\section{Introduction}

\noindent
Any smooth projective Fano variety in characteristic zero is rationally
connected and hence contains a very free rational curve. In positive
characteristic a smooth projective Fano variety is rationally chain
connected. However, it is not known whether such varieties are separably
rationally connected, or equivalently, whether they have a very free
rational curve. This is an open question even for nonsingular
Fano hypersurfaces. See \cite{kollar} as well as \cite{debarre}.

\medskip\noindent
In this paper we consider the degree $5$ Fermat hypersurface
$$
X \quad:\quad X_0^5 + X_1^5 + X_2^5 + X_3^5 + X_4^5 + X_5^5 = 0
$$
in $\mathbb{P}^5$ over an algebraically closed field $k$ of characteristic $2$.
Note that $X$ is a nonsingular projective Fano variety.

\begin{theorem}
\label{theorem-main}
Any free rational curve $\varphi : \mathbb{P}^1 \to X$ has degree $\geq 8$
and there exists a free rational curve of degree $8$. Any very free rational
curve $\varphi : \mathbb{P}^1 \to X$ has degree $\geq 9$ and there exists a
very free rational curve of degree $9$.
\end{theorem}

\noindent
This result, although perhaps expected, is interesting for several
reasons. First, it is known that $X$ is unirational,
see \cite[Page 52]{debarre} (the corresponding rational map
$\mathbb{P}^4 \dasharrow X$ is inseparable). Second, in
\cite{beauville} it is shown that every nonsingular hyperplane section of $X$
is isomorphic to a Fermat hypersurface of dimension 3 and this property
characterizes Fermat hypersurfaces among all hypersurfaces of degree 5
in characteristic $2$. We believe that these facts single out the Fermat as a
likely candidate for a counter example to the conjecture below; instead
our theorem shows that they are evidence for it.

\begin{conjecture}
\label{conjecture-fano-src}
Nonsingular Fano hypersurfaces have very free rational curves.
\end{conjecture}

\noindent
A little bit about the method of proof. In
Section \ref{section-setup}
we translate the geometric question into an algebra question which is
computationally more accessible. In
Sections \ref{splittingrelation}, \ref{section-numerology},
and \ref{section-degree-4-5} we exclude low degree solutions by theoretical
methods. Finally, in
Sections \ref{section-degree-8} and \ref{section-degree-9}
we explicitly describe some curves which
are free and very free in degrees 8 and 9 respectively.





\section{The overall setup}
\label{section-setup}

\noindent
In the rest of this paper $k$ will be an algebraically closed field of
characteristic $2$ and $X$ will be the Fermat hypersurface of
degree $5$ over $k$. Let $\varphi : \mathbb{P}^1 \to X$
be a nonconstant morphism. We will repeatedly use that every vector
bundle on $\mathbb{P}^1$ is a direct sum of line bundles, see
\cite{grothendieck}. Thus we can choose a splitting
$$
\varphi^*T_X =
\mathcal{O}_{\mathbb{P}^1}(a_1) \oplus
\mathcal{O}_{\mathbb{P}^1}(a_2) \oplus
\mathcal{O}_{\mathbb{P}^1}(a_3) \oplus
\mathcal{O}_{\mathbb{P}^1}(a_4)
$$
Recall that $\varphi$ is said to be a {\it free curve} on $X$ if
$a_i \geq 0$ and $\varphi$ is said to be {\it very free} if $a_i > 0$.
Consider the following commutative diagram
\begin{equation}
\label{equation-diagram}
\vcenter{
\xymatrix{
& 0 \ar[d] & 0 \ar[d] \\
& \mathcal{O}_X \ar@{=}[r] \ar[d] & \mathcal{O}_X \ar[d] \\
0 \ar[r] &
E_X \ar[d] \ar[r] &
\mathcal{O}_X(1)^{\oplus 6} \ar[d] \ar[r] &
\mathcal{O}_X(5) \ar@{=}[d] \ar[r] &
0 \\
0 \ar[r] &
T_X \ar[r] \ar[d] &
T_{\mathbb{P}^5}|_X \ar[r] \ar[d] &
N_{X/\mathbb{P}^5} \ar[r] &
0 \\
& 0 & 0
}
}
\end{equation}
with exact rows and columns as indicated. We will call $E_X$ the
{\it extended tangent bundle of $X$}. The left vertical exact sequence
determines a short exact sequence
$$
0 \to \mathcal{O}_{\mathbb{P}^1} \to \varphi^*E_X \to \varphi^*T_X \to 0
$$
The splitting type of $\varphi^*E_X$ will consistently be denoted
$(f_1, f_2, f_3, f_4, f_5)$ in this paper. Since
$\text{Hom}_{\mathbb{P}^1}(\mathcal{O}_{\mathbb{P}^1}(f),
\mathcal{O}_{\mathbb{P}^1}(a)) = 0$
if $f > a$ we conclude that
\begin{enumerate}
\item If $f_i \geq 0$ for all $i$, then $\varphi$ is free.
\item If $f_i > 0$ for all $i$, then $\varphi$ is very free.
\end{enumerate}
For the converse, note that the map
$\mathcal{O}_{\mathbb{P}^1} \to \varphi^*E_X$ has image contained
in the direct sum of the summands with $f_i \geq 0$. Hence if $f_i < 0$
for some $i$, then $\varphi$ is not free. Finally, suppose that
$f_i \geq 0$ for all $i$. If there are at least two $f_i$ equal to $0$,
then we see that $\varphi$ is free but not very free. We conclude that
\begin{enumerate}
\item[(3)] If $\varphi$ is free, then $f_i \geq 0$ for all $i$.
\item[(4)] If $\varphi$ is very free, then either (a) $f_i > 0$ for all $i$,
or (b) exactly one $f_i = 0$ and all others $> 0$.
\end{enumerate}
We do not know if (4)(b) occurs.

\medskip\noindent
{\bf Translation into algebra.}
Here we work over the graded $k$-algebra $R = k[S,T]$. As usual we let
$R(e)$ be the graded free $R$-module whose underlying module is $R$ with
grading given by $R(e)_n = R_{e + n}$. A {\it graded free $R$-module}
will be any graded $R$-module isomorphic to a finite direct sum of $R(e)$'s.
Such a module $M$ has a {\it splitting type}, namely the sequence of integers
$u_1, \ldots, u_r$ such that $M \cong R(u_1) \oplus \ldots \oplus R(u_r)$.

\medskip\noindent
We will think of a degree $d$ morphism
$\varphi: \mathbb{P}^1 \to \mathbb{P}^5$ as a $6$-tuple
$(G_0, \ldots, G_5)$ of homogeneous elements in $R$ of degree $d$
with no common factors. Then $\varphi$ is a morphism into $X$ if and only if
$G_0^5 + \ldots + G_5^5 = 0$. In this situation we define two graded
$R$-modules. The first is called the {\it pullback of the cotangent bundle}
$$
\Omega_X(\varphi) =
\text{Ker}(\tilde\varphi : R^{\oplus 6}(-d) \longrightarrow R)
$$
where the map $\tilde\varphi$ is given by
$(A_0, \ldots, A_5) \mapsto \sum A_iG_i$. The second is called the
{\it the pullback of the extended tangent bundle}
$$
E_X(\varphi) = \text{Ker}(R^{\oplus 6}(d) \longrightarrow R(5d))
$$
where the map is given by $(G_0^4, \ldots, G_5^4)$.
As the kernel of a map of graded free $R$-modules is a graded free
$R$-module, both $\Omega_X(\varphi)$ and $E_X(\varphi)$
are themselves graded free $R$-modules of rank $5$.

\begin{lemma}
\label{lemma-splitting-type}
The splitting type of $\varphi^*E_X$ is equal to the splitting type
of the $R$-module $E_X(\varphi)$.
\end{lemma}

\begin{proof}
Recall that $\mathbb{P}^1 = \text{Proj}(R)$. Thus a finitely generated
graded $R$-module corresponds to a coherent sheaf on $\mathbb{P}^1$, see
\cite[Proposition 5.11]{Hartshorne}.
Under this correspondence the module $R(e)$ corresponds to
$\mathcal{O}_{\mathbb{P}^1}(e)$. The lemma follows if we show that
$\varphi^*E_X$ is the coherent sheaf associated to $E_X(\varphi)$.
Diagram (\ref{equation-diagram}) shows that $\varphi^*E_X$ is the kernel
of a map
$\mathcal{O}_{\mathbb{P}^1}(d)^{\oplus 6} \to \mathcal{O}_{\mathbb{P}^1}$
given by substituting $(G_0, \ldots, G_5)$ into the partial derivatives
of the polynomial defining $X$. Since the equation is
$X_0^5 + \ldots + X_5^5$, the derivatives are $X_i^4$, and substituting
we obtain $G_i^4$ as desired.
\end{proof}




