In a 1983 paper by William Webb (link below), the author gives a version of the Selberg sieve for function fields and uses it to prove that for a fixed $K \in \mathbb{F}_q[t]$, the number $\mathcal{N}(n,K)$ of monic irreducible $P$ of degree $\leq n$ such that $P + K$ is also irreducible satisfies $\mathcal{N}(n,K) \leq c\frac{q^n}{n^2}$ ($c$ depending on $K$).
Webb, William A., Sieve methods for polynomial rings over finite fields, J. Number Theory 16, 343-355 (1983). ZBL0517.10049.
As per convention, we set $\lvert F \rvert = q^{\deg(F)}$.
Define
$$\mathscr{P} = \{P \textrm{ monic irreducible} : \lvert P \rvert \leq N^{1/2}\}$$
and
$$\mathscr{D} = \{D \textrm{ squarefree} : \forall P \mid D, \ P \in \mathscr{P}\}.$$
In the proof (Theorem 4 in the paper), it is claimed that $$\sum_{\substack{D \in \mathscr{D} \\ \lvert D \rvert \leq N^{1/4} \\ (D,K) = 1}} \frac{2^{\omega(D)}}{\lvert D \rvert} \geq c_1 \sum_{\substack{P \in \mathscr{P} \\ \lvert P \rvert \leq N^{1/4} \\ P \nmid K}} \left(1 – \frac{2}{\lvert P \rvert}\right)^{-1} \geq c_2 \log^2(N).$$
I suspect that the middle term in the previous inequality was supposed to be a product. Otherwise each term of the sum is $\geq 1$, and the sum should be $\gg \frac{N^{1/4}}{\log(N)}$ by counting irreducibles. So assuming a typo, I am trying to see why
$$\sum_{\substack{D \in \mathscr{D} \\ \lvert D \rvert \leq N^{1/4} \\ (D,K) = 1}} \frac{2^{\omega(D)}}{\lvert D \rvert} \geq c_1 \prod_{\substack{P \in \mathscr{P} \\ \lvert P \rvert \leq N^{1/4} \\ P \nmid K}} \left(1 – \frac{2}{\lvert P \rvert}\right)^{-1}.$$
Ignoring the coprimality condition with $K$ for the moment, I can see that
$$\prod_{\substack{P \in \mathscr{P} \\ \lvert P \rvert \leq N^{1/4}}} \left(1 – \frac{2}{\lvert P \rvert}\right)^{-1} \asymp \prod_{\substack{P \in \mathscr{P} \\ \lvert P \rvert \leq N^{1/4}}} \left(1 + \frac{2}{\lvert P \rvert}\right) \leq \prod_{\substack{P \in \mathscr{P} \\ \lvert P \rvert \leq N^{1/2}}} \left(1 + \frac{2}{\lvert P \rvert}\right) = \sum_{D \in \mathscr{D}} \frac{2^{\omega(D)}}{\lvert D \rvert},$$
but it seems that the terms in the right-most sum with $\lvert D \rvert \leq N^{1/4}$ would need to dominate the sum, which I have not been able to prove. Any suggestions?
Best Answer
I am not sure exactly what Webb had in mind, but here is my answer. There is no real difference between the Selberg sieve in integers and in the polynomial ring $\mathbb{F}_q[T]$. So you can consult the book "Multiplicative Number Theory I" by Montgomery and Vaughan where the same upper bound is established in integers, and adapt their argument to $\mathbb{F}_q[T]$. Specifically see page 94, where a sum analogous to the one you ask about is estimated.
In more detail, the required estimate is $$\sum_{\substack{D \in \mathscr{D} \\ \lvert D \rvert \leq N^{1/4} \\ (D,K) = 1}} \frac{2^{\omega(D)}}{\lvert D \rvert} \geq c \log^2(N).$$ For given $d\ge 0$ with $q^d \le N^{1/4}$, I claim that $$(\star)\quad\sum_{\substack{D \in \mathscr{D} \\ \deg(D)=d \\ (D,K) = 1}} \frac{2^{\omega(D)}}{\lvert D \rvert} =q^{-d}\sum_{\substack{\deg(D)=d \\(D,K)=1}} 2^{\omega(D)} \mu^2(D) \ge c' d$$ if $d$ is sufficiently large (in terms of $K$). The constant $c'$ depends on $K$. Summing this over $d$ gives the needed bound.
To establish $(\star)$ write the arithmetic function $f(D)=2^{\omega(D)}\mu^2(D)\mathbf{1}_{(D,K)=1}$ as a convolution of the divisor function $\tau(D)$ with a simple multiplicative $g$. Since $\sum_{\deg(D)=d} \tau(D) = q^ d (d+1)$, the problem reduces to showing $\sum_{D}g(D)/\lvert D \rvert$ converges to a positive constant and $\sum_{D} g(D) \deg(D)/\lvert D \rvert$ converges absolutely. This is easily verified by considering the Dirichlet series of $g$ (as in page 94 of the aforementioned book).