Irreducibility of $1+x+\dots+x^{n-1}$ over $\mathbb{F}_2[x]$ Can someone provide a reference of the proof (or the proof itself) of this statement?
The polynomial $1+x+\dots+x^{n-1}$ is irreducible over $\mathbf{F}_2[x]$ if and only if $n$ is an odd prime and $2$ is a primitive element of $\mathbf{F}_{n}$.
 A: The polynomial in question is irreducible if and only if the smallest field of characteristic $2$, which admits a $n$-th root of unity, is $\mathbb F_{2^{n-1}}$.
This is equivalent of saying $n | 2^k-1$ holds for $k = n-1$ but not for $0< k < n-1$. This implies that $n$ is prime and that $2$ is a primitive root of $(\mathbb Z/n\mathbb Z)^*$.
Let me clarify the first part of the proof (I think the second part is clear):
$f = 1+ x + \dotsb + x^{n-1}$ is irreducible iff it is the minimal polynomial of its roots. Its roots are exactly the non-trivial $n$-th roots of unity. Let $\zeta$ be such a $n$-th root of unity. Then $f$ is the minimal polynomial of $\zeta$ iff $[\mathbb F_2(\zeta):\mathbb F_2]=n-1$, which is the same as saying $\mathbb F_2(\zeta) = \mathbb F_{2^{n-1}}$.
By the way: It is a priori clear that $n$ must be prime, because if $n$ is not prime, then $f = 1+ x + \dotsb + x^{n-1}$ is even reducible over $\mathbb Z$. So we can assume $n$ to be prime and only have to care about the primitive root part. This makes life a little bit easier.
