +3R_1 
R_2^2\Delta+(R_2^3\Delta+3R_1^2 R_2)\sqrt{\Delta}$, hence
$$4R_3^3=R_2^3\Delta+3R_1^2 R_2 \hbox{ and } (12su-4s^3)R_3^3=R_1^3+3R_1 
R_2^2\Delta.$$

We now show that $R_3$ is a constant.
Assume that $S$ is an irreducible divisor of $R_3$.
Then $S$ divides $R_1^2-R_2^2\Delta$ by the first equation, hence $S$ 
divides $3R_1^3-3R_1R_2^2\Delta$. The third equation implies
that $S$ divides $R_1^3+3R_1R_2^2\Delta$, 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 
$3R_2^2\Delta$. Consequently, $S$ divides $R_2^2$ (even if 
$S=\Delta$), so
$S$ divides $R_2$. This is impossible since $R_1,R_2,R_3$ are relatively prime.
Hence, one can assume that $R_3=1$.

\medskip

We then have the equations $$64u^3=(R_1^2-R_2^2\Delta)^3, 
4=R_2^3\Delta+3R_1^2R_2, 12su-4s^3=R_1^3+3R_1R2^2\Delta.$$

The second equation then implies that $R_2$ and $R_2^2\Delta+3R_1^2$ 
are non zero constant polynomials.
In particular $R_1^2$ and $\Delta$ has same degree in $u$. This gives 
a contradiction since these degrees don't have the same parity.

\enddemo

\bigskip

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

\bigskip

Let $t$ be an indeterminate over $k$, let $\overline{k(t)}$ be an 
algebraic closure of $k(t)$. Let $i$ be the composite $k\to 
k(t)\to\overline{k(t)}$, where the first map is the natural 
inclusion, and $k(t)\to\overline{k(t)}$ is a fixed $k$-linear 
morphism which maps $t$ to itself. In the sequel we set 
$k'=\overline{k(t)}$.
Let $\lambda\in k'$ such that $j(C_\lambda)=t$ and consider the 
functor $\FF_\lambda$. By Lemma 1.2, we have 
$$\ed_k(i^*\FF_\lambda)=\ed_{k'}(\FF_\lambda)+\trdeg(k':k).$$
By Proposition 5.1, we have $\ed_{k'}(\FF_\lambda)=2$. Moreover we have 
$\trdeg(k':k)=1$, so we get
$$\ed_k(i^*\FF_\lambda)=3.$$
Thus there exists a field extension $L/k$ with $\trdeg(L:k)=3$ and 
and an element $x=(f,[C])\in (i^*\FF_\lambda)(L/k)$ which can not be 
defined over a subextension of $L$ of a smaller transcendence degree. 
By definition of $i^*\FF_\lambda(L/k)$, the extension $L/k$ is the 
composite $f\circ i$. Since $[C]\in F_\lambda(L/k')$, then in 
particular $[C]\in\Cub_3^+(L/k')$, so $[C]$ can be viewed as an 
element of $\Cub_3^+(L/k)$ via $i$.

\bigskip

We now proceed to show that $\ed([C])=3$ in $\Cub_3^+$.


\bigskip

Assume that there exists a subextension $K'/k$ of $L/k$ with 
$\trdeg(K':k)\leq 2$ and $[C']\in\Cub_3^+(L)$ such that $[C']_L=[C]$.
\smallskip
Let $\varphi:k\freccia L,\psi:k\freccia K'$ and $\theta:K'\freccia L$.
Since $K'/k$ is a subextension of $L/k$, we then have $\theta\circ\psi=\varphi$.
\smallskip

Since $[C]\in\FF_\lambda(L/k')$, we have 
$j(C'_L)=j(C)=j({C_\lambda}_L)=f(t)$, hence $j(C'_L)$ is 
transcendental over $k$.
Consequently, $j(C')\in K'$ is transcendental over $k$ and we can 
define a morphism of $k$-extensions $\beta:k(t)\to K'$ by $\beta(t)=j(C')$ and $\beta_{\vert k}=\psi$.


\bigskip

We now check that the composite maps $\eta_1:k(t)\freccia k'\freccia L$ and $\eta_2:k(t)\freccia K'\freccia L$ are 
the same.
\smallskip
Let $\alpha\in k$.  
We have $\eta_1(\alpha)=f(\alpha)$ and $\eta_2(\alpha)=\theta(\psi(\alpha))=\varphi(\alpha)$.
By definition of $L/k$, we have $\varphi=f\circ i$ (since $L/k$ factors through $i$).
Hence $\eta_2(\alpha)=f(i(\alpha))=f(\alpha)$.
Moreover, since we have $\eta_1(t)=f(t)$ and $\eta_2(t)=\psi(j(C'))=j(C'_L)$, the maps $\eta_1$ and $\eta_2$ coincide.

\smallskip

Hence we can define the compositum $E/k(t)$ of $k'/k(t)$ 
and $K'/k(t)$ in $L/k(t)$. 
\bigskip

By definition of $E/k(t)$, the composite map $K'\freccia E\freccia L$ is equal to 
$\theta$, so we have $([C']_E)_L=[C']_L=[C]$.
Now consider the extension $E/k$ defined by $k\freccia 
k'\freccia E$.
By construction, this extension factors through $i$. Let $g:k'\freccia E$.
We then have $(g,[C']_E)\in (i^*\FF_\lambda)(E/k)$.
Moreover, by definition of $E/k(t)$ again, the map $k'\freccia E\freccia L$ is precisely $f$, hence $(g,[C']_E)_L=(f,[C])$.

Since $\trdeg(K':k)\leq 2$ and $\trdeg(k(t):k)=1$, we have 
$\trdeg(K':k(t))\leq 1$.
Moreover, $\trdeg(k':k(t))=0$, hence $\trdeg(E:k(t))\leq 
1$, so $\trdeg(E:k)\leq 2$.
Consequently, $x=(f,[C])$ is defined over a subextension of $L/k$ of 
transcendence degree at most $2$, which is impossible by choice of 
$x$.

\bigskip

We then get $\ed_k(\Cub_3^+)\geq\ed([C])=3$, which concludes the 
proof of Theorem 5.1.


\vskip1cm

{\bf REFERENCES}

\vskip1cm


\smallskip
[BeF]G.Berhuy, G. Favi Essential dimension: a functorial point of 
view. Preprint (2003) Available on ...
\smallskip
[BK] E. Brieskorn, H. Kn\"{o}rrer, Plane Algebraic Curves, 
Birkha\"{u}ser Verlag, Basel, Boston,Stuttgart (1986)
\smallskip
[BOI] M.-A. Knus, A.S. Merkurjev, M. Rost, J.-P. Tignol, The Book of 
Involutions, AMS Coll. Pub. vol. 44  (1998)
\smallskip
[Re] Z. Reichstein, On a Theorem of Hermite and Joubert.  Canadian J. 
of Mathematics, {\bf 51}(1), 69-95 (1999)
\bigskip


Gr\'{e}gory Berhuy

 Ecole Polytechnique F\'{e}d\'{e}rale de Lausanne


 MA-Ecublens


 CH-1015 Lausanne SWITZERLAND 

\medskip

Email:gregory.berhuy\@epfl.ch 

\bigskip

Giordano Favi

Institut de Math\'{e}matiques

Universit\'{e} de Lausanne

Dorigny


CH-1015 Lausanne SWITZERLAND
\medskip

Email: giordano.favi\@ima.unil.ch

\end

