\section{Relating the Splitting Types}
\label{splittingrelation}
\noindent
Observe that $\Omega_X(\varphi)$ is a graded free module of rank 5 and so has a splitting type, which we denote using $e_1,\dots,e_5$. In this section, we relate the splitting type of $\Omega_X(\varphi)$ to the splitting type of $E_X(\varphi)$. If $(A_0,\dots,A_5) \in \Omega_X(\varphi)$, then $A_0G_0 + \cdots + A_5G_5 = 0$ so that $A_0^4G_0^4 + \cdots + A_5^4G_5^4 = 0$ by the Frobenius endomorphism in characteristic 2. Let $\mathcal{T} = \{(A_0^4,\dots,A_5^4) \mid (A_0,\dots,A_5) \in \Omega_X(\varphi)\} \subset E_X(\varphi)$. We denote the $R$-module generated by $\mathcal{T}$ as $R\langle\mathcal{T}\rangle$.

\begin{lemma} In the notation as above, $E_X(\varphi) = R\langle\mathcal{T}\rangle$.
\end{lemma}

\begin{proof}
Let $(B_0,\dots,B_5)$ be a homogeneous element of $E_X(\varphi)$ where $B_i$ is a homogeneous polynomial of degree $b$. Assume $b \equiv 0 \mod 4$. Observe that we can rewrite each monomial term of $B_i$ as $(c^{1/4}S^{\ell}T^k)^4S^iT^{4-i}$ or $(c^{1/4}S^{\ell}T^k)^4$ where $\ell,k$ are integers, $c \in k$, and $0 < i <4$. After collecting terms and applying the Frobenius endomorphism, we obtain $B_i = a_{i1}^4 + a_{i2}^4S^3T + a_{i3}^4S^2T^2 + a_{i4}^4ST^3$ where the $a_{ij}$ are polynomials in $S,T$.  Then, $\displaystyle B_0G_0^4 + \cdots + B_5G_5^4 = 0$, so we obtain the following after substituting our expression for the $B_i$'s and apporpirately applying Frobenius:
\begin{eqnarray}
(\sum_{i=0}^5 a_{i1}G_i)^4 + (\sum_{i=0}^5a_{i2}G_i)^4S^3T + (\sum_{i=0}^5a_{i3}G_i)^4S^2T^2 + (\sum_{i=0}^5a_{i4}G_i)^4ST^3  &=& 0 \nonumber \
\end{eqnarray}
The sums $\sum_{i=0}^5 a_{ij}G_i$ are each homogeneous polynomials. But since the degree of $T$ in each term in the preceeding equation is distinct modulo 4, the equation $\sum_{i=0}^5 a_{ij}G_i = 0$ implies that $(a_{0j},\dots,a_{5j}) \in \Omega_X(\varphi)$, so $(a_{0j}^4,\dots,a_{5j}^4) \in \mathcal{T}$ for $1 \le j \le 4$. The cases for $d \equiv 1,2,3 \mod 4$ are similar.

\medskip
\noindent
Hence, every homogeneous element of $E_X(\varphi)$ is contained in the submodule generated by $\mathcal{T}$. Since the reverse containment is trivial, it follows that $E_X(\varphi) = R\langle\mathcal{T}\rangle$. 
\end{proof}

\begin{proposition}
If $x_i = (x_{i0},\dots,x_{i5})$ for $1 \le i \le 5$ form a basis for $\Omega_X(\varphi)$, then $y_i = (x_{i0}^4,\dots,x_{i5}^4)$ for $1 \le i \le 5$ form a basis for $E_X(\varphi)$.
\end{proposition}

\begin{proof}
If $x_i \in \Omega_X(\varphi)$, then $y_i  \in \mathcal{T}$ and every element of $\mathcal{T}$ is a $R$-linear combination of the $y_i$'s. 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 $E_X(\varphi)$. Moreover, $E_X(\varphi)$ is a free module of rank 5 over a domain, so the generators $y_i$ for $E_X(\varphi)$ must also be linearly independent and hence form a basis.  
\end{proof}
\noindent 

\begin{corollary}
\label{splittingformula}
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{corollary}