Posted by Jason Polak on 20. December 2017 · Write a comment · Categories: commutative-algebra · Tags: ,

Over a finite field, there are of course only finitely many irreducible monic polynomials. But how do you count them? Let $q = p^n$ be a power of a prime and let $N_q(d)$ denote the number of monic irreducible polynomials of degree $d$ over $\F_q$. The key to finding $N_q(d)$ is the following fact: the product of all the monic, irreducible polynomials of degree $d$ with $d \mid n$ in the finite field $\mathbb{F}_q$ is the polynomial
$$x^{q^n} – x.$$
So let's say $f_1,f_2,\dots, f_k$ are all the irreducible monic polynomials of degree $d$ with $d\mid n$. By taking degrees on both sides of the equation $x^{q^n} -x = f_1f_2\cdots f_k$, we get the formula
$$q^n = \sum_{d\mid n} dN_q(d).$$
Hey this is pretty good! For example, if $q = 2$ and $n = 3$ then the formula reads
$$8 = N_2(1) + 3N_2(3)$$
Now, $N_q(1)$ is always easy to figure out. All monic linear polynomials are irreducible so $N_q(1) = q$. Therefore, $N_2(3) = 2$. In fact, these two polynomials are: $x^3 + x + 1$ and $x^3 + x^2 + 1$. Okay, what about if $q = 3$ and $n = 6$? Then our formula tells us that
$$3^6 = N_3(1) + 2N_3(2) + 3N_3(3) + 6N_3(6).$$
So we now have to recursively compute: if we did that we would get that $N_3(1) = 1$, and this gives $2N_3(2) = 6$. Finally, $3N_3(3) = 24$. Therefore, we would calculate that $N_3(6) = 116$. It would not be too hard to write such a recursive algorithm and I encourage the reader to try it.
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