I need to show the equivalence:
$x^{p-1}+x^{p-2}+\cdots+x+1$ is irreducible over $\mathbb{F}_2$ if and only if $2$ generates $\mathbb{F}_p^{*}$.
Maybe I should clarify the meaning of "generates": $\mathbb{F}_p^{*}$ is a cyclic group (as the mulplicitive group of a finite field) composed of the elements $1, 2, ...., p-1$, and $2$ generates $\mathbb{F}_p^{*}$ means that all elements are its powers.
I tried what seems to be, intuitively the easy direction. Here's my attempt:
Suppose $1+x+...+x^{p-1}$ is irreducible. Let $\alpha\in \overline{\mathbb{F}}_2$ be a root of the polynomial. Then $[\mathbb{F}_2[\alpha]: \mathbb{F}_2]=p-1$ thus $\mathbb{F}_2[\alpha]\cong \mathbb{F}_{2^{p-1}}$.
Now I'm stuck. I think it might have something to do with the multiplcitive group of this field, since it's order is $2^{p-1}-1$ who's divisable be $p$, but maybe I'm wrong.
Any ideas?
if $\alpha$ is a $p$th root of unity, then $\alpha^{n} = \alpha$ iff $n \equiv 1 \bmod p$.
if the order of $2$ modulo $p$ is $D$, then $\alpha^{2^m} = \alpha$ iff $D | m$.
let $r$ be the least integer such that $\alpha \in \mathbb{F}_{2^r}$. Then $\sigma(a) = a^2$ is the Frobenius automorphism and generates $Gal(\mathbb{F}_{2^r}/\mathbb{F}_{2})$, which means that the minimal polynomial of $\alpha$ is $$\phi(x) = \prod_{k=1}^d (x-\sigma^k(a))=\prod_{k=1}^d (x-a^{2^k})$$ where $d = [\mathbb{F}_{2}(\alpha):\mathbb{F}_{2}]$, which is also the least integer such that $\alpha = \alpha^{2^d}$ i.e. the order of $2$ modulo $p$.
if $d = p-1$, since $\alpha$ is a root of $\varphi(x) = \sum_{m=0}^{p-1} x^m$, then $\phi = \varphi$. Conversely, if $d \ne p-1$ then $\phi | \varphi$ which is not irreducible.