Representation of cyclotomic polynomial over finite field with prime characteristic If I have a finite field $F_p$ with characteristic $p$ prime (and size $p$), the cyclotomic polynomial over $F_p$ is defined as $$\Phi_r(X) = \prod\limits_{i=1}^k (X - \eta_i)$$ where $\eta_i$ is a primitive $r$th root of unity (maybe not an element of $F_p$ but certainly an element from the cyclotomic field). Is it true that $\Phi(X)$ can always be represented over $F_p[X]$? I mean, are the coefficients of $F_p[X]$ always elements of $F_p$?
I know there are lots of question regarding factorization of cyclotomic polynomials over finite fields, but I couldn't find the answer to this question. I don't really care if it's irreducible or not, just if it can be represented in the field.
Thanks!
 A: Your definition of $\Phi_r(X)$ a priori makes sense over some finite field $\mathbb{F}_{p^k}$ which is a splitting field of the polynomial $x^r - 1$ over $\mathbb{F}_p$, so its coefficients a priori live in $\mathbb{F}_{p^k}$. To check whether or not they live in $\mathbb{F}_p$ it's necessary and sufficient to check whether or not they're fixed by the Frobenius map $x \mapsto x^p$, whose fixed points on any field extension of $\mathbb{F}_p$ are exactly $\mathbb{F}_p$ (this is an easy special case of Galois theory). So we need to check whether the primitive $r^{th}$ roots of unity are permuted by $x \mapsto x^p$.
If $\gcd(r, p) = 1$ this is clear because $p$ is invertible $\bmod r$, so multiplication by $p$ is an isomorphism $(\mathbb{Z}/r\mathbb{Z})^{\times} \to (\mathbb{Z}/r\mathbb{Z})^{\times}$ and sends generators to generators.
If $p \mid r$ then there aren't any primitive $r^{th}$ roots of unity, because there aren't any primitive $p^{th}$ roots of unity.
This argument suggests but does not prove that the polynomial you've defined is actually the reduction $\bmod p$ of the cyclotomic polynomial $\Phi_r(X)$ in the usual sense (defined over $\mathbb{C}$ a priori, whose coefficients live in $\mathbb{Z}$). This is also true if $\gcd(r, p) = 1$, but if $p \mid r$ then $\Phi_r(X) \bmod p$ has repeated roots. For example
$$\Phi_p(X) = \frac{X^p - 1}{X - 1} \equiv (X - 1)^{p-1} \bmod p.$$
