\section{Degree 4 and 5 morphisms into X}
\label{section-degree-4-5}

\noindent
We will show that there are no free morphisms of degrees $4$ and $5$ into $X$.
A morphism $\varphi = (G_0, ..., G_5)$, where $G_i = \sum_{0 \leq j \leq d}a_{ij}S^{d - j}T^{j}$ are homogeneous polynomials of degree $d$, gives us a $6 \times (d+1)$ matrix $(a_{ij})$. We will denote this matrix as $M_{\varphi}$. 

\begin{lemma}
\label{lemma-matrices-have-max-rank}
If $\varphi$ is a degree $4$ or $5$  free morphism into $X$, then $M_{\varphi}$ has maximal rank.
\end{lemma}

\begin{proof} This follows from Remark \ref{remark-mike} (4) and (5), and the observation that for a degree $d$ morphism into $X$ the transpose of $M_{\varphi}$ is the matrix of the $k$-linear map $\tilde{\varphi}_d: (R(-d)^{\oplus 6})_d \rightarrow R_d$.
\end{proof}

\begin{lemma}
\label{lemma-morphisms-degree 4 and 5}
We have the following: 
\begin{enumerate}
\item[(a)] There are no degree $4$ free morphisms into $X$.
\item[(b)] There are no degree $5$ free morphisms into $X$.
\end{enumerate}
\end{lemma}

\begin{proof} (a) Assume a degree $4$ free morphism $\varphi = (G_0, ..., G_5)$ exists. By the previous lemma, the $6 \times 5$ matrix $M_{\varphi} = (a_{ij})$ has maximal rank. Since permuting the $G_i's$ does not affect the splitting type of $E_X(\varphi)$, we can assume that the first $5$ rows of $M_{\varphi}$ are linearly independent over $k$. Then det$\big{(}(a_{ij})_{i \leq 4}\big{)} \neq 0$.
Now consider the matrix $\overline{M_{\varphi}} = (a^4_{ij})$. By the Frobenius endomorphism on $k$, det$\big{(}(a^4_{ij})_{i \leq 4}\big{)}$ = $\big{(}$det$((a_{ij})_{i \leq 4})\big{)}^4 \neq 0$, proving that $\overline{M_{\varphi}}$ has maximal rank.\\

\noindent
Since $G^5_0 + ... + G^5_5 = 0$, computing the coefficients of $G^5_0 + ... + G^5_5 $, for $j = 0,1,2,3,4$, we get
\begin{equation}
\sum\nolimits_{0 \leq i \leq 5}a^4_{ij}a_{i1} = 0
\quad\text{and}\quad
\sum\nolimits_{0 \leq i \leq 5}a^4_{ij}a_{i3} = 0
\end{equation}
The kernel of the map $k^6 \rightarrow k^5$ given by right multiplication by the matrix $\overline{M_{\varphi}}$ has dimension $1$, because rank$\big{(}\overline{M}_{\varphi}\big{)} = 5$. By (5.2.1), $(a_{01}, a_{11}, ...,a_{51})$, $(a_{03}, a_{13}, ..., a_{53})$ $\in$ ker$(k^6 \rightarrow k^5)$, and as  these $6$-tuples are columns of $M_{\varphi}$, they are linearly independent over $k$. Then dim$_k$\big{(}ker$(k^6 \rightarrow k^5)\big{)}$ $\geq$ $2$, a contradiction.\\

\noindent
(b) Assume $\varphi = (G_0, ..., G_5)$ is a degree $5$ free morphism. By the previous lemma, the matrix $M_{\varphi} = (a_{ij})$ has maximal rank, and is invertible. Thus $\overline{M_{\varphi}} = (a^4_{ij})$ is invertible by the same argument as above.
Since $G^5_0 + ... + G^5_5 = 0$, computing the coefficients of the polynomial $G^5_0 + \cdots + G^5_5$ we get
$\sum_{0 \leq i \leq 5}a^4_{ij}a_{i2} = 0$, for $j = 0, 1, 2, 3, 4, 5$. 
So the product of the row matrix $(a_{02}, a_{12}, ... , a_{52})$ and the matrix $\overline{{M}_{\varphi}}$ is $0$, which is impossible because $(a_{02}, a_{12}, ... , a_{52}) \neq \vec{0}$ and $\overline{{M}_{\varphi}}$ is invertible.
\end{proof}
