1\section{Computations for the degree $8$ free curve}

\noindent Let $\varphi:\mathbb{P}_k^1\to\mathbb{P}_k^5$ be a morphism given by the 6-tuple

\begin{align*}
G_0 &= &    &      &       &\hspace{12.5pt}S^5T^3 &       &       &+S^2T^6 & & \\
G_1 &= &S^8 &      &       &       &       &+S^3T^5 &+S^2T^6 & &+T^8 \\
G_2 &= &S^8 &+S^7T &       &+S^5T^3 &       &       & &+ST^7 & \\
G_3 &= &    &      &S^6T^2 &       &       &+S^3T^5 &+S^2T^6 & &+T^8 \\
G_4 &= &S^8 &      &+S^6T^2 &       &+S^4T^4 &       &+S^2T^6 &+ST^7 & \\
G_5 &= &S^8 &+S^7T &+S^6T^2 &+S^5T^3 &+S^4T^4 &+S^3T^5 &+S^2T^6 &. &
\end{align*}

\noindent One can check by computer or by hand that this curve lies on the Fermat hypersurface $X\subset\mathbb{P}_k^5$.

\begin{proposition}
\label{proposition-degree-8}
The degree $8$ rational curve $\varphi$ on the Fermat hypersurface is free.
\end{proposition}

\begin{proof} Due to twisting, the domain of the map $\tilde{\varphi}:R(-8)^{\oplus 6}\to R$ has its first nontrivial graded piece in dimension $8$. The $G_i$ are linearly independent, hence the kernel is trivial in dimension $8$. The matrix for the map $\tilde{\varphi}_9:R(-8)_{9}^{\oplus 6}\to R_9$ is
$$
\begin{pmatrix}
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0
\end{pmatrix}
$$

\noindent where each direct summand of the domain has a basis $\left\{(S,0),(T,0)\right\}$, of which we take six copies (for total dimension $12$), and the range has basis given by the degree $9$ monomials in $S$ and $T$, ordered by increasing $T$-degree (for total dimension $10$). This matrix has rank $10$, which means that the map in degree $9$ is surjective. By rank-nullity, two dimensions of the kernel live in degree $9$; denote the generators by $g_1,g_2$. Surjectivity of $\tilde{\varphi}$ in degree $9$ implies surjectivity in all higher degrees. A second application of rank-nullity gives $\dim_R\Omega(\varphi)_{10}=7$. Four of the generators are inherited from the previous degree, taking the forms
$$
g_1S,g_2S,g_1T,g_2T.
$$

\noindent We conclude that there are three additional generators in degree $10$. Therefore, the splitting type of $\Omega(\varphi)$ is $e_1,...,e_5=-10,-10,-10,-9,-9$, which translates to a splitting type for $E_X(\varphi)$ of $f_1,...,f_5=0,0,0,4,4$, hence the curve is free.
\end{proof}

\end{document}