Prime Number Theorem in $\mathbb{F}_p[x]$

Question

What is the probability that a randomly chosen monic polynomial of large degree $n$ in $\mathbb{F}_p[x]$ is irreducible? We can interpret this probability as $\displaystyle\lim_{n\to\infty}\frac{N_p(n)}{p^n},$ where $$N_p(n)=\frac{1}{n}\sum_{d|n}p^d\mu\left(\frac{n}{d}\right)$$ is the number of monic irreducibles of degree $n$ in $\mathbb{F}_p[x]$. I'm having trouble seeing how one would go about evaluating/describing the asymptotic behavior of this limit, any help is appreciated.

Answer 1

The limit will be zero due to the factor of $\frac{1}{n}.$ Notice that $$\sum_{d|n} p^n\mu\left(\frac{n}{d}\right)=p^n+O\left(np^{n/2}\right),$$ since there are at most $n$ additional terms, each of which is at most $p^{n/2}$ in size. Thus $$\frac{nN_p(n)}{p^{n}}=1+O\left(n^2p^{-n/2}\right),$$ which implies that

$$\lim_{n\rightarrow\infty}\frac{nN_p(n)}{p^{n}}=1.$$

This is the version of the prime number theorem for finite fields, and as you can see, it is significantly easier to prove than the PNT for the integers.

Alternate proof that $N_p(n)/p^n\rightarrow 0$: As $|\mu(d)|\leq 1$, we have the crude upper bound $$N_p(n)\leq \frac{1}{n}\left(1+p+p^2+\cdots +p^n\right)\leq \frac{1}{n}\cdot \left(1-\frac{1}{p}\right)^{-1}p^n.$$ From this it follows that $$\frac{N_p(n)}{p^n}\leq \frac{1}{n}\left(1-\frac{1}{p}\right)^{-1},$$ and so the limit is $0$.