# Irreducible Polynomials over a Finite Field

I am motivated by this question, and want to find a solution to the following problem.

Question: How many monic, irreducible polynomials of degree $n$ are there over the finite field $\mathbb{F}_q$ for some prime $q$?

The solution provided in the original question pivots on two central claims:

Claim 1: $\mathbb{F}_{q^n}$ is the splitting field of the polynomial $g(x)=x^{q^n}−x$

Claim 2: Every monic irreducible polynomial of degree $n$ divides $g$.

Claim 1 I am happy with, but in the case of Claim 2 I cannot see why it is true. Could someone please explain.

## 2 Answers

The point is that every element of ${\mathbb F}_{q^n}$ is a root of $g(x)$, so $$g(x) = \prod_{\alpha \in {\mathbb F}_{q^n}} (x - \alpha).$$ Now an irreducible polynomial $h(x)$ over ${\mathbb F}_q$ of degree $n$ splits in distinct linear factors over ${\mathbb F}_{q^n}$, so all its factors are also factors of $g(x)$. Therefore $h(x)$ divides $g(x)$.

• Thanks, it's all so clear now – Mathmo May 2 '14 at 13:24

Denote by $N_q(n)$ the number of monic, irreducible polynomials in $\mathbb{F}_q[x]$ of degree $n$. Here $q$ can be any prime power. Then Claim $1$ and $2$ give $$q^m =\sum_{n\mid m}nN_q(n),$$ for all $m\ge 1$, and this gives the formula $$N_q(n)=\frac{1}{n}\sum_{d\mid n} \mu \left(\frac{n}{d}\right)q^d$$ by the Möbius inversion formula.