\begin{aligned}
\pi \colon \phantom{aaaaat}\F^k_q\phantom{aaaaa} &\longrightarrow \phantom{aaa} R/Rg\\
    (c_0,c_1,\dots,c_{k-1})&\longmapsto c_0+c_1x+\cdots c_{k-1}x^{k-1}\\
\end{aligned}$$ with $R=\F_q[x;\theta]$. Then, we can see that $\pi(P\tau^i )=x^i$ with $P=(1,0,\dots,0)$, for all $i\in\Z_{\geq 0}$. Thus, we have $$\begin{aligned}
\pi( a_0P+\cdots+a_{k-1}P\tau^{k-1}+P\tau^k)
&=\pi(a_0P)+\cdots+\pi(a_{k-1}P\tau^{k-1})+\pi(P\tau^k)\\
&=a_0\pi(P)+\cdots+a_{k-1}\pi(P\tau^{k-1})+\pi(P\tau^k)\\
&=a_0 (1)+a_1(x)+\cdots+a_{k-1}(x^{k-1})+(x^k)\\
&=g(x) = 0 \in R/Rg(x),\end{aligned}$$ i.e. $a_0P+a_1P\tau+\cdots+a_{k-1}P\tau^{k-1}+P\tau^k=(0,\dots,0)=\vec{0}$. Furthermore, as $g(x)$ is a right divisor of $x^n-\alpha$, we have $$\pi(P\tau^n-\alpha P)=\pi(P\tau^n)-\alpha\pi(P)=x^n-\alpha=0\in R/Rg(x),$$ and this implies that $P\tau^n-\alpha P=(0,0,\dots,0)$, that is, $P\tau^n=\alpha P$.\
Take $H=[(P_0)_t,(P_1)_t,\dots,(P_{n-1})_t]$, $P_i=P\tau^i$, $P_0=P=(1,0,\dots,0)$ and $\tau=\Theta \circ T_g$.

 \
“$\Leftarrow$” Note that $\vec{g}=(a_0,a_1,\dots,a_{k-1},1,0,\dots,0)\in\F_q^n$ is an element of $\mathscr{C}$, since $\vec{g}H_t=\vec{0}$. Furthermore, we get $\vec{g}\tau^iH_t=\vec{0}$ for every $i=0,\dots,n-k-1$, i.e. $\{ \vec{g},\vec{g}\tau,\dots,\vec{g}\tau^{n-k-1}\}$ are $n-k$ linear independent elements of $\mathscr{C}$. Thus $\mathscr{C}$ has a parity check matrix $G$ given by $$\label{MG}
\begin{pmatrix}
a_0     &  a_1  &  \cdots  &  a_{k-1}&1  & 0          & 0         & \cdots  & 0  \\
0         &  \theta(a_0)     &  \theta(a_1)     &  \cdots     & \theta(a_{k-1}) &1& 0         &  \cdots & 0  \\
\vdots   &  \ddots &  \ddots  &  \ddots    & \ddots    & \ddots &  \ddots  &\ddots  & \vdots \\
0         &  0        &   0         &  \theta^{n-k-1}(a_0)       &  \theta^{n-k-1}(a_1)      &   \cdots & \cdots  & \theta^{n-k-1}(a_{k-1})&1 \\
\end{pmatrix}.$$

As $P\tau^n=\alpha P$, by Theorem \[GMaruta:2\] we see that $H$ is a parity check matrix of a skew 