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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!

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You're using $q$ as the number of elements of $\mathbb F_q$ and also as a generator of the group of units, which is pretty confusing. Also, do you want the number of elements of your field to be relatively prime to $p$? –  Dylan Yott Nov 23 '13 at 23:16
    
@Dylan: $q$ is a fixed prime power, $p$ is another prime, and one assumes that $[q]$ is some generator of $\mathbb{F}_p^*$. So there is just one $q$ here! –  Martin Brandenburg Nov 23 '13 at 23:20
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@Lullaby: Do you know something about galois theory and cyclotomic polynomials? –  Martin Brandenburg Nov 23 '13 at 23:21
    
@Martin, unfortunately, we haven't covered these topics in the class yet, but thank you for the information, i'll read in Wikipedia about cyclotomic polynomials. –  Lullaby Nov 23 '13 at 23:24
    
You can find a simpler expression for that summation. –  Hurkyl Nov 23 '13 at 23:27

1 Answer 1

up vote 2 down vote accepted

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.

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+1: The same argument in other words is that the other roots of the minimal polynomial of $x$ are the Frobenius iterate $x^q$, $x^{q^2}$, $x^{q^3}$ et cetera. But $x$ is a primitive $p$th root of unity, and $q$ generates $\Bbb{Z}_p^*$, so those conjugates of $x$ are all the non-trivial powers of $x$. –  Jyrki Lahtonen Nov 24 '13 at 6:50
    
Thank you very much!! :) –  Lullaby Nov 24 '13 at 9:41

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