We have the following mod $p$ technique to detemine the Galois group of a polynomial in $\mathbb{Z}[x]$:
Assume $f\in\mathbb{Z}[x]$ is irreducible, $p$ is a prime, and $\bar{f}$ is the polynomial obtained by $f$ mod $p$. (Or more precisely, $\bar{f}$ is the image of the homomorphism $\phi: \mathbb{Z}[x]\to F_p[x]$, induced by $\pi: \mathbb{Z}\to F_p$). If $\bar{f}$ is separable over $F_p$ and it factors in $F_p$ into irreducible polynomials as $f=f_1\cdots f_l$, with ${\rm deg}f_i=d_i$, then ${\rm Gal}(f,\mathbb Q)$ contains a permutation which is the product of $l$ disjoint cycles, each with length $d_1,\cdots,d_l$.
I wonder is the converse true? Or, if ${\rm Gal}(f,\mathbb Q)$ contains a permutation which is the product of $l$ disjoint cycles, each with length $d_1,\cdots,d_l$, then does there always exist a prime $p$, s.t. the corresponding $\bar{f}$ is separable over $F_p$ and it factors in $F_p$ into irreducible polynomials as $f=f_1\cdots f_l$, with ${\rm deg}f_i=d_i$? If this is incorrect for the identity element, then is it correct for the non-identity elements in ${\rm Gal}(f,\mathbb Q)$?
(Attempt: For the simplest case $f(x)=x^2+ax+b$ and the identity element in $S_2$, it is equivalent to determine if exists $u,v\in\mathbb Z$ and $p$ prime s.t. $u+v\equiv a \mod p$ and $uv\equiv b \mod p$. This already seems hard for me.)