I want help in the following counting problem:
We know how many irreducible Monic polynomial of degree 3 are there in $\mathbb{F}_{q}[x]$. Now what I want to find is number of monic irreducible monic polynomials of degree 3 in $\mathbb{F}_{q}[x]$, with constant term -1. Assume that $q$ is a odd-prime power
I use the following method to solve this. To find such irreducible monic polynomials, we observe that the product of roots is 1. If $\alpha\in \mathbb{F}_{q^{3}}^{*}\setminus \mathbb{F}_{q}^{*}$. Then the other roots are given by the images of $\alpha$ under Galois automorphism and those are $\alpha^{q}$ and $\alpha^{q^{2}}$ resp. Hence we have $\alpha^{1+q+q^{2}}=1$. There are $1+q+q^{2}$ number of elements in $\mathbb{F}_{q^{3}}^{*}$ satisfying this. Now we have to find how many elements are there in $\mathbb{F}_{q}^{*}$, which satisfy this? This is where I get stuck into. If I assume there are $l$ number of such elements, then the final answer should be $\frac{1+q+q^{2}-l}{3}$. I hope this approach is fine, but I am stuck.
Thanks in advance for Any kind of help!
Best Answer
You made very good progress with your problem!
To find $\ell$ all you need to do is to observe that if $\alpha\in\Bbb{F}_q^*$ then $\alpha^q=\alpha$, and also $\alpha^{q^2}=\alpha$. Implying that $\alpha^{1+q+q^2}=\alpha^3$.
Because $\Bbb{F}_q^*$ is cyclic of order $q-1$ it has, by basic properties of cyclic groups, $\ell=\gcd(3,q-1)$ elements $\alpha$ such that $\alpha^3=1$.
As a final check I invite you to verify that $1+q+q^2-\gcd(q-1,3)$ is divisible by three for all values of $q$. This part also follows from the fact that your formula is bound to produce an integer as the answer, but it is a nice extra check :-)