To prove that a polynomial is irreducible in $\mathbb{F}_p[x]$.

Question

Actually, I am studying a paper where I didn't understand a line. Can anyone please help me.

Let $\bar f_p(x)=x^p-x^{p-1}-...-x-1$ be a polynomial in $\mathbb{F}_p[x]$ where $p$ is prime. Now they are proving that $\bar f_p(x)$ is irreducible in $\mathbb{F}_p[x]$.

For that, they are claiming that $\bar f_p(x)$ is irreducible in $\mathbb{F}_p[x]$ iff

(1) gcd($\bar f_p(x),x^p-x)=1$

(2) $\bar f_p(x)|(x^{p^p}-x)$.

So, I am trying to understand the backward implication. Here if (1) is true then $\bar f_p(x)$ has no roots in $\mathbb{F}_p$, but how the additional condition (2) implies $\bar f_p(x)$ is irreducible in $\mathbb{F}_p$?

And, also how $\bar f_p(x)$ is irreducible implies (2)?

Thanks!

Answer 1

We only need that $\overline{f}_p \in \mathbb{F}_p[x]$ is of degree $p$ (prime).

Let $\alpha_i$ be the roots of $\overline{f}_p$ and $\mathbb{F}_p(\alpha_i) =\mathbb{F}_{p^{n_i}}$. If $\overline{f}_p(x) | x^{p^p}-x$ then for each $i$ : $\alpha_i \in \mathbb{F}_{p^p}$, thus $n_i = 1$ or $p$. And $gcd(x^p -x , \overline{f}_p) = 1$ means $n_i \ne 1$ ie. $n_i = p$ and $\overline{f}_p$ is irreducible.

Conversely if $\overline{f}_p$ is irreducible then $n_i = p$.