Question:
Let $f,g$ be two irreducible polynomials over finit field $\mathbb{F}_q$ such that $\text{ord}(f)=\text{ord}(g)$. Prove that $\deg(f)=\deg(g)$.
Let $f$ be a polynomial over Finite field $\mathbb{F}_q$. The order of $f$, denoted by $\text{ord}(f)$, is defined by the least positive integer e such that $f \mid x^e−1$.
A lemma From here
Let $f$,$g$ be two square-free polynomials over finite field $\mathbb{F}_q$. Then the splitting field of $f$ is that of $g$, up to isomorphism, if and only if $\text{ord}(f)=\text{ord}(g)$.
Attempt: Denote $r=\deg(f)$ and $s=\deg(g)$. From the above lemma, we know that the splitting field of $f$ equals that of $g$, since $\text{ord}(f)=\text{ord}(g)$. Note that the splitting field of $f$ is $\mathbb{F}_{q^r}$ and the splitting field of $g$ is $\mathbb{F}_{q^s}$. Since $\mathbb{F}_{q^r}=\mathbb{F}_{q^s}$, we have $r=s$.
Am I right ? Thanks for any replies.