If $p$ is a prime, prove there are exactly $\frac{p^3-p}{3}$ monic irreducible cubic polynomials in $\mathbb{Z}_p[x]$

Question

If $p$ is a prime, prove there are exactly $\frac{p^3-p}{3}$ monic irreducible cubic polynomials in $\mathbb{Z}_p[x]$

I am looking some notes here but don't know in general how to approach this problem.

Answer 1

More General:

Number of monic irreducible poly. of degree $n$ over $\mathbb Z_p$ is $$\frac{1}{n}\sum_{d|n}\mu(d)p^{n/d}$$ where $\mu$ is the Mobius Function defined by $$\mu(n)=\begin{cases}1 & \text{ , if } n=1\\0 & \text{ , if $n$ is perfect square}\\(-1)^{\text{number of distinct prime factors of $n$}} & \text{ , otherwise }\end{cases}$$

Answer 2

The count follows from the fact that each irreducible cubic will have three distinct roots in $\mathbb{F}_{p^3} - \mathbb{F}_{p}$, and moreover each element of $\mathbb{F}_{p^3} - \mathbb{F}_{p}$ has a unique monic cubic minimal polynomial. The theorem that @S.Panja-1729 cited can be seen by similar analysis, only you need to keep track of the intermediate subfields of $\mathbb{F}_{p^n}$.

Answer 3

This is not so different from the other approaches, but it can be shown that the product of all monic irreducibles of degree $1$ and $3$ is $x^{p^3}-x$, because the irreducible factors of this polynomial are precisely those whose roots lie in the finite field of $p^3$ elements.

Dividing out by the $p$ degree $1$ irreducibles gives a degree $p^3-p$ polynomial, and the result follows.