Let $f(x)$ be a polynomial of degree $d$ with integer coefficients. Let $G\_p^+$ be the Galois group of the polynomial $f(x)-y$ over $\overline{\mathbb{F}}\_p(y)$ and $G\_p$ be the Galois group of the same polynomial over $\mathbb{F}\_p(y)$. It is known (see B. Birch and H. Swinnerton-Dyer, "Note on a problem of Chowla", Lemma 1) that for large enough $p$ all the groups $G\_p^+$ are isomorphic and also isomorphic to the Galois group over $\mathbb C(y)$. My question is, what happens with $G\_p$ when $p$ varies?


More precisely, for all large enough $p$ fix the order of roots of $f(x)-y$. Then $G\_p^+$ and $G\_p$ are subgroups of $S\_d$. Let us fix the order in a compatible way: so that $G\_p^+$ is always the same group $G\subseteq S\_d$. Let $H$ be a subgroup in $S\_d$, denote by
$$
\pi\_f(x;H)
$$
the number of $p\leq x$ such that $G\_p=H$. Then I'm interested in asymptotic formulas for $\pi\_f(x;H)$ and their dependence on $f$. Is it true that the main term actually depends only on $G=G\_p^+$?



The distribution of certain Galois groups
Denote by $\Omega$ the splitting field of $f(x)-y$ over $\mathbb{Q}$, by $k/\mathbb{Q}$ the maximal constant extension inside $\Omega$, and set $G:=Gal(\Omega/\mathbb{Q}(y))$, $G^+ =Gal(\Omega/k(y))$. Up to avoiding finitely many primes $p$, one has "good constant reduction" and can obtain the picture over $\mathbb{F}\_p(y)$ by reducing given defining polynomials of $\Omega/k(y)$ (i.e., $f(x)-y$) and of $k/\mathbb{Q}$ individually modulo $p$; in other words, if $x\_p\in G/G^+$ is a representative of the Frobenius class of $p$ in $k/\mathbb{Q}$ (well-defined up to conjugation in $G/G^+$), then $G\_p$ is the subgroup of $G$ mapping onto $\langle x\_p\rangle$ under projection $G \to G/G^+$, and these subgroups occur with frequency described by Chebotarev's density theorem (applied to $k/\mathbb{Q}$); this should give the "main term" in the question.
So then, if I understand the last question correctly, it basically amounts to asking whether the normal subgroup $G^+$ uniquely determines $G$ (independently of the concrete shape of $f$). This would not be true for polynomials of arbitrary shape, but due to the form $f(x)-y$, it should be true; indeed, it is known (cf. Lemma 3.4 in <https://arxiv.org/pdf/math/0109071.pdf>) that $G\le S\_d$ is generated by $G^+$ together with the normalizer (in $G$) of a $d$-cycle $\sigma$ (the latter being the inertia group generator at infinity); on the other hand, the so-called branch cycle lemma implies that this inertia group generator must be conjugate in $G$ to all powers $\sigma^k$ ($k$ coprime to $d$), i.e., $G$ must contain the full symmetric normalizer (of order $d\cdot \varphi(d)$) of this $d$-cycle. So then, once $G^+$ is given **and the $d$-cycle has been fixed**, there is really no freedom anymore in the choice of $G$, meaning that the answer to the last question should be "yes".


**EDIT:** Shortly after confidently typing away the above, I'm noticing the following crucial **gap**: It is not clear (to me at least) that, after fixing $G^+$, the ``fixing the $d$-cycle" is something that can be done without loss of generality; i.e., there are groups $H\le S\_d$ that have several classes of $d$-cycles $\sigma$, such that $H\trianglelefteq \langle H, N\_{S\_d}(\sigma)\rangle$, but the groups on the right can have different orders! This is not yet sufficient for these $d$-cycles to actually belong to the monodromy of some polynomial $f(x)-y$ over $\mathbb{Q}$, but if they did, it would actually give different $G$ with the same $G^+$, i.e., it would be a counterexample to the above claim!


