\section{The case of flexes} \label{flexcar3}  \label{open}

The heuristic results and computations of \cite{FOR} tend to suggest that a random plane smooth quartic over $\FF_q$ has an asymptotic probability of about $0.63$ to have at least one rational flex when $q$ tends to infinity. We describe how to turn the heuristic strategy into a proof.\\

Let $\PP^{14}$ be the linear system of all plane quartic curves over a field $K$ and $I_0=\{(P,l), \; P \in l\} \subset \PP^2 \times {\PP^2}^*$. Let $I_4 \subset \PP^{14} \times I_0$ be the locus
$$I_4=\{(C,(P,l)), \; C \; \textrm{is smooth and} \; P \; \textrm{is a flex of} \, C\, \textrm{with tangent line} \; l\}.$$   

 Harris proved the following result using monodromy arguments.
\begin{theorem}[{\cite[p.698]{harris-enu}}]
The Galois group of the cover $I_4 \to \PP^{14}$ over $\CC$ is the full symmetric group $S_{24}$.
\end{theorem}
\noindent Let us assume for a moment that this result is still valid over finite fields. Then, using a general Chebotarev density theorem for function fields like in \cite[Th.7]{serre-chebo}, this would mean that the probability of finding a rational flex is within $O(1/\sqrt{q})$
of the probability that a random permutation 
of $24$ letters has a fixed point, which is 
$$p_{24}:=1-\frac{1}{2!} + \frac{1}{3!} - \ldots -\frac{1}{24!} \approx 1-\exp(-1) \approx 0.63.$$
Unfortunately, it is not easy to transpose Harris's proof over any field. And actually, Harris's result is not true in characteristic $3$, as the following proposition implies.

\begin{proposition}
Let $C\,:\, f(x_1,x_2,x_3)=0$ be a smooth plane quartic defined over an algebraically closed field $K$ of characteristic $3$. The flexes of $C$ are  the intersection points of $C$ with a certain curve $H_C : h_C=0$ of degree less than or equal to $2$. The curve $H_C$ can be degenerate as in the case of the Fermat quartic where $h_C = 0$.
\end{proposition}

\begin{proof}
We use the method to compute flexes of a plane curve of degree $d$ described  in the appendix of \cite{FOR} (see also \cite[Th.0.1]{stvo2}). Indeed, when the characteristic of the field divides $2(d-1)$, one cannot use the usual Hessian and one should proceed  as follows.\\
Let  $C : f=0$ be the generic plane quartic over $K$
\begin{eqnarray}
f(x,y,z) &:=& a_{00}y^4+y^3(a_{10}x+a_{01}z)+y^2(a_{20}x^2+a_{11}xz+a_{02}z^2) \nonumber\\
	& & +y(a_{30}x^3+a_{21}x^2z+a_{12}xz^2+a_{03}z^3) \nonumber\\
	& & +(a_{40}x^4+a_{31}x^3z+a_{22}x^2z^2+a_{13}xz^3+a_{04}z^4),	\nonumber
\end{eqnarray}
We define as in \cite{FOR} 
\begin{eqnarray*}
2 \bar{h} &= & 2f_1 f_2 f_{12} - f_1^2 f_{22}-f_2^2 f_{11}, \\
&= &  f_1 ( f_2  f_{12} - f_1 f_{22})+ f_2 ( f_1 f_{12} - f_2 f_{11})
\end{eqnarray*}
where $f_i$ or $f_{ij}$ are the partial derivatives with respect to $i$th variable (or to $i$th and $j$th variables).

It is then easy to check  via a computer algebra system, see

{\small 
\begin{center}
\url{http://iml.univ-mrs.fr/~ritzenth/programme/flex-char3.mw},
\end{center}
}
\noindent that
$$
2\bar{h} -a_{20}f^2-f \cdot  (ax^3+by^3+cz^3)z = \tilde{h}_{C} \cdot z^2, $$
with

\begin{eqnarray}
a &:= & a_{40}a_{11}+a_{21}a_{30}-2a_{20}a_{31},  \nonumber\\
b	&:= & a_{10}a_{11}+a_{00}a_{21}-2a_{20}a_{01},\nonumber \\
c	&:= & 2a_{12}^2+a_{13}a_{11}+a_{20}a_{04}+a_{03}a_{21}+a_{02}a_{22},\nonumber
\end{eqnarray}

where $\tilde{h}_{C}$ is a homogeneous polynomial in $K\left[x,y,z\right]$ of degree $6$, for which nonzero coefficients appear  only for the monomials $ x^6,y^6,z^6,x^3y^3,x^3z^3$ and $y^3z^3$.
  Since the map $u\mapsto u^3$ is an isomorphism of $K$, there is a polynomial $h_{C}\in K\left[x,y,z\right]$ satisfying $\tilde{h}_{C}=h_{C}^3$. If we suppose that there is no flex at infinity, the flexes are the intersection points of  $\bar{h}=0$ and $f=0$, so they are also the intersection points of $h_C=0$ and $f=0$ and $h_C=0$ is the equation of a (possibly degenerate) conic $H_C$. \qed
\end{proof}
\begin{remark}
As $\tilde{h}_C=h_C^3$, we see that the weight of a flex which is not a hyperflex is $3$ in characteristic $3$.
\end{remark}

\begin{corollary}
The Galois group of the cover $I_4 \to \PP^{14}$ over $\bar{\FF}_3$ is the full symmetric group $S_{8}$.
\end{corollary}
\begin{proof}
Note that the Galois group $G$ of the cover is the Galois group of the $x$-coordinate of the $8$ flexes of the general quartic. Hence, $G$ is included in $S_8$. To show that $G$ is exactly $S_8$, we are going to specialize the general quartic to smooth quartics over finite fields with $8$ distinct flexes having different arithmetic patterns. More precisely, 
to generate $S_8$ we need to produce (see \cite[Lem.4.27]{milne-galois}):
\begin{itemize}
\item a smooth quartic with $8$  Galois conjugate flexes over $\FF_3$:
$$2 x^4 + 2 x^3 y + x^3 z + 2 x^2 z^2 + x y^3 + 2 x y^2 z + y^3 z + y z^3=0 ;$$
\item a smooth quartic with one rational flex and $7$ Galois conjugate flexes over $\FF_3$:
$$x^3 y + x^2 z^2 + 2 x y^3 + x y^2 z + 2 x y z^2 + 2 x z^3 + y^4 + 2 y z^3=0;$$
\item a smooth quartic with two quadratic conjugate flexes and $6$ rational flexes over $\FF_9$: 
\begin{eqnarray*}
a^6 x^4 &+& a x^3 y  +  a^7 x^3 z + a^6 x^2 y^2 + a^2 x^2 z^2 + a^7 x y^3 + a^7 x y^2 z \\ &+& x y z^2 
 +  a^5 x z^3 + a^5 y^4 + a^3 y^3 z + a^5 y^2 z^2 + 2 y z^3 + a^7 z^4=0,
\end{eqnarray*}
where $a^2-a-1=0$. \qed
\end{itemize} 

\end{proof}
\begin{corollary}
Let $C$ be a smooth plane quartic over $\FF_{3^n}$. The probability that $C$ has a rational flex tends to 
$$p_8:=1-\frac{1}{2!} + \frac{1}{3!} - \ldots -\frac{1}{8!}  \approx 0.63$$
when $n$ tends to infinity.
\end{corollary}
\begin{remark}
The fact that the Galois group $G$ in characteristic $3$ is $S_8$ and not $S_{24}$ was unnoticed in our computations in \cite{FOR} because $|p_{24}-p_8| \leq 10^{-5}$.
\end{remark} 
General reduction arguments show that the Galois group $G$ remains $S_{24}$  almost all $p$. 
We  conjecture that $p=3$ is the only exceptional case.
\begin{conjecture}
The Galois group of the cover $I_4 \to \PP^{14}$ over $\bar{\FF}_p$ is the full symmetric group $S_{24}$ if $p \ne 3$ and $S_8$ otherwise.
\end{conjecture}
{"config": "arxiv", "file": "1006.0873/Flexes_char_3.tex"}