1

I'm reading the following theorem about square-free values of polynomials over a finite field $\mathbb{F}_q$.

Suppose $q=p^e$ is a prime power, $k>0$ is an integer, and $m,n > 0$ are integers with $m\gg \log_q n \log_q \log_q n$ and $m \to \infty$. Suppose $f \in \mathbb{F}_q[t,x]$ is a square-free polynomial with $\deg_x f \leq k, \deg_t f \leq n$. Let $c_f$ be defined as $c_f = \prod_{P\in \mathcal{P}}\left(1-\frac{\rho_f(P^2)}{|P|^2}\right).$ Then $$\#\{a\in \mathbb{F}_q[t]:\deg a < m, \text{f(a) is square-free}\} = c_fq^m+o(c_fq^m)).$$

Could anyone provide me with an example of a polynomial for which this would hold, and how this then would be used?

Update

As per advice from the comment. We take $f(t,x) = tx+1$, with $q=2, k=1, n=1$ and $m=5$. We see that $m = 5 \gg \log_2 1 \log_2\log_2 1 = 0.$ In the end we have to let $m\to \infty$. Furthermore, we see that $\deg_x f = 1 = k$ and $\deg_t f = 1 = n$. So the assumptions there are met.

We define $c_f$ as usual. The theorem says that $$\#\{a \in \mathbb{F}_2[t] : \deg a < 5, f(a) \text{ is square free} = c_f2^5 + o(c_f2^5)$$.

Still, I'm not quite sure how to proceed. $a\in \mathbb{F}_2[t]$ is a polynomial with coefficients $0$ or $1$ and $\deg <5$, and then? Any thoughts?

Whizkid95
  • 759
  • 1
    Please try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Jun 07 '18 at 18:55
  • 1
    @Shaun My apologies, I understand your point. – Whizkid95 Jun 07 '18 at 19:00
  • Let $f(t,x)=tx+1$ for example, with $q=2$, $k=1$, $n=1$ and $m=5$. Now write down everything explicitly. Of course, $m$ needs to go to $\infty$ in the end. What do you get? – Dietrich Burde Jun 07 '18 at 19:04
  • Hmm. When you write $f(a)$ do you mean $f(t,a)$ or what? You introduced $f$ as a bivariate polynomial, so... – Jyrki Lahtonen Jun 08 '18 at 11:53
  • @DietrichBurde I'm still struggling, but edited the question. Any further thoughts? – Whizkid95 Jun 13 '18 at 12:43
  • 1
    There are $q^n-q^{n-1}$ square-free monic polynomials $a\in \Bbb{F}_2[t]$ of degree $n$, see here. Now you can compute your number explicitly, with $q=2$ and $deg(a)=n<5$. – Dietrich Burde Jun 13 '18 at 13:00
  • @DietrichBurde Thanks a lot, really helpful! One question, if I know how many square-free monic polynomials $a$ there are, how then does this help me to know explicitly when $f(a)$ is square-free? – Whizkid95 Jun 13 '18 at 13:37
  • What is $f(a)$ for $a\in \Bbb{F}_2[t]$? See Jyrki's comment. – Dietrich Burde Jun 13 '18 at 13:39
  • Well, $a$ is a polynomial with coefficients in $\mathbb{Z}/2\mathbb{Z}$, and $f(t,x) = tx+1$, so $f(a) = ax+1$? – Whizkid95 Jun 13 '18 at 13:42
  • @DietrichBurde But to be honest, I'm not quite sure. That is one of the reasons why I don't understand how to compute this. I'm looking at theorem 2.1 in the article https://arxiv.org/pdf/1605.07765.pdf – Whizkid95 Jun 13 '18 at 15:23
  • 1
    Judging from a quick look at the arXiv paper $f(a)$ means $f(t,a)$. This fits the bigger picture. In the purely number theoretic version $f$ is a polynomial from $\Bbb{Z}[x]$, $a$ takes values from $\Bbb{Z}$, and the question is whether $f(a)$ is square-free or not. When moving to the function field $A=\Bbb{F}_q[t]$ takes the role of $\Bbb{Z}$. So $f\in A[x]$ and $a\in A$. Therefore my somewhat educated guess is that with $f(t,x)=tx+1$ the notation $f(a)$ means $ta+1$. – Jyrki Lahtonen Jun 14 '18 at 18:16
  • (cont'd) In a sense, all polynomials in $t$ are now "constants", and the question is how often $f(a)$ is divisible by the square of some non-constant polynomial $a\in\Bbb{F}_q[t]\setminus \Bbb{F}_q$. – Jyrki Lahtonen Jun 14 '18 at 18:19

0 Answers0