 and the Ore Polynomial ring $F_q[Y^{q_0}, \circ]$.* $\Box$

For a skew cyclic code over $\mathbb{F}_q$, it can also be described in terms of the $n$-th root of unity. By the above mapping $\phi$, one can verify that $\phi(x^n-1)=Y^{q_0^n}-Y$. Since $\sigma=\theta^d$, the fixed subfield is $\mathbb{F}_{p^d}=\mathbb{F}_{q_0}$. Let $\mathbb{F}_{q^s}$ be the smallest extension of $\mathbb{F}_q$ containing $\mathbb{F}_{q_0^n}$ as a subfield. Then $\mathbb{F}_{q^s}$ is the splitting field of $\phi(x^n-1)$ over $\mathbb{F}_q$. An element $\alpha \in \mathbb{F}_{q^s}$ is called a *right root* of $f\in R$ if $x-\alpha$ is a right divisor of $f$.

Let the extension of $\sigma$ to an automorphism of $\mathbb{F}_{q^s}$ be also denoted by $\sigma$. For any $\alpha\in \mathbb{F}_{q^s}$ define ${\mathcal N}_{\sigma, i}(\alpha)=\sigma^{i-1}(\alpha)\sigma^{i-2}(\alpha)\cdots \sigma(\alpha)\alpha,~i>0$, with ${\mathcal N}_{\sigma, 0}=1$.

[**Lemma 2.4** ]{}[@Jacobson2 Proposition 1.3.11] *Let $f(x)=\sum_{i=0}^ka_ix^i \in R$. Then\
(i)  The remainder $r$ on right division of $f(x)$ by $x-\alpha$ is given by $r=a_0{\mathcal N}_{\sigma, 0}(\alpha)+a_1{\mathcal N}_{\sigma, 1}(\alpha)+\cdots +a_k{\mathcal N}_{\sigma, k}(\alpha)$.\
(ii)  Let $\beta\in \mathbb{F}_{q^s}$. Then $(x-\beta)\mid_rf(x)$ if and only if $\sum_{i=0}^ka_i{\mathcal N}_{\sigma, i}(\beta)=0$.* $\Box$

Note that $\sigma(\alpha)=\theta^d(\alpha)=\alpha^{p^d}=\alpha^{q_0}$, and hence ${\mathcal N}_{\sigma, i}(\alpha)=\alpha^{\frac{q_0^i-1}{q_0-1}}$. The following result can also be found in [@Chaussade Lemma 4], here we give another proof by Lemma 2.3.

[**Lemma 2.5** ]{} *Let $f(x)\in R$, and $\mathbb{F}_{q^s}$ be the smallest extension of $\mathbb{F}_q$ in which $\phi(f(x))$ splits. Then a non-zero element $\alpha \in \mathbb{F}_{q^s}$ is a root of $\phi(f(x))$ if and only if $\alpha^{q_0}/\alpha$ is a right root of $f(x)$.*

*Proof* If $\alpha^{q_0}/\alpha$ is a right root of $f(x)$, then $x-\alpha^{q_0}/\alpha$ is a right divisor of $f(x)$. From Lemma 2.3, we h