I am trying to solve a problem about irreducible polynomials over a finite field and i would like to ask you for a little help or any idea how to make this proof. Here is the problem:
We have a finite field $\mathbb{F}_{q}$ and a prime number $p$. Let $q$ be the generator of the multiplicative group $\mathbb{F}_{p}^{\times }$. Prove that the polynomial $\sum_{i=0}^{p-1}X^{i}$ is irreducible over $\mathbb{F}_{q}$.
$q$ as a generator of the multiplicative group has order $p-1$, but what kind of information i can take from this? Can anybody help me with an idea, please?
Thank you in advance!
Best Answer
Suppose there is a root $x \in \Bbb F_{q^n}$ of this polynomial for some $n$. Then, since $x \neq 0$ and $x^p = 1$, $\Bbb F_{q^n}^\times$ contains a cyclic group of order $p$, so its order, $q^n-1$, has to be divisible by $p$.
Since you supposed that $q$ was a primitive root of $\Bbb F_p^\times$, $q^n \equiv 1 \pmod p \iff n \equiv 0 \pmod {p-1}$. This shows that if $x$ is a root of this polynomial then it lives in an extension of $\Bbb F_{q^{p-1}}$. Since the polynomial is of degree $p-1$, it is irreducible.