The following lemma is well known
Let $f$ be an irreducible polynomail over $\mathbb{F}_q$ of degree $m$ with $f(0)\ne 0$. Then the order of $f$ is equal the order of any rootof $f$ in the multiplicative group group $\mathbb{F}_{q^m}^*$.
My question is, given a positive integer $r$ such that $r \mid q^m-1$, how to construct an irreducible polynomail of order $r$ of degree $m$ over $\mathbb{F}_q$ ?
My Answer: If $r=1$, it is trivial. Denote $s=\frac{q^m-1}{r}$. Let $g$ be a primitive element of $\mathbb{F}_{q^m}$. Then $\alpha=g^s$ is an element of order $r$. Let $n$ be the smallest integer such that $q^n\equiv 1\pmod{r}$. Then define $$f(x)=\prod_{i=0}^{n-1}(x-\alpha^{q^i}).$$ Obviously $f(x)$ is an irreducible polynomail over $\mathbb{F}_q$ of degree $n$ and $n \mid m$. Moreover, $\alpha$ is a root of $f(x)$. So $\text{ord}(f)=\text{ord}(\alpha)=r$. That's $\text{ord}(f) \mid q^n-1$.
We claim that, there are no irreducible polynomail over $\mathbb{F}_q$ of order $r$ of degree differing from $n$, which follows from the conlusion here.
Am I right? Thanks for any replies.