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It can be shown that there are $q^n-q^{n-1}$ (monic) polynomials of degree $n$ in $\mathbb{F}_q[t]$ that are squarefree. The standard way to show this is using Zeta functions.

Suppose I wanted to count the number of (monic) polynomials of degree $n$ such that it has exactly $m<n$ linear factors over $\overline{\mathbb{F}_q}[t]$. This is a combinatorially much harder problem and writing a Dirichlet series with coefficients coming from these counts won't yield a function with nice Euler product.

Does anyone know how one could count such things?

daruma
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  • At least one could split into the number (monic) polynomials of degree $m$ that split completely into linear factors, times the number of polynomials of degree $n-m$ that have no linear factors; the two contributions are independent. – Greg Martin Sep 20 '24 at 16:27
  • For the benefit of the curious: Qiaochu Yuan posted a solution to that count for the square free polynomials in the early years of the site. – Jyrki Lahtonen Sep 20 '24 at 18:09
  • @Jyrki Lahtonen When I meant a zeta function argument, this is what I meant. – daruma Sep 20 '24 at 19:06
  • @Greg Martin, I am looking at the linear factors over the algebraic closure $\overline{\mathbb{F}}_q$. Sorry, if my overline was not long enough to be visible. – daruma Sep 20 '24 at 19:10
  • Ah so you want degree-$n$ polynomials with exactly $m$ distinct roots in the algebraic closure? – Greg Martin Sep 20 '24 at 21:58
  • @Greg Martin Yes, that's exactly right. Do you have any ideas? – daruma Sep 20 '24 at 23:00
  • For instance when $m=n-1$, this isn't as difficult as some of the other cases because I know that we have precisely one factor that is repeated with a multiplicity of $2$. We also know that linear factor is defined over $\mathbb{F}_q$. But you can imagine that this starts to become quite complicated very quickly as $n-m$ is bigger than $1$. – daruma Sep 20 '24 at 23:03
  • Yeah, it seems like one would have to sum over all the partitions of $n$ into $m$ parts, and then for each part size $s$ appearing in a partition, count the number of $s$th powers of squarefree polynomials of degree $d$ where $d$ is the number of times $s$ appears in the partition.... – Greg Martin Sep 20 '24 at 23:47
  • @Greg Martin I agree that one could take over a sum of all partition of $n-m$ and over each possible partition count the possible ways but this seems to be very far away from anything closed form. Counting number of partitions is hard but also we need to count the number of irreducibles polynomials of a given degree. This seems very messy. Even something asymptotic would be desirable at this point. – daruma Sep 21 '24 at 00:21

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This is equivalent to fixing the total degree of the irreducible factors (not counting multiplicity) of a polynomial; that is, if $f(x) = \prod f_i(x)^{m_i}$ is the irreducible factorization over $\mathbb{F}_q$, then the number of linear factors of $f(x)$ over $\overline{\mathbb{F}_q}$ is $\sum \deg f_i$. You can write a generating function for this but it's not very nice. First, recall that the Euler product for the zeta function of $\mathbb{F}_q[t]$ takes the form

$$\frac{1}{1 - qt} = \prod_{n \ge 1} \left( \frac{1}{1 - t^n} \right)^{M(q, n)}$$

where

$$M(q, n) = \frac{1}{n} \sum_{d \mid n} \mu(d) q^{\frac{n}{d}}$$

is the number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_q$; this is the cyclotomic identity. To modify this product so that it also counts the total degree of the irreducible factors we add a second variable $z$ and replace $\frac{1}{1 - t^n}$ (the generating function corresponding to the powers of a fixed monic irreducible polynomial of degree $n$) with

$$1 + z^n (t^n + t^{2n} + \dots) = 1 + \frac{z^n t^n}{1 - t^n}$$

which gives us a modified generating function

$$\boxed{ F(z, t) = \prod_{n \ge 0} \left( 1 + \frac{z^n t^n}{1 - t^n} \right)^{M(q, n)} }.$$

The coefficient of $z^m t^n$ is the number of monic polynomials of degree $n$ over $\mathbb{F}_q$ with $m$ distinct roots over $\overline{\mathbb{F}_q}$.

This is not terribly nice but it could be worse. For example if you wanted to compute the expected value of $m$ you could do it by taking the logarithmic derivative in $z$.

Qiaochu Yuan
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  • So, essentially we can make a generating function in two variables with the additional variable $z$ keep tracking of the number of distinct roots over $\overline{\mathbb{F}_q}$. Can we expect a rationality result to hold for $F(z,t)$ to be able to read off the coefficients of the generating series? – daruma Sep 21 '24 at 14:46
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    @daruma: I don't think so. – Qiaochu Yuan Sep 21 '24 at 18:45