# Calculating a strange algebraic limes

I have a problem with calculating a strange limes:

Let $m \in \mathbb{N}$ be fixed and $q=p^n$ (a variabel prime power) for $n \in \mathbb{N}$ and $p$ prime. We define $$c_m=|\left\lbrace f \in \mathbb{F}_q[X]; f \ \text{irreducible, monic, deg}(f)=m \right\rbrace|.$$

Now I have to calculate $\lim_{q \to \infty} \frac{c_m}{q^m}$. And motivated from this limes the question is: Which magnitude has $$c_m=|\left\lbrace f \in \mathbb{F}_q[X]; f \ \text{irreducible, monic, deg}(f)=m \right\rbrace|$$ for large $q$?

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Calculating $\lim_{q \to \infty} \frac{c_m}{q^m}$ means to solve $\lim_{q \to \infty} \frac{\frac{1}{m} \sum_{d|m}\mu(d)q^{\frac{m}{d}}}{q^m}$ because $c_m=\frac{1}{m} \sum_{d|m}\mu(d)q^{\frac{m}{d}}$. I think the result must be either $0$ or $1$. Otherwise I would be surprised. But I don't know how to handle such a limes. My idea was to find an upper and lower bound and then to argue with the sandwich theorem, but maybe there is a shorter solution. – math_space Apr 5 '14 at 13:45

$$\lim_{q \to \infty}\frac{c_m}{q^m} = \frac{1}{m}$$
The reason is that $$\begin{eqnarray}\frac{c_m}{q^m} &=& \frac{1}{q^m}\frac{1}{m}\sum_{d\mid m}\mu(d) q^{m/d} \\&=& \frac{1}{m}\left(1+\sum_{1<d\mid m}\mu(d)q^{m/d-m}\right).\end{eqnarray}$$ Now for $q\to \infty$ the sum has a constant number of summands, therefore $$\begin{eqnarray}\lim_{q\to\infty}\frac{c_m}{q^m} &=& \frac{1}{m}+ \frac{1}{m} \sum_{1<d\mid m}\mu(d)\lim_{q\to\infty}q^{m/d-m} \\&=&\frac{1}{m}.\end{eqnarray}$$
Concerning the magnitude of $c_m$ we can see from the above that for $q \to \infty$ $$c_m = \frac{1}{m}q^m-O(q^\frac{m}{2})$$.
Very nice! Could you explain how you get $c_m = \frac{1}{m}q^m-O(q^\frac{m}{2})$ from $\lim_{q \to \infty}\frac{c_m}{q^m} = \frac{1}{m}$ for a constant $C$? Is $c_m = \frac{1}{m}q^m-O(q^\frac{m}{2})$ equivalent to $\frac{1}{m}q^m-c_m \leq C q^\frac{m}{2}$? – math_space Apr 5 '14 at 16:56
The claim of the $O$-statement is that there is $C$ independent from $q$ such that $|\frac{q^m}{m}-c_m| \leq Cq^\frac{m}{2}$ for all $q$. We can see this as follows: We know that $$c_m = \frac{q^m}{m}+\frac{1}{m}\sum_{1<d\mid m}\mu(d)q^{m/d}$$ Now as $$\frac{1}{m}\sum_{1<d\mid m}\mu(d)q^{m/d} \leq \frac{m-1}{m}q^{m/2}$$ we know that $|\frac{q^m}{m}-c_m| \leq q^{m/2}$. – benh Apr 5 '14 at 17:06
I think I got it. So we can say $$c_m \leq \frac{q^m}{m}+q^{\frac{m}{2}}.$$ If I try to interpret this reslut in words: For large $q$ we have also a large number of irreducible polynomials and as higher the grade $m$ gets as larger $c_m$ gets. Well - this is not really surprising. Perhaps I have overlooked something? – math_space Apr 5 '14 at 17:58
@math_space Well, of course we expect some term increasing in $q$ and $m$, but the question is how this behavior looks like exactly. The statement $c_m = \frac{1}{m}q^m-O(q^\frac{m}{2})$ does not only give you the answer to your question on $c_m/q^m$ but also an explanation for how fast this convergence actually is. I mentioned it because results similar to the above (but WAY harder to prove) play a role in the theory of function fields over finite fields, for example in Bombieri's proof of the "Riemann Hypothesis of function fields over finite fields". – benh Apr 5 '14 at 18:57