\section{Relating the Splitting Types}

\noindent
Observe that $\Omega_X(\varphi)$ and $E_X(\varphi)$ are equal to $\text{ker}(R^{\oplus 6} \to R(6d))$ and $\text{ker}(R^{\oplus 6} \to R(-d))$ as $R$-modules respectively and we work over these equivalent modules for convenience in this section. If $(A_0,\dots,A_5) \in \Omega_X(\varphi)$, then $A_0G_0 + \cdots + A_5G_5 = 0$ so that $A_0^4G_0^5 + \cdots + A_5^4G_0^4 = 0$ by the Frobenius endomorphism in characteristic 2. Let $\mathcal{T} = \{(A_0^4,\dots,A_5^4) \mid (A_0,\dots,A_5) \in \text{ker}(R^{\oplus 6} \to R(6d))\}$. 

\begin{proposition} If $(B_0,\dots,B_5)$ is a homogeneous element of $E_X(\varphi)$, then $(B_0,\dots,B_5)$ is a $R$-linear combination of finitely many elements of $\mathcal{T}$.
\end{proposition}

\begin{proof}
Suppose that the degree of $B_i$ for $i = 0,\dots,5$ is $d$. In the case that $d \equiv 0 \mod 4$, $B_i = a_{i1}^4 + a_{i2}^4S^3T + a_{i3}^4S^2T^2 + a_{i4}^4ST^3$ where the $a_{ij}$ are polynomials in $S,T$ since factoring powers of 4 on each monomial component of $B_i$ yields the decomposition $f(S,T)^4S^iT^{4-i}$ for $0 < i <4$ or $f(S,T)^4$ where $f(S,T)$ are polynomials in $S,T$.  Then, $B_0G_0^4 + \cdots + B_5G_5^4 = 0 \implies \sum_{0 \le i \le 5}(a_{i1}G_i)^4 + (a_{i2}G_i)^4S^3T + (a_{i3}G_i)^4S^2T^2 + (a_{i4}G_i)^4ST^3 = (\sum_i a_{i1}G_i)^4 + (\sum_ia_{i2}G_i)^4S^3T + (\sum_ia_{i3}G_i)^4S^2T^2 + (\sum_ia_{i4}G_i)^4ST^3 = 0$. Observe that the sums $\sum_i a_{ij}G_i$ are all homogeneous. But then, the power of $T$ in each term in the sum $(\sum_i a_{i1}G_i)^4 + (\sum_ia_{i2}G_i)^4S^3T + (\sum_ia_{i3}G_i)^4S^2T^2 + (\sum_ia_{i4}G_i)^4ST^3 = 0$ is distinct modulo 4 so hence $\sum_i a_{ij}G_i = 0\implies (a_{0j},\dots,a_{5j}) \in \Omega_X(\varphi) \implies (a_{0j}^4,\dots,a_{5j}^4) \in \mathcal{T}$. The cases for $d = 1,2,3 \mod 4$ are similar.




\noindent
Hence, every homogeneous element of $E_X(\varphi)$ is contained in the submodule generated by $\mathcal{T}$, which we denote  $R\langle\mathcal{T}\rangle$. Since the reverse containment is trivial, it follows that the $E_X(\varphi) = R\langle\mathcal{T}\rangle$. Now if $x_i = (x_{i0},\dots,x_{i5})$ is a generating set for $\Omega_X(\varphi)$, then $y_i = (x_{i0}^4,\dots,x_{i5}^4) \in \mathcal{T}$ and moreover every element of $\mathcal{T}$ is a $R$-linear combination of the $y_i$'s. Then since $E_X(\varphi) = R\langle\mathcal{T}\rangle$, every element of $E_X(\varphi)$ is also an $R$-linear combination of the $y_i$'s so that the $y_i$'s generate $E_X(\varphi)$. Since $E_X(\varphi)$ is a free module of rank 5 over a domain, the generators $y_i$ for $E_X(\varphi)$ must also be linearly independent and hence form basis.  
\end{proof}

\noindent
As a consequence, the splitting type of $E_X(\varphi)$ relates to the splitting type of $\Omega_X(\varphi)$ as $f_i = 4(e_i+d)+d$ so that: 

\begin{proposition}
If $f_i$ denotes the splitting type of $E_X(\varphi)$ and $e_i$ denotes the splitting type of $\Omega_X(\varphi)$, then $f_i = 4e_i + 5d$ where $d$ is the degree of the morphism.
\end{proposition}