Let $gcd(n,q)= 1$
Consider $x^n - 1 \in \mathbb{F}_{q}[x]$, and let $\mathbb{F}_{q^t}$ be a splitting field for $x^n - 1$ over $\mathbb{F}_q$. Then, $\mathbb{F}_{q^t}$ contains a primitive $n^{th}$ root of unity, $\alpha$.
Let $\beta$ be a primitive element in $\mathbb{F}_{q^t}$, so it generates the multiplicative group $(\mathbb{F}_{q^t})^*$.
Now, it is easy to show that the minimal polynomial of $\beta^s$ over $\mathbb{F}_{q}$ is $$M_{\beta^s}(x) = \prod_{i \in C_s} (x - \beta^i)$$ where $C_s$ is the unique $(q)-$ cyclotomic coset modulo $q^t - 1$ containing $s$.
From this, it is easy to deduce the following;
Let $\frac{q^t - 1}{n} = d$ and let ${\{s_1, ..., s_t}\}$ be a complete set of representatives of $(q)-$ cyclotomic cosets modulo $n$. Let $M^{(j)}(x)$ denote the minimal polynomial of $\beta^j$ with respect to $\mathbb{F}_q$. Then $x^n - 1$ has the factorisation into monic irreducibles; $$ x^n - 1 = \prod_{i=1}^{t} M^{(ds_i)}(x)$$
Now, I understand this, but I keep seeing the following written in various different places;
The minimal polynomial of $\alpha^s$ over $\mathbb{F}_{q}$ is $$M_{\alpha^s}(x) = \prod_{i \in C_s} (x-\alpha^i)$$ where $C_s$ is the $(q)-$ cyclotomic coset of $s$ modulo $n$.
Hence, $$x^n - 1 = \prod_{s} M_{\alpha^s}(x)$$
All the theory I've done has been regarding primitive elements; $\beta$ in this case, but how have they just stated the same results for a non-primitive $\alpha$? \ I understand that $\alpha = \beta^d$, but this doesn't make anything clearer to me? How can we just go from working with a primitive $\beta$ an cyclotomic cosets modulo $q^t - 1$ to working with $\alpha$ non-primitive with cyclotomic cosets modulo $n$?