Number of polynomials with the Frobenius class in $C$.

Is there a theorem to estimate the number of integer polynomials with Frobenius class in $C$, where $C$ is a subset $G=S_n$ that is stable under conjugation? In other words, For a fixed prime $p$,

$\#\{ f(x) \in \mathbb{Z}[x]: \{\text{Frob}_{p}\} \subseteq C\} \ll$ ?

where $\{\text{Frob}_{p}\}$ is a conjugacy class in $S_n$, i.e., Frobenius class.

• What is $Frob_p$ (which depends on $f$) ? Is it the Frobenius $x \mapsto x^p$ in $\mathbb{F}_p[x]/(f(x))$ seen as a permutation of the roots of $f$ ? In that case it depends only on $\{ deg(f_i), f_i | f \bmod p\}$ the factorization-type obtained from $f(x) = \prod_{i=1}^j f_i(x) \bmod p$. And to count those I would use the zeta function of $\mathbb{F}_p[x]$. – reuns Jul 11 '17 at 5:18
• yes to your first question. Can you elaborate your last statement? – Tasmia Jul 11 '17 at 5:25
• $\mathbb{F}_p[x]$ is a UFD and $t^{deg(u)}$ is multiplicative so $\sum_{u \in \mathbb{F}_p[x]\text{ monic}} t^{deg(u)} = \prod_{v \in \mathbb{F}_p[x]\text{ irreducible}} \frac{1}{1-t^{deg(v)}}$ from which we obtain $\# \{ {v \in \mathbb{F}_p[x]\text{ irreducible}}, deg(v) = n\} = \frac{1}{n}\sum_{d | n} \mu(d) p^{n/d}$ from which you can find the number of polynomials of a given factorization-type. See this – reuns Jul 11 '17 at 5:34
• OK that makes sense. Thanks! – Tasmia Jul 11 '17 at 5:46