The genuine prime number theorem says that there are about $N/\log N$ prime numbers between 1 and N. The prime number theorem in function fields is rather about the number $a_n$ of irreducible monic polynomials of degree $n$ in the polynomial ring $A=\mathbb{F}_q[t]$ on a finite field $\mathbb{F}_q$. It is remarkable that the result is of the same shape $q^n/n$, which is $N/\log N$ if you let $N=q^n$ be the number of monic polynomials of degree $n$. But the proof, as often with the polynomial analogues of number theory, is much neater and even yields a nice exact formula for $a_n$

$\displaystyle{q^n=\sum_{d\mid n}da_d\text{ and by Moebius inversion } a_n=\frac{1}{n}\sum_{d\mid n}\mu(d)q^{n/d}}$

First proof

This one requires a good deal of knowledge of the theory finite fields.

We prove that $\Phi=T^{q^n}-T$ is the product $\Psi$ of all monic irreducible polynomials whose degree divide $n$, which immediately provides the above formula by comparison of the degrees. To prove this, notice that the roots of $\Phi$ are all the elements of the extension $\mathbb{F}_{q^n}$ of $\mathbb{F}_q$, so that $\Phi$, as an element of $\mathbb{F}_{q^n}[t]$ can be factored as $\prod_{\alpha\in \mathbb{F}_{q^n}}(T-\alpha)$. On the other hand, if $P$ is an irreducible polynomial of degree $m\mid n$, it splits into linear factors in $A/(P)\sim \mathbb{F}_{q^m}\leq \mathbb{F}_{q^n}$. So $P$ has a root $\alpha\in \mathbb{F}_{q^n}$ and all its other roots are images of $\alpha\in \mathbb{F}_{q^n}$ under the Frobenius element, more precisely,

$P=(T-\alpha)(T-\alpha^{q^{n/m}})\cdots (T-\alpha^{q^{n(m-1)/m}})$

so that $P\mid \Phi$ (at first sight in $\mathbb{F}_{q^n}[t]$, but also in $A$ because they are both in $A$ – it is well known that the gcd is preserved under field extensions). Moreover, all monic irreducible $P$ are coprimes, so that their product still divides $\Phi$. Finally, we need to prove the converse divisibility, either in $A$ or in $\mathbb{F}_{q^n}[t]$, which is equivalent. So let’s do it in the latter; take a factor $T-\alpha$ with $\alpha \in\mathbb{F}_{q^n}$. Let $m\mid n$ be the order of the Frobenius element on $\alpha$, so that $\alpha^{q^m}=1$ and no smaller integer has this property; then $(T-\alpha)\mid (T-\alpha)(T-\alpha^{q^{n/m}})\cdots (T-\alpha^{q^{n(m-1)/m}})\in A$ which is indeed an irreducible of degree $m\mid n$, hence a factor of $\Psi$. Given the factorization of $\Phi$ and the fact that the factors $T-\alpha$ are coprime, their product $\Phi$ still divides $\Psi$

Second proof

This one proves exactly what we want and not more and uses only power series. We need the notation $\vert f\vert=q^{\text{deg }f}$.

Let

$\displaystyle{\zeta_A(s)=\prod_{P\in A\text{ monic irreducible}}(1-\vert P\vert ^{-s})^{-1}=\sum_{f\in A\text{ monic}}\vert f\vert^{-s}}$

be the zeta function (for s of real part >1). It is easy to see that $\zeta_A=(1-q^{1-s})^{-1}$ by looking at the sum expression, and $\zeta_A=\prod_{d\geq 1}(1-q^{-ds})^{-a_d}$ by looking at the product expression. Putting $u=q^{-s}$, both sides become nice power series so that we get the equality

$\displaystyle{\frac{1}{1-qu}=\prod_{d=1}^{+\infty}(1-u^d)^{-a_d}}$

and doing classical operations (logarithmic derivation and expansion in power series followed by termwise identification in particular) on both sides yield the result.

This proof seems to use about zero input of algebra, so its success looks like a miracle. The only step where a (tiny) bit of knowledge seems to be used is the transformation of the sum into an Euler product – it needs the existence of unicity of the decomposition of a monic polynomial into product of irreducibles.

A kind of sieve method ?

It could be reasonable to hope that the sieve of Eratosthenes would work better in the realm of polynomial rings. We’ll try to adapt it here. Notice that in this case we have count primes up to a certain large degree, and not of exactly certain large degree (both quantities are far from equivalent in fact !).

Let’s define $\pi(n)$ to be the number of monic irreducible (let’s say primes) polynomials of degree at most n, and $\delta(m)=1_{m=1}$. We also use Moebius’ function for monic polynomials, also defined in terms of the number of prime factors. The Moebius inversion formula $\delta=\mu\star 1$ still holds in the polynomial world. We also write

$\displaystyle{\Phi=\prod_{P \text{mon. irred,deg} P\leq n/2}P}$

Thus

$\displaystyle{\pi(n)-\pi(n/2)+1=\sum_{f\text{ monic of deg}\leq d}\delta((f,\Phi))=\sum_{d\mid\Phi}q^{n-\mathrm{deg}d}\mu(d)1_{\mathrm{deg} d\leq n}}$

and if we forget about $1_{\mathrm{deg} d\leq n}$ we obtain exactly

$\displaystyle{q^n\prod_{P \text{mon. irred,deg} P\leq n/2}(1-\vert P\vert ^{-1})}$

this looks quite nice, but the error term, just like in the integer case, can’t easily be bounded by anything better than $O(2^{\pi(n/2)})$ which we know will be vastly bigger than the main term. By the way this product can very easily be shown to be $\ll 1/n$ but the exact asymptotic is not any easier than the PNT itself… So this sieve seems to have very little interest, given the convenience of the other methods.