\section{A Very Free Rational Curve of Degree 9}
\label{section-degree-9}

\noindent
Let $\varphi:\mathbb{P}^1 \rightarrow \mathbb{P}^5$ be the morphism into the Fermat hypersurface given by
\begin{align*}
G_0 &= S^4T^5 \\
G_1 &= S^9 + S^8T + S^5T^4 \\
G_2 &= S^9 + S^4T^5 + ST^8 \\  
G_3 &= S^9 + S^8T + S^4T^5 + S^3T^6 + S^2T^7 + ST^8 \\ 
G_4 &= S^9 + S^5T^4 + S^3T^6 +S^2T^7 + ST^8 + T^9 \\ 
G_5 &= S^7T^2 + S^6T^3 + S^5T^4 + S^3T^6 + S^2T^7 +ST^8 +T^9.
\end{align*}

\noindent
Let $e_1,...,e_5$ denote the splitting type of $\Omega_X (\varphi)$, where we have that $e_i \le -9$ $(1 \le i \le 5)$.  Since the $G_i$ are linearly independent over $k$, $\dim_k(\Omega_X(\varphi)_9) = 0$.  Next we claim that $\tilde{\varphi_{10}} : R_1^{\bigoplus6} \rightarrow R_{10}$ is surjective.  In fact it can be checked that the $\tilde{\varphi}(b_i)$ $(1 \le i \le 12)$, where the $b_i$ are distinct basis elements of $R_1^{\bigoplus6}$, span $R_{10}$.  It follows that $\tilde{\varphi_n} : R(-9)_n^{\bigoplus6} \rightarrow R_n$ is surjective for $n \ge 10$.  Hence 
\begin{align*}
\dim_k(\Omega_X(\varphi)_{10}) &= \dim_k(R_1^{\bigoplus6}) - \dim_k(R_{10}) = 1\\
\dim_k(\Omega_X(\varphi)_{11}) &= \dim_k(R_2^{\bigoplus6}) - \dim_k(R_{11}) = 6.
\end{align*}

\noindent
After reordering this yields $e_1,...,e_5 = -11,-11,-11,-11,-10$, which corresponds to the splitting type $1,1,1,1,5$ of $E_X(\varphi)$, showing that $\varphi$ is very free.  This completes the proof of Theorem \ref{theorem-main}.
